2nd Session - Dr. S. P. Asok

8/13/2019 2nd Session - Dr. S. P. Asok

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1.2 Advantages of CFD

Possible to see simultaneously the effect of various parameters

and variables on the behaviour of the system. To study the same in an

experimental setup is not only difficult and time-consuming but in

many cases, impossible.

It is much cheaper than setting up big experiments or building

prototypes of physical systems.

Numerical modeling is versatile. A large variety of problems with

different levels of complexity can be simulated on a computer.

Numerical experimentation allows models and is similar to

conducting experiments.


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In some cases, it is the only feasible substitute for experiments, for

example, modeling loss of coolant accident (LOCA) in nuclear

reactors, numerical simulation of spread of fire in a building and

modeling of incineration of hazardous waste.

However, all problems can not be solved by CFD. Experiments arestill required to get an insight into the phenomena that are not well

understood and also to check the validity of the results of computer

simulation of complex problems.


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1.3 Applications of Fluid Flow and Heat Transfer

Coupled and individual Fluid flow and heat transfer play a very

important role in nature.The various applications of fluid flow and heat

transfer are:

All methods of power production, e.g. thermal, nuclear, hydraulic,

wind, and solar power plants.

Heating and air-conditioning plants.

Chemical and metallurgical industries, e.g. furnaces, heat

exchangers, condensers and reactors.


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Design of IC engines.

Optimization of heat transfer from cooling fins.

Aircraft and spacecraft.

Design of electrical machinery and electrical circuits.

Cooling of computers.

Weather prediction and environmental pollution.

Materials processing such as solidification and melting, metal

cutting, welding, rolling, extrusion, plastics and food processing in

screw extruders, laser cutting of materials.

Oil exploration.

Production of chemicals such as cement and aluminium oxide.

Drying

Processing of solid and liquid wastes.

Bio-heat transfer as in human and animal bodies.


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We can find from classical textbook on fluid dynamics and heat

transfer that there are only a handful of analytical or exact solutions. In

actual situations, problems are lot more complex as in those involving non-

linear governing equations and / or boundary conditions, and irregular

geometry which do not allow analytical solutions to be obtained. Therefore, it

is necessary to use numerical techniques for most problems of practical

interest. Furthermore, to design and optimize thermal processes and

systems, numerical simulation of the relevant transport phenomena is a

must, since experimentation is usually too involved and expensive. However,

necessary experimentation must still be done in checking the accuracy and

validity of numerical results. Sometimes, numerical model can be refined by

input from results of a companion experimental set-up for the same problem.

1.4 Necessity of CFD in Fluid Flow and Heat Transfer


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Suppose we wish to obtain the temperature field in a domain. We

imagine that the domain is filled by a grid, and seek the values of

temperature at the grid points. Therefore, the energy equation which is the

governing differential equation for the problem is valid at all the grid points.

The governing differential equation is then transformed into a system of

difference equations resulting in a set of simultaneous algebraic equations

which means that if there are 100 grid points there will be 100 equations to

solve per variable. The simplification inherent in the use of algebraic

equations rather than differential equations is what makes numerical

methods so powerful and widely applicable.

1.5 Basic Approach in Solving a Problem by Numerical Method


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There are mainly two methods of discretising a given differential

equation as briefed below:

Finite Difference method: The usual procedure for deriving finite-

difference equations consists of approximating derivatives in the differential

equation via a truncated Taylor series. The method includes the assumption

that the variation of the unknown to be computed is somewhat like a

polynomial in x, y or z so that higher derivatives are unimportant. The

popularity of finite-difference methods is mainly due to their straight-

forwardness and relative simplicity.

1.6 Methods of Discretization


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Linear second-order partial differential equation in two independent

variables is further classified into three canonical forms elliptic, parabolic, and

hyperbolic. The general form of this class of equations is

where coefficients are either constant or functions of the independent

variables only. The three canonical forms are determined by the following

criteria:

b24ac < 0 elliptic

b24ac = 0 parabolicb24ac > 0 hyperbolic

2.2 Elliptic, Parabolic and Hyperbolic Equations

02

22

2

2

gfy

ex

dy

cyx

bx

a (2.5)


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(a) Elliptic PDE: Typical examples are

Note that in both of the equations (2.6) and (2.7), b = 0, a = 1, c = 1 which

makes b24 ac =-4 which is


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In elliptical problems, the function (x,y) must satisfy both, the differential

equation over a closed domain and the boundary conditions on the closed

boundary of the domain. This is depicted pictorially in Fig. 2.1

Fig. 2.1 Pictorial representation of an elliptic problem


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(b) Parabolic PDE: A typical example is

Where is a positive, real constant,

Note that in equation (2.8), b = 0, c = 0, a = which makes

b24 ac = 0. In parabolic problems, the solution advances outward

indefinitely from known initial values, always satisfying the known

boundary conditions as the solution progresses. This is also called

marching type of problem. The open-ended type of solution domain is

pictorially illustrated in fig. 2.2.

2

2

xt(Heat conduction or diffusion equation) (2.8)


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(c) Hyperbolic PDE: A typical example is

Fig. 2.2 Pictorial representation of a parabolic problem

2

22

2

2

tx(wave equation) (2.9)

Where 2 is a real constant (always positive).


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Note that in equation (2.9), b = 0, a = 2 , c = -1 which makes b24ac = 4 2

which

is > 0.

The solution domain of hyperbolic PDE (Fig. 2.3) has the same

open-ended nature as in parabolic PDE. However, two initial conditions are

required to start the solution of hyperbolic equations in contrast with

parabolic equations where only one initial condition is required.

Fig. 2.3 Pictorial representation of a hyperbolic problem


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The initial and boundary conditions must be specified to obtain unique

numerical solutions to the partial differential equations. This is explained

next using the one-dimensional transient heat condition equation.

2.3 Initial and Boundary Conditions

2

2

xT

tT (2.10)

Equation (2.10) is the mathematical representation of a physical problem in

which for example, temperature within a large solid slab having finite

thickness changes in the x-direction (thickness direction) as a function of

time till steady state (corresponding to t ) is reached.

Following are the three categories of boundary conditions.


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The values of the dependent variables are specified at the boundaries(Fig.2.4).

Boundary conditions (abbreviated as B.C.) of the first kind can be

expressed as

2.3.1 Dirichlet Conditions (First Kind)

B.C.1 T = f(t) or T1 at x = 0

B.C.2 T = T2 at x = L

Initial conditions (abbreviated as I.C.)

0 x L

T = f(x) at t = 0

or T = T0


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Fig. 2.4 Dirichlet condition

The derivative of the dependent variable is given as a constant or as a

function of the independent variable on one boundary.

2.3.2 Neumann Conditions (second kind)

For example 0x

Tat x=L and t 0


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This condition specifies that the temperature gradient at the right

boundary is zero (insulation condition).

Cauchy conditions: A problem which combines both Dirchlet and Neumann

conditions is considered to have Cauchy conditions (Fig. 2.5)

Fig. 2.5 Cauchy condition (Dirichlet and Neumann)


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The derivative of the dependent variable is given as a function of

the dependent variable on the boundary (see Fig. 2.6). For the heat

conduction problems, this may correspondent to the case of cooling of a

large steel slab of finite thickness L by water or oil, the heat transfer

coefficient. hbeing finite.

2.3.3 Robbins Conditions (Third kind)

Fig. 2.6 Robbins condition


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On the basis of their initial and boundary conditions, PDEsmay be further

classified into initial value or boundary value problems.

2.4 Initial and Boundary Value Problems

2.4.1 Initial Value Problems

In this case, at least one of the independent variables has an open

region. In the unsteady state conduction problems the time variable has the

range 0 t , where no condition has been specified at t = ; therefore thisis an initial value problem.


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When the region is closed for all independent variables and conditions are

specified at all boundaries, then the problem is of the boundary value type.

An example of this is the three-dimensional steady state heat conduction

(with no heat generation) problem mathematically represented by the

equation:

2.4.2 Boundary Value problems

with the boundary conditions given at all the boundaries.

02

2

2

2

2

2

z

T

y

T

x

T (2.11)


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Fluid flow is described mathematically by Navier-Stokes equations

and equation of continuity. A two-dimensional, laminar, incompressible flow

with constant viscosity is described by:

3.2 Governing Equations

xMomentum: y

uvx

uut

u

=x

p

y

u

x

u2

2

2

2

(3.5)


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Equations (3.5 and 3.6) are called the NavierStokes equations

and Eq. (3.7) is the equation of continuity. It should be noted that pdenotes

the difference between the total pressure and hydrostatic pressure. This

causes the body forces to cancel, as they are in equilibrium with the

hydrostatic pressure.

yMomentum:y

vv

x

vu

t

v

=y

p

y

v

x

v2

2

2

2

(3.6)

and continuity 0u

y

v

x (3.7)


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1. Nonlinearity: A close look at Eqns. (3.5) and (3.6) reveal that the

convective part of the momentum equations involve nonlinear terms. For

example, in Eqn. (3.5), the convection coefficient u is a function of the

dependent variable u. However, this can be treated like the conductivity k

being a function of temperature T. Starting with a guessed velocity field, one

could iteratively solve the momentum equation to arrive at the converged

solution for the velocity components. Therefore, nonlinearity poses no

problems as such. It only makes the computations more involved.

3.3 Difficulties in Solving the Navier Stokes (NS) Equations


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2. Pressure gradient: The main hurdle to overcome in the calculation ofvelocity field is the unknown pressure field. The pressure gradient behaves

like a source term for a momentum equation. But, there is no particular

difficulty in solving the momentum equations. So, the challenging task is to

determine the correct pressure distribution.

The pressure field is indirectly linked with the continuity equation.

When the correct pressure field is plugged into the momentum equations the

resulting velocity field satisfied the continuity equation. Therefore, the

calculation of pressure field is the main problem facing the numerical analyst

in solving fluid flow problems.


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3. Turbulent Flow: In turbulent flows, there will be higher wall shear stressand pressure drops. Turbulence delays the point of separation in external

flows and reduces form drag. The unsteady velocity fluctuations present

can generate flow induced oscillations leading even to structural failure.

Time dependent NS equations are needed to define the flow along with

the smallest time and length scales since the eddies formed in turbulent

flow are available in a variety of sizes. Grids should be refined needing

higher computational effort and higher round of errors can result. Direct

Numerical Simulation (DNS) type of analysis is more advisable.


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4. FINITE DIFFERENCE

METHOD


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4.1 Finite Difference Discretisation :

1. Replace the derivatives in the governing equations by algebraic

difference quotients, resulting in a system of algebraic equations. In the

below mesh, uniform spacing is followed. x need not be equal to y.

Non uniform spacing is covered under automatic grid generation.

y

x

i-1,j+1 i,j+1 i+1,j+1

i+1,j

i+1,j-1

i,j

i,j-1i-1,j-1

i-1,j P

y

x

Fig. 4.1 Finite Difference Discretion 2D


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2. Let ui,j = the x component velocity at P(i,j); using Taylor series:.........

6

)(

2

)( 3

,

3

32

,

2

2

,

,,1

x

x

ux

x

ux

x

uuu

jijiji

jiji (4.1)

;2

)(~2

,

2

2

,

,,1

x

x

ux

x

uuu

jiji

jiji Second order accurate eqn. (4.2)

;~,

,,1 xx

u

uuji

jiji First order accurate eqn. (4.3)

The neglected higher order terms Truncation errors.


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3. The truncation error for (4.2) and (4.3) are :

2 ,3 ,!

)(&;

!

)(

n

n

ji

n

n

n

n

ji

n

n

n

x

x

u

n

x

x

u

Lesser the magnitude of x, lesser is the Truncation error.

4. Rearrange terms on (4.1)

........6

)(2

2

,

3

3

,

2

2

,,1

,

xxux

xu

x

uu

xu

jiji

jiji

ji

)(,,1

,

xox

uu

x

u jiji

ji

, (4.4)

The symbol o ( x) terms of order of x and imply the order of magnitude of the

truncation error. (4) first order accurate difference representation ofjix

u

,

called

the first order forward difference


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5. In a similar way, write ui-1, jfrom ui, j:

..............6

)(

2

)()(

3

,

3

32

,

2

2

,

,,1

x

x

ux

x

ux

x

uuu

jijiji

jiji

)(,1,

,

xox

uu

x

u jiji

ji

, (4.5)

(5) = First order backward or rearward difference expression for ( u/ x) at (i,j).


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6. ..............3

)(2

3

,

3

3

,

,1,1

x

x

ux

x

uuu

jiji

jiji

2,1,1

,

)()(2

xox

uu

x

u jiji

ji

, (4.6)

(6) Second order central difference for ( u/ x)i,j

we can get,

2

2

,1,,1

,

2

2

)()(

2xo

x

uuu

x

u jijiji

ji

, (4.7)

(4.7) second order central difference form forji

x

u

,

2

2

in the same way, the

expressions for the y direction can also be got.


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8. Consider the Laplace equation2

2

2

2

2 0yT

xTT

0)(

2

)(

22

1,,1,

2

,1,,1

y

TTT

x

TTT jijijijijiji

Puty

x = mesh aspect ratio

0)1(2)( ,2

1,1,

2

,1,1 jijijijiji TTTTT

)2

1,1,

2

,1,1

,1(2

)( jijijijiji

TTTT

T


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4.2 Burgers Equation

1. Fluid mechanics problems are usually non linear. Consider an equation having

convective, diffusive and time dependent terms. Burgers introduced a simple nonlinear equation to meet the conditions:

,2

2

xxu

t= any transferable property (4.8)

2. On (4.8), if the r.h.s term is neglected, the Eulers equation (or) the inviscid

Burgers equation is got.

0x

ut

3. Conservative property: a finite difference equation passes conservative

property if it preserves integral conservation relation to the continuum.

Preserving the conservative property is of special importance in the finite

volume approach - a special form of finite difference equations.


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4.3 Upwind Scheme

1. 0x

ut

, discretizing it

011

xu

t

n

i

n

i

n

i

n

i , (4.9)

Applying the Von Neumann stability analysis, it can be found that both the finite

difference equations are unconditionally unstable. The remedy is the upwind scheme.

Equation (4.9) can be made stable by substituting the forward space difference by a

backward space difference, provided u is +ve.


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2. Considering the full Burgers equation,

2

2

xxu

t;

,0),(11

ufortermviscousx

uu

t

n

i

n

i

n

i

n

i

,0),(1 ufortermviscousx

uu n

i

n

i

The upwind method of discretization is very essential for convection dominated

problems. The upwind bias retains transporative property of flow equation.


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4.4 Upwind Differencing and Artificial Viscosity

1. For the Burgers equation

0...,..........11

uforx

uu

t

n

i

n

i

n

i

n

i

(4.10)

2. Using Taylors expansion:

n

i

n

i

n

i

n

it

t

tt

2

221

2

)(+ (4.11)

n

i

n

i

n

i

n

ix

x

xx

2

22

12

)(. (4.12)


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3. substitute (4.11) and (4.12) on (4.10)

,0),(11

ufortermviscousx

uu

t

n

i

n

i

n

i

n

i

Dropping the superscript n and subscript i

tdiffusivexx

x

xx

x

ut

tt

tt

t

3

2

223

2

22 )(0

2

)()(0)(

2

11

2

2

2

2

2

)(012

1x

xxx

tuxu

xu

t,

;2

2

2

2

termsorderhigherxxx

ut

e (4.13)

Where )1(2

1Cxue

C = Courant number =x

tu


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4. While deriving (4.13) ,2

2

t was taken as

2

22

xu . However, the non

physical coefficient e leads to a diffusion like term which is dependent on the

discretisation procedure, e is called artificial/numerical viscosity. C should be

less than 1, so that e is a non zero +ve quantity, Normally Cx= Cy . Though

upwinding schemes suffer from false diffusion, finite differences provide stable

solutions.


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4.5 Consistency

1. A finite difference representation of a PDE is consistent if: mesh Lt 0 (PDEFDE)

= meshLt(TE) 0,

Consider ,2

2

x

ua

t

u

Using the DufortFrankel FD scheme for the r.h.s

2

111

111

)(2 x

uuuua

t

uu nin

in

in

in

in

i ,

The TE = )(6

1)(

12 3

32

2

22

4

4

tt

u

x

t

t

uax

x

ua n

i

n

i

n

i. The TE will be meaningful

forx

ttx

x

t;0&0,0


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2.0,

)(

0,

)(

xt

TELt

xt

FDEPDELt= ,

2

22

t

ua This r.h.s term is not vanishing

when the mesh vanishes.

Now reconstitute the PDE from the FDE2

2

2

22

xua

tua

tu we have

started from a parabolic equation and ended up in a hyperbolic equation. Thus the Dufort

Frankel scheme is not consistent unless 0

x

ttogether with t, x 0.


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4.6 Explicit and Implicit Methods

1. We have:2

11

1

)(

2

x

uuua

t

uu n

i

n

i

n

i

n

i

n

i , using the forward time and central space

(FTCS) scheme, the unknown dependent variable uin+1

can be explicitly got from the

known values of ui+1n, ui

n, ui-1

n and this is a typical example for explicit finite

difference method.

2. Now let us try a different discretion of ut = a2u, express the spatial differences on the

r.h.s. in terms of averages between n and (n+1) time level:

2

1

1

1

1

1

1

1

1

)(

22

2 x

uuuuuua

t

uu nin

i

n

i

n

i

n

i

n

i

n

i

n

i (4.14)

This is called the Crank-Nicolson implicit scheme; i.e. the unknown u in+1

has been

expressed in terms of the known quantities at time level n and the unknown quantities

at time level n+1; hence equation (4.14) at a grid point i cannot itself result in a

solution for uin+1

. We will need the equation to be written for all grid points and later

solve simultaneously.


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4.7 Error and Stability Analysis

1. Discretization error = TE + any error introduced by the numerical treatment of the

b.cs. If A = Analytical solution, D = exact solution of the FDE and N = numerical

solution from a real computer with finite accuracy, then discretization error = A-D.

Round off error = = N-D. The solution will be stable if ivalues shrink or at least

stay the same, as the solution progresses from n to n+1. If ivalues grow larger, then

the solution becomes unstable for a solution to be stable : 11 n

i

n

i


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4 9 Example for Unstable Calculation


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4.9 Example for Unstable Calculation

1. Consider the heat condition equation2

2

x

ua

t

u;

The simple explicit finite difference representation is:

2

11

1

)(

)2(

x

uuua

t

uu n

i

n

i

n

i

n

i

n

i 211

1

)(;)21()(

x

taruruuru

n

i

n

i

n

i

n

i

It can be found that the solution is stable for r 1/2

2. Consider a case where

r1/2 say r =1, ui

n+1 =

1(100+100) + (1-2) 0 =

200C; which cannot be

correct.

Fig. 4.2 Unstable Calculation


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5. FINITE DIFFERENCE

APPLICATIONS

5.1 Heat Dissipation Through a Straight Fin


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p g g

1. The governing differential equations of the problem are approximated at the nodes to

generate finite difference algebraic equations for the nodal values of (x,y,z,t). In the

FV method the governing equation is integrated over chosen CVsaround each node

of the grid before the derivatives are approximated.

1 2 i-1 i i+1

N

N+1

x

x=L

Tbx

Tf

dT/dx=0

a

P

Fig. 5.1 Heat Dissipation Through a Straight Fin

Governing equation: ,0)(2

2

fTTKa

hP

dx

Tdneglecting the losses from the end

face x = L, the b.csare; T = Tbat x = 0, 0

dx

dtat x = L


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2. Consider a finite difference grid of N equal steps having N+1 grids, x = L/N.

For any mode i in the interior

,0)()(

22

11fi

iii TTKahP

xTTT 2 i N (5.1)

T1(2+A) T2+ T3= -ATf, i = 2, where A =2)( x

Ka

hP

T2(2+A) T3+T4= -ATf, i = 3, TN(2+A) TN+ TN+1 = -ATf, i = N;3. For the boundary nodes, nodal equations are derived from the b.c s : At the fi

base i=1, T1=Tb, the zero heat flux condition at the end i = N+1 should be

discretized by introducing a hypothetical node (N+2) called a image node. The

central difference expression for the derivativedxdT at the node (N+1) is:

NNNN TT

x

TT

dx

dT2

2 02

(5.2)


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4. Using eqn. (5.1) for the end node i=N+1:

TN(2+A) TN+1+ TN+2 = -ATf, use equ. (5.2) on this 2TN(2+A) TN+1= -ATf

5. The final system of algebraic equations for the fin problem for a grid of N=4 is:

f

f

f

f

b

AT

AT

AT

ATT

T

T

T

TT

A

A

A

A

5

4

3

2

1

)2(2000

1)2(100

01)2(10

001)2(100001


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T

h

Example:A slab of 0.1m thickness with K=40W/mC generates energy at the rate o

106W/m

3. The surface at x = 0 is insulated and the end at x = 0.1m is subjected to

convection with a heat transfer coefficient of 200W/m2C with Tatm= 150C. Obtain the

temperature distribution.

Solution

Take M=5, x = L/M = 0.1/5 = 0.02m 1 2 3 4 5 6

For the internal nodes m = 2 to 5:

;0)(

)2(

2

11K

gxTTT mmmm

040

10)02.0(2

62

11

xTTT mmm

5&4,3,2,0102 11 mforTTT mmm

At x = 0,

022 12

12K

gxTT

1,01022 12 mforTT


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At x = L,

02

2

22 221

2

1

2

TK

xh

k

gx

Tk

xh

T

M

MM

,015040

20002.02

40

1002.0

40

20002.0222

62

65 xxxx

Txx

T

60402.22 65 mforTT

40

1010

10

10

10

2.220000

121000012100

001210

000121

000022

6

5

4

3

2

1

T

TT

T

T

T

5 3 One Dimensional Steady Conduction Through Cylinder


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5.3 One Dimensional Steady Conduction Through Cylinder

1.

b

H

r

rT1

T1T2

T3Tm-1

TmTm+1

TMTM+1

1 2 3 m-1 m m+1 M M+1

r

2

2

r r

r =M

b

b

Fig. 5.3 One Dimensional Steady Conduction Through Cylinder


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2. Governing equation: broorgkdr

rdTr

dr

d

r,)(

1)(1

3. [Rate of heat entering by condition] + [rate of energy generation] = 0

Applying this for a cylindrical volume element and simplifying:

1,02

)(

2

11

2

11 1

2

1 Mmfork

grT

k

rhT

k

rh

MT

M

MMM


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Example:A 10cm dia. Solid chrome-Nickel rod with k= 20 W/mC is electricity heated

by passing an electric current generating heat at the rate of 107

W/m3

. The surface of the

rod is subjected to convection with a heat transfer coefficient h=200 W/mC into an

ambient at 30C. Determine the finite difference equations.

Solution:

b = 0.05m, k =20 W/mC, g = 107W/m3, T = 30C, h =200W/m2C, Take M=5

,01.05

05.0m

M

br 1.0

20

20001.0,50

20

10)01.0()( 722 x

k

rh

k

gr

the f.d. equation for the central node m =1: 1;0)(

)(4 12

12 mfork

grTT

.1;050)(4 12 mforTT


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Now we have six algebraic equations for the six nodes from m = 1 to m = 6:

-4T1 + 4T2= -50, for m = 1,

0.5T1 - 2T2+1.5 T3= -50, for m = 2,

0.75T2 - 2T3+1.25 T4 = -50, for m = 3,

0.8333T3 - 2T4+1.666T5= -50, for m = 4,

0.875T4 - 2T5+1.125T6= -50, for m = 5,

0.9T5 - T6= -28, for m = 6

in the matrix form:

28

50

50

50

50

50

19.00000

125.12875.0000

01666.128333.000

0025.1275.00

0005.125.0

000044

6

5

4

3

2

1

T

T

T

T

T

T


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Example:Steady one dimensional heat conduction is taking place in a slab of thickness

L=1 cm, where energy is generated at a construction rate of 7.2 x 107 W/m3. The

boundary surface of the slab at x=0 is maintained at fo=50C, while the boundary surface

at the other end dissipated heat by convection with a heat transfer coefficient h =

200W/m2C into the surroundings at T = 100C. Taking the K value of the slab as

18W/mC, determine the nodal temperatures, taking five subdivisions.

Solution:

cmM

LxLxg

kx

T2.0

5

1;0,0

12

2

LxathThTdx

xdTK

CTxatCxTCB s

,)(

,50;050)(:.1

LxatTThdx

dTk )(


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we know : ;0)(

22

11

k

gxTTT mmmm

for the interior nodes and for the node 6, use the following:

02)(2

22 221

2

12 T

k

xh

k

gxT

k

xhT MMM

Now,

,1618/)102.7()102()( 723

2

xxk

gx

0444.018/200)102(22 3

xk

xh


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,44.4100

0444.02

k

xhTSubstituting, getting the nodal equations and bringing to

the matrix form

44.20

16

16

16

66

044.22000

12100

01210

00121

00012

6

5

4

3

2

T

T

T

T

T

Solving: T2 = 119.045C, T3 = 172.09C, T4 = 209.135C, T5 = 230.18C,

T6= 235.225C


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ydiffusivitwhere

X

T

t

T2

2

One-diml. Transient Analysis

The Governing Equation for the transient

analysis is given by


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Explicit Scheme

The FD equation for 1D transient equation

is given by

2

11

1

211

1

)(

)21()(

)(2

x

tr

where

TrTTrT

xTTT

tTT

p

m

p

m

p

m

p

m

p

m

p

m

p

m

p

m

p

m

Example: A marble slab of k=2 w/mC =1 x 10-6

m2/s L = 2cm thick is initially at a


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Example:A marble slab of k 2 w/m C, 1 x 10 m /s, L 2cm thick is initially at auniform temperature Ti= 200C. One of its surfaces is suddenly lowered to 0C and the

other surface is insulated. Develop an one dimensional explicit f.d. scheme to determinethe temperature distribution in the slab as a function of position and time.

Solution: 0,0,2

2

tLxx

T

t

T

B.C: 0x

T, at x = 0, t > 0

T = 0C, at x = L, t > 0

I.C: T = 200C, for t = 0

For the insulated side h1= 0,

For the x = L side hM+1= , because there is a sudden heat transfer.

Insulated

K,

2 cm

Ti= 200C

0C


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Also T M+1= 0C

We know: Tmi+1= r Tm-1

i+ (1-2r)Tmi+ r imT 1 ; for m = 2, 3, . . . ., M and r = stability

criterion

1

1

i

mT = 0C, at m = M+1; Tm= 200C, for i = 0; m = 1, 2, . . . ., M+1

Taking M = 5; x = LM

= 25= 0.4cm = 4mm

Take r =1

2 r =

1

2 =

t

( x)2 t = 8 sec

Now the equation for 1imT becomes (Tm-1i+ Tm+1i), for m = 2, 3, 4, 5

In addition we have: T1i+1

= T2i+1

for m=1

T6i+1= 0C for m = 6 and Tm

= 200C for m = 1 to 6


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Now we march on time and obtain the following values:

m=1 2 3 4 5 6Time

stept sec

x=0 4mm 8mm 12mm 16mm 20mm

0 0 200 200 200 200 200 200

1 8 200 200 200 200 200 0

2 16 200 200 200 200 100 0

3 24 200 200 200 150 100 0

4 32 200 200 175 150 75 0

5 40 187.5 187.5 175 125 75 0

6 48 181.2 181.2 156.2 125 62.5 0

7 56 168.7 168.7 153.1 109.4 62.5 0

8 64 160.9 160.9 139.1 107.8 54.7 0

9 72 150 150 134.4 96.9 53.9 0

10 80 142.2 142.2 123.5 94.2 48.5 0


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The labyrinth seal should have a minimum gap not

below0.2 mmand achieve a high pressure drop ratio

(Pr) of above 10at a low rated flow of2.5 m3/hof water

at 70C

The seal length is restricted to be as small as

possible above a minimum stipulatedlength of 120 mm.

Limiting Cavitation to tolerable levels over the seal

length.

Straight Annular Portion

Outer Sleeve


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Fig.2 Annular Seal (AS)

b

c

d

D/2

Axis of the Labyrinth

CavityChamber

aV1

V2

Fig.4 Circular-grooved Squarecavity Labyrinth Seal (CSLS)

Fig.3 Circular-grooved square cavity labyrinth


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Fig.5 Domain discretisation of CSLS named LS1

Fig.6 Domain discretisation of CTLS 1


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Fig.7 Velocity distribution for AS at Re = 2500 predicted by FEA

Fig.8 Velocity distribution of CSLS - LS1 at Re = 2500 predicted by FEA


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Fig.9 Velocity distribution of CTLS 1 at Re = 2500 predicted by FEA


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Fig.10 Pressure distribution for AS at Re = 750 predicted by FEA

ig.11 Pressure distributions for CSLS named LS1 at Re = 750 predicted by

Fig.12 Pressure contours for CTLS 1 at Re = 750 predicted by FEA


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Fig.13 CFD flow visualisation for CCLS-LS6


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Fig.14 CFD predicted flow visualisation view for CCLS-LS7


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THANK

2nd Session - Dr. S. P. Asok

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Transcript of 2nd Session - Dr. S. P. Asok