Differential Transformation Method for Mechanical Engineering Problems

DIFFERENTIALTRANSFORMATIONMETHOD FORMECHANICALENGINEERINGPROBLEMS

MOHAMMAD HATAMI

DAVOOD DOMAIRRY GANJI

MOHSEN SHEIKHOLESLAMI

AMSTERDAM • BOSTON • HEIDELBERG • LONDON

NEW YORK • OXFORD • PARIS • SAN DIEGO

SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO

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ISBN: 978-0-12-805190-0

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DEDICATED TO:

All the Iranian fathers and mothers who resist and defended with wholeexistence against the assault to our homeland during the 1980e88 years thatbrought safety and comfort for us these days to live in love and peace.

PREFACE

Great Lord of Life and Wisdom! In thy NameWhich to transcend no flight of thought may claim!

Shahnameh, Abul-Qasim Ferdowsi (940e1020 C.E.)

Many phenomena in the nature need to be modeled mathematically ornumerically for better perception of its physic and enhance the researchersto solve the possible problems. Mechanical engineering, chemical engi-neering, petroleum, energy crisis, industrial heat exchangers, boilers,engines, etc., all contain some studies which need mathematical modeling.During most of these modelings, ordinary differential equations or partiallydifferential equations will appear and need powerful solution methods,numerically or analytically. Current book introduces the differentialtransformation method (DTM) as one of the most powerful mathematic/analytic methods for solving these differential equations. The contents ofthe current book are able to benefit engineers, researchers, and graduatestudents who want to develop their knowledge in basic phenomena of allsciences, especially mechanical engineering. In the introductory chapters(Chapters 1 and 2), DTM is introduced in simple and complicated versionsincluding all improvements and developments. In other chapters (Chapters3e8), application of DTM on various examples in solid and fluid mechanicsis demonstrated, and several examples of recently published papers fromhigh-quality journals are included to illuminate the subject. The authors arevery much pleased to receive the readers’ comments and amendments onthe materials of the book. Finally, we would like to express our sincerethanks to the staff of books publishing at Elsevier for their helpful support.We hope this book will be of great benefit to you.

Mohammad HatamiAssistant Professor of Mechanical Engineering Department, EsfarayenUniversity of Technology, Esfarayen, North Khorasan, Iran

xi

CHAPTER 1

Introduction to DifferentialTransformation Method

1.1 INTRODUCTION

Most of the problems in mathematics and nature are inherently nonlinear.For solving and analyzing these problems, analytical and numerical methodsmust be applied. homotopy analysis method (HAM), homotopy perturbationmethod (HPM), adomian decomposition method (ADM), weighted residualmethod (WRM), etc., are some common and classical analytical methodshave been presented in the literature for solving nonlinear problems [1e76].

The differential transformation method (DTM) is an alternative pro-cedure for obtaining an analytic Taylor series solution of differentialequations. The main advantage of this method is that it can be applieddirectly to nonlinear differential equations without requiring linearizationand discretization, and therefore, it is not affected by errors associated withdiscretization. The concept of DTM was first introduced by Zhou [77],who solved linear and nonlinear problems in electrical circuits. DTM dueto the following advantages has been used by many researchers and theytried to improve and increase its accuracy, which is discussed in this book.1. Unlike perturbation techniques, DTM is independent of any small or

large quantities. So, DTM can be applied no matter if governing equa-tions and boundary/initial conditions of a given nonlinear problemcontain small or large quantities, or not.

2. Unlike HAM, DTM does not need to calculate auxiliary parameter Z1,through h-curves.

3. Unlike HAM, DTM does not need initial guesses and auxiliary linearoperator, and it solves equations directly.

4. DTM provides us with great freedom to express solutions of a givennonlinear problem by means of Padé approximant and Ms-DTM orother modifications.This chapter introduces DTM generally and contains the following:

1.1 Introduction1.2 Principle of Differential Transformation Method

Differential Transformation Method for Mechanical Engineering ProblemsISBN 978-0-12-805190-0http://dx.doi.org/10.1016/B978-0-12-805190-0.00001-2

© 2017 Elsevier B.V.All rights reserved. 1

http://dx.doi.org/10.1016/B978-0-12-805190-0.00001-2

1.3 Multistep Differential Transformation Method1.4 Hybrid Differential TransformationMethod and Finite DifferenceMethod1.5 Differential Transformation Method Applying on Initial-Value Prob-

lems and Ordinary Differential Equations1.6 Two-Dimensional Differential Transformation Method for Partial Dif-

ferential Equations1.7 Differential Transformation MethodePadé Approximation1.8 Differential Transformation Method on Singular Two-Point Boundary

Value Problem

1.2 PRINCIPLE OF DIFFERENTIAL TRANSFORMATIONMETHOD

For understanding the method’s concept, suppose that x(t) is an analyticfunction in domain D, and ti represents any point in the domain. Thefunction x(t) is then represented by one power series whose center is locatedat ti. The Taylor series expansion function of x(t) is:

xðtÞ ¼PNk¼0

ðt � tiÞkk!

�dkxðtÞdtk

�t¼ti

ct ˛D (1.1)

The Maclaurin series of x(t) can be obtained by taking ti ¼ 0 in Eq. (1.1)expressed as [77]:

xðtÞ ¼PNk¼0

tk

k!

�dkxðtÞdtk

�t¼0

ct ˛D (1.2)

As explained in Ref. [77], the differential transformation of the functionx(t) is defined as follows:

XðkÞ ¼XNk¼0

Hk

k!

�dkxðtÞdtk

�t¼0

(1.3)

Where X(k) represents the transformed function, and x(t) is the originalfunction. The differential spectrum of X(k) is confined within the intervalt˛[0, H ], where H is a constant value. The differential inverse transform ofX(k) is defined as follows:

xðtÞ ¼XNk¼0

�tH

�k

XðkÞ (1.4)

2 Differential Transformation Method for Mechanical Engineering Problems

It is clear that the concept of differential transformation is based uponthe Taylor series expansion. The values of function X(k) at values ofargument k are referred to as discrete, i.e., X(0) is known as the zerodiscrete, X(1) as the first discrete, etc. The more discrete available, the moreprecise it is possible to restore the unknown function. The function x(t)consists of the T-function X(k), and its value is given by the sum of theT-function with (t/H)k as its coefficient. In real applications, at the rightchoice of constant H, the larger values of argument k the discrete ofspectrum reduce rapidly. The function x(t) is expressed by a finite series andEq. (1.4) can be written as:

xðtÞ ¼Xnk¼0

�tH

�k

XðkÞ (1.5)

Some important mathematical operations performed by DTM are listed inTable 1.1.

Example A: As shown in Fig. 1.1, a rectangular porous fin profile isconsidered. The dimensions of this fin are length L, width w, and thickness t.The cross section area of the fin is constant, and the fin has temperature-dependent internal heat generation. Also, the heat loss from the tip ofthe fin compared with the top and bottom surfaces of the fin is assumed tobe negligible. Since the transverse Biot number should be small for the finto be effective, the temperature variations in the transverse direction are

Table 1.1 Some Fundamental Operations of the Differential Transform Method [77]Origin Function Transformed Function

x(t) ¼ af(x) � bg(t) X(k) ¼ aF(k) � bG(k)

xðtÞ ¼ dmf ðtÞdtm

XðkÞ ¼ ðkþ mÞ!Fðkþ mÞk!

x(t) ¼ f(t)g(t)XðkÞ ¼Pk

l¼0FðlÞGðk� lÞ

x(t) ¼ tmXðkÞ ¼ dðk� mÞ ¼

�1; if k ¼ m;

0; if ksm.

x(t) ¼ exp(t) XðkÞ ¼ 1k!

x(t) ¼ sin(ut þ a) XðkÞ ¼ uk

k!sin

�kp2þ a

�x(t) ¼ cos(ut þ a) XðkÞ ¼ uk

k!cos

�kp2

þ a

�

Introduction to Differential Transformation Method 3

neglected. Thus heat conduction is assumed to occur solely in the longi-tudinal direction [12].

Energy balance can be written as:

qðxÞ � qðxþ DxÞ þ q*$A$Dx ¼ _mcp½TðxÞ � TN� þ hðp$DxÞ½TðxÞ � TN�(1.6)

The mass flow rate of the fluid passing through the porous material is:

_m ¼ r$Vw$Dx$w (1.7)

The passage velocity from the Darcy’s model is:

Vw ¼ gKb

nðT � TNÞ (1.8)

Substitutions of Eqs. (1.7) and (1.8) into Eq. (1.6) yield:

qðxÞ � qðxþ DxÞDx

þ q*$A ¼ r$cp$g$K$b$wy

½TðxÞ � TN�2

þ hp½TðxÞ � TN�(1.9)

As, Dx / 0 Eq. (1.9) becomes:

dqdx

þ q*$A ¼ r$cp$g$K$b$wy

½TðxÞ � TN�2 þ hp½T ðxÞ � TN� (1.10)

Also from Fourier’s Law of conduction:

q ¼ �keff AdTdx

(1.11)

Tb VW

Air Flow(h,T∞)

q*

t

w

L

X dX

Porousmedia

Heatgeneration Adia

batic

tip

Figure 1.1 Schematic of convective porous fin with temperature-dependent heatgeneration.


Where A is the cross-sectional area of the fin A ¼ (w$t) and keff is the effec-tive thermal conductivity of the porous fin that can be obtained fromfollowing equation:

keff ¼ 4$kf þ ð1� 4Þks (1.12)

where 4 is the porosity of the porous fin. Substitution Eq. (1.11) into Eq.(1.10) leads to:

d2Tdx2

� r$cp$g$K$b

t$keff $n½TðxÞ � TN�2 � h$p

keff $A½TðxÞ � TN� þ q*

keff¼ 0 (1.13)

It is assumed that heat generation in the fin varies with temperature asEq. (1.14):

q* ¼ q*Nð1þ εðT � TNÞÞ (1.14)

Where q*N is the internal heat generation at temperature TN.For simplifying the above equations some dimensionless parameters are

introduced as follows:

q ¼ ðT � TNÞðTb � TNÞ; X ¼ x

L; M 2 ¼ hPL2

k0A; Sh ¼ Da$x$Ra

kr

�Lt

�2

G ¼ q*NhPðTb � TNÞ; εG ¼ εðTb � TNÞ

(1.15)

where Sh is a porous parameter that indicates the effect of the permeabilityof the porous medium as well as buoyancy effect, so higher value of Shindicates higher permeability of the porous medium or higher buoyancyforces. M is a convection parameter that indicates the effect of surface con-vecting of the fin. Finally, Eq. (1.13) can be rewritten as:

d2qdX2

�M 2qþM 2Gð1þ εGqÞ � Shq2 ¼ 0 (1.16)

In this research we study finite length fin with insulated tip. For thiscase, the fin tip is insulated so that there will not be any heat transfer at theinsulated tip and boundary condition will be,

qð0Þ ¼ 1

dqdX

��x¼1

¼ 0(1.17)

Properties of Porous Materials are presented in Table 1.2.


Now we apply DTM from Table 1.1 into Eq. (1.16) to find u(t):

ðkþ 1Þðkþ 2ÞQðkþ 2Þ � Sh

Xkl¼0

QðlÞQðk� lÞ!

� ðM 2 � εGGM 2ÞQðkÞ þGM 2dðkÞ ¼ 0

(1.18)

Rearranging Eq. (1.18), we have

Qðkþ 2Þ ¼Sh

�Pkl¼0

QðlÞQðk� lÞ�þ ðM 2 � εGGM 2ÞQðkÞ �GM 2dðkÞðkþ 1Þðkþ 2Þ

(1.19)

and boundary condition transformed form is,

Qð0Þ ¼ 1; Qð1Þ ¼ a (1.20)

where a is an unknown coefficient that must be determined. By solvingEq. (1.19) and using boundary conditions, the DTM terms are obtained as

Qð2Þ ¼ �12GM 2 þ 1

2Sh þ 1

2M 2 � 1

2εGGM 2

Qð3Þ ¼ 13Shaþ 1

6M 2a� 1

6M 2aεGG

Qð4Þ ¼ �18ShεGGM 2 � 1

12ShGM 2 þ 1

12S2h þ

18ShM

2 � 112εGGM 4

þ 112Sha

2 � 124

GM 4 þ 124M 4 þ 1

24M 4

εGG2 þ 1

24ε2GG

2M 4

Qð5Þ ¼ � 120ShaGM 2 þ 1

12S2haþ

112ShaM

2 � 112ShaεGGM 2

þ 1120

M 4a� 160

aεGGM 4 þ 1120

aε2GG2M 4

(1.21)

and etc.

Table 1.2 Properties of Porous MaterialsPorous Material Ks (W/K�m) Cp (kJ/(kg�K)) r (kg/m3)

Aluminum 218 0.91 2700Si3N4 25 0.5 2300


Now by substituting Eq. (1.21) into Eq. (1.17) and using boundarycondition the “a” coefficient and then q(X) function will be obtained. ForSi3N4 with following coefficient,

G ¼ 0:4; M ¼ 1; Da ¼ 0:0001;Lt¼ 10; Ra ¼ 10000; εG ¼ 0:6;

L ¼ 1; ks ¼ 954;

(1.22)

Sh can be calculated. After five iterations in DTM series, temperaturedistribution will be obtained as,

qðxÞ ¼ 1� 0:2430983438xþ 0:1524109014x2 � 0:03280383661x3

þ 0:01079942879x4 � 0:001716344389x5 þ 0:0004283472660x6

� 0:00007114658871x7

(1.23)

Fig. 1.2 shows the validation of this solution.

1.3 MULTISTEP DIFFERENTIAL TRANSFORMATIONMETHOD

Multistep differential transformation method (Ms-DTM) due to someadvantages is applied in physical application. For example, Ms-DTM due tosmall time steps has a powerful accuracy especially for initial-value prob-lems. Also, because it’s based on DTM, does not need to small parameter,auxiliary function and parameter, discretization, etc., versus other analyticalmethods. For perception of Ms-DTM basic idea, consider a generalequation of n-th order ordinary differential equation [29],

f�t; y; y0;.; yðnÞ

¼ 0 (1.24)subject to the initial conditions

yðkÞð0Þ ¼ dk; k ¼ 0;.; n� 1: (1.25)

To illustrate the DTM for solving differential equations, the basic def-initions of differential transformation are introduced as follows. Let y(t) beanalytic in a domain D, and let t ¼ t0 represent any point in D. Thefunction y(t) is then represented by one power series, whose center islocated at t0. The differential transformation of the k-th derivative of afunction y(t) is defined as follows:

YðkÞ ¼ 1k!

�dkyðtÞdtk

�t¼t0

;ct ˛D: (1.26)


In Eq. (1.26), y(t) is the original function and Y(k) is the transformedfunction. As in [29] the differential inverse transformation of Y(k) is definedas follows:

yðtÞ ¼XNk¼0

YðkÞðt � t0Þk; ct ˛D. (1.27)

In fact, from Eq. (1.26) and Eq. (1.27), we obtain

yðtÞ ¼XNk¼0

ðt � t0Þkk!

�dkyðtÞdtk

�t¼t0

; ct ˛D. (1.28)

Eq. (1.28) implies that the concept of differential transformation isderived from the Taylor series expansion. From the definitions of Eq. (1.26)and Eq. (1.27), it is easy to prove that the transformed functions complywith the following basic mathematics operations (see Table 1.1).

In real applications, the function y(t) is expressed by a finite series andEq. (1.28) can be written as

yðtÞ ¼XNk¼0

YðkÞðt � t0Þk; ct ˛D. (1.29)

Eq. (1.29) implies thatPN

k¼Nþ1 YðkÞðt � t0Þk is negligibly small.Let [0, T ] be the interval over which we want to find the solution of the

initial-value problem. In actual applications of the DTM, the approximatesolution of the initial-value problem can be expressed by the finite series,

1.00

1

0.99

0.98

0.97

0.96

0.95θ

θ

0.94

0.93

0.92

0.91

0 0.2 0.4 0.6XX

0.8 1

0.98

0.96

0.94

0.92

0.90

0.88

0 0.2

NUM CM DTM LS NUM CM DTM LS

0.4 0.6 0.8 1

Figure 1.2 Temperature validation among Collocation, DTM, LS, and numericalmethods for (left) Si3N4 (right) AL.


yðtÞ ¼PNk¼0

bktk; t ˛ ½0;T � . (1.30)

Assume that the interval [0, T ] is divided into N subintervals [ti�1, ti],i ¼ 1, 2.,M of equal step size h ¼ T/M by using the nodes ti ¼ ih. Themain ideas of the multistep DTM are as follows. First, we apply the DTMto Eq. (1.24) over the interval [0, t1], we will obtain the followingapproximate solution,

y1ðtÞ ¼XNk¼0

b1ktk; t ˛ ½0; t1� (1.31)

using the initial conditions yðkÞ1 ð0Þ ¼ dk. For i � 2, at each subinterval [ti�1, ti]we will use the initial conditions yðkÞi ðti�1Þ ¼ yðkÞi�1ðti�1Þ and apply theDTM to Eq. (1.24) over the interval [ti�1, ti], where t0 in Eq. (1.26) isreplaced by ti�1. The process is repeated and generates a sequence ofapproximate solutions yi(t), i ¼ 1, 2,.,M for the solution y(t),

yiðtÞ ¼XNk¼0

bikðt � ti�1Þk; t ˛ ½ti�1; ti� (1.32)

In fact, the multistep DTM assumes the following solution,

yðtÞ ¼

8>>><>>>:y0ðtÞ; t ˛ ½0; t1�y1ðtÞ; t ˛ ½t1; t2�:::

yMðtÞ; t ˛ ½tM ; tMþ1�

(1.33)

Example A: Consider a particle that slides along a surface that has theshape of a parabola z ¼ cr2 (see Fig. 1.3). Following assumptions areconsidered for particles motion modeling:• Particle is at equilibrium.• The particle rotates in a circle of radius R.• The surface is rotating about its vertical symmetry axis with angular

velocity u.By choosing the cylindrical coordinates r, q, and z as generalized

coordinates. The kinetic and potential energies are [29],

T ¼ 12m�_r2 þ r2 _q

2 þ _z2

U ¼ mgz(1.34)


We have in this case some equations of constraints that we must takeinto account, namely

z ¼ cr2

_z ¼ 2c _rr(1.35)

and

q ¼ ut_q ¼ u

(1.36)

Inserting Eq. (1.36) and Eq. (1.35) in Eq. (1.34), we can calculate theLagrangian for the problem

L ¼ T � U ¼ 12mð _r2 þ 4c2r2 _r2 þ r2u2Þ � mgcr2 (1.37)

It is important to note that the inclusion of the equations of constraintsin the Lagrangian has reduced the number of degrees of freedom to only

z

ω

RParticle

r

θ

Par

abol

ic s

urfa

ce

Figure 1.3 Schematic view of a spherical particle on a rotating parabolic surface.


one, i.e., r. We now calculate the equation of motion using Lagrange’sequation

vLvr

¼ mð4c2r _r2 þ ru2 � 2gcrÞddt

vLvr

¼ mð€r þ 4c2r2€r þ 8c2r _r2Þ(1.38)

and

€rð1þ 4c2r2Þ þ _r2ð4c2rÞ þ rð2gc � u2Þ ¼ 0 (1.39)

It is considered that 2gc � u2 ¼ ε2 and so

€r þ 4c2€rr2 þ 4c2r _r2 þ ε2r ¼ 0 (1.40)

It’s considered that initial particle position is in radius A, and its initialvelocity is zero. So, its initial conditions are:

rð0Þ ¼ A; _rð0Þ ¼ 0 (1.41)

For solving the particle motion on a rotating parabola by an efficient,fast, and high accurate method, Eq. (1.40) is solved by Ms-DTM,

ðkþ 2Þðkþ 1ÞRjðkþ 2Þ

� 4c2Xkk1

Xk1l¼0

RjðlÞRjðk1� lÞðk� k1þ 1Þðk� k1þ 2ÞRjðk� k1þ 2Þ

þ ε2RjðkÞ þ 4c2

Xkk1¼0

Xk1l¼0

RjðlÞðk1� l þ 1ÞRjðk1� l þ 1ÞRjðk� k1þ 1Þ

ðk� k1þ 1Þ ¼ 0

(1.42)

With initial condition as,

R0ð0Þ ¼ A; R0ð1Þ ¼ 0

Rið0Þ ¼ ri�1ðtiÞ; Rið1Þ ¼ r 0i�1ðtiÞ; i ¼ 1; 2;.;K � M(1.43)

Since, the procedure of solving Eq. (1.34) is autonomous of constants A,ε, and c; so, for generalization and simplification of problem for future caseswith different physical conditions the constants, which represent physicalproperties are assumed to be as the following,

A ¼ c ¼ ε ¼ 1. (1.44)

After solving Eq. (1.34), and using initial conditions Eq. (1.25) andEq. (1.36), position of the particle, r(t), will be appeared as followingequation for each 0.25 s time step,


2 4 6 8 100

0.25

1 11 553 59363 10256209( ) 1 [0,0.25)10 3000 2250000 3150000000 7087500000000

( ) 1.006293- 0.050230616

( )

r t t t t t t t

r t

r t

= − − − − − ∈

=

=

2 3 4 5

6 7 8 9 10

- 0.1013895( - 0.25) - 0.003744513( - 0.25) - 0.00390231( - 0.25) - 0.38551 -3( - 0.25)

- 0.279966 -3( - 0.25) - 0.40481 - 4( - 0.25) - 0.23109 - 4( - 0.25) - 0.396 -5( - 0.25) - 0.18719 -5( - 0.25)

t t t t e te t e t e t e t e t

2 3 4 50.50

6 7 8

[0.25,0.5)( ) 1.025707 - 0.10188 - 0.105738( - 0.5) - 0.0079821( - 0.5) - 0.0046757( - 0.5) - 0.88127 -3( - 0.5)

- 0.398166 -3( - 0.5) - 0.99646 - 4( - 0.5) - 0.383726 - 4( - 0.5) - 0.10192 - 4( - 0.5

tr t t t t t e t

e t e t e t e t

∈

=9 10

2 3 4 50.75

6 7

) - 0.32931 -5( - 0.5) [0.5,0.75)( ) 1.05996 - 0.1565587 - 0.113642( - 0.75) - 0.013349( - 0.75) - 0.00621702( - 0.75) - 0.0016488( - 0.75)

- 0.65612 -3( - 0.75) - 0.206483 -3( - 0.75)

e t tr t t t t t t

e t e t

∈

=8 9 10

2 3 4 59.0

6

- 0.7254 - 4( - 0.75) - 0.2103 - 4( - 0.75) - 0.5254 - 5( - 0.75) [0.75,1.0)...

( ) -2.63269 0.374569 - 0.1812( -9) 0.053389( -9) - 0.024782( -9) 0.0101267( -9)

- 0.00325035( -9) - 0.15

e t e t e t t

r t t t t t tt

∈

= + + +7 8 9 10

2 3 4 59.25

27698 -3( -9) 0.001640656( -9) - 0.0020994( -9) 0.0020165285( -9) [9.0,9.25)( ) -1.88514 .29261 - 0.1490558( -9.25) 0.033957( -9.25) - 0.014987( -9.25) 0.00580704( -9.25)

e t t t t tr t t t t t t

+ + ∈

= + + +6 7 8 9 10

2 39.5

- 0.00229197( -9.25) 0.72248 -3( -9.25) - 0.8865 - 4( -9.25) - 0.14027 -3( -9.25) 0.19074 -3( -9.25) [9.25,9.5)( ) -1.2385 0.223613 - 0.1284215( -9.5) 0.0219744( -9.5) - 0.009517(

t e t e t e t e t tr t t t t t

+ + ∈

= + + 4 5

6 7 8 9 10

9.75

-9.5) 0.0032032( -9.5)

- 0.001265( -9.5) 0.43899 -3( -9.5) - 0.1384 -3( -9.5) 0.2617 - 4( -9.5) 0.85178 -5( -9.5) [9.5,9.75) ( ) -0.6551 0.162983 - 0.11507( -9

tt e t e t e t e t t

r t t t

+

+ + + ∈

= + 2 3 4 5

6 7 8 9 10

.75) 0.0141168( -9.75) - 0.0064938( -9.75) 0.001773667( -9.75)

- 0.70295 -3( -9.75) 0.22445 -3( -9.75) - 0.78165 - 4( -9.75) 0.22326 - 4( -9.75) - 0.5106 -5( -9.75) [9.75,10]

t t te t e t e t e t e t t

⎧⎪⎪⎪⎪⎪

⎨

+ +

+ + ∈

⎪⎪⎪⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(1.45)

12DifferentialTransform

ationMethod

forMechanicalEngineering

Problems

Eq. (1.45) is depicted in Fig. 1.4 and is compared with obtained resultfrom DTM. The values are presented in Table 1.3. Fig. 1.4 shows theparticle’s position for three different c constants when A ¼ ε ¼ 1.

1.4 HYBRID DIFFERENTIAL TRANSFORMATION METHODAND FINITE DIFFERENCE METHOD

The differential transform is defined as follows [78]:

XðkÞ ¼ 1k!

�dkxðtÞdtk

�t¼t0

(1.46)

where, x(t) is an arbitrary function, and X(k) is the transformed function.The inverse transformation is as follows

xðtÞ ¼XNk¼0

XðkÞðt � t0Þk (1.47)

Figure 1.4 Multistep Differential Transformation Method efficiency in particle motionanalysis, compared with Differential Transformation Method and numerical solution.


Substituting Eq. (1.46) into Eq. (1.47), we have

xðtÞ ¼XNk¼0

ðt � t0Þkk!

�dkxðtÞdtk

�t¼t0

The function x(t) is usually considered as a series with limited terms, andEq. (1.47) can be rewritten as:

xðtÞzXmk¼0

XðkÞðt � t0Þk (1.48)

where, m represents the number of Taylor series’ components. Usually,through elevating this value, we can increase the accuracy of the solution.

Table 1.3 Multistep Differential Transformation Method’s Valuesfor Position of Particle, Compared With Numerical Resultst r(t)

Num. Ms-DTM0 1.0000000 1.00000000.5 0.974766908 0.97476691831.0 0.896067130 0.89606715971.5 0.753057776 0.75305785862.0 0.5189251820 0.5189252782.5 0.13321789149 0.1332184543.0 �0.33404132749 �0.3337829263.5 �0.64064202117 �0.6398159244.0 �0.82788893947 �0.82668330254.5 �0.9393086789 �0.93776772805.0 �0.9926709981 �0.99078762345.5 �0.9947172580 �0.99244885546.0 �0.94568648462 �0.94294589886.5 �0.83945793817 �0.8360780697.0 �0.65951111277 �0.6551442107.5 0.36531058744 0.3591902508.0 0.09376192157 0.1006018468.5 0.49425716809 0.4972761309.0 0.73793715747 0.7384362249.5 0.88704545678 0.88581378410 0.97044161417 0.9677503517


Although the DTM series solution has a good approximate of the exactsolution, but this series is diverged for greater areas. For this reason, themultistep DTM is used. Based on this technique, the solution domain isdivided to some subdomains.

To solve a differential equation in the domain [0, T] using multistepDTM, this domain is divided to N sections. We suppose the subdomains areequal, and length of each subdomain will be H ¼ T/N. So, a separatefunction is considered for every subdomain as follows:

xðtÞ ¼

8>>>>>><>>>>>>:

x1ðtÞ ; t ˛ ½t1; t2�«

xiðtÞ ; t ˛ ½ti; tiþ1�«

xNðtÞ ; t ˛ ½tN ; tNþ1�

(1.49)

where ti ¼ (i � 1)H. Multistep DTM for every sub domain defined as:

XiðkÞ ¼ Hk

k!

�dkxiðtÞdtk

�t¼ti

(1.50)

The inverse multistep DTM is

xiðtÞ ¼XNk¼0

XiðkÞ�t � tiH

�k

To solve the partial differential equation u(y, t) in the domain t˛[0, T ]and y˛[yfirs,yend] using hybrid multi-step differential transformation method(MDTM) and FDM, we apply finite difference approximate on y-directionand take MDTM on t. The time domain is divided to Nt sections. Wesuppose the time subdomains are equal, and length of each subdomain isH ¼ T/Nt. So there is a separate function for every subdomain as follows:

uiðy; tÞ ¼

8>>>>>><>>>>>>:

u1ðj; tÞ; t ˛ ½t1; t2�; 1 � j � Ny þ 1

«

uiðj; tÞ; t ˛ ½ti; tiþ1�; 1 � j � Ny þ 1

«

uNtðj; tÞ; t ˛ ½tN ; tNtþ1�; 1 � j � Ny þ 1

(1.51)

where ti ¼ (i � 1)H and Ny is the number of cells in y-direction.


Example A: The fluid is assumed to be flowing between two infinitehorizontal plates located at the y ¼ �h planes. The upper plate moveswith a uniform velocity U0 while the lower plate is kept stationary. Thetwo plates are assumed to be electrically insulating and kept at twoconstant temperatures, T1 for the lower plate and T2 for the upper platewith T2 > T1. A constant pressure gradient is applied in the x-direction. Auniform magnetic field B0 is applied in the positive y-direction while theinduced magnetic field is neglected by assuming a very small magneticReynolds number (See Fig. 1.5). The Hall effect is taken into consider-ation and consequently a z-component for the velocity is expected toarise. The fluid motion starts from rest at t ¼ 0, and the no-slip conditionat the plates implies that the fluid velocity has neither a z- nor anx-component at y ¼ �h and y ¼ h. The initial temperature of the fluid isassumed to be equal to T1. Since the plates are infinite in the x- and z-directions, the physical quantities do not change in these directions andthe problem is essentially one-dimensional. The flow of the fluid isgoverned by the NaviereStokes equation, which has the two components[78],

rvuvt

¼ �dPdx

þ mv2uvy2

þ vm

vyvuvy

� sB20

1þ m2ðuþ mwÞ (1.52)

rvwvt

¼ mv2wvy2

þ vm

vyvwvy

� sB20

1þ m2ðw � muÞ (1.53)

where r is the density of the fluid, m is the viscosity of the fluid, v is thevelocity vector of the fluid ¼u(y, t)i þ w(y, t)j, s is the electric conductivityof the fluid, m is the Hall parameter given by m ¼ sbB0, and b is the Hall

Figure 1.5 Schematic of the problem.


factor [78]. The energy equation describing the temperature distribution forthe fluid is given by:

rcpvTvt

¼ v

vy

�kvTvy

�þ m

"�vuvy

�2

þ�vwvy

�2#þ sB2

0

ð1þ m2Þ ðu2 þ w2Þ

(1.54)

where T is the temperature of the fluid, cp is the specific heat at constant pres-sure of the fluid, and k is thermal conductivity of the fluid. The viscosity ofthe fluid is assumed to vary exponentially with temperature and is defined asm ¼ m0 f(T) ¼ m0 exp(�a(T � T1)). Also the thermal conductivity of thefluid is varying linearly with temperature ask ¼ k0g(T) ¼ k0[1 þ b(T � T1)].The problem is simplified by writing the equations in the nondimensionalform. To achieve this define the following nondimensional quantities [78],

by ¼ yh; bt ¼ tU0

h; bP ¼ P

rU20; ðbu; bwÞ ¼ ðu;wÞ

U0; q ¼ T � T1

T2 � T1; a ¼ �dbP

dbx(1.55)

f(q) ¼ e�aq, a is the viscosity parameter. g(q) ¼ 1 þ bq b is the thermal

conductivity parameter. Re ¼ rU0hm0

is the Reynolds number.

Ha2 ¼ sB20h

2

m0, Ha is Hartmann number. Pr ¼ m0cp

k0is Prandtl number and

Ec ¼ U20

cpðT2 � T1Þ is Eckert number.

In terms of the above nondimensional quantities, the velocity andenergy Eqs. (1.52)e(1.54) read;

vuvt

¼ aþ 1Re

f ðqÞ v2uvy2

þ 1Re

vf ðqÞvy

vuvy

� 1Re

Ha2


vwvt

¼ 1Re

f ðqÞ v2wvy2

þ 1Re

vf ðqÞvy

vwvy

� 1Re

Ha2

1þ m2ðw � muÞ (1.57)

vq

vt¼ 1

RePrgðqÞ v

2q

vy2þ 1RePr

�vgðqÞvy

��vq

vy

�þ EcRe

f ðqÞ"�

vuvy

�2

þ�vwvy

�2#þ EcHa2

Reð1þ m2Þ ðu2 þ w2Þ

(1.58)


The boundary and initial conditions for components of velocity andtemperature are:

IC’s/

8>>><>>>:uðy; 0Þ ¼ 0

wðy; 0Þ ¼ 0

qðy; 0Þ ¼ 0

BC’s/

8>>><>>>:uð�1; tÞ ¼ 0; uð1; tÞ ¼ 1

wð�1; tÞ ¼ 0; wð1; tÞ ¼ 0

qð�1; tÞ ¼ 0; qð1; tÞ ¼ 1

(1.59)

Once the values of the velocities and temperature are obtained, thefriction coefficient and Nusselt number will be determined. The local skinfriction coefficient at the lower wall is,

Cf ¼ 2Re

vUvy

��y¼�1

(1.60)

And the local Nusselt number for the lower wall is defined as,

Nux ¼ vq

vy

��y¼�1

(1.61)

The solution of the system of Eqs. (1.56)e(1.58) can be assumed as thefollowing form:

for 1 � i � Nt; 1 � j � Ny þ 1

uið j; tÞ ¼Xmk¼0

Uið j; kÞ�t � tiH

�k

t ˛ ½ti; tiþ1�

wið j; tÞ ¼Xmk¼0

Wið j; kÞ�t � tiH

�k

t ˛ ½ti; tiþ1�

qið j; tÞ ¼Xmk¼0

Qið j; kÞ�t � tiH

�k

t ˛ ½ti; tiþ1�

(1.62)


After taking the second order accurate central finite differenceapproximation with respect to y and applying MDTM on Eqs. (1.56)e(1.58), the following recurrence relations can be obtained:

for 1 � i � Nt; 1 � j � Ny � 1

Uið j; kþ 1Þ ¼H

kþ 1

(adðkÞ þ 1

ReDy2Xkr¼0

Fið j; k� rÞðUið j þ 1; rÞ � 2Uið j; rÞ þ Uið j � 1; rÞÞ

þ 14ReDy2

Xkr¼0

ðFið j þ 1; k� rÞ � Fið j � 1; k� rÞÞðUið j þ 1; rÞ � Uið j � 1; rÞÞ

� Ha2

Reð1þ m2Þ ðUið j; kÞ þ mWið j; kÞÞ

(1.63)

Wið j; kþ 1Þ ¼ Hkþ 1

(1

ReDy2Xkr¼0

Fið j; k� rÞðWið j þ 1; rÞ � 2Wið j; rÞ

þWið j � 1; rÞÞ

þ 14ReDy2

Xkr¼0

ðFið j þ 1; k� rÞ � Fið j � 1; k� rÞÞðWið j þ 1; rÞ �Wið j � 1; rÞÞ

� Ha2

Reð1þ m2Þ ðWið j; kÞ � mUið j; kÞÞ

(1.64)

Qið j; kþ 1Þ ¼ Hkþ 1

�1

Re PrDy2ðQið j þ 1; kÞ � 2Qið j; kÞ þQið j � 1; kÞÞ

þ bRe PrDy2

Xkr¼0

Qið j; k� rÞðQið j þ 1; rÞ � 2Qið j; rÞ þQið j � 1; rÞÞ

þ 14Re PrDy2

Xkr¼0

ðQið j þ 1; k� rÞ �Qið j � 1; k� rÞÞðQið j þ 1; rÞ �Qið j � 1; rÞÞ

þ Ec4ReDy2

Xkr¼0

Xr

s¼0

Fið j; sÞ½ðUið j þ 1; r � sÞ � Uið j � 1; r � sÞÞðUið j þ 1; k� rÞ

�Uið j � 1; k� rÞÞðWið j þ 1; r � sÞ �Wið j � 1; r � sÞÞðWið j þ 1; k� rÞ �Wið j � 1; k� rÞÞ�

þ EcHa2

Reð1þ m2ÞXkr¼0

½Uið j; rÞUið j; k� rÞ þWið j; rÞWið j; k� rÞ�)

(1.65)


where Fi( j, k) and Gi( j, k) are the differential transform of the functions f(q)and g(q), respectively. Applying MDTM on initial conditions in Eq. (1.59),we have:

for 1 � j � Ny þ 1

U1ð j; 0Þ ¼ 0; W1ð j; 0Þ ¼ 0; T1ð j; 0Þ ¼ 0(1.66)

The boundary conditions in Eq. (1.59) can be transformed as follow:

for 1 � j � Nt

BC’s for uðy; tÞ/(Uið1; kÞ ¼ 0; k � 0

UiðNy þ 1; 0Þ ¼ 1; UiðNy þ 1; kÞ ¼ 0; k � 1

(1.67)

BC’s for wðy; tÞ/�Wið1; kÞ ¼ 0; k � 0

WiðNy þ 1; kÞ ¼ 0; k � 0(1.68)

BC’s for qðy; tÞ/�Qið1; kÞ ¼ 0; k � 0

QiðNy þ 1; 0Þ ¼ 1; QiðNy þ 1; kÞ ¼ 0; k � 1

(1.69)

For solving the problem in whole of the time subdomains, we must usethe continuity condition in each time subdomain. These conditions can beexpressed as:

for 2 � j � Ny; 2 � i � Nt

Uið j; 0Þ ¼Xmk¼0

Ui�1ð j; kÞ; Wið j; 0Þ ¼Xmk¼0

Wi�1ð j; kÞ

Qið j; 0Þ ¼Xmk¼0

Qi�1ðj; kÞ

(1.70)

The accuracy of this method is shown in Fig. 1.6.

1.5 DIFFERENTIAL TRANSFORMATION METHOD APPLYINGON INITIAL-VALUE PROBLEMS AND ORDINARYDIFFERENTIAL EQUATIONS

In this section, applying the DTM on initial-value problems (IVPs) andordinary differential equations (ODE) are discussed [79]. The differentialequation for the initial-value problem can be described as


Figure 1.6 Comparison between Hybrid Differential Transformation Method withRef. [5] when a ¼ 5, Pr ¼ 1, Re ¼ 1, Ec ¼ 0.2, Ha ¼ 1, m ¼ 3, a ¼ 0.5, b ¼ 0.5.


dydt

¼ f ðt; yÞ; a � t � b (1.71)

with initial condition

yðaÞ ¼ a. (1.72)

A. Differential Transformation Method With Fixed Grid SizeThe objective of this section is to find the solution of Eq. (1.71) at theequally spaced grid points {t0, t1, t2,.,tN}, where

ti ¼ aþ ih; for each i ¼ 0; 1;.;N ; and h ¼ ðb� aÞ=N (1.73)

The domain of interest [a, b] is divided into N subdomains and theapproximation functions in each subdomain are yi(t); i ¼ 0, 1, 2,.,N�1,respectively. Taking the differential transformation of Eq. (1.71), the trans-formed equation describes the relationship between the spectrum of y(t) as

ðkþ 1ÞYðkþ 1Þ ¼ FðYðkÞÞ; (1.74)

where F($) denotes the transformed function of f(t, y(k)). From the initialcondition Eq. (1.72), it can be obtained that

Yð0Þ ¼ a. (1.75)

In the first subdomain, y(t) can be described by y0(t). From Eqs. (1.74)and (1.75), y0ðtÞ can be represented in terms of its nth order Taylorpolynomial about a, that is

y0ðtÞ ¼ Y0ð0Þ þ Y0ð1Þðt � aÞ þ Y0ð2Þðt � aÞ2 þ.þ Y0ðnÞðt � aÞn;(1.76)

where the subscript 0 denotes that the Taylor polynomial is expandedabout t0¼(a). Once the Taylor polynomial is obtained, y(t1) can be eval-uated as

yðt1Þz y0ðt1Þ¼ Y0ð0Þ þ Y0ð1Þðt � aÞ þ Y0ð2Þðt � aÞ2 þ.þ Y0ðnÞðt � aÞn

¼ Y0ð0Þ þ Y0ð1ÞðhÞ þ Y0ð2Þh2 þ Y0ðnÞhn

¼Pnj¼0

Y0ðjÞhj

(1.77)


The final value y0(t1) of the first subdomain is the initial value of thesecond subdomain, i.e., y1(t1) ¼ Y1(0) ¼ y0(t1). In a similar manner, y(t2)can be represented as:

yðt2Þz y1ðt1Þ¼ Y1ð0Þ þ Y1ð1Þðt � t1Þ þ Y1ð2Þðt � t1Þ2 þ.þ Y1ðnÞðt � t1Þn¼ Y1ð0Þ þ Y1ð1ÞðhÞ þ Y1ð2Þh2 þ Y1ðnÞhn

¼Pnj¼0

Y1ð jÞhj.

(1.78)

Hence, the solution on the grid points (tiþ1) can be obtained as follows:

yðtiþ1Þz yiðtiþ1Þ¼ Yið0Þ þ Yið1Þðtiþ1 � tiÞ þ Yið2Þðtiþ1 � tiÞ2 þ.þ YiðnÞðtiþ1 � tiÞn¼ Yið0Þ þ Yið1ÞðhÞ þ Yið2Þh2 þ YiðnÞhn

¼Pnj¼0

Yið jÞhj.

(1.79)

For illustrative purposes, the procedure for solving the initial-value problemof linear and nonlinear differential equations is demonstrated as follows.

Example A. Consider the nonlinear initial-value problem

_y ¼ �ðyþ 1Þðyþ 3Þ; 0 � t � 3; with yð0Þ ¼ �2: (1.80)

Let N ¼ 10 and h ¼ 0.3. Taking the differential transformation of Eq.(1.80), it can be obtained that

Yiðkþ 1Þ ¼ �½ðYiðkÞ þ dðkÞÞ5ðYi þ 3dðkÞÞ�=ðkþ 1Þ ¼�½YiðkÞ5YiðkÞ þ 4YiðkÞ þ 3dðkÞ�=ðkþ 1Þ (1.81)

With Y0(0) ¼ �2 the approximation of y(t) on the grid point can beobtained by Eq. (1.81). The exact solution of this problem isy(t) ¼�3 þ (1 þ e�2t)�1.

B. Differential Transformation Method With VaryingGrid SizeThe adaptive technique of the DTM can improve the efficiency ofcomputation by the use of varying grid size techniques in integration


approximation. The precondition is that complications from these methodsdo not influence the nature of the system. The truncation errors ofcomputation can be estimated according to different grid sizes and withoutapproximation of the higher derivatives of the function. These methods arecalled adaptive because they adapt the number and position of the gridpoints used, such that the truncation error is bounded within a specifiedbound [79]. Using the ideal difference equation to solve the initial-valueproblem of Eq. (1.71), the approximated solution is

wiþ1 ¼ wi þ hiFðti;wi; hiÞ for i ¼ 0; 1;.; n. (1.82)

where F is the incremental function of ti, wi and hi. Theoretically, if a toler-ance e is given, the global error would not exceed e for any grid points.Thus,

jyðtiÞ � wij � ε for i ¼ 0; 1;.; n. (1.83)

From error estimates of the Taylor transformation method, if sufficientdifferentiability conditions are satisfied, an nth-order Taylor transformationmethod will have local errorO(hnþ1) and global errorO(hn). The local errorcan be predicted by changing the order of the Taylor series expansion.Through this method, one can choose a proper grid size according to somecriterions, such that the global error is constrained within a specified bound[79].

Example B. Consider the system of the initial-value problem asfollows:

_u1ðtÞ ¼ 9u1ðtÞ þ 24u2ðtÞ þ 5 cos t � 13sin t; with u1ð0Þ ¼ 4

3;

_u2ðtÞ ¼ �24u1ðtÞ � 51u2ðtÞ � 9 cos t þ 13sin t; with u2ð0Þ ¼ 2

3;

(1.84)

The exact solutions of Eq. (1.84) are

u1ðtÞ ¼ 2e�3t � e�39t þ 13cos t;

u2ðtÞ ¼ �e�3t þ 2e�39t � 13cos t.

The differential equation of the system, for t˛[ti, tiþ1], can be repre-sented as


_u1�t* ¼ 9u1

�t*þ 24u2ðtÞ þ

�5 cos ti � 1

3sin ti

�cos t* �

�5 sin ti þ 1

3cos ti

�sin t*;

_u2�t* ¼ �24u1

�t*� 51u2ðtÞ þ

�� 9 cos ti þ 1

3sin ti

�cos t* þ

�9 sin ti � 1

3cos ti

�sin t*;

(1.85)

where t* ¼ t � ti. Taking the differential transformation of Eq. (1.85), thecorresponding transformed equation can be obtained as

ðkþ 1ÞU1iðkþ 1Þ

¼ 9U1iðkÞ þ 24U2iðkÞ þ 1k!

�5 cos ti � 1

3sin ti

�cos

�P

2k

��1k!

�5 sin ti þ 1

3cos ti

�sin

�P

2k

�;

ðkþ 1ÞU2iðkþ 1Þ

¼ �24U1iðkÞ � 51U2iðkÞ þ 1k!

�� cos ti þ 1

3sin ti

�cos

�P

2k

�1k!

�9 sin ti � 1

3cos ti

�sin

�P

2k

�

(1.86)

With U10ð0Þ ¼ 43 and U20ð0Þ ¼ 2

3 ; the approximation of y(t) onadaptive grid points can be obtained by Eq. (1.86). More details about theaccuracy and errors of examples can be found in [79].

1.6 TWO-DIMENSIONAL DIFFERENTIAL TRANSFORMATIONMETHOD FOR PARTIAL DIFFERENTIAL EQUATIONS

In this section, two-dimensional differential transform method of solutionof the initial-value problem for partial differential equations (PDEs) hasbeen studied. The basic definitions and fundamental theorems 1e13 of thetwo-dimensional transform are defined in [80] as follows:

W ðk; hÞ ¼ 1k!h!

"vkþhwðx; yÞvxkvyh

#ð0;0Þ

; (1.87)

where w(x,y) is the original function, and W(k, h) is the transformed func-tion. Letters represent the original and transformed functions respectively.The differential inverse transform of W(k, h) is defined as


(1.88)

and from Eqs. (1.87) and (1.88), the following can be concluded

wðx; yÞ ¼XNk¼0

XNh¼0

1k!h!

"vkþhwðx; yÞvxkvyh

#ð0;0Þ

xkyh (1.89)

Theorem 1. If w(x, y) ¼ u(x, y) � y(x, y), then W(k, h) ¼ U(k, h) �V(k, h).

Theorem 2. If w(x, y) ¼ lu(x, y), then W(k, h) ¼ lU(k, h). Here, l is aconstant.

Theorem 3. If w(x, y) ¼ vu(x, y)/vx, thenW(k, h) ¼ (k þ 1)U(k þ 1, h).Theorem 4. If w(x, y) ¼ vu(x, y)/vy, then W(k, h) ¼ (k þ 1)U(k,

h þ 1).Theorem 5. If w(x, y) ¼ vrþsu(x, y)/vxrvys, then

W ðk; hÞ ¼ ðkþ 1Þðkþ 2Þ.ðkþ rÞðhþ 1Þðhþ 2Þ.ðhþ sÞlUðkþ r; hþ sÞ.

Theorem 6. Ifw(x,y) ¼ u(x,y)y(x,y), then

W ðk; hÞ ¼Xkr¼0

Xhs¼0

Uðr; h� sÞV ðk� r; sÞ. (1.90)

Theorem 7. If w(x,y) ¼ xmyn,

W ðk; hÞ ¼ dðk� m; h� nÞ ¼ dðk� mÞdðh� nÞ. (1.91)

where

dðk� mÞ ¼�1; k ¼ m;&; h ¼ n

0; otherwise

Theorem 8. If

W ðx; yÞ ¼�vuðx; yÞ

vxvyðx; yÞ

vx

�;

then

W ðk; hÞ ¼Xkr¼0

Xhs¼0

ðr þ 1Þðk� r þ 1ÞUðr þ 1; h� sÞV ðk� r þ 1; sÞ.


From the above definition,

W ð0; 0Þ ¼�vuðx; yÞ

vxvyðx; yÞ

vx

�ð0;0Þ

¼ Uð1; 0ÞV ð1; 0Þ (1.92)

W ð1; 0Þ ¼ 11!0!

v

vx

�vuðx; yÞ

vxvyðx; yÞ

vx

�ð0;0Þ

¼ 11!0!

�v2uðx; yÞ

vx2vyðx; yÞ

vxþ vuðx; yÞ

vxv2yðx; yÞ

vx2

�ð0;0Þ

¼ 2Uð2; 0ÞV ð1; 0Þ þ 2Uð1; 0ÞV ð2; 0Þ;

(1.93)

W ð2; 0Þ ¼ 12!0!

v2

vx2

�vuðx; yÞ

vxvyðx; yÞ

vx

�ð0;0Þ

¼ 3Uð3; 0ÞV ð1; 0Þ þ 4Uð2; 0ÞV ð2; 0Þ þ 3Uð1; 0ÞV ð3; 0Þ;(1.94)

W ð0; 1Þ ¼ Uð1; 1ÞV ð1; 0Þ þ Uð1; 0ÞV ð1; 1Þ; (1.95)

W ð1; 1Þ ¼ 2Uð2; 1ÞV ð1; 0Þ þ 2Uð2; 0ÞV ð1; 1Þ þ 2Uð1; 1ÞV ð2; 0Þþ2Uð1; 0ÞV ð2; 1Þ;

(1.96)

W ð2; 2Þ ¼ 3Uð3; 2ÞV ð1; 0Þ þ 3Uð3; 1ÞV ð1; 1Þ þ 3Uð3; 0ÞV ð1; 2Þþ4Uð2; 2ÞV ð2; 0Þ þ 4Uð2; 1ÞV ð2; 1Þ þ 4Uð2; 0ÞV ð2; 2Þþ3Uð1; 2ÞV ð3; 0Þ þ 3Uð1; 1ÞV ð3; 1Þ þ 3Uð1; 0ÞV ð3; 2Þ.

(1.97)

Theorem 9. If

wðx; yÞ ¼ vuðx; yÞvy

vyðx; yÞvy

;

then

W ðk; hÞ ¼Xkr¼0

Xhs¼0

ðk� r þ 1Þðh� sþ 1ÞUðk� r þ 1ÞV ðr; h� sþ 1Þ.


W ð0; 0Þ ¼�vuðx; yÞ

vxvyðx; yÞ

vy

�ð0;0Þ

¼ Uð1; 0ÞV ð0; 1Þ; (1.98)


W ð1; 0Þ ¼ 11!0!

v

vx

�vuðx; yÞ

vyvyðx; yÞ

vy

�ð0;0Þ

¼ 11!0!

�v2uðx; yÞvxvy

vyðx; yÞvy

þ vuðx; yÞvy

v2yðx; yÞvxvy

�ð0;0Þ

Uð1; 1ÞV ð0; 1Þ þ Uð0; 1ÞV ð1; 1Þ;

(1.99)

W ð2; 0Þ ¼ 12!0!

v2

vx2

�vuðx; yÞ

vxvyðx; yÞ

vx

�ð0;0Þ

¼ Uð0; 1ÞV ð2; 1Þ þ Uð1; 1ÞV ð1; 1Þ þ Uð2; 1ÞV ð0; 1Þ;(1.100)

W ð0; 1Þ ¼ 2Uð0; 2ÞV ð0; 1Þ þ 2Uð0; 1ÞV ð0; 2Þ; (1.101)

W ð1; 1Þ ¼ 2Uð0; 2ÞV ð1; 1Þ þ 2Uð0; 1ÞV ð1; 2Þ þ 2Uð1; 2ÞV ð0; 1Þþ2Uð1; 1ÞV ð0; 2Þ;

(1.102)

W ð2; 2Þ ¼ 3Uð0; 3ÞV ð2; 1Þ þ 4Uð0; 2ÞV ð2; 2Þ þ 3Uð0; 1ÞV ð2; 3Þþ3Uð1; 3ÞV ð1; 1Þ þ 4Uð1; 2ÞV ð1; 2Þ þ 3Uð1; 1ÞV ð1; 3Þþ3Uð2; 3ÞV ð0; 1Þ þ 4Uð2; 2ÞV ð0; 2Þ þ 3Uð2; 1ÞV ð0; 3Þ.

(1.103)

Theorem 10. If

wðx; yÞ ¼ vuðx; yÞvx

vyðx; yÞvy

;

then

W ðk; hÞ ¼Xkr¼0

Xhs¼0

ðk� r þ 1Þðh� sþ 1ÞUðk� r þ 1; sÞV ðr; h� sþ 1Þ:


W ð0; 0Þ ¼ v

vx

�vuðx; yÞ

vxvyðx; yÞ

vy

�ð0;0Þ

¼ Uð1; 0ÞV ð0; 1Þ; (1.104)

W ð1; 0Þ ¼ 11!0!

v

vx

�vuðx; yÞ

vxvyðx; yÞ

vy

�ð0;0Þ

¼ 11!0!

�v2uðx; yÞ

vx2vyðx; yÞ

vyþ vuðx; yÞ

vxv2yðx; yÞvxvy

�ð0;0Þ

¼ 2Uð2; 0ÞV ð0; 1Þ þ Uð1; 0ÞV ð1; 1Þ;

(1.105)


W ð2; 0Þ ¼ 12!0!

v2

vx2

�vuðx; yÞ

vxvyðx; yÞ

vy

�ð0;0Þ

(1.106)

¼ 3Uð3; 0ÞV ð1; 0Þ þ 2Uð2; 0ÞV ð1; 1Þ þ Uð1; 0ÞV ð2; 1Þ; (1.107)

W ð0; 1Þ ¼ 2Uð1; 0ÞV ð0; 2Þ þ 2Uð1; 1ÞV ð0; 1Þ; (1.108)

W ð1; 1Þ ¼ 4Uð2; 0ÞV ð0; 2Þ þ 2Uð2; 1ÞV ð0; 1Þ þ 2Uð1; 0ÞV ð1; 2ÞþUð1; 1ÞV ð1; 1Þ;

(1.109)

W ð2; 2Þ ¼ 9Uð3; 0ÞV ð0; 3Þ þ 6Uð3; 1ÞV ð0; 2Þ þ 3Uð3; 2ÞV ð0; 1Þþ6Uð2; 0ÞV ð1; 3Þ þ 4Uð2; 1ÞV ð1; 2Þ þ 2Uð2; 2ÞV ð1; 1Þþ3Uð1; 0ÞV ð2; 3Þ þ 2Uð1; 1ÞV ð2; 2Þ þ Uð1; 2ÞV ð2; 1Þ.

(1.110)

Theorem 11. If w(x,y) ¼ u(x,y)y(x,y)u(x,y), then

W ðk; hÞ ¼Xkr¼0

Xk�r

t¼0

Xhs¼0

Xh�s

p¼0

Uðr; h� s� pÞV ðt; sÞUðk� r � t; PÞ.

From the definition of transform,

W ð0; 0Þ ¼ ½uðx; yÞyðx; yÞuðx; yÞ�ð0;0Þ ¼ Uð0; 0ÞV ð0; 0ÞUð0; 0Þ; (1.111)

W ð1; 0Þ ¼ 11!0!

v

vx½uðx; yÞyðx; yÞuðx; yÞ�ð0;0Þ

¼ Uð0; 0ÞV ð0; 0ÞUð1; 0Þ þ Uð0; 0ÞV ð1; 0ÞUð0; 0ÞþUð1; 0ÞV ð0; 0ÞUð0; 0Þ;

(1.112)

W ð2; 0Þ ¼ 12!0!

v2

vx2½uðx; yÞyðx; yÞuðx; yÞ�ð0;0Þ

¼ Uð0; 0ÞV ð0; 0ÞUð2; 0Þ þ Uð0; 0ÞV ð1; 0ÞUð1; 0ÞþUð0; 0ÞV ð2; 0ÞUð0; 0Þ þ Uð1; 0ÞV ð0; 0ÞUð1; 0ÞþUð1; 0ÞV ð1; 0ÞUð0; 0Þ þ Uð2; 0ÞV ð0; 0ÞUð0; 0Þ;

(1.113)

W ð0; 1Þ ¼ Uð0; 1ÞV ð0; 0ÞUð0; 0Þ þ Uð0; 0ÞV ð0; 1ÞUð0; 0ÞþUð0; 0ÞV ð0; 0ÞUð0; 1Þ; (1.114)


W ð0; 2Þ ¼ Uð0; 2ÞV ð0; 0ÞUð0; 0Þ þ Uð0; 1ÞV ð0; 0ÞUð0; 1ÞþUð0; 0ÞV ð0; 0ÞUð0; 2Þ þ Uð0; 1ÞV ð0; 1ÞUð0; 0ÞþUð0; 0ÞV ð0; 1ÞUð0; 1Þ þ Uð0; 0ÞV ð0; 2ÞUð0; 0Þ;

(1.115)

W ð1; 1Þ ¼ Uð0; 1ÞV ð0; 0ÞUð1; 0Þ þ Uð0; 0ÞV ð0; 0ÞUð1; 1ÞþUð0; 0ÞV ð0; 1ÞUð1; 0Þ þ Uð0; 1ÞV ð1; 0ÞUð0; 0ÞþUð0; 0ÞV ð1; 0ÞUð0; 1Þ þ Uð0; 0ÞV ð1; 1ÞUð0; 0ÞþUð1; 1ÞV ð0; 0ÞUð0; 0Þ þ Uð1; 0ÞV ð0; 0ÞUð0; 1ÞþUð1; 0ÞV ð0; 1ÞUð0; 0Þ.

(1.116)

Theorem 12. If

W ðx; yÞ ¼ uðx; yÞ vuðx; yÞvx

vyðx; yÞvx

;

then

W ðk; hÞ ¼Pkr¼0

Pk�r

t¼0

Phs¼0

Ph�s

p¼0ðt þ 1Þðk� r � t þ 1Þ

�Uðr; h� s� pÞV ðt þ 1; sÞUðk� r � t þ 1; PÞ.From the definition of transform,

W ð0; 0Þ ¼�uðx; yÞ vyðx; yÞ

vxvuðx; yÞ

vx

�ð0;0Þ

¼ þUð0; 0ÞV ð1; 0ÞUð1; 0Þ;

(1.117)

W ð1; 0Þ ¼ 11!0!

v

vx

�uðx; yÞ vyðx; yÞ

vxvuðx; yÞ

vx

�ð0;0Þ

¼ Uð1; 0ÞV ð1; 0ÞUð1; 0Þ þ 2Uð0; 0ÞV ð2; 0ÞUð1; 0Þ þ 2Uð0; 0ÞV ð1; 0ÞUð2; 0Þ;(1.118)

W ð2; 0Þ ¼ 12!0!

v2

vx2

�uðx; yÞ vyðx; yÞ

vxvuðx; yÞ

vx

�ð0;0Þ

¼ 3Uð0; 0ÞV ð1; 0ÞUð3; 0Þ þ 4Uð0; 0ÞV ð2; 0ÞUð2; 0ÞþU3ð0; 0ÞV ð3; 0ÞUð1; 0Þ þ 2Uð1; 0ÞV ð1; 0ÞUð2; 0Þþ2Uð1; 0ÞV ð2; 0ÞUð1; 0Þ þ Uð2; 0ÞV ð1; 0ÞUð1; 0Þ;

(1.119)

W ð0; 1Þ ¼ Uð0; 1ÞV ð1; 0ÞUð1; 0Þ þ Uð0; 0ÞV ð1; 0ÞUð1; 1ÞUð0; 0ÞV ð1; 1ÞUð1; 0Þ; (1.120)


W ð0; 2Þ ¼ Uð0; 2ÞV ð0; 0ÞUð0; 0Þ þ Uð0; 1ÞV ð0; 0ÞUð0; 1ÞþUð0; 0ÞV ð0; 0ÞUð0; 2Þ þ Uð0; 1ÞV ð0; 1ÞUð0; 0ÞþUð0; 0ÞV ð0; 1ÞUð0; 1Þ þ Uð0; 0ÞV ð0; 2ÞUð0; 0Þ;

(1.121)

W ð1; 1Þ ¼ 2Uð0; 1ÞV ð1; 0ÞUð2; 0Þ þ Uð0; 0ÞV ð1; 0ÞUð2; 1ÞþUð0; 0ÞV ð1; 1ÞUð2; 0Þ þ 2Uð0; 1ÞV ð2; 0ÞUð1; 0ÞþUð0; 0ÞV ð2; 0ÞUð1; 1Þ þ Uð0; 0ÞV ð2; 1ÞUð1; 0ÞþUð1; 1ÞV ð1; 0ÞUð1; 0Þ þ Uð1; 0ÞV ð1; 0ÞUð1; 1ÞþUð0; 1ÞV ð1; 1ÞUð1; 0Þ

(1.122)

Theorem 13. If

W ðx; yÞ ¼ uðx; yÞyðx; yÞ v2uðx; yÞvx2

;

then

W ðk; hÞ ¼Pkr¼0

Pk�r

t¼0

Phs¼0

Ph�s

p¼0ðK � r � t þ 1Þðk� r � t þ 1Þ

�Uðr; h� s� pÞV ðt; sÞUðk� r � t þ 2; pÞ.From the definition of transform,

W ð0; 0Þ ¼�uðx; yÞyðx; yÞ v

2uðx; yÞvx2

�ð0;0Þ

¼ Uð0; 0ÞV ð0; 0ÞUð2; 0Þ;

(1.123)

W ð1; 0Þ ¼ 11!0!

v

vx

�uðx; yÞyðx; yÞ v

2uðx; yÞvx2

�ð0;0Þ

¼ 6Uð0; 0ÞV ð0; 0ÞUð3; 0Þ þ 2Uð0; 0ÞV ð1; 0ÞUð2; 0Þþ2Uð1; 0ÞV ð0; 0ÞUð2; 0Þ;

(1.124)

W ð2; 0Þ ¼ 12!0!

v2

vx2

�uðx; yÞyðx; yÞ v

2uðx; yÞvx2

�ð0;0Þ

¼ 12Uð0; 0ÞV ð0; 0ÞUð4; 0Þ þ 6Uð0; 0ÞV ð1; 0ÞUð3; 0Þþ2Uð0; 0ÞV ð2; 0ÞUð2; 0Þ þ 6Uð1; 0ÞV ð0; 0ÞUð3; 0Þþ2Uð1; 0ÞV ð1; 0ÞUð2; 0Þ þ 2Uð2; 0ÞV ð0; 0ÞUð2; 0Þ;

(1.125)


W ð0; 1Þ ¼ 2Uð0; 1ÞV ð0; 0ÞUð2; 0Þ þ 2Uð0; 0ÞV ð0; 0ÞUð2; 1Þþ2Uð0; 0ÞV ð0; 1ÞUð2; 0Þ; (1.126)

W ð0; 2Þ ¼ 2Uð0; 2ÞV ð0; 0ÞUð2; 0Þ þ 2Uð0; 1ÞV ð0; 0ÞUð2; 1Þþ2Uð0; 0ÞV ð0; 0ÞUð2; 1Þ þ 2Uð0; 1ÞV ð0; 1ÞUð2; 0Þþ2Uð0; 0ÞV ð0; 1ÞUð2; 1Þ þ 2Uð0; 0ÞV ð0; 2ÞUð2; 0Þ;

(1.127)

W ð1; 1Þ ¼ Uð0; 1ÞV ð0; 0ÞUð1; 0Þ þ Uð0; 0ÞV ð0; 0ÞUð1; 1ÞþUð0; 0ÞV ð0; 1ÞUð1; 0Þ þ Uð0; 1ÞV ð1; 0ÞUð0; 0ÞþUð0; 0ÞV ð1; 0ÞUð0; 1Þ þ Uð0; 0ÞV ð1; 1ÞUð0; 0ÞþUð1; 1ÞV ð0; 0ÞUð0; 0Þ þ Uð1; 0ÞV ð0; 0ÞUð0; 1ÞþUð1; 0ÞV ð0; 1ÞUð0; 0Þ;

(1.128)

Example A. Consider the following linear PDE with the initial con-dition by Eqs. (1.130) and (1.131),

v2uvx2

� 3v2uvxvt

þ v2uvt2

¼ 0; (1.129)

I : C1 ðx; 0Þ ¼ x2; (1.130)

I : C2vuðx; 0Þ

vt¼ ex. (1.131)

Taking two-dimensional transform of Eq. (1.129), we obtain

ðkþ 1Þðkþ 2ÞUðkþ 2; hÞ � 3ðkþ 1Þðhþ 1ÞUðkþ 1; hþ 1Þ�4ðkþ 1Þðkþ 2ÞUðkþ 2; hÞ ¼ 0;

(1.132)

from Eq. (1.130),

Uð0; 0Þ ¼ 0 (1.133)

Uð1; 0Þ ¼ 0 (1.134)

Uð2; 0Þ ¼ 1 (1.135)

Uði; 0Þ ¼ 0; i ¼ 3; 4;.; n; (1.136)

and from Eq. (1.131),

Uð0; 1Þ ¼ 1; (1.137)

Uð1; 1Þ ¼ 1; (1.138)


Uð2; 1Þ ¼ 22!; (1.139)

Uði; 1Þ ¼ 2n!; i ¼ 3; 4;.; n. (1.140)

Substituting Eqs. (1.133)e(1.140) into Eq. (1.132), and by recursivemethod, the results corresponding to n /N are listed as

Uð0; 2Þ ¼ �18; (1.141)

Uð0; 2Þ ¼ 1396

; (1.142)

Uð1; 1Þ ¼ 1; (1.143)

Uð2; 1Þ ¼ 12. (1.144)

Substituting all U(k,h) into Eq. (1.88), we obtain the series solution formas follows:

uðx; tÞ ¼ t � 18t2 þ x2 þ xt þ 1

8x2t þ. (1.145)

and the analytical solution of the problem is given as follows for comparison:

uðx; tÞ ¼ 45

�eðxþðt=4ÞÞ � eðx�tÞþ x2 þ 1

4t2 (1.146)

Example B. Consider the following wave equation [81]:

v2wðx; tÞvt2

� 4v2wðx; tÞ

vx2¼ 0; 0 � x � 1; 0 < t (1.147)

with the boundary conditions

wð0; tÞ ¼ wð1; tÞ ¼ 0; 0 < t (1.148)

and initial conditions

wð0; tÞ ¼ sinðpxÞ; 0 � x � 1;

vwðx; 0Þvt

¼ 0; 0 � x � 1:(1.149)

Taking the differential transform of Eq. (1.147), then

ðkþ 2Þðkþ 1ÞW ði; kþ 2Þ ¼ 4ðiþ 2Þðiþ 1ÞW ðiþ 2; kÞ. (1.150)


From the initial condition given by Eq. (1.149),

wðx; 0Þ ¼XNi¼0

W ði; 0Þxi ¼ sinðpxÞ ¼XN

i¼1;3;.

ð� 1Þði�1Þ=2

i!pixi (1.151)

the corresponding spectra can be obtained as follows:

W ði; 0Þ ¼

8><>:0; for i is even;

ð� 1Þði�1Þ=2

i!pi; for i is odd

(1.152)

and from Eq. (1.149) it can be obtained that

vwðx; 0Þvt

¼XNi¼0

W ði; 1Þxi ¼ 0: (1.153)

Hence,

wði; 1Þ ¼ 0: (1.154)

Substituting Eqs. (1.153) and (1.154) to Eq. (1.150), all spectra can befound as

W ði; kÞ ¼

8><>:0; for i is even or k is odd;

2kð� 1Þðiþk�1Þ=2

i!k!piþk; for i is odd or k is even.

(1.155)

Therefore, the closed form of the solution can be easily written as

wðx; tÞ ¼PNi¼0

PNk¼0

W ði; kÞxitk ¼XNi¼0

XNk¼0

2k

k!i!ð�1Þðiþk�1Þ=2

piþkxitk

¼ PN

i¼1;3;.

1i!ð�1Þði�1Þ=2ðpxÞi

! XNk¼0;2;.

1k!ð�1ÞðkÞ=2ð2ptÞk

!¼ sinðpxÞcosð2ptÞ.

(1.156)

As seen in Fig. 1.7, when the solution of the PDE is calculated by usingDTM, it is clearly appears that the boundary conditions are provided, as theapproximate solutions remain close to the exact solutions.

Example C. Consider the one-dimensional unsteady heat conductionproblem as follows:

uxx ¼ 4ut (1.157)


with the initial and boundary conditions are

uð0; tÞ ¼ 0; uð2; tÞ ¼ 0; uðx; 0Þ ¼ 2 sinpx2. (1.158)

Taking the differential transform of Eq. (1.157), then

ðkþ 1Þðkþ 2ÞUðkþ 2; hÞ ¼ 4ðhþ 1ÞUðk; hþ 1Þand also

Uðk; hþ 1Þ ¼ ðkþ 1Þðkþ 2Þ4ðhþ 1Þ Uðkþ 2; hÞ (1.159)

Substituting U(k,h) in DTM principle, it can be obtained that the closedform of the solution is,

uðx; tÞ ¼ 2

�x� 1

3!x3 þ 1

5!x5 � 1

7!x7 þ 1

9!x9 �.::

��t

�1:31!

x� 2:53!

x3 þ 3:75!

x5 � 4:97!

x7 þ 5:119!

x9 �.::

�þ t2

�3:52!2

x� 3:5:73!4

x3 þ 3:7:95!2

x5 � 4:9:117!2

x7 þ.::

�þ.:

(1.160)

The analytic solution of this equation is

uðx; tÞ ¼ 2 sinpx2$e�p2t=16 (1.161)

and the related analytic graph is given by Fig. 1.8.

10.5

0

0

–0.5

–0.5

0.5x

1–1

–0.5

0

0.5

1

t

w

–1–1

Figure 1.7 The analytic solution of Example 1.


1.7 DIFFERENTIAL TRANSFORMATION METHODePADÉAPPROXIMATION

A Padé approximant is the ratio of two polynomials constructed from thecoefficients of the Taylor series expansion of a function u(x). The [L/M]Pade approximants to a function y(x) are given by [31,82].�

LM

�¼ PLðxÞ

QMðxÞ (1.162)

where PL(x) is polynomial of the degree of at most L, and QM(x) is a poly-nomial of the degree of at most M. The formal power series

yðxÞ ¼XNi¼0

aixi (1.163)

and

yðxÞ � PLðxÞQMðxÞ ¼ OðxLþMþ1Þ (1.164)

determine the coefficients of PL(x) and QM(x) by the equation. Since wecan clearly multiply the numerator and denominator by a constant andleave [L/M] unchanged, we imposed the normalization condition which is

QMð0Þ ¼ 1:0 (1.165)

Finally, PL(x) and QM(x) need to include noncommon factors. If thecoefficient of PL(x) and QM(x) are written as�

PLðxÞ ¼ p0 þ p1xþ p2x2 þ.þ pLx

L;

QMðxÞ ¼ q0 þ q1xþ q2x2 þ.þ qMxM ;(1.166)

2

0

–2

–5

–2.5

0

2.5

5

x

u 1

0.5

0

–0.5

–1

t

Figure 1.8 The approximate solution of Example C.


and using Eqs. (1.165) and (1.166), we may multiply (14) by QM(x), whichlinearizes the coefficient equations. Eq. (1.164) can be presented in moredetail as 8>>><>>>:

aLþ1 þ aLq1 þ.þ aL�Mþ1qM ¼ 0;

aLþ2 þ aLþ1q1 þ.þ aL�Mþ2qM ¼ 0;

«

aLþM þ aLþM�1q1 þ.þ aLqM ¼ 0;

(1.167)

8>>>>>><>>>>>>:

a0 ¼ p0;

a1 þ a0q1 ¼ p1;

a2 þ a1q1 þ a0q2 ¼ p2;

«

aL þ aL�1q1 þ.þ a0qL ¼ pL

(1.168)

To solve these equations, we start with Eq. (1.167), which is a set oflinear equations for all the unknown q0s. Once the q0s are known, thenEq. (1.168) gives an explicit formula for the unknown p0s, which completethe solution. If Eqs. (1.167) and (1.168) are nonsingular, then we can solvethem directly and obtain Eq. (1.169), where Eq. (1.169) holds, and if thelower index on a sum exceeds the upper, the sum is replaced by zero:

�LM

�¼

det

2666664aL�Mþ1 aL�Mþ2 / aLþ1

« « 1 «

aL aLþ1 / aLþMPLj¼M

aj�Mxj PL

j¼M�1aj�Mþ1x

j /PLj¼0

ajxj

3777775

det

26664aL�Mþ1 aL�Mþ2 / aLþ1

« « 1 «

aL aLþ1 / aLþM

xM xM�1 / 1

37775(1.169)

To obtain a diagonal Padé approximants of a different order such as[2/2], [4/4], or [6/6], the symbolic calculus software Maple is used.

Example A. We considered the heat transfer analysis in the unsteadytwo-dimensional squeezing nanofluid flow between the infinite parallelplates (Fig. 1.9). The two plates are placed at z ¼ �[(1 � at)1/2 ¼ �h(t). Fora > 0 , the two plates are squeezed until they touch t ¼ 1/a, and for a < 0


the two plates are separated. The viscous dissipation effect, the generation ofheat due to friction caused by shear in the flow, is retained. This effect is quiteimportant in the case when the fluid is largely viscous or flowing at a highspeed. This behavior occurs at high Eckert number (>>1). Further thesymmetric nature of the flow is adopted. The fluid is a water-based nanofluidcontaining Cu (copper) nanoparticles. The nanofluid is a two componentmixture with the following assumptions: incompressible; no chemical reac-tion; negligible viscous dissipation; negligible radiative heat transfer;nanosolid-particles; and the base fluid are in thermal equilibrium and no slipoccurs between them. The thermo-physical properties of the nanofluid aregiven in Table 1.4. The governing equations for momentum and energy inunsteady two-dimensional flow of a nanofluid are [31]:

vuvx

þ vvvy

¼ 0; (1.170)

rnf

�vuvt

þ uvuvv

þ vvuvy

�¼ � vp

vxþ mnf

�v2uvx2

þ v2uvy2

�; (1.171)

z

xy

D

2ℓ(1

– α

t)0.5

Figure 1.9 Nanofluid between parallel plates.

Table 1.4 Thermo-Physical Properties of Water and Nanoparticlesr(kg/m3) Cp(j/kgk) k(W/m$k)

Pure water 997.1 4179 0.613Copper (Cu) 8933 385 401


rnf

�vvvt

þ uvvvv

þ vvvvy

�¼ � vp

vyþ mnf

�v2vvx2

þ v2vvy2

�; (1.172)

vTvt

þ uvTvx

þ vvTvy

¼ knfðrCpÞnf

�v2Tvx2

þ v2Tvy2

�

þ mnf

ðrCpÞnf

4

�vuvx

�2

þ�vuvx

þ vuvy

�2!;

(1.173)

Here u and v are the velocities in the x- and y-directions respectively, Tis the temperature, P is the pressure, effective density (rnf), the effectivedynamic viscosity (mnf), the effective heat capacity (rCp)nf, and the effectivethermal conductivity knf of the nanofluid are defined as [31]:

rnf ¼ ð1� fÞrf þ frs; mnf ¼mf

ð1� fÞ2:5; ðrCpÞnf ¼ ð1� fÞðrCpÞf þ fðrCpÞs

knfkf

¼ ks þ 2kf � 2fðkf � ksÞks þ 2kf þ 2fðkf � ksÞ

(1.174)

The relevant boundary conditions are:

v ¼ vw ¼ dh=dt; T ¼ TH at y ¼ hðtÞ;v ¼ vu=vy ¼ vT=vy ¼ 0 at y ¼ 0:

(1.175)

We introduce these parameters:

h ¼ yhlð1� atÞ1=2

i; u ¼ ax½2ð1� atÞ� f 0ðhÞ;

v ¼ � alh2ð1� atÞ1=2

i f ðhÞ; q ¼ TTH

;

A1 ¼ ð1� fÞ þ frs

rf.

(1.176)


Substituting the above variables into Eqs. (1.171) and (1.172), and theneliminating the pressure gradient from the resulting equations give:

f iv � S A1ð1� fÞ2:5�hf 000 þ 3f 00 þ f 0f 00 � ff000 ¼ 0; (1.177)

Using Eq. (1.176), Eqs. (1.172) and (1.173) reduce to the followingdifferential equations:

q00 þ PrS

�A2

A3

�ðf q0 � hq0Þ þ Pr Ec

A3ð1� fÞ2:5�f 002 þ 4d2f 02

¼ 0; (1.178)

Here A2 and A3 are constants given by:

A2 ¼ ð1� fÞ þ fðrCpÞsðrCpÞf

; A3 ¼ knfkf

¼ ks þ 2 kf � 2 fðkf � ksÞks þ 2 kf þ 2 fðkf � ksÞ (1.179)

With these boundary conditions:

f ð0Þ ¼ 0; f 00ð0Þ ¼ 0;

f ð1Þ ¼ 1; f 0ð1Þ ¼ 0;

q0ð0Þ ¼ 0; qð1Þ ¼ 1:

(1.180)

where S is the squeeze number, Pr is the Prandtl number, and Ec is theEckert number, which are defined as:

S ¼ al2

2yf; Pr ¼ mf ðrCpÞf

rf kf; Ec ¼ rf

ðrCpÞf

�ax

2ð1� atÞ�2

; d ¼ lx;

(1.181)

Physical quantities of interest are the skin fraction coefficient andNusselt number, which are defined as:

Cf ¼mnf

�vuvy

�y¼hðtÞ

rnf v2w; Nu ¼

�lknf

�vTvy

�y¼hðtÞ

kTH. (1.182)

In terms of Eq. (1.176), we obtain

C*f ¼ l2=x2ð1� atÞRexCf ¼ A1ð1� fÞ2:5f 00ð1Þ;

Nu* ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi1� at

pNu ¼ �A3 q

0ð1Þ.(1.183)


Now DTM into governing equations has been applied. Taking thedifferential transforms of Eqs. (1.177)e(1.179) with respect to c andconsidering H ¼ 1 gives:

ðkþ 1Þðkþ 2Þðkþ 3Þðkþ 4ÞF½kþ 4� þ SA1ð1� fÞ2:5Pkm¼0

ðD½k� m� 1�ðmþ 1Þðmþ 2Þðmþ 3ÞF½mþ 3�Þ

� 3Sðkþ 1Þðkþ 2ÞF½kþ 2� � SA1ð1� fÞ2:5Pkm¼0

ððk� mþ 1ÞF½k� mþ 1�ðmþ 1Þðmþ 2ÞF½mþ 2�Þ

þ SA1ð1� fÞ2:5Xkm¼0

ðF½k� m�ðmþ 1Þðmþ 2Þðmþ 3ÞF½mþ 3�Þ ¼ 0;

D½m� ¼(1 m ¼ 1

0 ms 1(1.184)

F½0� ¼ 0; F½1� ¼ a1; F½2� ¼ 0; F½3� ¼ a2 (1.185)

ðkþ 1Þðkþ 2ÞQ½kþ 2� þ Pr$S$

�A2

A3

�Xkm¼0

ðF½k� m�ðmþ 1ÞQ½mþ 1�Þ

�Pr:S:

�A2

A3

�Xkm¼0

ðD½k� m�ðmþ 1ÞQ½mþ 1�Þ

þ Pr Ec

A3ð1� fÞ2:5Xkm¼0

ððk� mþ 1Þðk� mþ 2ÞF½k� mþ 2�ðmþ 1Þðmþ 2ÞF½mþ 2�Þ

þ 4Pr Ec

A3ð1� fÞ2:5 d2Xkm¼0

ððk� mþ 1ÞF½k� mþ 1�ðmþ 1ÞF½mþ 1�Þ;

D½m� ¼(1 m ¼ 1

0 ms 1

(1.186)

Q½0� ¼ a3; Q½1� ¼ 0 (1.187)


where F[k] andQ[k] are the differential transforms of f(h), q(h) and a1, a2, a3are constants, which can be obtained through boundary condition. Thisproblem can be solved as followed:

F½0� ¼ 0; F½1� ¼ a1; F½2� ¼ 0; F½3� ¼ a2;F½4� ¼ 0

F½5� ¼ 320

S A1ð1� fÞ2:5a2 þ 120

S A1ð1� fÞ2:5a1a2 þ 120a1a2;.

(1.188)

Q½0� ¼ a3; Q½1� ¼ 0;Q½2� ¼ �2Pr Ec

A3ð1� fÞ2:5d2a21;

Q½3� ¼ 0;

Q½4� ¼ 13

Pr Ec

A3ð1� fÞ2:5 Pr S�A2

A3

�a31 d

2 � 3Pr Ec

A3ð1� fÞ2:5 a22 � 2Pr Ec

A3ð1� fÞ2:5a1a2;

Q½5� ¼ 0;.

(1.189)

The above process is continuous. By substituting Eqs. (1.188) and(1.189) into the main equation based on DTM, it can be obtained that theclosed form of the solutions is:

FðhÞ ¼ a1hþ a2h3 þ�320

SA1ð1� fÞ2:5a2 þ 120

SA1ð1� fÞ2:5a1a2

þ 120a1a2

�h4 þ.

(1.190)

qðhÞ ¼ a3 þ

� 2Pr Ec

A3ð1� fÞ2:5d2a21

!h2

þ 13a31

Pr Ec

A3ð1� fÞ2:5 Pr S�A2

A3

�d2 � 3Pr Ec

Pr Ec

A3ð1� fÞ2:5a22

� 2Pr EcPr Ec

A3ð1� fÞ2:5a1a2!h4 þ.

(1.191)

by substituting the boundary condition from Eq. (1.180) into Eqs. (1.190)and (1.191), in point h ¼ 1, it can be obtained the values of a1, a2, a3. Bysubstituting obtained a1, a2, a3 into Eqs. (1.190) and (1.161), it can beobtained the expression of F(h) and Q(h). For example, for Cu-water


nanofluid, when Pr ¼ 6.2. Ec ¼ 0.05, d ¼ 0.1, S ¼ 0.1, and 4 ¼ 0.01following equations will be obtained:

f ðhÞ ¼ 1:4870h� 0:47373h3 � 0:01368h5 þ 0:0001428h6

þ 0:0002479h7 � 3:174� 10�7h8 (1.192)

qðhÞ ¼ 1:227� 0:0135h2 � 0:20073h4 � 0:00935h6 þ 0:000176h7

� 0:00374h8

(1.193)

by applying Padé approximation to Eqs. (1.192) and (1.193) (for Padé [6,6]accuracy), we have,

Pad�e ½6; 6�ðf ðhÞÞ ¼1:487026h� 0:024727h2 � 0:454146h3

þ 0:007668h4 � 0:0197056h5 þ 0:00043511h6

0:9999� 0:016628hþ 0:0131716h2 � 0:00014h3

þ 0:000146h4 � 0:00000133h5 þ 0:00000264h6

(1.194)

Pad�e ½6; 6�ðqðhÞÞ ¼1:22716þ 0:011948h� 0:2471506h2 þ 0:0004207h3

� 0:21051h4 � 0:002414h5 þ 0:03393h6

0:9999þ 0:009736h� 0:1903901h2 þ 0:0004h3�0:0100589h4 � 0:00037h5 þ 0:004023h6

(1.195)

Figs. 1.10 and 1.11 show the results of DTM and DTMePadé [6,6],respectively for solving Eqs. (1.177) and (1.178) in different Eckert and

Figure 1.10 Comparison of Differential Transformation Method and numerical resultsfor (a) q(h) when Pr ¼ 6.2, S ¼ 0.1, d ¼ 0.1, and 4 ¼ 0.01 (b) f(h) when Pr ¼ 6.2,Ec ¼ 0.05, d ¼ 0.1, and 4 ¼ 0.01.


squeeze numbers. As seen in these figures, DTM and DTMePadé havegood agreement with numerical method in wide range of Ec and Snumbers, also Table 1.5, which is designed for an especial case of these twofigures (Pr ¼ 6.2, Ec ¼ 0.05, d ¼ 0.1, S ¼ 0.1, 4 ¼ 0.01), confirms thatDTMePadé has an excellent congruity with the numerical procedure andis more reliable than DTM. Also it can be seen that by increasing theaccuracy of Padé from [3,3] to [6,6], DTM results have more agreementwith numerical results. Fig. 1.12 shows the effect of squeeze number (S) onnondimensional temperature and velocity profiles respectively. When twoplates move together, thermal boundary layer thickness increases as theabsolute magnitude of the squeeze number enhances. The positive andnegative squeeze numbers have different effects on the velocity profile. Forthe case of squeezing flow, the velocity increases due to an increase in theabsolute value of squeeze number when h < 0.5, while it decreases forh > 0.5. In other words, when S is minus, a is minus too and by decreasingits magnitude, its behavior is like squeezing flow so temperature profile isdecreased. The same treatment is observed in velocity profile. By increasingin positive values for S number, space between two plates decreased andconsequently velocity profile near the lower plate is decreased.

1.8 DIFFERENTIAL TRANSFORMATION METHOD ONSINGULAR TWO-POINT BOUNDARY VALUE PROBLEM

We consider the singular two-point boundary value problem (BVP) [83].

y00ðxÞ þ 1xy0ðxÞ þ qðxÞyðxÞ ¼ rðxÞ; 0 < x � 1 (1.196)

Figure 1.11 Comparison of Differential Transformation MethodePadé [6,6] andnumerical results for (a) q(h) when Pr ¼ 6.2, S ¼ 0.1, d ¼ 0.1, and 4 ¼ 0.01 (b) f(h) whenPr ¼ 6.2, Ec ¼ 0.05, d ¼ 0.1, and f ¼ 0.01.


Table 1.5 Comparison of Differential Transformation Method and Differential Transformation MethodePade Results for Cu-WaterNanofluid When Pr ¼ 6.2, Ec ¼ 0.05, d ¼ 0.1, S ¼ 0.1, and 4 ¼ 0.01

q(h) f(h)

h NUM DTMDTMePade[3,3]

DTMePade[6,6] NUM DTM

DTMePade[3,3]

DTMePade[6,6]

0.0 1.2293076 1.227169 1.227169 1.227169 0.00000 0.00000 0.0000 0.000000.1 1.2291512 1.227014 1.227010 1.227014 0.148563042 0.1482287 0.1482287 0.14822870.2 1.2284338 1.226307 1.225837 1.226307 0.294239491 0.2936109 0.2936109 0.29361090.3 1.2264082 1.224320 1.230774 1.224320 0.434130652 0.4332839 0.4332838 0.43328390.4 1.2218221 1.219828 1.228738 1.219828 0.565313773 0.5643523 0.5643513 0.56435230.5 1.2129029 1.211086 1.228413 1.211086 0.684830361 0.6838729 0.6838679 0.68387290.6 1.1973334 1.195796 1.228287 1.195796 0.789674916 0.788838 0.7888208 0.78883880.7 1.1722092 1.171051 1.228223 1.171051 0.876784230 0.8761652 0.8761122 0.87616520.8 1.1339720 1.133258 1.228185 1.133258 0.943027415 0.9426750 0.9425400 0.94267500.9 1.0783086 1.078024 1.228161 1.078027 0.985196818 0.9850864 0.9847780 0.98508641.0 1.0000000 0.999999 1.228144 1.000001 1.00000 0.9999999 0.9993538 1.000001

Introductionto

DifferentialTransform

ationMethod

45

Figure 1.12 Effect of squeeze number (S) on (a) Temperature profile and (b) Velocityprofile when Pr ¼ 6.2, Ec ¼ 0.5, d ¼ 0.1, and f ¼ 0.06.

subject to the boundary conditions�yð0Þ ¼ a1; yð1Þ ¼ b1; or

y0ð0Þ ¼ a2; yð1Þ ¼ b2(1.197)

where q(x) and r(x) are continuous functions on (0,1); and a1, a2, b1, and b2are real constants. These problems generally arise frequently in many areasof science and engineering, for example, fluid mechanics, quantum me-chanics, optimal control, chemical reactor theory, aerodynamics, reactionediffusion process, geophysics, etc.

Example A. Consider the following singular two-point BVP

y00ðxÞ þ 1xy0ðxÞ þ yðxÞ ¼ 5

4þ x2

16; 0 < x � 1 (1.198)

subject to the boundary conditions

y0ð0Þ ¼ 0; yð1Þ ¼ 1716

(1.199)

The exact solution of this problem is yðxÞ ¼ 1þ x2

16. The transformed

version of Eq. (1.198) isPkl¼0

dðl � 1Þðk� l þ 1Þðk� l þ 2ÞYðk� l þ 2Þ þ ðkþ 1ÞYðkþ 1Þþ

þPkl¼0

dðl � 1ÞYðk� lÞ � 54dðk� 1Þ � 1

16dðk� 3Þ ¼ 0

(1.200)


The transformed boundary conditions are

ðkþ 1ÞYðkþ 1Þ ¼ 0;PNk¼0

YðkÞ ¼ 1716

(1.201)

Using Eqs. (1.200) and (1.201) by taking N ¼ 6, the following systemfor k ¼ 0,., 5 is obtained:

Yð1Þ ¼ 0

4Yð2Þ þ Yð0Þ ¼ 54


16Yð4Þ þ Yð2Þ ¼ 116



(1.202)

Solving the above system and using the inverse transformation rule, weget the following series solution

yðxÞ ¼ 1þ x2

16(1.203)

Note that for N > 6 we evaluate the same solution, which is the exactsolution of Eq. (1.198) with the boundary conditions Eq. (1.199).

Example B. Consider the following singular two-point BVP�1� x

2

y00ðxÞ þ 3

2

�1x� 1

�y0ðxÞ þ

�x2� 1 yðxÞ

¼ 5� 29x2

þ 13x2

2þ 3x3

2� x4

2(1.204)

with following boundary conditions

yð0Þ ¼ 0; yð1Þ ¼ 0 (1.205)

The exact solution of this problem is y(x) ¼ x2 � x3. The transformedversion of Eq. (1.204) is


Xkl¼0

dðl � 1Þðk� l þ 1Þðk� l þ 2ÞY ðk� l þ 2Þ

� 12

Xkl¼0

dðl � 1Þðk� l þ 1Þðk� l þ 2ÞY ðk� l þ 2Þ þ 32ðkþ 1ÞY ðkþ 1Þþ

� 32

Xkl¼0

dðl � 1Þðk� l þ 1ÞY ðk� l þ 1Þ þ 12

Xkl¼0

dðl � 2ÞY ðk� lÞ

�Xkl¼0

dðl � 1ÞYðk� lÞ � 5dðk� 1Þ � 292dðk� 2Þ

� 132ðk� 3Þ � 3

2dðk� 4Þ þ dðk� 5Þ

2¼ 0

(1.206)

The transformed boundary conditions are

Yð0Þ ¼ 0;PNk¼0

YðkÞ ¼ 0 (1.207)

Using Eqs. (1.204) and (1.205) by taking N ¼ 6, the following systemfor k ¼ 0,., 5 is obtained:

Yð1Þ ¼ 0

�32Yð1Þ þ 5Yð2Þ þ Yð0Þ ¼ 5

�Yð1Þ þ 212Yð3Þ � 4Yð2Þ þ 1

2Yð0Þ ¼ �29

2

18Yð4Þ � 152Yð3Þ � Yð2Þ þ 1

2Yð1Þ ¼ 13

2552Yð5Þ � 12Yð4Þ þ 1

2Yð2Þ � Yð3Þ ¼ 3

2

39Yð6Þ � 352Yð5Þ þ 1

2Yð3Þ � Yð4Þ ¼ �1

2

(1.208)


REFERENCES[1] Hatami M, Ganji DD. Optimization of the longitudinal fins with different geometries

for increasing the heat transfer. In: ISER- 12th international conference on recenttrends in engineering and technology (ICRTET-2015), Kuala Lumpur, MalaysiaNovember 8, 2015; November 2015.

[2] Jalili Palandi S, Hatami M, Ghasemi SE, Ganji DD. Temperature distribution ofconvective fin with temperature-dependent internal heat generation and thermalconductivity using DTM. In: International conference on nonlinear modeling &optimization; August 2012.

[3] Jalili Palandi S, Ghasemi SE, Hatami M, Ganji DD. Efficient analytical approaches formotion of a spherical solid particle in plane couette fluid flow. In: International con-ference on nonlinear modeling & optimization; August 2012.

[4] Gorji M, Hatami M, Hasanpour A, Ganji DD. Nonlinear thermal analysis of solar airheater for the purpose of energy saving. January 2012. http://dx.doi.org/10.5829/idosi.ijee.2012.03.04.010.

[5] Ghasemi SE, Hatami M, Mehdizadeh Ahangar GHR, Ganji DD. Electro-hydrodynamic flow analysis in a circular cylindrical conduit using Least SquareMethod. Journal of Electrostatics January 2013;72(1). http://dx.doi.org/10.1016/j.elstat.2013.11.005.

[6] Hatami M, Ganji DD. Investigation of refrigeration efficiency for fully wet circularporous fins with variable sections by combined heat and mass transfer analysis. Inter-national Journal of Refrigeration January 2013;40. http://dx.doi.org/10.1016/j.ijrefrig.2013.11.002.

[7] Khavaji A, Ganji DD, Roshan N, Moheimani R, Hatami M, Hasanpour A. Slopevariation effect on large deflection of compliant beam using analytical approach.Structural Engineering & Mechanics November 2012;44(3). http://dx.doi.org/10.12989/sem.2012.44.3.405.

[8] Hatami M, Ganji DD. Heat transfer and flow analysis for SA-TiO2 non-Newtoniannanofluid passing through the porous media between two coaxial cylinders. Journalof Molecular Liquids December 2013;188:155e61. http://dx.doi.org/10.1016/j.molliq.2013.10.009.

[9] Hatami M, Hatami J, Ganji DD. Computer simulation of MHD blood conveying goldnanoparticles as a third grade non-Newtonian nanofluid in a hollow porous vessel.Computer Methods and Programs in Biomedicine November 2013;113(2). http://dx.doi.org/10.1016/j.cmpb.2013.11.001.

[10] Hatami M, Nouri R, Ganji DD. Forced convection analysis for MHD Al2O3-waternanofluid flow over a horizontal plate. Journal of Molecular Liquids November2013;187:294e301. http://dx.doi.org/10.1016/j.molliq.2013.08.008.

[11] Ghasemi SE, Hatami M, Ganji DD. Analytical thermal analysis of air-heating solarcollectors. Journal of Mechanical Science and Technology November 2013;27(11).http://dx.doi.org/10.1007/s12206-013-0878-0.

[12] Hatami M, Hasanpour A, Ganji DD. Heat transfer study through porous fins (Si3N4

and AL) with temperature-dependent heat generation. Energy Conversion andManagement October 2013;74:9e16. http://dx.doi.org/10.1016/j.enconman.2013.04.034.

[13] Ganji DD, Gorji M, Hatami M, Hasanpour A, Khademzadeh N. Propulsion andlaunching analysis of variable-mass rockets by analytical methods September2013;2(3):225e33. http://dx.doi.org/10.1016/j.jppr.2013.07.006.

[14] Hatami M, Ganji DD, Boubaker K. Temperature variations analysis forCondensed matter micro- and nanoparticles combustion burning in gaseous oxidizingmedia by DTM and BPES September 2013;2013(24). http://dx.doi.org/10.1155/2013/129571.


http://dx.doi.org/10.5829/idosi.ijee.2012.03.04.010

http://dx.doi.org/10.5829/idosi.ijee.2012.03.04.010

http://dx.doi.org/10.1016/j.elstat.2013.11.005

http://dx.doi.org/10.1016/j.elstat.2013.11.005

http://dx.doi.org/10.1016/j.ijrefrig.2013.11.002

http://dx.doi.org/10.1016/j.ijrefrig.2013.11.002

http://dx.doi.org/10.12989/sem.2012.44.3.405

http://dx.doi.org/10.12989/sem.2012.44.3.405

http://dx.doi.org/10.1016/j.molliq.2013.10.009


http://dx.doi.org/10.1016/j.cmpb.2013.11.001

http://dx.doi.org/10.1016/j.cmpb.2013.11.001


http://dx.doi.org/10.1007/s12206-013-0878-0

http://dx.doi.org/10.1016/j.enconman.2013.04.034


http://dx.doi.org/10.1016/j.jppr.2013.07.006

http://dx.doi.org/10.1155/2013/129571

http://dx.doi.org/10.1155/2013/129571

[15] Sheikholeslami M, Hatami M, Ganji DD. Analytical investigation of MHD nanofluidflow in a semi-porous channel. Powder Technology September 2013;246:327e36.http://dx.doi.org/10.1016/j.powtec.2013.05.030.

[16] Gorji M, Ganji DD, Hatami M, Hasanpour A, Hossein AK. Launching process forvariable-mass rockets by series solution. June 2013. http://dx.doi.org/10.18005/JMET0101003.

[17] Hatami M, Ganji DD. Thermal performance of circular convective-radiative porousfins with different section shapes and materials. Energy Conversion and ManagementDecember 2013;76:185e93. http://dx.doi.org/10.1016/j.enconman.2013.07.040.

[18] Hatami M, Sheikholeslami M, Ganji DD. Laminar flow and heat transfer of nanofluidbetween contracting and rotating disks by least square method. Powder TechnologyFebruary 2014;253:769e79. http://dx.doi.org/10.1016/j.powtec.2013.12.053.

[19] Hatami M, Ganji DD. Thermal and flow analysis of microchannel heat sink (MCHS)cooled by Cu-water nanofluid using porous media approach and least square method.Energy Conversion and Management February 2014;78:347e58. http://dx.doi.org/10.1016/j.enconman.2013.10.063.

[20] Hatami M, Ganji DD. Heat transfer and nanofluid flow in suction and blowing processbetween parallel disks in presence of variable magnetic field. Journal of MolecularLiquids February 2014;190:159e68. http://dx.doi.org/10.1016/j.molliq.2013.11.005.

[21] Sheikholeslami M, Hatami M, Ganji DD. Nanofluid flow and heat transfer in a rotatingsystem in the presence of a magnetic field. Journal of Molecular Liquids February2014;190:112e20. http://dx.doi.org/10.1016/j.molliq.2013.11.002.

[22] Hatami M, Ganji DD. Motion of a spherical particle in a fluid forced vortex by DQMand DTM. Particuology March 2014;16. http://dx.doi.org/10.1016/j.partic.2014.01.001.

[23] Hatami M, Domairry G. Transient vertically motion of a soluble particle in a New-tonian fluid media. Powder Technology February 2014;253:481e5. http://dx.doi.org/10.1016/j.powtec.2013.12.015.

[24] Hatami M, Ganji DD. Natural convection of sodium alginate (SA) non-Newtoniannanofluid flow between two vertical flat plates by analytical and numerical methods.Case Studies in Thermal Engineering March 2014;2:14e22. http://dx.doi.org/10.1016/j.csite.2013.11.001.

[25] Nouri R, Ganji DD, Hatami M. Unsteady sedimentation analysis of spherical particlesin Newtonian fluid media using analytical methods June 2014;3(2). http://dx.doi.org/10.1016/j.jppr.2014.05.003.

[26] Hatami M, Ganji DD. Thermal behavior of longitudinal convectiveeradiative porousfins with different section shapes and ceramic materials (SiC and Si3N4). CeramicsInternational June 2014;40(5):6765e75. http://dx.doi.org/10.1016/j.ceramint.2013.11.140.

[27] Sheikholeslami M, Hatami M, Ganji DD. Micropolar fluid flow and heat transfer in apermeable channel using analytical method. Journal of Molecular Liquids June2014;194:30e6. http://dx.doi.org/10.1016/j.molliq.2014.01.005.

[28] Hatami M, Ganji DD. MHD nanofluid flow analysis in divergent and convergentchannels using WRMs and numerical method. International Journal of NumericalMethods for Heat and Fluid Flow May 2014;24(5). http://dx.doi.org/10.1108/HFF-01-2013-0010.

[29] Hatami M, Ganji DD. Motion of a spherical particle on a rotating parabola usingLagrangian and high accuracy Multi-step Differential Transformation Method. PowderTechnology May 2014;258:94e8. http://dx.doi.org/10.1016/j.powtec.2014.03.007.

[30] Ahmadi AR, Zahmatkesh A, Hatami M, Ganji DD. A comprehensive analysis of theflow and heat transfer for a nanofluid over an unsteady stretching flat plate. PowderTechnology May 2014;258:125e33. http://dx.doi.org/10.1016/j.powtec.2014.03.021.


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[31] Domairry G, Hatami M. Squeezing Cu-water nanofluid flow analysis between parallelplates by DTM-Padé Method. Journal of Molecular Liquids May 2014;193:37e44.http://dx.doi.org/10.1016/j.molliq.2013.12.034.

[32] Ganji DD, Hatami M. Three weighted residual methods based on Jeffery-Hamel flow.International Journal of Numerical Methods for Heat Fluid Flow June 2014;24(3).http://dx.doi.org/10.1108/HFF-06-2012-0137.

[33] Hatami M, Sheikholeslami M, Hosseini M, Ganji DD. Analytical investigation ofMHD nanofluidflow in non-parallel walls. Journal of Molecular Liquids June2014;194:251e9. http://dx.doi.org/10.1016/j.molliq.2014.03.002.

[34] Hatami M, Sheikholeslami M, Ganji DD. Nanofluid flow and heat transfer in anasymmetric porous channel with expanding or contracting wall. Journal of MolecularLiquids July 2014;195:230e9. http://dx.doi.org/10.1016/j.molliq.2014.02.024.

[35] Hatami M, Sheikholeslami M, Domairry G. High accuracy analysis for motion of aspherical particle in plane Couette fluid flow by Multi-step Differential TransformationMethod. Powder Technology July 2014;260:59e67. http://dx.doi.org/10.1016/j.powtec.2014.02.057.

[36] Hatami M, Mehdizadeh Ahangar GHR, Ganji DD, Boubaker K. Refrigeration effi-ciency analysis for fully wet semi-spherical porous fins. Energy Conversion andManagement August 2014;84:533e40. http://dx.doi.org/10.1016/j.enconman.2014.05.007.

[37] Hatami M, Hosseinzadeh Kh, Domairry G, Behnamfar MT. Numerical study of MHDtwo-phase Couette flow analysis for fluid-particle suspension between moving parallelplates. Journal of the Taiwan Institute of Chemical Engineers September 2014;45(5).http://dx.doi.org/10.1016/j.jtice.2014.05.018.

[38] Ghasemi SE, Hatami M, Ganji DD. Thermal analysis of convective fin withtemperature-dependent thermal conductivity and heat generation. Case Studies inThermal Engineering November 2014;4. http://dx.doi.org/10.1016/j.csite.2014.05.002.

[39] Hatami M, Ganji DD, Jafaryar M, Farkhadnia F. Transient combustion analysis for Ironmicro-particles in a gaseous media by weighted residual methods (WRMs). CaseStudies in Thermal Engineering November 2014;4. http://dx.doi.org/10.1016/j.csite.2014.06.003.

[40] Mosayebidorcheh S, Sheikholeslami M, Hatami M, Ganji DD. Analysis of turbulentMHD Couette nanofluid flow and heat transfer using hybrid DTMeFDM. Partic-uology November 2014. http://dx.doi.org/10.1016/j.partic.2014.07.004S.

[41] Ghasemi E, Valipour P, Hatami M, Ganji DD. Heat transfer study on solid and porousconvective fins with temperature-dependent heat generation using efficient analyticalmethod. Journal of Central South University of Technology December 2014;21(12).http://dx.doi.org/10.1007/s11771-014-2465-7.

[42] Fakour M, Vahabzadeh A, Ganji DD, Hatami M. Analytical study of micropolar fluidflow and heat transfer in a channel with permeable walls. Journal of Molecular LiquidsJanuary 2015;204. http://dx.doi.org/10.1016/j.molliq.2015.01.040.

[43] Sheikholeslami M, Hatami M, Domairry G. Numerical simulation of two phase un-steady nanofluid flow and heat transfer between parallel plates in presence of timedependent magnetic field. Journal of the Taiwan Institute of Chemical EngineersJanuary 2015;46:43e50. http://dx.doi.org/10.1016/j.jtice.2014.09.025.

[44] Hatami M, Hatami J, Jafaryar M, Domairry G. Differential transformation method forNewtonian and Non-Newtonian fluids flow analysis: comparison with HPM andnumerical solution. Journal of the Brazilian Society of Mechanical Sciences andEngineering February 2015. http://dx.doi.org/10.1007/s40430-014-0275-3.










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[45] Rahimi-Gorji M, Pourmehran O, Hatami M, Ganji DD. Statistical optimization ofmicrochannel heat sink (MCHS) geometry cooled by different nanofluids using RSManalysis. European Physical Journal Plus February 2015;130(2). http://dx.doi.org/10.1140/epjp/i2015-15022-8.

[46] Hatami M, Ghasemi SE, Sarokolaie AK, Ganji DD. Study on Blood flow containingnanoparticles trough porous arteries in presence of magnetic field using analyticalmethods. Physica E Low-dimensional Systems and Nanostructures March 2015;70.http://dx.doi.org/10.1016/j.physe.2015.03.002.

[47] Dogonchi AS, Hatami M, Hosseinzadeh K, Domairry G. Non-spherical particlessedimentation in an incompressible Newtonian medium by Pade approximation.Powder Technology 2015;278:248e56.

[48] Dogonchi AS, Hatami M, Domairry G. Motion analysis of a spherical solid particle inplane Couette Newtonian fluid flow. Powder Technology April 2015;274. http://dx.doi.org/10.1016/j.powtec.2015.01.018.

[49] Pourmehran O, Rahimi-Gorji M, Hatami M, Sahebi SAR, Domairry G. Numericaloptimization of microchannel heat sink (MCHS) performance cooled by KKL basednanofluids in saturated porous medium. Journal of the Taiwan Institute of ChemicalEngineers May 2015;55. http://dx.doi.org/10.1016/j.jtice.2015.04.016.

[50] Atouei SA, Hosseinzadeh Kh, Hatami M, Ghasemi SE, Sahebi SAR, Ganji DD. Heattransfer study on convectiveeradiative semi-spherical fins with temperature-dependentproperties and heat generation using efficient computational methods. Applied Ther-mal Engineering June 2015;89. http://dx.doi.org/10.1016/j.applthermaleng.2015.05.084.

[51] Dogonchi AS, Hatami M, Hosseinzadeh K, Domairry G. Non-spherical particlessedimentation in an incompressible Newtonian medium by Padé approximation.Powder Technology July 2015;278. http://dx.doi.org/10.1016/j.powtec.2015.03.036.

[52] Mosayebidorcheh S, Hatami M, Mosayebidorcheh T, Ganji DD. Effect of periodicbody acceleration and pulsatile pressure gradient pressure on non-Newtonian bloodflow in arteries. Journal of the Brazilian Society of Mechanical Sciences and Engi-neering August 2015;38(3). http://dx.doi.org/10.1007/s40430-015-0404-7.

[53] Mosayebidorcheh S, Hatami M, Ganji DD, Mosayebidorcheh T,Mirmohammadsadeghi SM. Investigation of transient MHD Couette flow and heattransfer of Dusty fluid with temperature-dependent properties. Journal of AppliedFluid Mechanics September 2015;8(4).

[54] Ghasemi SE, Hatami M, Hatami J, Sahebi SAR, Ganji DD. An efficient approach tostudy the pulsatile blood flow in femoral and coronary arteries by Differential Quad-rature Method. Physica A: Statistical Mechanics and its Applications October 2015;443.http://dx.doi.org/10.1016/j.physa.2015.09.039.

[55] Hatami M, Ghasemi SE, Sahebi SAR, Mosayebidorcheh S, Ganji DD, Hatami J.Investigation of third-grade non-Newtonian blood flow in arteries under periodicbody acceleration using multi-step differential transformation method. AppliedMathematics and Mechanics November 2015;36(11). http://dx.doi.org/10.1007/s10483-015-1995-7.

[56] Mosayebidorcheh S, Hatami M, Mosayebidorcheh T, Ganji DD. Optimization analysisof convectiveeradiative longitudinal fins with temperature-dependent properties anddifferent section shapes and materials. Energy Conversion and Management November2015;106. http://dx.doi.org/10.1016/j.enconman.2015.10.067.

[57] Ghasemi SE, Zolfagharian A, Hatami M, Ganji DD. Analytical thermal study onnonlinear fundamental heat transfer cases using a novel computational technique.Applied Thermal Engineering December 2015;98:88e97. http://dx.doi.org/10.1016/j.applthermaleng.2015.11.120.


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[58] Ghasemi SE, Vatani M, Hatami M, Ganji DD. Analytical and numerical investigationof nanoparticle effect on peristaltic fluid flow in drug delivery systems. Journal ofMolecular Liquids March 2016;215:88e97. http://dx.doi.org/10.1016/j.molliq.2015.12.001.

[59] Rashidi MM. The modified differential transform method for solving MHDboundary-layer equations. Computer Physics Communications 2009;180(11):2210e7.

[60] Rashidi MM, Laraqi N, Sadri SM. A novel analytical solution of mixed convectionabout an inclined flat plate embedded in a porous medium using the DTM-Padé.International Journal of Thermal Sciences 2010;49(12):2405e12.

[61] Rashidi MM, Mohimanian Pour SA. A novel analytical solution of heat transfer of amicropolar fluid through a porous medium with radiation by DTM-Padé. HeatTransferdAsian Research 2010;39(8):575e89.

[62] Rashidi MM, Parsa AB, Bég OA, Shamekhi L, Sadri SM, Bég TA. Parametric analysisof entropy generation in magneto-hemodynamic flow in a semi-porous channel withOHAM and DTM. Applied Bionics and Biomechanics 2014;11(1e2):47e60.

[63] Rashidi MM, Keimanesh M. Using differential transform method and padé approx-imant for solving mhd flow in a laminar liquid film from a horizontal stretching surface.Mathematical Problems in Engineering 2010;2010.

[64] Erfani E, Rashidi MM, Parsa AB. The modified differential transform method forsolving off-centered stagnation flow toward a rotating disc. International Journal ofComputational Methods 2010;7(04):655e70.

[65] Rashidi MM, Laraqi N, Basiri Parsa A. Analytical modeling of heat convection inmagnetized micropolar fluid by using modified differential transform method. HeatTransferdAsian Research 2011;40(3):187e204.

[66] Rashidi MM, Hayat T, Basiri Parsa A. Solving of boundary-layer equations withtranspiration effects, governance on a vertical permeable cylinder using modified dif-ferential transform method. Heat TransferdAsian Research 2011;40(8):677e92.

[67] Rashidi MM, Rahimzadeh N, Ferdows M, Uddin MJ, Bég OA. Group theory anddifferential transform analysis of mixed convective heat and mass transfer from a hor-izontal surface with chemical reaction effects. Chemical Engineering Communications2012;199(8):1012e43.

[68] Rashidi MM, Hayat T, Keimanesh T, Yousefian H. A study on heat transfer in asecond-grade fluid through a porous medium with the modified differential transformmethod. Heat TransferdAsian Research 2013;42(1):31e45.

[69] Rashidi MM, Chamkh AJ, Keimanesh M. Application of multi-step differentialtransform method on flow of a second-grade fluid over a stretching or Shrinking Sheet.American Journal of Computational Mathematics 2011;6:119e28.

[70] Rashidi MM, Abelman S, Freidooni Mehr N. Entropy generation in steady MHDflow due to a rotating porous disk in a nanofluid. International Journal of Heat andMass Transfer 2013;62:515e25.

[71] Keimanesh M, Rashidi MM, Chamkha AJ, Jafari R. Study of a third grade non-Newtonian fluid flow between two parallel plates using the multi-step differentialtransform method. Computers &Mathematics with Applications 2011;62(8):2871e91.

[72] Rashidi MM, Mohimanian Pour SA, Laraqi N. A semi-analytical solution of micro-Polar flow in a porous channel with mass Injection by using differential transformmethod. Nonlinear Analysis Modelling and Control 2010;15(3):341e50 [ISI].

[73] Rashidi MM, Shahmohamadi H, Domairry G. Variational iteration method for solvingthree-dimensional NaviereStokes equations of flow between two stretchable disks.Numerical Methods for Partial Differential Equations 2011;27(2):292e301.

[74] Rashidi MM, Erfani E. The modified differential transform method for investigatingnano boundary-layers over stretching surfaces. International Journal of NumericalMethods for Heat and Fluid Flow 2011;21(7):864e83.




[75] Basiri Parsa A, Rashidi MM, Anwar Bég O, Sadri SM. Semi-Computational simulationof magneto-hemodynamic flow in a semi-porous channel using optimal homotopy anddifferential transform methods. Computers in Biology and Medicine 2013;43(9):1142e53.

[76] Rashidi MM, Sadri SM. Solution of the laminar viscous flow in a semi-porous channelin the presence of a uniform magnetic field by using the differential transform method.International Journal of Contemporary Mathematical Sciences 2010;5(15):711e20.

[77] Zhou K. Differential transformation and its applications for electrical circuits. Wuhan,China: Huazhong University Press; 1986.

[78] Mosayebidorcheh S, Sheikholeslami M, Hatami M, Ganji DD. Analysis of turbulentMHD Couette nanofluid flow and heat transfer using hybrid DTMeFDM. Partic-uology 2016;26:95e101.

[79] Jang M-J, Chen C-L, Liy Y-C. On solving the initial-value problems using the dif-ferential transformation method. Applied Mathematics and Computation2000;115(2):145e60.

[80] Ayaz F. On the two-dimensional differential transform method. Applied Mathematicsand Computation 2003;143(2):361e74.

[81] Bildik N, Konuralp A, Bek FO, Küçükarslan S. Solution of different type of the partialdifferential equation by differential transform method and Adomian’s decompositionmethod. Applied Mathematics and Computation 2006;172(1):551e67.

[82] Bervillier C. Status of the differential transformation method. Applied Mathematics andComputation 2012;218(20):10158e70.

[83] Ravi Kanth ASV, Aruna K. Solution of singular two-point boundary value problemsusing differential transformation method. Physics Letters A 2008;372:4671e3.


CHAPTER 2

Differential TransformationMethod in Advance

2.1 INTRODUCTION

Many phenomena in viscoelasticity, fluid mechanics, biology, chemistry,acoustics, control theory, psychology, and other areas of science can besuccessfully modeled by the use of fractional-order derivatives. That isbecause of the fact that, a realistic modeling of a physical phenomenonhaving dependence not only at the time instant, but also the previous timehistory can be successfully achieved by using fractional calculus. In previouschapter the principle of Differential Transformation Method (DTM) wasintroduced, which can be applied in most of mechanical engineeringproblems [1]. But in some mechanical problems such as eigenvalue prob-lem, higher-order initial problems, fractional integro-differential equations,etc., the governing equations are some complicated and cannot be solvedby the traditional DTM. This chapter introduces DTM for advanceproblems and contains the following sections:2.1 Introduction2.2 Differential Transformation Method for Higher-Order Initial Value

Problems2.3 Fractional Differential Transform Method2.4 Differential Transformation Method for Integro-Differential Equation2.5 Differential Transformation Method for Eigenvalue Problems2.6 Two-Dimensional Differential Transformation Method for Fractional

Order Partial Differential Equations2.7 Reduced Differential Transform Method2.8 Modified Differential Transformation Method

2.2 DIFFERENTIAL TRANSFORMATION METHOD FORHIGHER-ORDER INITIAL VALUE PROBLEMS

A. Consider the Second-Order Initial Value Problem [2]y00ðtÞ � 2y0ðtÞ þ 2yðtÞ ¼ expð2tÞsinðtÞ; 0 � t � 1 (2.1)



http://dx.doi.org/10.1016/B978-0-12-805190-0.00002-4

with initial conditions

yð0Þ ¼ �0:4 (2.2)

y0ð0Þ ¼ �0:6: (2.3)

As above, with u1(t) ¼ y(t) and u2(t) ¼ y0(t). Eq. (2.1) transformed intothe system of the first-order differential equation

u01ðtÞ ¼ u2ðtÞ; (2.4)

u02 ¼ expð2tÞsinðtÞ þ 2u2ðtÞ (2.5)

and the initial conditions Eqs. (2.2) and (2.3) become

u1ð0Þ ¼ �0:4 (2.6)

u2ð0Þ ¼ �0:6 (2.7)

Let h ¼ 0.1 and N ¼ 10. The differential equations of the systemEqs. (2.4) and (2.5) between ti and tiþ1 can be represented as

u01�t*� ¼ u2

�t*�

(2.8)

u02�t*� ¼ exp

�2t* þ 2ti

�sin�t* þ ti

�þ 2u1�t*�þ 2u2

�t*�; (2.9)

where t* ¼ t � ti. Taking the differential transformation of Eqs. (2.8) and(2.9), respectively, we get

U1iðkþ 1Þ ¼ 1ðkþ 1ÞU2iðkÞ (2.10)

and

U2iðkþ 1Þ ¼ 1ðkþ 1Þ

(expð2tiÞcosðtiÞ

Xkl¼0

2l

ðl!Þðk� lÞ! sin�p2

�ðk� lÞ

þ expð2tiÞsinðtiÞXkl¼0

2l

ðl!Þðk� lÞ! cos�p2

�ðk� lÞ

� 2u1iðkÞ þ 2u2iðkÞ);

(2.11)

with

U10ð0Þ ¼ �0:4; U20ð0Þ ¼ �0:6; (2.12)


The approximation of u1(t) ¼ y(t) on the grid points can be obtainedfrom Eqs. (2.10) and (2.8). The actual solution of Eqs. (2.1)e(2.3) is

yðtÞ ¼ u1ðtÞ ¼ 0:2 expð2tÞ þ sinðtÞ; (2.13)

Fig. 2.1, shows errors involved with different order of DTM, along withthe result obtained by the RungeeKutta fourth-order method.

B. Consider the Third-Order Initial Value Problemy000 ðtÞ þ 2y00ðtÞ � y0ðtÞ � 2yðtÞ ¼ expðtÞ; 0 � t � 3 (2.14)


yð0Þ ¼ 1: (2.15)

y0ð0Þ ¼ 2 (2.16)

y00ð0Þ ¼ 0 (2.17)

With U1(t) ¼ y(t), U2(t) ¼ y0(t), and U3(t) ¼ y00(t). Eq. (2.14) trans-formed into the system of the first-order differential equation

u01ðtÞ ¼ u2ðtÞ. (2.18)

u02ðtÞ ¼ u3ðtÞ; (2.19)

u03ðtÞ ¼ �2u3ðtÞ þ u2ðtÞ þ 2u1ðtÞ þ expðtÞ (2.20)

5.0E–7

0.0E+00.0 0.1 0.2 0.3 0.4

Time (t)0.5 0.6 0.7 0.8 0.9 1.0

1.0E–6

1.5E–6

2.5E–6

3.0E–6

3.5E–6

SolidDottedCircleStarSquare

RKM of order4DTM of order4DTM of order5DTM of order10DTM of order15

4.0E–6

4.5E–6

5.0E–6

2.0E–6

Err

or

Figure 2.1 Computational errors corresponding to different order of DifferentialTransformation Method for case A.

Differential Transformation Method in Advance 57

U1ð0Þ ¼ 1 (2.21)

U2ð0Þ ¼ 2 (2.22)

U3ð0Þ ¼ 0 (2.23)

Let h ¼ 0:2 and N ¼ 15. The differential equation of the systemEqs. (2.18)e(2.20) between ti and ti þ 1 can be represented as

u01�t*� ¼ u2

�t*�. (2.24)

u02�t*� ¼ u3

�t*�

(2.25)

u03�t*� ¼ �2u3

�t*�þ u2

�t*�þ 2u1

�t*�þ exp

�t* þ ti

�; (2.26)

where t* ¼ t � tiTaking the differential transformation of Eqs. (2.24)e(2.26),respectively, we get

u1iðkþ 1Þ ¼ 1ðkþ 1Þu2iðkÞ (2.27)

u3iðkþ 1Þ ¼ 1ðkþ 1Þ

�� 2u3iðkÞ þ u2iðkÞ þ 2u1iðkÞ þ 1

k!expðtiÞ

�; (2.28)

and

u2iðkþ 1Þ ¼ 1ðkþ 1Þu3iðkÞ (2.29)

with

u10ð0Þ ¼ 0;

u20ð0Þ ¼ 0; (2.30)

u30ð0Þ ¼ 0:

The approximation of u1t ¼ y(t) on the grid points can be obtained fromEq. (2.27). The actual solution of Eqs. (2.1)e(2.3) is

yðtÞ ¼ u1t ¼ 4336

expðtÞ þ 14expð�tÞ � 4

9expð�2tÞ þ 1

6½t expðtÞ�. (2.31)

yðtÞ ¼ u1t ¼ 4336

expðtÞ þ 14expð�tÞ � 4

9expð�2tÞ þ 1

6½t expðtÞ�. (2.32)

Fig. 2.2 shows errors involved with different order of DTM, along withthe result obtained by the RungeeKutta fourth-order method. As indi-cated in Figs. 2.1 and 2.2, the computational error decreases as the order of


Taylor series increases. The order of computational error corresponding tothe RungeeKutta method and the DTM of the same order is the same.

2.3 FRACTIONAL DIFFERENTIAL TRANSFORM METHOD

There are several approaches to the generalization of the notion of dif-ferentiation to fractional orders. The fractional differentiation inRiemanneLiouville sense is defined by [3].

Dqx0f ðxÞ ¼ 1

Gðm� qÞdm

dxm

24Z x

x0

f ðtÞðx� tÞ1þq�m dt

35 (2.33)

for m � 1 � q < m, m ˛ Zþ, x > x0. Let us expand the analytical andcontinuous function f(x) in terms of a fractional power series as follows:

f ðxÞ ¼XNk¼0

f ðkÞðx� x0Þka (2.34)

where a is the order of fraction and F(k) is the fractional differential trans-form off(x). Concerning the practical applications encountered in various

SolidDottedStarCircleSquare

RKM of order4DTM of order4DTM of order5DTM of order10DTM of order15

Time (t)0.0 0.5 1.0 1.5 2.0 2.5 3.0

Err

or

2.0E–4

0.0E+0

4.0E–4

6.0E–4

8.0E–4

1.0E–3

1.2E–3

1.4E–3

1.6E–3

1.8E–3

2.0E–3

Figure 2.2 Computational errors corresponding to different order of DifferentialTransformation Method for case B.


branches of science, the fractional initial conditions are frequently not avail-able, and it may not be clear what their physical meaning is. Therefore, thedefinition in Eq. (2.33) should be modified to deal with integer orderedinitial conditions as follows:

Dqx0

"f ðxÞ �

Xm�1

k¼0

1k!ðx� x0Þkf ðkÞð0Þ

#

¼ 1Gðm� qÞ

dm

dxm

8>>>>><>>>>>:Z x

0

26664f ðtÞ �

Xm�1

k¼0

1k!ðt � x0Þkf ðkÞð0Þ

ðx� tÞ1þq�m dt

377759>>>>>=>>>>>;

(2.35)

Since the initial conditions are implemented to the integer order de-rivatives, the transformation of the initial conditions is defined as follows:

FðkÞ ¼

8>>>>>>><>>>>>>>:

ifka˛zþ/

1�ka

!

"d

kaf ðxÞdx

ka

#x¼x0

ifka;zþ/0

for k ¼ 0; 1; 2;.:ðnf� 1Þ

(2.36)

where, n is the order of fractional differential equation considered. UsingEqs. (2.33) and (2.34), the theorems of fractional differential transformmethod (FDTM), are introduced below.

Theorem 1. If f(x) ¼ g(x) � h(x), then F(k) ¼ G(k) � H(k).Theorem 2. If f ðxÞ ¼ gðxÞhðxÞ; then FðKÞ ¼PK

t¼0GðlÞHðk� 1ÞTheorem 3. If f(x) ¼ g1(x)g2(x),.,gn�1(x)gn(x), then

FðkÞ ¼Xkkn�1

Xkn�1

kn�2

.Xk3k2¼0

Xk2k1¼0

G1ðK1ÞG2ðK2 � K1Þ.

Gn�1ðkn�1 � kn�2ÞGNðK � KN�1Þ

Theorem 4. If f(x) ¼ (x � x0)p, then F(k) ¼ d(k � ap), where,

dðkÞ ¼�1/if ; k ¼ 0

0/if ; ks0


Theorem 5. If ðxÞ ¼ Dqx0 ½gðxÞ�; then FðkÞ ¼

G

�qþ 1þ

ka

G

�1þ

ka

GðK þ aqÞ

Theorem 6. For the production of fractional derivatives in the most

general form, if f ðxÞ ¼ dq1

dxq1½g1ðxÞ� d

q2

dxq2½g2ðxÞ�:: d

qn�1

dxqn�1½gn�1ðxÞ�

dqn

dxqn½gnðxÞ�; then:

FðKÞ ¼Xkkn�1¼0

Xkn�1

kn�2¼0

.Xk3k2¼0

Xk2k1¼0

Gðq1 þ 1þ k1=aÞGð1þ k1=aÞ

Gðq2 þ 1þ k2 � k1=aÞGð1þ k2 � k1=aÞ

.Gðqn�1 þ 1þ kn�1 � kn�2=aÞ

Gð1þ k2 � k1=aÞ � Gðqn þ 1þ ðk� kn�1Þ=aÞGð1þ ðk� kn�1Þ=aÞ

G1ðK1 þ aq1ÞG2ðK2 � K1 þ aq2Þ.GN�1ðKN�1 � KN�2 þ aqn�1Þ�GN�1ðK � KN�1 þ aqnÞ

where, aqi ˛ zþ, for, i ¼ 1, 2, 3,.,n.Proofs of all above theorems are presented in Ref. [3]. Following some

examples are discussed.Example A. Now, let us consider the BagleyeTorvik equation that

governs the motion of a rigid plate immersed in a Newtonian fluid.

Ad2xdt2

þ Bd32x

dx32þ cx ¼ f ðtÞ (2.37)

It is considered the case f(t) ¼ C(1 þ t), A ¼ 1; B ¼ 1; and C ¼ 1, withthe following boundary conditions:

xð0Þ ¼ 1; and; x0ð0Þ ¼ 1 (2.38)

Selecting the order of fraction as alpha ¼ 2, the boundary conditions aretransformed by using Eq. (2.36) as follows:

xð0Þ ¼ 1; xð1Þ ¼ 0; xð2Þ ¼ 1; and xð3Þ ¼ 0 (2.39)

By using theorems 4 and 5, the transform of Eq. (2.37) leads to thefollowing recurrence relation:

Xðkþ 4Þ¼�Gð5=2þ k=2ÞXðkþ 3ÞþGð1þ k=2Þ½XðkÞ� dðk� 2Þ�dðkÞ�Gð3þ k=2Þ

(2.40)


Using Eqs. (2.39) and (2.40), X(k) is evaluated up to a certain number ofterms and then using the inverse transformation rule, x(t) is evaluated asfollows:

XðtÞ ¼ 1þ t (2.41)

Example B. Consider the composite fractional oscillation equation [3]

d2udt2

� adaudta

� bu ¼ 8/t_0; 03a � 2 (2.42)

with the initial conditions

Uð0Þ ¼ 0 and u0ð0Þ ¼ 0 (2.43)

Taking a ¼ b ¼ �1 and using theorems 4 and 5, Eq. (2.42) can betransformed as follows:

uðkþ 2bÞ ¼ �Gðaþ 1þ k=bÞuðkþ baÞ þ Gð1þ k=bÞ½uðkÞ � 8dðkÞ�Gð3þ k=bÞ

(2.44)

where b is the unknown value of the fraction. The conditions in Eq. (2.43)can be transformed by using Eq. (2.36) as follows:

uðkÞ ¼ 0; for k ¼ 0; 1;.; 2b� 1 (2.45)

Example C. We consider the equation [3].

D2:2xðtÞ þ 1:3D1:5xðtÞ þ 2:6xðtÞ ¼ sinð2tÞ (2.46)

with the initial conditions

xð0Þ ¼ x0ð0Þ ¼ x00ð0Þ ¼ 0 (2.47)

By choosing a ¼ 10 and using theorem 4, Eq. (2.46) can be transformedas follows:

xðkþ 22Þ ¼ Gð1þ 0:1kÞ½sðkÞ � 2:6xðkÞ� � 1:3Gð2:5þ 0:1kÞxðkþ 15ÞGð3:2þ 0:1kÞ

(2.48)

where S(k) is the fractional differential transform of sin(2t) that can be eval-uated using Eq. (2.34) as

sðkÞ ¼XNi¼0

ð�1Þi22iþ1

ð2iþ 1Þ! d½k� 10ð2iþ 1Þ� (2.49)


The conditions in Eq. (2.47) can be transformed by using Eq. (2.36) as

XðkÞ ¼ 0 k ¼ 0; 1; 2;.21 (2.50)

Using the inverse transformation rule in Eq. (2.34) the following seriessolution is obtained:

xðtÞ ¼ 285613600000

t6 þ 2Gð21=6Þt

16=5 � 135Gð49=10Þt

39=10

þ 16950Gð28=5Þt

23=5 � 8Gð21=5Þt

26=5� 2197500Gð63=10Þt

53=10

� 265Gð32=5Þt

27=5þ 525Gð69=10Þt

59=10þ.

(2.51)

By using a mathematical software package, x(t) is evaluated up toN ¼ 1000 and plotted in Fig. 2.3.

Example D. Lastly, the fractional Ricatti equation is considered that isfrequently encountered in optimal control problems

dbydtb

¼ 2y� y2 þ 1; 03b � 1 (2.52)

with the initial condition

Yð0Þ ¼ 0 (2.53)

by using theorems 2, 4, and 5 will be as follows:

Yðkþ abÞ ¼ Gð1þ k=aÞGðbþ 1þ k=aÞ

"2YðkÞ �

Xkk1¼0

Yðk1ÞYðk� k1Þ þ dðkÞ#

(2.54)

0–0.4

–0.3

–0.2

–0.1

0

0.1

0.2

0.3

2 4 6t

x (t)

8 10

Figure 2.3 Time response of x(t) for N ¼ 1000 terms.


The condition in Eq. (2.53) is transformed by using Eq. (2.36) asfollows:

yðkÞ ¼ 0; for; k ¼ 0; 1;.:;ab� 1 (2.55)

For the values of b ¼ 1/2 and a ¼ 2, k ¼ 6y(t) is obtained as follows:

yðtÞ ¼ 2ffiffiffip

p t1=2 þ 2t þ 16ðp� 1Þ3p3=2

t3=2 þ p� 4p

t2 � 32ð3p2 þ 44p� 32Þ45p5=2

t5=2

þ�1289p2

� 719p

� 374

t3 þ/

(2.56)

2.4 DIFFERENTIAL TRANSFORMATION METHOD FORINTEGRO-DIFFERENTIAL EQUATION

Consider the fractional-order integro-differential equation of the form [4].

DqyðxÞ ¼ f ðxÞ þZ x

ak1ðx; tÞyðtÞdt þ

Z b

ak2ðx; tÞyðtÞdt (2.57)

With the nonlocal boundary conditions,Xmj¼1

�gijy

ðj�1ÞðaÞ þ hijyðj�1ÞðbÞ�þ li

Z b

aHiðtÞyðtÞdt ¼ di; i ¼ 1; 2;.;m

(2.58)

where Dq denotes a differential operator with fractional order q, f(x) andki(x, t) (i ¼ 1, 2) are holomorphic functions, Hi(t) is a continuous function.gij, hij, li, and di(i ¼ 1, 2,.,m) are constants, and y(x) is a function of classC (a class of functions that are piecewise continuous on J 0 ¼ (0, N) andintegrable on any finite subinterval J ¼ [0, N)). There are various typesof definition for the fractional derivative of order q > 0; the mostcommonly used definitions among various definitions of fractional deriva-tives of order q > 0 are the RiemanneLiouville and Caputo formulas, oneswhich use fractional integrations and derivatives of the whole order. Thedifference between the two definitions is in the order of evaluation.RiemanneLiouville fractional integration of order q is defined as

J qx0 f ðxÞ ¼1

GðqÞZ x

x0

ðx� tÞq�1f ðtÞdt; q > 0; x > 0 (2.59)


The following equations define RiemanneLiouville and Caputo frac-tional derivatives of order q, respectively:

Dqx0f ðxÞ ¼ dm

dxm�Jm�qx0

f ðxÞ� (2.60)

Dq*x0

f ðxÞ ¼ Jm�qx0

dm

dxmf ðxÞ

�(2.61)

where m � 1 � q < m and m ˛ N. From Eqs. (2.59) and (2.60), we have

Dqx0f ðxÞ ¼ 1

Gðm� qÞdm

dxm

Z x

x0

ðx� tÞm�q�1f ðtÞdt; x > x0 (2.62)

Consider the fractional-order integro-differential Eq. (2.57) with thenonlocal boundary conditions (2.58), where we assumed that the functionski(x, t), (i ¼ 1, 2) and f(x) are holomorphic and y(x) is a function of class C;hence ki(x,t), (i ¼ 1,2) can be approximated by separable functions.Therefore one can write kiðx; tÞ ¼

Pnj¼0 uijðxÞyijðtÞ; ði ¼ 1; 2Þ and then

Z b

akiðx; tÞyðtÞdt ¼

Xnj¼0

Z b

auijðxÞyðtÞyðtÞdt ¼

Xnj¼0

uijðxÞZ b

ayijðtÞyðtÞdt

(2.63)

By using FDTM, Eq. (2.57) is transformed to the following recurrencerelation:

Yðkþ aqÞ¼G

�1þ k

a

G

�1þ qþ k

a

0@FðkÞ þ RðkÞ þ

Xnj¼0

U2jðkÞZ b

ay2jðtÞyðyÞdt

1A;

k ¼ 0; 1;.;N

(2.64)


where N is an arbitrary natural number, R(k) is the transformationof

R xak1ðx; tÞyðtÞdt, and Y(0) ¼ y(a), Y(a) ¼ y0(a),

.Y

�ðm� 1Þa ¼ yðm�1ÞðaÞ

ðm�1Þ!

are the unknowns to be determined. If we set,

bj ¼Z b

ay2jðtÞyðtÞdt (2.65)

then Eq. (2.64) implies that

Yðkþ aqÞ ¼ LðkÞ AðkÞ þ

Xnj¼0

U2jðkÞbj

!; k ¼ 0; 1;.;N (2.66)

where

LðkÞ ¼G

�1þ k

a

G

�1þ qþ k

a

; AðkÞ ¼ FðkÞ þ RðkÞ (2.67)

By using inverse transformation on both sides of Eq. (2.66), y(x) isobtained as

yðxÞ ¼XaqþN

k¼0

YðkÞðx� aÞka

¼Xm�1

k¼0

Y ðkÞ

k!ðx� aÞk þ

XNk¼0

AðkÞ þ

Xnj¼0

U2jðkÞbj

!LðkÞðx� aÞaqþk

a

(2.68)

Substituting y(x) in the boundary conditions, we obtain, for i ¼ 1,2,.,m


yðaÞ24gi1 þ hi1 þ li

Z b

aHiðtÞdt

35þ y0ðaÞ24gi2 þ ðb� aÞhi1 þ hi2 þ li

Z b

aHiðtÞðt � aÞdt

35þ.

þ yðm�1ÞðaÞ"gim þ hi1

ðb� aÞðm�1Þ

ðm� 1Þ! þ hi2ðb� aÞðm�2Þ

ðm� 2Þ! þ hi3ðb� aÞðm�3Þ

ðm� 3Þ! þ.

þ him�1ðb� aÞ þ him þ li

Z b

aHiðtÞ ðt � aÞðm�1Þ

ðm� 1Þ! dt

35þ b0

"hi1

XNk¼0

U20ðkÞLðkÞðb� aÞaqþka þ hi2

XNk¼0

�aqþ k

a

U20ðkÞLðkÞðb� aÞaqþk

a �1 þ.

þ him

XNk¼0

�aqþ k

a

�aqþ k

a� 1

�aqþ k

a� mþ 2


a �mþ1

þ li

Z b

aHiðtÞ

XNk¼0

U20ðkÞLðkÞðt � aÞaqþka dt

35þ b1

"hi1

XNk¼0

U21ðkÞLðkÞðb� aÞaqþka þ hi2

XNk¼0

�aqþ k

a


a �1 þ.

þ him

XNk¼0

�aqþ k

a

�aqþ k

a� 1

.

�aqþ k

a� mþ 2


a �mþ1

þ li

Z b

aHiðtÞ

XNk¼0

U21ðkÞLðkÞðt � aÞaqþka dt

35þ.

þ bn

"hi1

XNk¼0

U2nðkÞLðkÞðb� aÞaqþka þ hi2

XNk¼0

�aqþ k

a

U2nðkÞLðkÞðb� aÞaqþk

a �1 þ.

þ him

XNk¼0

�aqþ k

a

�aqþ k

a� 1

.

�aqþ k

a� m� 2

U2nðkÞLðkÞðb� aÞaqþk

a �mþ1

þ li

Z b

aHiðtÞ

XNk¼0

U2nðkÞLðkÞðt � aÞaqþka dt

35¼ �

hi1

XNk¼0

LðkÞAðkÞðb� aÞaqþka þ hi2

XNk¼0

�aqþ k

a

LðkÞAðkÞðb� aÞaqþk

a�1 þ.

þ him

XNk¼0

�aqþ k

a

�aqþ k

a� 1::

�aqþ k

a� mþ 2

LðkÞAðkÞðb� aÞaqþk

a �mþ1

þ li

Z b

aHiðtÞ

XNk¼0

AðkÞLðkÞðt � aÞaqþka

35þ di

(2.69)

On the other hand; bj ¼R ba y2iðtÞyðtÞdt. Substituting y(x) from

Eq. (2.68) and rearranging the terms, for j ¼ 0, 1, 2,.,n yields


bj

0@Z b

ay2jðtÞ

XNk¼0

U2jðkÞLðkÞðt � aÞaqþka dt � 1

1A

þXNl¼0lsj

bl

Z b

ay2jðtÞ

XNk¼0

LðkÞU21ðkÞðt � aÞaqþka dt

!

¼ �Z b

ay2jðtÞ

Xm�1

k¼0

yðkÞðaÞk!

ðt � aÞk þXNk¼0

AðkÞLðkÞðt � aÞaqþka

!dt

(2.70)

Example A. Consider the following linear fractional integro-differential equation with the given nonlocal condition

D12yðxÞ ¼ �x2

ex

3yðxÞ � 1

2x2 þ 1

G

�32

x12 þ ex

Z x

0tyðtÞdt þ

Z 1

0x2yðtÞdt

(2.71)

yð0Þ þ yð1Þ � 3Z 1

0tyðtÞdt ¼ 0 (2.72)

where the order of fraction is a ¼ 2 . If we set b0 ¼R 10 yðtÞdt, then

Eq. (2.71) is transformed to

Yðkþ 1Þ ¼G

�1þ k

2

G

�32þ k2

� 13

Xkn¼0

Xnj¼0

dðj � 4ÞEðn� jÞYðj � nÞ

� 12dðk� 4Þ þ 1

G

�32

dðk� 1Þ

þ 2Xki¼2

1j

Xi�2

m¼0

dðm� 2ÞYðj � m� 2ÞEðk� jÞ þ dðk� 4Þb0

!.

where E(k) denotes the transformation of y(x) that can be expressed by us-ing Section 2.3 as follows:

EðkÞ ¼

8>>><>>>:0;

ka;Zþ

1�ka

!

;ka˛Zþ (2.73)


Thus LðkÞ ¼G

�1þ k

2

G

�32þ k2

; U20 ¼ ðkÞ ¼ dðk� 4Þ;

and

AðkÞ ¼ � 13

Xkn¼0

Xnj¼0

dðj � 4ÞEðn� jÞYðk� nÞ � 12dðk� 4Þ

þ 1

G

�32

dðk� 1Þ

þ2Xki¼2

1j

Xj�2

m¼0

dðm� 2ÞYðj � m� 2Þ � Eðk� jÞ:

Now, we solve the system

12yð0Þ þ 2

3G

�72

b0 þXNk¼0

k� 15þ k

LðkÞAðkÞ ¼ 0

yð0Þ þ

0BBB@ 4

7G

�72

� 1

1CCCAb0 þXNk¼0

23þ k

LðkÞAðkÞ ¼ 0

(2.74)

and obtain the following approximation for Y(0) ¼ y(0) and b0. ForN ¼ 20: Y(0) ¼ y(0) ¼ �0.96658 � 10�6 and b0 ¼ 0.5 þ 4.2 � 10�6.The exact values of y(0) and b0 are 0 and 0.5, respectively, and the exactsolution is y(x) ¼ x. For N ¼ 20, the approximate solution is

yðxÞ ¼ �0:96658� 10�6 þ xþ 0:24306� 10�5x52 � 0:831� 10�7x

72 � 0:36933� 10�7x

92

� 0:16178� 10�6x5 � 0:11192� 10�7x112 � 0:10295� 10�6x6 � 0:25827� 10�8x

132

þ 0:157� 10�6x7 þ 0:37576� 10�7x152 � 0:10389� 10�7x8 þ 0:64758� 10�7x

172

þ 0:89992� 10�8x9 þ 0:13509� 10�7x192 þ 0:73501� 10�7x10 þ.:

(2.75)


Example B. Consider the following linear fractional integro-differential equation with the given nonlocal condition

D13yðxÞ ¼ 3

2x32

G

�32

� 1þ ex2 � x2ex

2 þZ x

0x2extyðtÞdt (2.76)

yð0Þ þ 2yð1Þ þ 3Z 1

0tyðtÞdt ¼ 3 (2.77)

with exact solution y(x) ¼ x. We have

kðx; tÞzXMi¼0

x2þiti

i!(2.78)

By using FDTM, Eq. (2.76) is transformed to the following recurrencerelation

Yðkþ 1Þ ¼G

�1þ k

3

G

�43þ k3

0B@ 3

2G

�23

dðk� 2Þ � dðkÞ � E1ðkÞ �Xkk1¼0

dðk1 � 6Þ

�E1ðk� k1Þ þ 3Xkk1¼3

1k1Yðk1 � 3Þdðk� k1 � 6Þ þ 3

Xkk1¼3

Xk1�3

k2¼0

�XMk3¼1

1k1Yðk1 � k2 � 3Þ

�1k3!

dðk2 � 3k3Þdðk� k1 � 3ðk3 � 2ÞÞ1CA

(2.79)

where E1(k) denotes to the transformation of ex2that can be expressed by

using Section 2.3 as follows:

E1ðkÞ ¼

8>>><>>>:0;

k2a

;Zþ

1�k2a

!

;k2a

˛Zþ (2.80)

Thus we have LðkÞ ¼G

�1þ k

3

G

�43þ k3

and


AðkÞ ¼

3

2G

�32

dðk� 2Þ � dðkÞ þ E1ðkÞ �Xkk1¼0

dðk1 � 6ÞE1ðk� k1Þ

þ3Xkk1¼3

1k1Yðk1 � 3Þdðk� k1 � 6Þ þ 3

Xkk1¼3

Xk1�3

k2¼0

XMk3¼1

� 1k1Yðk1 � k2 � 3Þ

�1k3!

dðk2 � 3k3Þdðk� k1 � 3ðk3 þ 2ÞÞ!(2.81)

By solving the following equation we get the value of Y(0) ¼ y(0):

92yð0Þ þ

XNk¼0

�2kþ 237þ k

LðkÞAðkÞ ¼ 3 (2.82)

For N ¼ 30 and M ¼ 3, we have y(0) ¼ 0.14247 � 10�4. For N ¼ 20,the approximate solution is

yðxÞ ¼ 1:4247� 10�5 þ 9:9997� 10�1xþ 0:92308� 10�5x103 � 0:89711

� 10�5x133

(2.83)

2.5 DIFFERENTIAL TRANSFORMATION METHOD FOREIGENVALUE PROBLEMS

Following StrumeLiouville problem is used to illustrate how to solveeigenvalues and eigenfunctions by DTM [5]

ddx

pðxÞ dyðxÞ

dx

�þ ½qðxÞ þ lwðxÞ�yðxÞ ¼ 0 (2.84)

Boundary conditions:

yð0Þ þ ay0ð0Þ ¼ 0 (2.85)

yð1Þ þ by0ð1Þ ¼ 0 (2.86)

where a and b are constants.


Taking differential transformation of Eq. (2.84) and using DTM prin-ciple, we haveXk

l¼0

ðl þ 1ÞPðl þ 1Þðk� l þ 1ÞYðk� l þ 1Þ

þXkl¼0

PðlÞðk� l þ 1Þðk� l þ 2ÞYðk� l þ 1Þ

þXkl¼0

½lW ðlÞ �QðlÞ�Yðk� 1Þ ¼ 0

(2.87)

where P(k), Q(k), W(k), and Y(k) are T-function of p(x), q(x), w(x), andy(x), respectively.

DTM transformation of boundary condition (2.85) becomes

Yð0Þ þ aYð1Þ ¼ 0 (2.88)

and boundary condition (2.89) becomesXnk¼0

ð1þ bkÞYðkÞ ¼ 0 (2.89)

Let

Yð0Þ ¼ c. (2.90)

From Eq. (2.88) we have

Yð1Þ ¼ � ca. (2.91)

Substituting Eqs. (2.90) and (2.91), and k ¼ 0 into Eq. (2.87), we have

Yð2Þ ¼ c2Pð0Þ

pð1Þa

þQð0Þ � lW ð0Þ�

(2.92)

Substituting Eqs. (2.90)e(2.92), and k ¼ 1 into Eq. (2.87), we have

Yð3Þ ¼ c6Pð0Þ

�2Pð2Þa

� 2ðPð1ÞÞ2aPð0Þ

�þ

l

W ð0Þa

�W ð1Þ þ 2Pð1ÞW ð0ÞPð0Þ

�� 2Pð1ÞQð0Þ

Pð0Þ �Qð0Þa

þQð1Þ(2.93)


Following the same recursive procedure, we calculate up to the nthterm Y(n), and n is decided by the convergence of the eigenvalue, asdescribed later. Substituting Y(1) e Y(n) into Eq. (2.89), we have

c�f ðnÞðlÞ� ¼ 0 (2.94)

where f(n)(l) is a polynomial of a corresponding to n. For nontrivial solu-tions of eigenfunctions, we have c s 0, and

f ðnÞðlÞ ¼ 0 (2.95)

Solving Eq. (2.95), we get

l ¼ lðnÞi ;where i ¼ 1; 2;. (2.96)

lðnÞi is the ith estimated eigenvalue corresponding to n, and n is decided

by the following equation ��lðnÞi þ lðn�1Þi

�� ε (2.97)

where lðn�1Þi is the ith estimated eigenvalue corresponding to n � 1 and ε is

a small value, we set. From Eq. (2.24) we have two cases to discuss.

Case 1:If Eq. (2.97) is satisfied, then l

ðnÞi is the ith eigenvalue li. Substituting li into

Y(0) � Y(n) and using:

yiðxÞ ¼Xnk¼0

xkYliðkÞ (2.98)

where YliðkÞ is Y(k), whose l is substituted by li, and yi(x) is the eigen func-tion corresponding to eigenvalue li. For comparison with the analytic so-lution as shown later, the ith normalized eigenfunction is defined as follows:

byiðxÞ ¼ yiðxÞR 10 jyiðxÞjdx

(2.99)

Case 2:If Eq. (2.97) is not satisfied, then repeat the following steps until the itheigenvalue and the ith normalized eigenfunction are found.

Step 1: substituting n þ 1 for n.


Step 2: following the same procedure as shown in Eqs. (2.94)e(2.99).At first glance, the method introduced in this section looks very

involved in computation; in fact, however, these algebraic computationscan be finished very fast by symbolic computational software.

Example A. Consider the following equation,

y00 þ ly ¼ 0 (2.100)

yð0Þ � y0ð0Þ ¼ 0 (2.101)

yð1Þ þ y0ð1Þ ¼ 0 (2.102)

Taking differential transform of Eq. (2.100), we obtain

Yðkþ 2Þ ¼ � lYðkÞðkþ 1Þðkþ 2Þ (2.103)

Boundary condition (2.101) becomes

Yð0Þ � Yð1Þ ¼ 0 (2.104)

Boundary condition (2.102) becomesXnk¼0

ð1þ kÞYðkÞ ¼ 0 (2.105)

1. Solving the first eigenvalue and eigenfunction. For ease of demonstra-tion, list the computation and result corresponding to n ¼ 6.

Let

Yð0Þ ¼ c (2.106)

and from Eq. (2.104), we have

Yð1Þ ¼ c (2.107)

Substituting Eqs. (2.106) and (2.107), and k ¼ 0 into Eq. (2.103), wehave

Yð2Þ ¼ � c2l (2.108)

Following the same recursive procedure, we have

Yð3Þ ¼ � c6l (2.109)


Yð4Þ ¼ � c24l2 (2.110)

Yð5Þ ¼ � c120

l2 (2.111)

Yð6Þ ¼ � c720

l3 (2.112)

Substituting Eqs. (2.106)e(2.112) into Eq. (2.105), we have

f ð6ÞðlÞ ¼ 3� 136lþ 31

120l2 � 7

720l3 ¼ 0 (2.113)

Solving Eq. (2.113), we have

l ¼ 1:71; 12:43� 5:08i. (2.114)

Take real root

lð6Þ1 ¼ 1:71 (2.115)

When n ¼ 5, by the same way, we have

lð5Þ1 ¼ 1:75 (2.116)

From Eqs. (2.115) and (2.116), we have��lð6Þ1 � lð5Þ1

�� ¼ 0:04 � ε (2.117)

where ε is a small value we set. From Eq. (2.117) we have l1 ¼ 1.71and l1 is the first eigenvalue. Substituting l1 into Y(0) � Y(6) andDTM principle, we obtain the first eigenfunction.

y1ðxÞ ¼ ð1þ x� 0:855x2 � 0:285x3

þ 0:121838x4 þ 0:0243675x5 � 0:00694474x6Þc (2.118)

By Eq. (2.99), the first normalized eigenfunction is shown as follows

y_1ðxÞ ¼ 0:853835ð1þ x� 0:855x2 � 0:285x3 þ 0:121838x4

þ 0:0243675x5 � 0:00694474x6Þ (2.119)

we have the first eigenvalue and eigenfunction as follows

lðaÞ1 ¼ 1:71 (2.120)

yðaÞ1 ðxÞ ¼ 1:30767 cosð1:30767xÞ þ sinð1:30767xÞ (2.121)


After normalization, Eq. (2.121) becomes

y_ðaÞ1 ðxÞ ¼ 0:652999½1:30767 cosð1:30767xÞ þ sinð1:30767xÞ� (2.122)

Besides l1 ¼ lðaÞ1 , the calculated results from Eq. (2.119) are

compared closely with the analytic results from Eq. (2.122) as shownin Fig. 2.4.

2. Solving the second eigenvalue and eigenfunction: List the result corre-sponding to n ¼ 12.

Following the same procedure as shown earlier, we solve the equa-tion f (12)(l) ¼ 0, take the real root and get

lð12Þ1 ¼ 1:71 (2.123)

lð12Þ2 ¼ 13:49 (2.124)

Note that lð12Þ1 ¼ lð6Þ1 . Due to

��lð12Þ2 � lð11Þ2

�� ε, we have the sec-ond eigenvalue l2 ¼ 13.49. Following the same procedure shownabove, we get the second normalized eigenfunction as follows,

y_2ðxÞ ¼ 1:40347ð1þ x� 6:745x2 � 2:24833x3 þ 7:5825x4

þ 1:5165x5 � 3:4096x6 � 0:487086x7 þ 0:821348x8

þ 0:0912609x9 � 0:123111x10 � 0:0111919x11 þ 0:0125816x12Þ(2.125)

1.1

1.05

1

1st norm

aliz

ed e

igenfu

nction

0.95

0.9

0.850 0.2 0.4 0.6

x

analytic

present

0.8 1

Figure 2.4 Comparison of the calculated results of the first normalized eigenfunctionwith the analytic results after normalization.


By the analytic method, we have the second eigenvalue and normal-ized eigenfunction as follows,

lðaÞ2 ¼ 13:49 (2.126)



compared closely with the analytic results from Eq. (2.127) as shownin Fig. 2.5.

3. Solving the third eigenvalue and eigenfunction: List the result corre-sponding to n ¼ 22. Following the same procedure as shown earlier,we solve the equation f(22)(l) ¼ 0, take the real root and get

lð22Þ1 ¼ 1:71 (2.128)

lð22Þ2 ¼ 13:49 (2.129)

lð22Þ3 ¼ 43:36 (2.130)

Note that lð22Þ1 ¼ l

ð12Þ1 ¼ l

ð6Þ1 and l

ð22Þ2 ¼ l

ð12Þ2 . Due to��lð22Þ3 � l

ð21Þ3

�� ε, we have the third eigenvalue l3 ¼ 43.36. Following

the same procedure shown earlier, we have the third normalized eigen-function as follows,

2nd n

orm

aliz

ed e

igenfu

nction

1.5

1

–1

0.5

0

0 0.2 0.4

x

0.6 0.8 1

–0.5

–1.5

analytic

present

Figure 2.5 Comparison of the calculated results of the second normalized eigen-function with the analytic results after normalization.


y_3ðxÞ ¼ 1:51383ð1þ x� 21:68x2 � 7:22667x3 þ 78:3371x4

þ 15:6674x5 � 113:223x6 � 16:1747x7 þ 87:6671x8

þ 9:74073x9 � 42:2361x10 � 3:83964x11 þ 13:8739x12

þ 1:06722x13 � 3:30534x14 � 0:220356x15 þ 0:597165x16

þ 0:351274x17 � 0:0846179x18 � 0:00445358x19 þ 0:00965535x20

þ 0:000459779x21 � 0:000906182x22Þ(2.131)

By the analytic method, we have the third eigenvalue and normal-ized eigenfunction as follows,

lðaÞ3 ¼ 43:36 (2.132)



compared closely with the analytic results from Eq. (2.133) as shownin Fig. 2.6.The convergence of eigenvalues l1 � l3 is shown in Fig. 2.7, where l1,

l2, and l3 converge to 1.71, 13.49, and 43.36, respectively.Example B. IH AbdeleHalim [6] applied this method to solve the

eigenvalue problem in the shape of

3rd

norm

aliz

ed e

igen

func

tion

analyticpresent

2

1

1.5

0.5

–0.5

–1.5

–2

–1

0

0 0.2 0.4x

0.6 0.8 1

Figure 2.6 Comparison of the calculated results of the third normalized eigenfunctionwith the analytic results after normalization.


ddy

pðxÞ dy

dx

�þ ½lwðxÞ � qðxÞ�yðxÞ ¼ 0 (2.134)

with boundary conditions

yð0Þ þ ay0ð0Þ ¼ 0; (2.135)

yð0Þ þ by0ð1Þ ¼ 0; (2.136)

where a and b are constants, and illustrated the results by solving thisproblem

y00 þ lx2y ¼ 0 (2.137)


yð0Þ ¼ 0; yð1Þ ¼ 0 (2.138)

Consider the regular SturmeLiouville eigenvalue problem

ddy

pðxÞ dyðxÞ

dx

�þ ½lrðxÞ � qðxÞ�yðxÞ ¼ 0 (2.139)


a1yð0Þ þ b1y0ð0Þ ¼ 0; (2.140)

a2yð1Þ þ b2y0ð1Þ ¼ 0; (2.141)

convergence of eigenvalue1st eigenvalue

2nd eigenvalue

3rd eigenvalueeig

envalu

e

60

55

50

45

40

35

30

25

20

15

10

5

00 5 10

n

15 20 25

Figure 2.7 The convergence of eigenvalues l1 � l3.


where p(x) > 0, r(x) > 0, and p(x), r(x), q(x) and p0(x) are continuous on theclosed interval [0, 1], and ai � 0, bi � 0, and ai þ bi � 0 for i ¼ 1, 2. Takingdifferential transformation of Eq. (2.139) and using Eqs. (2.133)e(2.137)we obtainXk

[¼0

ð[þ 1Þpð[þ 1Þðk� [þ 1ÞYðk� [þ 1Þ þXkl¼0

pð[Þðk� [þ 1Þ

�ðk� [þ 2ÞYðk� [þ 2Þ þXk[¼0

½lRð[Þ �Qð[Þ�Yðk� [Þ ¼ 0;

(2.142)

where P(k), Q(k), R(k), and Y(k), are transformed functions of p(x), r(x),q(x), and y(x), respectively. Using Eq. (2.134), the boundary condition(2.140) becomes

a1Yð0Þ � b1Yð1Þ ¼ 0; (2.143)

the boundary condition (2.141) becomesXnk¼0

ða2 þ b2kÞYðkÞ ¼ 0 (2.144)

Put

yð0Þ ¼ c (2.145)

from Eq. (2.143) we find that

Yð1Þ ¼ a1cb1

(2.146)

At k ¼ 0, and substituting from Eqs. (2.145) and (2.146) intoEq. (2.142), we have

Yð2Þ ¼ � c2pð0Þ

a1pð1Þb1

þ lRð0Þ �Qð0Þ�. (2.147)

At k ¼ 1, and substituting from Eqs. (2.145)e(2.147) into Eq. (2.142),we have

Yð3Þ ¼ c6pð0Þ

�� 2a1pð2Þ

b1þ 2a1p

2ð1ÞPð0Þb1

þ l

2pð1ÞRð0Þ

b1� Rð1Þ

�� 2pð1ÞQð1Þ

pð0Þ þ a1Qð1Þb1

þQð1Þ�

(2.148)


Following the same procedure, we calculate up to the nth term Y(n) andsubstituting from Y(1) to Y(n) into Eq. (2.144), we obtain

c�f ðnÞðlÞ� ¼ 0 (2.149)

where f (n)(l) is a polynomial of k corresponding to n, and for c s 0, wehave

f ðnÞðlÞ ¼ 0 (2.150)

Solving Eq. (2.150), we li ¼ lni ; i ¼ 1; 2; 3;.; where lðnÞi is the nthestimated eigenvalue corresponding to n, and n is indicated by��lðnÞi � l

ðn�1Þi

�� x; (2.151)

where in lðn�1Þi i is the ith estimated eigenvalue corresponding to n�1 and x

is a small value we set and then we have two cases the same as Section 2.5described.

Example C: Consider the equation of

xy00ðxÞ þ l2yðxÞ ¼ 0; (2.152)

Boundary conditions:

yð0Þ ¼ 0; (2.153)

yð1Þ ¼ 0: (2.154)

Taking DTM, we obtainXk[¼0

dð[� 1Þðk� [þ 1Þðk� [þ 2ÞYðk� [þ 2Þ ¼ �l2YðkÞ. (2.155)

Using DTM, the boundary condition (2.153) becomes

Yð0Þ ¼ 0: (2.156)

and the boundary condition (2.154) becomesXnk¼0

YðkÞ ¼ 0. (2.157)

(I): Solving the first eigenvalue and eigenfunction: putting

Yð1Þ ¼ c. (2.158)

Substituting Eqs. (2.156) and (2.158) at k ¼ 1 into Eq. (2.155), we have

Yð2Þ ¼ � c2l2. (2.159)


Following the same procedure, Y(3) ¼ Y(7) can be solved as follows

Yð3Þ ¼ c12l4; (2.160)

Yð4Þ ¼ � c144

l6; (2.161)

Yð5Þ ¼ c2880

l8; (2.162)

Yð6Þ ¼ � c86400

l10; (2.163)

Yð7Þ ¼ � c3628800

l12. (2.164)

Substituting Eqs. (2.156), (2.158)e(2.164) into Eq. (2.157), we have

f ð7ÞðlÞ ¼ 1� l2

2þ l4

12� l6

144þ l8

28800� l10

86400þ l6

3628800¼ 0: (2.165)

Solving Eq. (2.165), and taking the real roots, we have

l1ð7Þ ¼ �1:9159: (2.166)

When n ¼ 6, by the same way, we have

lð6Þ1 ¼ �1:9143: (2.167)

From Eqs. (2.166) and (2.167), we have��l1ð7Þ � l1ð6Þ�� ¼ 0:01 � x. (2.168)

From Eq. (2.166) we take l1 ¼ 1.92 as the first eigenvalue. Substitutingl1 into Eqs. (2.158)e(2.164) and using DTM principle, we obtain the firsteigenfunction,

y1ðxÞ ¼ ðx� 1:8432x2 þ 1:1324x3 � 0:347892x4 þ 0:0641235x54� 10�4x7

�0:00787949x6 þ 6:91594� 10�4x7Þc.(2.169)

the first normalized eigenfunction is

y1^ ðxÞ ¼ 9:19243ðx� 1:8432x2 þ 1:1324x3 � 0:347892x4

þ 0:0641235x54� 10�4x7Þ.� 0:00787949x6 þ 6:91594� 10�4x7Þ.

(2.170)


By the analytical method, the first eigenvalue and eigenfunction are

l1ðaÞ ¼ 1:916; (2.171)

y1ðaÞ ¼ffiffiffix

pJ1�3:832

ffiffiffix

p �. (2.172)

After normalization, Eq. (2.172) becomes

y^ðaÞ1 ðxÞ ¼ 4:7589

ffiffiffix

pJ1�3:832

ffiffiffix

p �. (2.173)

From Eqs. (2.166) and (2.171) we deduce that l1 ¼ l1ðaÞ, and the

calculated results from Eq. (2.166) are compared closely with the analyticresults from Eq. (2.173) as shown in Fig. 2.8.

(II): Solving the second eigenvalue and eigenfunction: Listing the resultcorresponding to n ¼ 11, and following the same procedure as above, weget l2 ¼ 3.507502 and the second normalized eigenfunction as

y2^ ðxÞ ¼ � 41:0133ðx� 6:15303x2 þ 12:6199x3 � 12:9418x4 þ 6:96316x5

� 3:26649x6 þ 0:95088x7 � 0:210321x8 þ 0:0359475x9

� 0:00491526x10 þ 5:49886� 10�4x11Þ:(2.174)

0.0

present

analytic

–0.4

–0.2

–0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

0.1 0.2 0.3 0.4 0.5x

1st n

orm

aliz

ed e

igen

fuct

ion

0.6 0.7 0.8 0.9 1.0

Figure 2.8 Comparison of the calculated results of the first normalized eigenfunctionwith the analytic results after normalization.


By the analytical method, we have the second eigenvalue andnormalized eigenfunction as follows:

l2ðaÞ ¼ 3:508 (2.175)

y^ðaÞðxÞ ¼ �11:69

ffiffiffix

pJ1�7:016

ffiffiffix

p �. (2.176)

The calculated results from Eq. (2.174) are compared closely with theanalytic results from Eq. (2.176) as shown in Fig. 2.9. Regarding the thirdnormalized eigenfunction, we find that the comparison results are the samefrom Fig. 2.10.

The convergence of eigenvalues of the problem confirms that l1, l2,and l3 converge to 1.92, 3.51, and 5.09, respectively, as shown in Fig. 2.11.

2.6 TWO-DIMENSIONAL DIFFERENTIAL TRANSFORMATIONMETHOD FOR FRACTIONAL ORDER PARTIALDIFFERENTIAL EQUATIONS

There are several definitions of a fractional derivative of order a > 0. TheCaputo fractional derivative is defined as [7].

Daa f ðxÞ ¼ Jm�a

a Dmf ðxÞ; (2.177)

0.0–3.0

–2.4

–1.8

–1.2

–0.6

0.0

0.6

1.2

1.8

2.4

3.0

3.6

0.1 0.2 0.3 0.4 0.5x

2nd

norm

aliz

ed e

igen

func

tion

0.6 0.7 0.8 0.9 1.0

present

analytic

Figure 2.9 Comparison of the calculated results of the second normalized eigen-function with the analytic results after normalization.


where m� 13a � m: Here Dm is the usual integer differential operator oforder m and Jma is the RiemanneLiouville integral operator of order m > 0,defined by

Jma f ðxÞ ¼1

GðmÞZ x

aðx� tÞm�1f ðtÞdt; x_0: (2.178)

The Caputo fractional derivative is considered here because it allowstraditional initial and boundary conditions to be included in the formula-tion of the problem [4].

Consider a function of two variables u(x, y), and suppose that it can berepresented as a product of two single-variable functions, i.e., u(x, y) ¼ f(x)g(y). On the basis of the properties of generalized two-dimensional dif-ferential transform, the function u(x, y) can be represented as

uðx; yÞ ¼XNk¼0

FaðkÞðx� x0ÞkaXNh¼0

GbðhÞðy� y0Þhb

¼XNk¼0

XNh¼0

Ua;bðk; hÞðx� x0Þkaðy� y0Þhb;(2.179)

where 03a;b � 1; Ua,b(k, h) ¼ Fa(k)Gb(h) is called the spectrum of u(x, y).The generalized two-dimensional differential transform of the function u(x, y)is as follows:

Ua;bðk; hÞ ¼ 1Gðakþ 1ÞGðbhþ 1Þ

�Da

x0

�k�Db

y0

�huðx; yÞ

�ðx0;y0Þ

; (2.180)

0.0–6.0

–4.8

–3.6

–2.4

–1.2

0.0

1.2

2.4

3.6

4.8

0.1 0.2 0.3

present

analytic

0.4 0.5x

3rd

norm

aliz

ed e

igen

func

tion

0.6 0.7 0.8 0.9 1.0

Figure 2.10 Comparison of the calculated results of the third normalized eigenfunc-tion with the analytic results after normalization.


where�Da

x0

�k ¼ Dax0D

ax0.Da

x0, k times. In this work, the lowercase u(x, y)represents the original function while the uppercase Ua,b(k, h) stands for thetransformed function. On the basis of the definitions (I) and (II), we havethe following results:

Theorem I. Suppose thatUa,b(k, h), Va,b(k, h), andWa,b(k, h) are the dif-ferential transformations of the functions u(x,y), v(x,y), and w(x,y),respectively;

ðaÞ if uðx; yÞ ¼ yðx; yÞ � uðx; yÞ; then Ua;bðk; hÞ ¼ Va;bðk; hÞ �Wa;bðk; hÞ;ðbÞ if uðx; yÞ ¼ ayðx; yÞ; a ˛R; then Ua;bðk; hÞ ¼ aVa;bðk; hÞ;ðcÞ if uðx; yÞ ¼ yðx; yÞuðx; yÞ; then Ua;bðk; hÞ ¼

Xk

r¼0

Xh

s¼0Va;bðr; h� sÞWa;bðk� r; sÞ;

ðdÞ if uðx; yÞ ¼ ðx� x0Þnaðy� y0Þmb; then Ua;bðk; hÞ ¼ dðk� nÞdðh� mÞ;

ðeÞ if uðx; yÞ ¼ Dax0yðx; yÞ; 03a � 1; then Ua;bðk; hÞ ¼ Gðaðk þ 1Þ þ 1Þ

Gðak þ 1Þ Ua;bðk þ 1; hÞ.

Theorem II. If u(x, y) ¼ f(x)g(y) and the function f(x) ¼ x_h(x), wherel > �1, h(x) has the generalized Taylor series expansionhðxÞ ¼PN

n¼0 anðx� x0Þak; and

01.01.31.61.9

2.2

2.52.8

3.13.4

3.74.04.3

4.6

4.95.2

5.55.8

2 3 5 6 8 9 11 12 14n

convergence of eigenvalue to problem3

3rd eigenvalue

2nd eigenvalue

1st eigenvalue

eige

nval

ue

15 17 18 20 21 23 24 26

Figure 2.11 Convergence of the eigenvalues l1 � l3, where l1, l2 and l3 converge to1.92, 3.51, and 5.09, respectively.


ðaÞ b3lþ 1 and a is arbitrary; or

ðbÞ b3lþ 1; a is arbitrary and an ¼ 0 for n ¼ 0; 1;.m� 1;

where� 13b � m.

Then the generalized differential transform becomes

Ua;bðk; hÞ ¼ 1Gðakþ 1ÞGðbhþ 1Þ

�Da

x0

�k�Db

y0

�huðx; yÞ

�ðx0;y0Þ

; (2.181)

Theorem III. If v(x, y) ¼ f(x)g(y), the function f(x) satisfies the condi-tions given in Theorem 2.2, and Dg

x0yðx; yÞ then

Ua;bðk; hÞ ¼ Gðaðkþ 1Þ þ gÞGðakþ 1Þ Va;bðkþ g=a; hÞ (2.182)

Following two examples are presented that demonstrate the perfor-mance and efficiency of the generalized DTM for solving linear partialdifferential equations with time- or space-fractional derivatives.

Example A. Consider the following linear inhomogeneous time-fractional equation:

vauvta

þ xvuvx

þ v2xvx2

¼ 2ta þ 2xþ 2;Ua;1ðk; 0Þ ¼ dðk� 2Þ. (2.183)

where 03a � 1; subject to the initial condition

uðx; 0Þ ¼ x2. (2.184)

Suppose that the solution u(x,t) can be represented as a product ofsingle-valued functions. Selecting b ¼ 1 and applying the generalized two-dimensional differential transform to both sides of Eq. (2.183), the linearinhomogeneous time-fractional Eq. (2.183) transforms to"Xk

r¼0

Xhs¼0

dðr � 1Þdðh� sÞðk� r þ 1ÞUa;1ðk� r þ 1; sÞ

Ua;1ðk; hþ 1Þ ¼ Gðakþ 1ÞGðaðhþ 1Þ þ 1Þ

�ðkþ 1Þðkþ 2ÞUa;1ðkþ 2; hÞ þ 2dðk� 2ÞdðhÞ þ 2dðkÞdðhÞ# (2.185)


The generalized two-dimensional differential transform of the initialcondition Eq. (2.184) is

Ua;1ðk; 0Þ ¼ dðk� 2Þ. (2.186)

Utilizing the recurrence relation Eq. (2.185) and the transformed initialcondition Eq. (2.186), we get Ua,1(2, 1) ¼ 1, Ua;1ð0; 2Þ ¼ 2 Gðaþ 1Þ

Gð2aþ 1Þ, andUa,1(k,h) ¼ 0 for k s 2, h s 2. Therefore, according to above sections, thesolution of Eq. (2.183) is given by

uðx; tÞ ¼ x2 þ 2Gðaþ 1ÞGð2aþ 1Þt

2a; (2.187)

which is the exact solution of the linear inhomogeneous time-fractionalEq. (2.183).

Example B. Consider the following linear space-fractional telegraphequation

v1:5uvx1:5

¼ v2uvt2

þ vuvt

þ u; x_0; (2.188)

subject to the initial conditions

uð0; tÞ ¼ expð�tÞ; uxð0; tÞ ¼ expð�tÞ . (2.189)

Suppose that the solution u(x, t) can be represented as a product ofsingle-valued functions, u(x, t) ¼ v(x)w(t) where the function v(x) satisfiesthe conditions given in theorem II. Selecting a ¼ 1, b ¼ 0.5 and applyingthe generalized two-dimensional differential transform to both sides ofEq. (2.188), the linear space-fractional telegraph Eq. (2.188) transforms to

��ðhþ 1Þðhþ 2ÞU1;1=2ðk; hþ 2ÞU1;1=2ðk; hþ 1Þ þ U1;1=2ðk; hÞ�.

U1;1=2ðkþ 3; hÞ ¼ Gðk=2þ 1ÞGðk=2þ 5=2Þ

(2.190)

The generalized two-dimensional differential transforms of the initialconditions Eq. (2.189) are given by

U1;1=2ð0; hÞ ¼ ð�1Þh�h!;U1;1=2ð1; hÞ ¼ 0;

U1;1=2ð2; hÞ ¼ ð�1Þh�h!;Utilizing the recurrence relation Eq. (2.190) and the transformed initial

conditions, the first few components of U1,1/2(k, h) are calculated and givenin Table 2.1 in Ref. [7].


Therefore, the approximate solution of the linear space-fractionaltelegraph Eq. (2.188) can be derived as

uðx; tÞ ¼�1� t þ 1

2!t � 1

3!t3 þ 1

4!t4 � 1

5!t5

þ�1� t þ 1

2!t2 � 1

3!t3 þ 1

4!t4 � 1

5!t5x

þ�1� t þ 1

2!t2 � 1

3!t3 þ 1

4!t4 � 1

5!t5

x1:5

Gð5=2Þ

þ�1� t þ 1

2!t2 � 1

3!t3 þ 1

4!t4 � 1

5!t5

x2:5

Gð5=2Þ þ.

(2.191)

that is

uðx; tÞ ¼ expð�tÞ�1þ xþ x1:5

Gð5=2Þ þx2:5

Gð7=2Þ þx3

Gð4Þþ x4

Gð5Þ þx4:5

Gð11=2Þ þ.

;

(2.192)

Table 2.1 The Fundamental Operations of Reduced Differential Transform MethodFunctional Form Transformed Form

u(x, t)Ukðx; tÞ ¼ 1

k!

hvk

vxk uðx; tÞit¼0

w(x,t) ¼ u(x, t) � v(x,t) Wk(x) ¼ Uk(x) � Vk(x)w(x,t) ¼ au(x,t) Wk(x) ¼ aUk(x) (a is constant)w(x,t) ¼ xmtn Wk(x) ¼ xmd(k � n)w(x,t) ¼ xmtnu(x,t) Wk(x) ¼ xmUk�n

w(x,t) ¼ u(x,t)v(x,t)WkðxÞ ¼

Xkr¼0

VrðxÞUk�rðxÞ ¼

Xkr¼0

UrðxÞVk�rðxÞ

wðx; tÞ ¼ vk

vtk uðx; tÞWkðxÞ ¼ ðkþ 1Þ.ðkþ rÞUkþrðxÞ

¼ ðkþ rÞ!k!

UkþrðxÞ

wðx; tÞ ¼ vvx uðx; tÞ

WkðxÞ ¼ vvxUkðxÞ


2.7 REDUCED DIFFERENTIAL TRANSFORM METHOD

The reduced differential transform method (RDTM) was first proposed bythe Turkish mathematician Keskin [8] in 2009. It has received muchattention since it has applied to solve a wide variety of problems by manyauthors. In this section after introducing this method, it is applied to solvetwo examples of NLPDEs [9]. Consider a function of two variables u(x,t) andsuppose that it can be represented as a product of two single-variable func-tions, i.e., u(x,t) ¼ f(x)g(t). Based on the properties of one-dimensional dif-ferential transform, the function u(x,t) can be represented as follows [9]:

uðx; tÞ ¼ XN

i¼0

FðiÞxi! XN

j¼0

GðjÞtj!

¼XNk¼0

UkðxÞtk (2.193)

where Uk(x) is called t-dimensional spectrum function of u(x, t). The basicdefinitions of RDTM are introduced as follows:

Definition I. If function u(x,t) is analytic and differentiated continu-ously with respect to time t and space x in the domain of interest,then let

Ukðx; tÞ ¼ 1k!

"vk

vxkuðx; tÞ

#t¼0

(2.194)

where the t-dimensional spectrum function Uk(x) is the transformedfunction. In this paper, the lowercase u(x, t) represents the original func-tion, while the uppercase Uk(x) stands for the transformed function.Definition II. The differential inverse transform of Uk(x) is defined asfollows:

uðx; tÞ ¼XNk¼0

UkðxÞtk (2.195)

Then, combining Eqs. (2.194) and (2.195) we write

uðx; tÞ ¼XNk¼0

1k!

"vk

vtkuðx; tÞ

#t¼0

tk (2.196)

From the above definitions, it can be found that the concept of theRDTM is derived from the power series expansion. To illustrate the basicconcepts of the RDTM, consider the following nonlinear partial differentialequation written in an operator form

Luðx; tÞ þ Ruðx; tÞ þNuðx; tÞ ¼ gðx; tÞ (2.197)


with initial condition

uðx; 0Þ ¼ f ðxÞ (2.198)

where L ¼ vvt, R is a linear operator which has partial derivatives, Nu(x, t) is

a nonlinear operator and g(x,t) is an inhomogeneous term.According to the RDTM, we can construct the following iteration

formula:

ðkþ 1ÞUkþ1ðx; tÞ ¼ GkðxÞ � RUkðxÞ �NUkðxÞ (2.199)

where Uk(x), RUk(x), NUk(x), and Gk(x) are the transformations of thefunctions Lu(x,t), Ru(x,t), Nu(x,t), and g(x,t) respectively. From initial con-dition Eq. (2.198), we write

U0ðxÞ ¼ f ðxÞ (2.200)

Substituting Eq. (2.200) into Eq. (2.199) and by straightforward iterativecalculation, we get the following Uk(x) values. Then, the inverse trans-formation of the set of values fUkðxÞgnk¼0 gives the n-terms approximationsolution as follows

unw ðx; tÞ ¼

Xnk¼0

UkðxÞtk (2.201)

Therefore, the exact solution of the problem is given by

uðx; tÞ ¼ limn/N

unw ðx; tÞ (2.202)

The fundamental mathematical operations performed by RDTM can bereadily obtained and are listed in Table 2.1.

Following, the RDTM is applied to solve two nonlinear partial dif-ferential equations (PDEs), namely, generalized DrinfeldeSokolov (gDS)equations and KaupeKupershmidt (KK) equation.

Example A. We consider the generalized DrinfeldeSokolov (gDS)equations:

ut þ uxxx � 6uux � 6ðvaÞx ¼ 0

vt � 2vxxx þ 6uvx ¼ 0(2.203)


uðx; 0Þ ¼ �b2 � 4k4

4k2þ 2k2 tanh2ðkxÞ

vðx; 0Þ ¼ b tanhðkxÞ(2.204)


where a is a constant. According to the RDTM and Table 2.1, the differ-ential transform of Eq. (2.203) reads

ðkþ 1ÞUkþ1ðxÞ ¼ � v3

vx3UkðxÞ þ 6AkðxÞ þ 6BkðxÞ

ðkþ 1ÞVkþ1ðxÞ ¼ 2v3

vx3VkðxÞ � 6CkðxÞ

(2.205)

where the t-dimensional spectrum functions Uk(x), Vk(x) are the trans-formed functions. Ak(x), Bk(x), and Ck(x) are transformed form of thenonlinear terms. For the convenience of the reader, the first few nonlinearterms are as follows

A0 ¼ U0v

vxU0;A1 ¼ U1

v

vxU0 þ U0

v

vxU1;

A2 ¼ U2v

vxU0 þ U1

v

vxU1 þ U0

v

vxU2;

A3 ¼ U3v

vxU0 þ U2

v

vxU1 þ U1

v

vxU2 þ U0

v

vxU3

(2.206)

B0 ¼ v

vx

�V a

0

�;B1 ¼ v

vx

�aV a�1

0 V1

�;

B2 ¼ v

vx

�aða� 1Þ

2V a�2

0 V 21 þ aV a�1

0 V2

;

B3 ¼ v

vx

�aV a�1

0 V3 þ aða� 1ÞV a�20 V1V2

þ 16aða� 1Þða� 2ÞV a�3

0 V 31

(2.207)

C0 ¼ U0v

vxV0;C1 ¼ U1

v

vxV0 þ U0

v

vxV1;

C2 ¼ U2v

vxV0 þ U1

v

vxV1 þ U0

v

vxV2;

C3 ¼ U3v

vxV0 þ U2

v

vxV1 þ U1

v

vxV2 þ U0

v

vxV3

(2.208)

from initial conditions Eq. (2.204), we write

U0ðxÞ ¼ �b2 � 4k4

4k2þ 2k2 tanh2ðkxÞ

V0ðxÞ ¼ b tanhðkxÞ(2.209)


Substituting Eq. (2.209) into Eq. (2.205) (when a ¼ 2) and bystraightforward iterative steps, we can obtain

U1ðxÞ ¼ 2kð4k4 þ 3b2ÞsinhðkxÞcoshðkxÞ3 ;

U2ðxÞ ¼ �12

�2 coshðkxÞ2 � 3

�ð4k4 þ 3b2Þ2coshðkxÞ4 ;

U3ðxÞ ¼ 13

sinhðkxÞ�coshðkxÞ2 � 3�ð4k4 þ 3b2Þ3

coshðkxÞ5 ;

U4ðxÞ ¼ � 124

ð4k4 þ 3b2Þ4�15þ 2 coshðkxÞ4 � 15 coshðkxÞ2�coshðkxÞ6

«

(2.210)

and

V1ðxÞ ¼ 12bð4k4 þ 3b2ÞcoshðkxÞ2k ;

V2ðxÞ ¼ �14bð4k4 þ 3b2Þ2 sinhðkxÞ

coshðkxÞ3k2 ;

V3ðxÞ ¼ 124

�2 coshðkxÞ2 � 3

�bð4k4 þ 3b2Þ3

coshðkxÞ4k3 ;

V4ðxÞ ¼ � 148

sinhðkxÞð4k4 þ 3b2Þ4�coshðkxÞ2 � 3�b

coshðkxÞ5k4 .

«

(2.211)

and so on, in the same manner, the rest of components can be obtained byusing MAPLE software.

Taking the inverse transformation of the set of values fUkðxÞgnk¼0 andfVkðxÞgnk¼0 gives n-terms approximation solutions as follows

unw ðx; tÞ ¼

Xnk¼0

UkðxÞtk ¼ �b2 � 4k4

4k2þ 2k2 tanh2ðkxÞ þ 2kð4k4 þ 3b2ÞsinhðkxÞ

coshðkxÞ3 t þ.

þ 1n!

vn

vtn

��b2 � 4k4

4k2þ 2k2 tanh2

�kxþ ð4k4 þ 3b2Þ

2kt

�t¼0

tn

(2.212)


vnw ðx; tÞ ¼

Xnk¼0

VkðxÞtk ¼ b tanhðkxÞ þ 12bð4k4 þ 3b2ÞcoshðkxÞ2k t þ.

þ 1n!

vn

vtn

�b tanh


2kt

�t¼0

tn(2.213)

Therefore, the exact solution of the problem is readily obtained asfollows

uðx; tÞ ¼ limn/N

unw ðx; tÞ ¼ �b2 � 4k4

4k2þ 2k2 tanh2


2kt

vðx; tÞ ¼ lim

n/Nvnw ðx; tÞ ¼ b tanh


2kt

(2.214)

To examine the accuracy of the RDTM solution, the absolute errors ofthe 4-terms approximate solutions are plotted in Fig. 2.12.

Example B: Consider the nonlinear KaupeKupershmidt (KK)equation:

ut ¼ uxxxxx þ 5uuxxx þ 252uxuxx þ 5u2ux (2.215)

subject to the initial condition

uðx; 0Þ ¼ �2k2 þ 24k2

1þ ekx� 24k2

ð1þ ekxÞ2 (2.216)

10.8

0.60.4

0.20

1. × 10–24

2. × 10–24

3. × 10–24

10

5

0xt

–5

–10

(a)

1. × 10–23

8. × 10–24

6. × 10–24

4. × 10–24

2. × 10–24

10.8

0.60.4

0.20

10

5

0 xt

–5

–10

(b)

Figure 2.12 The absolute error of (a) u4w ðx; tÞ (b) v4w ðx; tÞ when k ¼ 0.01 and b ¼ 0.001.


where k is an arbitrary constant. By taking the differential transform ofEq. (2.215), we obtain

ðkþ 1ÞUkþ1ðxÞ ¼ v5

vx5UkðxÞ þ 5AkðxÞ þ 25

2BkðxÞ þ 5CkðxÞ (2.217)

where the t-dimensional spectrum function and Uk(x) is the transformedfunction. Ak(x), Bk(x), and Ck(x) are the transformed form of the nonlinearterms. For the convenience of the reader, the first few nonlinear terms areas follows

A0 ¼ U0v3

vx3U0;A1 ¼ U1

v3

vx3U0 þ U0

v3

vx3U1;

A2 ¼ U2v3

vx3U0 þ U1

v3

vx3U1 þ U0

v3

vx3U2;

A3 ¼ U3v3

vx3U0 þ U2

v3

vx3U1 þ U1

v3

vx3U2 þ U0

v3

vx3U3

(2.218)

B0 ¼ v

vxU0

v2

vx2U0;B1 ¼ v

vxU0

v2

vx2U1 þ v

vxU1

v2

vx2U0þ;

B2 ¼ v

vxU0

v2

vx2U2 þ v

vxU1

v2

vx2U1 þ v

vxU2

v2

vx2U0;

B3 ¼ v

vxU0

v2

vx2U3 þ v

vxU1

v2

vx2U2 þ v

vxU2

v2

vx2U1 þ v

vxU3

v2

vx2U0

(2.219)

C0 ¼ U20

v

vxU0;C1 ¼ 2U0U1

v

vxU0 þ U2

0

v

vxU1;

C2 ¼ 2U0U2v

vxU0 þ U2

1

v

vxU0 þ 2U0U1

v

vxU1 þ U2

0

v

vxU2;

C3 ¼ 2U0U3v

vxU0 þ 2U1U2

v

vxU0 þ U2

1

v

vxU1 þ 2U0U2

v

vxU1

þ 2U0U1v

vxU2 þ U2

0

v

vxU3

(2.220)

from the initial conditions Eq. (2.216), we write

U0ðxÞ ¼ �2k2 þ 24k2

1þ ekx� 24k2

ð1þ ekxÞ2 (2.221)


Substituting Eq. (2.29) into Eq. (2.25) and by straightforward iterativesteps, yields

U1ðxÞ ¼ � 264k7ekxð�1þ ekxÞð1þ ekxÞ3

;

U2ðxÞ ¼ � 1452k12ekxð4ekx � e2kx � 1Þð1þ ekxÞ4

;

U3ðxÞ ¼ 5324k17ekxð1� 11ekx þ 11e2kx � e3kxÞð1þ ekxÞ5

;

U4ðxÞ ¼ 14641k22ekxð1� 26ekx þ 66e2kx � 26e3kx þ e4kxÞð1þ ekxÞ6

«

(2.222)

and so on.Taking the inverse transformation of the set of values fUkðxÞgnk¼0 gives

n-terms approximation solution as

unw ðx; tÞ ¼

Xnk¼0

UkðxÞtk ¼ �2k2 þ 24k2

1þ ekx� 24k2

ð1þ ekxÞ2� 264k7ekxð�1þ ekxÞ

ð1þ ekxÞ3t þ.

þ 1n!

"vn

vtn

� 2k2 þ 24k2

1þ ekxþ11k5t� 24k2

ð1þ ekxþ11k5tÞ2!#

t¼0

tn

(2.223)

Therefore, the exact solution of the problem is readily obtained asfollows

uðx; tÞ ¼ limn/N

unw ðx; tÞ ¼ �2k2 þ 24k2

1þ ekxþ11k5t� 24k2

ð1þ ekxþ11k5tÞ2 (2.224)

2.8 MODIFIED DIFFERENTIAL TRANSFORMATION METHOD

In recent years, glorious developments have been presented by researcherson DTM to improve its accuracy. One of these developments is ModifiedDifferential Transformation Method or MDTM [10]. Let us consider thedifferential transform for,

u3ðx; tÞ ¼Xkr¼0

Xk�r

q¼0

Xhs¼0

Xh�s

p¼0

Ua;1ðr; h� s� pÞUa;1ðq; sÞUa;1ðk� r � q; pÞ

(2.225)


where involves four summations. Thus it is necessary to have a lot ofcomputational work to calculate such differential transform Ua,1(k, h) forthe large number of (k, h). Since, DTM is based on the Taylor series forall variables. To reduce the complexity in DTM, we introduce the DTMwith respect to the specific variable for the function u(x,t) known as themodified version of DTM. Assume that the specific variable is the variablet then, we have the Taylor series expansion of the function u(x, t) at t ¼ t0 asfollows,

uðx; tÞ ¼XNh¼0

1Gðahþ 1Þ

vahuðx; t�

vtah

!t¼t0

ðt � t0Þah (2.226)

The modified differential transform Ua,1(x, h) of u(x, t) with respect tothe variable t at t0 is defined by

Ua;1ðx; hÞ ¼ 1Gðahþ 1Þ

vahuðx; t�

vtah

!t¼t0

(2.227)

The modified differential inverse transform Ua,1(x, h) with respect tothe variable t at t0 is defined by

uðx; tÞ ¼XNh¼0

Ua;1ðx; hÞðt � t0Þah (2.228)

Since the MDTM results from the Taylor’s series of the function withrespect to the specific variable, it is expected that the corresponding alge-braic equation from the given problem is much simpler than the resultobtained by the standard DTM. The fundamental mathematical operationsperformed by two-dimensional DTM are listed in Tables 2.2 and 2.3.

Example A. Consider the nonlinear fractional KleineGordon equation

vauðx; tÞvta

� v2uðx; tÞvx2

þ u2ðx; tÞ ¼ 0 (2.229)

subject to the initial conditions

uðx; 0Þ ¼ 1þ sin x (2.230)

DTM transferred is

Gðaðhþ 1Þ þ 1ÞGðahþ 1Þ Ua;1ðk; hþ 1Þ � ðkþ 1Þðkþ 2ÞUa;1ðkþ 2; hÞ

þXkr¼0

Xhs¼0

Ua;1ðr; h� sÞUa;1ðk� r; sÞ ¼ 0

(2.231)


Table 2.2 The Operations for the Two-Dimensional Differential Transform MethodOriginal Function Transformed Function

w(x,t) ¼ u(x, t) � v(x,t) Wa,1(k,h) ¼ Ua,1(k, h) � Va,1(k, h)w(x,t) ¼ mu(x,t) Wa,1(k,h) ¼ mUa,1(k, h)

wðx;tÞ ¼ vuðx;tÞvx

Wa,1(k,h) ¼ (k þ 1)Ua,1(k þ 1, h)

wðx;tÞ ¼ Da*t0uðx; tÞ; 0 < a � 1 Wa;1ðk;hÞ ¼ Gðaðh þ 1Þþ 1Þ

Gðah þ 1Þ Ua;1ðk; hþ 1Þw(x,t) ¼ (x � x0)

m(t � t0)na

Wa;1ðk;hÞ ¼ dðk� m; ha� nÞ ¼�1 k ¼ m; h ¼ n

0 otherwisew(x,t) ¼ u2(x,t)

Wa;1ðk;hÞ ¼Xkm¼0

Xhn¼0

Ua;1ðm; h� nÞUa;1ðk� m; nÞw(x,t) ¼ u3(x,t)

Wa;1ðk;hÞ ¼Xkr¼0

Xk�r

q¼0

Xhs¼0

Xh�s

p¼o

Ua;1ðr; h� s� pÞUa;1ðq; sÞUa;1ðk� r � q; pÞ


ationMethod


Problems

Table 2.3 The Operations for the Modified Differential Transform MethodOriginal Function Transformed Function

w(x,t) ¼ u(x, t) � v(x, t) Wa,1(x, h) ¼ Ua,1(x, h) � Va,1(x, h)w(x,t) ¼ mu(x,t) Wa,1(x,h) ¼ mUa,1(x,h)

wðx;tÞ ¼ vuðx;tÞvx

Wa;1ðx;hÞ ¼ vUa;1ðx;hÞvx

wðx;tÞ ¼ Da*t0uðx; tÞ; 0 < a � 1 Wa;1ðx;hÞ ¼ Gðaðhþ1Þþ1Þ

Gðahþ1Þ Ua;1ðx; hþ 1Þw(x,t) ¼ (x � x0)

m(t � t0)na Wa,1(x,h) ¼ (x � x0)

md(ha � n)w(x,t) ¼ u2(x,t)

Wa;1ðx;hÞ ¼Xhm¼0

Ua;1ðx;mÞUa;1ðx; h� mÞw(x,t) ¼ u3(x,t)

Wa;1ðx;hÞ ¼Xhm¼0

Xml¼0

Ua;1ðx; h� mÞUa;1ðx; lÞUa;1ðx;m� lÞ

DifferentialTransform

ationMethod

inAdvance

99

The transformed of Eq. (2.230) is

Ua;1ðk; 0Þ ¼

8>>>>><>>>>>:

1; k ¼ 0

1k!; k ¼ 1; 5; 9.

�1k!; k ¼ 3; 7; 11.

(2.232)

By substituting Eq. (2.232) into Eq. (2.231) and using DTM inverse, wehave

uðx; tÞ ¼�1þ x� x3

3!þ x5

5!� x7

7!þ.

þ�1þ 3xþ x2 � x3

2� x4

3þ.

ta

Gðaþ 1Þ

þ�11xþ 12x2 þ x3

6� 4x4 þ.

t2a

Gð2aþ 1Þ þ.

(2.233)

For applying MDTM to Eq. (2.229) we have

Gðaðhþ 1Þ þ 1ÞGðahþ 1Þ Ua;1ðx; hþ 1Þ � v2Ua;1ðx; hÞ

vx2

þXhm¼0

Ua;1ðx;mÞUa;1ðx; h� mÞ ¼ 0

(2.234)

The transformed of Eq. (2.230) is

Ua;1ðx; 0Þ ¼ 1þ sin x (2.235)

Finally after substituting the result is

uðx; tÞ ¼ 1þ sin x� 1Gðaþ 1Þ ð1þ 3 sin xþ sin2 xÞta

þ 1Gð2aþ 1Þ ð11 sin xþ 12 sin2 xþ 2 sin3 xÞt2a

(2.236)


Aruna and Kanth [10] showed that MDTM can improve the results, forexample, they showed when the equation is

v2:5uðx; tÞvt2:5

� v2uðx; tÞvx2

� uðx; tÞ þ u3ðx; tÞ ¼ 0 (2.237)

With initial condition of

uðx; 0Þ ¼ �sec hx (2.238)

DTM and MDTM have a large difference as shown in Fig. 2.13.

REFERENCES[1] Joneidi AA, Ganji DD, Babaelahi M. Differential Transformation Method to determine

fin efficiency of convective straight fins with temperature dependent thermal con-ductivity. International Communication in Heat and Mass Transfer 2009;36:757e62.

[2] Hassan IHA-H. Differential transformation technique for solving higher-order initialvalue problems. Applied Mathematics and Computation 2004;154(2):299e311.

[3] Arikoglu A, Ozkol I. Solution of fractional differential equations by using differentialtransform method. Chaos, Solitons & Fractals 2007;34(5):1473e81.

[4] Nazari D, Shahmorad S. Application of the fractional differential transform method tofractional-order integro-differential equations with nonlocal boundary conditions.Journal of Computational and Applied Mathematics 2010;234(3):883e91.

[5] Chen C-K, Ho S-H. Application of differential transformation to eigenvalue problems.Applied Mathematics and Computation 1996;79(2):173e88.

[6] Hassan IHA-H. On solving some eigenvalue problems by using a differential trans-formation. Applied Mathematics and Computation 2002;127(1):1e22.

[7] Odibat Z, Momani S. A generalized differential transform method for linear partialdifferential equations of fractional order. Applied Mathematics Letters 2008;21(2):194e9.

–0.2–0.4

u[x,t] –0.6–0.8–1.0

–0.2–0.4

u[x,t] –0.6–0.8–1.0

–2 –2

0 0

2 20.000

0.005 t

0.010

0.000

DTM MDTM

(a) (b)

0.005 t

0.010

x x

Figure 2.13 Difference between Modified Differential Transformation Method andDifferential Transformation Method for u(x,t).


[8] Keskin Y, Oturanc G. Reduced differential transform method for partial differentialequations. International Journal of Nonlinear Sciences and Numerical Simulation2009;10(6):741e9.

[9] Al-Amr MO. New applications of reduced differential transform method. AlexandriaEngineering Journal 2014;53(1):243e7.

[10] Aruna K, Ravi Kanth ASV. Two-dimensional differential transform method andmodified differential transform method for solving nonlinear fractional KleineGordonequation. National Academy of Science Letters MarcheApril 2014;37(2):163e71.


CHAPTER 3

DTM for Heat Transfer Problems

3.1 INTRODUCTION

Most of the problems in mechanical engineering include the heat transferphenomena. Industrial engineering, cooling process, oil industry andmelting, shaping and deformations, automobile industry and many otherprocess have a heat transfer and researchers need to control it by increase/decrease devices. Fins are the most effective instrument for increasing therate of heat transfer, as we know they increase the area of heat transfer andcause an increase in the transferred heat volume. A complete review on thistopic is presented by Krause et al. [1]. Fins are widely used in the manyindustrial applications such as air conditioning, refrigeration, automobile,chemical processing equipment, and electrical chips. Also cooling by suit-able fluids such as nanofluids is another application of heat transfer discussedin the present chapter, which contains following sections:3.1 Introduction3.2 Longitudinal Fins With Constant Profile3.3 Natural Convection Flow of a Non-Newtonian Nanofluid3.4 Two-Dimensional Heat Transfer in Longitudinal Rectangular and

Convex Parabolic Fins3.5 Thermal Boundary Layer on Flat Plate3.6 FalknereSkan Wedge Flow3.7 Free Convection Problem

3.2 LONGITUDINAL FINS WITH CONSTANT PROFILE

Consider a longitudinal fin with constant rectangular profile, section areaA, length L, perimeter P, thermal conductivity k, and heat generation q*.The fin is attached to a surface with a constant temperature Tb and lossesheat to the surrounding medium with temperature TN through a constantconvective heat transfer coefficient h. In the problem we assume that thetemperature variation in the transfer direction is negligible, so heat con-duction occurs only in the longitudinal direction (x-direction). A schematicof the geometry of described fin and other properties is shown in Fig. 3.1.



http://dx.doi.org/10.1016/B978-0-12-805190-0.00003-6

For this problem, the governing differential equation and boundary con-dition can be written as [2]

d2Tdx2

� hPkA

ðT � TNÞ þ q*

k¼ 0 (3.1)

x ¼ 0;dTdx

¼ 0 (3.2)

x ¼ L; T ¼ Tb (3.3)

This problem is solved in two main cases using Differential Trans-formation Method (DTM). In the following subsections, the governingequations for these two cases are introduced.

A. Fin With Temperature-Dependent Internal HeatGeneration and Constant Thermal ConductivityIn the first case, we assume that heat generation in the fin varies withtemperature as given in Eq. (3.4) and the thermal conductivity is constant k0.

q* ¼ q*Nð1þ εðT � TNÞÞ (3.4)

where q*N is the internal heat generation at temperature TN. With theintroduction of following dimensionless quantities:

q ¼ ðT � TNÞðTb � TNÞ; X ¼ x

L; N2 ¼ hPL2

k0A

G ¼ q*NhPðTb � TNÞ; εG ¼ εðTb � TNÞ

(3.5)

Figure 3.1 Schematic of the fin geometry with heat generation source.


Eq. (3.1)e(3.3) can be rewritten as

d2qdX2

�N2qþN2Gð1þ εGqÞ ¼ 0 (3.6)

X ¼ 0;dqdX

¼ 0 (3.7)

X ¼ 1; q ¼ 1 (3.8)

Now we apply DTM from Chapter 1 into Eq. (3.6) to find q(x).

ðkþ 1Þðkþ 2ÞQðkþ 2Þ �N2QðkÞ þN2GðdðkÞ þ εGQðkÞÞ ¼ 0 (3.9)

Rearranging Eq. (3.9), a simple recurrence relation is obtained as follows

Qðkþ 2Þ ¼ N2QðkÞ �N2GðdðkÞ þ εGQðkÞÞðkþ 1Þðkþ 2Þ (3.10)

where

dðkÞ ¼�1 if k ¼ 0

0 if ks 0(3.11)

Similarly, the transformed form of boundary conditions can be writtenas

Qð0Þ ¼ a; Qð1Þ ¼ 0 (3.12)

By solving Eq. (3.10) and using boundary conditions (Eq. 3.12) theDTM terms are obtained as

Qð2Þ ¼ 12N2Qð0Þ � 1

2N2Gð1þ εGQð0ÞÞ

Qð3Þ ¼ 16N2Qð1Þ � 1


Qð4Þ ¼ 112N2Qð2Þ � 1


Qð5Þ ¼ 120N2Qð1Þ � 1


:

:

:

(3.13)

DTM for Heat Transfer Problems 105

Now by applying DTM principle equation, into Eq. (3.13), and byusing Eq. (3.3) the constant parameter “a” will be obtained, so the tem-perature distribution equation will be estimated.

B. Fin With Temperature-Dependent Internal HeatGeneration and Temperature-Dependent ThermalConductivityIn the second case, we assume that thermal conductivity of fin as well asinternal heat generation is temperature-dependent. We consider it to varylinearly with temperature, then we have

k ¼ k0½1þ bðT � TNÞ� (3.14)

The dimensionless form of Eq. (3.15) is

kk0

¼ ½1þ εcq� (3.15)

where

εc ¼ bðTb � TNÞ (3.16)

Eq. (3.6) for this condition becomes

ddX

�ð1þ εGqÞ dqdX

��N2qþN2Gð1þ εGqÞ ¼ 0 (3.17)

Where its boundary conditions are given by Eqs. (3.7) and (3.7). Nowwe must apply DTM to this governing equation.

ðkþ 2Þðkþ 1ÞQðkþ 2Þ þ εC

Xkm¼0

fðkþ 1� mÞQðkþ 1� mÞðmþ 1ÞQðmþ 1Þgþ

εC

Xkm¼0

fðk� mÞQðk� mÞðmþ 2ÞQðmþ 2Þ �N2QðkÞ þN 2GðdðkÞ þ εGQðkÞÞ)

¼ 0

(3.18)

Rearranging Eq. (3.17), a simple relation is obtained as follows

Qðkþ 2Þ ¼ �1ðkþ 2Þðkþ 1Þ ðεC

Xkm¼0

fðkþ 1� mQðkþ 1� mÞðmþ 1ÞQðmþ 1Þgþ

εC

Xkm¼0

fðk� mÞQðk� mÞðmþ 2ÞQðmþ 2Þ �N 2QðkÞ þN2GðdðkÞ þ εGQðkÞÞ)

(3.19)


Where the boundary conditions of this case is the same as that of theprevious case Eq. (3.12). By solving Eq. (3.19) and using boundary con-ditions Eq. (3.12), the DTM terms for this case can be as

Qð2Þ ¼ N2Qð0Þ �N2GεGQð0Þ �N2G2ð1þ εCQð0ÞÞ

Qð3Þ ¼ εCQð2ÞQð1Þ þ 16N2Qð1Þ � 1

6N2GεGQð1Þ

Qð4Þ ¼ � 112N2GεGQð2Þ � 3

4εCQð3ÞQð1Þ � 2

3εCQ

2ð2Þ þ 112N2Qð2Þ

:

:

:

(3.20)

Finally, by applying DTM into Eq. (3.20) and using Eq. (3.3) theconstant parameter “a” will be obtained and the temperature distributionequation will be calculated.

A. Case1: Fin With Temperature-Dependent Internal HeatGeneration and Constant Thermal ConductivityTemperature distribution in case 1 (temperature-dependent heat genera-tion and constant thermal conductivity) is shown in Figs. 3.2e3.5. It iscommon in fin design that the N parameter is considered to be 1. Fig. 3.1shows temperature distribution for this state and εG ¼ G ¼ 0.2,εG ¼ G ¼ 0.4, and εG ¼ G ¼ 0.6. This choice of parameters represents afin with moderate temperature-dependent heat generation and the thermalconductivity variation of 20% between the base and the surroundingcoolant temperatures that are often used in nuclear rods. As we see in thefigure by increasing εG and G the temperature of the fin is increasedbecause of increasing heat generation. By comparing the results withnumerical method, it was observed that DTM has a good efficiency andaccuracy, error of DTM is plotted in Fig. 3.1 and it reveals this fact. As seenin Fig. 3.1 the maximum error occurs in the tip of the fin. Fig. 3.4 showscomparison results, which pertain to N ¼ 0.5 (this choice is used incompact heat exchanger fin design), and this figure illustrates that fintemperature in this condition is greater than N ¼ 1 state. Fig. 3.5 shows theerrors for N ¼ 1 and εG ¼ G ¼ 0.2, εG ¼ G ¼ 0.4, and εG ¼ G ¼ 0.6. As


Figure 3.2 Temperature distribution in the fin with temperature-dependent internalheat generation and constant thermal conductivity for N ¼ 1.

Figure 3.3 Error of differential transformation method in comparison by numericalmethod for case 1 and N ¼ 1.


0 0.2 0.4 xεG = 0.4

θ

εG = 0.4εG = 0.2G = 0.2G = 0.4 G = 0.6-

+

0.6 0.8 1

1

0.98

0.97

0.96

0.95

0.94

0.93

0.92

0.99

Figure 3.4 Temperature distribution in the fin with temperature-dependent internalheat generation and constant thermal conductivity for N ¼ 0.5.

Err

or (%

)

1.2 × 10-6

1. × 10-6

8. × 10-7

6. × 10-7

4. × 10-7

2. × 10-7

0

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

εG = 0.2G = 0.2

εG = 0.4G = 0.4

εG = 0.4G = 0.6-

x

Figure 3.5 Error of differential transformation method in comparison by numericalmethod for case 1 and N ¼ 0.5.


already seen in Fig. 3.5 maximum error occurs in the tip of the fin, thisresult also occurred in Fig. 3.5. The range of the errors reveals that DTMhas a good agreement with numerical results.

B. Case 2: Fin with Temperature-Dependent Internal HeatGeneration and Temperature-Dependent ThermalConductivityFigs. 3.6e3.9 show the temperature distribution in case 2. As alreadymentioned, in case 2, thermal conductivity and heat generation are tem-perature dependent. Fig. 3.6 illustrates the temperature distribution withN ¼ 1, εG ¼ G ¼ 0.4, and εC increased from 0 to 0.6 with intervals 0.2. Asseen in Fig. 3.6, when εC increases, the local fin temperature increasesbecause the ability of the fin to conduct heat increases. Fig. 3.7 shows theerror of DTM in comparison with numerical method for N ¼ 1,εG ¼ G ¼ 0.4 and low maximum error in this figure emphasis on accuracyand efficiency of DTM. In Fig. 3.8 the N parameter is decreased to 0.5 andtemperature distribution is depicted. In Fig. 3.9, the error of DTM forN ¼ 0.5, G ¼ 0.4, εG ¼ 0.4 is depicted.

Figure 3.6 Temperature distribution in the fin with temperature-dependent internalheat generation and temperature-dependent thermal conductivity for N ¼ 1, G ¼ 0.4,εG ¼ 0.4.


Figure 3.7 Error of differential transformation method in comparison by numericalmethod for case 2 and N ¼ 1.

Figure 3.8 Temperature distribution in the fin with temperature-dependent internalheat generation and temperature-dependent thermal conductivity for N ¼ 0.5,G ¼ 0.4, εG ¼ 0.4.


Finally, by a comparative assessment of figures introduced for this caseand pervious case, it can be found that the local fin temperature increases asthe parameters G, εG, and εC increase. The increase in parameter εG impliesthat the heat generation is increased, and hence it causes to produce ahigher temperature in the fin. An increase in εC means the thermal con-ductivity of the fin is increased, and it makes more heat conducting throughthe fin and local temperature will increase.

3.3 NATURAL CONVECTION FLOW OF A NON-NEWTONIANNANOFLUID

A schematic theme of the problem is shown in Fig. 3.10. It consists of twovertical flat plates separated by a distance 2b apart. A non-Newtonian fluidflows between the plates due to natural convection. The walls at x ¼ þband x ¼ �b are held at constant temperatures T2 and T1, respectively,where T1 > T2. This difference in temperature causes the fluid near thewall at x ¼ �b to rise and the fluid near the wall at x ¼ þb to fall. The fluidis a water-based nanofluid containing Cu. It is assumed that the base fluidand the nanoparticles are in thermal equilibrium and no slip occurs between

Figure 3.9 Error of differential transformation method in comparison by numericalmethod for case 2 and N ¼ 0.5.


them. The thermophysical properties of the nanofluid are given inTable 3.1 [3].

The effective density rn f, the effective dynamic viscosity mn f, the heatcapacitance (rCp)n f, and the thermal conductivity kn f of the nanofluid aregiven as

rn f ¼ rf ð1� fÞ þ rsf (3.21)

mn f ¼mf

ð1� fÞ2:5 (3.22)

ðrCpÞn f ¼ ðrCpÞf ð1� fÞ þ ðrCpÞsf (3.23)

kn f

kf¼ ks þ 2kf � 2fðkf � ksÞ

ks þ 2kf þ fðkf � ksÞ (3.24)

here, f is the solid volume fraction.

Figure 3.10 Geometry of the problem.

Table 3.1 Thermophysical Properties of Water and Nanoparticlesr (kg/m3) Cp (j/kgk) k (W/m$k) b � 105 (K�1)

Pure water 997.1 4179 0.613 21Copper(Cu) 8933 385 401 1.67


Following, we define the similarity variables:

V ¼ y

V0;h ¼ x

b; q ¼ T � Tm

T1 � T2; (3.25)

Under these assumptions and following the nanofluid model proposedby MaxwelleGarnetts (MG) model, the NaviereStokes and energy equa-tions can be reduced to the following pair of ordinary differential equations:

d2Vdh2

þ 6dð1� fÞ2:5�dVdh

�2d2Vdh2

þ q ¼ 0; (3.26)

d2qdh2

þ EcPr

ð1� fÞ�2:5

A1

!�dVdh

�2

þ 2dEcPr

�1A1

��dVdh

�2

¼ 0: (3.27)

where Prandtl number (Pr), Eckert number (Ec), dimensionless non-Newtonian viscosity (d), and A1 have following forms:

Ec ¼ rf V20

ðrCpÞf ðT1 � T2Þ; Pr ¼ mf ðrCpÞfrf kf

; d ¼ 6b3V 20

mf b2(3.28)

A1 ¼ knfkf

¼ ks þ 2kf � 24ðkf � ksÞks þ 2kf þ 24ðkf � ksÞ (3.29)

The appropriate boundary conditions are

h ¼ �1 : V ¼ 0; q ¼ 0:5

h ¼ þ1 : V ¼ 0; q ¼ �0:5(3.30)

The average Nusselt number is defined as

Nuave ¼ � knfkf

�q0ð1Þ þ q0ð�1Þ

2

�(3.31)

Now DTM will be applied to the governing equations. Taking thedifferential transform of equations with respect to c, and considering H ¼ 1gives

ðkþ 1Þðkþ 2Þw½kþ 2� þQ½k� þ

6að1� fÞ2:5 Xk

m¼0

Xmr¼0

ðk� mþ 1Þw½k� mþ 1�ðm� r þ 1Þ

�w½k� r þ 1�ðr þ 1Þðr þ 2Þw½r þ 2�Þ!!

(3.32)


w½0� ¼ a1; w½1� ¼ a2 (3.33)

ðkþ 1Þðkþ 2ÞQ½kþ 2� þ EcPr

�ð1� fÞ2:5A1

�

�Xkm¼0

ððmþ 1Þw½mþ 1�ðk� mþ 1Þw½k� mþ 1�Þ

þ 2dEcPr

�1A1

�Xkm¼0

Xmr¼0

Xr

s¼0

ðk� mþ 1Þw½k� mþ 1�ðm� r þ 1Þ

� w½m� r þ 1�ðr � sþ 1Þw½r � sþ 1�ðr þ 1Þw½r þ 1�Þ(3.34)

Q½0� ¼ b1; Q½1� ¼ b2 (3.35)

where w(k) and Q(k) are the differential transforms of V(h) and q(h), alsoa1, a2, b1, and b2 are constants which can be obtained through boundaryconditions, Eqs. (3.26) and (3.27). This problem can be solved as follows:

w½0� ¼ a1; w½1� ¼ a2

w½2� ¼ � 12

b11þ 6dð1� fÞ2:5a22

w½3� ¼ � 16

12 b21a2 dð1� fÞ2:5 þ b2 þ 12b2dð1� fÞ2:5a22 þ 36b2�dð1� fÞ2:5�2a42�

1þ 6dð1� fÞ2:5a22�3

;.

(3.36)

Q½0� ¼ b1; Q½1� ¼ b2

Q½2� ¼ � 12EcPr

ð1� fÞ�2:5

A1

!a22 � EcPr

�1A1

�da42

Q½3� ¼ 13

EcPr

ð1� fÞ�2:5

A1

!b11þ 9dð1� fÞ2:5a22 þ 6

�dð1� fÞ2:5�2a42 � dð1� fÞ2:5a2b1

�1þ 6dð1� fÞ2:5a22

�2;.

(3.37)

The above process is continuous. By substituting Eqs. (3.36) and (3.37)into the main equation based on DTM, it can be obtained that the closedform of the solutions is


wðhÞ ¼ a1 þ a2hþ

� 12

b11þ 6dð1� fÞ2:5a22

!h2þ

� 16

12 b21a2 dð1� fÞ2:5 þ b2 þ 12b2dð1� fÞ2:5a22 þ 36b2�dð1� fÞ2:5�2a42�

1þ 6dð1� fÞ2:5a22�3

!h3 þ.

(3.38)

QnðhÞ ¼ b1 þ b2hþ

� 12EcPr

ð1� fÞ�2:5

A1

!a22 � EcPr

�1A1

�da42

!h2 þ

0BBBB@13

EcPr

ð1� fÞ�2:5

A1

!b11þ 9dð1� fÞ2:5a22 þ 6

�dð1� fÞ2:5�2a42 � dð1� fÞ2:5a2b1

�1þ 6dð1� fÞ2:5a22

�21CCCCAh3 þ.

(3.39)

By substituting the boundary condition from Eq. (3.30) into Eqs. (3.38)and (3.39) the values of a1, a2, b1, and b2 can be obtained.

wð�1Þ ¼ 0

wðþ1Þ ¼ 0

Qð�1Þ ¼ 0:5

Qðþ1Þ ¼ �0:5

(3.40)

By solving Eq. (3.40) the values of a1, a2, b1, and b2 are given. Bysubstituting obtained a1, a2, b1, and b2 into Eqs. (3.38)e(3.39), theexpression of w(k) and Q(k) can be obtained.

For example, when Ec ¼ 1, d ¼ 0.5, f ¼ 0, Pr ¼ 6.2, a1, a2, b1 and b2are obtained as follows:

a1 ¼ �1:719612304; a2 ¼ 0:2493778738e� 1; b1 ¼ 2:256751261;

b2 ¼ �0:6092964931

In this example, natural convection of a non-Newtonian nanofluidbetween two infinite parallel vertical flat plates has been investigated. Theseequations are solved analytically using DTM. To verify the accuracy of thepresent results, we have compared these results with a numerical method(the fourth-order RungeeKutta method). Comparison between numericalresults and DTM solutions for different values of effective parameter isshown in Fig. 3.11. It shows that the results obtained by DTM are in good


agreement with those carried out by the numerical solution obtained usingfourth-order Runge-Kutta method.

This accuracy gives high confidence to us about the validity of thismethod and reveals an excellent agreement of engineering accuracy. This

-1

-0.03

0

0.03

0.06

0.09

-0.5

(a)

0

Cu-Water

φ = 0.1, δ = 4, Ec

φ = 0, δ = 0.5,

φ = 0.1, δ = 0.5,

Ec

E

η0.5

NMDTMNMDTMNMDTM

c = 4

c = 0.5

Ec = 6

1

V (η

)θ

(η)

-1

-0.4

-0.2

0

0.2

0.4

(b)

-0.5

Cu-Wat

φ

φ

φ

0 0.5

ter

= 0.1, δ = 4, Ec = 4

= 0, δ = 0.5,

= 0.1, δ = 0.5,

Ec = 0.5

Ec = 6

η1

NMDTMNMDTMNMDTM

Figure 3.11 Comparison between numerical results and differential transformationmethod solution for different values of effective parameter.


investigation is completed by depicting the effects of important parameterssuch as nanoparticle volume fraction (f), dimensionless non-Newtonianviscosity (d), and Eckert number (Ec) to evaluate how these parametersinfluence this fluid. Fig. 3.12 shows the effect of nanoparticle volumefraction (f) on: (a) velocity profiles V(h); (b) temperature profiles q(h),

Figure 3.12 Effect of nanoparticle volume fraction (f) on: (a) velocity profiles V(h);(b) temperature profiles q(h), when d ¼ 1, Ec ¼ 1, and Pr ¼ 6.2.


when d ¼ 1, Ec ¼ 1, and Pr ¼ 6.2. When the volume fraction of thenanoparticles increases from 0 to 0.1, the velocity also increases. Also, wecan see that, with increasing volume fraction of the nanoparticles, thethermal boundary layer thickness decreases, and heat transfer rate increasesat the surface (Nusselt number). Fig. 3.13 displays the effect of dimen-sionless non-Newtonian viscosity (d) on: (a) velocity profiles V(h);

(a)

-1 -0.5 0 0.5 1

-0.02

0

0.02

0.04 δ = 0δ = 20δ = 50δ = 100

Cu-Water

V (η)

η(b)

Figure 3.13 Effect of Dimensionless non-Newtonian viscosity (d) on: (a) velocity pro-files V(h); (b) temperature profiles q(h), when f ¼ 0.1, Ec ¼ 1, and Pr ¼ 6.2.


(b) temperature profiles q(h), when f ¼ 0.1, Ec ¼ 1, and Pr ¼ 6.2. Thedimensionless non-Newtonian viscosity indicates the relative significance ofthe inertia effect compared to the viscous effect. The magnitudes of bothvelocity and temperature decrease as d increases, and in turn the magnitudeof the skin friction coefficient and the Nusselt number increase. Also, it canbe seen that at jhj ¼ 0:6 the maximum value of velocity is observed. Effectsof Eckert number (Ec) on: (a) velocity profiles V(h); (b) temperature profilesq(h), when f ¼ 0.1, d ¼ 1, and Pr ¼ 6.2 are shown in Fig. 3.14. It can befound that velocity and temperature increase due to increase in Eckertnumber. When we neglect viscous dissipation, minimum amount ofvelocity and temperature are obtained. Also, it can be seen that at jhj ¼ 0:5maximum value of velocity and at middle point/surface of two plates, andat (h ¼ 0) maximum value of temperature observed.

Fig. 3.15 depicts the effects of nanoparticle volume fraction (f),dimensionless non-Newtonian viscosity (d), and Eckert number (Ec) onNusselt number when (a) Ec ¼ 1; (b) d ¼ 1 and Pr ¼ 6.2. The Nusseltnumber has a direct relationship with dimensionless non-Newtonianviscosity parameter and volume fraction of the nanoparticles (whend < 20), but it has reverse relationship with Eckert number and volumefraction of the nanoparticles (when d > 20).

3.4 TWO-DIMENSIONAL HEAT TRANSFER INLONGITUDINAL RECTANGULAR AND CONVEXPARABOLIC FINS

We consider a one-dimensional longitudinal fin of an arbitrary profile F(X)and cross-sectional area Ac is shown in Fig. 3.16. The perimeter of the fin isdenoted by P and its length by L. The fin is attached to a fixed primesurface of temperature Tb and extends to an ambient fluid of temperatureTa. The fin thickness is given by d and the base thickness is db. We assumethat the fin is initially at ambient temperature. At time t ¼ 0, the tem-perature at the base of the fin is suddenly changed from Ta to Tb, and theproblem is to establish the temperature distribution in the fin for all t � 0.Based on the one-dimensional heat conduction, the energy balanceequation is then given by (see e.g., Ref. [4])

rcvTvt

¼ v

vx

�vb

2FðxÞKðTÞ vT

vX

�� PAC

HðTÞðT � TaÞ; 0 � X � L

(3.41)


-1 -0.5 0 0.5 1

0

0.04

(a)

(b)

0.08

Ec = 0Ec = 2Ec = 5Ec = 6

Cu-Water

V(η)

η

-1 -0.5 0 0.5 1

-0.4

-0.2

0

0.2

0.4 Ec = 0Ec = 2Ec = 5Ec = 6

Cu-Water

θ(η)

η

Figure 3.14 Effect of Eckert number (Ec) on: (a) velocity profiles V(h); (b) temperatureprofiles q(h), when f ¼ 0.1, d ¼ 1, and Pr ¼ 6.2.


where K and H are the nonuniform thermal conductivity and heat transfercoefficients depending on the temperature (see e.g., Ref. [4]), q is the den-sity, c is the specific heat capacity, T is the temperature distribution, t is thetime, and X is the space variable. Assuming that the fin tip is adiabatic

0 20 40 60 80 100

0.663

0.6635

0.664

0.6645

(a)

(b)

φ = 0φ = 0.05φ = 0.1

Cu-Water

δ

1 2 3 4 5 60.52

0.54

0.56

0.58

0.6

0.62

0.64

0.66φ = 0φ = 0.05φ = 0.1

Cu-Water

Ec

Nu a

veN

u ave

Figure 3.15 Effects of nanoparticle volume fraction (f), dimensionless non-Newtonianviscosity (d), and Eckert number (Ec) on Nusselt number, when (a) Ec ¼ 1; (b) d ¼ 1 andPr ¼ 6.2.


(insulated) and the base temperature is kept constant, then the boundaryconditions are given by

Tðt;LÞ ¼ Tb andvTvX

��X¼0

¼ 0 (3.42)

Initially the fin is kept at the ambient temperature,

Tð0;XÞ ¼ Ta (3.43)

Introducing the following dimensionless variables:

x ¼ XL; s ¼ Kat

rcvL2; q ¼ T � Ta

Tb � Ta; h ¼ H

kb; K ¼ K

Ka;

M 2 ¼ PhbL2

ACKa; f ðxÞ ¼ db

2FðXÞ

(3.44)

reduces Eq. (3.41) into

vq

vs¼ v

vX

�f ðxÞkðqÞ vq

vx

��M 2hðqÞq; 0 � x � 1 (3.45)

Figure 3.16 Schematic representation of a longitudinal fin of an arbitrary profile.


The prescribed boundary conditions are given by

qðs; 1Þ ¼ 1 (3.46)

vq

vxjx¼0 ¼ 0 (3.47)

and the initial condition becomes

qð0; xÞ ¼ 0 (3.48)

The dimensionless variable M is the thermo-geometric fin parameter, his the dimensionless temperature, x is the dimensionless space variable, k isthe dimensionless thermal conductivity, ka is the thermal conductivity ofthe fin at ambient temperature, hb is the heat transfer coefficient at the finbase. For most industrial application, the heat transfer coefficient maybegiven as a power law [4].

HðTÞ ¼ hb

�T � Ta

Tb � Ta

�n

(3.49)

where n and hb are constants. The constant nmay vary between �6.6 and 5.However, in most practical applications it lies between �3 and 3. Theexponent n represents laminar film boiling or condensation when n ¼ 1/4, laminar natural convection when n ¼ 1/4, turbulent natural convectionwhen n ¼ 1/3, nucleate boiling when n ¼ 2, radiation when n ¼ 3, andn ¼ 0 implies a constant heat transfer coefficient. Exact solutions may beconstructed for the steady-state one-dimensional differential equationdescribing temperature distribution in a straight fin when the thermal con-ductivity is a constant and the exponent of the heat transfer coefficient isgiven by n ¼ �1; 0; 1 or 2. Furthermore, exact solution for the steady statemay be constructed when thermal conductivity is a differential consequenceof the term involving heat transfer coefficient (see e.g., Ref. [4]), that is ifthe nonlinear ordinary differential equation is linearizable.

In dimensionless variables we have h(q) ¼ qn. Also, for many engi-neering applications, the thermal conductivity may depend linearly ontemperature, that is

KðTÞ ¼ Ka½1þ gðT � TaÞ� (3.50)

The dimensionless thermal conductivity given by the linear function oftemperature is k(q) ¼ 1 þ bq, where the thermal conductivity gradient isb ¼ m(Tb � Ta). As such the governing equation is given by

vq

vs¼ v

vx

�f ðxÞð1þ bqÞ vq

vx

��M 2qnþ1 (3.51)


We will employ the two-dimensional DTM to construct the seriessolutions to Eq. (3.51), subject to the initial and boundary conditionsEqs. (3.46)e(3.48). A comparison on temperature distribution in longitu-dinal rectangular and convex parabolic fins is established.

A. Heat Transfer in Fins of Rectangular ProfileHere the governing equation becomes

vq

vs¼ v

vx

�ð1þ bqÞ vq

vx

��M 2qNþ1 (3.52)

Taking the two-dimensional differential transform of Eq. (3.51) forn ¼ 1, we obtain the following recurrence relation,

ðK þ 1ÞQðK þ 1þHÞ ¼ ðH þ 1ÞðH þ 2ÞQðK ;H þ 1ÞQðK ;H þ 2Þ

þ bPKI¼0

PHJ¼0

QðK � I ; JÞðH þ 1� JÞðH þ 2� JÞQðI ;H þ 2� JÞ

þ bPKI¼0

PHJ¼0

ðJ þ 1ÞQðK � I ; J þ 1ÞðH þ 1� JÞQðI ;H þ 1� JÞ

�M 2XKI¼0

XHJ¼0

QðI ;H � JÞQðK � I ; JÞ

(3.53)

where Q(K,H) is the differential transform of q(s,X). Taking the two-dimensional differential transform of the initial condition Eq. (3.48) andboundary condition Eq. (3.47) we obtain the following transformations,respectively,

Qð0;HÞ ¼ 0;H ¼ 0; 1; 2; :: (3.54)

QðK ; 1Þ ¼ 0;K ¼ 0; 1; 2;. (3.55)

We consider the other boundary condition as follows:

QðK ; 1Þ ¼ a; a˛R; k ¼ 1; 2; 3 (3.56)

where the constant a can be determined from the boundary Eq. (3.46)at each time step after obtaining the series solution. SubstitutingEqs. (3.54)e(3.56) into (3.53) we obtain the following,

Qð1; 2Þ ¼ a (3.57)


Qð2; 2Þ ¼ 1=2ð3a� 2ba2 þ a2M 2Þ (3.58)

Qð3; 2Þ ¼ 1=2ð4a� 5ba2 þ 2b2a3 þ 2a2m2 � ba3m2 (3.59)

Qð1; 4Þ ¼ 1=12ð3a� 2ba2 þ a2m2Þ (3.60)

Qð2; 4Þ ¼ 1=24ð12a� 33ba2 þ 10b2a3 � 5ba3m2 (3.61)

Substituting Eqs. (3.57)e(3.61) into Eq. (3.53), we obtain the followinginfinite series solution given by

qðs; xÞ ¼ asþ as2 þ asx2 þ 1=2ð3a� 2ba2 þ a2m2Þs2x2 þ as3

þ 1=2�4a� 5ba2 þ 2b2a3 þ 2a2m2 � ba3m2

�s3x2

þ as4 þ 1=12ð3a� 2ba2 þ a2m2Þsx4

þ 1=24�12a� 33ba2 þ 10b2a3 þ 10a2m2 � 5ba3m2

�s2x4

(3.62)

The constant a can be determined from the boundary condition (3.46)at each time step. To obtain the value of a, we substitute the boundarycondition Eq. (3.46) into Eq. (3.62) at the point x ¼ 1. Thus, we have,

qðs; 1Þ ¼ asþ as2 þ asþ 1=2ð3a� 2ba2 þ a2m2Þs2 þ as3

þ 1=2�4a� 5ba2 þ 2b2a3 þ 2a2m2 � ba3m2

�s3

þ as4 þ 1=12ð3a� 2ba2 þ a2m2Þsþ 1=24

�12a� 33ba2 þ 10b2a3 þ 10a2m2 � 5ba3m2

�s2

(3.63)

We then obtain the expression for h(x) upon substituting the obtainedvalue of a into Eq. (3.62). Using the first 40 terms of the power seriessolution, we plot the solution Eq. (3.62) for special parameters as shown inFig. 3.17.

B. Heat Transfer in Fins of Convex Parabolic ProfileIn this case, the governing equation becomes

vq

vs¼ v

vx

�x1=2ð1þ bqÞ vq

vx

�� m2qnþ1 (3.64)

The transformation y ¼ x1/2 reduces Eq. (3.64) into

4yvq

vs¼ v

vy

�ð1þ bqÞ vq

vy

�� 4ym2qnþ1 (3.65)


Eq. (3.65) is more susceptible to DTM analysis. Following the techniquesabove we construct series solutions. When n ¼ 2, the solution will be

qðs; xÞ ¼ asþ as2 þ 4a3sx3=2 � 2a=3ð� 3þ 2ab� am2Þs2x3=2 þ as3

þ 2a=3�4� 5abþ 2a2b2 þ 2am2 � a2bm2

�s3x3=2 þ as4 þ.

(3.66)

Which is depicted in Fig. 3.18.

3.5 THERMAL BOUNDARY LAYER ON FLAT PLATE

Consider the flow of a viscous fluid over a semiinfinite flat plate, as shownin Fig. 3.19. The temperature of the wall, Tw, is uniform and constant and isgreater than the free stream temperature, TN. It is assumed that the freestream velocity, UN, is also uniform and constant.

Further, assuming that the flow in the laminar boundary layer is two-dimensional, and that the temperature changes resulting from viscousdissipation are small, the continuity equation and the boundary layerequations may be expressed as [5]

dudx

þ dn

dx¼ 0; (3.67)

Figure 3.17 Transient temperature distribution in a longitudinal rectangular fin. Heren ¼ 1; b ¼ 1; M ¼ 6.


ududx

þ ndudy

¼ nv2uvy2

; (3.68)

udtdx

þ ndtdy

¼ av2tvy2

; (3.69)

where u and v are the velocity components in x- and y-direction of thefluid, a is the thermal diffusivity of the fluid, T is the temperature distribu-tion in the vicinity of the plate, and the boundary conditions are given by

at y ¼ 0 : u ¼ n ¼ 0 and t ¼ tw (3.70)

at y/N : u/uN and t ¼ tN (3.71)

Figure 3.18 Transient temperature distribution in a rectangular and convex parabolicfin with linear thermal conductivity, n ¼ 2; M ¼ 6; b ¼ 1; s ¼ 1.2.

Figure 3.19 Velocity and thermal boundary layers on a flat plate.


at x/0 : u ¼ uN and t ¼ tN. (3.72)

A stream function, j(x, y), is introduced such that

u ¼ djdy

and n ¼ � djdx

: (3.73)

In addition to the physical considerations which require the introduc-tion of this function, the mathematical significance of its use is that theequation of continuity, i.e., Eq. (3.67), is satisfied identically, and themomentum equation becomes

dj

dyd2j

vxdy� dj

dxd2j

dy2¼ n

d3j

dy3(3.74)

Integrating Eq. (3.73) and introducing a similarity variable yield

f ðhÞ ¼ jffiffiffiffiffiffiuN

pnx

; (3.75)

h ¼ffiffiffiffiffiffiuNnx

r. (3.76)

Substituting Eqs. (3.68) and (3.69) into Eq. (3.67) gives

d3f ðhÞdh3

þ 12f ðhÞ d

2f ðhÞdh2

¼ 0: (3.77)

The boundary conditions of f(h) are given by

at h ¼ 0 f ð0Þ ¼ df ð0Þdh

¼ 0: (3.78)

at h/N :df Nð Þdm

¼ 1: (3.79)

A dimensionless temperature parameter is defined as follows:

q ¼ t � twtN � tw

. (3.80)

If Eq. (3.73) is substituted into Eq. (3.69), the boundary layer energyequation then becomes


d2qðhÞdh2

þ 12Prf ðhÞ dqðhÞ

dh¼ 0; (3.81)

with the following boundary conditions:

at h ¼ 0 : q ¼ 0; (3.82)

at h ¼ 0/N : q ¼ 1; (3.83)

where Pr is the Prandtl number, which is equal to the ratio of themomentum diffusivity of the fluid to its thermal diffusivity (i.e., Pr ¼ v

a).

The boundary value problems (Eqs. 3.70e3.72) can be reduced to a pairof initial value problems by means of a group of transformations [5]. Theinitial value problems are given by

d2qðhÞdh2

þ 12f ðzÞ d

2f ðzÞdz2

¼ 0; (3.84)

With initial conditions of

z ¼ 0 : f ð0Þ ¼ df ð0Þdz

¼ 0;d2f ð0Þd2z

¼ 1; (3.85)

and by,d3f ðhÞdh3

þ 12f ðhÞ d

2f ðhÞdh2

¼ 0; (3.86)

with initial conditions of

h ¼ 0 : f ð0Þ ¼ df ð0Þdh

¼ 0;d2f ð0Þdh2

¼

2664

1df ðNÞdz

3775

32

(3.87)

These equations suggest a transformation of the form:

f ðzÞ ¼ l�13 f ðhÞ; z ¼ l

13h; l ¼

2664

1df ðNÞdz

3775

32

. (3.88)

The DTM is then used to solve the pair of initial value problems(Eqs. 3.84e3.87). The following expression is initially defined:

yðzÞ ¼ df ðzÞdz

; (3.89)


and

zðzÞ ¼ dyðzÞdz

¼ d2f ðzÞdz2

¼ 0 (3.90)

Thereafter, the third-order ordinary differential equation (Eq. 3.86) isreduced to a first-order ordinary differential equation with the followingform:

dzðzÞdz

þ 12f ðzÞzðzÞ ¼ 0: (3.91)

The initial conditions become

z ¼ 0 : f ð0Þ ¼ yð0Þ ¼ 0; zð0Þ ¼ 1: (3.92)

Eqs. (3.89)e(3.91) undergo the differential transformation to give thefollowing:

kþ 1hi

Fiðkþ 1Þ ¼ YiðkÞ; (3.93)

kþ 1hi

Yiðkþ 1Þ ¼ ZiðkÞ; (3.94)

kþ 1hi

Ziðkþ 1Þ þ 12Fi � Ziðkþ 1Þ þ 1

2

Xki¼0

Fiðk� lÞZiðlÞ ¼ 0: (3.95)

The third-order ordinary differential equation (i.e., Eq. 3.86) becomes afirst-order ordinary differential equation with the following form [5]:

dyðhÞdm

þ 12f ðhÞyðhÞ ¼ 0 (3.96)


h ¼ 0 : f ð0Þ ¼ uð0Þ ¼ 0; yð0Þ ¼ l�32 . (3.97)

More details about the solution are presented in Ref. [5].Since the solutions of Eq. (3.70) can be established from the previous

calculations, f(h) is also known and can be substituted into Eq. (3.74) tosolve the boundary layer energy equation. Eq. (3.74) is a linear second-order ordinary differential equation with variable coefficients. The


solution of this energy equation can be obtained by using the method ofsuperposition. The following relationship is established.

qðhÞ ¼ cðhÞ þ sDðhÞ. (3.98)

Substituting Eq. (3.98) into Eq. (3.74) and separating the resultingequations into a group of terms give two initial value problems, i.e.,

D2CðhÞDh2

þ 12Prf ðhÞ dcðhÞ

dh¼ 0; (3.99)


h ¼ 0 : cð0Þ ¼ 0;dcð0Þdh

¼ 1; (3.100)

and

d2DðhÞdh2

þ 12Prf ðhÞ dDðhÞ

dh¼ 0; (3.101)


h ¼ 0 : Dð0Þ ¼ 0;dDð0Þdh

¼ �1: (3.102)


dqð0Þdh

¼ 1� s. (3.103)

The parameter “s” in Eq. (3.98) can be calculated by using the boundarycondition given in Eq. (3.74). This yields

s ¼ 1� cðNÞDðNÞ . (3.104)

The solutions of the pair of linear second-order ordinary differentialequations (Eqs. 3.99e3.102) can be obtained from the DTM. Initially, thefollowing relationship is defined:

wðhÞ ¼ dcðhÞdh

. (3.105)

Substituting Eq. (3.105) into Eq. (3.99) gives

dwðhÞdh

þ 12Prf ðhÞwðhÞ ¼ 0: (3.106)



h ¼ 0 : cð0Þ ¼ 0; wð0Þ ¼ 1: (3.107)

By applying DTM, Eqs. (3.105) and (3.106) undergo the differentialtransformation to give the following:

kþ 1h

Ciðkþ 1Þ ¼ WiðkÞ; (3.108)

kþ 1hi

Ciðkþ 1Þ þ 12fiðkÞ �WiðkÞ

¼ kþ 1hi

Wiðkþ 1Þ þ 12

Xmi¼0

fiðk� lÞWiðlÞ ¼ 0:(3.109)

More details of solutions and boundary conditions can be found inRef. [5].

The variation of the values of f(h) and its derivatives are plotted inFig. 3.20. Since f(h) is known, the boundary layer energy equation (Eq.3.74) can be solved numerically for various values of Prandtl number. Thetemperature distributions in the thermal boundary layer over the flat plateare shown in Fig. 3.21 for a range of Prandtl numbers.

3.6 FALKNEReSKAN WEDGE FLOW

Consider the flow of an incompressible viscous fluid over a wedge, asshown in Fig. 3.22. The temperature of the wall, Tw, is uniform and

Figure 3.20 Solutions of Eq. (3.70) for function f(h) and its derivatives.


constant and is greater than the free stream temperature, T1. It is assumedthat the free stream velocity, U1, is also uniform and constant. Further,assuming that the flow in the laminar boundary layer is two-dimensional,and that the temperature changes resulting from viscous dissipation aresmall, the continuity equation and the boundary layer equations may beexpressed as [6]

vuvx

þ vy

vy¼ 0; (3.110)

uvuvx

þ yvuvy

¼ UdUdx

þ yv2uvy2

; (3.111)

Figure 3.21 Temperature profiles in the laminar boundary on a flat plate.

Figure 3.22 Velocity and thermal boundary layers for the FalknereSkan wedge flow.


uvTvx

þ yvTvy

¼ þav2Tvy2

; (3.112)

where u and v are the respective velocity components in the x- and y-direction of the fluid flow, m is the viscosity of the fluid, and U is the refer-ence velocity at the edge of the boundary layer and is a function of x. a isthe thermal diffusivity of the fluid, T is the temperature in the vicinity ofthe wedge, and the boundary conditions are given by

at y ¼ 0 : u ¼ y ¼ 0; and T ¼ Tw (3.113)

at y/N : u/U xð Þ ¼ UN x=Lð Þm;and T ¼ TN;

(3.114)

at x ¼ 0 : u ¼ UN and T ¼ TN; (3.115)

where UN is the mean stream velocity, L is the length of the wedge, m isthe FalknereSkan power-law parameter, and x is measured from the tip ofthe wedge. A stream function, j(x,y), is introduced such that

u ¼ vj

vyand y ¼ � vj

vx: (3.116)

In addition to the physical considerations which require the introduc-tion of this function, the mathematical significance of its use is that theequation of continuity, i.e., Eq. (3.110), is satisfied identically. The mo-mentum equation becomes

vj

vyv2j

vxvy� vj

vxv2j

vy2¼ U

dUdx

þ yv3j

vy3. (3.117)

Integrating Eq. (3.116) and introducing a similarity variable yield

f ðhÞ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ m2

Lm

yUN

r �j xð1þmÞ=2�; (3.118)

f ðhÞ ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ m2

UN

yLm

r �y xð1�mÞ=2� (3.119)


d3f ðhÞdh3

þ f ðhÞ d2f ðhÞdh2

þ b

"1�

�df ðhÞdh

�2#¼ 0; (3.120)


which is known as the FalknereSkan boundary layer equation [6]. Theboundary conditions of f(h) are given by

at h ¼ 0 : f ð0Þ ¼ df ð0Þdh

¼ 0; (3.121)

at h/Ndf Nð Þdh

¼ 1: (3.122)

Note that in the equations above, parameters b and m are relatedthrough the expression b ¼ 2 m/(1 þ m). A dimensionless temperature isdefined as follows:

q ¼ T � TN

TN � Tw(3.123)

If Eq. (3.123) is substituted into Eq. (3.112), the boundary layer energyequation then becomes

d2qðhÞdh2

þ Prf ðh; bÞ dqðhÞdh

¼ 0 (3.124)


at h ¼ 0 : q ¼ 0 (3.125)

at h/N : q ¼ 1 (3.126)

where Pr is the Prandtl number, which is equal to the ratio of themomentum diffusivity of the fluid to its thermal diffusivity (i.e., Pr ¼ y

a).

Eqs. (3.120) and (3.124) present a system of ordinary differential equationsfor the FalknereSkan boundary layer problem. Simultaneous solution ofthese two equations yields the velocity and temperature profiles for theflow of a viscous fluid passing a wedge. To solve the FalknereSkan bound-ary layer equation for a family of values of b, it is first necessary to define adependent variable, g(h), i.e.,

gðhÞ ¼ vf ðhÞvb

. (3.127)

Differentiating Eqs. (3.120)e(3.122) with respect to b gives

d3gðhÞdh3

þ f ðhÞ d2gðhÞdh2

þ gðhÞ d2gðhÞdh2

þ"1�

�df ðhÞdh

�2#

� 2bdf ðhÞdh

dgðhÞdh

¼ 0;

(3.128)


The boundary conditions are given by

gð0Þ ¼ dgð0Þdh

¼ 0;dgðNÞdh

¼ 0 (3.129)

The method of superposition is used together with a group of trans-formations to solve the boundary layer equation given in Eq. (3.128).

gðhÞ ¼ PðhÞ þ C1QðhÞ; (3.130)

where C1 is a constant to be determined. Substituting Eq. (3.130) intoEq. (3.128) gives the following pair of initial value problems:

d3PðhÞdh3

þ f ðhÞ d2PðhÞdh2

þ PðhÞ d2gðhÞdh2

þ�2bdf ðhÞdh

dPðhÞdh

¼�df ðhÞdh

�2

� 1; (3.131)


Pð0Þ ¼ dPð0Þdh

¼ d2PðhÞdh2

¼ 0; (3.132)

and

d3QðhÞdh3

þ f ðhÞ d2QðhÞdh2

þQðhÞ d2f ðhÞdh2

þ�2bdf ðhÞdh

dQðhÞdh

¼ 0; (3.133)


Qð0Þ ¼ dQðNÞdh

¼ 0;d2QðhÞdh2

¼ 1: (3.134)

Details of this method can be found in Ref. [6]. The solution of thisenergy equation can be obtained by using the method of superposition.The following relationship is established:

qðhÞ ¼ CðhÞ þ C2DðhÞ. (3.135)

Substituting Eq. (3.135) into Eq. (3.124), and separating the resultingequations into a group of terms gives two initial value problems, i.e.,

d2CðhÞdh2

þ Prf ðhbÞ dCðhÞdh

¼ 0; (3.136)



h ¼ 0 : Cð0Þ ¼ 0;dCðhÞdh

¼ 1 (3.137)

and

d2DðhÞdh2

þ Prf ðhbÞ dDðhÞdh

¼ 0; (3.138)


h ¼ 0 : Dð0Þ ¼ 0;dDð0Þdh

¼ �1: (3.139)


dqð0Þdh

¼ 1� C2: (3.140)

The parameter “C2” in Eq. (3.135) can be calculated by using theboundary condition given in Eq. (3.126). This yields

C2 ¼ 1� CðNÞDðNÞ . (3.141)

By solving Eqs. (3.136)e(3.139) then gives C(h), D(h), and theirderivatives. The value of C2, the values of q(h) are derived from Eq.(3.135). Hence, we have been determined the solutions of the FalknereSkan wedge flow.

The DTM is then used to solve the pair of initial value problems (whenb ¼ 0). Initially, the following expressions are defined [6]:

yðzÞ ¼ dFðzÞdz

; (3.142)

and

zðzÞ ¼ dyðzÞdz

¼ d2FðzÞdz2

: (3.143)

Thereafter, the third-order ordinary differential equation is reduced to afirst-order ordinary differential equation with the following form:

dzðzÞdz

þ FðzÞzðzÞ ¼ 0: (3.144)



z ¼ 0 : Fð0Þ ¼ yð0Þ ¼ 0; zð0Þ ¼ 1: (3.145)

By a process of inverse differential transformation, performing differ-ential transformation of Eqs. (3.142)e(3.144) gives the following:

kþ 1Hi

Fiðkþ 1Þ ¼ YiðkÞ; (3.146)

kþ 1Hi


kþ 1Hi

Ziðkþ 1Þ þ FiðkÞ � ZiðkÞ

¼ kþ 1Hi

Ziðkþ 1Þ þXkl¼1

ðk� 1ÞZiðLÞ ¼ 0(3.148)

By changing the third-order ordinary differential equation into a first-order ordinary differential equation the solution can be found [6]. By thecurrent method, solution of other governing equation (energy) is obtainedand presented [6]. Fig. 3.23 plots the variation in the values of f(h) and itsderivatives for various values of b. Fig. 3.24 plots the dimensionless tem-perature distributions of the FalknereSkan boundary layer problem for thePrandtl number range of 0.001e10,000.

Figure 3.23 Numerical results of f(g) and its derivatives for various values of b.


3.7 FREE CONVECTION PROBLEM

Consider the flow of an incompressible viscous fluid over a vertical plate.Assuming that the flow in the laminar boundary layer is two-dimensional,the continuity equation and the boundary layer equations may beexpressed as [7],

vuvx

þ vvvy

¼ 0 (3.149)

uvuvx

þ vvuvy

¼ vv2uvy2

þ gTw � TN

TN

q (3.150)

uvq

vxþ v

vq

vy¼ a

v2q

vy2(3.151)

where u and n are the velocity components in the x- and y-direction of thefluid flow, respectively, n is the viscosity of the fluid, a is the thermal diffu-sivity of the fluid, g is the gravitational acceleration of the plate, Tw is thetemperature of the wall and is assumed to be uniform and constant, andTN is the free stream temperature (Fig. 3.25).

Figure 3.24 Dimensionless temperature profiles for b ¼ 0 and various Prandtlnumber.


A dimensionless temperature can be defined as follows:

q ¼ T � Tw

TN � Tw(3.152)

where T is the temperature distribution in the vicinity of the plate, and theboundary conditions are given by

at y ¼ 0 : u ¼ n ¼ 0; and q ¼ 1 (3.153)

at y/N : u ¼ 0; and q ¼ 0 (3.154)

A stream function, J(x,y), is introduced such that

u ¼ vj

vyand n ¼ �vj

vx(3.155)

Introducing the similarity variables yields

h ¼ Cy x1=4;j ¼ 4nCx3=4FðhÞ; C ¼ ½gðTw � TNÞ=4n2TN�1=4 (3.156)

Figure 3.25 Velocity and temperature profiles in free convection flow over a verticalplate.


Eqs. (3.149)e(3.151) then become a coupled system of differentialequations expressed in terms of F(h) and q(h), i.e.,

d3FðhÞdh3

þ 3FðhÞ d2FðhÞdh2

þ 2

�dFðhÞdh

�2

þ qðhÞ ¼ 0 (3.157)

d2qðhÞdh2

þ 3PrFðhÞ dqðhÞdh

¼ 0 (3.158)


at h ¼ 0 : Fð0Þ ¼ 0;dFð0Þdh

¼ 0; qð0Þ ¼ 1; (3.159)

at h/N :dFðNÞdh

¼ 0; qðNÞ ¼ 0; (3.160)

where Pr is the Prandtl number, which is given by the ratio of the mo-mentum diffusivity of the fluid to its thermal diffusivity (i.e., Pr ¼ n/a).Eqs. (3.157) and (3.158) represent a coupled system of ordinary differentialequations for the free convection problem. Simultaneous solution of thesetwo equations yields the velocity and temperature profiles for the flow of aviscous fluid passing the vertical plate. To solve the free convection equa-tions for a family of values of Pr, it is first necessary to define the dependentvariables, g(h) and S(h), i.e.,

gðhÞ ¼ vFðhÞvPr

; SðhÞ ¼ vqðhÞvPr

. (3.161)

Differentiating Eqs. (3.157) and (3.158) with respect to Pr gives

d3gðhÞdh3

þ 3FðhÞ d2gðhÞdh2

� 4dFðhÞdh

dgðhÞdh

þ 3d2FðhÞdh2

gðhÞ þ SðhÞ ¼ 0

(3.162)

d2SðhÞdh2

þ 3FðhÞ dqðhÞdh

þ 3PrgðhÞ dqðhÞdh

þ 3PrFðhÞ dSðhÞdh

¼ 0 (3.163)

The boundary conditions are given by

at h ¼ 0 : gð0Þ ¼ 0;dgð0Þdh

¼ 0; Sð0Þ ¼ 0; (3.164)

at h/N :dgðNÞdh

¼ 0; SðNÞ ¼ 0: (3.165)


The boundary layer equations given in Eqs. (3.162) and (3.163) aresolved by means of the superposition method and a group of trans-formations. Initially, the following expressions are defined in terms of thetwo parameters l and m:

gðhÞ ¼ g1ðhÞ þ lg2ðhÞ þ mg3ðhÞ; (3.166)

SðhÞ ¼ S1ðhÞ þ lg2ðhÞ þ mS3ðhÞ. (3.167)

Substituting these expressions into Eqs. (3.162)e(3.165) and thenseparating the resulting equations give the three sets of initial value prob-lems presented below

d3g1ðhÞdh3

þ 3FðhÞ d2g1ðhÞdh2

� 4dFðhÞdh

dg1ðhÞdh

þ 3d2FðhÞdh2

g1ðhÞ þ S1ðhÞ ¼ 0

(3.168)

d2S1ðhÞdh2


þ 3Prg1ðhÞ dqðhÞdhþ 3PrFðhÞ dS1ðhÞ

dh¼ 0

(3.169)


g1ð0Þ ¼ dg1ð0Þdh

¼ d2g1ð0Þdh2

¼ 0 (3.170)

S1ð0Þ ¼ dS1ð0Þdh

¼ 0 (3.171)

and

d3g2ðhÞdh3


� 4dFðhÞdh

dg2ðhÞdh

þ 3d2FðhÞdh2


(3.172)

d2S2ðhÞdh2


dh¼ 0 (3.173)



¼ d2g2ð0Þdh2

¼ 0; (3.174)

S2ð0Þ ¼ 0;dS2ð0Þdh

¼ 1 (3.175)


and

d3g3ðhÞdh3


� 4dFðhÞdh

dg3ðhÞdh

þ 3d2FðhÞdh2


(3.176)

d2S3ðhÞdh2


dh¼ 0 (3.177)



¼ 0;d2g3ð0Þdh2

¼ 1; (3.178)

S3ð0Þ ¼ dS3ð0Þdh

¼ 0; (3.179)

Substituting the boundary conditions at infinity from Eq. (3.165) intoEqs. (3.166) and (3.167) gives the values of parameters l and m as,

l ¼ S1ðNÞdg3ðNÞ=dh� S3ðNÞdg1ðNÞ=dhS3ðNÞdg2ðNÞ=dh� S2ðNÞdg3ðNÞ=dh ; (3.180)

m ¼ S2ðNÞdg1ðNÞ=dh� S1ðNÞdg2ðNÞ=dhS3ðNÞdg2ðNÞ=dh� S2ðNÞdg3ðNÞ=dh . (3.181)

More details about the solution process can be found in Ref. [7]. Tosolve Eqs. (3.157) and (3.158) using the DTM, it is first necessary to solvethese equations for Pr ¼ 1. The boundary conditions used are

at h ¼ 0 : Fð0Þ ¼ 0;dFð0Þdh

¼ 0;d2Fð0Þdh2 ¼ 0:6421;

qð0Þ ¼ 1;dqð0Þdh

¼ �0:5671

(3.182)

The DTM is then used to solve the system of initial value problems forPr ¼ 1. Initially, the following expressions are defined:

yðhÞ ¼ dFðhÞdh

; zðhÞ ¼ dyðhÞdh

¼ d2FðhÞdh2

(3.183)

and

sðhÞ ¼ dqðhÞdh

(3.184)


Thereafter, the third-order and second-order ordinary differentialequations (Eqs. 3.157 and 3.158) are reduced to the following first-orderordinary differential equations:

dzðhÞdh

þ 3FðhÞzðhÞ � 2y2ðhÞ þ qðhÞ ¼ 0; (3.185)

dsðhÞdh

þ 3FðhÞsðhÞ ¼ 0; (3.186)


h ¼ 0 : Fð0Þ ¼ yð0Þ ¼ 0; zð0Þ ¼ 0:6421

qð0Þ ¼ 1; sð0Þ ¼ �0:5671(3.187)

Performing differential transformation of Eqs. (3.185)e(3.187) yieldsthe following:

kþ 1Hi

Fðkþ 1Þ ¼ YiðkÞ; kþ 1Hi


kþ 1Hi

Qiðkþ 1Þ ¼ SiðkÞ; (3.189)

kþ 1Hi

Ziðkþ 1Þ þ 3Xki¼0

Fðk� iÞZiðiÞ � 2Xki¼0

Yðk� iÞYðiÞ þQðkÞ ¼ 0

(3.190)

kþ 1Hi

Siðkþ 1Þ þ 3Xki¼0

Fðk� iÞSiðiÞ ¼ 0: (3.191)

various values of Fi(k), Yi(k), Zi(k), Qi(k), and Si(k) are obtained by usingEqs. (3.188)e(3.191) and the transformed initial conditions.

To solve Eqs. (3.168)e(3.171) using the DTM, the following expres-sions are first defined:

AðhÞ ¼ dg1ðhÞdh

; BðhÞ ¼ dAðhÞdh

¼ d2g1ðhÞdh2

(3.192)

and

CðhÞ ¼ dS1ðhÞdh

. (3.193)


Hence, the third-order and second-order ordinary differential equations(Eqs. 3.168 and 3.169) become two first-order ordinary differential equa-tions with the following forms:

dBðhÞdh

þ 3FðhÞBðhÞ � 4dFðhÞdh

AðhÞ þ 3d2FðhÞdh2


(3.194)

dCðhÞdh


þ 3Prg1ðhÞ dqðhÞdhþ 3PrFðhÞ þ CðhÞ ¼ 0 (3.195)


g1ð0Þ ¼ Að0Þ ¼ Bð0Þ ¼ 0 (3.196)

S1ð0Þ ¼ Cð0Þ ¼ 0 (3.197)

As in the previous procedure, a process of inverse differential trans-formation is used to yield the following solutions:

g1iðhÞ ¼Xmk¼0

�h

Hi

�k

G1iðkÞ; AiðhÞ ¼Xmk¼0

�h

Hi

�k

AiðkÞ;

BiðhÞ ¼Xmk¼0

�h

Hi

�k

BiðkÞ;(3.198)

S1iðhÞ ¼Xmk¼0

�h

Hi

�k

S1iðkÞ; CiðhÞ ¼Xmk¼0

�h

Hi

�k

CiðkÞ;

where 0 � h � Hi;

(3.199)

where, as before, i ¼ 0,1,2,.,n indicates the ith subdomain, k ¼ 0,1,2,.,mrepresents the number of terms of the power series. From the initial condi-tions (Eqs. 3.195 and 3.196) and the solution equations (Eqs. 3.198 and3.199), it can be shown that

G1ð0Þ ¼ Að0Þ ¼ Bð0Þ ¼ 0 (3.200)

S1ð0Þ ¼ Cð0Þ ¼ 0 (3.201)

Eqs. (3.192)e(3.195) undergo a process of differential transformation togive the following:

kþ 1Hi

G1iðkþ 1Þ ¼ AiðkÞ; kþ 1Hi

Aiðkþ 1Þ ¼ BiðkÞ (3.202)


kþ 1Hi

S1iðkþ 1Þ ¼ CiðkÞ; (3.203)

kþ 1Hi

Biðkþ 1Þ þ 3Xki¼0

Biðk� iÞFiðiÞ � 4Xki¼0

Aiðk� iÞYðiÞ

þ 3Xki¼0

G1iðk� iÞZðiÞS1iðkÞ ¼ 0

(3.204)

kþ 1Hi

Ciðkþ 1Þ þ 3Xki¼0

Siðk� iÞFiðiÞ þ 3PrXki¼0

Siðk� iÞG1iðiÞ

þ 3PrXki¼0

Ciðk� iÞFiðkÞ ¼ 0:

(3.205)

As in the previous procedures, Eqs. (3.172)e(3.175) can be solved usingthe DTM. The following expressions are defined:

DðhÞ ¼ dg2ðhÞdh

; EðhÞ ¼ dDðhÞdh

¼ d2g2ðhÞdh2

(3.206)

and

HðhÞ ¼ dS2ðhÞdh

(3.207)


dEðhÞdh

þ 3FðhÞEðhÞ � 4dFðhÞdh

DðhÞ þ 3d2FðhÞdh2


(3.208)

dHðhÞdh

þ 3Prg2ðhÞ dqðhÞdhþ 3PrFðhÞHðhÞ ¼ 0 (3.209)


g2ð0Þ ¼ Dð0Þ ¼ Eð0Þ ¼ 0 (3.210)

S2ð0Þ ¼ 0;Hð0Þ ¼ 1 (3.211)


As in the previous procedure, a process of inverse differential trans-formation is used to yield the following solutions:

g2iðhÞ ¼Xmk¼0

�h

Hi

�k

G2iðkÞ; DiðhÞ ¼Xmk¼0

�h

Hi

�k

DiðkÞ;

EiðhÞ ¼Xmk¼0

�h

Hi

�k

EiðkÞ(3.212)

S2iðhÞ ¼Xmk¼0

�h

Hi

�k

S2iðkÞ; HiðhÞ ¼Xmk¼0

�h

Hi

�k

HiðkÞ;

where 0 � h � Hi

(3.213)

From the initial conditions (Eqs. 3.210 and 3.211) and the solutionequations (Eqs. 3.212 and 3.213), it can be shown that

G2ð0Þ ¼ Dð0Þ ¼ Eð0Þ ¼ 0; (3.214)

S2ð0Þ ¼ 0; Hð0Þ ¼ dðkÞ; where dðkÞ ¼�1 k ¼ 0

0 ks 0(3.215)


kþ 1Hi

G2iðkþ 1Þ ¼ DiðkÞ; kþ 1Hi

Diðkþ 1Þ ¼ EiðkÞ (3.216)

kþ 1Hi

S2iðkþ 1Þ ¼ HiðkÞ; (3.217)

kþ 1Hi

Eiðkþ 1Þ þ 3Xki¼0

Eiðk� iÞFðiÞ � 4Xki¼0

Diðk� iÞYðiÞ

þ 3Xki¼0

G2iðk� iÞZðiÞ þ S2iðkÞ ¼ 0

(3.218)

kþ 1Hi

Hiðkþ 1Þ þ 3PrXki¼0

Hiðk� iÞFðiÞ þ 3PrXkt¼0

Siðk� iÞG2iðiÞ ¼ 0

(3.219)

As in the previous procedures, Eqs. (3.176)e(3.179) can be solved usingthe DTM. The following expressions are defined:

IðhÞ ¼ dg3ðhÞdh

; JðhÞ ¼ dIðhÞdh

¼ d2g3ðhÞdh2

(3.220)


and

KðhÞ ¼ dS3ðhÞdh

(3.221)


dJðhÞdh

þ 3FðhÞJðhÞ � 4dFðhÞdh

IðhÞ þ 3d2FðhÞdh2

g3ðhÞ þ s3ðhÞ ¼ 0 (3.222)

dKðhÞdh

þ 3Prg3ðhÞ dqðhÞdhþ 3PrFðhÞKðhÞ ¼ 0 (3.223)


g3ð0Þ ¼ Ið0Þ ¼ 0; Jð0Þ ¼ 1 (3.224)

S3ð0Þ ¼ 0;Kð0Þ ¼ 0: (3.225)

As in the previous procedure, inverse differential transformation is usedto yield the following solutions:

g3iðhÞ ¼Xmk¼0

�h

Hi

�k

G3iðkÞ; IiðhÞ ¼Xmk¼0

�h

Hi

�k

I iðkÞ;

JðhÞ ¼Pmk¼0

�h

Hi

�k

J iðkÞ;(3.226)

S3iðhÞ ¼Xmk¼0

�h

Hi

�k

S3iðkÞ; KiðhÞ ¼Xmk¼0

�h

Hi

�k

KiðkÞ;

where 0 � h � Hi

(3.227)

From the initial conditions (Eqs. 3.224 and 3.225) and the solutionequations (Eqs. 3.226 and 3.227), it can be shown that

G3ð0Þ ¼ Ið0Þ ¼ 0; Jð0Þ ¼ dðkÞ; where dðkÞ ¼�1 k ¼ 0

0 ks 0(3.228)

S3ð0Þ ¼ Kð0Þ ¼ 0: (3.229)


kþ 1Hi

G3iðkþ 1Þ ¼ I iðkÞ; kþ 1Hi

I iðkþ 1Þ ¼ J iðkÞ; (3.230)


kþ 1Hi

S3iðkþ 1Þ ¼ KiðkÞ; (3.231)

kþ 1Hi

J iðkþ 1Þ þ 3Xki¼0

J iðk� 1ÞFiðiÞ � 4Xkt¼0

I iðk� iÞYðiÞ

þ 3Xki¼0

G3iðk� iÞZðiÞ þ S3iðkÞ ¼ 0

(3.232)

kþ 1Hi

Kiðkþ 1Þ þ 3PrXki¼0

Kiðk� iÞFiðiÞ þ 3PrXki¼0

Siðk� iÞG3iðiÞ ¼ 0

(3.233)

Fig. 3.26 plots the velocity distributions obtained by the present methodfor free convection over a vertical plate for various values of Prandtlnumber. It is observed that the maximum values of the velocity profilesoccur at larger values of g as the value of Pr decreases, and the velocityvalues decrease with increasing Prandtl number. Fig. 3.27 plots the tem-perature distributions obtained by the present method for free convectionboundary layer flow over a vertical plate for various values of Prandtlnumber.

Figure 3.26 Dimensionless velocity distributions for free convection flow over avertical plate for various values of Prandtl number.


REFERENCES[1] Kraus AD, Aziz A, Welty JR. Extended surface heat transfer. New York: John Wiley;

2002.[2] Ghasemi SE, Hatami M, Ganji DD. Thermal analysis of convective fin with

temperature-dependent thermal conductivity and heat generation. Case Studies inThermal Engineering 2014;4:1e8.

[3] Domairry D, Sheikholeslami M, Ashorynejad HR, Gorla RSR, Khani M. Naturalconvection flow of a non-Newtonian nanofluid between two vertical flat plates.Proceedings of the Institution of Mechanical Engineers, Part N: Journal of Nano-engineering and Nanosystems 2011;225(3):115e22.

[4] Ndlovu PL, Moitsheki RJ. Application of the two-dimensional differential transformmethod to heat conduction problem for heat transfer in longitudinal rectangular andconvex parabolic fins. Communications in Nonlinear Science and Numerical Simula-tion 2013;18(10):2689e98.

[5] Kuo B-L. Thermal boundary-layer problems in a semi-infinite flat plate by the dif-ferential transformation method. Applied Mathematics and Computation2004;150(2):303e20.

[6] Kuo B-L. Heat transfer analysis for the FalknereSkan wedge flow by the differentialtransformation method. International Journal of Heat and Mass Transfer 2005;48(23):5036e46.

[7] Kuo B-L. Application of the differential transformation method to the solutions of thefree convection problem. Applied Mathematics and Computation 2005;165(1):63e79.

Figure 3.27 Dimensionless temprature distributions for free convection flow over avertical plate for various values of Prandtl number.


CHAPTER 4

DTM for Fluids Flow Analysis

4.1 INTRODUCTION

Studies of fluid transport in biological organisms often concern the flow ofa particular fluid inside an expanding or contracting vessel with permeablewalls. For a valve vessel exhibiting deformable boundaries, alternating wallcontractions produce the effect of a physiological pump. The flow behaviorinside the lymphatic exhibits a similar character. In such models, circulationis induced by successive contractions of two thin sheets that cause thedownstream convection of the sandwiched fluid. Seepage across permeablewalls is clearly important to the mass transfer between blood, air, and tissue[1]. Therefore, a substantial amount of research work has been invested inthe study of the flow in different geometries in both Newtonian and non-Newtonian form. This chapter introduces Differential TransformationMethod (DTM) to solve these problems which contains the followingsections:4.1 Introduction4.2 Two-Dimensional Viscous Flow4.3 Magnetohydrodynamic Boundary Layer4.4 Nanofluid Flow Over a Flat Plate4.5 Non-Newtonian Fluid Flow Analysis

4.2 TWO-DIMENSIONAL VISCOUS FLOW

Consider the laminar, isothermal, and incompressible flow in a rectangulardomain bounded by two permeable surfaces that enable the fluid to enteror exit during successive expansions or contractions. A schematic diagram ofthe problem is shown in Fig. 4.1. The walls expand or contract uniformly ata time-dependent rate a•. At the wall, it is assumed that the fluid inflowvelocity Vw is independent of position. The equations of continuity andmotion for the unsteady flow are given as follows [2]:

vu*

vx*þ vv*

vy*¼ 0; (4.1)



http://dx.doi.org/10.1016/B978-0-12-805190-0.00004-8

vu*

vtþ u*

vu*

vx*þ v*

vu*

vy*¼ �1

r

vp*

vx*þ y

�v2u*

vx*2þ v2u*

vy*2

�; (4.2)

vv*

vtþ u*

vv*

vx*þ v*

vv*

vy*¼ �1

r

vp*

vy*þ y

�v2v*

vx*2þ v2v*

vy*2

�. (4.3)

In the above equations u* and v* indicate the velocity components in xand y directions; p* denotes the dimensional pressure; r; y, and t are thedensity, kinematic viscosity, and time, respectively. The boundary condi-tions will be:

y* ¼ aðtÞ : u* ¼ 0; v* ¼ �Vw ¼ �a•

c;

y* ¼ 0 :vu*

vy*¼ 0; v* ¼ 0;

x* ¼ 0 : u* ¼ 0$

(4.4)

where c ¼ a•Vw

is the wall presence or injection/suction coefficient, which isa measure of wall permeability. The stream function and mean flowvorticity can be introduced by putting,

u* ¼ vj*

vy*; v* ¼ vj*

vx*; x* ¼ vv*

vx*� vu*

vy*

vx*

vtþ u*

vx*

vx*þ v*

vx*

vy*¼ y

"v2x*

vx*2þ v2x*

vy*2

#.

(4.5)

0

x

y

a(t)

dadt

1 2 3 4 5 6 7 8 9Figure 4.1 Two-dimensional domain with expanding or contracting porous walls.


Due to mass conservation, a similar solution can be developed withrespect to x*. Starting with:

j* ¼ vx*f *ðy; tÞa

; u* ¼ vx*f *ya2

; v* ¼ �vf *ðy; tÞa

;

y ¼ y*

a; f *y h

vf *

vy.

(4.6)

Substitution of Eq. (4.6) into Eq. (4.5) yields,

u*y*t þ u*u*y*x* þ v*u*y*y* ¼ vu*y*y*y* (4.7)

To solve Eq. (4.7), one uses the chain rule to obtain,

f *yyyy þ a�yf *yyy þ 3f *yy

�þ f *f *yyy � f *y f

*yy � a2y�1f *yyt ¼ 0; (4.8)

With the following boundary conditions:

at y ¼ 0 : f * ¼ 0; f *yy ¼ 0;

at y ¼ 1 : f * ¼ Re; f *y ¼ 0;(4.9)

where aðtÞhaa•yis the nondimensional wall dilation rate defined positive for

expansion and negative for contraction. Furthermore, Re ¼ aVwyis the perme-

ation Reynolds number defined positive for injection and negative for suctionthrough the walls. Eqs. (4.6), (4.8), and (4.9) can be normalized by putting,

j ¼ j*

aa•; u ¼ u*

a•; v ¼ v*

a; f ¼ f *

Re; (4.10)

And so,

j ¼ xfc; u ¼ xf 0

c; v ¼ �f

c; c ¼ a

Re; (4.11)

f IV þ a�yf

000 þ 3f 00�þ Re f f

000 � Re f 0 f 00 ¼ 0 (4.12)

The boundary conditions (4.9) will be:

y ¼ 0 : f ¼ 0; f 00 ¼ 0

y ¼ 1 : f ¼ 1; f 0 ¼ 0(4.13)

DTM for Fluids Flow Analysis 155

The resulting Eq. (4.12) is the classic Berman’s formula [2], with a ¼ 0(channel with stationary walls).

After the flow field is found, the normal pressure gradient can be ob-tained by substituting the velocity components into Eqs. (4.1)e(4.3).Hence it is,

py ¼ �½Re�1f 00 þ f f 0 þ aRe�1ðf þ yf 0Þ�;

p ¼ p*

rV 2w

:(4.14)

We can determine the normal pressure distribution, if we integrate Eq.(4.14). Let pc be the centerline pressure, hence,

Z pðyÞ

pc

dp ¼Z y

0�½Re�1f 00 þ ff 0 þ a Re�1ð f þ yf 0Þ�; (4.15)

Then using ff 0 ¼ ð f 2Þ0=2 and�f þ yf 0

� ¼ ðyf Þ0, the resulting normalpressure drop will be,

Dpn ¼ Re�1f 0ð0Þ ��Re�1f 0 þ f 2

2þ a Re�1 yf

�: (4.16)

Another important quantity is the shear stress. The shear stress can bedetermined from Newton’s law for viscosity:

s* ¼ m�v*x* þ u*y*

�¼ rv2x*f *

00

a3: (4.17)

Introducing the nondimensional shear stress s ¼ s*rV 2

ww, we have,

s ¼ xf 00

Re: (4.18)

Now DTM into governing equations has been applied. Taking thedifferential transform of Eqs. (4.12) and (4.13) with respect to c andconsidering H ¼ 1 gives,


ðkþ 1Þðkþ 2Þðkþ 3Þðkþ 4ÞF½kþ 4�

þ aXkm¼0

ðd½m�ðk� mþ 1Þðk� mþ 2Þðk� mþ 3ÞF½k� mþ 3�Þ

þ 3aðkþ 1Þðkþ 2ÞF½kþ 2�

þ ReXkm¼0

ðF½k� m�ðmþ 1Þðmþ 2Þðmþ 3ÞF½mþ 3�Þ

� ReXkm¼0

ððk� mþ 1ÞF½k� mþ 1�ðmþ 1Þðmþ 2ÞF½mþ 2�Þ ¼ 0;

d½m� ¼(1 m ¼ 1

0 ms1

(4.19)

F½0� ¼ 0; F½1� ¼ a0; F½2� ¼ 0; F½3� ¼ a1 (4.20)

where F(k) are the differential transforms of f(h) and a0, a1 are constants,which can be obtained through boundary conditions from Eq. (4.13).This problem can be solved as follows:

F½0� ¼ 0; F½1� ¼ a0; F½2� ¼ 0; F½3� ¼ a1; F½4� ¼ F½6� ¼ F½8� ¼ 0

F½5� ¼ � 320

a a0

F½7� ¼ 3280

a2a1 þ 170

Re a21 þ1140

Re a0a1 a

F½9� ¼ � 12240

a3a1 � 1560

a Re a21 �1

1120Re a0 a1 a

2 � 11260

a0 a1 a2

� 11260

a0 Re a21 �

12520

Re a20 a a1

:

:

:

(4.21)

The above process is continuous. By substituting Eq. (4.21) into themain equation based on DTM, it can be obtained that the closed form ofthe solutions is,


FðhÞ ¼ a0hþ a1h3 þ

�� 320

h5

þ�

3280

a2a1 þ 170

Re a21 þ1140

Re a0a1 a

h7

þ�� 12240

a3a1

�� 1560

a Re a21 �1

1120Re a0 a1 a

2 � 11260

a0 a1 a2

� 11260

a0 Re a21 �

12520

Re a20 a a1

h9 þ.

(4.22)

By substituting the boundary conditions from Eq. (4.13) into Eq. (4.22)in point h ¼ 1 the values of a0, a1 can be obtained.

Fð1Þ ¼ a0 þ a1 þ�� 320

þ�

3280

a2a1 þ 170

Re a21 þ1140

Re a0a1 a

þ�� 12240

a3a1 � 1560

a Re a21 �1

1120Re a0 a1 a

2 � 11260

a0 a1 a2

� 11260

a0 Re a21 �

12520

Re a20 a a1

þ. ¼ 1

(4.23)

F 0ð1Þ ¼ a0hþ 3a1 þ 5

�� 320

þ 7

�3280

a2a1 þ 170

Re a21 þ1140

Re a0a1 a

þ 9

�� 12240

a3a1 � 1560

a Re a21 �1

1120Re a0 a1 a2 � 1

1260a0 a1 a2

� 11260

a0 Re a21 �

12520

Re a20 a a1

þ. ¼ 0

(4.24)


Solving Eqs. (4.23) and (4.24) gives the values of a0, a1. By substitutingobtained a0, a1 into Eq. (4.22), it can be obtained the expression of F(h).

The objective of the present example was to apply DTM compared tohomotopy perturbation method (HPM) and optimal homotopy asymptoticmethod (OHAM) to obtain an explicit analytic solution of laminar,isothermal, incompressible viscous flow in a rectangular domain boundedby two moving porous walls that enable the fluid to enter or exit duringsuccessive expansions or contractions (Fig. 4.1). Also, error percentage isintroduced as follows:

%Error ¼ f ðhÞNM � f ðhÞa

f ðhÞNM

� 100 (4.25)

where f(h)a is a value obtained using different analytical methods. Fig. 2.2shows the comparison between numerical method and other analytical so-lutions results for f(y) when a ¼ Re ¼ 1. It verifies that, there is an accept-able agreement between the numerical solution obtained by fourth-orderRungeekutta method and these methods. Tables 4.1 and 4.2 confirmthe last conclusion. Comparison between obtained results showed thatHPM and DTM are more accurate and acceptable than two other methods,as can be seen in Fig. 4.3 and Tables 4.3 and 4.4. After this validiation, re-sults are given for the velocity profile, normal pressure distribution, and wallshear stress for various values of permeation Reynolds number and nondi-mensional wall dilation rate.

1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

00 0.1 0.2 0.3

y

f(y)

0.4 0.5 0.6 0.7 0.8 0.9 1

Figure 4.2 Comparison between numerical method and other analytical solutions forf(y), when a ¼ Re ¼ 1.


Fig. 4.4 illustrates the behavior of f 0ðyÞ (or uc=x) for different perme-ation Reynolds number, over a range of nondimensional wall dilation rate.For every level of injection or suction, in the case of an expanding wall,increasing a leads to higher axial velocity near the center and the loweraxial velocity near the wall. The reason is that the flow toward the centerbecomes greater to make up for the space caused by the expansion of thewall and as a result, the axial velocity also becomes greater near the center.

The pressure distribution in the normal direction for various permeationReynolds numbers over a range of nondimensional wall dilation rates isplotted in Fig. 4.5. Fig. 4.5 shows that for every level of injection or suction,

Table 4.1 Constant Values With Different Nondimensional ParametersRe a a0 a1

1 1 1.586849 �0.681131 2 1.696949 �0.932642 1 1.585146 �0.67823

Table 4.2 The HPM, Differential Transformation Method (DTM), OHAM andNumerical Solution Results for f(y), When a ¼ 2, Re ¼ 1

f(y)

y NM HPM DTM OHAM

0 0 0 0 00.05 0.086267 0.086267 0.084731 0.0974310.1 0.17176 0.17176 0.168765 0.1937640.15 0.255721 0.255721 0.251416 0.287920.2 0.337421 0.337421 0.332018 0.3788680.25 0.416176 0.416176 0.409935 0.4656370.3 0.491359 0.491359 0.484572 0.5473450.35 0.562408 0.562408 0.555383 0.6232120.4 0.628838 0.628838 0.621875 0.6925810.45 0.690244 0.690244 0.683621 0.7549340.5 0.746307 0.746307 0.740259 0.8099060.55 0.796792 0.796792 0.791502 0.85730.6 0.841553 0.841553 0.837137 0.8970930.65 0.880522 0.880522 0.87703 0.9294480.7 0.91371 0.91371 0.911122 0.9547120.75 0.9412 0.9412 0.939433 0.9734250.8 0.963137 0.963137 0.962059 0.9863090.85 0.979721 0.979721 0.979164 0.9942670.9 0.991199 0.991199 0.990982 0.9983680.95 0.997854 0.997854 0.997811 0.999831 1 1 1 1


the absolute pressure change in the normal direction is lowest near the centralportion. Furthermore, by increasing nondimensional wall dilation rates theabsolute value of pressure distribution in the normal direction increases. The

wall shear stress�s ¼ xf 00ð1Þ

Re

�for permeation Reynolds number Re ¼ �1

and Re ¼ 1 over a range of nondimensional wall dilation rates is plotted inFig. 4.6. We can observe from Fig. 4.6 that the absolute shear stress along thewall surface increases in proportion to x. Furthermore, by increasingnondimensional wall dilation rates the absolute value of shear stress increases.

4.3 MAGNETOHYDRODYNAMIC BOUNDARY LAYER

Let us consider the Magnetohydrodynamic (MHD) flow of an incom-pressible viscous fluid over a stretching sheet at y ¼ D. The fluid is elec-trically conducting under the influence of an applied magnetic field B(x)normal to the stretching sheet. The induced magnetic field is neglected.The boundary layer equations are as follows [3]:

ux þ vy ¼ 0 (4.26)

uux þ vuy ¼ yuyy � sB2ðxÞu=r (4.27)

where u and n are the velocity components in the x and y directions,respectively, y is the kinematic viscosity, r is the fluid density, and s is

0

0

0.15

0.3%

Erro

r

0.2 0.4 0y 0.6 0.8

OHAM

HPM

DTM

1

Figure 4.3 Comparison between error percentages of HPM, DTM, and OHAM for f(y),when a ¼ Re ¼ 1.


the electrical conductivity of the fluid. In Eq. (4.27), the external electricfield and the polarization effects are negligible [3].

BðxÞ ¼ B0xðn�1Þ=2 (4.28)

The boundary conditions corresponding to the nonlinear stretching of asheet are:

uðx; 0Þ ¼ cxn vðx; 0Þ ¼ 0

uðx; yÞ/0 as y/0(4.29)

Using the similarity variables,

t ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffifficðnþ 1Þ

2y

rxn�12 y; u ¼ cxnf 0ðxÞ

v ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffifficyðnþ 1Þ

2y

rxn�12

�f ðtÞ þ n� 1

nþ 1tf 0ðtÞ

� (4.30)

Table 4.3 The Results of HPM, DTM, OHAM, and Numerical Solution for f(y), Whena ¼ 1, Re ¼ 2

f(y)

y NM HPM DTM OHAM

0 0 0 0 00.05 0.080088 0.080088 0.079173 0.086080.1 0.159633 0.159633 0.157837 0.171440.15 0.238097 0.238097 0.235491 0.2553680.2 0.314954 0.314954 0.311636 0.3371740.25 0.389691 0.389691 0.385788 0.4161930.3 0.461819 0.461819 0.457477 0.4917990.35 0.530872 0.530872 0.52625 0.5634120.4 0.596412 0.596412 0.591678 0.6305060.45 0.658037 0.658037 0.653355 0.6926210.5 0.715375 0.715375 0.710901 0.7493660.55 0.768097 0.768097 0.76397 0.8004320.6 0.815908 0.815908 0.812245 0.8455950.65 0.858559 0.858559 0.855444 0.8847230.7 0.895837 0.895837 0.893323 0.9177820.75 0.927572 0.927572 0.925674 0.9448430.8 0.953638 0.953638 0.952328 0.9660790.85 0.973944 0.973944 0.973157 0.9817720.9 0.988441 0.988441 0.988071 0.9923110.95 0.997119 0.997119 0.997021 0.9981891 1 1 1 1


Eqs. (4.26)e(4.29) are transformed into,

f000 ðtÞ þ f ðtÞf 00ðtÞ � bf 0ðtÞ2 �Mf 0ðtÞ ¼ 0 (4.31)

f ð0Þ ¼ 0; f 0ð0Þ ¼ 1; f 0ðþNÞ ¼ 0 (4.32)

where the primes denote differentiation with respect to t and

b ¼ 2n1þ n

; M ¼ 2sB20

rcð1þ nÞ (4.33)

We shall solve this nonlinear differential equation using the DTM, theDTMePade and the numerical methods (by using a fourth-orderRungeeKutta and shooting method).

Table 4.4 The Comparison Among Error Percentages of HPM, DTM, and OHAM forf(y), when a ¼ Re ¼ 1

f(y)

y HPM DTM OHAM

0 0 0 00.05 3.35E-06 0.016579 0.0865790.1 1.19E-06 0.008088 0.0980880.15 1.13E-06 0.00559 0.105590.2 1.26E-06 0.02375 0.133750.25 1.12E-06 0.045427 0.1654270.3 9.41E-07 0.069424 0.1994240.35 8.64E-07 0.094337 0.2243370.4 7.35E-07 0.118603 0.2486030.45 6.08E-07 0.140557 0.2705570.5 5.19E-07 0.158494 0.2884940.55 4.14E-07 0.170769 0.3007690.6 3.66E-07 0.175905 0.3059050.65 4.3E-07 0.172741 0.3027410.7 2.51E-07 0.160611 0.2906110.75 2.52E-07 0.139584 0.2695840.8 2.62E-07 0.110756 0.2407560.85 1E-07 0.076631 0.1966310.9 1.42E-07 0.041614 0.1516140.95 4.18E-07 0.012641 0.0926411 2E-07 2E-08 0


Taking differential transform of Eq. (4.31), one can obtain

ðkþ 1Þðkþ 2Þðkþ 3ÞFðkþ 3Þ

þXkr¼0

½ � bðr þ 1Þðk� r þ 1ÞFðr þ 1ÞFðk� r þ 1Þ

þ ðk� r þ 1Þðk� r þ 2ÞFðrÞFðk� r þ 2Þ��Mðkþ 1ÞFðkþ 1Þ ¼ 0

(4.34)

by using the DTM, the boundary conditions (Eq. 4.32) are transformedinto a recurrence equation that finally leads to the solution of a system ofalgebraic equations. After finding the DTM solutions, the Pade approxim-ant must be applied. We can consider the boundary conditions (Eq. 4.32) asfollows:

f ð0Þ ¼ 0; f 0ð0Þ ¼ 1; f 00ð0Þ ¼ a (4.35)

The differential transform of the boundary conditions are as follows:

Fð0Þ ¼ 0; Fð1Þ ¼ 1; Fð2Þ ¼ a (4.36)

Moreover, substituting Eq. (4.36) into Eq. (4.34) and by recursivemethod we can calculate other values of F(k). Hence, substituting all F(k)into Eq. (4.33), we have series solutions. After finding the series solutions,the Pade approximation must be applied, by using asymptotic boundarycondition ð f 0ðþNÞ ¼ 0Þ we can obtain a. For an analytical solution, theconvergence analysis was performed and the i value is selected equal to 20.The order of Pade approximation [L, M], [10, 10] has sufficient accuracy;on the other hand if the order of Padé approximation increases, the ac-curacy of the solution increases. For b ¼ 1.5 the analytical solutions are asfollows:

f ðtÞ½10;10�;M¼0 ¼ ðt þ 0:952076t2 þ 0:469526t3 þ 0:141262t4

þ 0:0289081t5 þ 0:00399352t6 þ 0:000382091t7

þ 0:0000221858t8 þ 8:00793� 10�7t9 � 1:77343� 10�12t10Þ=ð1þ 1:52943t þ 1:10254t2 þ 0:491682t3 þ 6:67233� 10�8t10Þ

(4.37)


f ðtÞ½10;10�;M¼1 ¼ ðt þ 0:682855t2 þ 0:271323t3 þ 0:054809t4 þ 0:0086483t5

þ 0:00011291t6 � 0:000121408t7 þ 0:0000478684t8

� 2:34566� 10�7t9 � 1:12636� 10�7t10Þ=ð1þ 1:44545t þ 0:95696t2 þ 0:372964t3 þ 0:089476t4

þ 0:0119558t5 þ 0:000181381t6 � 0:000251727t7

� 0:0000467392t8 � 3:73478� 10�6t9 þ 3:19695� 10�8t10Þ(4.38)

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

0

0.4

0.8

1.2

1.6(a)

(b)

0

0.4

0.8

1.2

1.6

α=0

α=−0.5

α=+0.5

y

α=0

α=−0.5

α=+0.5

f' (y)

f' (y)

y

Figure 4.4 f 0ðyÞ changes shown over a range of a at (a) Re ¼ �5 (b) Re ¼ 5.


f ðtÞ½10;10�;M¼5 ¼ ðt þ 0:398393t2 þ 0:134866t3 � 0:0046991t4

� 0:00243265t5 � 0:00941575t6 þ 0:000501479t7

þ 0:000144631t8 þ 0:0000729533t9 þ 0:0000112928t10Þ=ð1þ 1:65644t þ 1:13542t2 þ 0:3631t3 þ 0:00941431t4

0:0349225t5 � 0:0121299t6 � 0:000911216t7þ0:000530204t8 þ 0:00018197t9 þ 0:0000273471t10Þ

(4.39)

0 0.2 0.4 0.6 0.8 1-2.4

-2

-1.6

-1.2

-0.8

-0.4

0(a)

(b)

α=0

α=−0.5

α=+0.5

y

0 0.2 0.4 0.6 0.8 1

0

0.4

0.8

1.2

1.6

α=0

α=−0.5

α=+0.5

Pres

sure

dro

pPr

essu

re d

rop

y

Figure 4.5 The pressure drop in the normal direction (Dpn) changes shown over arange of a at (a) Re ¼ �1 (b) Re ¼ 1.


f ðtÞ½10;10�;M¼10 ¼ ðt þ 0:0799135t2 þ 0:440267t3 þ 0:157782t4

þ 0:034105t5 � 0:0206912t6 � 0:00395085t7

� 0:000579711t8 � 0:000251482t9 þ 0:000045754t10Þ=ð1þ 2:48228t þ 2:70166t2 þ 1:63052t3 þ 0:535353t4

þ 0:046855t5 � 0:0374833t6 � 0:0176568t7

� 0:00331496t8 � 0:000172074t9 þ 0:0000417687t10Þ(4.40)

0 0.2 0.4 0.6 0.8 10

0.8

1.6

2.4

3.2(a)

(b)

α=0

α=−0.5

α=+0.5

x

0 0.2 0.4 0.6 0.8 1-3.2

-2.4

-1.6

-0.8

0

α=0

α=−0.5

α=+0.5

τ wτ w

x

Figure 4.6 Shear stress changes shown over a range of a at (a) Re ¼ �1 (b) Re ¼ 1.


f ðtÞ½10;10�;M¼50 ¼ ðt � 22:6935t2 � 20:8798t3 � 31:0628t4 � 19:2098t5 � 0:841719t6

þ 16:6574t7 þ 1:82323t8 þ 5:78142t9 � 0:00715648t10Þ=ð1� 19:1112t � 97:9261t2 � 202:307t3 � 222:828t4 � 119:697t5

þ 11:8529t6 þ 70:5985t7 þ 55:4051t8 þ 22:1859t9 þ 4:17935t10Þ(4.41)

f ðtÞ½10;10�;M¼100 ¼ ðt � 0:441265t2 þ 0:3802484t3 þ 1:27347t4 � 6:29308t5

þ 4:93044t6 � 0:418915t7 � 4:56918t8 þ 3:63737t9 � 3:7257t10Þ=ð1þ 4:59754t þ 6:62968t2 � 0:266082t3 � 9:99615t4

� 7:66542t5 þ 4:24617t6 þ 8:29156t7

1:73129t8 � 3:29106t9 � 2:16751t10Þ(4.42)

f ðtÞ½10;10�;M¼500 ¼ ðt � 2:5452t2 þ 14:9318t3 � 34:3945t4 þ 21:1412t5 � 3:17934t6

þ� 106:258t7 þ 312:014t8 � 172:341t9 þ 190:748t10Þ=ð1þ 8:65t þ 28:1869t2 þ 26:501t3 � 89:1766t4

�315:503t5 � 307:616t6 þ 390:192t7 þ 1505:88t8

þ1883:72t9 þ 1023:08t10Þ(4.43)

f ðtÞ½10;10�;M¼1000 ¼ ðt � 3:67488t2 þ 31:4012t3 � 103:767t4 þ 135:989t5�170:603t6 � 530:6t7 þ 2608:13t8 � 2330:81t9 þ 3216:62t10Þ=ð1þ 12:147t þ 56:6735t2 þ 86:5044t3 � 278:191t4 � 1561:55t5

2423:89t6 þ 2469:64t7 þ 16765:5t8 þ 29253t9 þ 20837:3t10Þ(4.44)

The system of Eq. (4.34) with transformed boundary conditions wassolved analytically using the DTMePadé and numerically using the fourth-order RungeeKutta and shooting method. It was shown in Fig. 4.7 theanalytical and the exact solution of f(t) for different values of a magneticparameter. It is clear that as the magnetic parameter increases, the thickness


of boundary layer decreases, and so does the accuracy of the DTM for largevalues of t. It seems that the DTM solutions are only valid for small values ofindependent variable (t).

4.4 NANOFLUID FLOW OVER A FLAT PLATE

Consider a nanoliquid film flow and heat transfer in the neighborhood of athin elastic sheet. To understand more, the related physical model has beendepicted in Fig. 4.8. The Cartesian coordinate (x,y) is chosen in a mannerthat the x-axis is measured in the direction of wall stretching and the y-axis

β = 1, M = 50 β = 1, M = 500

0.2

0.15

0.1

0.05

00.5 1 1.50

f(t)

t

DTM, i=20

ExactDTM-Padé[10,10]

DTM, i=20

ExactDTM-Padé[10,10]0.05

0.04

0.03

0.02

0.01

00.1 0.2 0.3 0.40

f(t)

t

DTM, i=20

Exact

1.2

1

0.8

0.6

0.4

0.2

00 2

β = 1, M = 0 β = 1, M = 1

4t t

f(t)

f(t)

6 8 00

0.25

0.75

1

0.5

1 2 3 4 5

DTM-Padé[10,10]

DTM, i=20

ExactDTM-Padé[10,10]

Figure 4.7 The analytical solution of f(t) obtained by the differential transformationmethod (DTM), the DTMePade, and the exact solution.


is normal to the wall. The continuous surface at y ¼ 0 is stretched with thevelocity defined as:

Uw ¼ bx1� at

(4.45)

In the above equation b and a are constants with dimensions t�1. Afterthat the temperature distribution on the sheet is given as follows:

Ts ¼ T0 � Tr

�bx2

2vf

ð1� atÞ�3=2 (4.46)

In Eq. (4.46), T0 is the temperature at the slit, Tr can be consideredeither as a constant reference temperature or a constant temperature dif-ference, and finally vf is the kinematic viscosity of the base fluid. Supposethat the film is uniform and stable, and the gravity and end effects arenegligible. Therefore, the governing differential equations for this problemare expressed as follows:

vuvx

þ vvvy

¼ 0;vuvt

þ uvuvx

þ vvuvy

¼ mnf

rnf

v2uvy2

;vTvt

þ uvTvx

þ vvTvy

¼ anfv2Tvy2

(4.47)

Figure 4.8 The schematic diagram of the physical model.


In Eq. (4.47), u and v are the velocity components along x and y di-rections, T is defined as the temperature, mnf, rnf and then anf are theviscosity, the density, and the thermal diffusivity of nanofluid which areexpressed as follows:

anf ¼ knfðrcpÞnf

; mnf ¼mf

ð1� fÞ5=2; rnf ¼ ð1� fÞrf þ frs (4.48)

Afterward:

ðr cpÞnf ¼ ð1� fÞðr cpÞf þ fðr cpÞs;knfkf

¼ ðks þ 2kf Þ � 2fðkf � ksÞðks þ 2kf Þ þ fðkf � ksÞ

(4.49)

In the aforementioned equations f is the solid volume fraction of thenanofluid, knf and (rcp)nf are the thermal conductivity and the heatcapacitance of the nanofluid, respectively. Then, the related boundaryconditions for the differential equation governing on the mentioned systemare defined as:

when y ¼ 0/u ¼ Uw; v ¼ 0; T ¼ Ts (4.50)

and then,

at y ¼ hðtÞ/vuvy

¼ 0;vTvy

¼ 0; v ¼ dhdt

(4.51)

In the above equation, h(t) is the thickness of the film and x is assumed tobe a nonnegative quantity. The boundary layer thickness dðxÞ is propor-tional to (x vf/Uw)

0.5, so that the similarity variable h is defined as follows:

h ¼ y

ffiffiffiffiffiffiffiffiUw

x vf

s¼ y

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib

ð1� atÞvf

s(4.52)

In accordance with this explanation and substituting U(x) ¼ Uw, it ispossible to define x ; h and jðx; yÞ. As regards the above explanations, thefollowing new variables are introduced as [4]:

j ¼ b

�vf b

1� at

�1=2x f ðhÞ (4.53)

T ¼ T0 � Tr

�bx2

2vf

ð1� atÞ3=2qðhÞ (4.54)


h ¼ 1b

�b

vf ð1� atÞ�1=2

y (4.55)

where jðx; yÞ is the stream function explained in the usual way, take for

example u ¼ vjvy, v ¼ �vj

vx, and b > 0 is the dimensionless film thickness

defined by b ¼ (hb/vf)(1 � at)�1/2. Note that for the limiting casesb ¼ 0 and b ¼ N this transformation is no longer effective and particularapproaches are needed to give solutions. As a result, the velocity compo-nents u and v can be explicitly defined as:

u ¼ vj

vy¼�

bx1� at

f 0ðhÞ; v ¼ �vj

vx¼ �b

�vf b

1� at

1=2

f ðhÞ (4.56)

After substituting the obtained similarity variables from Eqs. (4.53)e(4.55)into Eq. (4.47), the continuity equation is automatically satisfied and themomentum and energy equations are reduced to,

ε1 f000 þ b2

hf f 00 � ðf 0Þ2 � S

�f 0 þ h

2f 00�i

¼ 0 (4.57)

ε2

Prq00 � b2

�S2ð3qþ hq0Þ þ 2qf 0 � f q0

�¼ 0 (4.58)

And then, the relevant boundary conditions for this problem can beexpressed as follows:

at h ¼ 0/ f ð0Þ ¼ 0; f 0ð0Þ ¼ 1; qð0Þ ¼ 1

at h ¼ 1/ f 00ð1Þ ¼ 0; q0ð1Þ ¼ 0; f ð1Þ ¼ S2

(4.59)

In the aforementioned equations, Pr ¼ (vf/af) is the Prandtl number,S ¼ (a/b) is the unsteadiness parameter, and finally ε1 and ε2 are twoconstants explained in the following form:

ε1 ¼ 1

ð1� fÞ2:5½ð1� fÞ þ f rs=rf �; ε2 ¼ ðknf =kf Þ�

1� fþ fðrcpÞs ðrcpÞf �

(4.60)

To understand more, it is better to indicate that the physical quantitiesfor this problem are the skin friction coefficient Cf and the Nusseltnumber Nu that are as follows:

Cf ¼ swrf ðUwÞ2

; Nu ¼ qwxkf ðTs � T0Þ (4.61)


In the above equation, the skin friction at the surface and the heat fluxfrom the surface are defined in the following form:

sw ¼ mnf

�vj

vy

y¼0

; qw ¼ �knf

�vTvy

y¼0

(4.62)

Therefore, after substituting Eq. (4.62) into Eq. (4.61), we will have thefinal form of the physical quantities as follows:

CfxRe�1=2x ¼ 1

bð1� fÞ5=2f 00ð0Þ; NuxRe

1=2x ¼ �knf

kf

1bq0ð0Þ (4.63)

It is notable that in Eq. (4.63) Rex ¼ Uwx/vf is the local Reynoldsnumber. For the current nanoliquid film flow, it is found that thesimilarity solutions are available in the same range of S. Beyond thisregion, no solutions can be found. On the other hand, it is citable thatthe film thickness b decreases monotonically as S increases for both theNewtonian fluids ðf ¼ 0Þ and the nanofluids ðfs0Þ as shown inTable 4.5.

This issue refers that the constants a and b as well as the wall stretchingvelocity Uw have significant effects on b. For a fixed value of b, the larger a,the smaller value of b. Moreover, it is concluded from Table 4.5 that thedecaying rate for b between any two prescribed values of S remains almostthe same for all of the considered nanofluids, take, for example, thedecaying rate for b between S ¼ 0.6 and S ¼ 0.8 is 31.2838% and this ratebetween S ¼ 1.0 and S ¼ 1.8 is 76.2191%. On the basis of the aboveexplanations, a linear formula for evaluating the variation of b in terms of

Table 4.5 The Dimensionless Film Thickness ðbÞ for Various Values of S and f,When Pr ¼ 1Types of Fluid S f ¼ 0.0 f ¼ 0.1 f ¼ 0.2

0.6 3.31710 2.66586 2.571090.8 2.15199 1.83187 1.766761.0 1.54361 1.31399 1.26729

Cu-water 1.2 1.12778 0.96002 0.925891.4 0.82103 0.69890 0.674051.6 0.57617 0.49046 0.473031.8 0.35638 0.30337 0.29259


nanoparticle volume fraction (f) for three different kinds of nanofluids isused from Ref. [4] and depicted in Fig. 4.9:

In accordance with the Chapter 1 explanations, DTM has been appliedto solve the presented problem as follows:

DTM1 ¼ ε1ðkþ 1Þðkþ 2Þðkþ 3Þðkþ 4ÞFðkþ 4Þ

þ b2

" Xkl¼0

ðl þ 1ÞFðl þ 1Þðkþ 1� lÞðkþ 2� lÞFðkþ 2� lÞ!

þ Xk

l¼0

FðlÞðkþ 1� lÞðkþ 2� lÞðkþ 3� lÞFðkþ 3� lÞ!

� 2

Xkl¼0

ðl þ 1ÞFðl þ 1Þðkþ 1� lÞðkþ 2� lÞFðkþ 2� lÞ!

� S

��32

ðkþ 1Þðkþ 2ÞFðkþ 2Þ

þ 12

Xkl¼0

ðdðl � 1Þðkþ 1� lÞðkþ 2� lÞðkþ 3� lÞFðkþ 3� lÞÞ!#

¼ 0

(4.64)

1.60 0.1 0.2 0.3

1.7

1.8

1.9

2

2.1

2.2

2.3

Al2O3TiO2Cu2.4

2.5

2.6

β

φ

Figure 4.9 The result of varying b in terms of f for three different kinds of nano-particles, when S ¼ 1.2.


DTM2 ¼ ε2

Prðkþ 1Þðkþ 2ÞQðkþ 2Þ

� b2

S2

3 QðkÞ þ

Xkl¼0

dðl � 1Þðkþ 1� lÞQðkþ 1� lÞ!

þ 2

Xkl¼0

QðlÞðkþ 1� lÞFðkþ 1� lÞÞ

�Xkl¼0

FðlÞðkþ 1� lÞQðkþ 1� lÞ!

¼ 0

(4.65)

where F and Q represent the DTM transformed form of f and q, respec-tively. After that the given boundary conditions can be transformed asfollows:

Fð0Þ ¼ 0; Fð1Þ ¼ 1; Fð2Þ ¼ a; Fð3Þ ¼ b;Qð0Þ ¼ 1; Qð1Þ ¼ c (4.66)

where a , b and c are constant coefficients that can be determined after spec-ifying FðhÞ and QðhÞ and applying the three remained boundary condi-tions from Eq. (4.59) on the obtained solutions. When our base fluid ispure water and the nanoparticles are copper and the constant coefficientsof the governing equations are assumed to be b ¼ 0.5, ε1 ¼ �0.332,ε2 ¼ 8.35, Pr ¼ 1 and finally S ¼ 1, we will have,

a ¼ �0:6481199115; b ¼ 0:06070499715; c ¼ �0:07341467053 (4.67)

In this step, to avoid the repeated mathematical operation for its ease ofunderstanding, we omit the detailed DTM procedure and only mention thefinal solution in the following form:

f ðhÞ ¼ h� 0:6481199115h2 þ 0:06070499715h3 þ 0:1016754379h4

� 0:01397202970h5 � 0:0002884940148h6

(4.68)

and then

qðhÞ ¼ 1� 0:07341467053hþ 0:05239520958h2 � 0:01385237686h3

þ 0:001460996324h4 þ 0:001072271513h5 � 0:0001706601508h6

(4.69)


To be sure about the precision of the achieved solutions, a comparisonbetween numerical method (RungeeKutta) and DTM has been presentedin Table 4.6.

To start, the obtained solution of the given set of differential equationsconsisted of Eq. (4.68) and Eq. (4.69) should be substituted into the mainset of differential equations , g(h) and h (h), and then it is necessary to depictthe yielded equations in Cartesian coordinates. Afterward, the yielded er-rors of the given set of differential equations can be observed from theobtained charts. To understand more, reading the following lines is rec-ommended. Consider a set of differential equations in the following form:�

gðhÞ ¼ gðf ðhÞ; f 0ðhÞ;.ÞhðhÞ ¼ hðqðhÞ; q0ðhÞ;.Þ (4.70)

And the answer of the aforementioned set of equations is assumed to bea function of h in the form of:

f ¼ h1ðhÞ; q ¼ h2ðhÞ (4.71)

Thus by substituting Eq. (4.71) into Eq. (4.70), the computational errorof the obtained solution by each analytical or semianalytical method can beachieved as follows:

gðhÞ ¼ gð f ðh1ðhÞÞ; f 0ðh2ðhÞÞ;.ÞhðhÞ ¼ hðqðh1ðhÞÞ; q0ðh2ðhÞÞ;.Þ (4.72)

Eventually, with regard to the given physical values and by substitutingthe yielded solutions using DTM, which are Eqs. (4.68)e(4.69) into Eqs.(4.57)e(4.58), the computational errors of DTM are depicted in the formsof Figs. 4.10e4.11.

The aforementioned figures (Figs. 4.10 and 4.11) show that the yieldedsolutions by DTM on the basis of the given physical values such as Pr ¼ 1 inthe specified domain are appropriate approximations for solving the pre-sented problem. In this case study, Cu-water nanofluid is chosen as aconvenient example for illustration and attempts have been made to discussthe effects of some physical parameters such as Pr, S andf on the velocityand temperature distribution as follows:

It is clear from Figs. 4.12 and 4.13 that by increasing the amount ofunsteadiness parameter S, the values of f(h) increases but vice versa, thetemperature profile diminishes smoothly with h in the specified domain. Inaccordance with Fig. 4.14, the velocity profile f 0ðhÞ decreases uniformly


Table 4.6 A Comparison Between the Obtained Values by Differential Transformation Method and Numerical Solution and the RelatedErrors for f(h) and q(h) in the Specified Domain, When b ¼ 0.5, ε1 ¼ �0.332, ε2 ¼ 8.35, Pr ¼ 1 and Finally S ¼ 1

f(h) q(h)

h NUM DTM The Error of DTM (%) NUM DTM The Error of DTM (%)

0 0.0 0.0 0 0.999999999999 1 00.1 0.09345612174 0.0935895334 0.1427 0.993171519127 0.9931687892 0.000270.2 0.17425663412 0.1747190347 0.2653 0.987307832315 0.9873047251 0.000310.3 0.24321770321 0.2440976516 0.3617 0.982331096286 0.9823314688 0.0000370.4 0.30136110613 0.3026445699 0.4258 0.978170169099 0.9781784958 0.00085120.5 0.34989036992 0.3514717280 0.4519 0.974761540350 0.9747820742 0.00210.6 0.39016599675 0.3918663230 0.4357 0.972050131446 0.9720861224 0.00370.7 0.42368002460 0.4252731100 0.3760 0.969989968411 0.9700429419 0.00540.8 0.45203012617 0.4532764928 0.2757 0.968544729388 0.9686138292 0.00710.9 0.47689340438 0.4775824081 0.1444 0.967688168589 0.9677695631 0.00841 0.49999999999 0.4999999998 0.00000004 0.967404418106 0.9674907698 0.0089

DTM

forFluids

FlowAnalysis

177

with h for different amounts of unsteadiness parameter and it is notable thatthe larger value of S, the higher velocity profile is. The mentioned variationwill be clearer while increasing the amount of h. Because the unsteadinessparameter is dependent to the a and b (S ¼ a/b), it is observed that the wall

–0.2

0.2 0.4 0.6 0.8 10

–0.15

–0.1

–0.05

0

The

erro

r of f

( )η

η

Figure 4.10 The achieved computational error for f(h) by differential transformationmethod.

–0.002

–0.004

–0.0060 0.2 0.4 0.6 0.8 1

0

The

erro

r of

( )η

η

θ

Figure 4.11 The resulted computational error for q(h) by differential transformationmethod.


0.6

0.4

0.2

00.2 0.4 0.6 0.8 1

0.8

1

f( )η

Symbol DTM

S = 0.5S = 1S = 1.5S = 2

Line Numerical Method

η

Figure 4.12 A comparison between the obtained results by differential transformationmethod and numerical solution in terms of varying f(h) for different values of un-steadiness parameter(S).

S = 0.5S = 1S = 1.5S = 2

0.650 0.2 0.4 0.6 0.8

0.7

0.75

0.8

0.85

0.9

0.95

1

θη(

)

η

Symbol DTMLine Numerical Method

Figure 4.13 Comparing the achieved results by differential transformation methodand numerical solution in terms of varying q(h) for different amounts of unsteadinessparameter(S).


stretching velocity is an important factor for determining the velocityprofile. As regards Fig. 4.15 which describes the variation of temperaturedistribution for any considered value of unsteadiness parameter, q(h) en-larges along with increasing f. This issue refers to the fact that nanofluidplays a vital role on the heat transfer properties. Afterward, the effects ofPrandtl number have been investigated on f(h) and q(h) completely and theresults are shown in Figs. 4.16 and 4.17. Regarding to the given figures, Prdoes not have any effect on f(h) and it enhances uniformly by increasing h,but the temperature distribution decreases significantly by growing thevalues of Prandtl number and it vanishes at the free surface (h ¼ 1) for largeamount of Prandtl number. Therefore, it is citable that the temperature atthe free surface is equal to the ambient temperature.

Moreover, the variation of f(h) in terms of various amounts of nano-particle volume fraction has been analyzed in Fig. 4.18. Based on thegraphical results, it is revealed that f(h) enhances in the specified domain ofh and for the effect of f, we can indicate that f(h) decreases monotonicallyby increasing f.

In this step, the effects of solid volume fraction of the nanofluid on thevelocity distribution have been presented in Fig. 4.19. In accordance with

S = 0.5S = 0.75S = 1S = 1.25S = 1.5S = 1.75

0

0

0.2

0.4

0.6

0.8

1

0.2 0.4 0.6 0.8 1

dfdη

η

Figure 4.14 The variation of velocity profile for some values of unsteadinessparameter.


the given information by increasing the values of f, f 0ðhÞ decreases in termsof h˛f0; 0:5g but this tendency inverses for h˛f0:5; 1g ,which means thevelocity profile enhances by increasing the values of f. Eventually, thevariations of local skin friction coefficient and also the local Nusselt number

Figure 4.15 The result of varying temperature distribution for four different values ofsolid volume fraction of the nanofluid (f).

Pr = 2Pr = 4Pr = 6Pr = 8Pr = 10

f ( )η

η

0.2

0.3

0.4

0.5

0.1

0 0.2 0.4 0.6 0.8 1

Figure 4.16 Comparing the variation of f(h) in terms of different values of PrandtlNumber.


presented as the physical quantities of the mentioned problem in terms of fare graphically represented in Figs. 4.20 and 4.21.

It is observed that by enlarging the unsteadiness parameter the amountof local skin friction coefficient enhances significantly, and the local skin

η

Pr = 2Pr = 4Pr = 6Pr = 8Pr = 10

0 0.20.75

0.8

0.85

0.9

0.95

1

0.4 0.6 0.8 1

ηθ(

)

Figure 4.17 The result of varying q(h) in terms of various amounts of Prandtl Number.

Figure 4.18 The result of varying f(h) in terms of various amounts of solid volumefraction of the nanofluid (f).


friction coefficient reduces uniformly with increasing f in the specifieddomain. This issue proves that nanofluids are sufficiently useful forreducing the drag force of fluid flow. Then, the variation of local Nusseltnumber with solid volume fraction of the nanofluid for different amounts

Figure 4.19 Comparing the variation of velocity profile in terms of four differentvalues of (f).

S = 0.5S = 1S = 1.5S = 1.9

–2

–2.5

–1.5

–0.5

0.5

0

0 0.01 0.02 0.03 0.04 0.05 0.06

–1

–3

Cfx

Re x

–1 2

φ

Figure 4.20 The obtained graphical results for the variation of local skin frictioncoefficient in terms of four different amounts of S.


of S is presented in Fig. 4.21. It can easily be seen that the local Nusseltnumber increases continuously with f defined from 0 to 0.6. This itemindicates that more the value of f, the more convective heat transfer ofthe nanofluids. Furthermore, it is necessary to mention that the localNusselt number enlarges by increasing the unsteadiness parameter S.Therefore, the obtained results indicate that the suspended nanoparticlesmainly enhance the heat transfer rate at any given values of S and Prandtlnumber.

To deeply understand the above procedure, the variation of local skinfriction coefficient and the local Nusselt number in terms of f for differentvalues of unsteadiness parameter has been presented in the form of nu-merical data in Tables 4.7 and 4.8.

4.5 NON-NEWTONIAN FLUID FLOW ANALYSIS

Consider an unsteady, incompressible, non-Newtonian fluid (such asblood) as a third-grade fluid in an artery. Table 4.9 shows the properties ofthe blood as considered non-Newtonian fluid. The flow is considered totake place axially through the circular tube of radius R under periodic body

S = 0.5S = 1S = 1.5S = 1.9

Nu x

Re x

1 2

02.5

3

3.5

4

4.5

5

5.5

6

6.5

0.01 0.02 0.03 0.04 0.05 0.06φ

Figure 4.21 Investigating the effect of (f) on the local Nusselt number in the specifieddomain.


acceleration and a pulsatile pressure gradient. Cauchy stress in a third-gradefluid is given by:

s ¼ �pI þ mA1 þ a1A2 þ a2A21 þ b1A3 þ b2ðA1A2 þ A2A1Þ

þ b3

�TrA2

1

�A1 (4.73)

where �pI shows the spherical stress due to the restraint of incompressibilityand a1, a2, b1, b2, and b3 are the material modules and are considered to befunctions of temperature generally. In Eq. (4.73), the kinematical tensorsA1, A2, and A3 can be defined by following equations [5]:

A1 ¼ ðVV Þ þ ðVV Þt (4.74)

An ¼ dAn�1

dtþ An�1ðVV Þ þ ðVV ÞtAn�1; n ¼ 2; 3; (4.75)

where V ¼ [0,0,v(r)] denotes the velocity field, superscript t stands for ma-trix transposition and D

Dt is the material time derivative, which is defined by

Dð:ÞDt

¼ vð:Þvt

þ ½gradð:Þ�V (4.76)

Table 4.8 The Variation of NuxRe12x in Terms of f for Different Values of Parameter S

4 S [ 0.5 S [ 1 S [ 1.5 S [ 1.9

0 4.1166 4.6547 5.1325 5.48350.01 4.1617 4.7158 5.2027 5.56040.02 4.2178 4.7772 5.2733 5.63750.03 4.2186 4.8388 5.3441 5.71500.04 4.3196 4.9007 5.4153 5.79280.05 4.3709 4.9630 5.4868 5.87100.06 4.4224 5.0256 5.5588 5.9496

Table 4.7 The Variation of CfxRe�12x in Terms of f for Different Values of Parameter S

4 S [ 0.5 S [ 1 S [ 1.5 S [ 1.9

0 �1.4566 �1.1022 �0.6073 �0.12950.01 �1.5094 �1.1459 �0.6325 �0.13500.02 �1.5631 �1.1902 �0.6580 �0.14050.03 �1.6177 �1.2349 �0.6837 �0.14610.04 �1.6733 �1.2802 �0.7096 �0.15180.05 �1.7299 �1.3262 �0.7359 �0.15750.06 �1.7875 �1.3729 �0.7625 �0.1633


The momentum equation for an incompressible, unsteady, axisym-metric (with z-axis as the axis of symmetry) and fully developed flow in acylindrical polar coordinate (r, q, z) is,

rvuvt

¼ vpvz

þ rG þ 1rv

vr½rsrz� (4.77)

where r; u; p; srz; t, and G denote density, axial velocity, pressure, shearstress, time, and body acceleration in axial direction, respectively. The shearstress srz for a third-grade fluid in an axisymmetric and thermodynamicallycompatible flow situation can be written as,

srz ¼"mþ b

�vuvr

2#

vuvr

(4.78)

In human beings the pressure gradient ðvpvzÞ produced by the pumping

action of the heart takes the approximate form [5].

�vpvz

¼ A0 þ A1 cosupt (4.79)

where A0, A1, up ¼ 2pfp and fp are respectively the constant component ofthe pressure gradient, the amplitude of the fluctuating component (givingrise to the systolic and diastolic pressures), the circular frequency, and thepulse frequency. The body acceleration G is assumed to be:

G ¼ Ag cosðubt þ fÞ (4.80)

where Ag, is the amplitude ub ¼ 2pfb, fb is the frequency, and f is the leadangle of G with respect to the heart action. By using Eqs. (4.78)e(4.80),equation of motion (4.77) can be written as,

rvuvt

¼ A0 þ A1 cosupt þ rAg cosðubt þ fÞ

þ 1rv

vr

(r

"mþ b

�vuvr

2#vuvr

) (4.81)

With the corresponding to initial and boundary conditions:

r ¼ R u ¼ 0

r ¼ 0vuvr

¼ 0

t ¼ 0 u ¼ 0

(4.82)


The initial condition is essential for the numerical scheme adapted toestimate the time at which the pulsatile steady state sets in. The nondi-mensional form of Eqs. (4.81)e(4.82) are respectively

a2

2pvuvT

¼B1ð1þ e cos 2pTÞ

þ B2 cosð2purT þ fÞ þ 1rv

vr

(rvuvr

"1þ B

�vuvr

2#) (4.83)

And boundary conditions:

r ¼ 1 u ¼ 0

r ¼ 0vuvr

¼ 0

T ¼ 0 u ¼ 0

(4.84)

where

a2 ¼ rupR2

mB1 ¼ A0R

2

mu0B2 ¼ rAgR

2

mu0

B ¼ bu20mR2 ur ¼ ub

upu ¼ u

u0r ¼ r

R

T ¼ upt2p

e ¼ A1

A0u0 ¼ A0R

2

8m

(4.85)

Here u0 is the cross-sectional average velocity of flow under steady statepressure gradient A0. The second-order nonlinear partial differential Eq.(4.83) with the boundary conditions (4.85) should be solved by efficientanalytical or numerical methods.

In this example analytical hybrid-DTM is used for the nondimensionaltime (T) and the r directions for obtaining the solution of Eq. (4.83) withboundary condition (4.85). Hybrid-DTM which is the combination offinite difference method (FDM) and multi-step differential transformationmethod (Ms-DTM) can solve the PDEs problem easily. To proceed with itscomputations, FDM is applied to uniform points in the T (nondimensionaltime) and Ms-DTM is based on r (nondimensional radius) directions, andsolutions are performed. Obtained results by hybrid-DTM are comparedwith CrankeNicolson method (CNM) [5] in Fig. 4.22, which shows anexcellent agreement between the analytical and numerical methods. In this


example, two different arteries are considered, femoral and coronary ar-teries. Femoral artery and coronary arteries have diameters 1 and 0.3 cmrespectively. Also, the constant component of the pressure gradient, A0, forcoronary and femoral arteries are taken as 698.65 and 32 dyne/cm,respectively as presented in the literature. In whole solution, physicalproperties of blood are considered constant as shown in Table 4.9 and g isthe acceleration due to gravity which considered being 10. Comparisonbetween hybrid-DTM and CNM [5] is depicted in Fig. 4.22 for both

Figure 4.22 Comparison between hybrid differential transformation method (presentstudy) with CrankeNicolson method, when f ¼ 0, e ¼ ur ¼ 1, Ag ¼ g and fb ¼ 1.2 Hzfor (a) femoral and (b) coronary arteries.


femoral and coronary arteries, when f ¼ 0, e ¼ ur ¼ 1, Ag ¼ g andfb ¼ 1.2 Hz in various times. As seen, hybrid-DTM has excellent agreementwith previous method. This is necessary to inform that due to the fluctu-ating flow, increasing the time makes the velocity take negative values andincreasing it further will increase the velocity and it reaches positive valuesagain. This process occurs continuously. For a better perception, Fig. 4.23 is

Table 4.9 Some Properties of Non-Newtonian Blood [5]Specification r (kg/m3) Cp (J/kg$k) k (W/m$k) m b

Blood 1060 3617 0.52 0.003 0.001

Figure 4.23 Velocity profiles versus time in different radii for (a) femoral and(b) coronary arteries, when f ¼ 0, e ¼ ur ¼ 1, Ag ¼ g, fb ¼ 1.2.


depicted for coronary and femoral arteries. This figure shows the velocityversus time in different radii, as seen, the maximum velocities occur in thecenter of tube (r ¼ 0), and in the whole domain, it has a fluctuating nature.Also, it’s obvious that when the velocity becomes negative, the maximumvalue is seen in the region beyond the center and near the tube wall, andmaximum values take place in lower time duration compared to positivevalues. Fig. 4.24 which is depicted for femoral and coronary arteries,

Figure 4.24 Velocity profiles versus radius in different times for showing the effect ofAg when f ¼ 0, e ¼ ur ¼ 1, and fb ¼ 1.2 Hz for (a) femoral and (b) coronary arteries.


demonstrate the Ag effects on velocity profiles. It can be concluded thatincreasing the amplitude, Ag makes higher velocity profiles in both negativeand positive values. These effects on velocity and wall shear stress during thetime viewpoint are presented in Fig. 4.25 for femoral artery when f ¼ p/6,e ¼ 20, ur ¼ 1, and fb ¼ 1.2 Hz. As Ag increases, the maximum magnitudeof the velocity along the axis increases and assumes both positive andnegative values during a cycle. Also, the maximum wall shear stress increaseswhen Ag increases. These effects are also observed in coronary artery as

Figure 4.25 Effect of Ag for femoral artery on (a) center velocity profile and (b) wallshear stress values, when f ¼ p/6, e ¼ 20, and fb ¼ 1.2 Hz.


shown in Fig. 4.26 by the difference that they are shapely near to parabolicor convex/concave profiles.

Finally, the effects of ur on the center velocity profile and wall shearstress for femoral and coronary arteries are depicted in Figs. 4.27 and 4.28,respectively. As seen, changing this value makes a mutation in velocityprofiles and completely changes the amplitude, shape, maximum values,etc. Reducing the ur from 1 to 0.5 makes a shift to right hand side inmaximum points in both velocity and shear stress profiles. Also, it makes a

Figure 4.26 Effect of Ag for coronary artery on (a) center velocity profile and (b) wallshear stress values, when f ¼ p/6, e ¼ 1, and fb ¼ 1.2 Hz.


reduction for maximum values of positive velocity and increase in itsnegative values. Its treatment for shear stress is completely vice versa (i.e.,increase in its positive maximum values and decrease in its negativevalues). Furthermore, when ur ¼ 0.5 all the graphs have a relativemaximum/minimum point and an absolute optimum point versus whenur ¼ 1.0.

Figure 4.27 Effect of ur for femoral artery on (a) center velocity profile and (b) wallshear stress values, when f ¼ p/3, e ¼ 20, and fb ¼ 1.2 Hz.


REFERENCES[1] Chang HN, Ha JS, Park JK, Kim IH, Shin HD. Velocity field of pulsatile flow in a

porous tube. Journal of Biomechanics 1989;22:1257e62.[2] Dinarvand S, Mehdi Rashidi M, Doosthoseini A. Analytical approximate solutions for

two-dimensional viscous flow through expanding or contracting gaps with permeablewalls. Central European Journal of Physics 2009;7(4):791e9.

[3] Rashidi MM. The modified differential transform method for solving MHD boundary-layer equations. Computer Physics Communications 2009;180(11):2210e7.

Figure 4.28 Effect of ur for coronary artery on (a) center velocity profile and (b) wallshear stress values, when f ¼ p/3, e ¼ 1, and fb ¼ 1.2 Hz.


[4] Ahmadi AR, Zahmatkesh A, Hatami M, Ganji DD. A comprehensive analysis of theflow and heat transfer for a nanofluid over an unsteady stretching flat plate. PowderTechnology 2014;258:125e33.

[5] Mandal PK, Chakravarty S, Mandal A, Amin N. Effect of body acceleration on un-steady pulsatile flow of non-Newtonian fluid through a stenosed artery. AppliedMathematics and Computation 2007;189(1):766e79.


CHAPTER 5

DTM for Nanofluids andNanostructures Modeling

5.1 INTRODUCTION

Nanofluids are produced by dispersing the nanometer-scale solid particlesinto base liquids with low thermal conductivity such as water, ethyleneglycol (EG), oils, etc. The term “nanofluid” was first coined by Choi [1] todescribe this new class of fluids. The presence of the nanoparticles in thefluids noticeably increases the effective thermal conductivity of the fluid andconsequently enhances the heat transfer characteristics. Therefore, numer-ous methods have been taken to improve the thermal conductivity of thesefluids by suspending nano/micro-sized particle materials in liquids. Alsonanostructures such as nanobeam, nanotube, etc., have many applicationsin mechanical engineering. This chapter focuses on the solving problems inthis field and contains the following sections:5.1 Introduction5.2 Nanofluid in Divergent/Convergent Channels5.3 MHD Couette Nanofluid Flow5.4 Nanofluid Between Parallel Plates5.5 Vibration Analysis of Nanobeams5.6 Buckling Analysis of a Single-Walled Carbon Nanotube

5.2 NANOFLUID IN DIVERGENT/CONVERGENT CHANNELS

Consider a system of cylindrical polar coordinates (r, q, z) which steady two-dimensional flow of an incompressible conducting viscous fluid from a sourceor sink at channel walls lie in planes, and intersect in the axis of z. Assumingpurely radial motion which means that there is no change in the flowparameter along the z-direction. The flow depends on r and q, and furtherassumes that there is no magnetic field in the z-direction (See Fig. 5.1). Thereduced form of continuity, NaviereStokes and Maxwell’s equations are [2]:

rnf

rvðruðr; qÞÞ

vrðruðr; qÞÞ ¼ 0 (5.1)

Differential Transformation Method for Mechanical Engineering ProblemsISBN 978-0-12-805190-0http://dx.doi.org/10.1016/B978-0-12-805190-0.00005-X


http://dx.doi.org/10.1016/B978-0-12-805190-0.00005-X

uðr; qÞ vuðr; qÞvr

¼ � 1rnf

vPvr

þ ynf

�v2uðr; qÞ

vr2þ 1

rvuðr; qÞ

vrþ 1r2

v2uðr; qÞvq2

� uðr; qÞr2

�� sB2

0

rnf r2uðr; qÞ

(5.2)

1rnf r

vPvq

� 2ynfr2

vuðr; qÞvq

¼ 0 (5.3)

where B0 is the electromagnetic induction, snf is the conductivity of thefluid, u(r) is the velocity along radial direction, P is the fluid pressure,ynf is the coefficient of kinematic viscosity, and rnf is the fluid density.

Figure 5.1 Schematic of the problem (MHD JefferyeHamel flow with nanofluid).


The effective density rnf, the effective dynamic viscosity mnf, and kinematicviscosity ynf of the nanofluid are given as:

rnf ¼ rf ð1� fÞ þ rsf; mnf ¼mf

ð1� fÞ2:5; ynf ¼mf

rnf;

snf

sf¼ 1þ

3

�ss

sf� 1

�f�

ss

sfþ 2

��ss

sf� 1

�f

(5.4)

Here, f is the solid volume fraction. Considering uq ¼ 0 for purelyradial flow, one can define the velocity parameter as:

f ðqÞ ¼ ruðrÞ (5.5)

Introducing the x ¼ qaas the dimensionless degree, the dimensionless

form of the velocity parameter can be obtained by dividing that to itsmaximum value as:

FðxÞ ¼ f ðqÞfmax

(5.6)

Substituting Eq. (5.5) into Eqs. (5.2) and (5.3), and eliminating P, onecan obtain the ordinary differential equation for the normalized functionprofile as [2]:

F000 ðxÞ þ 2a Re$A*ð1� fÞ2:5FðxÞF 0ðxÞ

þ �4� ð1� fÞ1:25B* Ha�a2F 0ðxÞ ¼ 0

(5.7)

where A* is a parameter. Reynolds number (Re) and Hartmann number (Ha)based on the electromagnetic parameter are introduced as following form:

A* ¼ ð1� fÞ þ rs

rff

B* ¼ 1þ3

�ss

sf� 1

�f�

ss

sfþ 2

��ss

sf� 1

�f

(5.8)

Re ¼ fmaxa

yf¼ Umaxra

yf

�divergent � channel : a > 0; fmax > 0

convergent � channel : a < 0; fmax < 0

�(5.9)

DTM for Nanofluids and Nanostructures Modeling 199

Ha ¼ffiffiffiffiffiffiffiffiffisf B2

0

rf yf

s(5.10)

With the following reduced form of boundary conditions

Fð0Þ ¼ 1; F 0ð0Þ ¼ 0; Fð1Þ ¼ 0 (5.11)

Physically these boundary conditions mean that maximum values ofvelocity are observed at centerline (x ¼ 0) as shown in Fig. 5.1, and weconsider fully developed velocity profile, thus the rate of velocity is zero at(x ¼ 0). Also, in fluid dynamics, the no-slip condition for fluid states that ata solid boundary, the fluid will have zero velocity relative to the boundary.The fluid velocity at all fluidesolid boundaries is equal to that of the solidboundary, so we can see that the value of velocity is zero at (x ¼ 1). Nowwe apply Differential Transformation Method (DTM) into Eq. (5.7). Takingthe differential transform of Eq. (5.7) with respect to x and consideringH ¼ 1 gives:

ðkþ 1Þðkþ 2Þðkþ 3ÞFðkþ 3Þ

þ 2aRe A*ð1� fÞ2:5Xkr¼0

½ðk� r þ 1ÞFðrÞFðk� r þ 1Þ�

þ �4�Hað1� fÞ1:25�a2ðkþ 1ÞFðkþ 1Þ ¼ 0

(5.12)

Fð0Þ ¼ 1; Fð1Þ ¼ 0 ; Fð2Þ ¼ b; (5.13)

where F(k) is the differential transformation of F(x), and b is a constant,which can be obtained through boundary condition, Eq. (5.11)

f ð1Þ ¼ 0 orXNk¼0

FðkÞ ¼ 0 (5.14)

This problem can be solved for different values of H,

Fð0Þ ¼ 1; Fð1Þ ¼ 0; Fð2Þ ¼ b; Fð3Þ ¼ 0

Fð4Þ ¼ �16

a Re A*ð1� fÞ2:5b� 13a2bþ 1

12a2bHað1� fÞ1:25

Fð5Þ ¼ 0

.

(5.15)


The above process is continuous. Substituting Eq. (5.15) into the mainequation based on DTM. It can be obtained that the closed form of thesolutions is:

FðxÞ ¼ 1þ bx2 þ��1

6a Re A*ð1� fÞ2:5b� 1

3a2b

þ 112a2bHað1� fÞ1:25

�x4 þ.

(5.16)

To obtain the value of b, we substitute the boundary condition fromEq. (5.11) into Eq. (5.16) in point x ¼ 1. This value is too long that are notshown in this paper. By substituting obtained b into Eq. (5.16). We can findthe expressions of F(x). For example, when Re ¼ 50, Ha ¼ 2000,f ¼ 0.04, and a ¼ 3 degrees for a channel filled with Cu-water nanofluid,F(x) function will be found as,

FðxÞ ¼ 1� 0:936173x2 þ 0:0298422x4 � 0:0914287x6

þ 0:00373416x8 � 0:00597469x10(5.17)

Subsequently, applying Padé approximation to Eq. (5.17) (for Padé [4,4]accuracy), we have,

Pad�e ½4; 4�ðFðxÞÞ ¼ 0:999999996� 0:9271132x2 � 0:07601269x4

0:999999996þ 0:00905979x2 � 0:09737341x4(5.18)

In this example the accuracy of three analytical methods named DTM,DTMePadé [4,4] compared with least square method (LSM) for obtain-ing the velocity profile of the MHD JefferyeHamel flow with nanofluid(Fig. 5.1) is investigated. Fig. 5.2 displays plots of F(x) (nondimensionalvelocity profile for MHD JefferyeHamel flow) for different cases of a, Ha,and Re numbers for a divergent and convergent channel where Cu-water isselected as nanofluid from Table 5.1. This figure compared three describedmethods with those of the numerical method. The numerical solution thatis applied to solve the present case is the fourth order RungeeKutta pro-cedure. As in the diagrams of Fig. 5.2, applied methods, specially DTMePadé [4,4] and LSM, show a good agreement with the numerical solution.Tables 5.2 and 5.3 show the values of F(x) when Re ¼ 50, Ha ¼ 2000,f ¼ 0.04, and a ¼ 3 degrees for a divergent and convergent channelrespectively, which are derived from different applied methods for showingvalidity of them. From these two tables the absolute errors of methods werecalculated and confirmed the accuracy and reliability of them. Also tables


reveal that the LSM has the lower error and higher accuracy among theother methods.

The effect of Hartmann number for a divergent and convergent channelis demonstrated in Fig. 5.3(a) and (b). Effect of Hartmann number on skinfriction coefficient is presented as contour plot in Fig.5.3(c). The velocitycurves show that the rate of transport is considerably reduced with theincrease of Hartmann number. This clearly indicates that the transversemagnetic field opposes the transport phenomena. Because the variation ofHa leads to the variation of the Lorentz force due to magnetic field, and theLorentz force produces more resistance to transport phenomena. As seen inthis figure increasing the Ha makes an increase in velocity profile so, by theincreasing of Hartmann number, the flow reversal disappears. IncreasingHartmann number leads to decrease in skin friction coefficient.

Table 5.1 Thermophysical Properties of Nanofluids and NanoparticlesMaterial Density (kg/m3) Electrical Conductivity, s ((Um)�1)

Silver 10,500 6.30 � 107

Copper 8933 5.96 � 107

Ethylene glycol 1113.2 1.07 � 10�4

Drinking water 997.1 0.05

1

0.8

0.4

F(x)

x xF(x)

0.2

00 0.2

DTM-Pade[4,4]Re =50, Ha=2000, α = 7

Re =50, Ha=2000, α = 3Re =50, Ha=0, α = 3

LSM DTM Num DTM-Pade[4,4]Re =50, Ha=2000, α = –7

Re =50, Ha=2000, α = –3Re =50, Ha=0, α = –3

LSM DTM Num

0.4 0.6 0.8

(a) (b)

1

0.6

1

0.8

0.4

0.2

00 0.2 0.4 0.6 0.8 1

0.6

Figure 5.2 Comparison of differential transformation method (DTM)ePadé [4,4], leastsquare method, DTM, and numerical results for nondimensional velocity in differentvalues of parameters for (a) divergent channel and (b) convergent channel, filled withCu-water nanofluid with f ¼ 0.04.


Table 5.2 Comparison of F(x) Values and Errors (%) of Applied Methods for a Divergent Channel, When Re ¼ 50, Ha ¼ 2000, f ¼ 0.04,and a ¼ 3 degrees With CueWater Nanofluid

x

Num DTM DTMePadé [4,4] LSM DTM DTMePadé [4,4] LSM

F(x) Values Errors (%)

0.0 1.00 1.00 1.00 1.00 0.00 0.00 0.000.1 0.9906413 0.9906411 0.9906411 0.9906669 2.02E-05 2.02E-05 0.002580.2 0.9625957 0.9625949 0.9625949 0.9627143 8.31E-05 8.31E-05 0.012320.3 0.9159215 0.9159197 0.9159196 0.9161175 0.000207 0.000197 0.02140.4 0.8506068 0.8506036 0.8506033 0.8507710 0.000411 0.000376 0.01930.5 0.7664072 0.7664020 0.7663992 0.7664192 0.001044 0.000678 0.001570.6 0.6626139 0.6626061 0.6625885 0.6624385 0.003833 0.001177 0.0264710.7 0.5377426 0.5377303 0.5376469 0.5374677 0.017797 0.002287 0.0511210.8 0.3891106 0.3890901 0.3887641 0.3888896 0.089049 0.005268 0.0567960.9 0.2122444 0.2122145 0.2111039 0.2121622 0.537352 0.014088 0.0387291.0 0.00 0.00 0.00 0.00 0.00 0.00 0.00

DTM

forNano

fluidsand

Nanostructures

Modeling

203

Table 5.3 Comparison of F(x) Values and Errors (%) of Applied Methods for a Convergent Channel, When Re ¼ 50, Ha ¼ 2000, f ¼ 0.04and a ¼ �3 degrees With CueWater Nanofluid

x

Num DTM DTMePadé [4,4] LSM DTM DTMePadé [4,4] LSM

F(x) Values Errors (%)

0.0 1.00 1.00 1.00 1.00 0.00 0.00 0.000.1 0.9957323 0.9957536 0.9957536 0.9956519 0.00214 0.00214 0.0080740.2 0.9824096 0.9824974 0.9824974 0.9820070 0.00894 0.00894 0.0409810.3 0.9584170 0.9586246 0.9586246 0.9576683 0.02166 0.02166 0.0781180.4 0.9208756 0.9212703 0.9212713 0.9200429 0.04297 0.04286 0.0904250.5 0.8653567 0.8660262 0.8660365 0.8647706 0.07856 0.07737 0.0677290.6 0.7854792 0.7865345 0.7866062 0.7852691 0.14348 0.13435 0.0267480.7 0.6724157 0.6739705 0.6743519 0.6723980 0.28795 0.23123 0.0026320.8 0.5143816 0.5164428 0.5181024 0.5142390 0.72335 0.40071 0.0277230.9 0.2962961 0.2983765 0.3045470 0.2959939 2.78468 0.70214 0.1019931.0 0.00 0.00 0.00 0.00 0.00 0.00 0.00


ationMethod


Problems

Fig. 5.4(a) and (b) displays the effect of Reynolds number for a divergentand convergent channel with a slope of 5 degrees respectively. Fig. 5.4(c)and (d) shows the effect of Reynolds number on skin friction coefficient.These figures reveal that increasing Reynolds number makes a decrease invelocity profile in divergent channels, also for higher Reynolds number, theflow moves reversely and a region of backflow near the wall is observed (seeFig. 5.4(a) for Re ¼ 200). As shown, in Fig. 5.4(b) for convergent channel,results were inversed and by increasing Reynolds number, velocity profileswere increased and no backflow was observed. Also for large Reynoldsnumbers velocity profile was approximately constant near the centerline andsuddenly reached to zero near the wall. Skin friction coefficient increaseswith increase of Reynolds number.

Fig. 5.5 displays the effect of nanoparticles volume fraction, f, whenRe ¼ 100, Ha ¼ 1000 for a divergent and convergent channel with

Figure 5.3 Effect of Ha on Cuewater velocity profile, when Re ¼ 100, f ¼ 0.05 for(a) divergent channel (a ¼ 5 degrees) and (b) convergent channel (a ¼ �5 degrees);(c) contour plot of f00(0), when Pr ¼ 6.2 (CueWater) and f ¼ 0.06, Re ¼ 5. Ha, Hart-mann number; Re, Reynolds number.


Figure 5.4 Effect of Re on Cuewater velocity profile, when Ha ¼ 1000, f ¼ 0.05 for(a) divergent channel (a ¼ 5 degrees) and (b) convergent channel (a ¼ �5 degrees);(c) and (d) contour plots of f00(0), when Pr ¼ 6.2(CueWater) and f ¼ 0.06. Ha, Hart-mann number; Re, Reynolds number.

Figure 5.5 Effect of nanoparticles volume fraction, f, on velocity profile, whenRe ¼ 100, Ha ¼ 1000 for (a) divergent channel (i ¼ 5 degrees) and (b) convergentchannel (a ¼ �5 degrees). Ha, Hartmann number; Re, Reynolds number.


5 degrees slope. It is assumed that the base fluid and the nanoparticles (Cu-water) are in thermal equilibrium and no slip occurs between them. It canbe seen that increasing nanoparticles volume fraction in divergent channelleads to decrease in velocity profile and the backflow may be started athigh Re.

Finally we considered four different and common structures of nano-fluid from Table 5.1 and their nondimensional velocity profiles, F(x), aredepicted in Figs. 5.6(a) and (b) for a divergent and convergent channelrespectively. As seen in this figure, for a divergent channel, when nanofluidincludes copper (as nanoparticles) or EG (as fluid phase) in its structure, F(x)values are greater than the other structures, but this treatment of nanofluidsstructure is completely vice versa for convergent channels.

5.3 MHD COUETTE NANOFLUID FLOW

Turbulent CuOewater nanofluid flow and heat transfer between twoinfinite horizontal plates located at the y ¼ �h planes are investigated. Theupper plate moves with a uniform velocity U0 while the lower plate is keptstationary. The two plates are assumed to be electrically insulating and keptat two constant temperatures T1 for the lower plate and T2 for the upperplate with T2 > T1. A constant pressure gradient is applied in the x-direction.A uniform magnetic field B0 is applied in the positive y-direction while theinduced magnetic field is neglected by assuming a very small magnetic

Figure 5.6 Effect of nanofluids structure on dimensionless velocity profile, whenRe ¼ 100, Ha¼ 1000, f ¼ 0.05 for (a) divergent channel (a ¼ 5 degrees) and(b) convergent channel (a ¼ �5 degrees). Ha, Hartmann number; Re, Reynolds number.


Reynolds number (see Fig. 5.7). The Hall effect is taken into considerationand consequently a z-component for the velocity is expected to arise. Theuniform suction implies that the y-component of the velocity is constant.Thus, the nanofluid velocity vector is given by v_ðy; tÞ ¼ u_ðy; tÞi þv0j þ w_ðy; tÞk. The nanofluid motion starts from rest at t ¼ 0, and theno-slip condition at the plates implies that the fluid velocity has neither az- nor an x-component at y ¼ �h and y ¼ h. The initial temperature of thefluid is assumed to be equal to T1. The thermophysical properties of thenanofluid are given in Table 5.4 [3].

The flow of the nanofluid is governed by the NaviereStokes equation,which has the two components

rnf

vu_

v t_ þ v0

vu_

vy_

!¼ � dP

_

dx_þ ðmnf þ mtÞ

v2u_

vy_2 �

snf B20

1þ mnf2

�u_ þ mnf w

_�(5.19)

rnf

vw_

v t_ þ v0

vw_

vy_

!¼ ðmnf þ mtÞ

v2w_

vy_2 �

snf B20

1þ m2nf

�w_ � mnf u

_� (5.20)


Table 5.4 Thermophysical Properties of Water and NanoparticlesP (kg/m3) Cp (j/kgk) K (W/m$k) dp (nm) s (U)

Pure water 997.1 4179 0.613 - 0.05CuO 6500 540 18 29 1 � 10�12


where m is the Hall parameter given by mnf ¼ snfb1B0 and b1 is the Hall

factor, and mt is turbulent viscosity given by mt ¼ rnf ‘2m

vu_v y_

. The energy

equation describing the temperature distribution for the fluid is given by

ðrcpÞnf�vTvt

þ v0vTvy

�¼ knf

v2T

vy_2 þ ðmnf þ mtÞ

" vu_

vy_

!2

þ vw_

vy_

!2#

þ snf B20�

1þ mnf2� �u_2 þ w_

2�(5.21)

where T is the temperature of the fluid. The effective density (rnf) and heatcapacitance (rCp)nf of the nanofluid are defined as,

rnf ¼ rf ð1� fÞ þ rsf

ðrCpÞnf ¼ ðrCpÞf ð1� fÞ þ ðrCpÞsf(5.22)

Also effective electrical conductivity of nanofluid was presented byMaxwell as below:

snf

sf¼ 1þ 3ðss=sf � 1Þf

ðss=sf þ 2Þ � ðss=sf � 1Þf (5.23)

In this study Brownian motion impact on thermal conductivity (knf ) andviscosity of nanofluid (mnf ) has been considered

mnf ¼ mstatic þ mBrownian

mstatic ¼mf

ð1� fÞ2:5

mBrownian ¼ 5� 104frf b

ffiffiffiffiffiffiffiffikbTrpdp

sf ðT ;fÞ

f ðT ;fÞ ¼ ð�6:04fþ 0:4705ÞT þ ð1722:3f� 134:63Þ;

b ¼8<:

0:0137ð100fÞ�0:8229f < 0:01

0:0011ð100fÞ�0:7272f > 0:01

(5.24)


keff ¼ kstatic þ kBrownian

kstatickf

¼ 1þ3

�kpkf� 1

�f�

kpkfþ 2

��kpkf� 1

�f

kBrownian ¼ 5� 104bfðrCpÞpffiffiffiffiffiffiffiffikbTrpdp

sf ðT ;fÞ; kb ¼ 1:385� 10�23

(5.25)

The problem is simplified by writing the equations in the nondimen-sional form. To achieve this define the following nondimensional quantities,

y ¼ y_

h; t ¼ t

_U0

h; P ¼ P

_

rf U20

; ðu;wÞ ¼�u_;w_

�U0

; q ¼ T � T1

T2 � T1; a ¼ �dP

_

d x_

(5.26)

Prandtl number (Pr), Reynolds number (Re), Suction parameter (S),Hartmann number (Ha), Eckert number (Ec), Hall parameter (m), Turbulentparameter (Lt), and Turbulent Eckert number (Ect) for base fluid areintroduced as follows:

Pr ¼ U0ðrCpÞf hrf kf

; Re ¼ rU0hmf

; S ¼ rv0hmf

Ha2 ¼ sf B20h

2

mf; Ec ¼ rf U

20 h

ðrCpÞf hðT2 � T1Þ; m ¼ sf b1B0

Lt ¼�‘mh

�2

; Ect ¼�‘mh

�2 rf U20

ðrCpÞf ðT2 � T1Þ(5.27)

In terms of the above nondimensional variables and parameters Eqs.(5.19)e(5.21) are,respectively, written as (where the hats are dropped forconvenience):

vuvt

þ SRe

vuvy

¼ aþ 1Re

A2

A1

v2uvy2

þ Ltv2uvy2

vuvy

� A3

A1

1Re

Ha2

1þ ðm A3Þ2ðuþ m A3wÞ

(5.28)vwvt

þ SRe

vwvy

¼ 1Re

A2

A1

v2wvy2

þ Ltv2wvy2

vuvy

� A3

A1

1Re

Ha2

1þ ðm A3Þ2ðw � m A3uÞ

(5.29)


vq

vtþ SRe

vq

vy¼ 1

PrA5

A4

v2q

vy2þ A2

A4Ec

"�vuvy

�2

þ�vwvy

�2#

þ A3

A1

EcHa2

Re�1þ ðm A3Þ2

� ðu2 þ w2Þ þ A1

A4Ect

"�vuvy

�3

þ vuvy

�vwvy

�2# (5.30)

where Ai (i ¼ 1.5) are defined as follows:

A1 ¼rnf

rf;A2 ¼

mnf

mf;A3 ¼ snf

sf;A4 ¼

ðrCpÞnfðrCpÞf

;A5 ¼ knfkf

(5.31)

The boundary and initial conditions for components of velocity andtemperature are:

u ¼ w ¼ q ¼ 0 for t � 0 and

u ¼ w ¼ q ¼ 0 at y ¼ �1; for t > 0

w ¼ 0; u ¼ q ¼ 1 at y ¼ 1

(5.32)

To solve the coupled nonlinear partial equations (Eqs. 5.28e5.30) in thedomain t ˛ [0, T] and y ˛ [�1, 1] using hybrid DTM and finite differencemethod,weapplyfinite difference approximationon y-direction and takeDTMon t. The following finite difference scheme is used based on a uniform mesh.The length in direction of y is divided intoNy equal intervals. The y-coordinatesof the grid points can be obtained by yj ¼ j(Dy), j ¼ 0:Ny, whereDy is themeshsize. After taking the second order accurate central finite difference approxi-mation with respect to y and applying DTM on Eqs. (5.28)e(5.30) for timedomain, the following recurrence relations can be obtained:

for 1 � j � Ny

Uðj; kþ 1Þ ¼ Hkþ 1

� SRe

�Uðj þ 1; kÞ � Uðj � 1; kÞ

2Dy

�þ adðkÞ

þ 1Re

A2ðj; kÞA1

5

�Uðj þ 1; kÞ � 2Uðj; kÞ þ Uðj � 1; kÞ

Dy2

�

þ Lt

�Uðj þ 1; kÞ � 2Uðj; kÞ þ Uðj � 1; kÞ

Dy2

�$


2Dy

�

� A3

A1

1Re

Ha2

1þ ðmA3Þ2ðUðj; kÞ þ mA3W ðj; kÞÞ

)

(5.33)


W ðj; kþ 1Þ ¼ Hkþ 1

� SRe

�W ðj þ 1; kÞ �W ðj � 1; kÞ

2Dy

�

þ 1Re

A2ðj; kÞA1

5

�W ðj þ 1; kÞ � 2W ðj; kÞ þW ðj � 1; kÞ

Dy2

�

þ Lt

�W ðj þ 1; kÞ � 2W ðj; kÞ þW ðj � 1; kÞ

Dy2

�$


2Dy

�

� A3

A1

1Re

Ha2

1þ ðmA3Þ2ðW ðj; kÞ � mA3Uðj; kÞÞ

)

(5.34)

Qðj; kþ 1Þ ¼ Hkþ 1

� SRe

�Qðj þ 1; kÞ �Qðj � 1; kÞ

2Dy

�

þ 1Pr

A5ðj; kÞA4

5

�Qðj þ 1; kÞ � 2Qðj; kÞ þQðj � 1; kÞ

Dy2

�

þ EcA4

A2ðj; kÞ5��

Uðj þ 1; kÞ � Uðj � 1; kÞ2Dy

�5


2Dy

�

þ�W ðj þ 1; kÞ �W ðj � 1; kÞ

2Dy

�5

�W ðj þ 1; kÞ �W ðj � 1; kÞ

2Dy

��

þ A3

A1

1Re

EcHa2

1þ ðmA3Þ2ðUðj; kÞ5Uðj; kÞ þW ðj; kÞ5W ðj; kÞÞ

þ EctA1

A4

"�Uðj þ 1; kÞ � Uðj � 1; kÞ

2Dy

�3

þ�Uðj þ 1; kÞ � Uðj � 1; kÞ

2Dy

��W ðj þ 1; kÞ �W ðj � 1; kÞ

2Dy

�2#)

(5.35)

where the symbol “5” denotes the convolution operation, A2ðj; kÞ andA5ðj; kÞ are the differential transform of the functions A2(q) and A5(q),respectively.

Applying DTM on initial conditions in Eq. (5.14), we have:

for 0 � j � Ny

Uðj; 0Þ ¼ 0; W ðj; 0Þ ¼ 0; Qðj; 0Þ ¼ 0:(5.36)


The boundary conditions in Eq. (5.32) can be transformed as follows:

BC0s for uðy; tÞ/Uð0; kÞ ¼ 0; k � 0

UðNy; 0Þ ¼ 1;UðNy; kÞ ¼ 0; k � 1(5.37)

BC 0s for wðy; tÞ/W ð0; kÞ ¼ 0; k � 0

W ðNy; kÞ ¼ 0; k � 0(5.38)

BC0s for qðy; tÞ/Qð0; kÞ ¼ 0; k � 0

QðNy; 0Þ ¼ 1;QðNy; kÞ ¼ 0; k � 1(5.39)

As shown in Fig. 5.8, they are in a very good agreement. After thisvalidity, effects of nanoparticle volume fraction, Reynolds number, Hallparameter, Hartmann number, Eckert number, turbulent parameter, andturbulent Eckert number on flow and heat transfer characteristics areexamined.

Fig. 5.9 shows the effect of volume fraction of nanofluid on the velocityand temperature profiles. The velocity components of nanofluid increase asa result of an increase in the energy transport in the fluid with the increasing

Figure 5.8 Comparison between Hybrid Differential Transformation Method in Ref. [3],when f ¼ 0.0, S ¼ 1, a ¼ 5, Re ¼ 1, Ec ¼ 0.2, Ha ¼ 3, m ¼ 3, Pr ¼ 1, Ect ¼ 0.0, Lt ¼ 0.0.


volume fraction. Thus, the skin friction coefficient increases with increasingvolume fraction of nanofluid. The sensitivity of thermal boundary layer thick-ness to volume fraction of nanoparticles is related to the increased thermalconductivity of the nanofluid. In fact, higher values of thermal conductivityare accompanied by higher values of thermal diffusivity. The high values ofthermal diffusivity cause a drop in the temperature gradients and accordinglyincrease the boundary thickness. This increase in thermal boundary layerthickness reduces the Nusselt number; however, the Nusselt number is amultiplication of temperature gradient and the thermal conductivity ratio(conductivity of the nanofluid to the conductivity of the base fluid). Sincethe reduction in temperature gradient due to the presence of nanoparticles ismuch smaller than thermal conductivity ratio, therefore an enhancement inNusselt is taken place by increasing the volume fraction of nanoparticles.

Effect of Hall parameter on velocity and temperature profiles is shownin Fig. 5.10. Hall parameter has a very important role in MHD Couetteflows, because this parameter causes the start of the secondary flow in di-rection z. As Hall parameter increases, velocity profiles also increase thusskin friction coefficient increases with increase of m. Thermal boundarylayer thickness decreases with the increase of Hall parameter.

Figure 5.9 Effect of volume fraction on velocity and temperature profiles, when S ¼ 1,a ¼ 5, Pr ¼ 6.2, Ec ¼ 0.2, Re ¼ 1, Ha ¼ 10, m ¼ 5, Ect ¼ 0.2, Lt ¼ 0.1.


Fig. 5.11 shows that the effect of nanoparticle volume fraction andHartmann number on Nusselt number over the upper and lower plates.Nusselt number over the upper and lower plates increase with increaseof nanoparticle volume fraction while they decrease with augment ofHartmann number. Effects of Reynolds number and turbulent Eckertnumber on Nusselt number over the upper and lower plates are shown in

Figure 5.10 Effect of Hall parameter on velocity and temperature profiles, when S ¼ 1,a ¼ 5, Pr ¼ 6.2, Ec ¼ 0.2, Re ¼ 1, Ha ¼ 10, m ¼ 5, Ect ¼ 0.2, L t¼ 0.1.

Figure 5.11 Effect of nanoparticle volume fraction and Hartmann number on Nusseltnumber over the upper and lower plates S ¼ 1, a ¼ 5, Pr ¼ 6.2, Ec ¼ 0.2, Re ¼ 1,Ect ¼ 0.2, Lt ¼ 0.1.


Fig. 5.12. It is worth to mention that the Reynolds number indicates therelative significance of the inertia effect compared to the viscous effect.Thus, thermal boundary layer thickness over the lower plate decreases as Reincreases and in turn increasing Reynolds number leads to increase inNusselt number. But, an opposite trend is observed for thermal boundarylayer thickness over the upper. Effects of turbulent Eckert number onNusselt number over the upper and lower plates are similar to that ofReynolds number. In the case in which Nusselt number has negativevalues, temperature of the surface near the wall has higher than that of onthe wall, so thermal gradient has a negative value. This phenomenon is dueto increase of viscous dissipation with increase of turbulent Eckert number.

Table 5.5 illustrates the effects of turbulent parameter and Hallparameter on Nusselt number over the upper and lower plates. As turbulentparameter increases, absolute values of Nusselt number over the upper andlower plates decreases. Nusselt number over the lower plate increases with

Figure 5.12 Effects of Reynolds number and turbulent Eckert number on Nusseltnumber over the upper and lower plates S ¼ 1, a ¼ 5, Pr ¼ 6.2, m ¼ 1, Ec ¼ 0.2,Ha ¼ 10, Lt ¼ 0.1, f ¼ 0.02.

Table 5.5 Effect of Turbulent Parameter and Hall Parameter on Nusselt Number Over the Upperand Lower Plates S ¼ 1, a ¼ 5, Pr ¼ 6.2, Ec ¼ Ect ¼ 0.2, Ha ¼ 10, f ¼ 0.02

m

Lt ¼ 0 Lt ¼ 0.1 Lt ¼ 0.2 Lt ¼ 0.3

NuLow NuUp NuLow NuUp NuLow NuUp NuLow NuUp

0 0.1151 �5.8986 0.1155 �5.8999 0.11580 �5.9047 0.1160 �5.91702 0.1566 �3.9305 0.1562 �3.4387 0.1557 �3.0751 0.1553 �2.79523 0.2040 �2.7836 0.2025 �2.2411 0.2011 �1.8450 0.1998 �1.5394 0.2634 �1.9319 0.2604 �1.4038 0.2575 �1.0156 0.2548 �0.71425 0.3321 �1.3311 0.3268 �0.8318 0.3219 �0.4618 0.3175 �0.1711


increase of Hall parameter while absolute value of Nusselt number over theupper plate decreases with augment of Hall parameter.

5.4 NANOFLUID BETWEEN PARALLEL PLATES

We consider the flow and heat transfer analysis in the unsteady two-dimensional squeezing flow of an incompressible nanofluid between theinfinite parallel plates. The two plates are placed at z ¼ �‘(1 � at)1/2 ¼�h(t) (‘ is distant of plate at t ¼ 0 and a is squeezed parameter). For a > 0 ,the two plates are squeezed until they touch t ¼ 1/a and for a < 0, the twoplates are separated. The viscous dissipation effect, the generation of heatdue to fraction caused by shear in the flow, is neglected. Further thesymmetric nature of the flow is adopted [4] and shown in Fig. 5.13.

It is also assumed that the time variable magnetic field�B!¼ Bey!; B ¼

B0ð1� atÞ� is applied, where ey! is unit vector in the Cartesian coordinatesystem. The electric current J and the electromagnetic force F are definedby J ¼ s

�V!� B

!�and F ¼ s

�V!� B

!�� B!, respectively. Also a heat

source (Q ¼ Q0/(1 � at)) is applied between two plates.The nanofluid is a two component mixture with the following as-

sumptions: incompressible; no-chemical reaction; negligible radiative heattransfer; nanosolid particles and the base fluid are in thermal equilibrium

Figure 5.13 Geometry of problem.


and no slip occurs between them. The thermophysical properties of thenanofluid are given in Table 5.6 [4].

The governing equations for conservative momentum and energy inunsteady two-dimensional flow of a nanofluid are:

vuvx

þ vvvy

¼ 0 (5.40)

rnf

�vuvt

þ uvuvx

þ vvuvy

�¼ � vp

vxþ mnf

�v2uvx2

þ v2uvy2

�� snf B

2u (5.41)

rnf

�vvvt

þ uvvvx

þ vvvvy

�¼ � vp

vyþ mnf

�v2vvx2

þ v2vvy2

�(5.42)

vTvt

þ uvTvx

þ vvTvy

¼ knfðrCpÞnf

�v2Tvx2

þ v2Tvy2

�þ QðrCpÞnf

T (5.43)

Here u and v are the velocities in the x and y directions respectively, T isthe temperature, P is the pressure, effective density (rnf), the effective heatcapacity (rCp)nf, and electrical conductivity (snf) of the nanofluid aredefined as:

rnf ¼ ð1� fÞ rf þ f rp;

ðrCpÞnf ¼ ð1� fÞðrCpÞf þ fðrCpÞp

snf

sf¼ 1þ

3

�sp

sf� 1

�f�

sp

sfþ 2

��sp

sf� 1

�f

(5.44)

The Brownian motion has a significant impact on the effective thermalconductivity. The effective thermal conductivity is

keff ¼ kstatic þ kBrownian (5.45)

Table 5.6 Thermophysical Properties of Water and Nanoparticles at RoomTemperature

P (kg/m3) Cp (j/kgk) K (W/m$k) dp (nm)

Pure water 997.1 4179 0.613 eAl2O3 3970 765 25 47CuO 6500 540 18 29


kstatickf

¼ 1þ3

�kpkf� 1

�f�

kpkfþ 2

��kpkf� 1

�f

(5.46)

where, kstatic is the static thermal conductivity based on Maxwell classicalcorrelation. The enhanced thermal conductivity component generated bymicroscale convective heat transfer of a particle’s Brownian motion andaffected by ambient fluid motion is obtained via simulating Stokes’ flowaround a sphere (nanoparticle). By introducing two empirical functions(b and f ), the interaction between nanoparticles in addition to the temper-ature effect in the model, leading to:

kBrownian ¼ 5� 104bfrf cp;f


sf ðT ;fÞ (5.47)

By introducing a thermal interfacial resistance Rf ¼ 4 � 10�8 km2/W, theoriginal kp in Eq. (5.46) was replaced by a new kp,eff in the form:

Rf þ dpkp

¼ dpkp;eff

(5.48)

For different based fluids and different nanoparticles, the functionshould be different. Only water-based nanofluids are considered in thecurrent study. For Al2O3ewater nanofluids and CuOewater nanofluids,this function follows the format:

g0ðT ;f; dpÞ ¼�a1 þ a2 lnðdpÞ þ a3 lnðfÞ þ a4 lnðfÞlnðdpÞ þ a5 lnðdpÞ2

�lnðT Þ

þ �a6 þ a7 lnðdpÞ þ a8 lnðfÞ þ a9 lnðfÞlnðdpÞ þ a10 lnðdpÞ2�

(5.49)

with the coefficients ai (i ¼ 0.10) are based on the type of nanoparticesand also with these coefficients, Al2O3ewater nanofluids and CuOewaternanofluids have an R2 of 96% and 98%, respectively (Table 5.7). Finally, theKKL (KooeKleinstreuereLi) correlation is written as:

kBrownian ¼ 5� 104frf cp;f


sg0ðT ;f; dpÞ (5.50)

For the effective viscosity due to micromixing in suspensions, its proposed:

meff ¼ mstatic þ mBrownian ¼ mstatic þkBrowniankf

� mf

Prf(5.51)


where mstatic ¼ mf

ð1 � fÞ2:5 is viscosity of the nanofluid, as given originally byBrinkman.

The relevant boundary conditions are:

v ¼ vw ¼ dh=dt; T ¼ TH at y ¼ hðtÞ;v ¼ vu=vy ¼ vT=vy ¼ 0 at y ¼ 0:

(5.52)

We introduce these parameters:

h ¼ yhlð1� atÞ1=2

i; u ¼ ax½2ð1� atÞ�f

0ðhÞ;

v ¼ � alh2ð1� atÞ1=2

i f ðhÞ; q ¼ TTH

.(5.53)

Substituting the above variables into Eqs. (5.41) and (5.42), and theneliminating the pressure gradient from the resulting equations gives:

f iv � SðA1=A4Þ�hf

000 þ 3f 00 þ f 0f 00 � ff000��Ha2ðA5=A4Þf 00 ¼ 0; (5.54)

Using Eq. (5.53), Eq. (5.43) reduces to the following differential equations:

q00 þ Pr S

�A2

A3

�ðf q0 � hq0Þ þ Hs

A3q ¼ 0 (5.55)

Here A1, A2, A3, A4, and A5 are dimensionless constants given by:

A1 ¼rnf

rf; A2 ¼

ðrCpÞnfðrCpÞf

; A3 ¼ keffkf; A4 ¼

meff

mf; A5 ¼ snf

sf(5.56)

Table 5.7 The Coefficient Values of Al2O3eWater Nanofluids and CuOeWaterNanofluidsCoefficient Values Al2O3eWater CuOeWater

a1 52.813488759 �26.593310846a2 6.115637295 �0.403818333a3 0.6955745084 �33.3516805a4 4.17455552786E-02 �1.915825591a5 0.176919300241 6.42185846658E-02a6 �298.19819084 48.40336955a7 �34.532716906 �9.787756683a8 �3.9225289283 190.245610009a9 �0.2354329626 10.9285386565a10 �0.999063481 �0.72009983664


With these boundary conditions:

f ð0Þ ¼ 0; f 00ð0Þ ¼ 0;

f ð1Þ ¼ 1; f 0ð1Þ ¼ 0;

q0ð0Þ ¼ 0; qð1Þ ¼ 1:

(5.57)

where S is the squeeze number, Pr is the Prandtl number, Ha is the Hart-mann number, and Hs is the heat source parameter, which are defined as:

S ¼ a‘2

2yf; Pr ¼ mf ðrCpÞf

rf kf; Ha ¼ ‘B0

ffiffiffiffiffisf

mf

r; Hs ¼ Q0‘

2

kf(5.58)

Physical quantities of interest are the skin fraction coefficient andNusselt number, which are defined as:

Cf * ¼mnf

�vuvy

�y¼hðtÞ

rnf v2w; Nu* ¼

�lknf

�vTvy

�y¼hðtÞ

kf TH. (5.59)

In terms of (5.56), we obtain

Cf ¼ jðA1=A4Þf 00ð1Þj;Nu ¼ jA3 q

0ð1Þj. (5.60)

Now DTM has been applied into governing equations (Eqs. 5.54 and5.55). Taking the differential transforms of Eqs. (5.54) and (5.55) withrespect to c and considering H ¼ 1 gives:

ðkþ 1Þðkþ 2Þðkþ 3Þðkþ 4ÞF½kþ 4� þ SðA1=A4ÞXkm¼0

ðD½k� m� 1�ðmþ 1Þðmþ 2Þðmþ 3ÞF½mþ 3�Þ

� 3SðA1=A4Þðkþ 1Þðkþ 2ÞF½kþ 2� � SðA1=A4ÞXkm¼0


þ SðA1=A4ÞXkm¼0


�Ha2ðA5=A4Þðkþ 1Þðkþ 2ÞF½kþ 2� ¼ 0;

D�m� ¼

(1 m ¼ 1

0 m s 1

(5.61)


F½0� ¼ 0; F½1� ¼ a1; F½2� ¼ 0; F½3� ¼ a2 (5.62)

ðkþ 1Þðkþ 2ÞQ½kþ 2� þ Pr$S$

�A2

A3

�Pkm¼0

ðF½k� m�ðmþ 1ÞQ½mþ 1�Þ

�Pr$S$

�A2

A3

�Pkm¼0

ðD½k� m�ðmþ 1ÞQ½mþ 1�Þ

þ HsA3

Q½k� ¼ 0;

D�m� ¼

(1 m ¼ 1

0 m s 1

(5.63)

Q½0� ¼ a3;Q½1� ¼ 0 (5.64)

where F[k] and Q[k] are the differential transforms of f(h), q(h), and a1, a2,a3 are constants which can be obtained through boundary condition. Thisproblem can be solved as followed:

F½0� ¼ 0; F½1� ¼ a1; F½2� ¼ 0; F½3� ¼ a2;F½4� ¼ 0

F½5� ¼ 320

S a2 ðA1=A2Þ þ 120

Sa1a2ðA1=A2Þ þ 120a1a2 þ 1

20Ha2a2;.

(5.65)

Q½0� ¼ a3; Q½1� ¼ 0;Q½2� ¼ �12HsA3

a3;Q½3� ¼ 0:0;

Q½4� ¼ 112

PrS

�A2

A3

�HsA3

a3a1;Q½5� ¼ 0;.

(5.66)

The above process is continuous. By substituting Eqs. (5.64) and (5.66)into the main Eq. (5.61) based on DTM, it can be obtained that the closedform of the solutions is:

FðhÞ ¼ a1hþ a2h3 þ

�320

S a2ðA1=A2Þ þ 120

Sa1a2ðA1=A2Þ

þ 120a1a2 þ 1

20Ha2a2

�h4 þ.

(5.67)

qðhÞ ¼ a3 þ�� 12HsA3

a3

�h2 þ

�112

Pr S

�A2

A3

�HsA3

a3a1

�h4 þ. (5.68)


by substituting the boundary condition from Eq. (5.53) into Eqs. (5.67) and(5.68) in point h ¼ 1, it can be obtained the values of a1, a2, a3. Bysubstituting obtained a1, a2, a3 into Eqs. (5.67) and (5.68), it can be obtainedthe expression of F(h) and Q(h).

The results of this method are compared with previous work inFig. 5.14. This companion indicates that DTM has a good accuracy to solvesuch problem. The kind of nanoparticle is a key factor for heat transferenhancement. Fig. 5.14 shows the effect of nanofluid volume fraction onskin friction coefficient and Nusselt number. As nanofluid volume fractionincrease, Nusselt number also increases while skin friction coefficientdecreases.

Fig. 5.15 shows the effect of the squeeze number on the velocity andtemperature profiles. It is important to note that the squeeze number (S)describes the movement of the plates (S > 0 corresponds to the platesmoving apart, while S < 0 corresponds to the plates moving together theso-called squeezing flow). Vertical velocity decreases with increase ofsqueeze number while horizontal velocity has different behavior. It meansthat horizontal velocity decrease with increase of S when h < 0.5 whileopposite trend is observed for h > 0.5. Thermal boundary layer thicknessincreases with increase of squeeze number. Effect of the Hartmann numberon velocity and temperature profiles is shown in Fig. 5.16. Effects ofHartmann number on velocity profiles are similar to that of squeezenumber. While increasing Hartmann number leads to decrease in thermalboundary layer thickness.

Figure 5.14 Effect of volume fraction of nanofluid on skin friction coefficient andNusselt number, when S ¼ 1, Hs ¼ �1, Pr ¼ 6.2 (CuOewater).


Fig. 5.17 depicts the effects of Hartmann number and squeeze numberon skin friction coefficient and Nusselt number. This figure shows thatHartmann number has direct relationship with both of the skin frictioncoefficient and Nusselt number. While squeeze number has direct rela-tionship with skin friction coefficient and reverse relationship with Nusseltnumber. Effects of heat source parameter on temperature profile and Nusseltnumber is shown in Fig. 5.18. As heat source parameter increases, temper-ature boundary layer thickness decrease and in turn Nusselt number increases.

5.5 VIBRATION ANALYSIS OF NANOBEAMS

In this example, nonlocal EulereBernoulli beam theory is employed forvibration analysis of functionally graded (FG) size-dependent nanobeams.Fig. 5.19 shows the coordinate system for an FG nanobeam with length L,

Figure 5.15 Effect of the squeeze number on velocity and temperature profiles, whenHa ¼ 1, Hs ¼ �1, Pr ¼ 6.2 (CuOeWater).


Figure 5.16 Effect of the Hartmann number on velocity and temperature profiles,when S ¼ 1, Hs ¼ �1, Pr ¼ 6.2 (CuOeWater).

Figure 5.17 Effects of Hartmann number and squeeze number on skin friction coef-ficient and Nusselt number, when f ¼ 0.04, Hs ¼ �1, Pr ¼ 6.2 (CuOeWater).


width b, and thickness h. It is assumed that bottom surface (z ¼ �h/2) offunctionally graded beam is pure metal, whereas the top surface (z ¼ h/2) ispure ceramics. One of the most favorable models for functionally gradedmaterials (FGMs) is the power-law model, in which material properties ofFGMs are assumed to vary according to a power law about spatial co-ordinates. The FG nanobeam is assumed to be composed of ceramic andmetal, and effective material properties (Pf ) of the FG beam such asYoung’s modulus Ef, shear modulus Gf, and mass density rf are assumed tovary continuously in the thickness direction (z-axis direction) according toa power function of the volume fractions of the constituents. According tothe rule of mixture, the effective material properties, P, can be expressed as[5],

Pf ¼ PcVc þ PmVm (5.69)

where Pm, Pc, Vm, and Vc are the material properties, and the volume frac-tions of the metal and the ceramic constituents related by:

Vc þ Vm ¼ 1 (5.70a)

Figure 5.18 Effects of heat source parameter on temperature profile and Nusseltnumber, when f ¼ 0.04, Hs ¼ �1, Pr ¼ 6.2 (CuOeWater).

Figure 5.19 Typical functionally graded beam with Cartesian coordinates.


The volume fraction of the ceramic and metal constituents of thebeam in both power law and MorieTanaka models are assumed to begiven by:

Vc ¼�zhþ 12

�p

;Vm ¼ 1��zhþ 12

�p

(5.70b)

Here p is the nonnegative variable parameter (power-law exponent),which determines the material distribution through the thickness of thebeam. Therefore, from Eqs. (5.69) and (5.70), the effective material prop-erties of the FG nanobeam can be expressed as follows:

Pf ðzÞ ¼ ðPc � PmÞ�zhþ 12

�p

þ Pm (5.71)

Additionally, in this study, MorieTanaka homogenization technique isalso employed to model the effective material properties of the FG nano-beam. According to MorieTanaka homogenization technique the localeffective material properties of the FG nanobeam such as effective local bulkmodulus Ke and shear modulus me can be calculated by:

Ke � Km

Kc � Km¼ Vc

1þ VmðKc � KmÞ=ðKm þ 4mm=3Þ(5.72a)

me � mm

mc � mm¼ Vc

1þ Vmðmc � mmÞ=½mm þ mmð9Km þ 8mmÞ=ð6ðKm þ 2mmÞÞ�(5.72b)

Therefore from Eq. (5.72), the effective Young’s modulus (E), Poisson’sratio (y), and mass density (r) based on MorieTanaka scheme can beexpressed by:

EðzÞ ¼ 9Keme

3Ke þ me(5.73a)

yðzÞ ¼ 3Ke � 2me

6Ke þ 2me(5.73b)

rðzÞ ¼ rcVc þ rmVm (5.73c)

It is worth mentioning that the chosen material gradations demonstratethe principle and are employed for demonstration purposes. Upon the


EulereBernoulli beam model, the displacement field at any point of thebeam can be written as:

uxðx; z; tÞ ¼ uðx; tÞ � zvwðx; tÞ

vx(5.74a)

uzðx; z; tÞ ¼ wðx; tÞ (5.74b)

where t is time, u and w are displacement components of the midplanealong x and z directions, respectively. By assuming the small deformations,the only nonzero strain of the EulereBernoulli beam theory is:

εxx ¼ ε0xx � zk0; ε0xx ¼

vuðx; tÞvx

; k0 ¼ v2wðx; tÞvx2

(5.75)

where ε0xx is the extensional strain and k0 is the bending strain. Based on theHamilton’s principle, which states that, the motion of an elastic structureduring the time interval t1 < t < t2 is such that the time integral of the totaldynamics potential is extremum:Z t

0dðU � T þWextÞdt ¼ 0 (5.76)

Here U is strain energy, T is kinetic energy, and Wext is work done byexternal forces. The virtual strain energy can be calculated as:

dU ¼ZvsijdεijdV ¼

ZvðsxxdεxxÞdV (5.77)

Substituting Eq. (5.75) into Eq. (5.77) yields:

dU ¼Z L

0

�N�dε0xx

��Mðdk0Þ�dx (5.78)

In which N, M are the axial force and bending moment respectively.These stress resultants used in Eq. (5.78) are defined as:

N ¼ZAsxxdA;M ¼

ZAsxxzdA (5.79)

The kinetic energy for EulereBernoulli beam can be written as:

T ¼ 12

Z L

0

ZArðzÞ

�vxxvt

�2

þ�vuzvt

�2!dAdx (5.80)


Also the virtual kinetic energy is:

dT ¼Z L

0

�I0

�vuvt

vduvt

þ vwvt

vdwvt

�� I1

�vuvt

v2dwvtvx

þ vduvt

v2wvtvx

�

þ I2v2wvtvx

v2dwvtvx

�dx

(5.81)

Where (I0, I1, I2) are the mass moment of inertias, defined as follows:

ðI0; I1; I2Þ ¼ZArðzÞð1; z; z2ÞdA (5.82)

The first variation of external forces work of the beam can be written inthe form:

dWext ¼Z L

0ðf ðxÞduþ qðxÞdwÞdx (5.83)

where f(x) and q(x) are external axial and transverse loads distribution alonglength of beam, respectively. By Substituting Eqs. (5.78), (5.81), and (5.83)into Eq. (5.76), and setting the coefficients of du, dw, and dvw/vx to zero,the following EulereLagrange equation can be obtained:

vNvx

þ f ¼ I0v2uvt2

� I1v3wvxvt2

(5.84a)

v2Mvx2

þ q ¼ I0v2wvt2

þ I1v3uvxvt2

� I2v4w

vx2vt2(5.84b)

Under the following boundary conditions:

N ¼ 0 or u ¼ 0 at x ¼ 0 and x ¼ L (5.85a)

vMvx

� I1v2uvt2

þ I2v3wvxvt2

¼ 0 or w ¼ 0 at x ¼ 0 and x ¼ L (5.85b)

M ¼ 0 orvwvx

¼ 0 at x ¼ 0 and x ¼ L (5.85c)

By using nonlocal elasticity model for FG nanobeam the explicit rela-tion of the nonlocal normal force can be derived by [5].

N ¼ Axxvuvx

� Bxxv2wvx2

þ m

�I0

v3uvxvt2

� I1v4w

vx2vt2� vfvx

�(5.86)


Also the explicit relation of the nonlocal bending moment can bederived by [5].

M ¼ Bxxvuvx

� Cxxv2wvx2

þ m

�I0v3wvt2

þ I1v3uvxvt2

� I2v4w

vx2vt2� q

�(5.87)

The nonlocal governing equations of EulereBernoulli FG nanobeam interms of the displacement can be derived by substituting for N and M fromEqs. (5.86) and (5.87), respectively, into Eq. (5.84) as follows:

Axxv2uvx2

� Bxxv3wvx3

þ m

�I0

v4uvx2vt2

� I1v5w

vx3vt2� v2fvx2

�

� I0v2uvt2

þ I1v3wvt2vx

þ f ¼ 0

(5.88a)

Bxxv3uvx3

� Cxxv4wvx4

þ m

�I0

v4wvt2vx2

þ I1v5u

vt2vx3� I2

v6wvt2vx4

� v2qvx2

�

� I0v2wvt2

� I1v3uvt2vx

þ I2v4w

vt2vx2þ q ¼ 0

(5.88b)

In this section to solve the free vibration problem of the FG nanobeamwith various boundary conditions, the DTM is employed. A sinusoidalvariation of u (x, t) and w (x, t) with a circular natural frequency u isassumed and the functions are approximated as,

uðx; tÞ ¼ uðxÞeiut (5.89a)

wðx; tÞ ¼ wðxÞeiut (5.89b)

Substituting Eqs. (5.89a) and (5.89b) into Eqs. (5.88a) and (5.88b),equations of motion can be rewritten as follows:

Axxv2uvx2

� Bxxv3wvx3

þ m

��I0u

2v2u

vx2þ I1u

2v3wvx3

� v2fvx2

�

þ I0u2u� I1u

2vwvx

þ f ¼ 0

(5.90a)

Bxxv3uvx3

� Cxxv4wvx4

þ m

��I0u

2v2wvx2

� I1u2v

3uvx3

þ I2u2v

4wvx4

� v2qvx2

�

þ I0u2w þ I1u2vuvx

� I2u2v2wvx2

þ q ¼ 0

(5.90b)


According to the basic transformation operations introduced in Chapter1, the transformed form of the governing Eqs. (5.90a) and (5.90b) aroundx0 ¼ 0 may be obtained as:

Axxðkþ 1Þðkþ 2ÞU ½kþ 2� � Bxxðkþ 1Þðkþ 2Þðkþ 3ÞW ½kþ 3�� I0u

2ð�U ½k� þ mðkþ 1Þðkþ 2ÞU ½kþ 2�Þ� I1u

2ð�mðkþ 1Þðkþ 2Þðkþ 3ÞW ½kþ 3� þ ðkþ 1ÞW ½kþ 1�Þ ¼ 0

(5.91)

Bxxðkþ 1Þðkþ 2Þðkþ 3ÞU ½kþ 3�� Cxxðkþ 1Þðkþ 2Þðkþ 3Þðkþ 4ÞW ½kþ 4�� I0u2ð�W ½k� þ mðkþ 1Þðkþ 2ÞW ½kþ 2�Þ� I1u2ð�ðkþ 1ÞU ½kþ 1� þ mðkþ 1Þðkþ 2Þðkþ 3ÞU ½kþ 3�Þ� I2u2ð�mðkþ 1Þðkþ 2Þðkþ 3Þðkþ 4ÞW ½kþ 4�þ ðkþ 1Þðkþ 2ÞW ½kþ 2�Þ ¼ 0

(5.92)

where U[k$]and W [k$] are the transformed functions of u and w respec-tively. Additionally, applying DTM to Eqs. (5.85a)e(5.85c), the variousboundary conditions are given as follows:

Simply supportedesimply supported:

W ½0� ¼ 0; W ½2� ¼ 0; U ½0� ¼ 0PNk¼0

W ½k� ¼ 0;PNk¼0

kðk� 1ÞW ½k� ¼ 0;PNk¼0

kU ½k� ¼ 0(5.93a)

Clampedeclamped:

W ½0� ¼ 0;W ½1� ¼ 0;U ½0� ¼ 0PNk¼0

W ½k� ¼ 0;PNk¼0

kW ½k� ¼ 0;PNk¼0

U ½k� ¼ 0(5.93b)

Clampedesimply supported:

W ½0� ¼ 0;W ½1� ¼ 0;U ½0� ¼ 0PNk¼0

W ½k� ¼ 0;PNk¼0


kU ½k� ¼ 0(5.93c)

Clamped-Free:

W ½0� ¼ 0;W ½1� ¼ 0;U ½0� ¼ 0PNk¼0


kðk� 1Þðk� 2ÞW ½k� ¼ 0;PNk¼0

kU ½k� ¼ 0(5.93d)


By using Eqs. (5.91) and (5.92) together with the transformed boundaryconditions, one arrives at the following eigenvalue problem:2

64A11ðuÞ A12ðuÞ A13ðuÞA21ðuÞ A22ðuÞ A23ðuÞA31ðuÞ A32ðuÞ A33ðuÞ

375½C� ¼ 0 (5.94a)

where [C] corresponds to the missing boundary conditions at x ¼ . For thenontrivial solutions of Eq. (5.94a), it is necessary that the determinant of thecoefficient matrix is equal to zero:2

64A11ðuÞ A12ðuÞ A13ðuÞA21ðuÞ A22ðuÞ A23ðuÞA31ðuÞ A32ðuÞ A33ðuÞ

375 ¼ 0 (5.94b)

Solution of Eq. (5.94b) is simply a polynomial root finding problem. Inthis example, the NewtoneRaphson method is used to solve the governingequation of the nondimensional natural frequencies.

Solving Eq. (5.94b), the ith estimated eigenvalue for nth iteration�u ¼ u

ðnÞi

�may be obtained and the total number of iterations is related to

the accuracy of calculations, which can be determined by the followingequation: uðnÞ

i � uðn�1Þi

Dε (5.95)

In this study ε ¼ 0.0001 considered in procedure of finding eigenvalueswhich results in four digit precision in estimated eigenvalues. Further aMatlab program has been developed according to DTM rule stated above, tofind eigenvalues. As mentioned before, DTM implies an iterative procedureto obtain the high-order Taylor series solution of differential equations. TheTaylor series method requires a long computational time for large orders,whereas one advantage of employing DTM in solving differential equationsis a fast convergence rate and a small calculation error. To show the results,functionally graded nanobeam is composed of steel and alumina (Al2O3),where its properties are given in [5]. The bottom surface of the beam is puresteel, whereas the top surface of the beam is pure alumina. The beam geo-metry has the following dimensions: L (length) ¼ 10,000 nm, b (width) ¼1000 nm, and h (thickness) ¼ 100 nm. Relation described in Eq. (5.96) isperformed to calculate the nondimensional natural frequencies.

u ¼ uL2ffiffiffiffiffiffiffiffiffiffiffiffiffiffirA=EI

p(5.96)

where I ¼ bh3/12 is the moment of inertia of the cross section of the beam.


Fig. 5.20 demonstrates the variation of first four fundamental fre-quencies of power law FG nanobeam with varying material distribution andnonlocality parameter at L/h ¼ 100. As can be noted, the first four dimen-sionless frequency of simply supported FG nanobeam decrease acutely asthe material index parameter increases from 0 to 10. It can be observed that,the first and second frequencies reduce with a high rate, where the powerexponent in range from 0 to 5 than the power exponent in range between5 and 10. While the third and fourth frequencies reduce and they havehigh rate in range from 0 to 2. Figs. 5.21 and 5.22 demonstrate the varia-tion of mode number with changing of the nonlocality parameter atconstant slenderness ratio (L/h ¼ 50) of FG nanobeam with simply sup-ported edge conditions and different material distribution for power lawand MorieTanaka models, respectively. As presented, the influence of

7.540

35

30

25

20

15

(a) (b)

(c) (d)

6.5

5.5

Dim

ensi

onle

ss F

requ

ency

( )

5.0

0

90

80

70

60

50

40

30

2 4Power Index (p)

Power Index (p)

Power Index (p)6 8 10

160

140

120

100

80

60

40

0 2 4 6 8 10

0 2 4 6 8 10

Power Index (p)0 2 4 6 8 10

7.0

6.0

Dim

ensi

onle

ss F

requ

ency

( ) ω

Dim

ensi

onle

ss F

requ

ency

( ) ω

Dim

ensi

onle

ss F

requ

ency

( ) ω

ω

Figure 5.20 The variation of the (a) first, (b) second, (c) third, and (d) fourth dimen-sionless frequency of simply supported FG nanobeam with material graduation fordifferent nonlocality parameter (L/h ¼ 100).


nonlocality parameter on the nondimensional frequency increased by thegrowing in mode number. Also, it can be deduced that, the influence ofnonlocality parameter on the frequencies is unaffected with the materialdistribution. More results and discussion on accuracy of method can befound in [5].

μ=0×10μ=1×10

μ=3×10μ=4×10μ=5×10

μ=2×10

μ=0×10μ=1×10

μ=3×10μ=4×10μ=5×10

μ=2×10

μ=0×10μ=1×10

μ=3×10μ=4×10μ=5×10

μ=2×10

p=0 p=0.1

p=0.5

p=5 p=10

p=1

250

200

150

100

Dim

ensi

onle

ss F

requ

ency

( )

Dim

ensi

onle

ss F

requ

ency

( )

Dim

ensi

onle

ss F

requ

ency

( )

Dim

ensi

onle

ss F

requ

ency

( )

Dim

ensi

onle

ss F

requ

ency

( )

50

0

200

150

100

50

0

150150

100

50

0

100

50

0

140

120

100

80

60

40

20

0

Dim

ensi

onle

ss F

requ

ency

( )

140

120

100

80

60

40

20

0

1 2 3 4 5

1 2 3Mode Number

Mode Number

1 2 3 4 5Mode Number




4 5

ω ω

ωω ω

ω

Figure 5.21 The effect of nonlocality parameter on dimensionless frequency of powerlaw FG nanobeam for various mode numbers and with different material graduationindexes (p ¼ 0, 0.1, 0.5, 1, 5, 10) (L/h ¼ 50).


p=0.5

p=5

p=1

p=10

μ=0×10μ=1×10

μ=3×10μ=4×10μ=5×10

μ=2×10

μ=0×10μ=1×10

μ=3×10μ=4×10μ=5×10

μ=2×10

μ=0×10μ=1×10

μ=3×10μ=4×10μ=5×10

μ=2×10

μ=0×10μ=1×10

μ=3×10μ=4×10μ=5×10

μ=2×10

μ=0×10μ=1×10

μ=3×10μ=4×10μ=5×10

μ=2×10

μ=0×10μ=1×10

μ=3×10μ=4×10μ=5×10

μ=2×10

p=0250

200

150

100

Dim

ensi

onle

ss F

requ

ency

( )

Dim

ensi

onle

ss F

requ

ency

( )

Dim

ensi

onle

ss F

requ

ency

( )

50

0

150

100

50

0

Dim

ensi

onle

ss F

requ

ency

( )

140

120

100

40

60

80

20

0

Dim

ensi

onle

ss F

requ

ency

( )

140

120

100

40

60

80

20

0

Dim

ensi

onle

ss F

requ

ency

( )

150

100

50

0

200

150

100

50

01 2 3

Mode Number Mode Number4 5

1 2 3Mode Number

4 5

1 2 3Mode Number

4 5

1 2 3Mode Number

4 5

1 2 3Mode Number

4 5

1 2 3 4 5

p=0.1ω ω

ω ω

ω ω

Figure 5.22 The effect of nonlocality parameter on dimensionless frequency ofMorieTanaka FG nanobeam for various mode numbers and with different materialgraduation indexes (p ¼ 0, 0.1, 0.5, 1, 5, 10) (L/h ¼ 50).


5.6 BUCKLING ANALYSIS OF A SINGLE-WALLEDCARBON NANOTUBE

Carbon nanotubes (CNTs) are an allotrope of carbon. They take the formof cylindrical carbon molecules and have novel properties that make thempotentially useful in a wide variety of applications in nanotechnology, elec-tronics, optics, and other fields of materials science. In the recent few years,CNTs and single-walled carbon nanotubes (SWCNTs) have been one ofthe most promising studies in the field of mechanics, physics, chemistry,materials science, and so on, which motivated the researchers to work on[6e8]. Fig. 5.23 show a sample of CNT.

In this example, we considered the EulereBernoulli beam model usingstress gradient approach for the buckling analysis of SWCNT with nonlocaleffect. The governing differential equation for the buckling is

EId4wdx4

þ Pd2

dx2

�w � ðe0aÞ2 d

2wdx2

�¼ 0 (5.97)

where w ¼ (x$t) is the transverse beam deflection, t, x, are the spatial coor-dinate and the time; E is the Young modulus of elasticity; P is the bucklingload, I is the moment of inertia of the beam cross section, e0 is a constantappropriate to each material, a is an internal characteristic length. The e0a isdetermined by matching the dispersion curves based on the atomic models.For the buckling analysis, Eq. (5.97) can be nondimensionalised using L(length of the beam) and rewritten as

d4wdx4

þ PL2

EId2

dx2

�w � ðe0a=LÞ2 d

2wdx2

�¼ 0 (5.98)

Fluid InCNT

L

kw

kp

hFluid Out

ShearLayer

Figure 5.23 Single-walled carbon nanotube embedded in a Pasternak-type founda-tion model.


By applying DTM to above equation, following expression can bewritten easily

W ðkþ 4Þ ¼ �rW ðkþ 2Þðkþ 3Þðkþ 4Þ (5.99)

where

r ¼ PL2=EI�1� ðe0aÞ2ðP=EIÞ

� (5.100)

The boundary conditions are different for different conditions.A. Clampedeclamped

In this case, consider the SWCNT supported by clamped at both theends and the boundary conditions defined as

wð0Þ ¼ 0;w0ð0Þ ¼ 0;wðLÞ ¼ 0;w0ðLÞ ¼ 0 (5.101)

And DTM transformed shape is

W ð0Þ ¼ 0;W ð1Þ ¼ 0;XNk¼0

W ðkÞ ¼ 0;XNk¼0

kW ðkÞ ¼ 0 (5.102)

B. Simply supportedThe boundary conditions for the case of simply supported SWCNT

at both the ends are defined as

wð0Þ ¼ 0;w00ð0Þ ¼ 0;wðLÞ ¼ 0;w00ðLÞ ¼ 0 (5.103)


W ð0Þ ¼ 0;W ð2Þ ¼ 0;XNk¼0

W ðkÞ ¼ 0;XNk¼0

kðk� 1ÞW ðkÞ ¼ 0 (5.104)

C. Clampedesimply supportedFor the SWCNT supported by clamped at one end and simply

supported in the other end, the boundary conditions are defined as

wð0Þ ¼ 0;w0ð0Þ ¼ 0;wðLÞ ¼ 0;w00ðLÞ ¼ 0 (5.105)


W ð0Þ ¼ 0;W ð1Þ ¼ 0;XNk¼0

W ðkÞ ¼ 0;XNk¼0

kðk� 1ÞW ðkÞ ¼ 0 (5.106)


D. Clampedefree SWCNTFor the SWCNT supported by clamped at one end and free in the

other end, the boundary conditions are defined as

wð0Þ ¼ 0;w0ð0Þ ¼ 0;MNLðLÞ ¼ 0;VNLðLÞ ¼ 0

MNL ¼ �EI

�d2wdx2

þ ðe0aÞ2 d4wdx4

�

VNL ¼ �EI

�d3wdx3

þ ðe0aÞ2 d5wdx5

�� P

dwdx

(5.107)

where MNL and VNL are the nonlocal moment and nonlocal shearforce. And DTM transformed shape is

W ð0Þ ¼ 0;W ð3Þ ¼ 0PNk¼0

kðk� 1ÞW ðkÞ�1þ ðe0aÞ2ðk� 2Þðk� 3Þ� ¼ 0

PNk¼0

kðk� 1Þðk� 2ÞW ðkÞ�1þ ðe0aÞ2ðk� 3Þðk� 4Þ�þ ðrÞðkÞW ðkÞ ¼ 0

(5.108)

REFERENCES[1] Choi SUS. Enhancing thermal conductivity of fluids with nanoparticles. In: Siginer DA,

Wang HP, editors. Developments and applications of non-Newtonian flows, FED-vol.231/MD-vol. 66. New York: ASME; 1995. p. 99e105.

[2] Hatami M, Sheikholeslami M, Hosseini M, Ganji DD. Analytical investigation of MHDnanofluid flow in non-parallel walls. Journal of Molecular Liquids 2014;194:251e9.

[3] Mosayebidorcheh S, Sheikholeslami M, Hatami M, Ganji DD. Analysis of turbulentMHD Couette nanofluid flow and heat transfer using hybrid DTMeFDM. Partic-uology 2016;26:95e101.

[4] Sheikholeslami M, Ganji DD. Nanofluid flow and heat transfer between parallel platesconsidering Brownian motion using DTM. Computer Methods in Applied Mechanicsand Engineering 2015;283:651e63.

[5] Ebrahimi F, Salari E. Size-dependent free flexural vibrational behavior of functionallygraded nanobeams using semi-analytical differential transform method. Composites PartB: Engineering 2015;79:156e69.

[6] Senthilkumar V. Buckling analysis of a single-walled carbon nanotube with nonlocalcontinuum elasticity by using differential transform method. Advanced Science LettersSeptember 2010;3(3):337e40 (4).

[7] Valipour P, Ghasemi SE, Khosravani MR, Ganji DD. Theoretical analysis on nonlinearvibration of fluid flow in single-walled carbon nanotube. Journal of Theoretical andApplied Physics. http://dx.doi.org/10.1007/s40094-016-0217-9.

[8] Ahmadi Asoor AA, Valipour P, Ghasemi SE. Investigation on vibration of single-walledcarbon nanotubes by variational iteration method. Applied Nanoscience 2016;6:243e9.


http://dx.doi.org/10.1007/s40094-016-0217-9

CHAPTER 6

DTM for Magnetohydrodynamic(MHD) and Porous MediumFlows

6.1 INTRODUCTION

In recent years the effect of magnetic field and porous medium in differentengineering applications such as the cooling of reactors and many metal-lurgical processes involves the cooling of continuous tiles has been moreconsiderable. And also, in several engineering processes, materials manu-factured by extrusion processes and heat treated materials traveling betweena feed roll and a wind up roll on convey belts possess the characteristics of amoving continuous surface, oil industry and combustion, penetration etc.,you can find some applications of Magnetohydrodynamic (MHD) andporous medium [1]. Although in previous chapters some examples werepresented, which contained these two important topics, but in this separatechapter some other examples are discussed due to its importance in thefollowing sections:6.1 Introduction6.2 Magnetohydrodynamic Couette Fluid Flow Between Parallel Plates6.3 Micropolar Fluid in a Porous Channel6.4 Magnetohydrodynamic Viscous Flow Between Porous Surfaces

6.2 MAGNETOHYDRODYNAMIC COUETTE FLUID FLOWBETWEEN PARALLEL PLATES

The fluid is assumed to be flowing between two infinite horizontal plateslocated at the y ¼ �h planes. The upper plate moves with a uniformvelocity U0 while the lower plate is kept stationary. The two plates areassumed to be electrically insulating and kept at two constant temperaturesT1 for the lower plate and T2 for the upper plate with T2 > T1. A constantpressure gradient is applied in the x-direction. A uniform magnetic field B0

is applied in the positive y-direction while the induced magnetic field is



http://dx.doi.org/10.1016/B978-0-12-805190-0.00006-1

neglected by assuming a very small magnetic Reynolds number (SeeFig. 6.1). The Hall effect is taken into consideration and consequently az-component for the velocity is expected to arise. The fluid motion startsfrom rest at t ¼ 0, and the no-slip condition at the plates implies that thefluid velocity has neither a z nor an x-component at y ¼ �h and y ¼ h. Theinitial temperature of the fluid is assumed to be equal to T1. Since the platesare infinite in the x and z directions, the physical quantities do not changein these directions and the problem is essentially one dimensional [2].

The flow of the fluid is governed by the NaviereStokes equation whichhas the two components.

rvuvt

¼ �dPdx

þ mv2uvy2

þ vm

vyvuvy

� sB20


rvwvt

¼ mv2wvy2

þ vm

vyvwvy

� sB20

1þ m2ðw � muÞ (6.2)

where r is the density of the fluid, m is the viscosity of the fluid, v is thevelocity vector of the fluid ¼ u(y, t)i þ w(y, t)j, s is the electric conductivityof the fluid, m is the Hall parameter given by m ¼ sbB0, and b is the Hallfactor. The energy equation describing the temperature distribution for thefluid is given by Ref. [2].

rcpvTvt

¼ v

vy

�kvTvy

�þ m

"�vuvy

�2

þ�vwvy

�2#þ sB2

0

ð1þ m2Þ ðu2 þ w2Þ

(6.3)

where T is the temperature of the fluid, cp is the specific heat at constantpressure of the fluid, and k is thermal conductivity of the fluid. The viscosityof the fluid is assumed to vary exponentially with temperature and isdefined as m ¼ m0f(T) ¼ m0exp(�a(T � T1)). Also the thermal conductivity

Moving wall

Couette flow

z

y

x

Stationary wall

U0

Magnetic field (B0)

T2

2h

T1

T2>T1



of the fluid is varying linearly with temperature as k ¼ k0gðT Þ ¼k0½1þ bðT � T1Þ]. The problem is simplified by writing the equations inthe nondimensional form. To achieve this define the following nondimen-sional quantities,

by ¼ yh;bt ¼ tU0

h; bP ¼ P

rU20; ðbu; bwÞ ¼ ðu;wÞ

U0; q ¼ T � T1

T2 � T1;a ¼ �dbP

dbx (6.4)

f ðqÞ ¼ e�aq, a is the viscosity parameter. g(q) ¼ 1 þ bq b is the thermal con-ductivity parameter. Re ¼ rU0h

m0is the Reynolds number. Ha2 ¼ sB2

0h2

m0, Ha is

Hartmann number. Pr ¼ m0cpk0

is Prandtl number, and Ec ¼ U20

cpðT2�T1Þ is Eckertnumber. In terms of the above nondimensional quantities the velocity and en-ergy Eqs. (6.1)e(6.3) read,

vuvt

¼ aþ 1Re

f ðqÞ v2uvy2

þ 1Re

vf ðqÞvy

vuvy

� 1Re

Ha2


vwvt

¼ 1Re

f ðqÞ v2wvy2

þ 1Re

vf ðqÞvy

vwvy

� 1Re

Ha2

1þ m2ðw � muÞ (6.6)

vq

vt¼ 1

RePrgðqÞ v

2q

vy2þ 1RePr

�vgðqÞvy

��vq

vy

�þ EcRe

f ðqÞ"�

vuvy

�2

þ�vwvy

�2#

þ EcHa2

Reð1þ m2Þ ðu2 þ w2Þ

(6.7)

The boundary and initial conditions for components of velocity andtemperature are,

IC’s/

8>><>>:uðy; 0Þ ¼ 0

wðy; 0Þ ¼ 0

qðy; 0Þ ¼ 0

BC’s/

8>><>>:uð�1; tÞ ¼ 0; uð1; tÞ ¼ 1

wð�1; tÞ ¼ 0; wð1; tÞ ¼ 0

qð�1; tÞ ¼ 0; qð1; tÞ ¼ 1

(6.8)

Once the values of the velocities and temperature are obtained, thefriction coefficient and Nusselt number will be determined. The local skinfriction coefficient at the lower wall is [2],

DTM for Magnetohydrodynamic (MHD) and Porous Medium Flows 241

Cf ¼ 2Re

vUvy

��y¼�1

(6.9)

And the local Nusselt number for lower wall is defined as,

Nux ¼ vq

vy

��y¼�1

(6.10)

To solve the partial differential equation u(y, t) in the domain t˛ ½0;T �and y˛ ½yfirst; yend� using hybrid modified differential transformation methodand finite difference method (FDM), we apply finite difference approxi-mate on y-direction and take modified differential transformation method(MDTM) on t. The time domain is divided to Nt sections. We suppose thetime subdomains are equal and length of each subdomain is H ¼ T/Nt. Sothere is a separate function for every subdomain as follows:

uiðy; tÞ ¼

8>>>>>><>>>>>>:

u1ð j; tÞ; t˛ ½t1; t2�; 1 � j � Ny þ 1

«

uið j; tÞ; t˛ ½ti; tiþ1�; 1 � j � Ny þ 1

«

uNtð j; tÞ; t˛ ½tN ; tNtþ1�; 1 � j � Ny þ 1

(6.11)

where ti ¼ (i � 1)H and Ny is the number of cells in y-direction. The so-lution of the system of Eqs. (6.5) to (6.8) can be assumed as the followingform:

for 1 � i � Nt; 1 � j � Ny þ 1

uið j; tÞ ¼Xmk¼0

Uið j; kÞ�t � tiH

�k

t˛ ½ti; tiþ1�

wið j; tÞ ¼Xmk¼0

Wið j; kÞ�t � tiH

�k

t˛ ½ti; tiþ1�

qið j; tÞ ¼Xmk¼0

Qið j; kÞ�t � tiH

�k

t˛ ½ti; tiþ1�

(6.12)

After taking second-order accurate central finite difference approxi-mation with respect to y and applying MDTM on Eqs. (6.5) to (6.8), thefollowing recurrence relations can be obtained:


for 1 � i � Nt; 1 � j � Ny � 1

Uið j; kþ 1Þ ¼ Hkþ 1

(adðkÞ þ 1

ReDy2Xk

r¼0

Fið j; k� rÞðUið j þ 1; rÞ

� 2Uið j; rÞ þ Uið j � 1; rÞÞ þ 14ReDy2

Xk

r¼0

ðFið j þ 1; k� rÞ

�Fið j � 1; k� rÞÞðUið j þ 1; rÞ � Uið j � 1; rÞÞ

� Ha2

Reð1þ m2Þ ðUið j; kÞ þ mWið j; kÞÞ�

ð6:13Þ

Wið j; kþ 1Þ ¼ Hkþ 1

(1

ReDy2Xk

r¼0

Fið j; k� rÞðWið j þ 1; rÞ

� 2Wið j; rÞ þWið j � 1; rÞÞ þ 14ReDy2

Xk

r¼0

ðFið j þ 1; k� rÞ

�Fið j � 1; k� rÞÞðWið j þ 1; rÞ �Wið j � 1; rÞÞ

� Ha2

Reð1þ m2Þ ðWið j; kÞ � mUið j; kÞÞ�

ð6:14Þ

Qið j; kþ 1Þ ¼ Hkþ 1

�1

RePrDy2ðQið j þ 1; kÞ � 2Qið j; kÞ þQið j � 1; kÞÞ

þ bRePrDy2

Xk

r¼0

Qið j; k� rÞðQið j þ 1; rÞ � 2Qið j; rÞ þQið j � 1; rÞÞ

þ 14RePrDy2

Xk

r¼0

ðQið j þ 1; k� rÞ �Qið j � 1; k� rÞÞ

ðQið j þ 1; rÞ �Qið j � 1; rÞÞ þ Ec4ReDy2

Xk

r¼0

Xr

s¼0

Fið j; sÞ

½ðUið j þ 1; r � sÞ � Uið j � 1; r � sÞÞðUið j þ 1; k� rÞ � Uið j � 1; k� rÞÞðWið j þ 1; r � sÞ �Wið j � 1; r � sÞÞðWið j þ 1; k� rÞ �Wið j � 1; k� rÞÞ�

þ EcHa2

Reð1þ m2ÞXk

r¼0

½Uið j; rÞUið j; k� rÞ þWið j; rÞWið j; k� rÞ�)

ð6:15Þ


where Fi( j, k) and Gi( j, k) are the differential transform of the functions f(q)and g(q), respectively. Applying MDTM on initial conditions in Eq. (6.8),

for 1 � j � Ny þ 1

U1ð j; 0Þ ¼ 0;W1ð j; 0Þ ¼ 0;T1ð j; 0Þ ¼ 0(6.16)

The boundary conditions in Eq. (6.8) can be transformed as follows:

for 1 � j � Nt

BC’s for uðy; tÞ/(Uið1; kÞ ¼ 0; k � 0

UiðNy þ 1; 0Þ ¼ 1;UiðNy þ 1; kÞ ¼ 0; k � 1

(6.17)

BC’s for wðy; tÞ/(Wið1; kÞ ¼ 0; k � 0

WiðNy þ 1; kÞ ¼ 0; k � 0(6.18)

BC’s for qðy; tÞ/(Qið1; kÞ ¼ 0; k � 0

QiðNy þ 1; 0Þ ¼ 1;QiðNy þ 1; kÞ ¼ 0; k � 1(6.19)

For solving the problem in whole of the time subdomains, we must usethe continuity condition in each time subdomain. These conditions can beexpressed as,

for 2 � j � Ny ; 2 � i � Nt

Uið j; 0Þ ¼Xmk¼0

Ui�1ð j; kÞ; Wið j; 0Þ ¼Xmk¼0

Wi�1ð j; kÞ

Qið j; 0Þ ¼Xmk¼0

Qi�1ð j; kÞ

(6.20)

In this study all calculations are based on main data for a ¼ 5, Pr ¼ 1,Re ¼ 1, Ec ¼ 0.2, Ha ¼ 1, m ¼ 3, a ¼ 0.5, b ¼ 0.5 for Eq. (6.7), and in theparametric study for effect of each parameter, other parameters are assumedto be constant and the effect of the main parameter in an acceptable range isinvestigated. In the first step, the accuracy of hybrid differential trans-formation method (DTM) is investigated comparing to FDM. As seen inFig. 6.2, which presents the velocity and temperature distributions asfunctions of y for various values of time (0.1, 0.5, and 2 s), hybrid-DTM iscompletely accurate and efficient. This figure reveals that temperature inCouette flow can exceed from hot plate temperature in large times(Fig. 6.2(c), t ¼ 2s). As seen hybrid-DTM has an excellent agreement with


FDM numerical technique, furthermore its solution is obtained using asimple iterative procedure, also it reduces the execution time and memoryrequirements for large scale computations. Following a parametric study forshowing the effect of constant numbers appeared in the mathematicalequations is investigated by hybrid-DTM.

Fig. 6.3 shows the effect of Hartmann number (Ha) on velocities andtemperature versus time in the center of channel (y ¼ 0). Fig. 6.3(a) con-firms that increasing Ha decreases u as it increases the damping force on u.Fig. 6.3(b) presents an interesting phenomenon, which is, increasing Haincreases w for small t and decreases w for large t and steady state condition.Fig. 6.3(c) shows that the effect of Ha on the temperature q depends on t. IfHa is small 0 < Ha < 1, then increasing Ha increases q as a result ofincreasing the Joule dissipation. However, for large values of t, increasingHa decreases q due to the reduction in the Joule and viscous dissipations.

Fig. 6.4 demonstrates the effect of Reynolds number (Re) on velocitiesand temperature distribution in the center of channel. These figures reveal

(a) (b)

(c)

0

0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.5

1

1.5

2

2.5

3

3.5

4

0 0.2–0.2

y

y y

u w

θ

–0.4–0.6–0.8–1 0.4 0.6 0.8 1

0 0.2–0.2–0.4–0.6–0.8–1 0.4 0.6 0.8 1 0 0.2–0.2–0.4–0.6–0.8–1 0.4 0.6 0.8 1

0.2

0.4

0.6

0.8

1

1.2

PresentAttia

PresentAttia

PresentAttia

t = 0.1,0.5,2

t = 0.1,0.5,2

t = 0.1,0.5,2

Figure 6.2 Comparison between hybrid differential transformation method, whena ¼ 5, Pr ¼ 1, Re ¼ 1, Ec ¼ 0.2, Ha ¼ 1, m ¼ 3, a ¼ 0.5, b ¼ 0.5. (a) u velocity, (b) wvelocity and, (c) temperature.


that the velocity components, u and w, and temperature q reach steady statequickly for low Re numbers but their steady states value is great for large Renumbers. Also these figures confirm that u reaches steady state faster than w.This is expected as u is the source of w.

Effect of Hall parameter (m) is depicted in Fig. 6.5. It is evident that uincreases with m for all times (See Fig. 6.5(a)). This is due to the fact that anincrease in m decreases the effective conductivity, and hence the magneticdamping decreases which increases u. Fig. 6.5(b) illustrates effect of m on w,velocity component. The effect of m on q is shown in Fig. 6.5(c). For allvalues for m, increasing m decreases q slightly. This is because an increase inm results in an increase in u but a decrease in w, so the Joule dissipation thatis proportional also to (1/(1 þ m2)) decreases.

According to Eqs. (6.9) and (6.10), local Nusselt number and skinfriction coefficient for lower plate are calculated and results are presented

00

0.2

0.4

0.6

0.8

1

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

0.5 1 1.5 2 2.5 3 3.5 4 54.50

00.5

0.5

1.5u w

θ

tt

t

2.5

2

1

1 1.5 2 2.5 3 3.5 4 54.5

0 0.5 1 1.5 2 2.5 3 3.5 4 54.5

(a) (b)

(c)

Ha = 8Ha = 6Ha = 4Ha = 2

Ha = 8Ha = 6Ha = 4Ha = 2

Ha = 8Ha = 6Ha = 4Ha = 2

Figure 6.3 Effect of Hartmann number (Ha) in center (y ¼ 0), when a ¼ 5, Pr ¼ 1,Re ¼ 1, Ec ¼ 0.2, m ¼ 3, a ¼ 0.5, b ¼ 0.5. (a) u velocity, (b) w velocity and, (c)temperature.


through Figs. 6.6 and 6.7 respectively. As seen in Fig. 6.7, increasing inReynolds number and Hall parameter makes high Nusselt number, butincreasing the Hartmann number decreases Nusselt number and heattransfer. Fig. 6.7 reveals that increasing the Hall parameter, Hartmann andReynolds number have a similar behavior on skin friction coefficientcompared to Nusselt number in Fig. 6.6, which is due to relevancy be-tween temperature and velocity gradients.

6.3 MICROPOLAR FLUID IN A POROUS CHANNEL

A micropolar fluid is the fluid with internal structures in which couplingbetween the spin of each particle and the macroscopic velocity field is takeninto account. We consider the steady laminar flow of a micropolar fluidalong a two-dimensional channel with porous walls through which fluid isuniformly injected or removed with speed v0. The lower channel wall has a

Figure 6.4 Effect of Reynolds number (Re) in center (y ¼ 0), when a ¼ 5, Pr ¼ 1,Ha ¼ 1, Ec ¼ 0.2, m ¼ 3, a ¼ 0.5, b ¼ 0.5. (a) u velocity, (b) w velocity and, (c)temperature.


solute concentration C1 and temperature T1 while the upper wall has soluteconcentration C2 and temperature T2 as shown in Fig. 6.8. Using Cartesiancoordinates, the channel walls are parallel to the x-axis and located aty ¼ �h, where 2h is the channel width. The relevant equations governingthe flow are [3]:

vuvx

þ vvvy

¼ 0; (6.21)

r

�uvuvx

þ vvuvy

�¼ � vP

vxþ ðmþ kÞ

�v2uvx2

þ v2uvy2

�þ k

vNvy

; (6.22)

r

�uvvvx

þ vvvvy

�¼ � vP

vyþ ðmþ kÞ

�v2vvx2

þ v2vvy2

�� k

vNvx

; (6.23)

r

�uvNvx

þ vvNvy

�¼ � k

j

�2N þ vu

vy� vvvx

�þ�ms

j

��v2Nvx2

þ v2Nvy2

�;

(6.24)

Figure 6.5 Effect of Hall parameter (m) in center (y ¼ 0), when a ¼ 5, Pr ¼ 1, Ha ¼ 1,Ec ¼ 0.2, R ¼ 1, a ¼ 0.5, b ¼ 0.5. (a) u velocity, (b) w velocity and, (c) temperature.


r

�uvTvx

þ vvTvy

�¼ k1

cp

v2Tvy2

; (6.25)

r

�uvCvx

þ vvCvy

�¼ D*v

2Cvy2

$ (6.26)

where u and v are the velocity components along the x- and y-axis respec-tively, r is the fluid density, m is the dynamic viscosity, N is the angular ormicrorotation velocity, P is the fluid pressure, T and cp are the fluid temper-ature and specific heat at constant pressure respectively, C is the speciesconcentration, k1 and D* are the thermal conductivity and molecular

Figure 6.6 Nusselt number variations versus Reynolds number and (a) Hartmannnumber (b) Hall parameter (m), when a ¼ 5, Pr ¼ 1, Ec ¼ 0.2, a ¼ 0.5 and b ¼ 0.5 forsteady state condition.


Figure 6.7 Skin friction coefficient variations versus Reynolds number and (a)Hartmann number (b) Hall parameter (m), when a ¼ 5, Pr ¼ 1, Ec ¼ 0.2, a ¼ 0.5 andb ¼ 0.5 for steady state condition.

Figure 6.8 Geometry of problem.


diffusivity respectively, j is the microinertia density, k is a material param-eter, and ys ¼

�mþ k

2

�j is the micro rotation viscosity.

The appropriate boundary conditions are:

y ¼ �h : v ¼ u ¼ 0;N ¼ �svuvy;

y ¼ þh : v ¼ 0; u ¼ v0xh;N ¼ v0x

h2$

(6.27)

where s is a boundary parameter and indicates the degree to which the mi-croelements are free to rotate near the channel walls. The case s ¼ 0 repre-sents concentrated particle flows in which microelements close to the wallare unable to rotate. Other interesting particular cases that have beenconsidered in the literature include s ¼ 1/2, which represents weak con-centrations and the vanishing of the antisymmetric part of the stress tensor,and s ¼ 1, which represents turbulent flow. We introduce the followingdimensionless variables:

h ¼ yh; j ¼ �v0xf ðhÞ; N ¼ v0x

h2gðhÞ;

qðhÞ ¼ T � T2

T1 � T2; fðhÞ ¼ C � C2

C1 � C2;

(6.28)

where T2 ¼ T1 � Ax, C2 ¼ C1 � Bx with A and B as constants. The

stream function is defined in the usual way; u ¼ vj

vy; v ¼ �vj

vx$

Eqs. (6.21)e(6.27) reduce to the coupled system of nonlinear differ-ential equations:

ð1þN1Þ f IV �N1g � Re�ff

000 � f 0f 00� ¼ 0; (6.29)

N2g00 þN1ðf 00 � 2gÞ �N3Re ðfg0 � f 0gÞ ¼ 0; (6.30)

q00 þ Peh f 0q� Peh f q0 ¼ 0; (6.31)

f00 þ Pem f 0f� Pem f f0 ¼ 0; (6.32)

subject to the boundary conditions:

h ¼ �1 : f ¼ f 0 ¼ g ¼ 0; q ¼ f ¼ 1;

h ¼ þ1 : f ¼ q ¼ f ¼ 0; f 0 ¼ �1; g ¼ 1:(6.33)

The parameters are the buoyancy ratio N, the Peclet numbers for thediffusion of heat Peh and mass Pem, respectively, the Reynolds number Re


where for suction Re > 0 and for injection Re < 0 and the Grashof numberGr given by:

N1 ¼ k

m;N2 ¼ ns

mh2; N3 ¼ j

h2; Re ¼ v0

yh;

Pr ¼ yrcpk1

; Sc ¼ y

D*;Gr ¼ gbTAh

4

y2;

Peh ¼ PrRe; Pem ¼ ScRe;

(6.34)

where Pr is the Prandtl number, Sc is the generalized Schmidt number, N1

is the coupling parameter, and N2 is the spin-gradient viscosity parameter.In technological processes, the parameters of particular interest are the localNusselt and Sherwood numbers. These are defined as follows:

Nux ¼q00y¼�hx

ðT1 � T2Þk1 ¼ �q0ð� 1Þ; (6.35)

Shx ¼ m00y¼�hx

ðC1 � C2ÞD*¼ �f0ð� 1Þ; (6.36)

where q00 and m00 are local heat flux and mass flux respectively. Now DTMinto governing equations has been applied. Taking the differential transformof Eqs. (6.29) and (6.32) with respect to c, and considering H ¼ 1 gives:

ð1þN1Þðkþ 1Þðkþ 2Þðkþ 3Þðkþ 4ÞF½kþ 4� �N1 G½k�

�ReXk

m¼0


þReXk

m¼0

ððk� mþ 1ÞF½k� mþ 1�ðmþ 1Þðmþ 2ÞF½mþ 2�Þ ¼ 0;

(6.37)

F½0� ¼ a0; F½1� ¼ a1; F½2� ¼ a2; F½3� ¼ a3; (6.38)

N2ðkþ 1Þðkþ 2ÞG½kþ 2� þN1ðkþ 1Þðkþ 2ÞF½kþ 2� � 2N1G½k�

�N3ReXk

m¼0

ððmþ 1ÞG½mþ 1�F½k� m�Þ

þN3ReXk

m¼0

ððmþ 1ÞF½mþ 1�Gk� m�Þ ¼ 0; ð6:39Þ

G½0� ¼ b0;G½1� ¼ b1; (6.40)


ðkþ 1Þðkþ 2Þw½kþ 2� þ PehXk

m¼0

ððmþ 1ÞF½mþ 1�w½k� m�Þ

�PehXk

m¼0

ððmþ 1Þw½mþ 1�F½k� m�Þ ¼ 0;

(6.41)

w½0� ¼ c0; w½1� ¼ c1; (6.42)

ðkþ 1Þðkþ 2Þ4½kþ 2� þ PemXk

m¼0

ððmþ 1ÞF½mþ 1�4½k� m�Þ

�PemXk

m¼0

ððmþ 1Þ4½mþ 1�F½k� m�Þ ¼ 0;

(6.43)

4½0� ¼ d0;4½1� ¼ d1; (6.44)

where FðkÞ;GðkÞ;wðkÞ, and 4ðkÞ are the differential transforms off ðhÞ; gðhÞ; qðhÞ, and fðhÞ, and a0, a1, a2, a3, b0, b1, c0, c1, d0, d1 are constantswhich can be obtained through boundary condition. Using Eqs. (6.29) and(6.32) this problem can be solved as followed:

F½0� ¼ a0; F½1� ¼ a1; F½2� ¼ a2; F½3� ¼ a3;

F½4� ¼ 112ð1þN1Þ

�N1

12N2

ð�N3Reb0a1 � 2N1a2 þ 2N1b0

þ N3Rea0 b1Þ þ 3ReN1N4 � ReN2N3

�;

(6.45)

.

G½0� ¼ b0;G½1� ¼ b1;

G½2� ¼ 12N2

ð�N3Reb0a1 � 2N1a2 þ 2N1b0 þN3Rea0b1Þ;

G½3� ¼ 16b2

ð� 2N3ReN2b0a2 � 6N1a3N2 þ 2N1N2b1 �N23Re

2a0a1b0

�2N3ReN1a0a2 þ 2N3N1Rea0b0 þN23Re

2a20b1Þ; ð6:46Þ.

w½0� ¼ c0;w½1� ¼ c1;

w½2� ¼ � 12Peha1c0 þ 1

2Pehc1a0;

w½3� ¼ � 13Peha2c0 � 1

6a0Pe

2ha1c0 þ

16Pe2ha

20c1;

(6.47)


.

4½0� ¼ d0;4½1� ¼ d1;

4½2� ¼ � 12Pema1d0 þ 1

2Pemd1a0;

4½3� ¼ � 13Pema2d0 � 1

6a0Pe

2ma1d0 þ

16Pe2ma

20d1.

(6.48)

.The above process is continuous. By substituting Eqs. (6.45) to (6.48)

into the main equation based on DTM, it can be obtained that the closedform of the solutions is:

FðhÞ ¼ a0 þ a1hþ a3h2 þ

�1

12ð1þN1Þ�N1

12N2

��N3Reb0a1 � 2N1a2

�þ 2N1b0 þN3Rea0 b1

�þ 3ReN1N4 � ReN2N3

�h3 þ/;

(6.49)

GðhÞ ¼ b0 þ b1hþ 12N2

ð�N3Reb0a1 � 2N1a2 þ 2N1b0 þN3Rea0b1Þh2

þ�

16b2

�� 2N3ReN2b0a2 � 6N1a3N2 þ 2N1N2b1 �N2

3Re2a0a1b0

� 2N3ReN1a0a2 þ 2N3N1Rea0b0 þN23Re

2a20b1�h3/;

(6.50)

wðhÞ ¼ c0 þ c1hþ�� 12Peha1c0 þ 1

2Pehc1a0

�h2

þ�� 13Peha2c0 � 1

6a0Pe2ha1c0 þ

16Pe2ha

20c1

�h3/;

(6.51)

4ðhÞ ¼ d0 þ d1hþ�� 12Pema1d0 þ 1

2Pemd1a0

�h2

�� 13Pema2d0 � 1

6a0Pe

2ma1d0 þ

16Pe2ma

20d1

�h3/;

(6.52)

By substituting the boundary conditions into Eqs. (6.49) to (6.52) inpoint h ¼ �1 and h ¼ 1 the values of a0, a1, a2, a3, b0, b1, c0, c1, d0, d1 can beobtained.


Fð�1Þ ¼ F 0ð�1Þ ¼ Gð�1Þ ¼ 0; wð�1Þ ¼ fð�1Þ ¼ 1;

Fð1Þ ¼ wð1Þ ¼ fð1Þ ¼ 0; F 0ð�1Þ ¼ �1;Gð1Þ ¼ 1:(6.53)

By solving Eq. (6.53) the values of a0, a1, a2, a3, b0, b1, c0, c1, d0, d1 willbe given. By substituting obtained a0, a1, a2, a3, b0, b1, c0, c1, d0, d1 into Eqs.(6.50)e(6.53), the expression of FðhÞ;GðhÞ;wðhÞ, and 4ðhÞ can be ob-tained. For DTM solving, constant values with different nondimensionalparameters are shown in Table 6.1. q(h) values in different steps of DTM atN1 ¼ N2 ¼ N3 ¼ Re ¼ 1 and Peh ¼ Pem ¼ 1 are shown in Table 6.2. Thistable shows that this method is converged in step 7 and error has beenminimized. In this table, %error is introduced as followed:

%Error ¼��f ðhÞNM � f ðhÞDTM

f ðhÞNM

�� (6.54)

Also, it can also be seen that the maximum error for the DTM occursnear the middle of the interval h ¼ 0. The results that obtained by DTMwere well matched with the results carried out by the numerical solutionobtained using four-order Rungeekutta method as shown in Fig. 6.9. Afterthis validity, the influence of significant parameters such as Reynoldsnumbers, microrotation/angular velocity, and Peclet number on the flowand heat transfer characteristics is discussed. Fig. 6.10 shows the effects ofmoderate Reynolds numbers on the f and g, when N1 ¼ N2 ¼N3 ¼ Peh ¼ Pem ¼ 1.

It is worth to mention that the Reynolds number indicates the relativesignificance of the inertia effect compared to the viscous effect. Thus, ve-locity boundary layer thickness decreases as Re increases. The Reynoldsnumber has little effect on the temperature and concentration fields. In ageneral manner, there is an increment in velocity profiles from suction toinjection. At higher Reynolds numbers the maximum velocity ( f ) pointshifts to the upper wall where shear stress becomes larger as the Reynoldsnumber grows. Also from this figure we observe that with an increase in thevalue of the Reynolds number the point at which minimum rotation ( g)occurs moves away from the origin of the channel to upper wall.

Fig. 6.11 depicts the effect of coupling parameter (N1) on the (a) f (b) g,when N2 ¼ N3 ¼ Re ¼ Peh ¼ Pem ¼ 1. The values of velocity profile ( f )increases with increase in N1, but the values of microrotation profile ( g)decreases with increasing N1. Fig. 6.12 displays the effects of spin-gradient viscosity parameter (N2) on the (a) f (b) g, whenN1 ¼ N3 ¼ Re ¼ Peh ¼ Pem ¼ 1. The values of velocity profile decreases


Table 6.1 Constant Values With Different Nondimensional ParametersN1 [ N2 [ N3 Re Peh [ Pem a0 a1 a2 a3 b0 b1 c0 c1 d0 d1

0.5 1 1 0.252 0.250 �0.254 �0.250 0.072 0.161 0.619 �0.540 0.619 �0.5401 1 1 0.262 0.250 �0.274 �0.250 0.076 0.173 0.623 �0.544 0.076 �0.5441 0.5 1 0.262 0.250 �0.274 �0.250 0.076 0.173 0.623 �0.544 0.623 �0.5441 1 0.5 0.259 0.250 �0.268 �0.250 0.067 0.160 0.557 �0.521 0.557 �0.522


ationMethod


Problems

with increase in N2. Similarly, with the range N2 > 1, the angular velocityincreases with N2. However, when N2 < 1 the behavior of the angularvelocity is oscillatory and irregular. The parameter N3 was found to have aneffect only on the angular velocity ( g) and as shown in Fig. 6.13, increasingN3 leads to decrease in the angular velocity.

The topographical effects of the Peclet number on the fluid temperatureand solute concentration are shown in Fig. 6.14. The fluid temperature andconcentration increases with increasing Peclet number. The maximumvalue of temperature profile occurred in the middle of the channel.However, the Peclet number was found to have no effect on the velocityand the microrotation vectors. Effects of Re and Peh ¼ Pem on Nusseltnumber and Sherwood number when N1 ¼ N2 ¼ N3 ¼ 1 are shown inFigs. 6.15 and 6.16, respectively. For both suction and injection it can befound that Reynolds number has direct relationship with Nusselt numberand Sherwood number, but Peclet number has reverse relationship withthem.

Table 6.2 %Error of q(h) Values in Different Steps of Differential TransformationMethod at: N1 ¼ N2 ¼ N3 ¼ Re ¼ Peh ¼ Pem ¼ 1h Step[ 4 Step [ 5 Step[ 6 Step [ 7

�1 0 0 0 0�0.9 0.084731 0.086267 0.003871 0.000468�0.8 0.168765 0.17176 0.005964 0.001576�0.7 0.251416 0.255721 0.006769 0.002984�0.6 0.332018 0.337421 0.006684 0.004466�0.5 0.409935 0.416176 0.006017 0.005876�0.4 0.484572 0.491359 0.005001 0.007128�0.3 0.555383 0.562408 0.003801 0.008171�0.2 0.621875 0.628838 0.002522 0.00898�0.1 0.683621 0.690244 0.001226 0.0095390 0.740259 0.746307 6.02E-05 0.0098380.1 0.791502 0.796792 0.00133 0.0098650.2 0.837137 0.841553 0.002582 0.0096040.3 0.87703 0.880522 0.003805 0.009040.4 0.911122 0.91371 0.004962 0.0081580.5 0.939433 0.9412 0.005968 0.006960.6 0.962059 0.963137 0.006676 0.0054760.7 0.979164 0.979721 0.006846 0.0037890.8 0.990982 0.991199 0.006135 0.0020720.9 0.997811 0.997854 0.004063 0.0006371 1 1 0 0


6.4 MAGNETOHYDRODYNAMIC VISCOUS FLOW BETWEENPOROUS SURFACES

Consider the steady flow of an electrically conducting fluid between twohorizontal parallel plates when the fluid and the plates rotate togetheraround the axis, which is normal to the plates with an angular velocity [4].A Cartesian coordinate system is considered as followed: the x-axis is alongthe plate, the y-axis is perpendicular to it, and the z-axis is normal to the xyplane (see Fig. 6.17). The origin is located on the lower plate, and the platesare located at y ¼ 0 and h. The lower plate is being stretched by two equaland opposite forces so that the position of the point (0,0,0) remains un-changed. A uniform magnetic flux with density B0 is acting along y-axisabout which the system is rotating. The upper plate is subjected to a

-1 -0.5 0 0.5 10

0.15

0.3

0.45

f

η-1 -0.5 0 0.5 10

0.2

0.4

0.6

0.8

1

g

η

-1 -0.5 0 0.5 10

0.4

0.8

1.2

N = N = N = 0.1 ,Re = 1,Pe = Pe = 2

N = N = N = 1 ,Re = 1,Pe = Pe = 1

N = N = N = 1 ,Re = -0.5,Pe = Pe = 1.5

N = N = N = 0.1 ,Re = 0.1,Pe = Pe = 0.1

θ

η-1 -0.5 0 0.5 10

0.4

0.8

1.2

φ

η

(a) (b)

(c) (d)N

Figure 6.9 Comparison between numerical and differential transformation methodsolution results.


constant wall injection with velocity v0. The governing equations of mo-tion in a rotating frame of reference are:

vuvx

þ vvvy

þ vwvz

¼ 0 (6.55)

uvuvx

þ vvuvy

þ 2Uw ¼ �1r

vp*

vxþ y

v2uvx2

þ v2uvy2

� sB2

0

ru; (6.56)

uvvvy

¼ �1r

vp*

vyþ y

v2vvx2

þ v2vvy2

; (6.57)

-1 -0.5 0 0.5 10

0.1

0.2

0.3

0.4Re = -5Re = 1Re = 5Re = 10

f

η

-1 -0.5 0 0.5 10

0.3

0.6

0.9Re = -5Re = 1Re = 5Re = 10

g

η

(a)

(b)

Figure 6.10 Effects of moderate Reynolds number on the (a) f (b) g, whenN1 ¼ N2 ¼ N3 ¼ 1.


uvwvx

þ vvwvy

� 2Uw ¼ y

v2wvx2

þ v2wvy2

� sB2

0

rw; (6.58)

where u, v, and w denote the fluid velocity components along the x, y, andz directions, y is the kinematic coefficient of viscosity, r is the fluid den-

sity, and p* is the modified fluid pressure. The absence ofvp*vz

in

Eq. (6.58) implies that there is a net cross-flow along the z-axis.The boundary conditions are:

u ¼ ax; v ¼ 0; w ¼ 0 at y ¼ 0 (6.59)

-1 -0.5 0 0.5 10

0.1

0.2

0.3

0.4

f

η

N = 0.51

N = 0.1

N = 5N = 1

1

11

-1 -0.5 0 0.5 1

-0.3

0

0.3

0.6

0.9

g

η

N = 0.51

N = 0.1

N = 5N = 1

1

11

(a)

(b)

Figure 6.11 Effects of N1 on the (a) f (b) g, when N2 ¼ N3 ¼ Re ¼ Peh ¼ Pem ¼ 1.


u ¼ 0; v ¼ �v0; w ¼ 0 at y ¼ þh

Nondimensional variables are introduced as followed:

h ¼ yh; u ¼ axf 0ðhÞ; v ¼ �ah f ðhÞ; w ¼ ax gðhÞ (6.60)

where a prime denotes differentiation with respect to h.Substituting Eq. (6.60) in Eqs. (6.55)e(6.58), we have:

� 1r

vp*

vy¼ a2x

f 0 � ff 00 � f

000

RþM

Rþ 2Kr

Rg

; (6.61)

� 1rh

vp*

vh¼ a2h

ff 0 þ 1

Rf 00; (6.62)

-1 -0.5 0 0.5 10

0.1

0.2

0.3

0.4

f

η

N = 0.52

N = 0.1

N = 5N = 1

2

2 2

-1 -0.5 0 0.5 1

-0.4

0

0.4

0.8

g

η

N = 0.52

N = 0.1

N = 5N = 1

2

2

2

(a)

(b)

Figure 6.12 Effects of N2 on the (a) f (b) g, when N1 ¼ N3 ¼ Re ¼ Peh ¼ Pem ¼ 1.


-1 -0.5 0 0.5 1

0

0.4

0.8

g

η

N = 0.53

N = 0.1

N = 5N = 1

3

33

Figure 6.13 Effects of N3 on the g, when N1 ¼ N2 ¼ Re ¼ Peh ¼ Pem ¼ 1.

-1 -0.5 0 0.5 10

0.5

1

1.5

2

2.5

θ

η

Pe =1h

Pe = 0.5Pe = 5Pe = 2.5h

h

h

-1 -0.5 0 0.5 10

4

8

12

φ

η

Pe = 5m

Pe = 2.5Pe = 7.5Pe = 7.5m

mm

(a)

(b)

Figure 6.14 Effects of (a) Peh on temperature profile at Pem ¼ 1 (b) Pem on concen-tration profile at Peh ¼ 1, when N1 ¼ N2 ¼ N3 ¼ Re ¼ 1.

g00 � Rðf 0g � fg0Þ þ 2Krf0 �Mg ¼ 0 (6.63)

and the nondimensional quantities are defined through in which R is theviscosity parameter, M is the magnetic parameter, and Kr is the rotationparameter.

R ¼ ah2

y;M ¼ sB2

0h2

ry;Kr ¼ Uh2

y(6.64)

Eq. (6.61) with the help of Eq. (6.62) can be written as,

-0.4 -0.2 0 0.2 0.4

0.39

0.405

0.42 Pe = Pe = 0.4Pe = Pe = 0.45Pe = Pe = 0.5

m h

m

Nu

Re

x

m

hh

Figure 6.15 Effects of Re and Peh ¼ Pem on Nusselt number, when N1 ¼ N2 ¼ N3 ¼ 1.

-0.4 -0.2 0 0.2 0.4

0.3

0.315

0.33Pe = Pe = 0.8Pe = Pe = 0.9Pe = Pe = 1

mh

m

Sh

Re

x

m

h h

Figure 6.16 Effects of Re and Peh ¼ Pem on Sherwood number, whenN1 ¼ N2 ¼ N3 ¼ 1.


f000 � R½f 02 � ff 00� � 2K2

r g �M 2f 0 ¼ A (6.65)

Differentiation of Eq. (6.65) with respect to h gives,

f iv � Rðf 0 f 00 � f f 00Þ � 2Krg0 �M f 00 ¼ 0 (6.66)

Therefore, the governing equations and boundary conditions for thiscase in nondimensional form are given by:

f iv � Rðf 0 f 00 � f f 00Þ � 2Krg0 �M f 00 ¼ 0 (6.67)

uvTvx

þ vvTvy

þ wvTvz

¼ kr cp

�v2Tvx2

þ v2Tvy2

þ v2Tvz2

�þ m4 (6.68)

f ¼ 2

"�vuvx

�2

þ�vvvy

�2

þ�vwvz

�2#þ�vvvx

þ vuvy

�2

þ�vwvy

þ vvvz

�2

þ�vwvx

þ vuvz

�2

� 23

�vuvx

þ vvvy

þ vwvz

�2

(6.69)

g00 � Rðf 0g � fg0Þ þ 2Krf0 �Mg ¼ 0 (6.70)

subject to the boundary conditions:

f ¼ 0; f 0 ¼ 1; g ¼ 0 at h ¼ 0

f ¼ l; f 0 ¼ 0; g ¼ 0 at h ¼ 1 l ¼ v0ah

(6.71)

Th

T0v0

uwuw

B

z

y

x

h

Ω

Figure 6.17 Geometry of the problem.


With replacing nondimensional variables and using similarity solutionmethod, by neglecting last term of viscous dissipation in the energyequation, we have the following energy equation:

q00 þ Pr½R f q0 þ Ecð4f 02 þ g2Þ þ Ecxðf 002 þ g02Þ� ¼ 0 (6.72)

subject to the boundary conditionsqð0Þ ¼ 1; qð1Þ ¼ 0 (6.73)

where Pr ¼ mCp

kis the Prandtl number, Ec ¼ a2h2

Cpðq0 � qhÞ is the Eckert

number, Ecx ¼ a2x2

Cpðq0 � qhÞ is the local Eckert number, and the nondimen-

sional temperature is defined as,

qðhÞ ¼ T � Th

T0 � Th(6.74)

where T0 and Th are temperatures at the lower and upper plates.Now we apply DTM into governing equations. Taking the differential

transform of these equations with respect to c and considering H ¼ 1 gives,

ðkþ 1Þðkþ 2Þðkþ 3Þðkþ 4ÞF½kþ 4�

�RXk

m¼0


þRXk

m¼0

ðF½k� m�ðmþ 1Þðmþ 2Þðmþ 3ÞF½mþ 3�Þ � 2Krðkþ 1ÞF½kþ 2�

�Mðkþ 1Þðkþ 2ÞF½kþ 2� ¼ 0 ð6:75ÞF½0� ¼ 0;F½1� ¼ 1;F½2� ¼ a;F½3� ¼ b (6.76)

ðkþ 1Þðkþ 2Þw½kþ 2� þ PrRXk

m¼0

ðF½k� m�ðmþ 1Þw½mþ 1�Þ

þ 4PrEcXk

m¼0

ððk� mþ 1ÞF½k� mþ 1�ðmþ 1ÞF½mþ 1�Þ

þPrEcXk

m¼0

ðG½k� m�G½m�Þ

þPrEcxXk

m¼0


þPrEcxXk

m¼0

ððk� mþ 1ÞG½k� mþ 1�ðmþ 1ÞG½mþ 1�Þ ¼ 0 ð6:77Þ

G½0� ¼ 0;G½1� ¼ g (6.78)


ðkþ 1Þðkþ 2Þw½kþ 2� þ PrRXk

m¼0

ðF½k� m�ðmþ 1Þw½mþ 1�Þ

þ 4PrEcXk

m¼0

ððk� mþ 1ÞF½k� mþ 1�ðmþ 1ÞF½mþ 1�Þ

þPrEcXk

m¼0

ðG½k� m�G½m�Þ

þPrEcxXk

m¼0


þPrEcxXk

m¼0

ððk� mþ 1ÞG½k� mþ 1�ðmþ 1ÞG½mþ 1�Þ ¼ 0 ð6:79Þ

w½0� ¼ 1;w½1� ¼ x (6.80)

whereFðkÞ;GðkÞ and wðkÞ are the differential transforms of f(h), g(h), and q(h)and a; b;g and x are constants, which can be obtained through boundaryconditions.

This problem can be solved as followed:

F½0� ¼ 0; F½1� ¼ 1

F½2� ¼ a; F½3� ¼ b

F½4� ¼ 112

Raþ 112Kr þ 1

12Ma

F½5� ¼ 120

Mbþ 130

Ra2 þ 130Kra

F½6� ¼ 1360

MKr þ 1360

M 2aþ 130

Rba� 1360

R2a� 1300

Raþ 160

ab

(6.81)G½0� ¼ 0

G½1� ¼ g

G½2� ¼ �Kr

G½3� ¼ � 23Kraþ 1

6Mg

G½4� ¼ � 112

MKr þ 112

RKr þ 112

Rag� 12bKr

G½5� ¼ � 115

MaKr þ 1120

M 2gþ 130

RaKr � 160

RMgþ 110

RbKr � 130K2

r

G½6� ¼ � 1360

M 2Kr þ 190

MaKr þ 1180

MRag� 130

MKrb

� 1120

R2Kr þ 160

RKrbþ 190

RKra2 þ 1

120RKra� 1

90K2

r a

(6.82)

w½0� ¼ 1

w½1� ¼ x

w½2� ¼ �2PrEc � 2PrEcxa2 � 12PrEcxg2

w½3� ¼ 23PrEcxgKr � 1

6PrRx� 8

3aPrEc � 4PrEcxba

w½4� ¼ � 13PrEcxRa2 � 1

3PrEcxKra� 1

3PrEcxMa2 � 3PrEcxb

2 � 112

PrRax

þ 13Pr2REc þ 1

3Pr2Ecxa2 þ 1

12PrEcxMg2 � 4

3PrEca2 � 2PrEcb

� 112

PrEcg2 þ 13PrEcxgaKr � 1

12PrEcxMg2 � 1

3PrEcxK2

r

w½5� ¼ � 45PrEcxaMb� 2

15PrEcxRa3 � 2

15PrEcxKra

2 � 35PrEcxRba

� 35PrEcxKrb� 1

20PrRbxþ 3

5Pr2REaþ 1

5Pr2Ra3Ecx þ 1

20Pr2g2RaEcx

� 110

pr2REcxKrgþ 140

Pr2R3xþ 35Pr2REcxba

� 215

ðPrEcRaþ PrEcKr þ PrEcMaÞ � 125PrEcbaþ 1

10PrEKrg

þ 215

PrEcxMag� 130

ðPrEcxRagþ PrEcxMag2Þ

þ 15PrEcxbag� 2

5PrEcxK2

r a

(6.83)

The above process is continuous. Substituting Eqs. (6.80)e(6.82) intothe main equation based on DTM. It can be obtained that the closed formof the solutions is,

FðhÞ ¼ 1þ ah2 þ bh3 þ�112

Raþ 112Kr þ 1

12Ma

�h4

þ�120

Mbþ 130

Ra2 þ 130Kra

�h5

þ�

1360

MKr þ 1360

M 2aþ 130

Rba� 1360

R2a� 1300

Raþ 160

ab

�h6 þ :::

(6.84)


GðhÞ ¼ gh� Krh2 þ

�� 23Kraþ 1

6Mg

�h3

þ�� 112

MKr þ 112

RKr þ 112

Rag� 12bKr

�h4

þ�� 115

MaKr þ 1120

M 2gþ 130

RaKr � 160

RMgþ 110

RbKr � 130K2

r

�h5

þ�� 1360

M 2Kr þ 190

MaKr þ 1180

MRag� 130

MKrb� 1120

R2Kr

þ 160

RKrbþ 190

RKra2 þ 1

120RKra� 1

90K2

r a

�h6 þ :::

(6.85)

wðhÞ ¼ 1þ xhþ�� 2PrEc � 2PrEcxa

2 � 12PrEcxg

2

�h2

þ�23PrEcxgKr � 1

6PrRx� 8

3aPrEc� 4PrEcxbaÞh3

þ�� 13PrEcxRa

2 � 13PrEcxKra� 1


2

� 112

PrRaxþ 13Pr2REc þ 1

3Pr2Ecxa

2 þ 112

PrEcxMg2 � 43PrEca2

� 2PrEcb� 112


12PrEcxMg2 � 1

3PrEcxK2

r

�h4 þ :::

(6.86)

To obtain the values of a;b;g and x we substitute the boundaryconditions into Eqs. (6.83)e(6.85) in point h ¼ 1. So we have,

Fð1Þ ¼ 1þ aþ bþ�112

Raþ 112Kr þ 1

12Ma

�þ�120

Mbþ 130

Ra2 þ 130Kra

�þ�

1360

MKr þ 1360

M 2aþ 130

Rba� 1360

R2a� 1300

Raþ 160

ab

�þ. ¼ l

(6.87)


F 0ð1Þ ¼ 1þ 2aþ 3bþ 4

�112

Raþ 112Kr þ 1

12Ma

�þ 5

�120

Mbþ 130

Ra2 þ 130Kra

�þ 6

�1360

MKr þ 1360

M 2aþ 130

Rba� 1360

R2a� 1300

Raþ 160

ab

�þ. ¼ 0

(6.88)

Gð1Þ ¼ g� Kr þ�� 23Kraþ 1

6Mg

�þ�� 112

MKr þ 112

RKr þ 112

Rag� 12bKr

�þ�� 115

MaKr þ 1120

M 2gþ 130

RaKr � 160

RMgþ 110

RbKr � 130K2

r

�þ�� 1360

M 2Kr þ 190

MaKr þ 1180

MRag� 130

MKrb� 1120

R2Kr

þ 160

RKrbþ 190

RKra2 þ 1

120RKra� 1

90K2

r a

�þ. ¼ 0

(6.89)

wð1Þ ¼ 1þ xþ�� 2PrEc � 2PrEcxa

2 � 12PrEcxg

2

�þ�23PrEcxgKr � 1

6PrRx� 8

3aPrEc � 4PrEcxba

�þ�� 13PrEcxRa

2 � 13PrEcxKra� 1


2

� 112

PrRaxþ 13Pr2REc þ 1

3Pr2Ecxa

2 þ 112

PrEcxMg2 � 43PrEca2

� 2PrEcb� 112


12PrEcxMg2 � 1

3PrEcxK

2r

�þ. ¼ 0

(6.90)


Solving Eqs. (6.86)e(6.89) gives the values of a; b;g and x. Bysubstituting obtained a; b;g and x into Eqs. (6.83e6.85), we can find theexpressions of FðhÞ;GðhÞ;wðhÞ.

The effects of acting parameters on velocity profile and temperaturedistribution are discussed. Constant values with different nondimensionalparameters are shown in Table 6.3. q(h) values in different steps of DTM atR ¼ 2, Kr ¼ 0.5, M ¼ 1, Pr ¼ 1, l ¼ 0.5, Ec ¼ Ecx ¼ 0.5 are shown inTable 6.4. Also it shows that this method is converged in step 16 and errorhas been minimized. There is an acceptable agreement between the nu-merical solution obtained by four-order Rungeekutta method and DTMas shown in Table 6.5. In these tables, error is introduced as followed:

Error ¼ ��f ðhÞNM � f ðhÞDTM

�� (6.91)

Figs. 6.18 and 6.19 show the magnetic field effect on nondimensionalvelocity component ðf and f 0Þ. The decrease of f curve is observed byapplying higher magnetic field intensity. Also f 0 values increase nearstretching sheet and decrease under porous sheet while at the middle pointthese values are constant.

Blowing velocity parameter (l) has noticeable effect on nondimensionalvelocity component, which by increasing l profile of f and f 0 becomenonlinear and the maximum amount of f and f 0 increase. Also velocitycomponent in x-direction increase severely as shown in Figs. 6.20 and 6.21.

Fig. 6.22 shows by increasing rotating parameter (Kr), values of trans-verse velocity component ( g) between two sheets increase and the locationof maximum amount of g approaching stretching sheet. Lorentz force hasan inverse effect on g in comparison with Coriolis force which means thatwith increasing magnetic field (M), transverse velocity component betweentwo plates decrease as shown in Fig. 6.23. The viscosity parameter (R)effects on g profile similar to magnetic field, however, with less intensitychanges. Also with increasing R the location of maximum amount of gapproaching stretching sheet indicates decreasing boundary layer thicknessnear stretching plate as shown in Fig. 6.24. Increasing the blowing velocityparameter leads to g increase, which shows blowing velocity parameter andmagnetic field effects on g are in opposite as shown in Fig. 6.25.

Fig. 6.26 shows that increasing viscosity parameter leads to increasingthe curve of temperature profile (q) and the decreasing of q values. Withincreasing rotating parameter near stretching plate a small amount of qincreasing has been observed. This effect decreases while we are


Table 6.3 Constant Values With Different Nondimensional ParametersR K M Pr l Ec Ecx a [ f 00 (0) b [ f 0 00 (0) g [ g 0 (0) x [ q 0 (0)

2 0.5 0.5 1 0.5 0.5 0.5 �1.137160 0.938104 0.297786 �0.5100600.5 2 0.5 1 0.5 0.5 0.5 �1.028720 �0.296040 1.260540 �0.2148700.5 0.5 1 1 0.5 0.5 0.5 �1.104230 0.637858 0.305824 �0.3130010.5 0.5 0.5 7 0.5 0.5 0.5 �1.063670 0.383391 0.315570 3.7556080.5 0.5 0.5 1 1 0.5 0.5 2.095631 �6.765270 0.536101 0.8453580.5 0.5 0.5 1 0.5 2 0.5 �1.063670 0.383391 0.315570 1.1756120.5 0.5 0.5 1 0.5 0.5 2 �1.063670 0.383391 0.315570 0.460925

DTM

forMagnetohydrodynam

ic(M

HD)and

PorousMedium

Flows

271

Table 6.4 q(h) Values in Different Steps of Differential Transformation Method at: R ¼ 2, Kr ¼ 0.5, M ¼ 1, Pr ¼ 1, l ¼ 0.5, Ec ¼ Ecx ¼ 0.5h NM n[ 4 Error n[ 8 Error n[ 12 Error N[ 16 Error

0 1 1 0 1 0 1 0 1 00.1 0.936372 0.927427 0.008944 0.937194 0.000822 0.936681 0.000311 0.936681 0.0003090.2 0.852373 0.834493 0.017880 0.854101 0.001728 0.853094 0.000721 0.853093 0.0007200.3 0.754207 0.727977 0.026229 0.756699 0.002493 0.755213 0.001006 0.755211 0.0010050.4 0.647177 0.613845 0.033332 0.650215 0.003038 0.648265 0.001089 0.648263 0.0010860.5 0.535656 0.497241 0.038415 0.539025 0.003370 0.536636 0.000980 0.536633 0.0009770.6 0.423083 0.382494 0.040590 0.426604 0.003521 0.423818 0.000735 0.423816 0.0007320.7 0.311987 0.273113 0.038875 0.315487 0.003499 0.312415 0.000428 0.312412 0.0004250.8 0.204023 0.171790 0.032232 0.207229 0.003207 0.204161 0.000139 0.204159 0.0001360.9 0.100014 0.080401 0.019613 0.102321 0.002307 0.099973 0.000040 0.099972 0.0000411 0 0 0 0 0 0 0 0 0


ationMethod


Problems

Table 6.5 Comparison Between Numerical Results and Differential Transformation Method Solution at: R ¼ 2, Kr ¼ 0.5, M ¼ 1, Pr ¼ 1,l ¼ 0.5, Ec ¼ Ecx ¼ 0.5

h

f g q

NM DTM Error NM DTM Error NM DTM Error

0 0 0 0 0 0 0 1 1 00.1 0.094302 0.094486 0.000184 0.024198 0.024248 0.000050 0.936372 0.936681 0.0003090.2 0.177851 0.178417 0.000566 0.039813 0.039882 0.000069 0.852373 0.853093 0.0007200.3 0.251398 0.252343 0.000945 0.048179 0.048233 0.000054 0.754207 0.755211 0.0010050.4 0.315423 0.316618 0.001195 0.05055 0.050562 0.000011 0.647177 0.648263 0.0010860.5 0.370172 0.371425 0.001254 0.048099 0.048057 0.000041 0.535656 0.536633 0.0009770.6 0.415676 0.416791 0.001115 0.041924 0.041837 0.000087 0.423083 0.423816 0.0007320.7 0.451786 0.452607 0.000821 0.033062 0.032952 0.000109 0.311987 0.312412 0.0004250.8 0.478190 0.478647 0.000458 0.022494 0.022394 0.000099 0.204023 0.204159 0.0001360.9 0.494445 0.494583 0.000139 0.011166 0.011107 0.000059 0.100014 0.099972 0.0000411 0.5 0.5 0 0 0 0 0 0 0

DTM

forMagnetohydrodynam

ic(M

HD)and

PorousMedium

Flows

273

approaching to porous plate as shown in Fig. 6.27 .The magnetic field doesnot have a noticeable effect on (q) as shown in Fig. 6.28.

Different effects of prandtl number on temperature profile in thepresence and absence of viscous dissipation are discussed in Figs. 6.29 and6.30 as follows. Increasing prandtl, in the presence of viscous dissipation,leads to increasing temperature between two plates, while in absence of

0 0.2 0.4 0.6 0.8 10

0.2

0.4

NMDTMNMDTMNMDTMNMDTM

f

η

M= 0

M= 50

M= 10

M= 5

Figure 6.18 Velocity component profile (f) for variable M at R ¼ 2, Kr ¼ 0.5, Pr ¼ 1,l ¼ 0.5, Ec ¼ Ecx ¼ 0.5.

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1NMDTMNMDTMNMDTMNMDTM

f

η

M= 0

M= 50

M= 10

M= 5

'

Figure 6.19 Velocity component profile ðf 0Þ for variable M at R ¼ 2, Kr ¼ 0.5, Pr ¼ 1,l ¼ 0.5, Ec ¼ Ecx ¼ 0.5.


viscous dissipation the changes are inverse. Increasing temperature betweentwo plates observed, which caused by increasing this effect, is more sensiblenear stretching plate. (It can be seen in Fig. 6.31). The effects of viscousdissipation for which the Eckert number (Ec) and the local Eckert number(Ecx) are responsible as shown in Figs. 6.32 and 6.33. It is obvious from thegraphs that by increasing Ec and Ecx the temperature increase near the

0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

2

f

η

λ= 0.5

λ= 2λ= 1.5λ= 1

Figure 6.20 Velocity component profile (f) for variable l at R ¼ 2, Kr ¼ 0.5, M ¼ 1,Pr ¼ 1, Ec ¼ Ecx ¼ 0.5.

0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

2

2.5

3

f

η

λ= 0.5

λ= 2λ= 1.5λ=1

'

Figure 6.21 Velocity component profile ðf 0Þ for variable l at R ¼ 2, Kr ¼ 0.5, M ¼ 1,Pr ¼ 1, Ec ¼ Ecx ¼ 0.5.


stretching wall. This is due to the fact that heat energy is stored in the fluiddue to the frictional heating. This phenomenon is more sensible on Ec thanEcx. In Figs. 6.18, 6.19, 6.22, and 6.26 these results are compared withnumerical method. It can be seen that there is good agreement betweennumerical solution and DTM.

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7NMDTMNMDTMNMDTMNMDTM

g

η

K

r

= 2

K

K

K r

r

r

= 0.5

= 6

= 4

Figure 6.22 Velocity component profile (g) for variable Kr at R ¼ 2, M ¼ 1, Pr ¼ 1,l ¼ 0.5, Ec ¼ Ecx ¼ 0.5.

0 0.2 0.4 0.6 0.8 10

0.02

0.04

0.06

g

η

M= 0

M= 50M= 10M= 5

Figure 6.23 Velocity component profile (g) for variable M at R ¼ 2, Kr ¼ 0.5, Pr ¼ 1,l¼0.5, Ec ¼ Ecx ¼ 0.5.


0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

g

η

R= 20R= 15R= 10R= 5

Figure 6.24 Velocity component profile (g) for variable R at Kr ¼ 4, M ¼ 1, Pr ¼ 1,l ¼ 0.5, Ec ¼ Ecx ¼ 0.5.

0 0.2 0.4 0.6 0.8 10

0.05

0.1

0.15

DTM λ= 0.50.2 DTM λ= 1

DTM λ= 1.5DTM λ= 2

g

η

Figure 6.25 Velocity component profile (g) for variable l at R ¼ 2, Kr ¼ 0.5, M ¼ 1,Pr ¼ 1, Ec ¼ Ecx ¼ 0.5.


0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1NMDTMNMDTMNMDTMNMDTM

θ

η

R= 5

R= 20

R= 15

R= 10

Figure 6.26 Temperature profile (q) for variable R at Kr ¼ 0.5, M ¼ 1, Pr ¼ 1, l ¼ 0.5,Ec ¼ Ecx ¼ 0.5.

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

θ

η

Kr= 0.5

Kr= 6Kr= 4Kr = 2

Figure 6.27 Temperature profile (q) for variable Kr at R ¼ 2, M ¼ 1, Pr ¼ 1, l ¼ 0.5,Ec ¼ Ecx ¼ 0.5.


0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

θ

η

M= 0

M= 50M= 10M= 5

Figure 6.28 Temperature profile (q) for variable M at R ¼ 2, Kr ¼ 0.5, Pr ¼ 1, l ¼ 0.5,Ec ¼ Ecx ¼ 0.5.

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

θ

η

Pr = 0.7

Pr = 40Pr = 10Pr = 7

Figure 6.29 Temperature profile (q) for variable Pr at R ¼ 2, Kr ¼ 0.5, M ¼ 1, l ¼ 0.5,Ec ¼ Ecx ¼ 0.


0 0.2 0.4 0.6 0.8 10

0.3

0.6

0.9

1.2

1.5

θ

η

Pr = 0.7

Pr = 40Pr = 10Pr = 7

Figure 6.30 Temperature profile (q) for variable Pr at R ¼ 2, Kr ¼ 0.5, M ¼ 1, l ¼ 0.5,Ec ¼ Ecx ¼ 0.5.

0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

2

2.5

3

θ

η

λ= 0.5

λ= 2λ= 1.5λ= 1

Figure 6.31 Temperature profile (q) for variable l at R ¼ 2, Kr ¼ 0.5, M ¼ 1, Pr ¼ 1,Ec ¼ Ecx ¼ 0.5.


REFERENCES[1] Ellahi R. The effects of MHD and temperature dependent viscosity on the flow of non-

Newtonian nanofluid in a pipe: analytical solutions. Applied Mathematical Modelling2013;37(3):1451e67.

[2] Mosayebidorcheh S, Sheikholeslami M, Hatami M, Ganji DD. Analysis of turbulentMHD Couette nanofluid flow and heat transfer using hybrid DTMeFDM. Particuology2016;26:95e101.

0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

θ

η

Ec = 0.5

Ec = 6Ec = 4Ec = 2

Figure 6.32 Temperature profile (q) for variable Ec at R ¼ 2, Kr ¼ 0.5, M ¼ 1, Pr ¼ 1,l ¼ 0.5, Ecx ¼ 0.5.

0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

θ

η

Ec = 0.5

Ec = 6Ec = 4Ec = 2

x

x

xx

Figure 6.33 Temperature profile (q) for variable Ecx at R ¼ 2, Kr ¼ 0.5, M ¼ 1, Pr ¼ 1,l ¼ 0.5, Ec ¼ 0.5.


[3] Sheikholeslami M, Ashorynejad HR, Ganji DD, Rashidi MM. Heat and mass transfer ofa micropolar fluid in a porous channel. Communications in Numerical Analysis2014;2014.

[4] Sheikholeslami M, Ashorynejad HR, Ganji DD, Kolahdooz A. Investigation of rotatingMHD viscous flow and heat transfer between stretching and porous surfaces usinganalytical method. Mathematical Problems in Engineering 2011;2011.


CHAPTER 7

DTM for Particles Motion,Sedimentation, and Combustion

7.1 INTRODUCTION

Many phenomena are existed in the environment in which particle’smotion can be observed on them, such as centrifugation, centrifugal filters,industrial hopper, etc. Surface of the motion has different shapes especiallyfor rotating application; it can be a circular, parabolic, or conical surface. It’snecessary for scientists to analyze the motion of the particles on thesesurfaces, so an analytical solution is usually the more preferred andconvenient method in engineering area because of less computational workas well as high accuracy which is widely used for predicting the motion ofparticles. Some of them are introduced in these sections:7.1 Introduction7.2 Motion of a Spherical Particle on a Rotating Parabola7.3 Motion of a Spherical Particle in Plane Couette Fluid Flow7.4 Nonspherical Particle Sedimentation7.5 Motion of a Spherical Particle in a Fluid Forced Vortex7.6 Combustion of Microparticles7.7 Unsteady Sedimentation of Spherical Particles7.8 Transient Vertically Motion of a Soluble Particle

7.2 MOTION OF A SPHERICAL PARTICLE ON A ROTATINGPARABOLA

Consider a particle slides along a surface that has the shape of a parabolaz ¼ cr2 (see Fig. 7.1). Following assumptions are considered for particlesmotion modeling [1]:• Particle is at equilibrium.• The particle rotates in a circle of radius R.• The surface is rotating about its vertical symmetry axis with angular ve-

locity u.



http://dx.doi.org/10.1016/B978-0-12-805190-0.00007-3

By choosing the cylindrical coordinates r, q, and z as generalized co-ordinates, the kinetic and potential energies can be given as,

T ¼ 12m�_r2 þ r2 _q

2 þ _z2�

U ¼ mgz(7.1)

We have in this case some equations of constraints that we must takeinto account, namely

z ¼ cr2

_z ¼ 2c _rr(7.2)

and

q ¼ ut_q ¼ u

(7.3)

Inserting Eqs. (7.3) and (7.2) in Eq. (7.1), we can calculate theLagrangian for the problem

z

ω

RParticle

r

θ

Par

abol

ic s

urfa

ce

Figure 7.1 Schematic view of a spherical particle on a rotating parabolic surface.

L ¼ T � U ¼ 12mð _r2 þ 4c2r2 _r2 þ r2u2Þ � mgcr2 (7.4)

It is important to note that the inclusion of the equations of constraintsin the Lagrangian has reduced the number of degrees of freedom to only


one, i.e., _r. We now calculate the equation of motion using Lagrange’sequation

vLvr

¼ mð4c2r _r2 þ ru2 � 2gcrÞddt

vLvr

¼ mð€r þ 4c2r2€r þ 8c2r _r2Þ(7.5)

and

€rð1þ 4c2r2Þ þ _r2ð4c2rÞ þ rð2gc � u2Þ ¼ 0 (7.6)

By considering 2gc � u2 ¼ ε2;

€r þ 4c2€rr2 þ 4c2r _r2 þ ε2r ¼ 0 (7.7)

It’s considered that initial particle position is in radius A, and its initialvelocity is zero. So, its initial conditions are

rð0Þ ¼ A; _rð0Þ ¼ 0 (7.8)

For solving the particle motion on a rotating parabola by an efficient,fast, and high accurate method, Eq. (7.7) is solved by multi-step differentialtransformation method (Ms-DTM),

ðkþ 2Þðkþ 1ÞRjðkþ 2Þ

�4c2Xkk1

Xk1l¼0

RjðlÞRjðk1� lÞðk� k1þ 1Þðk� k1þ 2ÞRjðk� k1þ 2Þ þ ε2RjðkÞ

þ4c2Xkk1¼0

Xk1l¼0

RjðlÞðk1� l þ 1ÞRjðk1� l þ 1ÞRjðk� k1þ 1Þðk� k1þ 1Þ ¼ 0

(7.9)


R0ð0Þ ¼ A; R0ð1Þ ¼ 0

Rið0Þ ¼ ri�1ðtiÞ; Rið1Þ ¼ r 0i�1ðtiÞ; i ¼ 1; 2;.;K � M(7.10)

Since the procedure of solving Eq. (7.9) is autonomous of constants A,ε, and c, for generalization and simplification of problem for future caseswith different physical conditions the constants that represent physicalproperties are assumed to be:

A ¼ c ¼ ε ¼ 1; (7.11)

After solving Eq. (7.9) and using initial condition Eqs. (7.10) and (7.11),position of the particle, r(t), will be appeared as following equation for each0.25-s time step,

DTM for Particles Motion, Sedimentation, and Combustion 285

2 4 6 8 100

0.25

1 11 553 59363 10256209( ) 1 [0,0.25]10 3000 2250000 3150000000 7087500000000

( ) 1.006293- 0.050230616

( )

r t t t t t t t

r t

r t

= − − − − − ∈

=

=

2 3 4 5

6 7 8 9 10

- 0.1013895( - 0.25) - 0.003744513( - 0.25) - 0.00390231( - 0.25) - 0.38551 -3( - 0.25)

- 0.279966 -3( - 0.25) - 0.40481 - 4( - 0.25) - 0.23109 - 4( - 0.25) - 0.396 -5( - 0.25) - 0.18719 -5( - 0.25)

t t t t e te t e t e t e t e t

2 3 4 50.50

6 7 8

[0.25,0.5]( ) 1.025707 - 0.10188 - 0.105738( - 0.5) - 0.0079821( - 0.5) - 0.0046757( - 0.5) - 0.88127 -3( - 0.5)

- 0.398166 -3( - 0.5) - 0.99646 - 4( - 0.5) - 0.383726 - 4( - 0.5) - 0.10192 - 4( - 0.5

tr t t t t t e t

e t e t e t e t

∈

=9 10

2 3 4 50.75

6 7

) - 0.32931 -5( - 0.5) [0.5,0.75]( ) 1.05996 - 0.1565587 - 0.113642( - 0.75) - 0.013349( - 0.75) - 0.00621702( - 0.75) - 0.0016488( - 0.75)

- 0.65612 -3( - 0.75) - 0.206483 -3( - 0.75)

e t tr t t t t t t

e t e t

∈

=8 9 10

2 3 4 59.0

6

- 0.7254 - 4( - 0.75) - 0.2103 - 4( - 0.75) - 0.5254 - 5( - 0.75) [0.75,1.0]...

( ) -2.63269 0.374569 - 0.1812( -9) 0.053389( -9) - 0.024782( -9) 0.0101267( -9)

- 0.00325035( -9) - 0.15

e t e t e t t

r t t t t t tt

∈

= + + +7 8 9 10

2 3 4 59.25

27698 -3( -9) 0.001640656( -9) - 0.0020994( -9) 0.0020165285( -9) [9.0,9.25]( ) -1.88514 .29261 - 0.1490558( -9.25) 0.033957( -9.25) - 0.014987( -9.25) 0.00580704( -9.25)

e t t t t tr t t t t t t

+ + ∈

= + + +6 7 8 9 10

2 39.5

- 0.00229197( -9.25) 0.72248 -3( -9.25) - 0.8865 - 4( -9.25) - 0.14027 -3( -9.25) 0.19074 -3( -9.25) [9.25,9.5]( ) -1.2385 0.223613 - 0.1284215( -9.5) 0.0219744( -9.5) - 0.009517(

t e t e t e t e t tr t t t t t

+ + ∈

= + + 4 5

6 7 8 9 10

9.75

-9.5) 0.0032032( -9.5)

- 0.001265( -9.5) 0.43899 -3( -9.5) - 0.1384 -3( -9.5) 0.2617 - 4( -9.5) 0.85178 -5( -9.5) [9.5,9.75] ( ) -0.6551 0.162983 - 0.11507( -9

tt e t e t e t e t t

r t t t

+

+ + + ∈

= + 2 3 4 5

6 7 8 9 10

.75) 0.0141168( -9.75) - 0.0064938( -9.75) 0.001773667( -9.75)

- 0.70295 -3( -9.75) 0.22445 -3( -9.75) - 0.78165 - 4( -9.75) 0.22326 - 4( -9.75) - 0.5106 -5( -9.75) [9.75,10]

t t te t e t e t e t e t t

⎧⎪⎪⎪⎪⎪

⎨

+ +

+ + ∈

⎪⎪⎪⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(7.12)


ationMethod


Problems

Eq. (7.12) is depicted in Fig. 7.2 and is compared with obtained resultfrom differential transformation method (DTM). The values are presentedin Table 7.1. Fig. 7.2 shows the particle’s position for three different cconstants when A ¼ ε ¼ 1. Influence of constant parameter (c) on stabilityand phase plane is investigated through Fig. 7.3.

7.3 MOTION OF A SPHERICAL PARTICLE IN PLANECOUETTE FLUID FLOW

The Vander Werff model for particle motion in Couette flow is adoptedin this modeling while the positive direction rotation of particle is clock-wise, and combined effects of inertia, gravity, and buoyancy forces areassumed to be negligible [2]. So, the inertia force in left-hand side offorce balance equation is the product of spherical particle mass by its ac-celeration [3]

T ¼ 4pa3

3rs _V ¼ 4pa3

3rsð€x;€y; 0Þ (7.13)

where a, r, and V denote the radius, density, and velocity of spherical par-ticle, respectively. _V is the first derivative of particle’s velocity, and €x and €y

Figure 7.2 Multi-step differential transformation method (Ms-DTM) efficiency in par-ticle motion analysis, compared with DTM and numerical solution. DTM, differentialtransformation method.


Table 7.1 Multi-step Differential Transformation Method’s (Ms-DTM) Values forPosition of Particle, Compared With Numerical Results

t

r(t)

Num. Ms-DTM

0 1.0000000 1.00000000.5 0.974766908 0.97476691831.0 0.896067130 0.89606715971.5 0.753057776 0.75305785862.0 0.5189251820 0.5189252782.5 0.13321789149 0.1332184543.0 �0.33404132749 �0.3337829263.5 �0.64064202117 �0.6398159244.0 �0.82788893947 �0.82668330254.5 �0.9393086789 �0.93776772805.0 �0.9926709981 �0.99078762345.5 �0.9947172580 �0.99244885546.0 �0.94568648462 �0.94294589886.5 �0.83945793817 �0.8360780697.0 �0.65951111277 �0.6551442107.5 0.36531058744 0.3591902508.0 0.09376192157 0.1006018468.5 0.49425716809 0.4972761309.0 0.73793715747 0.7384362249.5 0.88704545678 0.88581378410 0.97044161417 0.9677503517

Figure 7.3 Phase plane for particle in different constant using multi-step differentialtransformation method (Ms-DTM) when A ¼ ε ¼ 1.


are the second derivatives of particle motion in horizontal and vertical di-rections respect to time. To calculate the drag force, the velocities of thespherical particle are considered small adequately so that the Stokes lawcan be governed:

TDx ¼ 6pmaVrx ¼ 6pmað _x� ayÞ (7.14a)

TDy ¼ 6pmaVry ¼ 6pma _y (7.14b)

where m signifies the viscosity of fluid. The rotation and shear portion ofthe lift force is obtained as:

TRx ¼ 12pa3ra _y (7.15a)

TRy ¼ 12pa3raðay� _xÞ (7.15b)

TSx ¼ 0 (7.16a)

TSy ¼ 6:46a2ra1=2n1=2ðay� _xÞ (7.16b)

where n is the dynamic viscosity, ðmrÞ, and a is defined as positive propor-

tionality constant.An illustration of the spherical particle in plane Couette fluid flow and

applied forces on particle are shown in Fig. 7.4. The mass of particle isassumed in the center of sphere, and the forces caused from the rotation andshear fields and their interactions on drag and lift forces of particle areillustrated in Fig. 7.4(a) and (b), respectively. By forming the force balanceequation of the inertia force to the drag and lift forces; the equations ofmotion for the particle are driven as:

T ¼ TR þ TS � TD (7.17)

Eventually, by substituting Eqs. (7.14) and (7.16) into Eq. (7.17) thesystem of equation of motion for spherical particle in plane Couette flowyields:8>>><>>>:

4pa3

3rs€x ¼ 1

2pa3ra _y� 6pmað _x� ayÞ

4pa3

3rs€y ¼

�12pa3raþ 6:46a2ra1=2n1=2

�ðay� _xÞ � 6pma _y

(7.18)


For simplicity the governing equations have been expressed as:�€x ¼ A _y� Bð _x� ayÞ€y ¼ B _y� ðAþ CÞð _x� ayÞ (7.19)

where the coefficients AeC are defined as follows:

A ¼�3a8

��r

rS

�

B ¼�9n2a2

��r

rS

�

C ¼ 4:845

� ffiffiffiffiffian

pa

��r

rS

�(7.20)

An appropriate initial condition is required to avoid trapping the pro-cedure in nontrivial solution:

xðt ¼ 0Þ ¼ 0; _xðt ¼ 0Þ ¼ u0 (7.21)

yðt ¼ 0Þ ¼ 0; _yðt ¼ 0Þ ¼ n0

For applying Ms-DTM to the present problem, in first step traditionalDTM using Table 7.1 should be applied to Eq. (7.19),

Rotaryeffect

Rotaryeffect

mx, TRx

my, TRy,TS

TDx

TDyV = 0 V = 0

V = υ υ0 V = 0y y

x

(a)

Drag interactions Lift interactions

(b)

x

Figure 7.4 Schematic view of exerted forces on a spherical particle in Couette fluidflow. (a) Drag interactions; (b) lift interactions.


ðkþ 2Þðkþ 1ÞXjðkþ 2Þ � Aðkþ 1ÞYjðkþ 1ÞþBððkþ 1ÞXjðkþ 1Þ � aYjðkÞÞ ¼ 0

ðkþ 2Þðkþ 1ÞYjðkþ 2Þ þ Bðkþ 1ÞYjðkþ 1ÞþðAþ CÞððkþ 1ÞXjðkþ 1Þ � aYjðkÞÞ ¼ 0

(7.22)


X0ð0Þ ¼ 0; X0ð1Þ ¼ 1

Xið0Þ ¼ xi�1ðtiÞ; Xið1Þ ¼ x0i�1ðtiÞ; i ¼ 1; 2;.;K � M(7.23)

Y0ð0Þ ¼ 0; Y0ð1Þ ¼ 1

Yið0Þ ¼ yi�1ðtiÞ; Yið1Þ ¼ y0i�1ðtiÞ; i ¼ 1; 2;.;K � M

For example, if we divide the domain t ∊ [0, 8] to eight subintervals(each 1 s) for the first subinterval t ∊ [0, 1], we should obtain the DTMsolution of Eq. (7.22) using initial condition in Eq. (7.21) and for nextsubinterval, t ∊ [1, 2], DTM solution of Eq. (7.22) should be obtained as thefirst subinterval but with a new initial condition. This new initial condition,as previously mentioned, can be calculated from the obtained equationfrom the first subinterval by substituting t ¼ 1 in it. And for the thirdsubinterval, initial condition can be determined by substituting t ¼ 2 in theDTM solution obtained in the second step. This process should berepeated, which generates a sequence of approximate solutions for xi(t) andyi(t), i ¼ 1, 2,.,8 for the solution x(t) and y(t) after derivative with respectto time, u(t) and v(t) will be obtained.

Since the procedure of solving Eq. (7.22) is autonomous of constants A,B, and C, for generalization and simplification of problem for future caseswith different physical conditions the constants that represent physicalproperties are assumed to be as following:

A ¼ B ¼ a ¼ u0 ¼ n0 ¼ 1; C ¼ 0 (7.24)

After solving Eq. (7.22) using initial condition Eqs. (7.23) and(7.21), position of the particle, x(t) and y(t), will be appeared, and afterderivative, velocity profiles will be calculated as following for each 1-s timestep,


⎪⎪

0

1

( )

( )

( )

u t

u t

u t

= 2 3 4 5 6 71 1 1 1 1 112 3 8 30 144 840t t t t t t− + − + − + t [0,1]∈

=

=

2 3 42

2 3 41.104167 - 0.368254 0.992063 - 4( -1) 0.0613095( -1) - 0.030663( -1)t e t t t+ + + 5 0.0091997( -1)t -

( ) 0.947912 - 0.270994 0.0677689( - 2) - 0.13686 - 4( - 2) - 0.005641( - 2)u t t t e t t 50.0022569( - 2) -t

6 70.002044477( -1) 0.36509 -3( -1)t e t+ t [1, 2]∈

= + +60.56428 -3( - 2) 0.1074883 -3( - 2)e t e t+ 7

3

4

( )

( ) 0.384619 -

u t

u t

t [2,3]∈

= 2 3 40.647533- 0.149562 0.0498569( -3) - 0.0083108( -3) 6.9893 - 7( -3)t t t e t 50.415265 -3( -3) -e t+ + +6 70.138445 -3( -3) 0.2966841 - 4( -3)e t e t+ t [3, 4]∈

= 2 3 4

2 35

0.073370 0.0275137( - 4) - 0.0061140( - 4) 0.0007642( - 4)

( ) 0.20862 - 0.033743 0.0134968( -5) - 0.00337401( -5)

t t t t 52.0687 -8( - 4) -e t

u t t t t

+ + +6 70.00002548( - 4) 0.000007279586( - 4)t t+ t [4,5]∈

= + 4

2 3 46

0.56227 -3( -5) -

( ) 0.106183- 0.014897 0.0062069( - 6) - 0.0016551( - 6) 0.3103 -3( - 6) -

e t 50.562082 - 4( -5) -e t

u t t t t e t

+6 76.32323 -9( -5) 0.134009 -5( -5)e t e t+ t [5,6]∈

= + +

2 3 47 ( ) 0.051485 - 0.0063942 0.00274( - 7) - 0.00076115( - 7) 0.000152( - 7) -

0.413657 - 4e 5( - 6)t

u t t t t t6 70.254 -5( - 7) -1.81 - 7( - 7) e t e t

+6 70.34452 -5( - 6) 5.6481 -10( - 6)e t e t+ t [6,7] ∈

= + + 50.228 - 4( - 7)e t +

⎧⎪⎪⎪⎪

⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪

t [7,8]∈⎪⎪⎩

(7.25)


ationMethod


Problems

and velocity in y direction, v(t), will be:

2 3 4 5 6 70

2 3 41

5

3 2 5 1 7 1( ) 1 2 t [0,1]2 3 24 20 720 630

v ( ) 0.36766 - 0.36785 0.367956( -1) - 0.18399( -1) 0.061334( -1) -

0.015334( -1) 0.00306685(

( )

v t t t t t t t t

t t t t tt t

v t

= − + − + − + − ∈

= + +

+

=

6 7

2 3 42

5 6 7

-1) - 0.511148 -3( -1) t [1,2]( ) -0.135702 0.8212 - 4 0.067687( - 2) - 0.045138( - 2) 0.0169285( - 2) -

0.004514507( - 2) 0.940551 -3( - 2) - 0.161241 -3( - 2)

e tv t e t t t t

t e t e t

∈

= + + +

+2 3 4

35 6 7

4

t [2,3]( ) -0.24931 0.049865 - 0.83872 -5( -3) - 0.0083053( -3) 0.00415334( -3)

- 0.00124607( -3) 0.276913 -3( -3) - 0.49449 - 4( -3) t [3,4]( ) -0.201764 0.0366842

v t t e t t tt e t e t

v t

∈

= + +

+ ∈

= + 2 3 4

5 6 7

2 35

- 0.0091704( - 4) - 4.1374 - 7( - 4) 0.000764( - 4) -

0.000306( - 4) 0.0000764( - 4) - 0.000014559( - 4) t [4,5]( ) -0.128214 0.020244 - 0.006747( -5) 0.00112416( -5) 1

t t e t tt t t

v t t t t

+

+ ∈

= + + + 4

5 6 7

2 3 46

.897 - 7( -5) -

0.56284 - 4( -5) 0.18755 - 4( -5) - 0.40185 -5( -5) t [5,6]( ) -0.071998 0.009931 - 0.003724( - 6) 0.000827( - 6) - 0.00010336( - 6) -

2.3722 -8(

e te t e t e t

v t t t t te t

+ ∈

= + +5 6 7

2 3 47

5 6 7

- 6) 0.0000034531( - 6) -9.86028 - 7( - 6) t [6,7]( ) -0.037449 0.0045669 - 0.001826( - 7) 0.000457( - 7) - 0.76083 - 4( - 7)

0.76029 -5( - 7) 1.7969 -9( - 7) -1.81536 - 7( - 7)

t e tv t t t t e t

e t e t e t

+ ∈

= + +

+ + t [7,8]

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪

∈⎪⎪⎩

(7.26)

Eqs. (7.25) and (7.26) are depicted in Fig. 7.5 and are compared withthose obtained by homotopy perturbation method (HPM)ePade, alsotheir values are presented in Tables 7.2 and 7.3 for u(t) andv(t), respectively. Fig. 7.6 shows the acceleration of the particleobtained from Fig. 7.5. As seen in these figures and tables,

1

0.8

(a) (b)

0.6

0.4

u(t)

v(t)

0.2

0

1

0.8

0.6

0.4

0.2

01 2 3 4 5 6

Ms-DTM HPM-Pade, Ref. [9] Num.

7 8 1 2 3 4 5tt

6 7 8

Figure 7.5 Multi-step differential transformation method (Ms-DTM) efficiency in par-ticle velocity estimating, compared with homotopy perturbation methodePade [2].(a) velocity in x direction, u(t); (b) velocity in y direction, v(t).


Table 7.2 Comparison Between u(t)’s Values and Errors Obtained From Homotopy Perturbation Method (HPM)ePade and Multi-stepDifferential Transformation Method (Ms-DTM) When A ¼ B ¼ a ¼ 1, C ¼ 0t Num. HPMePade Ms-DTM % Error HPMePade % Error Ms-DTM

0 1.0 1.0 1.0 0 01 0.735758911174129 0.7357586420 0.7359126984 3.66E-07 0.000212 0.406005878786318 0.4059621586 0.4059247848 0.000108 0.00023 0.199148297016095 0.1985113312 0.1988459460 0.003198 0.0015184 0.0915782073185634 0.08815295207 0.09113681554 0.037403 0.004825 0.0404276801389359 0.02950000148 0.03991176120 0.270302 0.0127626 0.0173512487117868 �0.00813908291 0.01679880445 1.469078 0.0318397 0.0072950308693911 �0.04120493255 0.006725715286 6.648356 0.0780428 0.0030191406936549 �0.07720068776 0.002442357340 26.57042 0.191042


ationMethod


Problems

Table 7.3 Comparison Between v(t)’s Values and Errors Obtained From Homotopy Perturbation Method (HPM)ePade and Multi-stepDifferential Transformation Method (Ms-DTM) When A ¼ B ¼ a ¼ 1, C ¼ 0t Num. HPMePade Ms-DTM % Error HPMePade % Error Ms-DTM

0 1.0 1.0 1.0 0 01 �4.38556843 E�8 2.36772637 E�7 �1.984126984E-7 1.838492 4.5142 �0.1353353333998 �0.1352915191 �0.1355378794 0.000324 0.00153 �0.0995741854390 �0.09892589377 �0.09971382172 0.006511 0.00144 �0.0549469585114 �0.05140825685 �0.05502746765 0.064402 0.001475 �0.0269518219371 �0.01550417243 �0.02699360654 0.424745 0.001556 �0.0123937803694 0.01465314367 �0.01241399659 2.182298 0.001637 �0.0054712929795 0.04659627467 �0.005480594942 9.516501 0.00178 �0.0023482314014 0.08469760161 �0.002352349259 37.06868 0.00175

DTM

forParticles

Motion,Sedim

entation,andCom

bustion295

Ms-DTM is more accurate than HPMePade, especially when timeincreased. This may be due to perturbation, small parameter, p, and line-arization in HPMePade method. Also Ms-DTM with eight time steps hasexcellent agreement with numerical solution, but it is easily capable toincrease the time steps.

Figs. 7.7 and 7.8 demonstrate the velocity variation of the particle forother values of the A, B, C, and a. These two figures, with mathematicalpurpose, are presented to show that Ms-DTM is completely equal to

Figure 7.6 Multi-step differential transformation method (Ms-DTM) efficiency in par-ticle acceleration estimating, A ¼ B ¼ a ¼ 1, C ¼ 0. (a) Acceleration in x direction, ax(t);(b) acceleration in y direction, ax(t).

Figure 7.7 Particle velocity when A ¼ B ¼ C ¼ a ¼ 1. (a) Velocity in x direction, u(t);(b) velocity in y direction, v(t).


numerical solution for other values of constants appeared in the mathe-matical formulation. The particle positions in a 2D plane couette fluid flowwhen A ¼ B ¼ C ¼ 1 and A ¼ B ¼ 1, C ¼ 0 are presented graphically inFig. 7.9(a) and (b), respectively, for each 1-s time step. As these figuresreveal, when C ¼ 0, due to nonshear portion effect in y direction or Tsy,the rotation of particle in y direction is lower than that in the case whenC ¼ 1.

Eq. (7.19) can be also solved by differential transformation method withPadé approximation (DTMePade), for this aim it can be rewritten as;�

€x� A _yþ Bð _x� ayÞ ¼ 0€yþ B _yþ ðAþ CÞð _x� ayÞ ¼ 0

(7.27)

Figure 7.8 Particle velocity when A ¼ B ¼ C ¼ 1, a ¼ 0. (a) Velocity in x direction, u(t);(b) velocity in y direction, v(t).

–4 –3 –2X

X0.5 1 1.5 20

0

0.1

0.2YY

0.3–1

0.2

0 10

–0.2

–0.4

–0.6

–0.8

(a) (b)

Figure 7.9 Particle position in each 1-s time step using multi-step differentialtransformation method when (a) A ¼ B ¼ C ¼ 1, (b) A ¼ B ¼ 1, C ¼ 0.


where coefficients A to C are defined as:

A ¼�3a8

��rf

rs

�(7.28)

B ¼�9n2r2

��rf

rs

�(7.29)

C ¼ 1:542

�a1=2n1=2

r

��rf

rs

�(7.30)

As mentioned in the text, the nonzero initial conditions of equations ofmotion could be different for unlike situations. The following mightrepresent either injection of the particle into the fluid or statisticalfluctuations: �

x ¼ 0 ; _x ¼ u0 at t ¼ 0y ¼ 0 ; _y ¼ v0 at t ¼ 0

(7.31)

To simplify the solution, constants dependent on physical properties ofsolidefluid combination are considered to be A ¼ B ¼ C ¼a ¼ u0 ¼ v0 ¼ 1.

Now we apply the differential transformation method for Eq. (7.27),and taking the differential transform of Eq. (7.27) with respect to t, gives:8><>:

ðkþ 1Þðkþ 2ÞXðkþ 2Þ þ ðkþ 1ÞXðkþ 1Þ � ðkþ 1ÞYðkþ 1Þ � YðkÞ ¼ 0

ðkþ 1Þðkþ 2ÞYðkþ 2Þ þ ðkþ 1ÞYðkþ 1Þ � ðkþ 1ÞXðkþ 1Þ � YðkÞ ¼ 0

(7.32)

From a process of inverse differential transformation, it can be shownthat the solutions of each subdomain take n þ 1 term for the power series,i.e., 8>>>><

>>>>:xiðtÞ ¼

Xnk¼0

�tHi

�k

XiðkÞ; 0 � t � Hi

yiðtÞ ¼Xnk¼0

�tHi

�k

YiðkÞ; 0 � t � Hi

(7.33)

where k ¼ 0, 1, 2,.,n represents the number of term of the power series,i ¼ 0, 1, 2,. expresses the ith subdomain, and Hi is the subdomain interval.


From initial condition in Eq. (7.31), we have it in point t ¼ 0 and t ¼ 1,and exerting transformation 8><

>:Xð0Þ ¼ 0

Yð0Þ ¼ 0

(7.34)

8><>:

Xð1Þ ¼ 1

Yð1Þ ¼ 1

(7.35)

Accordingly, from a process of inverse differential transformation, in thisproblem we calculated X(k þ 2) and Y(k þ 2) from Eq. (7.32) as following:8><

>:Xð2Þ ¼ 0

Yð2Þ ¼ �1

(7.36)

8>><>>:

Xð3Þ ¼ �16

Yð3Þ ¼ 12

(7.37)

8>>>>><>>>>>:

Xð4Þ ¼ 112

Yð4Þ ¼ �16

(7.38)

8>>>>><>>>>>:

Xð5Þ ¼ � 140

Yð5Þ ¼ 124

(7.39)

8>>>>><>>>>>:

Xð6Þ ¼ 1180

Yð6Þ ¼ � 1120

(7.40)


The above process is continuous. Substituting Eqs. (7.36)e(7.40) intothe main equation based on DTM, the closed form of the solutions can beobtained as:

xðtÞ ¼ P22K¼0

XðKÞtK

¼ t � 16t3 þ 1

12t4 � 1

40t5 þ 1

180t6 � 1

1008t7 þ 1

6720t8 � 1

51840t9

þ 1453600

t10 � 14435200

t11 þ 147900160

t12 � 1566092800

t13

þ 17264857600

t14 � 1100590336000

t15 þ 11494484992000

t16

� 123712495206400

t17 þ 1400148356608000

t18

� 17155594141696000

t19 þ 1135161222676480000

t20

� 12688996956405760000

t21 þ 156200036388880384000

t22

(7.41)

yðtÞ ¼ P22K¼0

YðKÞtK

¼ t � t2 þ 12t3 � 1

6t4 þ 1

24t5 � 1

120t6 þ 1

720t7 � 1

5040t8 þ 1

40320t9

� 1362880

t10 þ 13628800

t11 � 139916800

t12 þ 1479001600

t13

� 16227020800

t14 þ 187178291200

t15 � 11307674368000

t16

þ 120922789888000

t17 � 1355687428096000

t18

þ 16402373705728000

t19 � 1121645100408832000

t20

þ 12432902008176640000

t21 � 151090942171709440000

t22

(7.42)

As it is obvious, solution of terms vary periodically and in each step moreduration of particle motion is covered. By increasing series terms, the accu-racy of DTM solution is improved and a larger period of acceleration motionof the particle is covered. Basically, by estimating the constants A � C foreach selected combinations of solidefluid, results can be derived easily.


After obtaining the result of 22nd iteration for DTM, it is seen fromgraphs and tables that the DTM for this problem do not have exact solu-tion; therefore we will apply the Pade approximation for variations ofvelocities of the particle as follows

uðtÞ½4=4� ¼ _xðtÞ½4=4�

¼31

58060800��

15378128512000

�t2 þ

�47

1354752000

�t3 �

�3539

1463132160000

�t4 þ

�67

217728000

�t

3158060800

þ�

67217728000

�t þ�

2112709504000

�t2 þ

�43

4064256000

�t3 þ

�143

209018880000

�t4

(7.43)

vðtÞ½4=4� ¼ _yðtÞ½4=4�

¼1457

870912000��

2089870912000

�t þ�

68998128512000

�t2 �

�481

3657830400

�t3 þ

�1763

209018880000

�t4

1457870912000

þ�

1111612160

�t þ�

19018128512000

�t2 þ

�1

32659200

�t3 þ

�907

487710720000

�t4

(7.44)

The results are compared with DTM and numerical solution. Fig. 7.10depicts the horizontal and vertical velocities versus time. It is observed thatthe DTMePade approximant solution is more accurate than DTM.Comparing DTMePade [4/4] and DTMePade [10/10], DTMePade [10/10] gives closer results to numerical solution. This fact is more pronouncedfor large values of time, i.e., t ¼ 10.

7.4 NONSPHERICAL PARTICLES SEDIMENTATION

Consider a rigid, nonspherical particle with sphericity f, equivalent volumediameter D, mass m, and density rs is falling in an infinite extent of anincompressible Newtonian fluid of density r and viscosity m (Fig. 7.11).

U represents the velocity of the particle at any instant time, t, and g is theacceleration due to gravity. The unsteady motion of the particle in a fluidcan be described by the BasseteBoussinesqeOseen (BBO) equation. For adense particle falling in light fluids (assuming r � rs), Basset History forceis negligible. Thus, the equation of particle motion is given as [4]

mdudt

¼ mg

�1� r

rs

�� 18pD2rCDu

2 � 112

pD3rdudt

(7.45)

where CD is the drag coefficient. In the right-hand side of Eq. (7.45), thefirst term represents the buoyancy effect, the second term corresponds to


drag resistance, and the last term is associated with the added mass effectwhich is due to acceleration of fluid around the particle.

The nonlinear terms due to nonlinearity nature of the drag coefficientCD is the main difficulty in solving Eq. (7.45). By substituting CD in Eq.(7.45), following expression is gained (assuming Re ¼ ruD

m):�

mþ 112

pD3r

�dudt

þ 18pD2r

�30mruD

þ 67:289e�5:03f

�u2 � mg

�1� r

rs

�¼ 0

(7.46)

Figure 7.10 Comparison of horizontal velocity (u) and vertical velocity (v) obtained bydifferential transformation method (DTM) and DTMePade when A ¼ B ¼ a ¼u0 ¼ v0 ¼ 1.


by rearranging parameters, Eq. (7.46) could be rewritten as follows:

adudt

þ buþ cu2 � d ¼ 0; uð0Þ ¼ 0 (7.47)

where

a ¼�mþ 1

12pD3r

�(7.48)

b ¼ 3:75pDm (7.49)

c ¼ 67:289e�5:03f

8pD2r (7.50)

d ¼ mg

�1� r

rs

�(7.51)

By the transforming of Eq. (7.47) considered by using the related def-initions of DTM, and considering a ¼ b ¼ c ¼ d ¼ 1, we have thefollowing:

ðkþ 1ÞUðkþ 1Þ þ UðkÞ þXkl¼0

UðlÞUðk� lÞ � dðkÞ ¼ 0 (7.52)

Figure 7.11 Schematic view of vertically falling particle in ethylene alcohol.


From a process of inverse differential transformation, it can be shown thatthe solutions of each subdomain takes n þ 1 term for the power series, i.e.,

uiðtÞ ¼Xnk¼0

�tHi

�k

UiðkÞ; 0 � t � Hi (7.53)

where k ¼ 0, 1, 2,.,n represents the number of terms of the power series,i ¼ 0, 1, 2,. expresses the ith subdomain, and Hi is the subdomain interval.From initial condition in Eq. (7.47), that we have it in point t ¼ 0, andexerting transformation,

Uð0Þ ¼ 0 (7.54)

Accordingly, from a process of inverse differential transformation, inthis problem we calculated U(K þ 1)U(k þ 1) from Eq. (7.52) as following:8>>>>>><

>>>>>>:

Uð0Þ ¼ 0;

Uð1Þ ¼ 1;

Uð2Þ ¼ �1=2;

Uð3Þ ¼ �1=6;

«

(7.55)

The above process is continuous. Substituting Eq. (7.55) into the mainequation based on DTM, the closed form of the solutions can be obtained as:

uðtÞ ¼X20K¼0

UðKÞtK

¼ t � 12t2 � 1

6t3 þ 7

24t4 � 1

24t5 � 17

144t6 þ 67

1008t7 þ 227

8064t8 þ.

(7.56)

In a similar manner, we will obtain another subdomain’s series solution,and we can present the solution of Eq. (7.47) accurately.

The calculations presented in this chapter adopt a value of n ¼ 20.Having determined the various values of U(K þ 1) from Eq. (7.52) with thetransformed initial condition of Eq. (7.54), the first subdomain solutions ofEq. (7.47) can be obtained by means of the inverse transformed equations ofEq. (7.53). The final values of the first subdomain, i.e., the solutions of theprevious calculation, are then taken as the initial condition of the secondsubdomain, which is subsequently calculated using the same procedure asdescribed above. By repeatedly adopting the final values of one subdomain


as the initial condition of the following subdomain, the differential equationcan be solved from its first subdomain to its final subdomain. Therefore, theproposed method enables the solutions of Eq. (7.47) to be solved over theentire time domain, but not exactly, therefore after obtaining the results,the Pade approximation is applied as follows:

uðtÞ½4=4� ¼125

6967296t þ 625

292626432t3

1256967296

þ 12513934592

t þ 62565028098

t2 þ 625585252864

t3 þ 6252341011456

t4

(7.57)

An analytical solution for velocity and acceleration of the nonsphericalparticle during the unsteady motion by DTMePade approximant is obtained.The results are compared with DTM and numerical method (NM). Fig. 7.12depicts the velocity versus time for the three methods. It is observed that theDTMePade approximate solution is more accurate than DTM. DTMePade[4/4] gives closer results to numerical solution. This fact is more pronounced forlarger values of time, i.e., t¼ 2. Moreover, this interesting agreement betweenDTMePade approximation and numerical solution is shown in Table 7.4.

As noted previously, DTMePade [4/4] gives closer results to numericalsolution; therefore, it will produce more acceptable results regarding

Figure 7.12 Velocity profiles of the particle (a ¼ b ¼ c ¼ d ¼ 1). DTM, differentialtransformation method.


Table 7.4 The Results of the Differential Transformation Method (DTM) and DTMePade and NM at a ¼ b ¼ c ¼ d ¼ 1t u(t)DTM�Padé[4/4] u(t)DTM order 20 u(t)NM ErrorDTM�Padé[4/4] ErrorDTM order 20

0 0 0 0 0 00.4 0.316007088 0.316007092 0.316007131 0.0000000434 0.00000003940.8 0.483836437 0.483835084 0.483837560 0.0000011223 0.00000247551.2 0.561131387 0.552128895 0.561150879 0.0000194924 0.00902198421.6 0.59429563 �2.319449392 0.594422825 0.0001271951 2.91387221662 0.607843137 �246.4477393 0.608320171 0.0004770338 247.0560594709

DTMePade, differential transformation method with Padé approximation; NM, numerical method.


ationMethod


Problems

accelerations of the particle during the falling process. This is confirmed bythe curves in Fig. 7.13.

Figs. 7.14 and 7.15 show the position of the particle in the sedimen-tation process for each time step equal to 0.02 s. The effect of particlesphericity on settling position is shown in Fig. 7.14. From this figure, it canbe concluded that the particle with larger sphericity moves faster thansmaller ones. Finally, Fig. 7.15 explains the material of particle’s effect onsettling position.

Figure 7.13 Acceleration variation of the particle (a ¼ b ¼ c ¼ d ¼ 1). DTM, differentialtransformation method.

Figure 7.14 Positions of falling particles for different sphericity of copper particleswith time interval ¼ 0.02 s.


7.5 MOTION OF A SPHERICAL PARTICLE IN A FLUIDFORCED VORTEX

Consider the polar coordinates r-q with pole at the origin r ¼ 0. Consideringthe 2D model for this problem, when rotating speed is approximately con-stant, is an acceptable assumption and simplification which is considered inthe literature [5]. So, in this study, motion of the particle in a plane is studiedaccording to Fig. 7.16. The mass of fluid is rotating in the

Figure 7.15 Positions of falling particles for different materials with timeinterval ¼ 0.02 s.

Ω

Ω

u=r.Ω

r

r

Particle

Forced vortex

Figure 7.16 Schematic view of spherical particles in a sample fluid forced vortex(industrial hopper).


counterclockwise direction around the origin (see Fig. 7.16). The streamlinesare concentric circles with common center at the origin. For forced vortex,the tangential velocity is given by uq ¼ U$r whereas for free vortex uq ¼ c/r, U and c are constants. In both cases the radial components of velocity offluid are zero. At the instant t ¼ 0, a small spherical particle is released in theflow field with initial radius r0; zero angular velocity; and u0 as radialcomponent of velocity. The fluid will drive the particle to rotation, exertingdrag force [5],

FD ¼ cDA12ru2 (7.58)

where u is the relative velocity between the particle and the flow; r is themass density of fluid; A is the projected area of sphere; and cD is the drag co-efficient and is a function of the Reynolds number. The equations of motionof the particle can be written as m$a ¼ F; m is the mass of the particle; a is theacceleration; and F is the exerted external force on particle. The equations inthe radial and tangential directions are given respectively [5],

m�€r � r _q

2� ¼ �cDA12r _r2 (7.59)

m�r€qþ 2 _r _q

� ¼ cDA12r�r _q� uqfluid

�2(7.60)

where uqfluid ¼ U$r for the forced vortex and uqfluid ¼ c/r for free vortex.The quantities to the left side are the radial and tangential components ofacceleration in polar coordinates. Now, by using U ¼ u

u0; U ¼ u

u0; R ¼ r

r0nondimensionalized equations will be:8>>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>>:

_RðsÞ � u0r0u0

UðsÞ ¼ 0

u0r0u0

_UðsÞ � RðsÞUðsÞ2 þ a2

�u0r0u0

�2

UðsÞ2 ¼ 0

RðsÞ _UðsÞ þ u0r0u0

2UðsÞUðsÞ � a2RðsÞ2

8>>>><>>>>:

ðUðsÞ � 1Þ2 Forced Vortex

UðsÞ � 1

RðsÞ2!2

Free Vortex¼ 0

(7.61)


The nondimensional parameter a2 ¼cDA

12rr0

m represents the drag toinertia ratio. Due to importance of particle motion analysis which wasintroduced in Section 7.1, this system of equation should be solved, andaccording to its rational function nonlinearity form, analytical and nu-merical methods should be applied. By obtaining the particle position andvelocities, its motion and treatment in vortices are completely under-standable. For solving the particle motion in a forced vortex by an efficient,fast, and high accurate method, system of Eq. (7.61) is solved byDTMePadé,8>>>>>>>>>>>>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>>>>>>>>>>>>:

ðkþ 1ÞRðkþ 1Þ � u0r0u0

UðkÞ ¼ 0

u0r0u0

ðkþ 1ÞUðkþ 1Þ �Xkk1

Xk1l¼0

RðlÞUðk1� lÞUðk� k1Þ

þa2

�u0r0u0

�2Xkl¼0

UðlÞUðk� lÞ ¼ 0

Pkl¼0

RðlÞðkþ 1� lÞUðk� l þ 1Þ þ 2u0r0u0

Xkl¼0

UðlÞUðk� lÞ

�a2

0BBBB@Pkk2¼0

Pk2k1¼0

Pk1l¼0

RðlÞRðk1� lÞUðk2� k1ÞUðk� k2Þ�

2Pkk1¼0

Pk1l¼0

RðlÞRðk1� lÞUðk� k1Þ þPkl¼0

RðlÞRðk� lÞ

1CCCCA ¼ 0

(7.62)

where UðkÞ; RðkÞ; UðkÞ are DTM transformed forms of U(k), R(k), U(k)in Eq. (7.61). With initial condition,

Uð0Þ ¼ 1; Rð0Þ ¼ 1; Uð0Þ ¼ 0 (7.63)

Since the procedure of solving Eq. (7.62) is autonomous of constants u0,r0, u0, and a2, for generalization and simplification of the problem forfuture cases with different physical conditions the constants that representphysical properties are assumed to be as following:

u0 ¼ r0 ¼ u0 ¼ a2 ¼ 1 (7.64)


After solving Eq. (7.62) by using initial condition Eq. (7.63), other terms ofparticle position R(s), radial velocity U(s), and angular velocity U(s) will beappeared as follows,

Rð1Þ ¼ u0r0u0

;Uð1Þ ¼ �a2u0r0u

;Uð1Þ ¼ a2

Rð2Þ ¼ �12u20a

2

r20u20

;Uð2Þ ¼ u20a4

r20u20

;Uð2Þ ¼ �12a2ðu0 þ 2a2r0u0Þ

r0u0

Rð3Þ ¼ �13u30a

4

r30u30

;Uð3Þ ¼ 13

a4�r40u

40 � 3a2u40

�r30u

30u0

;Uð3Þ

¼ 16

a2�6a4r20u

20 þ 6u20 þ 2a2u0r0u0 þ 3u20a

2�

r20u20

:::

(7.65)

For example, when using constant values in Eq. (7.64), final functionswill be calculated as:8>>>>>>>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>>>>>>>:

RðtÞ ¼ 1þ t � 12t2 þ 1

3t3 � 1

6t4 þ 1

15t5 þ 17

360t6 � 499

2520t7 þ 8147

20160t8 � 21079

30240t9

þ509813453600

t10 � 175051619979200

t11 þ 8053627129937600

t12 � 63553234631556755200

t13 þ 103136216031676505600

t14

UðtÞ ¼ 1� t þ t2 � 23t3 þ 1

3t4 þ 17

60t5 � 499

360t6 þ 8147

2520t7 � 21079

3360t8 þ 509813

45360t9

�17505161907200

t10 þ 805362712494800

t11 � 6355323463119750400

t12 þ 10313621603119750400

t13 � 552091929139916800

t14

UðtÞ ¼ t � 32t2 þ 17

6t3 � 11

2t4 þ 41

4t5 � 6641

360t6 þ 3398

105t7 � 1122019

20160t8 þ 2439641

25920t9

�28445111181440

t10 þ 45874001178200

t11 � 624104432914968800

t12 þ 10370857074591556755200

t13 � 2288364060278921794572800

t14

(7.66)

Eq. (7.66) is the DTM solution of Eq. (7.61), results are depicted inFigs. 7.17e7.19 and presented in Tables 7.5 and 7.6. Also the outcomeswere compared with numerical fourth order RungeeKutta and differentialquadrature method (DQM). As seen in this figure DTM solution after ashort agreement with numerical result, suddenly reaches to infinity (for allfunctions, R(t), U(t), and U(t)). To overcome this shortcoming andincreasing the convergence of this method, Padé approximation is applied.


2.0

1.8

1.6

1.4R

1.2

1.0

0.8

0 0.2

DTM-Pade DTM RK4 DQM

0.4τ

0.6 0.8 1

Figure 7.17 Position trajectory for the particle (R) obtained by different methods.DQM, differential quadrature method; DTM, differential transformation method;DTMePade, differential transformation method with Padé approximation.

0.50 0.2


0.4 0.6 0.8 1

0.6

0.7

0.8

U

τ

0.9

1.0

Figure 7.18 Radial velocity for the particle (U) obtained by applied methods. DQM,differential quadrature method; DTM, differential transformation method; DTMePade,differential transformation method with Padé approximation.


0 0.2 0.4 0.6 0.8 1


0.5

0.4

0.3

0.2

Ω

τ

0.1

0

Figure 7.19 Angular velocity for the particle (U) obtained by applied methods. DQM,differential quadrature method; DTM, differential transformation method; DTMePade,differential transformation method with Padé approximation.

Table 7.5 Comparison Between Applied Methods in Nonuniform and NondimensionalTime for Angular Velocity Resultss U(s)

RK4 DTMePadé[6,6] DQM

0.0000000000 0.000 0.000 0.0000.0301536896 0.0288632031758779 0.02886319921 0.03155900378703640.1169777784 0.100142863870931 0.1001428582 0.09956370644741570.2500000000 0.185921658986676 0.1859216559 0.1912457420765930.4131759114 0.267081663618700 0.2670816367 0.2700994189995330.5868240886 0.334822176009758 0.3348218891 0.3424470695101790.7500000000 0.386168789277334 0.3861657749 0.3902611734143340.8830222216 0.421185866197261 0.4211728745 0.4265683515874050.9698463104 0.441257247764227 0.4412281585 0.4463105657080621.0000000000 0.447764342038054 0.4477267112 0.453066160230719

DTM, differential transformation method; DTMePade, differential transformation method with Padéapproximation.


Using Padé with accuracy of [6,6] results in Eq. (7.66) and will be trans-formed as the following form,8>>>>>>>>><>>>>>>>>>:

RðtÞ ¼ ð1þ 5:5828532t þ 10:728458t2 þ 7:9406046t3 þ 1:7311221t4 þ 0:48787532t5 þ 0:27840808t6Þð1þ 4:5828532t þ 6:6456053t2 þ 3:2530926t3 þ 0:43988108t4 þ 0:15648097t5 þ 0:01235856t6Þ

UðtÞ ¼ ð1þ 3:6270954t þ 5:2449498t2 þ 3:6857199t3 þ 1:2403949t4 þ 1:4403629t5 þ 0:58851084t6Þð1þ 4:6270954t þ 8:8720452t2 þ 8:5973364t3 þ 3:7170831t4 þ 0:64910789t5 þ 0:3698456t6Þ

UðtÞ ¼ ð0:091822563t6 þ 0:44660102t5 þ 1:2839632t4 þ 2:4523533t3 þ 2:4126985t2 þ tÞð1þ 3:9126985t þ 5:4880679t2 þ 3:93008583t3 þ 2:06204617t4 þ 0:576084t5 þ 0:2009531t6Þ

(7.67)

In this chapter, for the first time, analytical approaches called DTMePadé and DQM have been successfully applied to find the most accurateanalytical solution for the motion of a particle in a forced vortex. Radialposition, angular velocity, and radial particle velocity were calculated anddepicted. As a main outcome from the present study, it is observed that theresults of DQM are in excellent agreement with numerical ones. Also thismethod is a simple, fast, powerful, and efficient technique for findingproblems solution in science and engineering with coupled nonlinear dif-ferential equations. Also, it reduces the size of calculations. Results showthat by passing time, particle recedes from the vortex center, and its radialvelocity decreases while its angular velocity increases.

Table 7.6 Comparison Between Applied Methods in Nonuniform andNonDimensional Time for Radial Velocity Results

s

U(s)

RK4 DTMePadé[6,6] DQM

0.0000000000 1.000 1.000 1.0000.0301536896 0.970737559394717 0.9707375589 0.9698607936576300.1169777784 0.895704744070214 0.8957047452 0.8953675797818970.2500000000 0.803455212921360 0.8034552207 0.8032069257292270.4131759114 0.720357770547646 0.7203578045 0.7231998898020200.5868240886 0.660545915322201 0.6605462648 0.6674546307593410.7500000000 0.625835951196769 0.6258387861 0.6347128496643660.8830222216 0.610250219603200 0.6102609542 0.6194288765556330.9698463104 0.605365864743497 0.6053880716 0.6139844755054961.0000000000 0.604553915451281 0.6045818949 0.612909563556282

DTM, differential transformation method; DTMePade, differential transformation method withPadé approximation.


7.6 COMBUSTION OF MICROPARTICLES

Consider a spherical particle which due to high reaction with oxygen willbe combusted. Since the thermal diffusivity of substance is large and theBiot number is small (BiH << 0.1), it is assumed that the particle isisothermal. In this state, a lumped system analysis is applicable. When thiscriterion is satisfied, the variation of temperature with location within theparticle will be slight and can be approximated as being uniform, so particlehas a spatially uniform temperature; therefore the temperature of particle isa function of time only, T ¼ T(t) and is not a function of radial coordinate,Ts T(r). The assumptions used in this modeling include the following [6].1. The spherical particle burns in a quiescent, infinite ambient medium,

and there are no interactions with other particles, also the effects offorced convection are ignored.

2. Thermophysical properties for the particle and ambient gaseous oxidizerare assumed to be constant.

3. The particle radiates as a gray body to the surroundings without contri-bution of the intervening medium.By these assumptions and considering the particle as a thermodynamic

system, and by using the principle of conservation of energy (first law ofthermodynamics), the energy balance equation for this particle can bewritten as (Fig. 7.20);

_Ein � _Eout þ _Egen ¼�dEdt

�p

(7.68)

where _Ein is the rate of energy entering the system which is owing to ab-sorption of total radiation incident on the particle surface from the

Figure 7.20 Iron microparticle and its energy balance.


surrounding, _Eout is the rate of energy leaving the system by mechanisms ofconvection on the particle surface and thermal radiation that emits from theouter surface of particle, _Egen is the rate of generation of energy inside theparticle due to the combustion process and equals to the heat released fromthe chemical reaction, and (dE/dt)p is the rate of change in total energy ofparticle. These energy terms can be calculated by:

_Ein ¼ assAsT4surr (7.69)

_Eout ¼ hconvAsðTs � TNÞ þ εssAsT 4s (7.70)

_Egen ¼ _Qcomb ¼ _RpAsDh+comb (7.71)�dEdt

�p

¼ rpVpcpdTs

dt(7.72)

By substituting of Eqs. (7.69)e(7.72) in Eq. (7.68),

assAsT4surr �

�hconvAsðTs � TNÞ þ εssAsT

4s

�þ _RpAsDh+comb ¼ rpVpcp

dTs

dt(7.73)

Three reasonable assumptions are used for improving Eq. (7.73):1. Both absorptivity and emissivity of the surface depend on the temper-

ature and the wavelength of radiation. Kirchhoff’s law of radiation statesthat the absorptivity and the emissivity of a surface at a given tempera-ture and wavelength are equal ðεsxasÞ.

2. The initial temperature of the particle at the beginning of combustioncan be regarded as the initial condition. This temperature is known asignition temperature (T(0) ¼ Tig).

3. The density of particle is a function of particle temperature, so it can beconsidered as a linear function (rp ¼ rp(T) ¼ rp,N[1 þ b(T � TN)]).By applying these assumptions, Eq. (7.73) will be converted to the

following:

rp;N½1þ bðT � TNÞ�VpcpdTs

dtþ hconvAsðTs � TNÞ þ εssAs

�T 4s � T 4

surr

�� _RpAsDh

+comb ¼ 0

(7.74)


For solving this nonlinear differential equation, it’s more suitable that allthe terms be converted to the dimensionless form. The following set ofdimensionless variables is defined:8>>>>><

>>>>>:

q ¼ TTig

; qN ¼ TN

Tig; qsurr ¼ Tsurr

Tig; ε1 ¼ bTig

s ¼ t�rp;NVpcphconvAs

�;j ¼_Qcomb

hconvAsTig; ε2 ¼

εssT3ig

hconv

(7.75)

Consequently, the nonlinear differential equation and its initial condi-tion can be expressed in the dimensionless form

ε1qdqds

þ ð1� ε1qNÞ dqds þ ε2

�q4 � q4surr

�þ q� j� qN ¼ 0 (7.76)

qð0Þ ¼ 1 (7.77)

By applying DTM from Chapter 1 principle, transformed form of Eq.(7.76) will be,

ðkþ 1ÞQðkþ 1Þ þ ε1

Xkl¼0

QðlÞ$ðkþ 1� lÞ$Qðkþ 1� lÞ

�ε1qNðkþ 1Þ$Qðkþ 1Þ þQðkÞþ

ε2

Xkk2¼0

Xk2k1¼0

Xk1l¼0

QðlÞ$Qðk2� k1Þ$Qðk� k2Þ$Qðk1� lÞ

�dðkÞ$�qN þ ε2q4surr þ j

� ¼ 0

(7.78)

where Q is transformed form of q and

dðkÞ ¼�1 k ¼ 0

0 ks1(7.79)

Transformed form of initial condition (Eq. 7.77) will be,

Qð0Þ ¼ 1 (7.80)

For example, for an iron particle with 20 mm diameter (see Table 2 inRef. [6]) solving Eq. (7.78) makes,

Qð0Þ ¼ 1;Qð1Þ ¼ 1:170326597;Qð2Þ ¼ �0:6324112173

Qð3Þ ¼ 0:2462459086;Qð4Þ ¼ �0:08656001439;.(7.81)


By substituting DTM transformed terms of T (Eq. 7.81) into trans-formed DTM equation, q(s) can be determined as (Figs. 7.21 and7.22),

qðsÞ ¼ 1þ 1:17033s� 0:632411s2 þ 0:246245s3 � 0:08656s4

þ 0:0324505s5 � 0:0130388s6 þ 0:00504942s7

� 0:0017s8 þ 0:00042358s9 � 0:114879e� 4s10(7.82)

7.7 UNSTEADY SEDIMENTATION OFSPHERICAL PARTICLES

For modeling the particle sediment phenomenon, consider a small, rigidparticle with a spherical shape of diameter D and mass of m and density of rsfalling in infinite extent filled water as an incompressible Newtonian fluid.Density of water, r, and its viscosity, m, are known. We considered thegravity, buoyancy, Drag forces, and added mass (virtual mass) effect onparticle. According to the BBO equation for the unsteady motion of theparticle in a fluid, for a dense particle falling in light fluids and by assuming

0

2.6

2.4

dp=100 µm

dp=60 µm

dp=20 µm

2.2

2

1.8θ

τ

1.6

1.4

1.2

10.1 0.2

NUM DTM

0.3 0.4 0.5

Figure 7.21 Comparison between DTM and numerical method in different particlediameters. DTM, differential transformation method.


r << rs, Basset History force is negligible. So by rewriting force balance forthe particle, the equation of motion is gained as follows [7],

mdudt

¼ mg

�1� r

rs

�� 18pD2rCDu

2 � 112

pD3rdudt

(7.83)

where CD is the drag coefficient. In the right-hand side of Eq. (7.83), thefirst term represents the buoyancy affect, the second term corresponds todrag resistance, and the last term is due to the added mass effect which isdue to acceleration of fluid around the particle. The main difficulty of solv-ing Eq. (7.83) is the nonlinear terms due to the nonlinearity nature of thedrag coefficient CD. By considering CD in range of Reynolds number,0 � Re � 105 as following equation

CD ¼ 24Re

�1þ 1

48Re

�(7.84)

Substituting Eq. (7.84) into Eq. (7.83), and mass of the spherical particle is

m ¼ 16pD3rs (7.85)

Eq. (7.83) can be rewritten as

adudt

þ buþ cu2 � d ¼ 0; uð0Þ ¼ 0 (7.86)

Figure 7.22 (a) Effect of ε1 on nondimensional temperature profile for micro- andnanoparticles. (b) Effect of ε2 on nondimensional temperature profile for micro- andnanoparticles.


where

a ¼ 112

pD3ð2rs þ rÞ (7.88)

b ¼ 3pD m (7.89)

c ¼ 116

pD2r (7.90)

d ¼ 16pD3gðrs � rÞ (7.91)

Eq. (7.86) is a nonlinear equation with an initial condition and it can besolved by numerical and analytical methods. In the present example, wechoose three different materials for solid particle, aluminum, copper, and leadand considered three different diameters (1, 3, and 5 mm) for them. Aschematic of described problem is shown in Fig. 7.23. Physical properties ofthe selected material are shown in Table 7.7, and the resulted coefficients a, b,c, and d from Eqs. (7.88)e(7.91) are listed in Table 7.8, and Eq. (7.86) as anonlinear equation, is solved by numerical method, DTMePadé approxi-mation compared to collocation method and Galerkin method. It is necessaryto inform that professional version of this problem with more complexity canbe solved by CFD methods that are available in the literature.

Figure 7.23 Schematic view of particles settling in water. It is assumed that particlesare falling freely such that no particleeparticle or particleewall interaction exists.


Applying the DTM transformation from Chapter 1 into Eq. (7.86) tofind u(t), we have:

aðkþ 1ÞUðkþ 1Þ þ bUðkÞ þ cXkl¼0

UðlÞUðk� lÞ � ddðkÞ ¼ 0; Uð0Þ ¼ 0

(7.92)

Rearranging Eq. (7.92), we have

Uðkþ 1Þ ¼ ddðkÞ � cPk

l¼0½UðlÞUðk� lÞ� � bUðkÞaðkþ 1Þ ; Uð0Þ ¼ 0 (7.93)

where

dðkÞ ¼�1 if k ¼ 0

0 if ks0(7.94)

By solving Eq. (7.93) and using the initial condition, the DTM terms areobtained as

Uð1Þ ¼ da

Uð2Þ ¼ �12bda2

Uð3Þ ¼ 16dðb2 � 2cdÞ

a3

Uð4Þ ¼ � 124

bdðb2 � 8cdÞa4

«

(7.95)

Now representing Eq. (7.95) in series form, u(t) function will be ob-tained. After five iterations in DTM series for a ¼ b ¼ c ¼ d ¼ 1, the u(t)function is obtained as

uðtÞ ¼ t � 12t2 � 1

6t3 þ 7

24t4 � 1

24t5 (7.96)

Table 7.7 Properties of the Selected MaterialsMaterial Density (kg/m3) Viscosity (kg/m s)

Aluminum 2702 eCopper 8940 eLead 11,340 eWater 996.51 0.001


Table 7.8 Coefficients in Eq. (7.86)

LiquidSolid(Particle)

Diameter(mm) a b c d

Water Aluminum 1 0.0000016756496 0.00000942477796 0.000195664281 0.0000087602563 0.0000452425392 0.00002827433389 0.001760978529 0.0002365269175 0.0002094562001 0.00004712388981 0.004891607024 0.0010950320241 0.0000049418587 0.00000942477796 0.000195664281 0.000040801768

Copper 3 0.0001334301866 0.00002827433389 0.001760978529 0.0011016477385 0.0006177323456 0.00004712388981 0.004891607024 0.0051002210101 0.0000061984958 0.00000942477796 0.000195664281 0.000053129377

Lead 3 0.0001673593872 0.00002827433389 0.001760978529 0.0014344931965 0.0007748119783 0.00004712388981 0.004891607024 0.006641172207


ationMethod


Problems

Subsequently, applying Padé approximation to Eq. (7.96) (for Padé [4,4]accuracy), we have,

Pad�e ½4; 4�ðuðtÞÞ ¼344113480

t4 � 2241011

t3 � 227674

t2 þ t

1þ 55337

t þ 9337

t2 þ 71685

t3 þ 11685

t4(7.97)

At first, a comparison between the three analytical methods andRungeeKutta methods is provided to select the best and reliable analyticalmethod for present problem. Eq. (7.86) is considered with all constants areequal to unity (i.e., a ¼ b ¼ c ¼ d ¼ 1). Results of the solutions are depictedin Fig. 7.24. From Fig. 7.24, it is revealed that GM and CM have a goodagreement with numerical result; but in DTM when time tends to infinity,even in high iterations (20 iterates), it cannot estimate a constant velocity as“terminal velocity” and its value suddenly reaches to zero. For solving thisproblem, Padé approximation in different orders such as [2,2], [4,4], [8,8](e.g., Eq. 7.97) is used, and convergence results are depicted in Fig. 7.25. AsFig. 7.25 shows, higher order of Padé approximation leads to obtain resultscloser to the numerical solution. Table 7.9 shows the velocity values versustime for applied analytical methods and are compared with numericalprocedure. The errors (%) of these methods, with respect to the numericalmethod, were listed in Table 7.10. As seen in Table 7.10, DTMePadé [8,8]is the best method for this equation. As GM also has a good agreement and

Figure 7.24 Comparison between numerical (NM), collocation (CM), Galerkin methods(GM) and DTM (in three different iterations) for Eq. (7.86) and a ¼ b ¼ c ¼ d ¼ 1. DTM,differential transformation method.


acceptable accuracy, it can be considered as an efficient method. In thefollowing section, these two methods are used for analysis of the practicalsettling motion of some spherical particles in water.

Afterward, to present some practical examples, aluminum, copper, andlead are selected in various sizes submerged in water. Physical properties ofthe materials and calculated coefficients for Eq. (7.86) in these practical ex-amples are listed in Tables 7.7 and 7.8, respectively. Figs. 7.26 and 7.27 depictthe variations of velocity and acceleration for aluminum in different di-ameters. Velocity variations for copper and lead are shown in Fig. 7.28 and7.29. The effect of the particle material on velocity and acceleration for thesethree metals, aluminum, copper, and lead, are investigated in Figs. 7.30 and7.31 for a constant diameter of 1 mm. These effects for larger diameters,D ¼ 3 mm and D ¼ 5 mm are shown in Figs. 7.32 and 7.33, respectively.Figs. 7.34 and 7.35 presented the position of the particle in sedimentationprocess for each time step equal to 0.02 s. Fig. 7.34 shows the effect of particlesize, and Fig. 7.35 explains the material of particle’s effect on settling position.

Generally speaking, the physical behavior of the settling particles is wellcaptured. For all the selected particles, terminal velocity is calculated andpresented in Table 7.11. It can be concluded from Table 7.11 thatDTMePade [8,8] estimated the terminal velocity excellently, although itserrors might increase compared with HPM when t tends to infinity.Outcomes reveal that the value of the terminal velocity increases

Figure 7.25 Convergence of the DTMePadé regarding to the method order, [L, M], forEq. (7.86) and a ¼ b ¼ c ¼ d ¼ 1. DTMePadé, differential transformation method withPadé approximation.


Table 7.9 Obtained Values for u(m/s) From Different Methods for Eq. (7.86) and a ¼ b ¼ c ¼ d ¼ 1Time(s)

NumericalMethod

CollocationMethod

GalerkinMethod

DTM(n ¼ 20)

DTMePade[2,2]

DTMePade[4,4]

DTMePade[8,8]

0 0 0 0 0 0 0 00.2 0.17911335972 0.1792419 0.17916766 0.17911334 0.1791044 0.1791133 0.1791133460.4 0.31600713125 0.3160935 0.31602345 0.31600709 0.3157894 0.3160070 0.3160070920.6 0.41502836932 0.4150925 0.41506767 0.41502829 0.4137931 0.4150281 0.4150282970.8 0.48383755960 0.4838791 0.48385852 0.48383508 0.4799999 0.4838364 0.4838374591.0 0.53032985719 0.5303603 0.53032540 0.53010503 0.5217391 0.5303244 0.5303297561.2 0.56115087915 0.5611641 0.56116612 0.55212889 0.5454545 0.5611313 0.5611507401.4 0.58132582197 0.5813853 0.58132775 0.38037826 0.5562913 0.5812713 0.5813256941.6 0.59442282455 0.5956222 0.59457135 e 0.5581395 0.5942956 0.5944226651.8 0.60287916150 0.6103034 0.60512051 e 0.5538461 0.6026196 0.6028790312 0.60832017092 0.6373060 0.62039362 e 0.5454545 0.6078431 0.608320058

DTM, differential transformation method; DTMePade, differential transformation method with Padé approximation.

DTM

forParticles

Motion,Sedim

entation,andCom

bustion325

Table 7.10 Calculated Errors (%) for Various Methods Solving Eq. (7.86) and a ¼ b ¼ c ¼ d ¼ 1Time (s) Collocation Galerkin DTM (n ¼ 20) DTMePade [2,2] DTMePade [4,4] DTMePade [8,8]

0 0 0 0 0 0 00.2 0.000303 0.000303 7.27454E-08 4.95894E-05 7.33037E-08 7.44203E-080.4 0.000274 5.17E-05 1.24524E-07 0.000688774 1.36866E-07 1.23258E-070.6 0.000155 9.47E-05 1.88481E-07 0.002976341 4.32802E-07 1.73302E-070.8 8.6E-05 4.33E-05 5.1164E-06 0.007931504 2.31877E-06 2.06488E-071.0 5.75E-05 8.39E-06 0.000423933 0.016198837 1.02894E-05 1.89122E-071.2 2.36E-05 2.72E-05 0.016077644 0.027971681 3.4736E-05 2.46742E-071.4 0.000102 3.32E-06 0.345671144 0.043064372 9.36325E-05 2.18938E-071.6 0.002018 0.00025 4.902019398 0.06103953 0.000213981 2.66898E-071.8 0.012315 0.003718 e 0.081331402 0.000430478 2.14978E-072 0.047649 0.019847 e 0.10334299 0.000784182 1.85468E-07

DTM, differential transformation method; DTMePade, differential transformation method with Padé approximation.


ationMethod


Problems

significantly with particle diameter and density. Moreover, the smallerparticle reaches to its terminal velocity earlier. Thus, the acceleration periodis shorter for smaller and lighter particles. From a physical point of view, itcan be concluded that larger particles reach zero acceleration (terminalvelocity) more slowly. Results also explain that when particle is massive(because of greater density or diameter) it has lower position in the sametime steps because of its greater terminal velocity.

00

0.1

0.2

u(m

/s)

0.3

0.4

0.5

0.10t

0.05

Num GM DTM-Pade[8,8]

0.15 0.20

D=5mm

D=3mm

D=1mm

Figure 7.26 Velocity variation for different particle diameters (aluminum).

Figure 7.27 Acceleration variation for different particle diameters (aluminum).


7.8 TRANSIENT VERTICALLY MOTION OF ASOLUBLE PARTICLE

For modeling the particle sediment phenomenon, consider a small particlewith a spherical shape of variable diameter D(t) and mass of m(t) and densityof rs, falling in infinite extent filled by an incompressible Newtonian fluid.

D=5mm

D=3mm

D=1mm

0.2

0

0 0.05 0.10

Num GM

t(s)

u(m

/s)

DTM-Pade[8,8]

0.15 0.20 0.25

0.4

0.6

0.8

1

Figure 7.28 Velocity variation for different particle diameters (copper).

u(m

/s)

0.1 0.2 0.30

0.2

0

0.4

0.6

0.8

1

t(s)Num GM DTM-Pade[8,8]

D=5mm

D=3mm

D=1mm

Figure 7.29 Velocity variation for different particle diameters (lead).


Density of fluid, r, and its viscosity, m, are known. We considered thegravity, buoyancy, drag forces, and added mass (virtual mass) effect onparticle. According to the BBO equation for the unsteady motion of theparticle in a fluid, for a dense particle falling in light fluids and by assuming

u(m

/s)

0.1

0

0 0.1 0.2 0.3

Aluminium

Copper

Lead

0.2

0.3

0.4

0.5

Num DTM-Pade[8,8]

t(s)

Figure 7.30 Comparison of velocity variation over time for different particle materialswhen D ¼ 1 mm.

9

8

7

6

5

4

3

2

1

00 0.1 0.2 0.3t(s)

Aluminum Copper Lead

Acc

(m/s

2 )

Figure 7.31 Comparison of acceleration variation for different particle materials(D ¼ 1 mm). DTMePadé, differential transformation method with Padé approximation.


0.2

0

0 0.1 0.2 0.3 0.4 0.5

0.4

0.6

0.8

1

Num DTM-Pade[8,8]

t(s)

u(m

/s)

Aluminium

Copper

Lead

Figure 7.32 Velocity variation for different particle materials when D ¼ 3 mm.DTMePadé, differential transformation method with Padé approximation

Aluminium

Copper

Lead

0

0

0.2

0.4

0.6

0.8

1

1.2

0.1 0.2 0.3 0.4 0.5 0.6

u(m

/s)

t(s)Num DTM-Pade[8,8]

Figure 7.33 Velocity variation for different particle materials when D ¼ 5 mm.DTMePadé, differential transformation method with Padé approximation


r << rs, Basset History force is negligible. So, by rewriting force balancefor the particle, the equation of motion is gained as follows [8],

mðtÞ duðtÞdt

¼ mðtÞg�1� r

rs

�� 18pDðtÞ2rCDuðtÞ2 � 1

12pDðtÞ3r duðtÞ

dt

(7.98)

Figure 7.34 Positions of falling particles for different size of aluminum particles, timeinterval ¼ 0.02 s.

Figure 7.35 Positions of falling particle for different materials, D ¼ 1 mm and timeinterval ¼ 0.02 s.


where CD is the drag coefficient, in the right-hand side of Eq. (7.98), thefirst term represents the buoyancy effect, the second term corresponds todrag resistance, and the last term is due to the added mass effect which isdue to acceleration of fluid around the particle. The main difficulty ofsolving Eq. (7.98) is nonlinear terms due to the nonlinearity nature ofthe drag force which comes from drag coefficient CD, u(t)

2 and D(t)2 terms.A correlation for CD of spherical particles which has good agreement withthe experimental data in a wide range of Reynolds number, 0 � Re � 105

is

CD ¼ 24Re

�1þ 1

48Re

�(7.99)

It’s necessary to inform that Eq. (7.99) is based on the nonslip conditionon the surface of the soluble particle. Substituting Eq. (7.99) into Eq. (7.98)and variable mass of the spherical particle is

mðtÞ ¼ 16pDðtÞ3rs (7.100)

Eq. (7.98) can be rewritten as

112

pDðtÞ3ð2rs þ rÞ duðtÞdt

þ 3pDðtÞmuðtÞ þ 116

pDðtÞ2ruðtÞ2

� 16pDðtÞ3gðrs � rÞ ¼ 0

(7.101)

Table 7.11 Terminal Velocity (m/s) for Particles Calculated by DTMePadé andCompared by Homotopy Perturbation Method (HPM) and Numerical MethodParticleMaterial

Diameter(mm) Numerical

DTMePadé[8,8] HPM

Aluminum 1 0.1888759074 0.1771953458 0.18887432123 0.3585504311 0.3581483507 0.35854256215 0.4683308257 0.4682662708 0.46832679581 0.4332008767 0.4336398188 0.4332014752

Copper 3 0.7829089418 0.7828929076 0.78289215315 1.0156731422 1.015681926 1.01567210431 0.4975607949 0.4837738933 0.4965829147

Lead 3 0.8944267047 0.8944040156 0.89441598565 1.1589079584 1.158902486 1.1589059863

DTMePade, differential transformation method with Padé approximation.


It is completely evident that in different industrial process and appli-cations, diameter of the particle maybe varied by a known function whichdepends on its solubility. In this study, it is considered that diameter variesthrough a linear function, so for other functions it can be solved easily tooby the same method.

DðtÞ ¼ D0 � _D$t (7.102)

where D0 is the initial diameter, and _D is the reduction rate of the diameterdue to particle solubility. By substituting to Eq. (7.101),

112

p�D0 � _D$ t

�3ð2rs þ rÞ duðtÞdt

þ 3p�D0 � _D$t

�muðtÞ

þ 116

p�D0 � _D$ t

�2ruðtÞ2

�16p�D0 � _D$ t

�3gðrs � rÞ ¼ 0

(7.103)

Eq. (7.103) is a nonlinear equation with an initial condition (u(0) ¼ 0)which can be solved by numerical and analytical methods. Here,DTMePade and RungeeKutta methods are presented for solving theproblem. A schematic of described problem is shown in Fig. 7.36.

Figure 7.36 Schematic view of soluble particle settling in Newtonian media while itsdiameter reduces by a linear function due to dissolve in fluid.


Because the procedure of solving Eq. (7.103) is autonomous of constantcoefficients 1

12pð2rs þ rÞ, 3pm, 116pr,

16pgðrs � rÞ, and D0, for general-

ization and simplification of the problem for future cases with differentphysical conditions, all the constants are assumed to be unity. The problemwill be solved by this assumption and for _D ¼ 0:4:0:2; 0:1 mm

s . For instancewhen _D ¼ 0:2 mm

s , Eq. (7.103) will be rewritten as,

duðtÞdt

� 0:6

�duðtÞdt

�t þ 0:12

�duðtÞdt

�t2 � 0:008

�duðtÞdt

�t3þ

uðtÞ � 0:2uðtÞt þ uðtÞ2 � 0:4uðtÞ2t þ 0:04uðtÞ2t2 � 1þ 0:6t

�0:12t2 þ 0:008t3 ¼ 0

(7.104)

by applying DTMePadé to Eq. (7.104),

ðkþ 1ÞUðkþ 1Þ þ UðkÞ � 0:6Pkl¼0

ðl þ 1ÞUðl þ 1Þdk�l�1

þ0:12Pkl¼0

ðl þ 1ÞUðl þ 1Þdk�l�2

�0:008Pkl¼0

ðl þ 1ÞUðl þ 1Þdk�l�3 � dk � 0:2Xkl¼0

UðlÞdk�l�1 þXkl¼0

UðlÞUðk� lÞ

�0:4Pkk1¼0

Pk1l¼0

UðlÞUðk1� lÞdk�k1�l þ 0:04Xkk1¼0

Xk1l¼0

UðlÞUðk1� lÞdk�k1�2

þ0:6dk�1 � 0:12dk�2 þ 0:008dk�3 ¼ 0

(7.105)

where U(k) is DTM transformed forms of u(t) in Eq. (7.103). After solvingEq. (7.105) using initial condition, particle’s velocity, u(t), will be appearedas fallows,

uðtÞ ¼ 1$ t � 0:5t2 � 0:3t3 þ 0:295t4 þ 0:0726t5 � 0:15577t6

þ0:00074381t7 þ 0:073487t8 � 0:01683t9 � 0:030966t10(7.106)

Eq. (7.106) is DTM solution of Eq. (7.103). For increasing theconvergency of DTM, Padé approximation with accuracy [6,6] is applied,and Eq. (7.106) is transformed to the following equation,

uðtÞ ¼ ð0:999999t � 1:35275t2 þ 884:7181t3 � 334:2273t4 þ 109:575t5 � 21:7561t6Þð0:99999� 0:85275t þ 884:5917t2 þ 107:52t3 þ 428:89t4 � 35:793t5 þ 14:698t6Þ

(7.107)

Results of DTMePadé [6,6] and numerical solution are presented inFig. 7.37.


As seen DTMePadé has an excellent agreement with numerical solu-tion. Table 7.12 reveals that DTMePadé results are completely acceptable;furthermore, it is possible to increase its accuracy by increasing the Padéorder. As seen in Fig. 7.37 for soluble particle, against the insoluble andrigid particle, no terminal velocity is observed and velocity increases to amaximum value and then starts to reduce until it reaches zero. According tothis figure, by increasing the solubility rate _D, maximum velocity increases,

Num.

D =0.4, 0.2, 0.1 mm/s

DTM-Pade [6,6]

00

0.1

0.2

0.3

0.4

0.5

2 4 6 8 10t

u(t)

Figure 7.37 Velocity function for particle in three different reduction rate of diameter�_Dðmm=sÞÞ with numerical an DTMePade [6,6]. DTM, differential transformation

method; DTMePadé, differential transformation method with Padé approximation.

Table 7.12 Comparison Between DTMePadé [6,6] and RungeeKutta NumericalResults for u(t)t Numerical RungeeKutta Method DTMePadé [6,6]

0 0.000 0.0000.5 0.356011026288972 0.35601094551.0 0.455673617777084 0.45566637571.5 0.424741800010445 0.42459438502.0 0.348494696237181 0.34740655532.5 0.260477668651496 0.25572160633.0 0.174182615316337 0.15901575553.5 0.099044601445866 0.060456979754.0 0.043185698190710 0.037995445524.5 0.010414005234102 0.01345128694

DTMePadé, differential transformation method with Padé approximation.


but occurs some later. Positions and sizes of the particle are depictedgraphically in Fig. 7.38 that demonstrates the falling process of the particleuntil it completely dissolves in Newtonian fluid media and disappears in thefluid.

REFERENCES[1] Hatami M, Ganji DD. Motion of a spherical particle on a rotating parabola using

Lagrangian and high accuracy Multi-step Differential Transformation Method. PowderTechnology 2014;258:94e8.

[2] Hatami M, Sheikholeslami M, Domairry G. High accuracy analysis for motion of aspherical particle in plane Couette fluid flow by Multi-step Differential TransformationMethod. Powder Technology 2014;260:59e67.

[3] Dogonchi AS, Hatami M, Domairry G. Motion analysis of a spherical solid particle inplane Couette Newtonian fluid flow. Powder Technology 2015;274:186e92.

[4] Dogonchi AS, Hatami M, Hosseinzadeh K, Domairry G. Non-spherical particlessedimentation in an incompressible Newtonian medium by Padé approximation.Powder Technology 2015;278:248e56.

[5] Hatami M, Ganji DD. Motion of a spherical particle in a fluid forced vortex by DQMand DTM. Particuology 2014;16:206e12.

[6] Hatami M, Ganji DD, Jafaryar M, Farkhadnia F. Transient combustion analysis for ironmicro-particles in a gaseous media by weighted residual methods (WRMs). Case Studiesin Thermal Engineering 2014;4:24e31.

[7] Nouri R, Ganji DD, Hatami M. Unsteady sedimentation analysis of spherical particlesin Newtonian fluid media using analytical methods. Propulsion and Power Research2014;3(2):96e105.

[8] Hatami M, Domairry G. Transient vertically motion of a soluble particle in a New-tonian fluid media. Powder Technology 2014;253:481e5.

Figure 7.38 Positions of falling particle for different rates of solubility, D0 ¼ 1 mm, andtime interval ¼ 1 s.


CHAPTER 8

DTM for Solid Mechanics,Vibration, and Deflection

8.1 INTRODUCTION

Most scientific problems in solid mechanics are inherently nonlinear bynature, and, except for a limited number of cases, most of them do not haveanalytical solutions. Accordingly, the nonlinear equations are usually solvedusing other methods including numerical techniques or by using analyticalmethods such as differential transformation method (DTM). Thereforeobtaining analytical limit state functions or using analytical techniques toobtain reliability index for nonlinear problems is almost impossible.Analytical methods which recently are widely used are one of the simpleand reliable methods for solving system of coupled nonlinear differentialequations. The most nonlinear problems in solid mechanics are vibrationanalysis, deflection and deformation of different beams, materials, or plates.In this chapter some of these problems are presented and solved by DTM,which are categorized in the following sections:8.1 Introduction8.2 Deflection Prediction of a Cantilever Beam8.3 Vibration Analysis of Stepped FGM Beams8.4 Piezoelectric Modal Sensors for Cantilever Beams8.5 Damped System With High Nonlinearity8.6 Free Vibration of a Centrifugally Stiffened Beam8.7 Deflections of Orthotropic Rectangular Plate8.8 Free Vibration of Circular Plates8.9 Vibration of Pipes Conveying Fluid8.10 Piezoelectric Modal Sensor for Nonuniform EulereBernoulli Beams

With Rectangular Cross Section8.11 Free Vibrations of Oscillators8.12 Composite Sandwich Beams With Viscoelastic Core



http://dx.doi.org/10.1016/B978-0-12-805190-0.00008-5

8.2 DEFLECTION PREDICTION OF A CANTILEVER BEAM

A cantilever beam OA is subjected to coplanar loading consisting of an axialcompressive force FA and of a transverse force QA (Fig. 8.1). FA and QA arefollower forces, i.e., they will rotate with the end section A of the beamduring the deformation, and they will at all times remain tangential andperpendicular, respectively, to the deformed beam axis. It is assumed thatthe effect of the material nonlinearity is negligible in the mathematicalderivation. Therefore at any point of coordinates x(s) and y(s) the externalmoment M is expressed as [1]:

M ¼ ðFA cos qA þQA sin qAÞðyA$yÞ þ ðFA sin qA þQA cos qAÞðxA$yÞ(8.1)

where x and y are the longitudinal and transverse coordinates, respec-tively; q is the slope of the normal to the beam cross section; and xA,yA, and qA denote the coordinates and the normal slope at the end sec-tion. The classical EulereBernoulli hypothesis assumes that the bendingmoment M at any point of the beam is proportional to the correspondingcurvature, i.e.,

M ¼ EI q0 (8.2)

where E is Young’s modulus and I is the area moment of inertia of thebeam cross section about the x-axis. By using the following relations,

dxds

¼ cos q;dyds

¼ sin q (8.3)

QA

QA

FA

L

O

y

s

x

FA

θ

Figure 8.1 The geometry of a cantilever beam subjected to nonconservative externalloading (follower forces).


and based on the trigonometric relations and by substituting Eqs. (8.2) and(8.3) into Eq. (8.1), the nonlinear differential equation governing the prob-lem is obtained as follows,

v2h

vs2þ FA

EIsinðhðsÞÞ þQA

EIcosðhðsÞÞ ¼ 0 (8.4)

The boundary conditions associated with the above equation are,

hð0Þ ¼ 0; h0ðLÞ ¼ 0 (8.5)

where

h ¼ q� qA (8.6)

Using only the two terms of a Taylor’s series expansion for cos(h(s)) andsin(h(s)), and substituting in Eq. (8.4) yields,

v2h

vs2þ FA

EI

�hðsÞ � 1

6h3ðsÞ

�þQA

EI

�1� 1

2h2ðsÞ

�¼ 0 (8.7)

Now DTM is applied from Chapter 1 into Eq. (8.7) to find h(s). So,

ðkþ 1Þðkþ 2ÞHðkþ 2Þ þ FA

EIHðkÞ

��FA

EI

��16

� Pkk1¼0

Pk1l¼0

HðlÞHðk1� lÞHðk� k1Þ

þQA

EId½k� �

�QA

EI

��12

�Xk

l¼0

HðlÞHðk� lÞ ¼ 0

(8.8)

where

d½k� ¼�1 if k ¼ 0

0 if ks0(8.9)

Similarly, the transformed form of boundary conditions (Eq. 8.5) can bewritten as,

Hð0Þ ¼ 0; Hð1Þ ¼ a (8.10)

in which the boundary condition of this case is the same as that of the pre-vious case. By solving Eq. (8.8) and using boundary conditions Eq. (8.10),

DTM for Solid Mechanics, Vibration, and Deflection 339

the DTM terms for this case for FA ¼ 300 KN, QA ¼ 350 KN, E ¼ (40)GPa, and I ¼ (1.6)(10�5) m4 can be,

Hð2Þ ¼ �0:2734375000

Hð3Þ ¼ �0:7812500000a

Hð4Þ ¼ 0:1068115234þ 0:02278645833a2

Hð5Þ ¼ �0:005645751950aþ 0:003906250000a3

(8.11)

etc.Now by applying Eq. (8.6) into Eq. (8.11) and using Eq. (8.10), the

constant parameter “a” will be obtained so that the slope parameterequation will be estimated,

hðsÞ ¼ a$s� 0:2734375000s2 � 0:07812500000a$s3

þð0:01068115234þ 0:02278645833a2Þs4þð�0:005645751950aþ 0:003906250000a3Þs5þð0:0005145867667� 0:003916422527a2Þs6þð0:0008974756514a� 0:0001828632660a3Þs7þð�0:00006135087461þ 0:0001701215902a2 þ 0:0001335144043a4Þs8

(8.12)

By using boundary condition in s ¼ 1, the “a” parameter will bedetermined as,

a ¼ 0:6310718570 (8.13)

and by substituting it into Eq. (8.12), h(s) will be found as,

hðsÞ ¼ 0:6310718570s� 0:2734375000s2 � 0:04930248883s3

þ 0:01975589785s4 � 0:002581135196s5 � 0:001045135118s6

þ 0:0005204134364s7 þ 0:00002757630897s8(8.14)

In this problem, deflection of a cantilever beam subjected to static coplanarloading is compared by two analytical methods called homotopy perturbationmethod (HPM) and DTM and fourth-order RungeeKutta numerical method.For showing the efficiency of these analytical methods, Fig. 8.2 is presented.

As seen in most cases of these figures, HPM and DTM have an excellentagreement with numerical solution but in some cases (for example, whenFA ¼ 100 KN, QA ¼ 100 KN, E ¼ 109 Pa, and I ¼ 10�4 m4, see Fig. 8.2)the accuracy of the DTM is greater than the HPM. Table 8.1 compared the


results of the HPM and DTM with numerical procedure and homotopyanalysis method which is presented in the literature when FA ¼ 300 KN,QA ¼ 350 KN, E ¼ (40) GPa, and I ¼ (1.6)(10�5) m4.

These data and calculated errors confirm that these two analyticalmethods are suitable and semi-exact for solving these kinds of problems. Inthe following step, effect of some parameters appeared in the mathematicalformulation such as area moment of inertia (I), Young’s modulus (E),transverse force (QA), and compressive force (FA) on slope variation areinvestigated. Selected values for showing the variation of these parametersare considered according to the mean value of stress limit state concludedfrom [1]. These mean values are presented in Table 8.2.

Effect of compressive force (FA) on slope parameter, h(s), is presented inFig. 8.3. As seen in this figure, slope parameter increases as well ascompressive force increases. Variation of slope parameter based on

00 0.2 0.4 0.6 0.8 1

0.1

0.2

0.3

η(s)

s

0.4

0.5

Numerical

I. II, III, IV

DTMHPM

0.6

Figure 8.2 Comparison of slope parameter for homotopy perturbation method (HPM),differential transformation method (DTM), and numerical solution when

I : FA ¼ 400 KN; QA ¼ 450 KN; E ¼ 40 GPa and I ¼ 7� 10�5 m4

II : FA ¼ 300 KN; QA ¼ 350 KN; E ¼ 40 GPa and I ¼ 1:6� 10�5 m4

III : FA ¼ 100 KN; QA ¼ 50 KN; E ¼ 109 Pa and I ¼ 10�4 m4

IV : FA ¼ 100 KN; QA ¼ 100 KN; E ¼ 109 Pa and I ¼ 10�4 m4


Table 8.1 Comparison Between Homotopy Perturbation Method (HPM), Differential Transformation Method (DTM), and NumericalSolution for FA ¼ 300 KN, QA ¼ 350 KN, E ¼ (40) GPa, and I ¼ (1.6)(10�5) m4

s h(s)Numerical h(s)HAM h(s)HPM h(s)DTM (%) ErrorHPM (%) ErrorDTM

0.1 0.06036251 0.06034326 0.06035369 0.06032545 0.00014625 0.000613970.2 0.11498708 0.11495123 0.11497007 0.11491317 0.00014793 0.000642740.3 0.16364459 0.16356398 0.16362082 0.16353411 0.00014527 0.000675080.4 0.20614594 0.20615968 0.20611719 0.20599929 0.00013944 0.000711390.5 0.24233792 0.24234267 0.24230515 0.24215566 0.00013523 0.000752070.6 0.27209923 0.27215896 0.27206024 0.27188220 0.00014331 0.000797630.7 0.29533656 0.29535986 0.29528238 0.29508623 0.00018348 0.000847600.8 0.31198116 0.31195631 0.31189042 0.31170062 0.00029085 0.000899210.9 0.32198607 0.32195632 0.32181657 0.32168184 0.00052643 0.000944861 0.32532409 0.32501035 0.32500059 0.32500948 0.00099440 0.00096706


ationMethod


Problems

transverse force (QA) is depicted in Fig. 8.4. By increasing the QA, slopeparameter is increased significantly. From these two figures, it is evident thatthe effect of QA for increasing the slope parameter is more than FA. It is dueto their direction applied to beam which as Fig. 8.1 shows, QA is a shearforce and make more effects on beam slope parameter.

8.3 VIBRATION ANALYSIS OF STEPPED FGM BEAMS

Two types of stepped FGM beams made from ceramicemetal phases arechosen to investigate their vibration behavior. The geometries and de-scriptions of the beam types are shown in Fig. 8.5. The beams shown in thisfigure are supported by elastic conditions at both ends including

Table 8.2 Mean Values for Parameters According toStress Limit State Presented in Ref. [1]Parameter Mean Value

Shear force (QA) 350 (KN)Normal force (FA) 300 (KN)Young’s modulus (E) 40 (GPa)Area moment of inertia (I) 2.6 E-5 (m4)

Figure 8.3 Effect of compressive force for QA ¼ 350 KN, E ¼ (40) GPa andI ¼ (1.6)(10�5) m4.


translational and rotational springs which are defined as the EeE beams. Itis known that FGMs are inhomogeneous spatial composite materials,typically composed of a ceramicemetal pair of materials. The materialcompositions are varied throughout the thickness direction from the top

Figure 8.4 Effect of transverse force for FA ¼ 300 KN, E ¼ (40) GPa andI ¼ (1.6)(10�5) m4.

kR1

W1x1

L1 h1

x2

w2

p2

p2

p1

p1

L2h2 kR2

kR2

kT2

kT2

L

kT1

kR1

kT1

ceramic

ceramic

ceramic/metal

ceramic/metal

FGM Type-I (p1=p2)

FGM Type-II (p1≠p2)ceramic

metal

metalmetalSection 1 Section 2

Figure 8.5 Two types of FGM beams under study.


surface (ceramic 100%emetal 0%) to the bottom one (ceramic 0%emetal100%). A schematic of these two beams is shown in Fig. 8.5 [2].

More assumptions of volume fractions of ceramic based on the powerlaw distribution of the stepped FGM beams can be found in Ref. [2].Consider classical beam theory: the partial differential equation used todescribe the free vibration in each section of the stepped FGM beams canbe expressed as:

v4wjðxj; tÞvx4j

þ Iojlj

v2wjðxj; tÞvt2

� 0; xj ˛ ½0;L� ðj ¼ 1; 2Þ (8.15)

It is defined that Ioj is the moment of inertia, lj ¼�D11 � B2

11A11

�is the

material stiffness coefficient in each beam section. For harmonic vibration,wj(xj, t) ¼Wj(xj)e

iwt is substituted into Eq. (8.15) to obtain a time-independent governing equation as follows.

d4WjðxjÞdx4j

� Iojlju2WjðxjÞ � 0 (8.16)

where u is a natural frequency. On the principle of the DTM, the governingdifferential equation and boundary condition equations as well as the conti-nuity conditions are transformed into a set of algebraic equations using trans-formation rule. The basic operations required in differential transformationfor the governing differential equations, boundary conditions, and the con-tinuity conditions are shown in Tables 2 and 3 in Ref. [2], respectively.The general function, fj(xj) is considered as the transverse displacementWj(xj):

Wj½r þ 4� ¼ Ioju2

ljðr þ 1Þðr þ 2Þðr þ 3Þðr þ 4ÞWj½r�. (8.17)

It can be seen that Eq. (8.17) is independent from boundary conditions.Therefore to obtain frequency results, the displacement function of Eq. (8.17)must be used to satisfy the corresponding boundary equations. To demon-strate the application of the DTM to vibration response of the stepped FGMbeam, let us consider the EeE beams as shown in Fig. 8.5 which have elasticboundary conditions at both ends. The governing equation in the form of therecurrence equation for Section 1 of the beams can be expressed as:

W1½r þ 4� ¼ I01u2

l1ðr þ 1Þðr þ 2Þðr þ 3Þðr þ 4ÞW1½r� (8.18)


For this case, the boundary conditions at the left end can be expressed as:

d3W1

dx31þ kT1

l1W1 ¼ 01

d2W1

dx21

kR1l1

dW1

dx1¼ 01 (8.19)

where kT1 and kR1 are the translational spring and rotational spring con-stants at the left end, respectively. Let the nonzero values of deflectionand slope at x1 ¼ 0 be C0 and C1, respectively. Applying the basic opera-tions of DTM for these nonzero quantities at x1 ¼ 0, one obtains:

W1½0� �W1ðx1Þ � C0 W1½1� � dW1ðx1Þdx1

� C1 (8.20)

The expression for nonzero values of bending moment and shear forceat x1 ¼ 0 can be written as:

W1½2� � kR12l1

C1; W1½3� � kR16l1

C0 (8.21)

To find W1(r) for all values of r, the components in Eqs. (8.20) and(8.21) are substituted into the recurrence equation in Eq. (8.18).

W1½4r� � u2r I r01lr1ð4rÞ!

C0 r ¼ 0; 1; 2; 3;. (8.22a)

W1½4r þ 1� � u2r I r01lr1ð4r þ 1Þ!C1 r ¼ 0; 1; 2; 3;. (8.22b)

W1½4r þ 2� � u2r I r01kR1

lðrþ1Þ1 ð4r þ 2Þ!

C1 r ¼ 0; 1; 2; 3;. (8.22c)

W1½4r þ 3� � u2r I r01kT1


C0 r ¼ 0; 1; 2; 3;. (8.22d)

Next procedure is given for considering Section 2 of the beam,therefore, the governing recurrence equation of this section is

W2½r þ 4� � I02u2

l2ðr þ 1Þðr þ 2Þðr þ 3Þðr þ 4ÞW2½r� (8.23)

To consider the boundary conditions at the right end (x2 ¼ 0), theconditions can be expressed as:

d3W2

dx32þ kT2

l2W2 � 0

d2W2

dx22� kR2

l2

dW2

dx2� 0 (8.24)


where kT2 and kR2 are the translational spring and rotational spring con-stants at the right end, respectively. The nonzero values of the deflectionand slope account for C2 and C3, respectively. Again, applying thebasic operations of DTM for the boundary condition at (x2 ¼ 0), oneobtains:

W2½0� ¼ W2ðx2Þ ¼ C2 W2½1� ¼ dW2ðx2Þdx2

¼ C3 (8.25)

The expression for nonzero values of bending moment and shear forceat x2 ¼ 0 can be written as:

W2½2� ¼ kR22l2

C3 W2½3� ¼ kR16l1

C2 (8.26)

Again by using Eqs. (8.25) and (8.26) with Eq. (8.23), the expressions ofW2(r) for all values of r can be written as follows.

W2½4r� ¼ u2r I r02lr2ð4rÞ!

C2 r ¼ 0; 1; 2; 3;. (8.27a)

W2½4r þ 1� ¼ u2r I r02lr2ð4r þ 1Þ!C3 r ¼ 0; 1; 2; 3;. (8.27b)

W2½4r þ 2� ¼ u2r I r02kR2


C3 r ¼ 0; 1; 2; 3;. (8.27c)

It is assumed that, for an FGM beam having discontinuous cross section,stress concentration at the interchange or at the step location of such beamis neglected. Hence, the continuity conditions are

W1ðL1Þ �W2ðL2Þ; dW1ðL1Þdx1

� dW2ðL2Þdx2

(8.28a)

d2W1ðL1Þd2x1

� l2

l1

d2W2ðL2Þd2x2

;d3W1ðL1Þ

d3x1� l2

l1

d3W2ðL2Þd3x2

(8.28b)

According to the principle of the DTM, the continuity conditions aretransformed into algebraic equations as well. The results of the trans-formation can be expressed as,

W1½r�LðrÞ1 �W2½r�LðrÞ

2 ¼ 0 (8.29)


W1½r�rLðr1Þ1 þW2½r�rLðr1Þ

2 ¼ 0 (8.30)

W1½r�rðr � 1ÞLðr2Þ1 � dW2½r�rðr � 1ÞLðr2Þ

2 ¼ 0 (8.31)

W1½r�rðr � 1ÞLðr3Þ1 þ dW2½r�rðr � 1ÞLðr3Þ

2 ¼ 0 (8.32)

where d ¼ l2l1. The components of Eqs. (8.22) and (8.27) are substituted into

the transformed continuity conditions in Eqs. (8.29)e(8.32). The results ofthis substitution can be arranged and presented in the matrix form asfollows:

26664e11 e12 e13 e14e21 e22 e23 e24e31 e32 e33 e34e41 e42 e43 e44

3777526664c0c1c2c3

37775 ¼ 0 (8.33)

where the elements in the matrix are

2 (4 ) 2 (4 3)01 1 01 1

11 1 ( 1)0 01 1

2 (4 1) 2 (4 2)01 1 01 1

12 1 ( 1)0 01 1

2 (4 ) 2 (4 3)02 2 02 1

13 1 ( 1)0 2 2

(4 )! (4 3)!

(4 1)! (4 2)!

(4 )! (4

r r r r r r

Tr rr r

r r r r r r

Rr rr r

r r r r r r

Tr rr

I L I Le kr r

I L I Le kr r

I L I Le kr r

ω ωλ λ

ω ωλ λ

ω ωλ λ

+∞ ∞

+= =

+ +∞ ∞

+= =

+∞

+=

= −+

= ++ +

= ++

∑ ∑

∑ ∑

∑0

2 (4 1) 2 (4 2)02 2 02 1

14 2 ( 1)0 02 2

2 (4 2) 2 (4 1)01 1 01 1

21 10 11 1

2 (4 ) 2 (4 1)01 1 01 1

22 10 1 1

3)!

(4 1)! (4 2)!

(4 2)! (4 1)!

(4 )!

r

r r r r r r

Rr rr r

r r r r r r

T r rr r

r r r r r r

Rr rr

I L I Le kr r

I L I Le kr r

I L I Le kr

ω ωλ λ

ω ωλ λ

ω ωλ λ

∞

=

+ +∞ ∞

+= =

+ −∞ ∞

= =

+∞

=

= −+ +

= ++ −

= +

∑

∑ ∑

∑ ∑

∑0

2 (4 2) 2 (4 1)02 2 02 2

23 2 ( 1)0 12 2

2 (4 ) 2 (4 1)02 2 02 2

24 2 ( 1)0 02 2

2 (4 1) 201 1 01

31 1 ( 1)0 1

(4 1)!

(4 2)! (4 1)!

(4 )! (4 1)!

(4 1)!

r

r r r r r r

T r rr r

r r r r r r

Rr rr r

r r r r

T rr

rI L I Le k

r rI L I Le k

r rI L Ie k

r

ω ωλ λ

ω ωλ λ

ω ωλ

∞

=

+ −∞ ∞

+= =

+∞ ∞

+= =

+∞

+=

+

= ++ −

= ++

= ++

∑

∑ ∑

∑ ∑

∑(4 2)1

1 12 (4 ) 2 (4 1)

01 1 01 132 1 ( 1)

0 11 1

(4 2)!

(4 )! (4 1)!

r r

rr

r r r r r r

R r rr r

Lr

I L I Le kr r

λ

ω ωλ λ

−∞

=

−∞ ∞

+= =

−

= +−

∑

∑ ∑

(8.34)


2 (4 1) 2 (4 2)02 2 02 2

33 2 ( 1)0 12 2

2 (4 ) 2 (4 1)02 2 02 2

34 2 ( 1)0 12 2

2 (4 ) 2 (4 3)01 1 01 1

41 1 ( 1)0 1 1

(4 1)! (4 2)!

(4 )! (4 1)!

(4 )!

r r r r r r

T r rr r

r r r r r r

R r rr r

r r r r r r

T r rr

I L I Le kr r

I L I Le kr r

I L I Le kr

ω ωδ δλ λ

ω ωδ δλ λ

ω ωλ λ

+ −∞ ∞

+= =

−∞ ∞

+= =

−∞

+=

= −+ −

= −−

= +

∑ ∑

∑ ∑

∑1

2 (4 2) 2 (4 1)01 1 01 1

42 1 ( 1)1 11 1

2 (4 ) 2 (4 3)02 2 02 2

43 2 ( 1)0 12 2

2 (4 2) 202 2 02

44 21 2

(4 3)!

(4 2)! (4 1)!

(4 )! (4 3)!

(4 2)!

r

r r r r r r

Rr rr r

r r r r r r

T r rr r

r r r r

Rrr

rI L I Le k

r rI L I Le k

r rI L Ie k

r

ω ωλ λ

ω ωδ δλ λ

ω ωδ δλ

∞

=

− −∞ ∞

+= =

−∞ ∞

+= =

−∞

=

−

= +− −

= +−

= +−

∑

∑ ∑

∑ ∑

∑(4 1)2

( 1)1 2 (4 1)!

r r

rr

Lrλ

−∞

+= −∑

(8.35)

To obtain a nontrivial solution, the determinant of coefficient matrix inEq. (8.33) could be set equal to zero. It is also noted that, for practicalcalculation, the finite number of r terms in each element of the matrix inEq. (8.34) should be used. Therefore a convergence study will be per-formed to determine an appropriate maximum number of r in the upperlimit of this equation. Mode shapes of the stepped FGM beams can beplotted by setting C0 to unity in Eq. (8.33), so that the remaining nonzeroconstants (C1, C2, C3) are solved. Thus, the mode shapes corresponding toany frequency can be expressed as the function of WjðxjÞ ¼

PNr¼0 x

rjWj½r�.

To plot the mode shapes with the whole length coordinate (x), forexample, in the case of EeE beam, its mode shape functions are

W1ðxÞ ¼XNr¼0

I r01u2r

lr1ð4rÞ!xð4rÞ þ C1

XNr¼0

I r01u2r

lr1ð4r þ 1Þ!xð4rþ1Þ

þC1kR1XNr¼0

I r01u2r


xð4rþ2Þ � kT1

XNr¼0

I r01u2r

lð4rþ1Þ1 ð4r þ 3Þ!

xð4rþ3Þx˛½0;L1�

W2ðL � xÞ ¼ C2

XNr¼0

I r02u2r

lr2ð4rÞ!ðL � xÞð4rÞ þ C3

XNr¼0

I r02u2r

lr2ð4r þ 1Þ!ðL � xÞð4rþ1Þ

þC3kR2XNr¼0

I r02u2r


ðL � xÞð4rþ2Þ

�C2kT2

XNr¼0

I r02u2r


ðL � xÞð4rþ3Þx˛½L1;L�

(8.36)


By following the same procedure, one can solve the vibration problemof the stepped FGM beams with other kinds of boundary conditions. Thematrix elements for other boundary conditions such as clampedeelasticsupported (CeE) and simply supportedeelastic supported (SeE) beams arepresented in Appendix A in Ref. [2]. In this section, the stepped FGMbeams made from alumina (Al2O3) and aluminum (Al) are chosen toinvestigate their vibration behavior. The mechanical properties of thematerials are:

For ceramic: (Al2O3), Ec ¼ 380 GPa, qc ¼ 3960 kg/m3, m ¼ 0.3.For metal: (Al), Em ¼ 70 GPa, qm ¼ 2702 kg/m3, m ¼ 0.3.The translational and rotational spring constants can be obtained by

using the following forms kTj ¼ bTjl1

L3 and kRj ¼ bRjl1

L in which bTj and bRjare the translational and rotational spring factors. To clearly illustrate thechanges of frequencies owing to the variations of spring constant factors,Fig. 8.6 plots the first to fourth frequency results of EeE beams against thespring constant factors (bT1 ¼ bR1 ¼ bT2 ¼ bR2). As can be observed, allfrequencies remain constant when the beams are supported by soft springs,using small values of spring constant factors. The frequencies increaseconsiderably as the increase of spring constant factors is in the range ofmoderate spring stiffness bT1 ¼ bR1 ¼ bT2 ¼ bR2 ¼ 102 / 104.

Mode shape functions for CeE, SeE, and EeE beams provided in thisproblem are used to plot and illustrate the vibration mode shapes of steppedFGM beams in Fig. 8.4. The first to fourth mode shapes shown in this

(a) (b)Ω1

Ω Ω

Ω2Ω3

80

70

60

50

40

30

20

10

0

70

60

50

40

30

20

10

0–6 –4 –2 2 4 60 –6 –4 –2 2 4 60

Ω4

Ω1Ω2Ω3

Ω4

log10 (T1 R1 R2T2β β β β )= = = log10 (

T1 R1 R2T2β β β β )= = =

Figure 8.6 Dimensionless frequencies of EeE beams with the variations of springconstant factors. (a) FGM Type-I. (b) FGM Type-II (L/h ¼ 20; L1 ¼ 0.5 L; n ¼ 0.5;N ¼ n ¼ 0.5).


figure are obtained from FGM Type-I beams, using bT1 ¼ bR1 ¼ bT2 ¼bR2 ¼ 10. Fig. 8.7(a) is for the mode shapes of CeE beams, (b) for SeEbeams, and (c) for EeE beams.

8.4 PIEZOELECTRIC MODAL SENSORS FORCANTILEVER BEAMS

In general, shaped piezoelectric sensors made of polyvinylidene fluoride(PVDF) are chosen since these add little loading to light structures and inaddition are easy to cut into desired shapes. The PVDF sensors are designedby shaping the surface electrode, whereby the output of the sensor can bemade sensitive to selected modal coordinates, other modal coordinates maybe filtered out. Consider a beam with length L, width b, and thickness h.A shaped PVDF film of constant thickness is attached onto the top surfaceand spanned across the entire length of the beam, as shown in Fig. 8.8.

Assuming that the PVDF sensor thickness is much smaller than beamthickness h, the mass and stiffness of the sensor is then negligible compared

–10

10

20

30

40

0

02

–2

4

–4

6

–6

8

3

2

1

–1

0

5

5 10 15 20

5 10 15 20

10x

x

x15 20

(c)

(a) (b)

W (x

)

W (x

)

W (x

)1st mode2nd mode

3rd mode4th mode

1st mode2nd mode

3rd mode4th mode

1st mode2nd mode

3rd mode4th mode

Figure 8.7 The first to fourth mode shapes of FGM Type-I beams. (a) Mode shapes ofCeE beams, (b) SeE beams, and (c) EeE beams.


to the properties of the beam. The output charge Q(t) of the PVDF sensorcan be expressed as [3]:

QðtÞ ¼ h2

ZL

0

FðxÞ$�e31

v2wðx; tÞvx2

�dx (8.37)

where e31 is the PVDF sensor stress/charge coefficient, w(x, t) is thedisplacement of the beam, and F(x) is the PVDF film shape function.The displacement distribution of the vibrating beam may be representedby the form of a modal expansion:

wðx; tÞ ¼XNn¼1

hnðtÞfnðxÞ (8.38)

where hn(t) and fnðxÞ are the nth modal coordinates and structural modeshape function. N is the index for the highest order structural mode.Substituting Eq. (8.38) into Eq. (8.37), and expressing in dimensionlessform, one obtains:

QðtÞ ¼ h2L

e31XNn¼1

hnðtÞZL

0

FðxÞ$v2fnðxÞvx2

dx (8.39)

where X ¼ xL ; FnðXÞ ¼ fnðxÞ. We make an approximation by expanding

F(X) as a linear function of the second derivative of the mode shape func-

tiond2fjðxÞdx2 .

FðxÞ ¼XNj¼1

Bjv2fjðxÞvX2

(8.40)

y

0

PVDF

F(x)

bx

h/2h/2 x

L

Beam

0

(a)

(b)

Figure 8.8 A shaped polyvinylidene fluoride (PVDF) film bonded on a beam(dimensions are not scaled). (a) Top view; (b) front view.


where Bj are the unknown shape coefficients for the PVDF sensor.Substituting Eq. (8.40) into Eq. (8.39), we obtain

QðtÞ ¼ h2L

e31XNn¼1

XNj¼1

Bk

Z1

0

v2fnðxÞvX2

$v2fjðxÞvX2

dX$hnðtÞ (8.41)

Eq. (8.41) can be simplified as a matrix form:

QðtÞ ¼ BTkh (8.42)

where B and g are the vectors. K is an N � N matrix with element.

kðn; jÞ ¼ h2L

e31

Z1

0

v2fnðxÞvX2

$v2fjðxÞvX2

dX (8.43)

To construct a shaped PVDF sensor that accurately measures the Jthstructural mode, the output signal of the sensor Q(t) should be directlyproportional to the Jth modal coordinate hj(t) and orthogonal to othermodes. For simplicity, we set Q(t) ¼ hj(t) to obtain

hJðtÞ ¼ BTkh (8.44)

From Eq. (8.44), the PVDF shape coefficients can be obtained:

BT ¼ k�1ðJ ; 1 :NÞ (8.45)

In accordance with the above analysis, to design a modal sensor, thesecond spatial derivative of the mode shapes must be obtained. This willbe solved by using the DTM technique. For this purpose, one particularcase, say, of a cantilever beam with intermediate support is considered, asshown in Fig. 8.9. The beam is divided into two sections with the twomirror systems of reference x1 and x2. The positive direction of the spatialcoordinate x1 is defined in the direction to the right for Section 1, and x2is defined in the direction to the left for Section 2. The ordinary dif-ferential equation describing the free vibration in each section is asfollows:

d4fjðxjÞdx4j

� msu2

EIfjðxjÞ ¼ 0 xj ˛ ½0$Lj�; ðj ¼ 1; 2Þ (8.46)

where subscripts j ¼ 1 and 2 denote Sections 1 and 2 of the beam, respec-tively. E is Young’s modulus, I is the cross-sectional moment of inertia of


the beam I ¼ bh312 , ms ¼ rbh is the mass per unit length, and r is the density

of the beam. Introduction of dimensionless variables,

Xj ¼ xjL; FjðXjÞ ¼

fjðxjÞL

(8.47)

can be rewritten in dimensionless form,

d4fjðxjÞdx4j

� U4fjðxjÞ ¼ 0 (8.48)

where U4 ¼ msu2L4

EI U is the dimensionless natural frequency and the nthnatural frequency is denoted as U(n). According to the DTM principle,fj(xj) in Eq. (8.48) can be expressed in differential transformationformulation:

fjðxjÞ ¼XMm¼0

Xmj fjðmÞ (8.49)

And the second spatial derivative of the mode shapes can be expressed as

d2fjðxjÞdx2j

¼XMm¼0

ðmþ 1Þðmþ 2ÞXmj fjðmþ 2Þ (8.50)

By applying the DTM to Eq. (8.48), and using the transformationoperations in Chapter 1 and some simplification, the following recurrenceequation can be obtained:

fjðmþ 4Þ ¼ U4

ðmþ 1Þðmþ 2Þðmþ 3Þðmþ 4ÞfjðmÞ (8.51)

w1(x,t)

x1

L1 L2

x2

w2(x,t)

L

k

0 0

Figure 8.9 The coordinate system for a cantilever beam with intermediate support.


To calculated2fjðxjÞdx2j

in Eq. (16), the differential transformation fjðmÞshould be solved. From Eq. (8.51), it can be seen fjðmÞ (m � 4) is a

function of fjð0Þ;fjð1Þ;fjð2Þ;fjð3Þ; and U. These nine unknown pa-

rameters, namely U and fjðsÞ (s ¼ 0, 1, 2, 3 and j ¼ 1, 2), can be deter-mined by the boundary condition equations of each section of the beamand the continuity conditions at intermediate support. By using boundarycondition of the cantilever beam shown in Fig. 8.8, we get at left end(clamped):

f1ðx1Þ ¼df1ðx1ÞdX1

¼ 0; ðX1 ¼ 0Þ (8.52)

At right end (free):

d2fðx2ÞdX2

2¼ d3fðx2Þ

dX32

¼ 0; ðX2 ¼ 0Þ (8.53)

The differential transformations of Eq. (8.51) are obtained with thedefinition Eq. (8.49) as

f1ð0Þ ¼ 0; f1ð1Þ ¼ 0 (8.54)

Similarly, for differential transformations of Eq. (8.53) can be expressed as:

f1ð2Þ ¼ 0; f1ð3Þ ¼ 0 (8.55)

For a beam with intermediate support, the continuity conditions indimensionless form are

f1ðR1Þ ¼ f2ðR2Þ; df1ðR1ÞdX1

¼ �df2ðR2ÞdX2

(8.56)

d2f1ðR1ÞdX2

1

¼ d2f2ðR2ÞdX2

2

;d3f1ðR1Þ

dX31

� Ksf1ðR1Þ ¼ � d3f2ðR2ÞdX3

2

(8.57)

where R1 ¼ L1L is denoted as dimensionless step location, R2 ¼ L2

L andR1 þ R2; Ks ¼ KsL3

EI are the dimensionless stiffness, and Ks is the stiffnessof the intermediate support. Substituting Eq. (8.49) into Eqs. (8.56) and(8.55), we obtain

XMm¼0

Rm1f1ðmÞ ¼

XMm¼0

Rm2 f2ðmÞ (8.58)


XMm¼0

ðmþ 1ÞRm1 f1ðmþ 1Þ ¼

XMm¼0

ðmþ 1ÞRm2 f2ðmþ 1Þ (8.59)

XMm¼0

ðmþ 1Þðmþ 2ÞRm1f1ðmþ 2Þ ¼

XMm¼0

ðmþ 1Þðmþ 2ÞRm2 f2ðmþ 2Þ

(8.60)

XMm¼0

ðmþ 1Þðmþ 2Þðmþ 3ÞRm2f2ðmþ 3Þ � KsR

m1f1ðmÞ

¼ �XMm¼0

ðmþ 1Þðmþ 2Þðmþ 3ÞRm2 f2ðmþ 3Þ (8.61)

Substituting Eqs. (8.54) and (8.55) into Eqs. (8.58)e(8.61) and thenrewriting in matrix form, we obtain:2

6664f11ðUÞ f12ðUÞ f13ðUÞ f14ðUÞf21ðUÞ f22ðUÞ f23ðUÞ f24ðUÞf31ðUÞ f32ðUÞ f33ðUÞ f34ðUÞf41ðUÞ f42ðUÞ f43ðUÞ f44ðUÞ

37775

2666664f1ð2Þf1ð3Þf2ð0Þf2ð1Þ

3777775 ¼ 0 (8.62)

where fij(U) is a linear function of f2ð0Þ; f1ð2Þ; f1ð2Þ and f2ð1Þ. FromEq. (8.62), the dimensionless natural frequencies U can be solved by

det

26664f11ðUÞ f12ðUÞ f13ðUÞ f14ðUÞf21ðUÞ f22ðUÞ f23ðUÞ f24ðUÞf31ðUÞ f32ðUÞ f33ðUÞ f34ðUÞf41ðUÞ f42ðUÞ f43ðUÞ f44ðUÞ

37775 ¼ 0 (8.63)

Substituting the solved U(n) into Eq. (8.51) and using Eq. (8.50), theclosed-form series solution for the second spatial derivative of the modeshapes d2f1ðx1Þ

dx21for Section 1 and d2f2ðx2Þ

dx22for Section 2 can be determined.

Rewriting the second spatial derivative of the mode shape functions with auniform coordinate X, we obtain:

v2fðxÞvX2

¼

8>>><>>>:

v2f1ðxÞvX2

v2f2ð1� xÞvX2

X ˛½0;R�;X ˛½R1; 1� (8.64)


It can be found that the solution of d2fðxÞdx2 using DTM is a continuous

function and not discrete numerical values at knot point by finite elementor finite difference methods.

To verify the proposed method to design the shaped PVDF modalsensor, an aluminum beam with dimensions of Lx ¼ 500 mm, ho ¼ 5 mm,bo ¼ 40 mm is considered. The intermediate support is located atR1 ¼ 0.85 with dimensionless stiffness ks ¼ 5. Assume that modal dampingis 0.01. A point force located at x0 ¼ Lx

20 20 is used as the excitation. Theexcellent numerical stability of the solution can also be observed inFig. 8.10.

For simplicity, the DTM solutions are truncated to M ¼ 20 in allthe subsequent calculations. Fig. 8.11 shows the first 10 mode shapesf(x) and the corresponding second spatial derivative of the mode

shapes d2fðxÞdx2 . For the case of a uniform bending beam with classical

boundary conditions (i.e., which are clamped, free, simply supported, orsliding), the second spatial derivative of the mode shapes form anorthogonal set.

Figure 8.10 The first five dimensionless natural frequencies X(n) with different numberof the series summation limit M.


8.5 DAMPED SYSTEM WITH HIGH NONLINEARITY

The equation of motion of a damped vibration system with high nonlin-earity can be expressed as follows [4]:

€xþ z _xþ xþ cxn ¼ 0; n ¼ 2pþ 1; p ¼ 0; 1; 2;. (8.65)

where the superposed dots (.) denote differentiation with respect to time,z is the damping coefficient, c is a constant parameter, and n is the degreeof nonlinearity. The initial conditions of x(t) are given by

xð0Þ ¼ 1; _xð0Þ ¼ 0 (8.66)

transformation domain.

Let yðtÞ ¼ _xðtÞ. (8.67)

Substituting Eq. (8.67) into Eqs. (8.65) and (8.66) yields

_yþ zyþ xþ cxn ¼ 0: (8.68)

(a)

(b)

–2

–1

–1

–0.5

0.5

1

0

0

0 0.1 0.2 0.3 0.4 0.5X

X

0.6 0.7 0.8 0.9 1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

1st

1st

10th

10th

1

2

Φ (X

)d2 Φ

(X)/(

X2 )

Figure 8.11 (a) The first 10 mode shapes and (b) the corresponding second spatialderivative of the mode shapes.


The initial conditions becomexð0Þ ¼ 1; yð0Þ ¼ 0: (8.69)

By a process of inverse differential transformation, the solutions of eachsubdomain take m þ 1 terms for the power series, i.e.,

xiðtÞ ¼Xmk¼0

�tHi

�k

XiðkÞ; 0 � t � Hi (8.70)

YiðtÞ ¼Xmk¼0

�tHi

�k

YiðkÞ; 0 � t � Hi (8.71)

where i ¼ 0, 1, 2,.,n indicates the ith subdomain, k ¼ 0, 1,2,.,m denotesthe term of the power series, Hi represents the subdomain interval, andXi(k) and Yi(k) are the transformed functions of xi(t) and yi(t), respectively.From the initial conditions (Eq. 8.69), it can be seen that

X0ð0Þ ¼ dðkÞ; where dðkÞ ¼�1k ¼ 0

0ks 0(8.72)

Y0ð0Þ ¼ 0: (8.73)

Eqs. (8.67) and (8.68) undergo differential transformation to yield thefollowing

kþ 1Hi

Xiðkþ 1Þ ¼ YiðkÞ; (8.74)

kþ 1Hi

Yiðkþ 1Þ þ zYiðkÞ þ XiðkÞ

þ cXk

l¼1

�nþ 1k

L � 1

�$XiðlÞXið0Þ $X

ni ðk� LÞ ¼ 0:

(8.75)

The present problem employs the DTM described above to generate anumber of numerical results for the response of a damped system with highnonlinearity. The responses of x(t) for different values of nonlinearity, n,and damping coefficient, z, are plotted in Fig. 8.12. It is noted that thepresent results are in excellent agreement with the numerical results ob-tained from the fourth-order RungeeKutta approach. Fig. 8.13 presentsthe displacement and velocity time responses of a damped system withnonlinearity orders of n ¼ 3. The results indicate that the amplitude decaysare more rapidly for higher values of nonlinearity, particularly at highervalues of the damping coefficient.


–1.00 0.4 0.8

t

t

t1.2 1.6 2.0

0 0.4

: Runge-Kutta solutions: Differential Transformation solutions



0.8 1.2 1.6 2.0

0 0.4 0.8 1.2 1.6 2.0

–0.5

0

0.5

1.0

–1.0

–0.5

0x

x x

0.5

1.0

–1.0

–0.5

0

0.5

1.0(a) (b)

(c)

= 0.02ξ

= 0.2ξ

= 1.0ξ

= 0.02ξ

= 0.02ξ = 0.2ξ

= 0.2ξ = 1.0ξ

= 1.0ξ

Figure 8.12 Comparison of differential transformation method and RungeeKuttanumerical methods. (a) n ¼ 3. (b) n ¼ 101. (c) n ¼ 1001.

0–1.0

0.0

1.0

–2.0

0.0

2.0

20

Displacement time response. Velocity time response.

40 60 80 100 0

(b)(a)

20

11

0.20.2

40 60 80 100

x

t t

x'

= 0.02ξ

= 0.02ξ

Figure 8.13 Response of damped system for different values of z with n ¼ 3.(a) Displacement time response and (b) velocity time response (c ¼ 1.0).


8.6 FREE VIBRATION OF A CENTRIFUGALLYSTIFFENED BEAM

A uniform rotating beam of doubly symmetric cross section is consideredand illustrated in Fig. 8.14.

Such a rotating beam vibrates both out-of-plane and in-plane; however,the out-of-plane and in-plane vibrations are uncoupled. The equation ofmotion for out-of-plane vibration of a centrifugally tensioned uniformEulereBernoulli beam is as follows [5];

EIzzv4yvx4

� rAu2y� v

vx

�TðxÞ vy

vx

�¼ 0 (8.76)

where T(x) is the centrifugal tension at a distance x from the origin and isgiven by

TðxÞ ¼ 0:5rAp2ðL2 þ 2R0L � 2R0x� x2Þ (8.76a)

x is the radial coordinate, y(x) is the transverse deflection, R is the radius ofhub, L is the length of the beam, A is the cross-sectional area, r is the massdensity, p is the angular rotational speed, u is the angular frequency, EIzz isthe flexural rigidity for bending in the xey plane. The dimensionless equa-tion of motion is as follows;

D4YðXÞ � 0:5vð1þ 2roÞD2YðXÞ þ vroD�XDYðXÞ�

þ 0:5vD�X2DYðXÞ�� mYðXÞ ¼ 0

(8.77)

o

2R0

p

y, Y

z, Z

x, X

Figure 8.14 A cantilever rotating beam.


where

X ¼ x=L; Dn ¼ dn=dXn

YðXÞ ¼ yðxÞ=L; v ¼ h2 ¼ rAp2L4=EIzzr0 ¼ R0=L; m ¼ U2 ¼ rAu2L4=EIzz

(8.78)

r0 is the dimensionless offset parameter, h is the dimensionless rotationalspeed, U is the dimensionless frequency, X is the dimensionless radial coor-dinate, and YðXÞ is the dimensionless deflection. V and m are the dimen-sionless rotational speed and the dimensionless frequency-relatedparameters. The dimensionless bending moment M(X) and shear forceQ(X) are accordingly

MðXÞ ¼ D2YðXÞ; QðXÞ ¼ �DMðXÞ þ bðXÞDYðXÞ (8.79)

where b(X) is the dimensionless tension at coordinate X and is given by

bðXÞ ¼ 0:5vð1þ 2ro � 2roX � X2Þ (8.80)

The dimensionless boundary conditions of the three most commonlyused boundaries are

Clamped boundary : YðXÞ ¼ 0; DYðXÞ ¼ 0;Free boundary : MðXÞ ¼ 0; QðXÞ ¼ 0;Pinned boundary : YðXÞ ¼ 0; MðXÞ ¼ 0:

(8.81)

From the definition and properties of DTM given in Chapter 1, theDTM of the equation of motion Eq. (8.76) is found as

ðkþ 1Þðkþ 2Þðkþ 3Þðkþ 4ÞYðkþ 4Þ � 0:5vð1þ 2roÞðkþ 1Þ�ðkþ 2ÞYðkþ 2Þ � mYðkÞ þ vr0ðkþ 1Þ2Yðkþ 1Þþ 0:5vkðkþ 1ÞYðkÞ ¼ 0

(8.82)

Rearranging Eq. (8.82), one has the following recurrence relation

Yðkþ 4Þ ¼ 0:5vð1þ 2r0Þðkþ 1ÞY ðkþ 2Þ þ mY ðkÞ � vr0ðkþ 1Þ2Y ðkþ 1Þ � 0:5vkðkþ 1ÞY ðkÞðkþ 1Þðkþ 2Þðkþ 3Þðkþ 4Þ

(8.83)

Combining Eq. (8.83) and the appropriate boundary conditions, oneobtains solutions to a free vibration problem. As an example, let us considera rotating cantilever beam. This is because turbine, propeller, and roboticmanipulators can all be modeled as cantilevered beams. Performing DTMto the boundary conditions, with origin chosen at the clamped end, one has


Yð0Þ ¼ 0 (8.84)

Yð1Þ ¼ 0 (8.85)

At the free end, that is, at X ¼ 1, one has

XNk¼0

ðkþ 1Þðkþ 2ÞYðkþ 2Þ ¼ 0 (8.86)

PNk¼0

ðkþ 1Þðkþ 2Þðkþ 3ÞYðkþ 3Þ

�Pnk¼0

0:5vð1þ 2r0Þðkþ 1ÞYðkþ 1Þ

þPnk¼0

vr0ðkÞYðkÞ þXn

k¼2

0:5vðk� 1ÞYðk� 1Þ ¼ 0

(8.87)

From Eqs. (8.83)e(8.85), it can be seen that Y(k) is a linear functionof Y(2) and Y(3). Eqs. (8.86) and (8.84) can then be written in matrixform as

f11 f12f21 f22

Yð2ÞYð3Þ

¼ 0 (8.88)

Setting the determinant of the coefficient matrix of Eq. (8.88) to zerogives the characteristic equation of the structure, from which thenatural frequencies are found. The natural frequencies corresponding tovarious hub offset and rotational speed are computed using Matlab.Though for comparison purpose, the natural frequencies are keptaccurate to the fourth decimal place, the precision of the natural fre-quencies can be as high as the machine precision of the computer used.Both the rotational speed and the offset are seen to affect the naturalfrequencies greatly. Results for three representative cases are presented inFig. 8.15.

• Case 1: Dimensionless offset r0 ¼ 0 and dimensionless rotational speedh ¼ 0.




8.7 DEFLECTIONS OF ORTHOTROPIC RECTANGULARPLATE

This example considers large deflections of the orthotropic rectangularplate. The plate has thickness “h” and edge lengths “a” and “b” in the x-and y-directions, respectively. Let Oxyz be a Cartesian coordinate systemwhich lies in the midplane of the plate and has its origin at the corner. Thelateral loading on the plate is denoted by p(t). Under these conditions, theequations of motion for the plate are given [6].

uxx þ wxwxx þ c1ðuyy þ wxwyyÞ þ c2ðyxy þ wywxyÞ ¼ 0 (8.89)

c1ðyxx þ wywxxyyÞ þ c3ðyyy þ wywyyÞ þ c2ðuxy þ wxwxyÞ ¼ 0 (8.90)

00

0.1

0.2

0.3

0.4

0.5

0.60.7

0.04

–0.08–0.06–0.04

0.04

–0.02

0.02

0.060.08

0

0.03

0.02

–0.02

–0.03

0.01

–0.01

–0.1

0

0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4

Mod

e sh

apes

Mod

e sh

apes

Mod

e sh

apes

xx

x

0.6 0.8 1

(a) (b)

(c)

Figure 8.15 Mode shapes of the first three modes corresponding to Case 1 (d), Case2 (- - -), and Case 3 (. . .). (a) First mode, (b) second mode, and (c) third mode.


wxxxx þ 2c4wxxyy þ c3wyyyy þ rhd1

wu þ d

d1wt � p

d1� 12

h2�wxx

ux þ y21yy þ 1

2w2x þ

12y21w2

y

þ c3wyy

y12uxx þ yy þ 1

2w2x

12w2y

þ 2c1½uy þ yx þ wxwy��

¼ 0;

(8.91)where c1 ¼ mg12

e1; c2 ¼ c1 þ y21; c3 ¼ e2

e1; c4 ¼ 2c1 þ y21; D ¼ e1h3

12m ; m ¼ 1�y21y12 and y12e2 ¼ y21e1. Furthermore, w(x, y, t) is the deflection of theneutral plane of the plate, u(x, y, t) and y(x, y, t) are the displacements ofthe neutral plane of the plate in the x- and y-directions, respectively. Addi-tionally, q is the density of the plate material, h is the plate thickness, and E1

and E2 are moduli of elasticity along the principal material axes x and y,respectively, G12 is the modulus of rigidity characterizing the change ofthe angle between the principal material directions, and y21 and y12 arePoisson’s ratios with the first subscript indicating the direction of the tensileforce and the second indicating the direction of contraction. d is the damp-ing coefficient of the plate material.

This problem considers two types of boundary conditions:1. The boundary conditions at all of the edges are clamped and immovable:

w ¼ vwvx

¼ u ¼ y ¼ 0 at x ¼ 0; a; (8.92)

w ¼ vwvy

¼ u ¼ y ¼ 0 at y ¼ 0; a; (8.93)

2. The boundary conditions at all of the edges are simply supported andimmovable:

w ¼ v2wvx2

¼ u ¼ y ¼ 0 at x ¼ 0; a; (8.94)

w ¼ v2wvy2

¼ u ¼ y ¼ 0 at y ¼ 0; a; (8.95)

To reduce the order of the system dynamic equation, this study in-troduces the following velocity function:

g ¼ vwvt

; (8.96)


The system dynamic equation is then given by

vwvt

¼ g; (8.97)

wxxxx þ 2c4wxxyy þ c3wyyyy þ rhd1wu þ d

d1wt � p

d1� 12

h2�wxx

ux þ y21yy þ 1

2w2x þ

12y21w

2y

þ c3wyy

y12uxx þ yy þ 1

2w2x

12w2y

þ 2c1½uy þ yx þ wxwy��

¼ 0;

(8.98)

Taking the differential transform of the system equation with respect totime, it can be shown that

kþ 1H

wðx; y; kþ 1Þ ¼ gðx; y; kÞ (8.99)

wðx; y; kÞxxxx þ 2c4wðx; y; kÞxxyy þ c3wðx; y; kÞyyyyþrhd1

kþ 1h

gðx; y; kÞ þ d

d1gðx; y; kÞ � p

d1� 12

h2�wðx; y; kÞxx5

uxðx; y; kÞ þ y21vðx; y; kÞy þ

12wðx; y; kÞx5wðx; y; kÞþx

þ 12y21wðx; y; kÞy5ðx; y; kÞy

��

�5c3wyy

y12uðx; y; kÞxx þ vðx; y; kÞy þ

12y12wðx; y; kÞy5wðx; y; kÞþx

�

þ�12wðx; y; kÞy5wðx; y; kÞy þ 2c1

�uðx; y; kÞy þ yðx; y; kÞx

þwðx; y; kÞx5wðx; y; kÞy�� ¼ 0;

(8.100)

where w(x, y, k); g(x, y, k); u(x, y, k); and v(x, y, k) are the differential trans-forms of w(x, y, T); g(x, y, T); u(x, y, T); and n(x, y, T), respectively.

wðx; y; kÞy5wðx; y; kÞX ¼XKl¼0

wðx; y; k� lÞxwðx; y; lÞxl. (8.101)


wðx; y; kÞxx5wðx; y; kÞX5wðx; y; kÞX¼

XKl¼0

wðx; y; k� lÞxxXl

m¼0

wðx; y; l � mÞxl;wðx; y; kÞX .(8.102)

By taking the differential transform and the finite difference approxi-mation, the associated boundary conditions are transformed to:1. All edges are clamped

wð1; j; kÞ ¼ 0; wði; 1; kÞ ¼ 0; wðm� 1; j; kÞ ¼ 0 wði; n� 1; kÞ ¼ 0;

(8.103)

uð1; j; kÞ ¼ 0; uði; 1; kÞ ¼ 0; uðm� 1; j; kÞ ¼ 0 uði; n� 1; kÞ ¼ 0;(8.104)

vð1; j; kÞ ¼ 0; vði; 1; kÞ ¼ 0; vðm� 1; j; kÞ ¼ 0 vði; n� 1; kÞ ¼ 0;(8.105)

2. All edges are simply supported

wð1; j; kÞ ¼ 0; wði; 1; kÞ ¼ 0; wðm� 1; j; kÞ ¼ 0 wði; n� 1; kÞ ¼ 0;

(8.106)

uð1; j; kÞ ¼ 0; uði; 1; kÞ ¼ 0; uðm� 1; j; kÞ ¼ 0 uði; n� 1; kÞ ¼ 0;(8.107)

vð1; j; kÞ ¼ 0; vði; 1; kÞ ¼ 0; vðm� 1; j; kÞ ¼ 0 vði; n� 1; kÞ ¼ 0;

(8.108)

wð2; j; kÞ � 2wð1; j; kÞ þ wð0; j; kÞh2x

¼ 0;wði; 2; kÞ � 2wði; 1; kÞ þ wði; 0; kÞ

h2y¼ 0;

(8.109)

wðm; j; kÞ � 2wðm� 1; j; kÞ þ wðm� 2; j; kÞh2x

¼ 0;

wði; n; kÞ � 2wði; n� 1; kÞ þ wði; n� 2; kÞh2y

¼ 0;(8.110)

In investigating the dynamic motion and large deflections of theorthotropic rectangular plate, this study initially considered the case of a


plate with EE ¼ 3; G

E ¼ 0:5; y21 ¼ 0:25; y12 ¼ 0:08333; d ¼ 10, andE1 ¼ 1, subjected to a variable lateral load of P. The dynamic motion andthe large deflections were solved using the proposed hybrid methodcombining the finite difference method and the differential transformmethod. Simulations were performed with H ¼ 0.01, k ¼ 5, a ¼ 1.0, andb ¼ 1.0. The edges of the plate were divided into 10 equal units such thatm ¼ n ¼ 11. It was assumed that the edges of the plate were all clamped andthat a lateral step force load of P ¼ 156.34 were applied.

Fig. 8.16 shows the dynamic motion of the center of the plate at x ¼ 0.5and y ¼ 0.5. The corresponding deflections of the center of the orthotropicplate are shown in Fig. 8.17. It can be seen that the deflection varies

w

1.5

0.8958

00 50 100 150

Time200 250 300

x0.01

Chia [2]

X: 289Y: 1.055

Figure 8.16 Deflection of center of plate under applied lateral load of P ¼ 100.

Figure 8.17 Loadedeflection curves for clamped rectangular plate under increasinguniform lateral load.


nonlinearly with the lateral force. Furthermore, it is apparent that thecurrent deflection results are in good general agreement with those ofRefs. [2] and [6].

8.8 FREE VIBRATION OF CIRCULAR PLATES

The governing differential equation of a thin circular plate undergoing freeharmonic vibration in a nondimensional form is as follows [7]:

V4w ¼ U2w (8.111)

where V4 is the biharmonic operator, W ¼ W(r, q) is dimensionlessdeflection, r is dimensionless coordinate along the radial axis of the plate,q is dimensionless coordinate along the tangential axis, and U is dimen-sionless frequency of vibration. From the classical plate vibration theory,deflection of a circular plate in polar coordinates may be expressed asfollows:

w ¼ f ðrÞcosðmqÞ; (8.112)

where m is the integer number of nodal diameters and f(r) is the radial modefunction. Substituting Eq. (8.112) into Eq. (8.111), the governing differen-tial equation of the circular plate becomes:

d4fdr4

þ 2rd3fdr3

� Br2

d2fdr2

þ Br3

dfdr

þ Ar4f ¼ U2f ; (8.113)

where

A ¼ m4 � 4m2 and B ¼ 2m2 þ 1: (8.114)

The boundary conditions at the outer edge (r ¼ 1) of the circular platemay be one of the following; simply supported, clamped, and free. Theseconditions may be written in terms of the radial mode function f(r) asfollows:

Simply supported:

f ðrÞjr¼1 ¼ 0;Mr jr¼1 ¼ �D

d2fdr2

þ n

�1rdfdr

þ m2

r2f

�¼ 0: (8.115)

Clamped:

f ðrÞjr¼1 ¼ 0;dfdr

��r¼1

¼ 0 (8.116)


Free:

Mr jr¼1 ¼ �D

d2fdr2

þ n

�1rdfdr

þ m2

r2f

�¼ 0;

Vr jr¼1 ¼d3fdr3

þ 1rd2fdr2

þ�m2v � 2m2 � 1

r2

�dfdr

þ�3m2 � m2v

r3

�f

¼ 0;

(8.117)

whereMr is the radial bending moment, Vr is the effective radial shear force,D is the flexural rigidity, and m is the Poisson’s ratio. It can easily be noticedthat, since Eq. (8.113) is a fourth-order differential equation, four initialconditions are required. One may obtain two of those from the boundaryconditions at the outer edge of the circular plate. However, remainingtwo conditions must be investigated within the regularity conditions atthe center of the plate.

Antisymmetric case:

f ðrÞjr¼0 ¼ 0; Mr jr¼0 ¼d2fdr2

��r¼0

¼ 0 for ðm ¼ 1; 3; 5;.Þ (8.118)

Symmetric case:

dfdr

��r¼0

¼ 0; Vr jr¼0 ¼d3fdr3

��r¼0

¼ 0 for ðm ¼ 0; 2; 4; 6;.Þ (8.119)

A. Transformation of Free Vibration EquationUsing the transformation operations defined in Chapter 1 and taking thedifferential transform of Eq. (8.113) at r0 ¼ 0, one may obtain:

AFk þ BXk

l¼0

dðl � 1Þðk� l þ 1ÞFk�lþ1

�BPkl¼0

dðl � 2Þðk� l þ 1Þðk� l þ 2ÞFk�lþ2 þ 2Xk

l¼0

dðl � 3Þðk� l þ 1Þ

*ðk� l þ 2Þðk� l þ 3ÞFk�lþ3

þ Pkl¼0

dðl � 4Þðk� l þ 1Þðk� l þ 2Þðk� l þ 3Þðk� l þ 4ÞFk�lþ4

¼ U2Xk

l¼0

dðl � 4ÞFk�l.

(8.120)


Simplifying Eq. (8.120) and using the last theorem, the equation ofmotion in Eq. (8.113) can be transformed into the following recurrenceequation:

Fkþ4 ¼ U2

A� Bðkþ 4Þðkþ 2Þ þ ðkþ 4Þðkþ 3Þ2ðkþ 2ÞFk. (8.121)

From Eq. (8.121), the following equations can be obtained for k ¼ 0, 1,2,.,n:

F4 ¼ U2F0

A� Bð4:2Þ þ ð4:32:2Þ ; F5 ¼ U2F1

A� Bð5:3Þ þ ð5:42:3Þ ; F6 ¼ U2F2

A� Bð6:4Þ þ ð6:52:4Þ ;

F7 ¼ U2F3

A� Bð7:5Þ þ ð7:62:5Þ ; F8 ¼ U2F4

A� Bð8:6Þ þ ð8:72:6Þ ; F9 ¼ U2F5

A� Bð9:7Þ þ ð9:82:7Þ ;/

and in general they can be formulated as follows:

F4k ¼ U2kF0Ykk¼1

�A� Bð4kÞð4k� 2Þ þ ð4kÞð4k� 1Þ2ð4k� 2Þ�

; (8.122)

F4kþ1 ¼ U2kF1Ykk¼1

�A� Bð4kþ 1Þð4k� 1Þ þ ð4kþ 1Þð4kÞ2ð4k� 1Þ�

; (8.123)


�A� Bð4kþ 2Þð4kÞ þ ð4kþ 2Þð4kþ 1Þ2ð4kÞ�

; (8.124)


�A� Bð4kþ 3Þð4kþ 1Þ þ ð4kþ 3Þð4kþ 2Þ2ð4kþ 1Þ�

;

(8.125)

In terms of the transforms appearing in Eqs. (8.122)e(8.125),Eq. (8.111) may be written as follows:

f ðrÞ ¼XNk¼0

r4kF4k þXNk¼0

r4kþ1F4kþ1 þXNk¼0

r4kþ2F4kþ2 þXNk¼0

r4kþ3F4kþ3.

(8.126)


B. Transformation of Boundary/Regularity ConditionsThe following equations are obtained by applying the theorems inChapter 1 to the boundary conditions at the outer edge (r ¼ 1);

Simply supported:

XNk¼0

F4k ¼ 0;XNk¼0

ðkðk� 1Þ þ vk� m2vÞFk ¼ 0: (8.127)

Clamped:

XNk¼0

Fk ¼ 0;XNk¼0

kFk ¼ 0. (8.128)

Free:

PNk¼0

ðkðk� 1Þ þ vk� m2vÞFk ¼ 0;

PNk¼0

ðkðk�1Þðk�2Þ þ kðk�1Þ þ ðm2v � 2m2 � 1Þkþ ð3m2 � m2vÞÞFk ¼ 0.

(8.129)

At the center of the circular plate (r ¼ 0), the boundary conditionswhich are derived from regularity conditions can be transformed asfollows:

Antisymmetric case:

F0 ¼ F2 ¼ F4 ¼ .F4K ¼ F4kþ2 ¼ 0 for ðm ¼ 1; 3; 5;.Þ. (8.130)

Symmetric case:

F1 ¼ F3 ¼ F5 ¼ .F4kþ1 ¼ F4kþ3 ¼ 0 for ðm ¼ 0; 2; 4; 6.Þ. (8.131)

To avoid the unnecessary repeating of derivations, we consider only oneparticular case, that is, a circular plate simply supported at the outer edgehaving symmetric modes (m ¼ 0, 2, 4,.). The boundary conditions atr ¼ 1 introduced in Eq. (8.127) can be written as follows:

XNk¼0

F4k þXNk¼0

F4kþ1 þXNk¼0

F4kþ2 þXNk¼0

F4kþ3 ¼ 0; (8.132)


XNk¼0

ðð4kÞð4k� 1Þ þ v4k� m2vÞF4k þXNk¼0

ðð4kþ 1Þð4kÞ þ vð4kþ 1Þ

� m2vÞF4kþ1 þXNk¼0

ðð4kþ 2Þð4kþ 1Þ þ vð4kþ 2Þ � m2vÞF4kþ2

þXNk¼0

ðð4kþ 2Þð4kþ 3Þð4kþ 2Þ þ vð4kþ 3Þ � m2vÞF4kþ3 ¼ 0

(8.133)

It must be noted that Eqs. (8.132) and (8.133) are valid for both symmetricand antisymmetric cases. For the symmetric modes, Eq. (8.131) must be takeninto account. Substituting Eqs. (8.122) and (8.124) into the boundary con-ditions in Eqs. (8.132) and (8.133), the following expressions are obtained:

XNk¼0

U2kF0Ykk¼1


þXNk¼0

U2kF2Ykk¼1


¼ 0;

(8.134)

XNk¼0

ðð4kÞð4k� 1Þ þ v4k� m2vÞU2kF0Ykk¼1


þXNk¼0

ðð4kþ 2Þð4kþ 1Þ þ vð4kþ 2Þ � m2vÞU2kF2Ykk¼1

�A� Bð4kþ 2Þð4kÞ þ ð4kþ 2Þð4k� 1Þ2ð4kÞ�

¼ 0:

(8.135)

Eqs. (8.134) and (8.135) can be grouped and rewritten as follows:

X ðnÞ11 ðUÞF0 þ X ðnÞ

12 ðUÞF2 ¼ 0;

X ðnÞ21 ðUÞF0 þ X ðnÞ

22 ðUÞF2 ¼ 0;(8.136)


where XðnÞ11 , X

ðnÞ12 , X

ðnÞ21 , and X ðnÞ

22 are closed-form polynomials of U corre-sponding to the nth term. It can be clearly seen that X ðnÞ

11 , XðnÞ12 and X ðnÞ

21 ,X ðnÞ22 terms represent the closed-form series expressions in Eqs. (8.134)

and (8.135). Eq. (8.136) can be expressed in the matrix form:"X ðnÞ

11 ðUÞ X ðnÞ12 ðUÞ

X ðnÞ21 ðUÞ X ðnÞ

22 ðUÞ

#�F0

F2

�¼

�0

0

�. (8.137)

The frequency equation of the circular plate is obtained by setting thedeterminant of the coefficient matrix of Eq. (8.137) equal to zero:��X

ðnÞ11 ðUÞ X ðnÞ

12 ðUÞX ðnÞ

21 ðUÞ X ðnÞ22 ðUÞ

�� ¼ 0: (8.138)

Performing the nontrivial solution in Eq. (8.138), we get U ¼ UðnÞj ,

where j ¼ 1, 2, 3,., n, where UðnÞj is the jth estimated eigenvalue corre-

sponding to n. The value of n can be obtained by the following equation:��UðnÞj � U

ðn�1Þj

�� x; (8.139)

where x is the tolerance parameter that is taken as x ¼ 0.00001 in thisexample. If Eq. (8.139) is satisfied, then we get jth eigenvalue Uj. The cor-responding eigen function, f(r), describing the instantaneous deflected shapeof the circular plate can be obtained by

f ðrÞ ¼ PNk¼0

r4kU2k

Ykk¼1


�

8>>>>><>>>>>:

PNk¼0

U2k

Ykk¼1


PNk¼0

U2k

Ykk¼1


9>>>>>=>>>>>;

*

8>>><>>>:

PNk¼0

r4kþ2U2k

Ykk¼1


9>>>=>>>;

(8.140)


Following the similar procedure, the frequency equations for othertypes of boundary and regularity conditions are derived as follows:

Simply supported and symmetric case:8>>><>>>:

PNk¼0

U2k

Ykk¼1


9>>>=>>>;

.

8>>><>>>:

PNk¼0

ðð4kþ 2Þð4kþ 1Þ þ vð4kþ 2Þ � m2vÞU2k

Ykk¼1


9>>>=>>>;

�

8>>><>>>:

PNk¼0

U2k

Ykk¼1


9>>>=>>>;

.

8>>><>>>:

PNk¼0

ðð4kÞð4k� 1Þ þ vð4kÞ � m2vÞU2k

Ykk¼1


9>>>=>>>;

¼ 0:

(8.141)

Simply supported and antisymmetric case:8>>><>>>:

PNk¼0

U2k

Ykk¼1


9>>>=>>>;

.

8>>><>>>:

PNk¼0


Ykk¼1


9>>>=>>>;

�

8>>><>>>:

PNk¼0

U2k

Ykk¼1


9>>>=>>>;

.

8>>><>>>:

PNk¼0

ðð4kþ 1Þð4kÞ þ vð4kþ 1Þ � m2vÞU2k

Ykk¼1


9>>>=>>>;

¼ 0:

(8.142)


Clamped and symmetric case:

8>>><>>>:

PNk¼0

U2k

Ykk¼1


9>>>=>>>;

.

8>>><>>>:

PNk¼0

ð4kþ 2ÞU2k

Ykk¼1


9>>>=>>>;

�

8>>><>>>:

PNk¼0

U2k

Ykk¼1


9>>>=>>>;

.

8>>><>>>:

PNk¼0

ð4kÞU2k

Ykk¼1


9>>>=>>>;

¼ 0:

(8.143)

Clamped and antisymmetric case:

8>>><>>>:

PNk¼0

U2k

Ykk¼1


9>>>=>>>;

.

8>>><>>>:

PNk¼0

ð4kþ 3ÞU2k

Ykk¼1


9>>>=>>>;

�

8>>><>>>:

PNk¼0

U2k

Ykk¼1


9>>>=>>>;

.

8>>><>>>:

PNk¼0

ð4kþ 1ÞU2k

Ykk¼1


9>>>=>>>;

¼ 0:

(8.144)


Free and symmetric case:8>>><>>>:

PNk¼0

ðð4kÞð4k� 1Þ þ v4k� m2vÞU2k

Ykk¼1


9>>>=>>>;

.

8>>><>>>:

PNk¼0

�ð4kþ 2Þð4kþ 1Þ2 þ ðm2v � 2m2 � 1Þð4kþ 2Þ þ ð3m2 � m2vÞ�U2k

Ykk¼1


9>>>=>>>;

�

8>>><>>>:

PNk¼0


Ykk¼1


9>>>=>>>;

.

8>>><>>>:

PNk¼0

�ð4kÞð4k� 1Þ2 þ ðm2v � 2m2 � 1Þ4kþ ð3m2 � m2vÞ�U2k

Ykk¼1


9>>>=>>>;

¼ 0:

(8.145)

Free and antisymmetric case:8>>><>>>:

PNk¼0

ðð4kþ 1Þð4kÞ þ vð4kþ 1Þ � m2vÞU2k

Ykk¼1


9>>>=>>>;

.

8>>><>>>:

PNk¼0


Ykk¼1


9>>>=>>>;

�

8>>><>>>:

PNk¼0

ðð4kþ 3Þð4kþ 2Þ þ vð4kþ 3Þ � m2vÞÞU2k

Ykk¼1


9>>>=>>>;

.

8>>><>>>:

PNk¼0


Ykk¼1


9>>>=>>>;

¼ 0:

(8.146)

In Fig. 8.18, the convergence of the first four natural frequencies withrespect to the number of terms considered is presented for the free endcondition and m ¼ 1. It is observed from this figure that evaluating the


higher natural frequencies needs more number of terms to be considered.Also note that the accuracy of the method increases dramatically with thenumber of terms taken into consideration.

Fig. 8.19 shows the first three mode shapes for the clamped, simplysupported, and free end conditions. Only two cases of the nodal diameter,m ¼ 0 and m ¼ 1, are considered as examples of symmetric and antisym-metric regularity conditions.

8.9 VIBRATION OF PIPES CONVEYING FLUID

Consider an elastic, straight, fluid-conveying pipe with length l between twoends, as shown in Fig. 8.20, the linear equation of motion is given by [8].

EIv4wvx4

þMU2v2wvx2

þ 2MUv2wvxvt

þ ðM þ mÞ v2wvt2

¼ 0; (8.147)

where EI is the flexural rigidity, M and m are the mass per unit length offluid and the pipe, respectively, U is the fluid flow velocity, w(x, t) is thetransverse deflection of the pipe, x is the horizontal coordinate along thecenterline of the pipe, and t is the time. Introducing the following nondi-mensional quantities

x ¼ xl; b ¼ M

M þ m; s ¼ t

L2

�EI

M þ m

�12

; u ¼ UL

�MEI

�12

. (8.148)

10

50

100

150

200

20 30N

1. Mode

2. Mode

3. Mode

Nat

ural

Fre

quen

cies

4. Mode

40 50 60

Figure 8.18 Convergence of the natural frequencies for m ¼ 1 and free end boundaryconditions.


Eq. (8.147) can be rewritten as a dimensionless form

v4h

vx4þ u2

v2h

vx2þ 2u

ffiffiffib

p v2h

vxvsþ v2h

vs2¼ 0: (8.149)

The dimensionless boundary conditions considered in this paper includethe following:1. Cantilevered pipe

hð0; sÞ ¼ h0ð0; sÞ ¼ 0;

h00ð1; sÞ ¼ h000 ð1; sÞ ¼ 0;

(8.150)

(e) (f)

(d)

(b)

(c)

(a)

0.25

0.250.2

0.1

0

0.15

0.05

–0.05–0.1

Mod

e S

hape

sM

ode

Sha

pes

Mod

e S

hape

s

Mod

e S

hape

sM

ode

Sha

pes

Mod

e Sh

apes

–0.25

0.75

–0.75

1

10.80.60.40.2

0–0.2–0.4

10.80.60.40.2

0–0.2–0.4 –0.1

0.1

0.2

0.3

0

0

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4x

x

x x

x

x0.6 0.8 1 0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

0

0

0.1

–0.1

0.2

–0.2

0.2 0.4 0.6 0.8 1

0.5

–0.5

1. M. S.2. M. S.3. M. S.

1. M. S.2. M. S.3. M. S.

1. M. S.2. M. S.3. M. S.

1. M. S.2. M. S.3. M. S.

1. M. S.2. M. S.3. M. S.

1. M. S.2. M. S.3. M. S.

Figure 8.19 Mode shapes for (a) simply supported and m ¼ 0; (b) simply supportedand m ¼ 1; (c) clamped and m ¼ 0; (d) clamped and m ¼ 1; (e) free and m ¼ 0; (f) freeand m ¼ 1.


2. Pinnedepinned pipe

hð0; sÞ ¼ h00ð0; sÞ ¼ 0;

hð1; sÞ ¼ h00ð1; sÞ ¼ 0;(8.151)

3. Clampedepinned pipe

hð0; sÞ ¼ h0ð0; sÞ ¼ 0;

hð1; sÞ ¼ h00ð1; sÞ ¼ 0;(8.152)

4. Clampedeclamped pipe

hð0; sÞ ¼ h0ð0; sÞ ¼ 0;

hð1; sÞ ¼ h0ð1; sÞ ¼ 0:(8.153)

The solution of Eq. (8.7) may be expressed as

hðx; sÞ ¼ wðxÞeus. (8.154)

Substituting Eq. (8.154) into Eq. (8.149) yields

v4w- ðxÞvx4

þ u2v2w- ðxÞvx2

þ 2uuffiffiffib

p vw- ðxÞvx

þ u2w- ðxÞ ¼ 0: (8.155)

(a)

(b)

(c)

U

U

U

U

(d)

Figure 8.20 Pipe conveying fluid with different boundary conditions. (a) Cantileveredpipe, (b) pinnedepinned pipe, (c) clampedepinned pipe, (d) clampedeclamped pipe.


From Chapter 1, the differential transformation form of Eq. (8.155) isfound as

ðkþ 1Þðkþ 2Þðkþ 3Þðkþ 4ÞW ðkþ 4Þ þ u2ðkþ 1Þðkþ 2ÞW ðkþ 2Þþ 2uu

ffiffiffib

pðkþ 1ÞW ðkþ 1Þ þ u2W ðkÞ ¼ 0:

(8.156)

Rearranging Eq. (8.156), a simple recurrence relation is obtained asfollows

W ðkþ 4Þ ¼ ��u2ðkþ 1Þðkþ 2ÞW ðkþ 2Þ þ 2uu

ffiffiffib

p ðkþ 1ÞW ðkþ 1Þ þ u2W ðkÞ�ðkþ 1Þðkþ 2Þðkþ 3Þðkþ 4Þ . (8.157)

Similarly, the differential transformation form of boundary conditionscan be written as:1. Cantilevered pipe

W ð0Þ ¼ W ð1Þ ¼ 0; (8.158)

XNk¼0

kðk� 1ÞW ðkÞ ¼ 0; (8.159)

XNk¼0

kðk� 1Þðk� 2ÞW ðkÞ ¼ 0; (8.160)

2. Pinnedepinned pipe

W ð0Þ ¼ W ð2Þ ¼ 0; (8.161)

XNk¼0

W ðkÞ ¼ 0; (8.162)

XNk¼0

kðk� 1ÞW ðkÞ ¼ 0; (8.163)

3. Clampedepinned pipe

W ð0Þ ¼ W ð1Þ ¼ 0; (8.164)

XNk¼0

W ðkÞ ¼ 0; (8.165)


XNk¼0

kðk� 1ÞW ðkÞ ¼ 0; (8.166)

4. Clampedeclamped

W ð0Þ ¼ W ð1Þ ¼ 0; (8.167)

XNk¼0

W ðkÞ ¼ 0; (8.168)

XNk¼0

kW ðkÞ ¼ 0: (8.169)

It ought to be pointed out that, in Ref. [8], the number of terms, N, ofDTM is chosen to be N ¼ 55, showing sufficiently accurate results. In thecalculations, convergence of the first four natural frequencies is shown inFig. 8.21. Here, it can be seen that with the increasing ofN, the precision of

(a) (c)

(b) (d)

Figure 8.21 Convergence of the first four dimensionless natural frequencies. (a)Cantilevered pipe, (b) pinnedepinned pipe, (c) clampedepinned pipe, and(d) clampedeclamped pipe.


DTM increases. Compared with the lower-order modes, the higher-ordernatural frequency requires more terms adopted in the DTM. In thefollowing analysis, therefore, we choose N ¼ 60. In the calculations basedon DQM, 17 sampling points were used to obtain the convergence solu-tions shown. It is found that even the sampling points are chosen more than17, the precision of DQM does not increase. If, however, the number ofterms (N) is chosen to be sufficiently large (e.g., N ¼ 60), the solution ofDTM would infinitely approach to the exact solution. In the case of us0,Fig. 8.22 represents the natural frequencies of pipes conveying fluid for asimply supported pipe conveying fluid. Very good agreement is foundbetween the results obtained by DTM and DQM.

8.10 PIEZOELECTRIC MODAL SENSOR FOR NONUNIFORMEULEReBERNOULLI BEAMS WITH RECTANGULARCROSS SECTION

Consider the free vibration of an EulereBernoulli beam with length Lx,varying thickness h(x) and varying width b(x). A shaped PVDF film ofconstant thickness is attached onto the top surface and spanned across theentire length of the beam, as shown in Fig. 8.23. Assuming that the PVDFsensor thickness is much smaller than beam thickness, the mass and stiffnessof the sensor is then negligible compared to the properties of the beam.The sensor thickness is typically 28e110 mm. As referred to Ref. [9], theoutput charge Q(t) of the PVDF sensor shown in Fig. 8.23 can beexpressed as:

QðtÞ ¼Z L

0

hðxÞ2

FðxÞ$�e31

d2wðx; tÞdx2

�dx (8.170)

where h(x) is the beam thickness function, e31 is the PVDF sensor stress/charge coefficient, w(x, t) is the displacement of the beam, and F(x) isthe PVDF film shape function. The displacement distribution of thevibrating beam may be represented by a series expansion:

wðx; tÞ ¼XNn¼1

hnðtÞfnðxÞ (8.171)

where Zn(t) and fn(x) are the nth modal coordinates and structural modeshape function. N is the index for the highest order structural mode.


4th mode

DTMDQM

DTMDQM

3rd mode

2nd mode

1st mode

0

010203040

–25

–20

–15

–10

–5

0

5

10

15

20

25

5060708090

100110120130140150160

2 4

u

Im(ω

)R

e(ω

)

u

6 8

1st and 2nd mode combined

1st and 2nd mode combined

1st mode

1st modedivergency

2nd modedivergency

Paidoussiscoupled-modeflutter

2nd mode 3rd and 4th mode

0 2 4 6 8

(a)

(b)

Figure 8.22 (a) The imaginary component and (b) the real component of thedimensionless frequency, as a function of fluid velocity, u, for the lowest four modes ofa simply supported pipe conveying fluid, b ¼ 0.1.


Substituting Eq. (8.171) into Eq. (8.170), and expressing in dimen-sionless form, one obtains:

QðtÞ ¼ e312Lx

XNn¼1

htðtÞZ 1

0hðXÞFðXÞ$d

2fnðxÞdX2

dX ¼ e312Lx

XNn¼1

hnðtÞPðnÞ

(8.172)

where X ¼ (x/L), fn(X) ¼ fn(x), pn ¼R 10 hðXÞFðXÞ$ðd2fnðXÞ=dX2ÞdX is

designated as the modal sensitivity [5,6]. To construct a sensor that accu-rately measures the target kth structural mode, a shaped PVDF film isused, the output signal of the sensor should be directly proportional tothe target kth modal information. According to Eq. (8.172), a sufficientcondition for QðtÞNhk is given by

Pn ¼Z 1

0hðXÞFðXÞ$d

2fnðxÞdX2

dX ¼�a n ¼ k

0 ns kfor n ¼ 1;.;N

(8.173)

where a is a nonzero scaling factor, a is given as unity for convenience. Todesign the appropriate PVDF film shape function F(X), which satisfies theabove modal sensing condition equation, we make an approximation byexpanding F(X) as a linear function of the second derivative of the modeshape function

d2fjðxÞdX2 dX ¼ dX and beam physical parameters, such as

FðXÞ ¼XNj¼1

BjEðXÞIðXÞ

hðXÞ $d2fjðXÞdX2

¼XNj¼1

BjEðXÞbðXÞh2ðXÞ

12

v2fjðXÞvx2

(8.174)

where Bj are the unknown shape coefficients for the PVDF sensor. E(X) isYoung’s modulus, I(X) ¼ b(X)h3(X)/12 is the cross section second momentof the beam.

BeamPVDFb(x)

h(x)

F(x)

Lx

Figure 8.23 A shaped polyvinylidene fluoride (PVDF) film bonded on a nonuniformbeam.


Substituting Eq. (8.174) into Eq. (8.173), the modal sensitivity Pn can berewritten as

Pn ¼XNj¼1

Bj

Z 1

0EðXÞIðXÞ d

2fjðXÞdX2

$d2fnðXÞdX2

dX ¼XNj¼1

BjKðn; jÞ

(8.175)

where

kðn; jÞ ¼Z 1

0EðXÞIðXÞ d

2fjðXÞdX2

$d2fnðXÞdX2

dX (8.176)

Using Eqs. (8.174)e(8.176), the PVDF sensor shape coefficient Bj canbe solved by:

Bj ¼ aK�1ðk; jÞ (8.177)

For a beam with classical boundary conditions (such as clamped, free,simply supported, or sliding), thanks to the orthogonality of the secondderivative of the mode shapes, we get

K�1ðk; jÞ ¼ 0 js k (8.178)

Thus, the modal sensor shape coefficients Bj and shape function F(X)under the classical boundary conditions for the kth structural mode can besimplified as:

Bj ¼

8><>:

1Kkk

j ¼ k

0 js kFðXÞ ¼ EðXÞbðXÞh2ðXÞ

12

v2fjðXÞvx2

(8.179)

In accordance with the above analysis, to design a modal sensor, thesecond spatial derivative of the mode shapes must be obtained. This will bediscussed in the next section by using the DTM technique. Consider the freevibration of a stepped EulereBernoulli beam consisting of J uniform sectionselastically restrained at both ends, as shown in Fig. 8.24. The stepped beam isdivided into J sections with the J mirror systems of referencexj( j ¼ 1,2,.,J ). The dimensionless ordinary differential equation describingthe free vibration in each section of the stepped beam is as follows.

d4fjðXjÞdX4

j

� U4j fjðXjÞ ¼ 0 Xj ˛½ 0 Rj � (8.180)


where subscript j denote the section-j of the stepped beamU4

j ¼ rjAju2L4

.EjIj; Rj ¼ ðLj=LÞ. I is the cross-section second order

area moment of the section-j, Ij ¼�bjh3J

�12�; Aj ¼ bjhj is the cross-

section area. Ej, rj, Lj, bj, and hj are the Young’s modulus, density, length,width and thickness of section-j, respectively.

clearly U4j ¼

rjAjL4

r1A1L4

E1I1EjIj

U4j ¼ mjU

4j (8.181)

where mj ¼ (rjAj/r1A1)(E1I1/EjIj), U1 is the dimensionless natural fre-quency, and the nth dimensionless natural frequency is denoted as U1(n).According to Eq. (8.179), fjðXjÞ in Eq. (8.180) can be expressed in differ-ential transformation formulation:

fjðXjÞ ¼XMm¼0

Xmj fjðmÞ (8.182)

And the second spatial derivative of the mode shapes can be expressed as

d2fjðXjÞdX2

j

¼XM � 2

m¼0

ðmþ 1Þðmþ 2ÞXmj fjðmþ 2Þ (8.183)

By applying the DTM to Eq. (8.180) and using the transformationoperations in Chapter 1 and using Eqs. (8.181) and (8.182), the followingrecurrence equation can be obtained

fjðmþ 4Þ ¼ mjU41

ðmþ 1Þðmþ 2Þðmþ 3Þðmþ 4ÞfjðmÞ (8.184)

Section 1 Section 2 Section j Section J

w1(x, t) w2(x, t) wj(x, t) wJ(x, t)

kR1 kRJ

kr1 krJ

L

L2L1

x2x1

Lj

xj

LJ

xJ00 0 0

Figure 8.24 The coordinate system for a multiple-stepped beam elastically restrainedat both ends.


To calculate d2fjðXjÞ=dX2j in Eq. (8.17), the differential transformation

fj(m) should be solved. From Eq. (18), it can be seen fj(m) (m � 4) are afunction of fjð0Þ; fjð1Þ; fjð2Þ; fjð3Þ and U1. These 2J þ 1 unknownparameters, namely U1 and fjðsÞ ðs ¼ 0; 1; 2; 3 and j ¼ 1wJÞ can bedetermined by the continuity conditions and the boundary conditionequations of each section of the beam. For a beam with discontinuous crosssections, stress concentrations at the step locations of the beam areneglected. Then, the continuity conditions in dimensionless form are

fjþ1ð0Þ ¼ fjðRjÞ;dfjþ1ð0ÞdXjþ1

¼ dfjðRjÞdXj

(8.185)

d2fjþ1ð0ÞdX2

jþ1

¼ EJIJEJþ1IJþ1

d2fjðRjÞ (8.186)

where Rj ¼ (Lj/L) is denoted as dimensionless length of the jth section andPjj¼1 Rj ¼ 1.Substituting Eq. (8.180) into Eqs. (8.182) and (8.183), the differential

transformation for the section-j (j � 2) can be written as

fjþ1ð0Þ ¼XMm¼0

Rmj fjðmÞ (8.187)

fjþ1ð1Þ ¼XMm¼0

ðmþ 1ÞRmj fjðmþ 1Þ (8.188)

fjþ1ð2Þ ¼ ð1=2Þ EjIjEjþ1Ijþ1

XMm¼0

ðmþ 1Þðmþ 2ÞRmj fjðmþ 2Þ (8.189)

fjþ1ð3Þ ¼ ð1=6Þ EjIjEjþ1Ijþ1

XMm¼0

ðmþ 1Þðmþ 2Þðmþ 3ÞRmj fjðmþ 3Þ

(8.190)

Notice that there are only five unknown parametersf1ð0Þ; f1ð1Þ; f1ð2Þ; f1ð3Þ and U1 in Eqs. (8.187)e(8.190) through arecursive way. It means that the mode shape of the Jth section of the beamcan be expressed as linear functions of f1ð0Þ; f1ð1Þ; f1ð2Þ; f1ð3Þ, such as

fjðmÞ ¼ f0mðU1Þf1ð0Þ þ f1mðU1Þf1ð1Þ þ f2mðU1Þf1ð2Þ þ f3mðU1Þf1ð3Þm ¼ 0; 1; 2; 3

(8.191)


The boundary conditions at the ends of the beam shown in Fig. 8.24can be expressed in dimensionless form

d2f1ð0ÞdX2

1

� KR1df1ð0ÞdX1

¼ 0;d3f1ð0ÞdX3

1

þ KT1f1ð0Þ ¼ 0 (8.192a)

d2fjðRjÞdX2

j

� KR1dfjðRjÞdXj

¼ 0;d3fjðRjÞdX3

j

þ KT1f1ðRjÞ ¼ 0 (8.192b)

where KR1 ¼ (KR1L/E1I1), KT1 ¼ (KT1L3/E1I1), KRj ¼ (KRjL/E1I1),

KTj ¼ (KTjL3/E1I1), andRj ¼ (Lj/L); KT1 and KTj are the stiffness of the

translational springs; and KR1 andKRj are the stiffness of the rotationalsprings at x1 ¼ 0 and xj ¼ Lj, respectively. The differential transformationsof Eq. (8.192) are obtained with the definition Eq. (8.179) as

2f1ð2Þ � KR1f1ð1Þ ¼ 0 6f1ð3Þ � KT1f1ð0Þ ¼ 0 (8.193)

2fJð2Þ � KRJfJð1Þ ¼ 0 6fJð3Þ � KTJfJð0Þ ¼ 0 (8.194)

Substituting Eq. (8.189) into Eq. (8.192) and then rewriting Eqs. (8.193)and (8.194) in matrix form, we obtain:

26664

0 �KR1 2 0

KT1 0 0 6

D31 D32 D33 D34

D41 D42 D43 D44

37775

2666664f1ð0Þf1ð1Þf1ð2Þf1ð3Þ

3777775 ¼ 0 (8.195)

where D3m ¼ kRj f1m(U1), D4m ¼ 6fm(U1) � kTj f0m(U1) with m ¼ 0, 1, 2, 3.From Eq. (8.195), the dimensionless frequency parameter U1 can besolved by:

det

26664

0 �KR1 2 0

KT1 0 0 6

D31 D32 D33 D34

D41 D42 D43 D44

37775 ¼ 0 (8.196)

Notice that the matrix in Eq. (8.196) is singular at each frequencyparameter U1, and the unknown parameters f1ðmÞ (m ¼ 0, 1, 2, 3) cannotbe directly determined. However, one may choose one quantity of f1ðmÞas the arbitrary nonzero constant, and the remaining three as functions ofthis arbitrary constant. Without loss of generality, one may choose


f1ð0Þ ¼ 1. Hence, the remaining three can be solved as functions by usingEq. (8.196): 2

640 0 6

D32 D33 D34

D42 D43 D44

375 ¼

2664f1ð1Þf1ð2Þf1ð3Þ

3775 ¼

264KTL

D31

D41

375 (8.197)

By using the solved f1ðmÞ, the closed-form series solution for thesecond spatial derivative of the mode shapes

�v2fJðXÞ�vX2

J

�for each

section can be obtained. The second spatial derivative of the mode shapefunction for the entire beam can be written as

df2ðXÞdX2

¼"df2

1ðXÞdX2

1

df22ðXÞdX2

2.df2

j ðXÞdX2

j

#(8.198)

Substituting (d2f(X)/dX2) into Eqs. (8.176) and (8.177), the PVDF sensorshape coefficientsBj are determined: Then substitutingBj into Eq. (8.174), theshape of the modal sensor can be obtained. It should be noticed that theproposedDTMcan be used to obtain the second spatial derivative of themodeshape function (d2f(X)/dX2) of beams consisting of an arbitrary number ofsteps through a recursiveway.Consequently, the complexity of the problem isreduced to the same order of a beam without any steps. (d2f(X)/dX2) can beobtained by solving a set of algebraic equations with only five unknowns. It iswell known that any type of nonuniform beam can be approximated by astepped beam with a suitable number of uniform sections. It means that theproposedmethod can be used to design the piezoelectricmodal sensors for anytype of nonuniform EulereBernoulli beams.

A. Simply Supported Beam With Two StepsIn this example, the shaped piezoelectric modal sensor of the two-stepsimply supported beam is studied using the DTM. The beam parametersare as shown in Fig. 8.25. For simply supported boundary condition, thestiffness of the translational and rotational springs in Eq. (8.192) can be set to1 � 109 and 0, respectively. It is important to check how rapidly thedimensionless natural frequencies U1(n) computed through the DTMconverge toward the exact value as the series summation limit M isincreased, because the closed-form series solution of the second spatialderivative of the mode shape functions (d2fj(Xj)/dXj

2) in Eq. (8.182) willhave to be truncated in numerical calculations. Fig. 8.26 shows the first five


dimensionless natural frequencies U1(n) as the function of the series sum-mation limit M. The dimensionless natural frequencies U1(1) � U1(5)converge to 3.16, 6.28, 9.68, 12.57, and 16.31 very quickly as the seriessummation limit M is increased. The excellent numerical stability of thesolution can also be observed in Fig. 8.26. For simplicity, the DTM solu-tions are truncated to M1/420 in Eq. (8.182) in all the subsequent calcula-tions. Fig. 8.27 shows the first four mode shapes and the correspondingsecond spatial derivative of the mode shapes. Fig. 8.28 shows the shapes of

25 10

M15 20

4

6

8

10

12

14

16

18

1st natural frequency2nd natural frequency3rd natural frequency

4th natural frequency5th natural frequency

Ω1

(n)

Figure 8.26 The first five dimensionless natural frequencies U1(n) as the function ofthe series summation limit M for the beam shown in Fig. 8.25.

Figure 8.25 A simply supported two-step beam with constant width and step varyingthickness when R1 ¼ 17/38, R2 ¼ 4/38, R3 ¼ 17/38, h2/h1 ¼ 2, h3/h1 ¼ 1 (dimensionsare not scaled).


(a)

(b)

2

1

0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0 0.1 0.2 0.3 0.4X

X

0.5 0.6 0.7 0.8 0.9 1

–1

–2

1

0

–1

–0.5

0.5

∂2 Φ(X

)/∂ X

2Φ

(X)

1st mode shape2nd mode shape

3rd mode shape4th mode shape

Figure 8.27 The first four (a) normalized mode shapes and (b) corresponding secondspatial derivative of the mode shapes for the beam shown in Fig. 8.25.

1

00 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

1

0

1

0

1

0

(a)

(b)

(c)

(d)

+

+

+

+

–

–

– + –

X

F(X

)

Figure 8.28 The modal sensor shapes of (a) the first; (b) the second; (c) the third;(d) the fourth mode for the beam shown in Fig. 8.25.


the PVDF modal sensors for the first four structural modes. It can be foundthat the second spatial derivative of the mode shape and modal sensor shapeof the third mode shown in Figs. 8.26(b) and 8.27(c) agree well with thoseshown in Ref. [9].

B. Tapered Beam With Abrupt Changes of Cross SectionIn this second, a tapered cantilever beam with abrupt changes of crosssection is considered to verify the present method, the beam cross sectionhas constant width but with linearly varying thickness at each segment. Thedimension of the beam is shown in Fig. 8.29. Figs. 8.30 and 8.31 show themode shape, the corresponding second spatial derivative of the modeshapes, and the modal sensor shapes for the first four structural modes.

8.11 FREE VIBRATIONS OF OSCILLATORS

It is known that the free vibrations of an autonomous conservative oscillatorwith inertia and static type fifth-order nonlinearities are expressed by [10].

d2uðtÞdt2

þ luðtÞ þ ε1u2ðtÞ d

2uðtÞdt2

þ ε1uðtÞ�duðtÞdt

�2

þ ε2u4ðtÞ d

2uðtÞdt2

þ 2ε2u3ðtÞ

�duðtÞdt

�2

þ ε3u3ðtÞ þ ε4u

5ðtÞ ¼ 0:

(8.199)

The initial conditions for Eq. (8.199) are given by u(0) ¼ A and du(0)/dt ¼ 0, where A represents the amplitude of the oscillation. Motion isassumed to start from the position of maximum displacement with zero

h1 h2

Lx/2 Lx/2

h3 h4

Figure 8.29 A stepped tapered cantilever beam (clamped at left end) with constantwidth when h2/h ¼ 0.8, h3/h1 ¼ 1, h4/h1 ¼ 0.8 (dimensions are not scaled).


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0–1

–0.5

0.5

0

0

2

1

–1

–2

1

0.1 0.2 0.3 0.4 0.5

X

X0.6 0.7 0.8 0.9 1

(a)

(b)

1st mode shape3rd mode shape

2nd mode shape4th mode shape

∂2 Φ(X

)/∂ X

2Φ

(X)

Figure 8.30 The first four (a) normalized mode shapes and (b) corresponding secondspatial derivative of the mode shapes for the beam shown in Fig. 8 29.

1

0

1

0

1

0

1

00 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

+

+ +

++

–

–

– –

(a)

(b)

(c)

(d)

F(X

)

X

Figure 8.31 The modal sensor shapes of (a) the first; (b) the second; (c) the third;(d) the fourth mode for the beam shown in Fig. 8.29.


initial velocity. Eq. (8.199) can be expressed as two simultaneous first-orderdifferential equations written in terms of u(t) and v(t), i.e.,

duðtÞdt

¼ yðtÞ (8.200)

dyðtÞdt

þ luðtÞ þ ε1u2ðtÞ dyðtÞ

dtþ ε1uðtÞy2ðtÞ þ ε2u

4ðtÞ dyðtÞdt

þ 2ε2u3ðtÞy2ðtÞ

þ ε3u3ðtÞ þ ε4u

5ðtÞ ¼ 0

(8.201)

Taking the differential transform of Eq. (8.201) with respect to time tgives

kþ 1H

Uðkþ 1Þ ¼ V ðkÞ (8.202)

where U(k) and V(k) are the differential transformations of functions u(t)and v(t), respectively. Taking the differential transform of Eq. (8.201)with respect to time t yields

kþ 1H

V ðkþ 1Þ þ lUðkÞ þ ε1

Xk

l¼0

kþ 1H

V ðk� lþ 1ÞXl

m¼0

Uðl� mÞUðmÞ

þ ε1

Xk

l¼0

Uðk� lÞXl

m¼0

V ðl� mÞV ðmÞ þ ε2

Xk

l¼0

kþ 1H

V ðk� lÞ

þ Plm¼0

Uðl� mÞPmp¼0

Uðm� pÞPpq¼0

Uðp� qÞUðqÞ þ 2ε2Xk

l¼0

Uðk� lÞ

þ Plm¼0

Uðl� mÞPmp¼0

Uðm� pÞPpq¼0

V ðp� qÞV ðqÞ

þ ε3

Xk

l¼0

Uðk� lÞXl

m¼0

Uðl� mÞUðmÞ þ ε4

Xk

l¼0

Uðk� lÞ

þXl

m¼0

Uðl� mÞXmp¼0

Uðm� pÞXp

q¼0

Uðp� qÞUðqÞ¼ 0 (8.203)


Eqs. (8.202) and (8.203) can be rewritten as follows:

Uðkþ 1Þ ¼ Hkþ 1

V ðkÞ (8.204)

V ðkþ 1Þ ¼ �Hðkþ 1Þ½1þ ε1U

2ð0Þ þ ε2U4ð0Þ�

�

26666666666666666666664

lUðkÞþ ε1

Xk

l¼1

kþ 1H

V ðk� lþ 1ÞXl

m¼0

Uðl� mÞUðmÞ

þε1Xk

l¼0

Uðk� lÞXl

m¼0

V ðl� mÞV ðmÞ

þ ε2

Xk

l¼1

kþ 1H

V ðk� lÞ þXl

m¼0

Uðl� mÞXmp¼0

Uðm� pÞXp

q¼0

Uðp� qÞUðqÞ

2ε2

Xk

l¼0

Uðk� lÞ þXl

m¼0

Uðl� mÞXmp¼0

Uðm� pÞXp

q¼0

V ðp� qÞV ðqÞ

þε3Xk

l¼0

Uðk� lÞXl

m¼0

Uðl� mÞUðmÞ

þ ε4

Xk

l¼0

Uðk� lÞ þXl

m¼0

Uðl� mÞXmp¼0

Uðm� pÞXp

q¼0

Uðp� qÞUðqÞ

37777777777777777777775

¼ 0

(8.205)

where the initial conditions are given by U(0) ¼ A and V(0) ¼ 0. The dif-ference equations presented in Eqs. (8.204) and (8.205) describe the free vi-brations of a conservative oscillator with inertia and static fifth-ordernonlinearities. From a process of inverse differential transformation, it canbe shown that the solutions of each subdomain take n þ 1 terms for thepower series of the DTM principle, i.e.,

uiðtÞ ¼Xn

k¼0

�tHi

�k

UiðkÞ; 0 � t � Hi (8.206)

yiðtÞ ¼Xn

k¼0

�tHi

�k

ViðkÞ; 0 � t � Hi (8.207)

where k ¼ 0, 1, 2,., n represents the number of terms of the power series;i ¼ 0, 1, 2,. expresses the ith subdomain; and Hi is the subdomain interval.


To verify the effectiveness of the proposed DTM, the fourth-orderRungeeKutta numerical method is used to compute the displacementresponse of the nonlinear oscillator for initial amplitude under four differentmodes. Fig. 8.32 compares the results for an example of amplitude con-ditions of the proposed DTM with those of the RungeeKutta method foreach of the four calculation modes. Note that Mode 1 results are given bylines 1 and 2, while the results for Modes 2, 3, and 4 are represented by lines3 and 4, 5 and 6, and 7 and 8, respectively.

8.12 COMPOSITE SANDWICH BEAMS WITHVISCOELASTIC CORE

The assumptions used to derive the kinematic relations and the governingequations are as follows [11]:1. The shear angle of the top and bottom face plates is neglected.2. The core layer is relatively soft and viscoelastic with a complex modulus.3. The contribution from the core layer is only by transverse shear stresses.4. Layers are assumed to be incompressible through the thickness.5. Transverse displacement does not change between the layers.6. The beam deflection is small.7. There is no slip between the layers.

Figure 8.32 Comparison of the differential transformation method (lines 2, 4, 6, 8) andRungeeKutta method (lines 1, 3, 5, 7) results with l ¼ 1 and A ¼ 1, for Modes 1e4.


The configuration of the sandwich beam is presented in Fig. 8.33. Usingthe geometry in Fig. 8.33, the kinematic relations are derived as follows:

u1 ¼ u0ðx; tÞ � h224ðx; tÞ �

�zþ h1

2

�v

vxwðx; tÞ; � h1

2� z � h1

2

(8.208)

u2 ¼ u0ðx; tÞ � z4ðx; tÞ; � h12� z � h1

2(8.209)

u3 ¼ u0ðx; tÞ þ h224ðx; tÞ �

�z� h3

2

�v

vxwðx; tÞ; � h3

2� z � h3

2

(8.210)

wi ¼ wðx; tÞ; i ¼ 1; 2; 3 (8.211)

where 4ðx; tÞ ¼ Fðx; tÞw0ðx; tÞ is the total angular displacement, u0(x, t) isthe longitudinal displacement, w(x, t) is the transverse displacement of thecentroid of viscoelastic core. Also, ui and wi correspond to the longitudinaland transverse displacements of the ith layer. The strainedisplacement rela-tions for the sandwich beam can be evaluated from Eqs. (8.208)e(8.211) asfollows:

εð1Þxx ¼ vu0

vx� h2

2v4

vx��zþ h1

2

�v2wvx2

(8.212)

gð2Þxz ¼ vw

vx� 4 (8.213)

Figure 8.33 Geometry and displacement field.


εð3Þxx ¼ vu0

vxþ h2

2v4

vx��z� h3

2

�v2wvx2

(8.214)

The stressestrain relations can be given as follows:

sð1Þxx ¼ E1ε

ð1Þxx (8.215)

sð2Þxz ¼ G*

2gð2Þxz (8.216)

sð3Þxx ¼ E3ε

ð3Þxx (8.217)

where G*2 ¼ G2ð1þ ihÞ is the complex shear modulus of the viscoelastic

core and E1, E3 corresponding to the Young’s moduli of the constraininglayer and base layer, respectively. Let us consider the face layers as ortho-tropic. Then, the Young’s moduli of these layers can be calculated asfollows:

Ei ¼ Q11 cos4ðqiÞ þQ22 sin

4ðqiÞ þ 2ðQ12 þ 2Q66Þsin2ðqiÞcos2ðqiÞ(8.218)

where qi is the angle of lamination of the ith layer and

Q11 ¼ E11

1� n12n21; Q12 ¼ n12E22

1� n12n21; Q22 ¼ E22

1� n12n21;

Q66 ¼ G12; n21 ¼ n12E22

E11

(8.219)

are the material properties of the composite face layers. For the free vibra-tions of the beam, Hamilton’s principle can be expressed as follows:Z T

oðdU � dKÞdt ¼ 0 (8.220)

where U and K correspond to the elastic strain energy and the kineticenergy, respectively. For the problem considered, Eq. (8.220) becomes:Z T

0

Z L

0

Z h1=2

�h1=2

sð1Þxx dε

ð1Þxx � r1

�vwvt

dvwvt

þ vu1vt

dvu1vt

�dzdxdt

þZ T

0

Z L

0

Z h2=2

�h2=2

sð2Þxz dg

ð2Þxz � r2

�vwvt

dvwvt

þ vu2vt

dvu2vt

�dzdxdt

þZ T

0

Z L

0

Z h3=2

�h3=2

sð3Þxx dε

ð3Þxx � r3

�vwvt

dvwvt

þ vu3vt

dvu3vt

�dzdxdt ¼ 0

(8.221)


From Eq. (8.221), the governing equations are obtained as follows:

ðE1h1 þ E3h3Þ v2u0vx2

þ h22ðE3h3 þ E1h1Þ v

24

vx2þ 12

�E3h

23 þ E1h

21

� v3wvx3

¼ ðh1r1 þ h2r2 þ h3r3Þv2u0vt2

þ h22ðh3r3 � h1r1Þ

v24

vt2

12

�h23r3 � h21r1

� v3wvt2vx

(8.222)

ðE3h3 þ E1h1Þ v2u0vx2

þ h22ðE1h1 þ E3h3Þ v

24

vx2þ 12

�E1h

21 þ E3h

23

� v3wvx3

þ 2G2

�vwvx2

� 4

�

¼ ðh3r3 � h1r1Þv2u0vt2

þ h26ð3h1r1 þ h2r2 þ 3h3r3Þ

v24

vt2

(8.223)

12

�h21r1 þ h23r3

� v3wvt2vx

þG2h2

�v4

vx� vwvx2

�þ 12

�E3h

23 � E1h

21

� v3u0vx3

þ h24

�E1h

21 þ E3h

23

� v34vx3

þ 13

�E1h

31 þ E3h

33

� v4wvx4

� 12

�h23r3 � h21r1

� v3u0v2tvx

� ðh1r1 þ h2r2 þ h3r3Þv2wvt2

þ h4

�h21r1 þ h23r3

� v34

v2tvxþ 13

�h31r1 þ h33r3

� v4wvx2vt2

(8.224)

Notice that, Eq. (8.222) that governs the axial motion of the sandwichbeam is uncoupled with the shear angle and the transverse displacement fora symmetrically sectioned beam, where the material and geometric prop-erties of layers 1 and 3 are identical. The boundary conditions are alsoevaluated as follows:

�N ð1Þ

x þN ð3Þx

�du0

�� L0 ¼ 0 (8.225)


h22

�N ð3Þ

x þN ð1Þx

�d4

�� L0 ¼ 0 (8.226)

h22

�Qð2Þ

x

v

vxM ð1Þ

x þ v

vxM ð3Þ

x þ h12

v

vxN ð1Þ

x � h32

v

vxN ð3Þ

x

�dw

�� L0 ¼ 0 (8.227)

�h32N ð3Þ

x � h12N ð1Þ

x �M ð1Þx �M ð3Þ

x

�d

�vwvx

�� L0 ¼ 0 (8.228)

where

N ðiÞx ¼

Zhi=2�hi=2

sðiÞxxdz; QðiÞ

x ¼Zhi=2

�hi=2

sðiÞxxdz;

M ðiÞx ¼

Zhi=2�hi=2

zsðiÞxxdz; i ¼ 1; 2; 3

(8.229)

For the problem considered, the sectional moment and forces are ob-tained from Eqs. (8.212)e(8.217) together with Eq. (8.229) as follows:

N ð1Þx ¼ E1

�h1vu0vx

� 12h1h2

v4

vx� 12h21v2wvx2

�(8.230)

N ð3Þx ¼ E3

�h3vu0vx

þ 12h2h3

v4

vx� 12h23v2wvx2

�(8.231)

Qð2Þx ¼ G2h2

�vwvx

� 4

�(8.232)

M ð1Þx ¼ � 1

12E1h

31

v2wvx2

(8.233)

M ð3Þx ¼ � 1

12E3h

33

v2wvx2

(8.234)

For the harmonic vibrations of the sandwich beam, the displacementfield can be assumed as follows:

u0ðx; tÞ ¼ u0ðx;uÞeiut;wðx; tÞ ¼ wðx;uÞeiut;4ðx; tÞ ¼ 4ðx;uÞeiut.

(8.235)


Then, Eqs. (8.22)e(8.24) become:

ðE1h1 þ E3h3Þu000 þ h22ðE3h3 � E1h1Þ400 þ 1

2

�E3h

23 � E1h

21

�w00

¼ �u2

ðh1r1 þ h1r2 þ h3r3Þu0 þ

h22ðh3r3 � h1r1Þ40 þ 1

2

�h23r3 � h21r1

�w0

(8.236)

ðE3h3 � E1h1Þu000 þ h22ðE1h1 � E3h3Þ400 þ 1

2

�E1h

21 � E3h

23

�w

000 þ 2G2ðw0 � 4Þ

¼ �u2

ðh3r3 � h1r1Þu0 þ

h26ð3h1r1 þ h2r2 þ 3h3r3Þ4þ 1

2

�h21r1 þ h23r3

�w0

(8.237)

G2h2ð40 � w00Þ þ 12

�E3h

23 � E1h

21

�u0000 þ h2

4

�E1h

21 þ E3h

23

�4

000

þ 13

�E1h

31 þ E3h

33

�win

¼ �u2

12

�h23r3 � h21r1

�u0 � ðh1r1 þ h2r2 þ h3r3Þw

þ h24

�h21r1 þ h23r3

�40 þ 1

3

�h31r1 þ h33r3

�w00

(8.238)

The resulting differential equation system that is presented in Eqs.(8.236)e(8.238) are transformed by using the basic rules of DTM,

Ukþ2 ¼ 14ðkþ 3Þðh1 � h3Þwkþ3

� 2G2ðE1h1 � E3h3Þ þ u2h1h3ðE1h3r3 � E3h1r1Þ4ðkþ 2ÞE1E3h1h3

wkþ1

þ 12G2ðE1h1 � E3h3Þ þ u2h2½E3h3�ð6h1r1 þ h2r2Þ � E1h1ðh2r2 þ 6h3r3Þ24ðkþ 1Þðkþ 2ÞE1E3h1h3

Jk

�u2½E3h3ð2h1r1 þ h2r2Þ þ E1h1ðh2r2 þ 2h3r3Þ�4ðkþ 1Þðkþ 2ÞE1E3h1h3

Uk

(8.239)


Jkþ2 ¼ 12G2ðE1h1 þ E3h3Þ � u2h2ð6E3h1h3r1 þ E1h1h2r2 þ E3h2h3r2 þ 6E1h1h3r3Þ12ðkþ 1Þðkþ 2ÞE1E3h1h2h3

Jk

�ðkþ 3Þðh1 � h3Þ2h2

wkþ3 � 2G2ðE1h1 þ E3h3Þ þ u2h1h3ðE3h1r1 þ E1h3r3Þ2ðkþ 2ÞE1E3h1h3

wkþ1

þu2½E3h3ð2h1r1 þ h2r2Þ � E1h1ðh2r2 þ 2h3r3Þ�2ðkþ 1Þðkþ 2ÞE1E3h2h3

Uk

(8.240)

Boundary conditions for the sandwich beam:

wkþ4 ¼ 1

2ðkþ 1Þðkþ 2Þðkþ 3Þðkþ 4Þ�E1h31 þ E3h

33

��f6u2h2ðh3 � h1Þr2ðkþ 1ÞUkþ1

þ2�6G2ðh1 þ 2h2 þ h3Þ � u2

�h31r1 þ h33r3

��ðkþ 1Þðkþ 2Þwkþ2

þ24u2ðh1r1 þ h2r2 þ h3r3Þwk

�u2h22ðh1 þ h3Þr2 � 12G2ðh1 þ 2h2 þ h3Þ

�ðkþ 1ÞJkþ1g

(8.241)

The boundary condition and matrix of solution can be found in detail inRef. [11]. The real and imaginary parts of normalized mode shapes for thetransverse displacement are presented in Fig. 8.34 for the first three modes.The convergence of the loss factor for the first four modes with increasingnumber of terms considered is presented in Fig. 8.35. The figure presents arapid convergence for the loss factors. Another important point is that, it isnecessary to consider more number of terms in the DTM calculations toevaluate higher modes.

0.00–1.0

–0.5

0.0

Re

w

x

0.5

1.0

0.05 0.10 0.15 0.00–1.0

–0.5

0.0

Im w

x

0.5

1.0

0.05 0.10 0.15

Figure 8.34 The real and imaginary parts of the first three mode shapes (-, first mode;–, second mode; -.- third mode).


REFERENCES[1] Hatami M, Vahdani S, Ganji DD. Deflection prediction of a cantilever beam subjected

to static co-planar loading by analytical methods. HBRC Journal 2014;10(2):191e7.[2] Suddoung K, Charoensuk J, Wattanasakulpong N. Vibration response of stepped FGM

beams with elastically end constraints using differential transformation method. AppliedAcoustics 2014;77:20e8.

[3] Mao Q. Design of shaped piezoelectric modal sensors for cantilever beams with in-termediate support by using differential transformation method. Applied Acoustics2012;73(2):144e9.

[4] Kuo B-L, Lo C-Y. Application of the differential transformation method to the so-lution of a damped system with high nonlinearity. Nonlinear Analysis: Theory,Methods & Applications 2009;70(4):1732e7.

[5] Mei C. Application of differential transformation technique to free vibration analysis ofa centrifugally stiffened beam. Computers & Structures 2008;86(11):1280e4.

[6] Yeh Y-L, Chi Wang C, Jang M-J. Using finite difference and differential trans-formation method to analyze of large deflections of orthotropic rectangular plateproblem. Applied Mathematics and Computation 2007;190(2):1146e56.

[7] Yalcin HS, Arikoglu A, Ozkol I. Free vibration analysis of circular plates by differentialtransformation method. Applied Mathematics and Computation 2009;212(2):377e86.

[8] Ni Q, Zhang ZL, Wang L. Application of the differential transformation method tovibration analysis of pipes conveying fluid. Applied Mathematics and Computation2011;217(16):7028e38.

[9] Mao Q. Design of piezoelectric modal sensor for non-uniform EulereBernoulli beamswith rectangular cross-section by using differential transformation method. MechanicalSystems and Signal Processing 2012;33:142e54.

[10] Chen S-S. Application of the differential transformation method to the free vibrationsof strongly non-linear oscillators. Nonlinear Analysis: Real World Applications2009;10(2):881e8.

[11] Arikoglu A, Ozkol I. Vibration analysis of composite sandwich beams with viscoelasticcore by using differential transform method. Composite Structures 2010;92(12):3031e9.

4

3

2

Loss

fact

or

1

015 20 25 30

N35 40

1. Mode2. Mode3. Mode4. Mode

Figure 8.35 Convergence of the modal loss factor with N.


INDEX

‘Note: Page numbers followed by “f” indicate figures, “t” indicate tables.’

BBasseteBoussinesqeOseen (BBO)

equation, 301e302

CCantilever beamdeflection prediction ofHPM and DTM, 340e341slope parameter, variation of,

341e343piezoelectric modal sensorsDTM solutions, 357mode shapes, second spatial derivative

of, 353e354PVDF sensor, 351e353, 357

Carbon nanotubes (CNTs), 236CrankeNicolson method (CNM),

187e192

DDifferential quadrature method (DQM),

311e314Differential transformation method

(DTM)advantage, 1cantilever beamdeflection prediction of, 338e343piezoelectric modal sensors, 351e357

concept of, 1eigenvalue problemsfirst eigenfunction, 74e76, 81e83second eigenfunction, 76e77, 83e84StrumeLiouville problem, 71third eigenfunction, 77e78

FDTMinverse transformation rule, 63Ricatti equation, 63RiemanneLiouville sense, 59e60theorems of, 60e61

fractional order partial differentialequations

Caputo fractional derivative, 84e85generalized two-dimensionaldifferential transform, 85e87

linear inhomogeneous time-fractionalequation, 87

linear space-fractional telegraph,88e89

free vibrationcentrifugally stiffened beam,361e363

circular plates, 369e378oscillators, 393e397

heat transfer problems. See Heat transferproblems

high nonlinearity, damped system with,358e359

higher-order initial value problemssecond-order initial value problem,55e57

third-order initial value problem,57e59

hybrid DTM-FDMfluid motion, 16e17Hall effect, 16e17inverse transformation, 13nondimensional quantities, 16e17Nusselt number, 18

integro-differential equation,64e71

IVPs and ODE, 20e25fixed grid size, 22e23varying grid size, 23e25

MDTM, 96e101mechanical problems, 55MHD boundary layeranalytical solutions, 164e169, 169fPade approximation, 164e168

405

Differential transformation method(DTM) (Continued)

Ms-DTMinitial-value problem, 8e9k-th derivative, 7Lagrangian equation, 10e11particles motion modeling, 9, 11

nanofluid flow, flat platecomputational errors, 176, 178fCu-water nanofluid, 176local Nusselt number, 182e184, 184fphysical model, 169e170, 170fPrandtl number, 176e180,181fe182f

non-newtonian fluid flow analysisfemoral artery and coronary arteries,187e192

higher velocity profiles, 187e192hybrid-DTM, 187e192non-Newtonian blood, properties of,187e192, 189t

nonuniform EulereBernoulli beamsbeam cross section, 393dimensionless natural frequencies,390e393

PVDF sensor, 383orthotropic rectangular plate, deflections

of, 364e369Padé approximationEckert and squeeze numbers, 43e44nanofluid, thermo-physical propertiesof, 37e39

viscous dissipation effect, 37e39PDEs for, two-dimensional differential

transformationanalytic solution, 34e35, 35fbasic definitions and fundamentaltheorems, 25e26

boundary conditions, 33initial conditions, 33e34

pipes conveying fluid, vibration of,378e383

porous medium flows. See Porousmedium flows

principle ofboundary conditions, 6e7Darcy’s model, 4

porous fin, 3e5, 4fporous materials, properties of, 5, 6tTaylor series expansion, 3

RDTMgeneralized DrinfeldeSokolov (gDS)equations, 91e92

KaupeKupershmidt (KK) equation,94e95

t-dimensional spectrum function, 90singular two-point BVP, 44e48stepped FGM beams, vibration analysis

ofmatrix form, 347e349mechanical properties, 350mode shape functions, 350e351nonzero values, 346e347translational and rotational springfactors, 350

types of, 343e345, 344ftwo-dimensional viscous flow

Berman’s formula, 156nondimensional wall dilation rate,155, 160e161

numerical method vs. analyticalsolutions, 159, 159f

permeation Reynolds numbers,160e161

shear stress, 156stream function and mean flowvorticity, 154

viscoelastic core, composite sandwichbeams with, 397e403

EEckert number, 17, 40EulereBernoulli beam theory,

224e228

FFinite difference method (FDM),

13e20, 187e192Fractional differential transform method

(FDTM)inverse transformation rule, 63Ricatti equation, 63RiemanneLiouville sense, 59e60theorems of, 60e61

406 Index

Free vibrationcentrifugally stiffened beam, 361e363circular platesantisymmetric case, 370boundary/regularity conditions,

transformation of, 372e378classical plate vibration theory, 369radial mode function, 369symmetric case, 370transformation operations, 370

oscillators, 393e397Functionally graded materials (FGMs),

224e226

HHartmann number, 17Heat transfer problemsconstant profile, longitudinal fins withtemperature-dependent internal heat

generation and constant thermalconductivity, 104e110

temperature-dependent internal heatgeneration and temperature-dependent thermal conductivity,106e107, 110e112

FalknereSkan wedge flow, 133e139flat plate, thermal boundary layer onfirst-order ordinary differential

equation, 131linear second-order ordinary

differential equations, 132Prandtl numbers, 133third-order ordinary differential

equation, 131free convection problemdimensionless temperature

distributions, 150, 151fdimensionless velocity distributions,

150, 150finitial value problems, 143e144velocity and temperature profiles,

140, 141fnon-Newtonian nanofluid, natural

convection flow ofdimensionless non-Newtonian

viscosity, effect of, 117e120, 118fEckert number, 117e120, 121f

fourth-order RungeeKutta method,116e117

MaxwelleGarnetts (MG) model, 114nanoparticle volume fraction, effectof, 117e120, 118f, 122f

thermophysical properties of,112e113, 113t

two-dimensional heat transferconvex parabolic profile, fins of,126e127

dimensionless thermal conductivity,124

rectangular profile, fins of, 125e126Homotopy perturbation method (HPM),

293e296

KKleineGordon equation, 97

LLeast square method (LSM), 201e202

MMagnetohydrodynamic (MHD),

161e162Hall parameter, effect of, 246, 248fHartmann number, effect of, 245, 246fhybrid-DTM, 244e245NaviereStokes equation, 240e241nondimensional quantities, 240e241Nusselt number and skin friction

coefficient, 246e247porous channel, micropolar fluid inCartesian coordinates, 247e251lower channel wall, 247e251microelements, 251Peclet number, topographical effectsof, 257

Reynolds number, 255velocity profile, values of, 255e257

porous surfaces, MHD viscous flowCartesian coordinate system,258e260

Coriolis force, 270different nondimensional parameters,constant values with, 270, 271t

DTM, 265e268

Index 407

Magnetohydrodynamic (MHD)(Continued)

Lorentz force, 270nondimensional form, 264e265temperature profile, 270e276, 278f,281f

velocity component profile, 270,274f, 277f

Reynolds number, effect of, 245e246,247f

MaxwelleGarnetts (MG) model, 114Modified differential transformation

method (MDTM), 96e101, 242Multistep differential transformation

method (Ms-DTM)initial-value problem, 8e9k-th derivative, 7Lagrangian equation, 10e11sparticles motion modeling, 9, 11

NNanofluiddivergent/convergent channelsanalytical methods, 201e202Hartmann number, effect of, 202MHD JefferyeHamel flow, 201e202nanoparticles volume fraction, effectof, 205e207, 206f

NaviereStokes and Maxwell’sequations, 197e199

Reynolds number, 205, 206fthermophysical properties of, 202tvelocity parameter, 199

MHD couette nanofluid flowBrownian motion, 209e210Hall parameter, 214, 215fnondimensional variables andparameters, 210e211

nonlinear partial equations, 211e212Nusselt number, 213e214thermophysical properties, 207e208,208t

Turbulent CuOewater nanofluidflow and heat transfer, 207e208

volume fraction, 213e214, 214fnanobeams, vibration analysis ofFGMs, 224e226

MorieTanaka models, 227nonlocal elasticity model, 229DTM rule, 232

parallel platesBrownian motion, 218e219squeeze number, 223e224thermophysical properties, 217e218,218t

viscous dissipation effect, 217single-walled CNTs, buckling analysis

of, 236e238Newton’s law, 156NewtoneRaphson method, 232Nonspherical particles sedimentationBBO equation, 301e302subdomain solutions, 304e305Nusselt numbernanoparticle volume fraction and

Hartmann number, effect of,215e216, 215f

Reynolds number and turbulent Eckertnumber, effect of, 215e216,216f

Turbulent parameter and Hall parameter,effect of, 216e217, 216t

PPartial differential equations (PDEs),

25e26Polyvinylidene fluoride (PVDF), 351Porous finporous materials, properties of, 5, 6ttemperature-dependent heat generation,

3e4, 4fPorous medium flowsporous channel, micropolar fluid in

Cartesian coordinates, 247e251lower channel wall, 247e251microelements, 251Peclet number, topographical effectsof, 257

Reynolds number, 255velocity profile, values of, 255e257

porous surfaces, MHD viscous flowCartesian coordinate system,258e260

Coriolis force, 270

408 Index

different nondimensional parameters,constant values with, 270, 271t

DTM, 265e268Lorentz force, 270nondimensional form, 264e265temperature profile, 270e276, 278f,

281fvelocity component profile, 270,

274f, 277fPrandtl number, 17, 40PVDF. See Polyvinylidene fluoride

(PVDF)

RReduced differential transform method

(RDTM)generalized DrinfeldeSokolov (gDS)

equations, 91e92KaupeKupershmidt (KK) equation,

94e95t-dimensional spectrum function, 90

Reynolds number, 17RungeeKutta fourth-order method,

57e59

SSingular two-point boundary value

problem (BVP), 44e48Spherical particle, motion offluid forced vortexDQM, 314DTM solution, 311e314nondimensional parameter, 310radial and tangential components of,

308e309

microparticles, combustion of,315e318

plane couette fluid flowDTMePade approximant, 301, 305HPMePade method, 293e296inverse differential transformation,298e299

nontrivial solution, 290particle positions, 296e297, 297fparticle velocity, 296e297,296fe297f

Vander Werff model,287e289

rotating parabolaLagrange’s equation, 284e285Ms-DTM, 285, 287, 288t,287f

particles motion modeling, 283soluble particle, transient vertically

motion of, 328e336unsteady sedimentation ofacceleration variation, 324, 327fBBO equation, 318e319, 321tPadé approximation, 323e324physical properties of, 320terminal velocity, 323e324velocity variations, 324, 327f

TTaylor transformation method, 24

VVander Werff model, 287e289

Index 409

Differential Transformation Method for Mechanical Engineering Problems

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