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서울대학교 화학생물공학부

Heat and Mass Transfer

19UNSTEADY-STATE

HEAT CONDUCTION

Supercritical Fluid Process Lab

Unsteady-State Heat Conduction

is of importance to engineers in many circumstances

It may control the rate at which process equipment is brought to stable operating conditions

It is also important in determining the processing time of many solid articles. For example, the curing time of objects made of molded plastic or rubber is often dependent on the time required to bring the center to some specifiedtemperature without causing thermal damage to the material at the surface.

There are also many applications of unsteady-state-conduction theory in the heat treating and casting of the metals.

A hot metal billet that is removed from a furnace

Re-entry of Columbia into the earth’s atmosphere

200oC 40oC

θ=20min

q1

200oC 40oC

θ=60min

q2

q1 = q2

200oC 40oC

θ=20min

q1

100oC 50oC

θ=60min

q2

q1 ≠ q2

(a) Steady (b) Unsteady

Steady vs. Unsteady State Heat Transfer


Unsteady problem typically arise when the boundary conditions of a system are changed.If the surface temperature of a system is altered, the temperature at each point in the systemwill also begin to change. The changes will continue to occur until a new steady-state temperature distribution is reached.

The first law of thermodynamics

Ein Eout

Control volumeEgenEacc

Ein – Eout + Egen = Eacc

Ein : Energy transport by the fluid into the control volume surface phenomenonEout : Energy transport by the fluid out of the control volume surface phenomenonEgen : Energy generation in the control volume volumetric phenomenonEacc : Energy accumulation in the control volume volumetric phenomenon

re temperatu:T time,:

heat Specific:C volume,:V density,:

)(

volume:

eunit volumper rate generationheat :

p

t

VTCE

V

g

VgE

ptacc

gen

ρ

ρ=

=

∂∂

&

&


qx

qy

qz

qx+Δx

qz+Δz qy+Δy

dz

dy

dx

General heat conduction equation

Energy Balance: In - Out + Generation = AccumulationEin - Eout + Egen =Eacc

Egen = gV (g: heat generation rate/volume)

Eacc = )( VTCpρθ∂∂

)(

)(

)(

)(

dxdydzgEztdxdykq

ytdxdzkq

xtdydzkq

gen

z

y

x

&=∂∂

−=

∂∂

−=

∂∂

−=

θρ

∂

⋅⋅∂=

∂∂

+=

∂

∂+=

∂∂

+=

Δ+

Δ+

Δ+

)( tdxdydzCE

dzz

qqq

dyy

qqq

dxxqqq

pacc

zzzz

yyyy

xxxx



General heat conduction equation

pp

p

CCk

tkg

zt

yt

xt

tCgztk

zytk

yxtk

x

ρρ

α

θα

ρθ

1

)(

2

2

2

2

2

2

=

∂∂

=+∂∂

+∂∂

+∂∂

∂∂

=+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

&

&

For constant properties (k, ρ, Cp)

: thermal diffusivity : thermal capacitance

Materials of large α will respond quickly to changes in their environment, taking shorter to reach a new equilibrium condition.


Fundamental Equations

2

2

2

2

2

2

2

2

2

2

2

2

2

2

xt

Ckt

zt

yt

xt

Ckt

zt

yt

xt

Ckt

ztu

ytu

xtu

p

p

pzyx

∂∂

=∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=∂∂

+∂∂

+∂∂

+∂∂

ρθ

ρθ

ρθ(8-11)

(8-12)

(19-1)

The basic differential equation for unsteady-state conduction in a solid(For constant thermal conductivity and no heat generation)

Unsteady-state conduction in only x direction

A differential energy balance


0

0

2

2

2

2

2

2

2

2

=∂∂

=∂∂

+∂∂

+∂∂

xt

zt

yt

xt

(19-2)

Steady-state conduction in only x direction

Laplace’s equation


The basic differential equation for steady-state conduction in a solid(For constant thermal conductivity and no heat generation)

Which is readily integrated to give a linear relation between temperature and distance. This was shown earlier in Eq. (18-3), obtained by integrating the Fourier conduction equation.


General Heat Transfer Equation


Cylindrical coordinates

zzryrx

===

θθ

sincos

zrrV ΔΔΔ=Δ θφrd

rΔzΔ

θΔr

θΔθ



θθρ

θ

θθ

θθ

θθ

θθθθθθ

′∂∂

ΔΔΔ

ΔΔΔ∂∂

ΔΔ−−∂∂

ΔΔ−=−

∂∂

ΔΔ−−∂∂

ΔΔ−=−

∂∂

ΔΔ−−∂∂

ΔΔ−=−

Δ+Δ+

Δ+Δ+

Δ+Δ+

trzrC

rzrSztzkr

ztrkrqq

rtzrk

rtzrkqq

rtzkr

rtzkrqq

p

zzzzzz

rrrrrr

)(

)(

)()(

)()(

)()(

Input-Output

r-direction:

θ-direction:

z-direction:

Generation:

Accumulation:



⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

=′∂

∂

′∂∂

=⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

′∂∂

=+Δ

∂∂

−∂∂

+Δ

∂∂

−∂∂

+Δ

∂∂

−∂∂

→ΔΔΔ

Δ+Δ+Δ+

2

2

2

2

2

20,,

11

11lim

ztt

rrtr

rrCkt

tCztk

ztk

rrtkr

rr

tCSz

ztk

ztk

rr

tktk

rrrtkr

rtkr

p

pzr

pzzzrrr

θρθ

θρ

θθ

θρ

θθθ

θ

θθθ

If k=constant

)( rzr ΔΔΔ÷ θ

0


⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

=′∂

∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=′∂

∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

+∂∂

=′∂

∂

rt

rrt

Ckt

trr

trr

tCkt

ztt

rrt

rrt

Ckt

p

p

p

1

11

11

2

2

2

2

22

2

2

2

2

2

22

2

ρθ

θρθ

θρθ(19-3)

No axial conduction


The basic differential equation for unsteady-state conduction in a solid(For constant thermal conductivity and no heat generation)


No axial conduction & angular conduction

RadialConduction

AxialConduction

θφθφθ

cossinsincossin

===

zryrxθΔr

φθ Δ⋅sinr

Δφ)Δθ, φΔr, θ(r +++

),,( φθr

rΔrrV Δ⋅Δ⋅Δ⋅=Δ θφθsin2

Spherical coordinates


θφθθρ

φθθ

φθθ

φθθ

θφθ

θφθ

φθθφθθ

φφφ

θθθθθθ

′∂∂

ΔΔΔ

ΔΔΔ

∂∂

ΔΔ−−∂∂

ΔΔ−=−

∂∂

ΔΔ−−∂∂

ΔΔ−=−

∂∂

ΔΔ−−∂∂

ΔΔ−=−

Δ+Δ+

Δ+Δ+

Δ+Δ+

trrC

rrS

tr

zkrtr

rkrqq

rtrkr

rtrkrqq

rtkr

rtkrqq

p

zzz

rrrrrr

)sin(

)sin(

sin1)(

sin1)(

)sin()sin(

)sin()sin(

2

2

22

Input-Output

r-direction:

θ-direction:

z-direction:

Generation:

Accumulation:




⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

=′∂

∂

′∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

′∂∂

=Δ

∂∂

−∂∂

+Δ

∂∂

−∂∂

+Δ

∂∂

−∂∂

→ΔΔΔ

Δ+Δ+Δ+

2

2

2222

2

2222

20,,

2

22

sin1sin

sin11

sin1sin

sin11lim

sinsinsin

sinsin

φθθθ

θθρθ

θρ

φφθθθ

θθ

θρ

θφθφφ

θθθ

θθ

θ

θ

φφφθθθ

tr

trr

trrrC

kt

tCtkr

tkrr

tkrrr

tCrr

tktk

rr

tktk

rrrtkr

rtkr

p

pzr

prrr

If k=constant

)sin( 2 φθθ ΔΔΔ÷ rr


Differential Energy Balance

gtr

trr

trrrC

k

tr

utr

urtut

p

r

&+⎥⎦

⎤⎢⎣

⎡∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

=

∂∂

+∂∂

+∂∂

+′∂

∂

2

2

2222

2 sin1sin

sin11

sin

φθθθ

θθρ

φθθθφθ


gztt

rrtr

rrCk

ztut

ru

rtut

pzr &+⎥

⎦

⎤⎢⎣

⎡∂∂

+∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

=∂∂

+∂∂

+∂∂

+′∂

∂2

2

2

2

2

11θρθθ

θ

gzt

yt

xt

Ck

ztu

ytu

xtut

pzyx &+⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=∂∂

+∂∂

+∂∂

+∂∂

2

2

2

2

2

2

ρθ

Rectangular coordinates



Energy Equations by ConductionConstant Properties

gtr

trr

trrrC

kt

p

&+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

=′∂

∂2

2

2222

2 sin1sin

sin11

φθθθ

θθρθ


gztt

rrtr

rrCkt

p

&+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

=′∂

∂2

2

2

2

2

11θρθ

gzt

yt

xt

Ckt

p

&+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=∂∂

2

2

2

2

2

2

ρθ



Rectangular coordinates

One-Dimensional Conduction in a Planar Medium with Constant Properties and No Generation

gzt

yt

xt

Ckt

p

&+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=∂∂

2

2

2

2

2

2

ρθ

2

2

xt

Ckt

p ∂∂

=∂∂

ρθ

gzt

yt

xt

Ck

ztu

ytu

xtut

pzyx &+⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=∂∂

+∂∂

+∂∂

+∂∂

2

2

2

2

2

2

ρθ

Boundary and Initial Conditions

Constant Surface Temperature

Constant Heat Flux

Applied Flux

Insulated Surface Convection

2

2

xt

Ckt

p ∂∂

=∂∂

ρθ

For transient conduction, heat equation is first order in time, requiring specification of an initial temperature distribution:

Since heat equation is second order in space, two boundary conditions must be specified. Some common cases:

st

),( θxt ),( θxt),( θxt

),( θxt

),0( θt

Boundary Conditions at the surface (x=0)

1. Constant Surface Temperature

stt =),0( θst

),( θxt

2. Constant Surface Heat Flux(a) Finite heat flux


Applied Flux

),( θxt

sx

qxtk ′′=∂∂

−=0

2. Constant Surface Heat Flux(b) Adiabatic or insulated surface


00

=∂∂

=xxt

),( θxt

3. Convection Surface Condition


),( θxt

),0( θt

ht ,∞

( )[ ]θ,00

tthxtk

x

−=∂∂

− ∞=


Ex. 19-1Problems of heating or cooling spheres are best solved using the differential energy balance written in spherical coordinates. Derive the equation for the case where there is no variation of temperature with angular position.

rt

r+dr

t+dt

Control volume

Rate of heat flow into control volume

Rate of heat flow out of control volume

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

−=

⎟⎠⎞

⎜⎝⎛

∂∂

−−

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

+∂∂

+∂∂

+∂∂

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

++−=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛∂∂

+∂∂

+−=

∂∂

−=

drrtrdr

rtrk

rtrk

drrtdrdr

rtrdrdr

rtr

rtdr

rtrdr

rtrk

drrt

rtdrrdrrk

rtd

rtdrrk

rtrk

24

)4(

224

)2(4

])(4[

)4(

2

22

2

2

22

2

2

2

2222

2

222

2

2

π

π

π

π

π

π

Net heat flow000

spherical coordinates

Ex. 19-1

rt

r+dr

t+dt

Control volume

The rate of accumulation energy in the control volume

Net heat flow = rate of accumulation

⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

+∂∂

ρ=

θ∂∂

θ∂∂

ρπ=⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

+∂∂

π

θ∂∂

ρπ

rt

rrt

Ckt

tCdrrdrrtrdr

rtrk

tCdrr

p

p

p

2

)4(24

)4(

2

2

22

22

2

spherical coordinates


The equation for unsteady conduction in a sphere

Modeling in Heat Transfer

Potato

100oC

PhysicalPhenomena

H2O

100oC

Actual Ideal(simplification)

Modeling

PhysicalPrinciples & laws

2

2

xtt

∂∂

=∂∂ αθ

ODE, PDE

Mathematicalformulation

Solution


Physical PhenomenaPhysical Phenomena

Solution of the Fundamental Equations

ODE, PDEODE, PDE

Numerical SolutionNumerical Solution

Modeling

Analytical SolutionAnalytical Solution

(1) Separation of variables(2) Similarity variable(3) Laplace, Fourier transform

- FDM, FEM


Similarity Variable

Conduction into a plate of infinite thickness

Analytical Solution

αθ

αθ

4

t t,at x :2 B.C.t t0,at x :1 B.C.t t0,θat : I.C.

0

s

0

2

2

xn

xtt

=

=∞=====

∂∂

=∂∂

Let

(19-1)

(19-4)


t0

tS

x

θ

∞0

tS

x

t0

Similarity VariableAnalytical Solution

2

2

2

2

2

2

23

41

41

41

41

2

41;

221

421

4

nt

xn

nt

xn

nt

nt

xt

nt

xn

nt

xt

ntnn

ntt

xnnxxn

∂∂

⋅=∂∂⋅

∂∂

⋅=∂∂⋅⎟⎠

⎞⎜⎝

⎛∂∂

⋅∂∂

=∂∂

∂∂

⋅=∂∂⋅

∂∂

=∂∂

∂∂

⋅−=∂∂

⋅∂∂

=∂∂

=∂∂

−=⋅−=⎟⎟⎠

⎞⎜⎜⎝

⎛−=

∂∂ −

αθαθαθ

αθ

θθθ

αθθθαθθ

αθ



02

41

2

41

2

2

2

2

2

2

2

2

2

=+

∂∂

⋅⋅=∂∂

⋅−

∂∂

⋅=∂∂

∂∂

⋅−=∂∂

dndtn

dntd

nt

ntn

nt

xt

ntnt

αθα

θ

αθ

θθ (19-5)

(19-6)

(19-7)



201

1

2

2

02

: variablenew

CdneCt

eCp

npdndp

dndtp

n n

n

+=

=

=+

=

∫ −

−

(19-8)

(19-9)

(19-10)

(19-7)

Two boundary conditions are needed. Supercritical Fluid Process Lab

022

2

=+dndtn

dntd


αθπ

π

π

θ

αθ

42

)(22

,n 0, , x:2 B.C.C ,t t0,n 0, x:1 B.C.

40

0

01

1

010

0

2s

2

2

xerfdnetttt

ttC

tC

tdneCt

ttt

xn

s

s

s

s

sn

s

==−−

−=

+=

+=

=∞→=∞→====

∫

∫

−

∞ −(19-11)

(19-12)

(19-13)

Gauss error function or probability functionSupercritical Fluid Process Lab

The error function (also called the Gauss error function) is a non-elementary function which occurs in probability, statistics and partial differential equations. One of its main purposes is for graphing a normal distribution. It is defined as:

By expanding the right-hand side in a Taylor series and integrating, one can express it in the form

for every real number x, or any complex number z = x. The error function is evidently odd.

Plot of the error function

Analytical Solution

αθπαθ

42 4

00

2 xerfdnetttt x

n

s

s ==−−

∫ −

Gauss error function (probability function)

erf(z)

z

zπ2erf(z) z, smallfor =


Analytical Solution

αθαθπ

αθπαθ

αθπαθ

018.0~42

01.0

42

401.0

42 4

00

2

=

≈=

==−−

∫ −

T

TT

xn

s

s

x

xxerf

xerfdnetttt

Thermal penetration depth, xT

x @ 1% of driving force


zπ2erf(z) z, smallfor =

Analytical Solution

)(

)(1

@4

24

0

0

0

0

ttCk

ttk

xtk

Aq

smallisxxxerftttt

sp

s

x

s

s

s

−⎟⎟

⎠

⎞

⎜⎜

⎝

⎛=

−⎟⎠

⎞⎜⎝

⎛ −−=

∂∂

−=

≈=−−

=

πθ

ρ

παθ

αθπαθ


x @ 1% of driving force

pCkρ : Thermal effusivity


Analytical Solution


과도기 상태에서는 가 중요하다.pCkρ

겨울에 나무 손잡이와 쇠 손잡이를 잡을 때더 빨리 차가움을 느끼는 것

뜨거운 공기와 뜨거운 물에 손을 담글 때화상을 입는 정도

100oC 스팀 사우나에서는 오랫동안 견디면서45oC 목욕탕 물 속에서는 잠깐도 견디기 어려운 것


Separation of Variables

Unsteady state heat conduction in a rubber sheet

Analytical Solution

hft

xtt

/0028.0Ck

F290 tand 0 x@ ?θ

0dt/dx 0,at x:2 B.C.

F292t t,1/4"xat x:1 B.C.

F70t t0,θat : I.C.

2

p

s0

0

2

2

=ρ

=α

===

==

====

===

∂∂

α=θ∂∂

o

o

o

(19-1)

x

x0=1/4”

292oF


292oF

Curing at 292oFfor 50min

Analytical Solution

2

2

20

02

2

002

20

0

20

0

2000

2

2

1)(1)(

1)(

)(

;;

nY

xtt

xn

nY

xtt

xt

nY

xtt

xn

nY

Yt

xt

Yx

ttYYtt

xxxn

ttttY

xtt

ss

s

s

s

s

∂∂

−=∂∂

∂∂

−=∂∂

∂∂

−=∂∂

∂∂

∂∂

=∂∂

∂∂

−=∂∂

∂∂

∂∂

=∂∂

==−−

=

∂∂

=∂∂

τα

θτ

τθ

αθτ

αθ (19-1)

Dimensionless variables

(1)

(1-1)

(1-2)



Analytical Solution

0,1,:2..

0,00,0:1..

1,0,0:..

1)()(

0

0

2

2

2

2

20

020

0

2

2

==→==

=∂∂

=→=∂∂

=

==→==∂∂

=∂∂

∂∂

−=∂∂

−

∂∂

=∂∂

YnttxxCBnYn

xtxCB

YttCInYY

nY

xttY

xtt

xtt

s

ss

τθτ

ατ

α

αθ (19-1)

(1-1) & (1-2) (19-1)

(1)

(2)



Analytical Solution


0,0@:B.C.2

0,1@:B.C.11,0@:I.C.

2

2

=∂∂

=

====

∂∂

=∂∂

nYn

YnY

nYY

τ

τ

0dt/dx 0, x@:2 B.C.t t,x x@:1 B.C.

tt0,θ @: I.C.

s0

0

2

2

====

==

∂∂

=∂∂

xtt α

θ

2000

;;xx

xnttttY

s

s αθτ ==−−

=

Dimensionless variables


Analytical Solution

2

2

2

2

2

2

2

2

11

)()(

dnNd

NddT

T

dnNdT

ddTN

dnNdT

nY

ddTNY

nNTY

=

=

=∂∂

=∂∂

⋅=

τ

τ

ττ

τSolution: (3)

(4)

(5)Supercritical Fluid Process Lab

Separation of variables:

Separation of VariablesT is a function of only τN is a function of only n

2

2

nYY

∂∂

=∂∂τ

Left is a function of only τRight is a function of only n

Analytical Solution

( )anCanCeCNTY

anCanCNNadn

Nd

eCTTaddT

adn

NdNd

dTT

a

a

cossin

cossin0

0

11

321

322

2

2

12

22

2

2

2

+=⋅=

+=→=+

=→=+

−==

−

−

τ

τ

τ

τ(5)

(10)

(6) (8)

(7) (9)

4 unknowns

Since time and distance are independent of each other. The left side of this equation is independent of distance and the right side is independent of time. Hence each side must be constant.



Analytical Solution

( )

ππππ

τθ

τ

τ

ττ

212,2/5,2/3,2/

cos0:2..

)A (wherecos

0

0sincos:1..

0,1,:2..

0,00,0:1..

1,0,0:..

2

2

22

31

2

21321

0

0

−=

=

==

=

==−=∂∂

==→==

=∂∂

=→=∂∂

=

==→==

−

−

−−

ia

aAeCB

CCanAeY

C

eCaCanaCanaCeCnYCB

YnttxxCBnYn

xtxCB

YttCI

a

a

aa

s

L

(11)


This equation contains four constants C1, C2, C3, and a: however C1 can be combined with C2 and C3, so that only three IC & BCs are needed to complete the solution.

n=0, sinan=0, cosan=1

(symmetrical system)


Analytical Solution

L

L

+⎟⎠⎞

⎜⎝⎛ −

+

+⎟⎠⎞

⎜⎝⎛+

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

−−

−

−−

nieA

neA

neAneAY

ii 2

)12(cos

25cos

23cos

2cos

2

2

22

)2/)12[(

)2/5(3

)2/3(2

)2/(1

π

π

ππ

τπ

τπ

τπτπ

GeneralSolution

(12)

The substitution of one of these values for a in Eq. (11) would meet the requirement of the second boundary condition, but it is not possible to represent an arbitrary temperature distribution in the slab by a cosine curve.



Analytical Solution

LL +⎟⎠⎞

⎜⎝⎛ −

++⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛= niAnAnA i 2

)12(cos2

3cos2

cos1 21πππ

(13)

Both side of this equation are multiplied byAnd integrated over the range of 0 to 1.

I.C.

ni π⎟⎠⎞

⎜⎝⎛ −

2)12(cos

i

ini

idnni )1(1

122

212sin1

122

212cos

1

0

1

0−

−−=⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −

−=⎟

⎠⎞

⎜⎝⎛ −

∫ ππ

ππ

( )

( ) 02

sin)22(1sin

2sin

)22(1sin

212cos

2cos

1

1

01

1

0 1

=⎥⎦

⎤⎢⎣

⎡+

−−

=

⎥⎦

⎤⎢⎣

⎡+

−−

=⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛

∫

ππ

ππ

ππ

ππππ

ii

iiA

ini

iniAdnninA



Analytical Solution

LL +⎟⎠⎞

⎜⎝⎛ −

++⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞


)12(cos2

3cos2

cos1 21πππ

(13)

Both side of this equation are multiplied byAnd integrated over the range of 0 to 1.

I.C.

ni π⎟⎠⎞

⎜⎝⎛ −

2)12(cos

2000

212

21

122

212)2sin(

41

212

21

122

212cos

1

0

1

0

21

ii

i

Aii

A

ninii

A

dnniA

=⎥⎦⎤

⎢⎣⎡ −−+

−−

=

⎥⎦⎤

⎢⎣⎡ −

+−

−=

⎟⎠⎞

⎜⎝⎛ −

∫

ππ

πππ

π



Analytical Solution

LL +⎟⎠⎞

⎜⎝⎛ −

++⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞


)12(cos2

3cos2

cos1 21πππ

(13)I.C.

i

i)1(1

122

−−

−π

= 0 + 0 + 0 + … + 2iA

L,54,

34,4

)12()1(4

2)1(1

122

321 πππ

π

π

=−==

−−−

=

=−−

−

AAA

iA

Ai

i

i

ii



Analytical Solution

[ ] niei

Y

ne

neneY

i

i

i

i

i

⎟⎠⎞

⎜⎝⎛ −

−−−

=

−⎟⎠⎞

⎜⎝⎛+

⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=

−−∞=

=

−

−−

∑ 2)12(cos

2/)12()1(2

25cos

54

23cos

34

2cos4

2

2

22

]2/)12[(

1

)2/5(

)2/3()2/(

ππ

ππ

ππ

ππ

τπ

τπ

τπτπ

L

(15)

The general solution is

(14)



Analytical Solution

L−π

+π

−π

=

=

−−

=

τπ−τπ−τπ− 222 )2/5()2/3()2/(

54

344009.0

0@70292

290292

eee

n

Y

The specific example:

The solution for τ must be obtained by trial and error.As a first approximation, only the first term on the right hand side will be considered. This gives

min)7.18(313.0)01.2(121

41

0028.011

01.24

)009.0)((ln2

220

20

2

hrxxk

C p =⎟⎠⎞

⎜⎝⎛=τ

α=τ

ρ=θ

=π

⎟⎠⎞

⎜⎝⎛π

−=τ


(18.7min)


Analytical Solution

L

L

)1040.1)(254.0()1050.3)(424.0()00694.0)(27.1(

54

344

01.2@

5420

1248.4497.4

−−

−−−

×+×−=

−π

+π

−π

=

=τ

eeeY

Check the relative magnitude of the terms in the series solution

The validity of the approximation which employed only the first term of the series is apparent.



Analytical Solution

Temperature distribution in rubber sheet

0 1 70 70 70 70

1 9.76x10-1 75 92 139 209 292

5 3.33x10-1 217 223 239 263 292

10 9.02x10-2 272 273 278 284.3 292

20 6.37x10-3 290.6 290.7 291 291.5 292

30 4.52x10-4 291.9 291.9 292 292 292

40 3.19x10-5 292 292 292 292 292

TimeElapsed(min)

τπ

π2)2/(4 −e

Temperature in sheet, oF

12

cos

0

=

⎟⎠⎞

⎜⎝⎛

=

πnn

924.02

cos

4/1

=

⎟⎠⎞

⎜⎝⎛

=

πnn

707.02

cos

2/1

=

⎟⎠⎞

⎜⎝⎛

=

πnn

383.02

cos

4/3

=

⎟⎠⎞

⎜⎝⎛

=

πnn

02

cos

1

=

⎟⎠⎞

⎜⎝⎛

=

πnn



Analytical Solution

Temperature profiles in rubber sheet

300

200

100

00 1/4 1/2 3/4 1surfacecenter

10min

5min

1min

Distance, n(=x/x0)Supercritical Fluid Process Lab

Tem

pera

tur e

, oF

292oF


Analytical Solution Conduction in Cylinder

sttRrrtrat

ttat

ztt

rrtr

rrt

==

=∂∂

=

==

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

=′∂

∂

,

0,0

,0

11

0

2

2

2

2

2

θ

θα

θ

0 0 (L>>R)

I.C.:

B.C.1:

B.C.2:

ts

0=∂∂rt


Analytical Solution Conduction in Cylinder


0,0 =∂∂

=rtr

sttRr == ,0,0 tt ==θ

),( θ= rft

Conduction in CylinderAnalytical Solution

22

2

2

2

2

20

2

2

111

)()(

11

1

addT

TdrdN

rNdrNd

N

ddTN

drdN

rT

drNdT

rNTY

YrY

rrY

ttttY

rt

rrtt

s

s

−==+

=+

⋅=

∂∂

=∂∂

+∂∂

−−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

=′∂

∂

θα

θα

θ

θα

αθ

Solution:

GE:



0

)exp(0

111

222

22

2

21

2

22

2

2

2

=++

×

−=→=+

−==+

=+

NardrdNr

drNdr

Nr

aCTTaddT

addT

TdrdN

rNdrNd

N

ddTN

drdN

rT

drNdT

θααθ

θα

θα



∫

∑

∞ −−

∞

=

+

=Γ

++Γ

⎟⎠⎞

⎜⎝⎛−

=

+=

=−++

0

1

0

2

222

22

)(

)1(!21)1(

)(

)()(

0)(

dtetz

kmm

xxJ

xBYxAJy

ykxdxdyx

dxydx

tz

m

kmm

k

kk

Bessel’s equation of order k

Solution:



constant sEuler' :5772.0ln131

211lim

)!(21)1(

2)(21ln2)(

)1(!21)1(

)(

)()(

00

12

2

00

0

2

0

00

222

22

LL =⎟⎠⎞

⎜⎝⎛ −++++=

⎟⎠⎞

⎜⎝⎛−

−⎥⎦

⎤⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛=

+Γ

⎟⎠⎞

⎜⎝⎛−

=

+=

==→=++

∞→

∞

=

∞

=

∑

∑

mm

m

ararJararY

mm

ararJ

arBYarAJN

arxandkNardrdNr

drNdr

m

m

mm

m

mm

γ

πγ

π

Bessel function of the 1st kind of order zero

Bessel function of the 2nd kind of order zero


( )

0,0:2..

0,:1..1,0:..

)()()()( 001

2

=∂∂

=

====

+=⋅= −

rYrCB

YRrCBYCI

arBYarAJecrNTY a

θ

θ θα



differentiation

)()()(

)()()(

1

1

xxYxkYxYdydx

xxJxkJxJdydx

kkk

kkk

αααα

αααα

+

+

−=

−=

)()()()(

)()()()(

1010

1010

araYarYdrdararYarY

drdr

araJarJdrdararJarJ

drdr

−=→−=

−=→−=



,

0,0@ =∂∂

=rYr

))()()(exp( 112

1 arBaYarAaJacrY

+−−=∂∂ θα

))0()0()(exp( 112

10 BaYAaJacrY

r +−−=∂∂

= θα

001 =→=∴ BBc


0

Supercritical Fluid Process Lab( ))()()()( 001

2

arBYarAJecrNTY a +=⋅= − θαθ

0,@ == YRr

,

)()exp(0 02

1 aRJaAc θα−=

0)( ofroot th theis a 0 =RaJ iιι

∑∞

=

−=0

02 )()exp(

iiii raJacY θα


Supercritical Fluid Process Lab( ))()()()( 001

2

arBYarAJecrNTY a +=⋅= − θαθ

1,0@ == Yθ

)(1 00

raJc ii

i∑∞

=

=

integralthen),(0 raxJa jj×

)()( 1,kwhen

function :)()1()(

)()(

11

1

xJxJ

xJxJ

xJxdxxJx

kk

k

kk

kk

αα

αα

ααα

−==

Γ−=

=

−

−

−∫


Left: )()( 10 1 RaRJdxxJx j

R

kk =∫ − αα

)()( 10 0 RaRJdrraxJa j

R

jj =∫

Supercritical Fluid Process Lab∑∞

=

−=0

02 )()exp(

iiii raJacY θα

right: ji ≠

∫ =−−

=R

jijijiji

ijii RaJRaJaRaJRaJa

aaRCdxxaJxaxJC

0 10102200 0)]()()()([)()(

ji =

∫ −−=R

iiiii

ii RaJRaJRaJRaCdxxaxJC0 11

20

20 )]()()([

2)(

)]()([2

)( 111 RaJRaJRaCRaRJ iiii

j −−=

)(2

)(2

11 RaJRaRaJRaC

iiiii =

−=∴

−


0

0

0


)()(

)/exp(2

)()exp()(

2

01 1

2

02

1 10

raJRaJa

CkaR

raJaRaJRatt

ttY

ii ii

pi

iii iis

s

∑

∑

∞

=

∞

=

−=

−=−−

=

ρθ

θα


Final Solution


)()(

)/exp(20

1 1

2

0

raJRaJa

CkaRtt

tti

i ii

pi

s

s ∑∞

=

−=

−− ρθ


Final Solution


0)( ofroot th theis a 0 =RaJ iιι

J0 and J1: Bessel function of the 1st kind (zero and first order)

(19-15)

The Fourier equation has been solved for many geometries and sets of conditions. A set of general solutions has been plotted for use in obtaining reasonably good solutions with less work. These are the Gurney-Lurie or Heisler Charts. Figures 19-3, 19-4, and 19-5 are simple versions of such a chart.

Use of the charts is restricted to cases where: 1. there is no internal heat source (generation) 2. the thermal diffusivity of the object is constant 3. the problem can be approximated as one-dimensional 4. the initial temperature of the object is uniform 5. the system is forced by a step change in temperature

of the surroundings (or of the surface, when 1/h=0)


Graphical Solution: Gurney-Lurie Chart

Graphical Solution: Gurney-Lurie ChartThe chart below is only an example of how a Gurney-Lurie chart might look and is not based on any actual data. The chart shows how four different dimensionless groups depend on each other.

Bihxkm

xxn

xCk

xX

ttttY

ps

s 1,,,00

20

200

=====−−

=ρ

θαθ

0ttttY

s

s

−−

=

20

20 xC

kx

Xpρθαθ

==


Gurney-Lurie Chart

To use the charts, some variables need to be defined. Various versions of the charts are slightly different in how the variables are defined:

resistance relative :1

position relative :

time)(relativenumber Fourier :

0

0

20

20

max0

Bihxkm

xxn

FoxC

kx

X

tt

ttttY

p

s

s

==

=

===

ΔΔ

=−−

=

ρθαθ

change tempshedunaccompli fractional the:


Fourier number (Fo)

Supercritical Fluid Process Labstored

conducted

QQ

LFo &

&== 2

αθ

Gurney-Lurie Chart for Plate


Gurney-Lurie Chart for Cylinder


Gurney-Lurie Chart for Sphere


Gurney-LurieChart for Sphere



Gurney-Lurie Chart for Sphere

Newman’s Rule :the technique for solving many problems of systems with finite dimensions in all direction

″

21

″

21

″

21

Finite dimension in all direction

If a brick-shape object is heated or cooled, the general solution describing temperature as a function of time and three distance variables, x, y, and z.

Y=YxYyYz

Yx=f1(x, θ) : x-dir. unsteady state conductionYy=f2(y, θ) : y-dir. unsteady state conductionYz=f3(z, θ) : z-dir. unsteady state conduction

70oF

292oF

290oF

208.0

009.070292

29029270292

292

0

=

=−−

=

−−

=−−

=

x

s

s

Y

tttttY


t = 290oF at center θ = ?

( )

min)7(113.044.673.0

48/10028.044.673.0

3-19 Fig. From.0,0

208.00090.0

220

00

3

h

x

hxkm

xxn

YY

YY

YYY

x

x

zyx

==θ

⎟⎟⎠

⎞⎜⎜⎝

⎛ θ=θ==

αθ

====

=→=

=

==

Newman’s Rule: Y=YxYyYz


208.0

009.070292

29029270292

292

0

=

=−−

=

−−

=−−

=

x

s

s

Y

tttttY

h→∞

Three dimensional conductionThree dimensional conduction

2

2

2

2

2

2

2

2

2

2

2

2

zY

yY

xYY

zt

yt

xt

Ckt

p ∂∂

+∂∂

+∂∂

=∂∂

→⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=∂∂

τρθ

2

2

2

2

2

2

2

2

2

2

2

2

1111zY

YyY

YxY

Y

zYYY

yY

YYxYYYYYY

z

z

y

y

x

x

zyx

yzx

xzyzyx

∂∂

+∂

∂+

∂∂

=∂∂

∂∂

+∂

∂+

∂∂

=∂∂

ττ

τ

τττττ

)()()( zYyYxYY zyx=τ

The basis for validity of Newman’s Rule

Solution:

(1)

(2)

Differentiating of Eq. (2)

(3)

Rearrangement yields


)(1

1

1

1

23

22

21

232

2

222

2

212

2

aaa

azY

Y

ayY

Y

axY

Y

z

z

y

y

x

x

++−=∂∂

−=∂∂

−=∂

∂

−=∂∂

ττ

τ

Newman’s RuleEach part of Eq(3) is a function of one of the four independent variables and hence, it is reasoned, each of the four terms is a constant.


( )

( ) ( )( )( )zaCzaCyaCyaC

xaCxaCeCY

zaCzaCY

yaCyaCYxaCxaCY

eC

aaa

z

y

x

aaa

37362524

13121

3736

2524

1312

1

cossincossincossin

cossin

cossincossin

23

22

21

23

22

21

+++=

+=

+=+=

=

++−

++−

τ

ττ

Newman’s Rule

These individual solutions are combined

These four expressions can be written as ordinary differential equationsfor which the solutions are

(4)

(5)

(6)

(7)

(8)

These general solution for three-dimensional conduction is consist of the products of each of the one-dimensional solutions.


Homework #2

PROBLEMSPROBLEMS

1919--111919--22

Due date: October 19Due date: October 19


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