1 CHAPTER 3 TWO-DIMENSIONAL STEADY STATE CONDUCTION 3.1 The Heat Conduction Equation Assume: Steady...

1

CHAPTER 3

TWO-DIMENSIONAL STEADY STATE CONDUCTION

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

2

• 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

3

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 with

variable coefficients

3.2 Method of Solution and Limitations

• Method: Separation of variables

• Basic approach: Replace PDE with sets of ODE

4

• 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.

variabledependent the iflinear is equation Anunitypower to raisedappear sderivative itsor

occurnot do productstheir if and

5

3.3 Homogeneous Differential Equations and Boundary Conditions

• Example: Eq. (3.1): Replace T by cT and divide

through by c

02

2

2

2

ck

q

x

T

k

Uc

y

T

x

T p

NH (a)

not isit if shomogeneou is condition buondaryor equation Anconstant a by multiplied is variabledependent the when altered

6

Example: Boundary condition

TThx

Tk (b)

Replace T by cT

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

7

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) )()()(

32

212

2

xaxadx

dxa

dx

dn

nn

0n

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

dxp

dx

dn

n

0

n

8

where

,exp)(1dxaxp ),()(

2xpaxq )()(

3xpaxw

(3.6)

NOTE:

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

• Equation (3.5) represents a set of n equations

corresponding to n values of n

• The solutions n are known as the characteristic

functions

9

• Important property of the Sturm-Liouville problem: orthogonality

• Two functions, n ,

mand are said to be orthogonal

in the interval a and b 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:

10

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

(2) BC at x = a and x = b are homogeneous of the form

0n (3.8a)

(3.8b) 0dx

dn

(3.8c) 0dx

d nn

11

Special Case: If p(x) = 0 at x = a or x = b , these conditions can be extended to include

)()( bann (3.9a)

and

dx

bd

dx

adnn

)()( (3.9b)

(3.9a) and (3.9b) are periodic boundary conditions.

• Physical meaning of (3.8) and (3.9)

12

• Relationship between the Sturm-Liouville problem, orthogonality, and the separation of variables method:

PDE + separation of variables 2 ODEOne of the 2 ODE = Sturm-Liouville problem

Solution to this ODE = n = orthogonal functions

13

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

• 4 BC are needed

• 3 BC are homogeneous

2 HBC

x

y

1.3Fig.

T = f(x)

T = 0 T = 0

T = 00

0

14

(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) Constant k

15

(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 HT = 02 HBC

x

y

1.3Fig.

T = f(x)

T = 0

T = 00

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

(2) ,0),( yLT H

16

(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

17

2

2

2

2

)(1

)(1

dy

YdyYdx

XdxX

(c)

)()( yGxF

where 2n is known as the separation constant

22

2

2

2

)(

1

)(

1ndy

Yd

yYdx

Xd

xX (d)

0)()(2

2

2

2

dy

YdxX

dx

XdyY

18

NOTE:

• The separation constant is squared

• The constant is positive or negative

• The subscript n. Many values: n ,...,,

321are

known as eigenvalues or characteristic values

• Special case: constant = zero must be considered

• Equation (d) represents two sets of equations

022

2

nn

n Xdx

Xd (e)

19

022

2

nn

n Ydy

Yd (f)

• The functionsn

Xn

Yand

or characteristic values

are known as eigenfunctions

(ii) Selecting the sign of the n terms

Select the plus sign in the equation representing the variable

with 2 HBC. Second equation takes the minus sign

20

022

2

nn

n Xdx

Xd (g)

022

2

nn

n Ydy

Yd (h)

• Important case: ,0n equations (g) and (h) become

02

02

dx

Xd (i)

21

020

2

dy

Yd(j)

(iii) Solutions to the ODE

xBxAxXnnnnn cossin)( (k)

yDyCyYnnnnn coshsinh)( (l)

xBAxX000

)( (m)

yDCyY000

)( (n)

22

NOTE: Each product is a solution

)()(),(000

yYxXyxT (p)

)()(),( yYxXyxTnnn

(o)

The complete solution becomes

100

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

nnyYxXyYxXyxT (q)

(iv) Applying Boundary Conditions

23

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

24

Therefore0

nB

Similarly, (r) and (m) give

00B

B.C. (2)

0)()(),( yYLXyLT nnn

Therefore

0)( LXn

25

Applying (k)

0sin LAnn

Therefore

0sin Ln (t)

Thus

,L

nn

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

Similarly, BC 2 and (m) give

0)(000 BLALX

26

Therefore0

0A

With ,000BA the solution corresponding to

0n vanishes.

BC (3)

0)0()()0,( nnn

YxXxT

0)0( n

Y

00cosh0sinh)0( nnn

DCY

27

Which gives

0n

D

Results so far: ,000

nn

DBBA therefore

and

Temperature solution:

1

))(sinh(sin),(n

nnnyxayxT (3.10)

xAxXnnnsin)(

yCyYnnnsinh)(

28

Remaining constant

nnnCAa

B.C. (4)

xWaxfWxTn

nnn

1

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

To determinen

a from (3.11) we apply orthogonality.

(v) Orthogonality

• Use eq. (7) to determinen

a

29

• The function xnsin in eq. (3.11) is the

solution 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

nn

n Xdx

Xd (g)

Compare with eq. (3.5a)

0)()()(3

2212

2

nn

nn xaxadx

dxa

dx

d

(3.5a)

30

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

31

xdxxwxfL

m0

sin)()(

xdxxwxWaL

mn

nnn

0 1

sin)(sin)sinh(

(3.12)

Interchange the integration and summation, and use

1)( xw

Introduce orthogonality (3.7)

0)()()( dxxwxxb

amn mn (3.7)

32

Apply (3.7) to (3.12)

xdxWaxdxxfn

L

nn

L

n

0

2

0

sinsinhsin)(

Solve for n

a

(3.13) xdxxfWL

aL

nn

n 0

sin)(sinh

2

(5) Checking

33

The solution is

1

))(sinh(sin),(n

nnnyxayxT (3.10)

Dimensional check

Limiting check

Boundary conditions

Differential equations

(6) Comments

Role of the NHBC

34

3.6 Cartesian Coordinates: Examples

Example 3.3: Moving Plate with Surface Convection

Outside furnace the plate exchanges heat by convection.

Determine the temperature distribution in the plate.

y

xU

0

Th,

L

L

3.3Fig.

HBC2

Th,

oT

Semi-infinite plate leaves a furnace at oT

35

Solution

(1) Observations

• Symmetry

• NHBC


(3) Formulation (i) Assumptions

(1) 2-D

(2) Steady state

• At 0x plate is at oT

36

(4) Uniform velocity

,k(3) Constantp

c and ,

(ii) Governing EquationsDefine TT

022

2

2

2

xyx

(a)

k

Ucp

2

(b)

(5) Uniform andhT

37

(iii) Independent variable with 2 HBC: y- variable

(iv) Boundary Conditions

(1) 0)0,(

y

x

(2) ),(),(

Lxhy

Lxk

(3) 0),( y

(4) TTy

0),0(

y

xU

0

Th,

L

L

Fig. 3.3

HBC2

Th,oT

38

(4) Solution


)()( yYxX

02 22

2 nn

nn Xdx

dX

dx

Xd (c)

(d) 022

2

nnn Y

dy

Yd

(ii) Selecting the sign of the 2n terms

39

02 22

2 nn

nn Xdx

dX

dx

Xd (e)

(f) 022

2 nn

n Ydy

Yd

For :0n

(g) 02 02

02

dx

dX

dx

Xd

(h) 020

2

dy

Yd

40


)(xXn

2222 expexp)exp( nnnn xBxAx

(i)

yDyCyYnnnnn cossin)( (j)

(k) 000

2exp)( BxAxX

And

(l) yDCyY000

)(

41

100

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

nnyYxXyYxXyx (m)

Complete solution:

(iv) Application of Boundary Conditions

BC (1)0

0DC

n

BC (2)

BiLLnn tan (n)

00C

42

0n

ABC (3)

With

,000DC

000YX

(3.17)

yxayxn

nnn

1

22 cosexp),(

BC (4)

10

cosn

nnyaTT (o)

43

(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-and0

21aa .1

3aLiouville equation with

44

L

n

L

n

n

ydy

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

sin20

(q)

45

(5) Checking

Dimensional check

Limiting check: If 00

),(,0 TyxTqTT

(6) Comments

.0UFor a stationary plate,

46

3.7 Cylindrical Coordinates: Examples

Example 3.5: Radial and Axial Conduction in a 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,

L0

2 HBC

Fig. 3.5

LTh, Th,

Th,

47

Find the interface temperature.

Solution

(1) Observations

• Symmetryor

r

z

Th,

L0

2 HBC

Fig. 3.5

LTh, Th,

Th,

• Interface frictional heat

= tangential force x velocity

• Transformation of B.C., TT


48

(3) Formulation

(i) Assumptions

(1) Steady

(2) 2-D

(4) Uniform interface pressure

(3) Constant ,k and

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


(5) Uniform

h and T

49

01

2

2

zrr

rr

(3.17)

(iii) Independent variable with 2 HBC: r- variable

(iv) Boundary

conditions (1) ,0),0(

r

zor ),0( z finite

50

(2) ),(),(

00 zrhr

zrk

(3) ),(),(

Lrhz

Lrk

(4),

2

0

( 0)( )

r Fk r f r

z r

(4) Solution


or

r

z

Th,

L0

2 HBC

Fig. 3.5

LTh, Th,

Th,

51

)()(),( zZrRzr (a)

01 2

R

dr

dRr

dr

d

r kk (b)

022

2

kk

k Zdz

Zd (c)

(ii) Selecting the sign of the 2k

r-variable has 2 HBC

52

0222

22

kkkk Rr

dr

dRr

dr

Rdr (d)

022

2

kk

k Zdz

Zd (e)

0020

2

dr

dR

dr

Rdr (f)

For :02 k

02

02

dz

Zd(g)

53


(d) is a Bessel equation:

,0BA 0n,1C ,k

D

Since 0n and D is real

rYBrJArRkkkkk

00)( (h)

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

zDzCzZkkkkk coshsinh)( (i)

54

000Bln rAR (j)

000DxCZ (k)

Complete solution

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

00zZrRzZrRzr

kk

k

(l)


BC (1)

)0(0

Y and 0lnNote:

55

00AB

k

BC (2)

)()(00,0

0

rhJrJdr

dk

kzrk

(m) BirJ

rJr

k

kk

)(

)(

00

010

00B

khrBi0

(n)

BC (2)

56

Since ,000BA 0n

LLBiL

LLBiLCD

kkk

kkkkk

cosh)(sinh

sinh)(cosh

1 cosh)(sinh

sinh)(cosh[sinh),(

k kkk

kkkkk LLBiL

LLBiLzazr

)](cosh0

rJzkk (3.20)

BC (3)

57

BC (4)

)()(0

1

rJakrfkk

kk

(o)

(v) Orthogonality

)(0

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

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

,/11

ra .13aLiouville equation with ,0

2a

Eq. (3.6):

rwp and 0q (p)

58

)(0

rJk and )(

0rJ

i are orthogonal with respect to

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

)(])()[

)()(2

)(

)()(

020

20

20

00

0

20

00

0

0

0

rJkhrrk

drrrJrf

drrrJk

drrrJrf

akk

r

kk

r

kk

r

k

k

(q)

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

59

)(cosh)(sinh

sinh)(cosh)0,(

01

rJLLBiL

LLBiLar

kk kkk

kkkk

(r)

(5) Checking

Dimensional check: Units ofk

a

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

(6) Comments

)(xw is not always equal to unity.

60

3.8 Integrals of Bessel Functions

0

0

2 )(

r

knndrrrJN (3.21)

61

Table 3.1 Normalizing integrals for solid cylinders [3]

Boundary Condition at

(3.8a):

(3.8b):

(3.8c):

0

0

2 )(

r

knndrrrJN 0

r

0)(0rJ

kn

0)(

0 dr

rdJkn

)()(

00 rhJ

dr

rdJk

knkn

2

02

20

)(

2

dr

rdJrkn

k

)()(2

10

22202

rJnrknk

k

)()()(2

10

2220

22

rJnrBiknk

k

62

3.9 Non-homogeneous Differential Equations

Example 3.6: Cylinder with Energy Generation

L

aT

oT

r

or q z0 q

0

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 .a

T

Find the steady state temperature distribution.

63

Solution

(1) Observations

• Energy generation leads to

NHDE

• Use cylindrical coordinates

L

aT

oT

ror q

z0q0

Fig. 3.6 • Definea

TT to make BC at surface homogeneous


64

(3) Formulation

(i) Assumptions

(1) Steady state

(2) 2-D

k(3) Constant and


Define

aTT

65

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

TTr 0

)0,(

66

(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)

67

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)

68

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

69

Thus

0)(

dz

Ld(d-1)

BC (4)

aTTr

0)0()0,(

Let(c-4) 0)0,( r

Thus

aTT

0)0( (d-2)

Solution to (d)

70

FEzzk

qz

2

2)( (e)


)()(),( zZrRzr (f)

(f) into (c), separating variables

(g) 022

2

kk

k Zdz

Zd

0222

22

kkkk Rr

dr

Rdr

dr

Rdr (h)

71

(ii) Selecting the sign of the 2k terms

(i) 022

2

kk

k Zdz

Zd

0222

22

kkkk Rr

dr

Rdr

dr

Rdr (j)

For ,0k

(k) 02

02

dz

Zd

72

0020

22

dr

Rdr

dr

Rdr (l)


(m) kkkkkBAzZ cossin)(

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

.iDk

(n) )()()(0

rKDrICzRkkkokk

73

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)

00CD

k

74

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

00BB

k

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

00A

Equation fork

,0cos Lk or

2

)12(

kL

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

With 000BA

000ZR

The solutions to ),( zr become

75

zrIazrk

kkk

10

sin)(),( (s)

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

TTF0

kLqE / and

)()/()/(22

)(0

22

TTLzLzk

Lqz (t)

BC (c-2)

)()/()/(22 0

22

TTLzLzk

Lq

zrIak

kkk

100

sin)(

(u)

76

(v) Orthogonality

zksin are solutions to equation (i).

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

.13aLiouville equation with ,0

21aa

Eq. (3.6) gives

0z and ,Lz 2 HBC at zksin are orthogonal

with respect to 1w

rwp and 0q

77

dzTTLzLzk

Lqk

L

sin)()/()/(22

00

22

L

kkkdzrIa

0

200

sin)(

Evaluating the integrals and solving for k

a

2

0/)(

)()(

2kao

okkk kqTT

rILa

(v)

78

Solution to ),( zr

(5) Checking

Limiting check: 0q and .0

TTa

Dimensional check: Units of kLq /2 and ofk

a in

solution (s) are in °C

(w) a

TzrTzr ),(),(

22

0)/()/(2

2)( LzLz

k

LqTT

a

zrIak

kkk

10

sin)(

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

80

Solution:),(),(),(),(),(

4321yxTyxTyxTyxTyxT

)( 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

1 CHAPTER 3 TWO-DIMENSIONAL STEADY STATE CONDUCTION 3.1 The Heat Conduction Equation Assume: Steady...

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Transcript of 1 CHAPTER 3 TWO-DIMENSIONAL STEADY STATE CONDUCTION 3.1 The Heat Conduction Equation Assume: Steady...