# sov conduction

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Separation of Variables for Multi-

dimensional Steady Systems Example 1: Laplaces equation in rectangular

coordinates (p.93):

See solution on page 93-99

),()()(),(

yxuYbXaYXT

++ =0

'''=+

+

k

g

TT yyxx

TL

TL

Ty=0

Tx=0

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Example 2: Laplaces equation in cylindricalcoordinates

r0 1

011

2 =++ UrUrU rrr

U(1,)=f()

1

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Use separation of variables:

1 becomes:

Also, it must be periodic in :

So:

This is satisfied only if =n where n=0,1,2,, so:

)()( = rRU

:or

cos,sin"'"

0"1

'1

"

22

2

==+

=++

RRr

RRr

Rr

Rr

R

)](cos[)](sin[

cossin

22 +++=+

BA

BA

)2()( +=

,...2,1,0cossin =+= nnBnA nnn ;

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R equation:

This is the equidimesional equation. Characteristicequation obtained substituting R=rn:

For n=0:

For n0:

0'"

22

=+ RnrRRr

2220 nn ==+ or-1)-(

rccRr

c

R

rcRrRRr

ln'

'

'

lnln'ln0'"

00

0

0

2

+==

==+

oror

or

n

n

n

nn rcrcR += '

rccR ln'00+=

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solution:

However, u

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Apply non homog. B.C.:

Use orthogonality:

First multiply by sin(m) and integrate in (0,2):

=

=++==

11

0 cossin)(),1(n

n

n

n nbnaafU

m

n

n

n

n

dnmbdnma

=

++

=

=

=

2

0

0

1

2

01

2

0

0

2

0

0

2

0

sin)(

cossinsinsin

sinsin)(

or

4444 34444 21

44 344 21

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Next multiply by cos(m) and integrate in (0,2)

Next multiply by 1 and integrate in (0,2)

General 2-D cylindrical coordinate problems:

mbdmf =2

0

cos)(

2)( 0

2

0

),(),(),( zTzrTrT

Equidimensionalequation in r:

sin and cos

Bessel functionin r, exponential

in z

Thin walled tubes,can reduce to

Cartesian coords.

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Example 3: Semi-infinite solid cylinder

Let = T - TThen:

r

r0

z

T

)(0),(

)()()0,(0),(

),0(0),0(

0

0

2

2

finiter

rFTrfrzr

finitezzr

zr

rr

r

or

=== =

==

=

+

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Let (r,z) = R(r)Z(z) Need eigenvalue problem in r-direction:

( )

0)(,0

and

0)(,)0(or0)0(

0

"'

1

2

2

2

0

2

2

==

===

=+

==

ZZ

dr

Zd

rRfiniteRdr

dR

rR

dr

dRr

dr

d

Z

ZrR

dr

d

Rr

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Rn = AnJo(nr) where n are determined from Jo(nro)

Also,

Apply the non-homogeneous condition:

Use orthogonality:

=

=

=

1

0 )(),(

)(

n

n

z

n

z

nn

rJeazr

eCzZ

n

n

===

1

0 )()()0,(

n

nn rJarFr

=0

0

0

2

0

0 0

)(

)()(

r

n

r

n

n

drrrJ

drrJrrFa

2

)( 02

1

2

0 rJr n=

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Example 4: Laplaces eqn. In spherical coordinates:

Axisymmetric no variation in direction

r

y

zT(r,,)

)(),(,),0(

0sinsin

12

fRTfiniteT

T

r

Tr

r

==

=

+

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Let T(r,) = R(r)() is the only possible homogeneous direction.

Choose the (-) sign to end up with the following:

Recast 1 as

finiteR

R

dr

dRr

dr

Rdr

dd

dd

=

=+

=+

)0(

02

0sinsin

1

2

22

and

1

2

0sin

cos1

sin

1 2

=+

d

d

d

d

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Let x = cos ; then and

This is the so called Legendres eqn. If = n:

0)1()1( 2 =++

43421

dx

dx

dx

d

dx

d

d

dx

dx

d

d

d []sin

[][]

==

3

)(),(: xQxP nn

Legendre functionsof first kind (polynomials

Of order n)

Legendre functionsof second kind (in finiteseries)

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Temperature Variation in a Solid Sphere

Eqn. 2 is equidimensional eqn. Characteristiceqn. obtained by setting R = r

K,2,1,0);(cos == nPA nn

+=+=

++=

+=

=+=+

)1(2

)12(1

2

)1(411

2

411

002)1(

2,1

2

n

nn

nn

oror

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Hence

Apply non-homog. B.C.:

=

+

=

==+=

0

)1(

)(cos),(

0)0(;)(

n

n

n

n

nnn

nn

nn

PrarT

DfiniteRrDrCrR

==

+=

=

=

1

1 0

0

0

0sin)(cos)(cos0)()(

sin)(cos)(2

12

)(cos)(

dPPdxxPxP

dPfn

Ra

PRaf

nnmn

n

n

n

n

n

n

n

or

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For example: if

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[ ]

K

+

+=

==

===

)(cos16

7

)(cos4

3

2

1),(

16

7

)35(2

1

2

7

04

5)13(

2

1

2

5

3

3

10

1

0 0

3

0

3

3

1

0

301

0

202

2

PR

r

PR

r

T

rT

TdxxxTRa

xxT

dxxT

Ra

.

.

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Legendre Polynomials

Consider the following linear second-order ODE withvariable coefficients:

Or equivalently:

0)1()1( 2 =++

ydx

dyx

dx

d

0)1(2)1( 2

22

=++ ydxdy

xdx

yd

x (5.109a)

(5.109b)

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This equation is known as Legendres diferential

equationand its two linearly independent solutionsare called Legendre functionsof the firstand secondkind, respectively.

When = n, where n is a positive integer or zero, theLegendre functions of the first kind becomes apolynomial of degree n.

The general solution of Legendres equation for

n = 0, 1, 2, can be written as

Pn(x) are polynomials called Legendre Polynomialsofdegreen, and Qn(x) are the Legendre functions of thesecond kind

)()()( 21 xQCxPCxy nn += (5.110)

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The Legendre polynomials are given by:

...!5

)4)(2)(3)(1(

!3

)2)(1()(

...!4

)3)(1)(2(

!2

)1(1)(

,...7,5,3),()1(642

531)1()(,)(

,...6,4,2),(642

)1(531)1()(,1)(

53

42

2/)1(

1

2/

0

++

++

=

++

++

=

=

==

===

xnnnn

xnn

xxV

xnnnn

xnn

xU

nxVn

nxPxxP

nxUn

nxPxP

n

n

n

n

n

n

n

n

(5.111a,b)

(5.111c,d)

(5.112a)

(5.112b)

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The first six Legendre Polynomials are readily foundto be:

Rodrigues Formula

)157063(81)()13(

21)(

)33035(8

1)()(

)35(2

1)(1)(

355

22

24

51

3

30

xxxxPxxP

xxxPxxP

xxxPxP

+==

+==

==

n

n

n

nn x

dx

d

nxP )1(

!2

1)(

2 =

(5.113)

(5.114)

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Recurrence formula:

The following property of Legendre Polynomials is

also frequently used:

xn

xPnxnPxP nnn

)12(

)()1()()( 11

+

++= +

,...3,2,1),()12(

11

=+= +

nxPndx

dP

dx

dPn

nn

(5.115)

(5.116)

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If |x| < 1, the Legendre functions of the second kindare given by

Notice that the Legendre Functions of the secondkind are infinite series, which are convergent when

|x| < 1, but diverges as x 1

,...7,5,3),(531

)1(642)1()(

),()(

,...6,4,2),()1(531

642)1()(

),()(

2/)1(

11

2/

00

=

=

=

=

=

=

+nxU

n

nxQ

xUxQ

nxVn

nxQ

xVxQ

n

n

n

n

n

n

(5.117a,b)

(5.117c,d)

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Legendre Functions of the second kind Qn(x) also

satisfy recurrence formulas (5.115) and (5.116)

When |x| < 1

Qn for any positive integer value of n:

xxxxQ 10 tanh

11ln

21)( =+=

(5.118)

)...(

)1(3

)52()(

1

)12()()()( 310 xP

n

nxP

n

nxQxPxQ nnnn

=

(5.119)

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If |x| < 1, the substitution x = cos transforms

Legendres differential equation from the form(5.109b) into the form:

or, equivalently

when = n. Hence the general solution to (5.120a) is

0)1(sinsin

1

=++

ynnd

dy

d

d

(5.120a)

0)1(cot2

2 =+++ ynnddy

dyd

(5.120b)

)(cos)(cos)(21 nn QCPCy += (5.121)

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It should be noted here that Qn(cos) is not finite

when cos = 1; that is when = k, k = 0, 1, 2, ,whereas Pn(cos) is merely a polynomial of degree nin cos. In particular we have

.

.

.

)cos33cos5(

8

1)cos3cos5(

2

1)(cos

)12cos3(4

1)1cos3(

2

1)(cos

cos)(cos

1)(cos

3

3

22

1

0

+==

+==

==

P

P

P

P

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When |x| < 1, the functions

are called the associated Legendre functionsofdegreen and orderm of the firstand second kinds,respectively. They can be shown to satisfy thedifferential equation

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

)()(,)(

)1()(

02/2

02/2

xQxQdx

xQdxxQ

xPxPdx

xPdxxP

nnmn

m

mmn

nnm

n

mmm

n

==

==

(5.122a)

(5.122b)

01

)1(2)1(2

2

2

22 =

++ y

x

mnn

dx

dyx

dx

ydx (5.123)

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Non-Homogeneous Problems

May result from non-homog. boundaries and/or non-homog. Governing equation

Example 5: Non-homog. Boundaries

The solutions can be added only when expressed ina common coordinate system. Sign for upper andlower B.C. depend on whether heat flux is in or out

2 = 0 = 21= 0 22= 0 23= 0+ +

F(y)

o

F(y)

0

0

0 0

0

= hy

k

"q

y

k =

11 =

hy

k

01 =

y 03 =

y"2 q

yk =

22 =

hy

k 33 =

hy

k

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Example 6: Non-homog. Term in the G.E. (use of

partial solutions)

L L

x

y

l

l 0 u

0

0

0

0

u

xy

=0

=0

y=0

x=0

0),(),()0,(),0(

0

"'

=====++

yLlxxy

k

u

yx

yyxx 1

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Assume soln. of following form:

With form given by 2 :

If u is function of x and y, try variation of parameters

)(),(),(or

)(),(),(

43

21

yyxyx

xyxyx

+=+= 2

3

(

2

21

111

2

1

2

2

1

2

22

2

2

),

0),()0,(),0(;0

0)(;0)0(;0"'

=

==

=

=

+

===+

lx

yLx

y

y

xyx

Ldx

dk

udx

d

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Two-Dimensional Transients

Example 7: Separation of variables (p134)

Let U = W(r,)()

4 becomes:

)()0,,(;0),,1(

),,(

12

rfrUU

rUU

UUrr

UU rrr

==

==++

where4

2

2

'111 =

=

++ Wr

Wr

WW

rrr

exp. decayIn time

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Also:

Let W = R(r)()

2= e

011 2

2

=+++ WWr

Wr

Wrrr

2222

2

2

"'"

0"1

'1

"

=

=++

=+++

rR

Rr

R

Rr

RRr

Rr

R

or

need periodicbehavior in

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So;

Now , so satisfies:

),(),,(:

0)('"

)2()(,cos,sin:

2222

rYrJR

RnrrRRr

n

nn

forwhere

=++

+==

must be absent forfinite temp. at r=0

0)(0),,1( == mnnJU

0)( =mnnJ where mn is the mth zero crossing of Jn

m = 1, 2, 3, ; n = 0, 1, 2,

eigenvalues in r-dir. eigenvalues in -dir.

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Unknown constants are again evaluated using

orthogonality:

first:

=

=

+=1 0

2)()cossin(),,(m n

mnnmnmnmnerJnBnArU

=

= +=

=

1 0)()cossin(

),()0,,(

m n

mnnmnmn rJnBnA

rfrU

44 344 21444 3444 21

m

mnnmn

nrF

=

= 2

0

2

1

'''

)',(

2

0'sin)('sin),(

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next:

Drop primes for convenience:

=

=

=

1

0 ''

2'''

1

01

1

0 '''''''''

)(

)()()()';(

drrrJA

nmnnm

m

nmnmnnmnnmn

{ }

1

0

2

11

2

1

0

)()()(2

);()(

= +

rJrJrJr

A

drnrFrrJ

mnnmnnmnnmn

mnn

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or

Determine Bmn similarly. See p. 138. Note n=0 hasto be treated separately (see p. 90)

{ }

= + )()()(2

1 211 mnnmnnmnnmn JJJA L.H.S.

0)(

2)(1 mnn

mn

mnn Jn

J

+= +

=

=

+

+

1

0

2

02

1

2

1

1

0

sin),()()(

2

)(2

1);()(

drdnrfrrJJ

A

mnn

mnn

mn

mnnmnmnn

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Example 8: Reduction of multi-dimensional to one-dimensional transients.

- Problem must be homog. and linear

- I.C. should be constant or expressible in

product form: e.g. Ri(r)Zi(z) or Xi(x)Yi(y),etc.

L

z

rto

to

to ro

0

0

2

2

2

2

)0,,(

0),,(0),0,(

0),,(

11

ttzrT

LrTrT

zrT

T

z

T

r

T

rr

T

i===

=

=

+

+

1

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Let

Substitute in 1 :

This will be satisfied if:

),(),( rZrRT= 2

0111

)(11

=

+

+

+=++

ZZRZRR

r

R

RZZRRZZRr

ZR

zzrrr

zzrrr

or

ZZ

RRrR

zz

rrr

1

11

=

=+

and

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B.C.:

These are satisfied when:

I.C.:

One choice that will satisfy this:

),(),(),,(

),0(),(),0,(

),(),(),,( 00

LZrRLrT

ZrRrT

zZrRzrT

==

=

0),(),0(0),( 0 == LZZrR ,,

)0,()0,()0,,(0 zZrRttzrT i ==

1)0,()0,( 00 == zZttrR i ,

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This reduces the originial problem into 2 one-dim.

Problems. Z is the solution to a 1-D transient in a slab: (p.84)

R is the solution to a 1-D transient in an infinitely long

cylinder: (p.120)

where m satisfy J0(mr0) = 0. Solution for T is obtained from 2

=

+

++

= 0/)12(

222

12

]/)12sin[(4

),(n

Ln

en

LznzZ

( )

=

=1

/)(

010

00

20

20

)()(

)(2),(

m

rr

mm

mi

merJr

rJttrR

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Many other problems can be handled in a similar

manner, e.g.

See other examples, also p.265 I&P

Can also include other homog. Conditions such asconvection. Cannot handle flux.

Multi-D I.C. must be of product form.

=

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Some Multi-D Steady Problems

Solvable By Separation of VariablesRectangular:

2D (s.s):

1 2 3

qf1(x)

f4(x)

f3(x)

f2(x) T0T0

T0 f2(x)

h, T0

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3D (s.s.):

A: 1 at T1, 2 at T2, all others at T0.B: 1 at T1, 2 at T2, all others convective conditions

C: 1 at T1, convection at all others

D: Region extends infinitely in the x-dirn

from X=0.Surface at x=0 at T1. Convection at other foursurfaces

x

yz

2

6

3

5

4

1

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Cylindrical:

2D:

(a) Short cylinder (r,z)

(b) Circumferential temperature variation

variation (r,) in a half infinite solidlong half cylinder: z > 0

z

r

f(r) = T-Tr

TS or h(for r=R or z=L)

f()

T0

r

z T

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(c) Disk shaped input B.C. (flux or temperature)

Can apply to bottom surface of a semi-infiniteregion or between two semi-infinite regions:

Spherical (r,); axisymmetric

(a) hollow sphere (b) solid hemisphere

T0f()

T0

f()

Multi Dimensional Transients Using

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Multi-Dimensional Transients Using

Separation of Variables Rectangular Regions:

(i) Assuming surface temperature

is T0 at all faces for t > 0 Ti(x,y,z)

2H

2W2L

zx

y

( ) ( ) ( )[ ]( ) dxdydz

H

zm

L

ym

W

xmTzyxTA

HpLnWm

H

zm

L

ym

W

xmAe

TtzyxT

W L H

imnp

mnp

m n p

mnp

tmnp

sinsinsin),,(

///

sinsinsin

),,,(

0 0 0 0

22222

1 1 1

0

2

=++=

=

=

=

=

=

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(ii)

Cylindrical Regions:

(i) (ii)

L T0x

y Ti(x,y)

h, T0

h, T0

Ti(r,) h(Ts- T)T0

T0

T0

Ti