Chapter 3 TWO-DIMENSIONAL STEADY STATE CONDUCTION

7/31/2019 Chapter 3 TWO-DIMENSIONAL STEADY STATE CONDUCTION

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CHAPTER 3

TWO-DIMENSIONAL STEADY STATECONDUCTION

3.1 The Heat Conduction Equation

Assume: Steady state, isotropic, stationary,

k = = constant

(3.1)02

2

2

2

k

q

x

T

k

Uc

y

T

x

T p


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Special case: Stationary material, no energy

generation

Cylindrical coordinates:

02

2

2

2

y

T

x

T (3.2)

(3.3)01

2

2

z

T

r

Tr

rr

Eq. (3.1), (3.2) and (3.3) are special cases of


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0)()(

)()()()(

01

2

2

2012

2

2

Tygy

Tyg

y

TygTxf

x

Txf

x

Txf

(3.4)

Eq. (3.4) is homogenous, second order PDE withvariable coefficients

3.2 Method of Solution and Limitations

Method: Separation of variables

Basic approach: Replace PDE with sets of ODE


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The number of sets depends on the number of

independent variables

Limitations:

(1) Linear equations

Examples: Eqs. (3.1)-(3.4) are linear

(2) The geometry is described by an orthogonal

coordinate system. Examples: Rectangles,

cylinders, hemispheres, etc.

variabledependenttheiflinearisequationAn unitypowertoraisedappearsderivativeitsoroccurnotdoproductstheirifand


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3.3 Homogeneous Differential Equations and

Boundary Conditions

Example: Eq. (3.1): Replace T bycTand divide

through byc

02

2

2

2

ck

q

x

T

k

Uc

y

T

x

T p NH (a)

notisitifshomogeneouisconditionbuondaryorequationAnconstantabymultipliedisvariabledependentthewhenaltered


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Example: Boundary condition

TThxTk (b)

Replace T bycT

c

TTh

x

Tk NH (c)

Simplest 2-D problem: HDE with 3 HBC and 1 NHBC

Separation of variables method applies to NHDE and

NHBC


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3.4 Sturm-Liouville Boundary Value

Problem: Orthogonality

PDE is split into sets of ODE. One such sets is known

as Sturm-Liouville equation if it is of the form

Rewrite as

(3.5a)

)()()(3

2

212

2xaxa

dx

dxa

dx

dn

nn 0n

(3.5b))()()( 2 xwxqdx

dxp

dx

dn

n 0n


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where

,exp)( 1dxaxp ),()( 2 xpaxq )()( 3 xpaxw

(3.6)

NOTE:

w(x) = weighting function, plays a special role

Equation (3.5) represents a set ofn equations

corresponding ton values of n

The solutionsn

are known as thecharacteristic

functions


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Important property of the Sturm-Liouville problem:

orthogonality

Two functions,n

,m

and are said to be orthogonal

in the intervala andb with respect to a weighting

function w(x), if

b

a mndxxwxx 0)()()( (3.7)

The characteristic functions of the Sturm-Liouville

problem are orthogonal if:


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(1)p(x) , q(x) and w(x) are real, and

(2) BC atx = a andx = b are homogeneous of the

form

0n

(3.8a)

(3.8b)0dx

dn

(3.8c)0dx

d nn


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Special Case: Ifp(x) = 0 atx = a orx = b , these

conditions can be extended to include

)()( bann

(3.9a)

and

dx

bd

dx

adnn

)()((3.9b)

(3.9a) and (3.9b) areperiodic boundary conditions.

Physical meaning of (3.8) and (3.9)


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Relationship between the Sturm-Liouville problem,

orthogonality, and the separation of variablesmethod:

PDE + separation of variables2 ODE

One of the 2 ODE =Sturm-Liouville problem

Solution to this ODE = n= orthogonal functions


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3.5 Procedure for the Application of

Separation of Variables Method

Example 3.1: Steady state 2-D conduction in

a rectangular plate

Find T (x,y)

(1) Observations

2-D steady state

4BC are needed

3 BC are homogeneous

2 HBC

x

y

1.3Fig.

T = f(x)

T = 0 T = 0

T = 0

00


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(2) Origin and Coordinates

Origin at the intersection of the two simplest BC

Coordinates are parallel to the boundaries

(3) Formulation

(i) Assumptions

(1) 2-D

(2) Steady

(3) Isotropic

(4) No energy generation

(5) Constantk


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(ii) Governing Equations

02

2

2

2

y

T

x

T(3.2)

(iii) Independent Variable with 2 HBC: x-

variable(iv) Boundary Conditions

(1) ,0),0( yT H

T = 02 HBC

x

y

1.3Fig.

T = f(x)

T = 0

T = 00

0(3) ,0)0,(xT H

(2) ,0),( yLT H


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(4) Solution

(i) Assumed Product Solution

)()(),( yYxXyxT (a)

(a) into eq. (3.2)

0)()()()(

2

2

2

2

y

yYxX

x

yYxX (b)

(4) ),(),( xfWxT non-homogeneous


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2

2

2

2

)(

1

)(

1

dy

Yd

yYdx

Xd

xX(c)

)()( yGxF

where 2n

is known as theseparation constant

2

2

2

2

2

)(

1

)(

1n

dy

Yd

yYdx

Xd

xX(d)

0)()(2

2

2

2

dy

YdxX

dx

XdyY


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NOTE:

The separation constant is squared

The constant is positive or negative

The subscriptn. Many values:n

,...,,321

are

known as eigenvalues orcharacteristic values

Special case: constant = zero must be considered

Equation (d) represents two sets of equations

022

2

nnn X

dx

Xd (e)


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022

2

nnn Y

dy

Yd(f)

The functionsn

Xn

Yand

or characteristic values

are known as eigenfunctions

(ii) Selecting the sign of then

terms

Select the plus sign in the equation representing the variable

with 2 HBC. Second equation takes the minus sign


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022

2

nnn X

dx

Xd(g)

02

2

2

nnn Y

dy

Yd(h)

Important case: ,0n

equations (g) and (h) become

02

0

2

dx

Xd (i)


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020

2

dy

Yd(j)

(iii) Solutions to the ODE

xBxAxX nnnnn cossin)( (k)

yDyCyYnnnnn

coshsinh)( (l)

xBAxX 000 )( (m)

yDCyY000

)( (n)


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NOTE: Each product is a solution

)()(),(000

yYxXyxT (p)

)()(),( yYxXyxT nnn (o)

The complete solution becomes

100

)()()()(),(n nn

yYxXyYxXyxT (q)

(iv) Applying Boundary Conditions


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NOTE: Each product solution must satisfy the BC

BC (1)

)()0(),0( yYXyT nnn

)()0(),0(000

yYXyT

Therefore

0)0(n

X (r)

0)0(0

X (s)

Applying (r) to solution (k)

00cos0sin)0(nnn

BAX


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Therefore

0n

B

Similarly, (r) and (m) give

00

B

B.C. (2)

0)()(),( yYLXyLT nnn

Therefore

0)(LXn


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Applying (k)

0sin LAnn

Therefore

0sin Ln

(t)

Thus

,L

nn

...3,2,1n (u)

Similarly, BC 2 and (m) give

0)(000

BLALX


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Therefore

00

A

With ,000

BA the solution corresponding to

0n

vanishes.

BC (3)

0)0()()0,(nnn

YxXxT

0)0(nY

00cosh0sinh)0(nnn

DCY


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Which gives

0n

D

Results so far: ,000 nn

DBBA therefore

and

Temperature solution:

1

))(sinh(sin),(n

nnnyxayxT (3.10)

xAxXnnn

sin)(

yCyYnnn

sinh)(


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Remaining constant

nnn CAaB.C. (4)

xWaxfWxT

n nnn1

sin)sinh()(),((3.11)

To determinen

a from (3.11) we applyorthogonality.

(v) Orthogonality

Use eq. (7) to determinen

a


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The function xn

sin in eq. (3.11) is thesolution to (g)

If (g) is a Sturm-Liouville equation with 2 HBC,orthogonality can be applied to eq. (3.11)

Return to (g)

022

2

nnn X

dx

Xd(g)

Compare with eq. (3.5a)

0)()()(3

2

212

2

nnnn xaxa

dx

dxa

dx

d

(3.5a)


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0)()(21

xaxa 1)(3

xaand

Eq. (3.6) gives

1)()( xwxp and 0)(xq

BC (1) and (2) are homogeneous of type (3.8a).

The characteristic functions xxXxnnn

sin)()(

are orthogonal with respect to 1)(xw

0x to

Multiply eq. (3.11) by ,sin)( xdxxwm

integrate from

Lx


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xdxxwxf

L

m0

sin)()(

xdxxwxWa

L

mn

nnn0 1

sin)(sin)sinh(

(3.12)

Interchange the integration and summation, and use

1)(xw

Introduce orthogonality (3.7)

0)()()( dxxwxx

b

amn

mn (3.7)


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Apply (3.7) to (3.12)

xdxWaxdxxfn

L

nn

L

n0

2

0

sinsinhsin)(

Solve for na

(3.13)xdxxf

WL

a

L

n

n

n

0

sin)(

sinh

2

(5) Checking


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The solution is

1))(sinh(sin),(

nnnn yxayxT (3.10)

Dimensional check

Limiting checkBoundary conditions

Differential equations

(6) Comments

Role of the NHBC


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3.6 Cartesian Coordinates: Examples

Example 3.3: Moving Plate with SurfaceConvection

Outside furnace the plate

exchanges heat by convection.

Determine the temperature distribution in the plate.

y

xU

0

Th,

L

L

3

.

3

Fig.

HBC2

Th,

oT

Semi-infinite plate leaves a

furnace atoT


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Solution

(1) Observations

Symmetry

NHBC


(3) Formulation

(i) Assumptions

(1) 2-D

(2) Steady state

At 0x plate is atoT


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(4) Uniform velocity

,k(3) Constantp

c and ,


Define TT

022

2

2

2

xyx

(a)

k

Ucp

2(b)

(5) Uniform andh T


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(iii) Independent variable with 2 HBC: y-

variable

(iv) Boundary Conditions

(1) 0)0,(

y

x

(2) ),(),(

Lxhy

Lxk

(3) 0),( y

(4) TTy0

),0(

y

x

U

0

Th,

L

L

Fig. 3.3

HBC2

Th,oT


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(4) Solution

(i) Assumed Product Solution)()( yYxX

02 22

2

nnnn X

dxdX

dxXd (c)

(d)0

2

2

2

nn

n

Ydy

Yd

(ii) Selecting the sign of the 2n terms


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022

2

2

nnnn X

dx

dX

dx

Xd(e)

(f)02

2

2

nnn Y

dy

Yd

For :0n

(g)02 02

0

2

dx

dX

dx

Xd

(h)02

0

2

dy

Yd


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)(xXn

2222 expexp)exp( nnnn xBxAx

(i)

yDyCyYnnnnn

cossin)( (j)

(k)000

2exp)( BxAxX

And

(l)yDCyY 000 )(


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100 )()()()(),(

nnn yYxXyYxXyx (m)

Complete solution:

(iv) Application of Boundary Conditions

BC (1)

00

DCn

BC (2)

BiLLnn

tan (n)

00

C


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0n

A

BC (3)

With ,000

DC

000

YX

(3.17)

yxayxn

nnn1

22 cosexp),(

BC (4)

10

cosn

nnyaTT (o)


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(v) Orthogonality

Characteristic Functions:

ynn

cos are solutions to equation (f).

Thus eq. (3.6) gives

1wp and 0q (p)

2 HBC at 0y and Ly

ynn

cos are orthogonal

Comparing (f) with eq. (3.5a) shows that it is a Sturm-

and021 aa .13aLiouville equation with


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L

n

L

n

nydy

ydyTT

a

0

2

00

cos

cos

to

Multiply both sides of (o) by ,cos xdxm

integrate

from 0x Lx and apply orthogonality

LLL

LTTa

nnn

nn cossin2

sin2 0 (q)


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(5) Checking

Dimensional check

Limiting check: If00

),(,0 TyxTqTT

(6) Comments

.0UFor a stationary plate,


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3.7 Cylindrical Coordinates: Examples

Example 3.5: Radial and Axial Conduction ina Cylinder

Two solid cylinders

are pressed co-axially with a force

F and rotated in

opposite directions.

Coefficient of friction is .

Convection at the outer surfaces.

or

r

z

Th,

L

0

2 HBC

Fig. 3.5

L

Th, Th,

Th,


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Find the interface temperature.

Solution

(1) Observations

Symmetry or

r

z

Th,

L

0

2 HBC

Fig. 3.5

LTh,

Th,

Th,

Interface frictional heat

= tangential forcex velocity

Transformation of B.C., TT



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(3) Formulation

(i) Assumptions

(1) Steady

(2) 2-D

(4) Uniform interface pressure

(3) Constant ,k and

(6) Radius of rod holding cylinders is smallcompared to cylinder radius


(5) Uniformh and T


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01

2

2

zr

r

rr

(3.17)

(iii) Independent variable with 2 HBC: r-

variable

(iv) Boundary conditions

(1) ,0),0(

r

z or ),0( z finite


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(2) ),(),(

00 zrh

r

zrk

(3) ),(),(

Lrhz

Lrk

(4),

2

0

( 0)( )

r Fk r f r

z r

(4) Solution


orr

z

Th,

L0

2 HBC

Fig. 3.5

LTh,

Th,

Th,


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)()(),( zZrRzr (a)

01 2R

dr

dRr

dr

d

r kk (b)

022

2

kkk Z

dz

Zd (c)

(ii) Selecting the sign of the 2k

r-variable has 2 HBC


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0222

22

kkkk Rr

dr

dRr

dr

Rdr (d)

022

2

kkk Z

dz

Zd(e)

0020

2

dr

dR

dr

Rdr (f)

For :02k

02

0

2

dz

Zd(g)


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(d) is a Bessel equation:

,0BA 0n,1C ,k

D

Since 0n andD

is real

rYBrJArRkkkkk 00

)( (h)

Solutions to (e), (f) and (g):

zDzCzZkkkkk

coshsinh)( (i)


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000BlnrAR (j)

000 DxCZ (k)

Complete solution

)()()()(),(1

00 zZrRzZrRzr kk

k(l)


BC (1)

)0(0

Y and 0lnNote:


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00

ABk

BC (2)

)()(00,0

0

rhJrJdr

dk

kzrk

(m)BirJ

rJr

k

kk )(

)(

00

010

00

B

khrBi 0 (n)

BC (2)


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Since ,000

BA 0n

LLBiL

LLBiLCD

kkk

kkkkk cosh)(sinh

sinh)(cosh

1cosh)(sinh

sinh)(cosh[sinh),(

k kkk

kkkkk LLBiL

LLBiLzazr

)](cosh0

rJzkk

(3.20)

BC (3)


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BC (4)

)()( 01

rJakrf kkk

k (o)

(v) Orthogonality

)(0

rJ k is a solution to (d) with 2HBC in .r

and

Comparing (d) with eq. (3.5a) shows that it is a Sturm-

,/11

ra .13

aLiouville equation with ,02

a

Eq. (3.6):

rwp and 0q (p)


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)(0

rJk

and )(0

rJi

are orthogonal with respect to

ka.)( rrw Applying orthogonality, eq. (3.7), gives

)(])()[

)()(2

)(

)()(

0

2

0

2

0

2

0

0

0

0

2

0

0

0

0

0

0

rJkhrrk

drrrJrf

drrrJk

drrrJrf

a

kk

r

kk

r

kk

r

k

k

(q)

Interface temperature: set 0z in eq. (3.20)


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)(cosh)(sinh

sinh)(cosh)0,(

0

1

rJLLBiL

LLBiLar

k

k kkk

kkkk

(r)

(5) Checking

Dimensional check: Units of ka

Limiting check: If 0 or 0 then TzrT ),(

(6) Comments

)(xw is not always equal to unity.


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3.8 Integrals of Bessel Functions

0

0

2 )(

r

knndrrrJN (3.21)


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Table 3.1 Normalizing integrals for solid cylinders [3]

Boundary Condition at

(3.8a):

(3.8b):

(3.8c):

0

0

2 )(

r

knn drrrJN0r

0)(0

rJkn

0)(

0

dr

rdJkn

)()(

00 rhJ

dr

rdJk

knkn

2

0

2

2

0)(

2dr

rdJrkn

k

)()(2

10

222

02rJnr

knk

k

)()()(2

10

222

0

2

2rJnrBi

knk

k


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3.9 Non-homogeneous Differential Equations

Example 3.6: Cylinder with EnergyGeneration

L

aT

oT

r

or q

z0 q0

Fig. 3.6

Solid cylinder generates

heat at a rate One end is at.q

0T while the other is insulated.

Cylindrical surface is at .aTFind the steady state temperature distribution.


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Solution

(1) Observations Energy generation leads to

NHDE

Use cylindrical coordinates

L

aT

oT

r

or qz0q

0

Fig. 3.6 Definea

TT to make

BC at surface homogeneous



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(3) Formulation

(i) Assumptions(1) Steady state

(2) 2-D

k(3) Constant and


Definea

TT


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01

2

2

k

q

zrr

rr(3.22)

(iii) Independent variable with 2 HBC

(iv) Boundary conditions

(1) ),0( z finite

(2) 0),(0

zr L

aT

oT

ror q

z0q0

Fig. 3.6

(3) 0),(

z

Lr

(4)

a

TTr0

)0,(


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(4) Solution

)(),(),( zzrzr (a)

Substitute into eq. (3.22)

Split (b): Let

(c)01

2

2

zrr

rr

Let:

01

2

2

2

2

k

q

zd

d

zrr

rr(b)


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Therefore

NOTE:

(d) is a NHODE for the )(z

Guideline for splitting PDE and BC:

),( zr (c) is a HPDE for

should be governed by HPDE and three HBC.

Let take care of the NH terms in PDE and BC)(z

),( zr

02

2

k

q

zd

d(d)


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BC (1)

),0( z finite(c-1)

BC (2)

)(),(0

zzr (c-2)

BC (3)

0)(),(

dz

Ld

z

Lr

Let

(c-3)0),(

z

Lr


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Thus

0)(

dz

Ld(d-1)

BC (4)

aTTr

0)0()0,(

Let

(c-4)0)0,(r

Thus

aTT0)0( (d-2)

Solution to (d)


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FEzzk

qz 2

2)( (e)


)()(),( zZrRzr (f)

(f) into (c), separating variables

(g)022

2

kkk Z

dz

Zd

0222

22

kkkk Rr

dr

Rdr

dr

Rdr (h)


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(ii) Selecting the sign of the 2k

terms

(i)022

2

kkk Z

dz

Zd

0222

22

kkkk Rr

dr

Rdr

dr

Rdr (j)

For,0k

(k)02

0

2

dz

Zd


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00

2

0

22

dr

Rdr

dr

Rdr (l)


(m)kkkkk

BAzZ cossin)(

(j) is a Bessel equation with ,0nBA ,1C

.iD k

(n))()()( 0 rKDrICzR kkkokk

S l i (k) d (l)


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Solution to (k) and (l)

(o)000)( BzAzZ

(p)000

ln)( DrCzR

Complete Solution:

100

)()()()()(),(k

kkzZrRzZrRzzr (q)


BC (c-1) to (n) and (p)

00

CDk

BC ( 4) t ( ) d ( )


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BC (c-4) to (m) and (o)

00

BBk

BC (c-3) to (m) and (o)

00

A

Equation for k

,0cos Lk

or2

)12( kL

k,...2,1k (r)

With 000 BA0

00ZR

The solutions to ),( zr become


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zrIazrk

kkk

10

sin)(),( (s)

BC (d-1) and (d-2)

TTF0

kLqE / and

)()/()/(22)( 02

2

TTLzLzk

Lqz (t)

BC (c-2)

)()/()/(22 0

2

2

TTLzLzkLq

zrIak

k

kk

1

00sin)(

(u)


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(v) Orthogonality

zksin are solutions to equation (i).

and

Comparing (i) with eq. (3.5a) shows that it is a Sturm-

.13

aLiouville equation with ,021

aa

Eq. (3.6) gives

0z and ,Lz2 HBC at zk

sin are orthogonal

with respect to 1w

rwp and 0q


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77

dzTTLzLz

k

Lqk

L

sin)()/()/(2

200

22

L

kkkdzrIa

0

2

00sin)(

Evaluating the integrals and solving fork

a

20

/)()()(

2kao

okkk kqTTrILa (v)

Solution to )(


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Solution to ),( zr

(5) Checking

Limiting check: 0q and .0

TTa

Dimensional check: Units of kLq /2 and of ka insolution (s) are in C

(w)a

TzrTzr ),(),(

22

0)/()/(2

2)( LzLz

k

LqTT

a

zrIa kk

kk1

0 sin)(


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79

3.10 Non-homogeneous Boundary Conditions

Method of Superposition: Decompose problem

Example:

)(yg

)(xfx

y

L

Woq

Th,

0

3.7Fig.

Problem with 4 NHBC is decomposed

into 4 problems each having one NHBC

DE

02

2

2

2

y

T

x

T

Solution:


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80

Solution:

),(),(),(),(),(4321

yxTyxTyxTyxTyxT

)(yg

)(xfx

y

L

Woq

Th,

0

x

y

L

Woq

0

0,h

0

0

x

y

L

W

0 0

0

Th,

4T

)(xf x

y

L

W

0

0,h

0

3T

)(yg

x

y

L

W

0

0,h

0

2T


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Chapter 3 TWO-DIMENSIONAL STEADY STATE CONDUCTION