sepvarcylss

8/10/2019 sepvarcylss

1/11

J. McKelliget, 2002

1/1

2D Separation of Variables in Cylindrical Coordinates

Consider the problem of a cylinder with a specified heat flux at the outer surface, andspecified temperature T=0at each end.

r

0( )q zT

r k

=

bT=0

T=0

h

Assuming azimuthal symmetry, T=T(r,z),the steady-state heat conduction equation incylindrical polar coordinates becomes

2 2

2 2

10

T T T

rr r z

+ + =

Postulate that

( , ) ( ) ( )T r z R r Z z =

The heat conduction equation becomes

2 22

2 2

1 1 1d R dR d Z

R r Zd r d r d z

+ = =

or,2

22

1 0d R dR Rrd r d r

+ =

22

20

d ZZ

d z+ =

As with the cartesian case we choose the direction where both boundary conditions are

homogeneous (separable). Thezdirection boundary conditions separate to


2/11

J. McKelliget, 2002

2/2

(0) 0

( ) 0

Z

Z h

==

We have seen this before. The eigenvalues are

; 1,2,3.....h n n = =

and the eigenfunctions are

sinn nn z

Z Ah

=

The equation forRis

2 2

21 0d R dR Rrd r d r

+ =

When re-expressed in terms of the independent variable,x= r , it reduces to the

modified Bessel equation of zero order

2

2

10

d R dRR

x d rd x+ =

The general solution is

0 0( ) ( )R CI r DK r = +

Since 0K as 0r we must setD=0to get

0( )R CI r=

The temperature solution now becomes

0( , ) ( )sin( ) ; 1,2,3.....n nn r n z

T r z A I nh h

= =

These functions satisfy the heat conduction equation, the zero temperature boundary

conditions in the z direction, and have zero derivative at the axis of symmetry. Each one,by itself, is incapable of satisfying the the specified flux boundary condition at the outer

surface of the cylinder. Previous experience, however, suggests that a linear combinationof these functions might be able to satisfy this boundary condition. Consider


3/11

J. McKelliget, 2002

3/3

01

( , ) ( )sin( )nn

n r n z T r z A I

h h

==

taking the derivative and using the derivative property of modified Bessel functions( '0 1I I= ),

11

( )sin( )nn

T n n r n z A I

r h h h

=

=

The boundary condition becomes

01

1

( ) sin( )n

n

qT n n b n z A I

r k h h h

=

= =

Again, this is a Fourier sine expansion of the function 0( )q z

kwith the term [] being the

fourier expansion coefficient.

01

0

2( ) sin( )

h

nqn n b n

A I z dzh h h k h

=

or

0

01

2sin( )

( )

h

nq n

A z dzn b k hn I

h

=

The analytic solution becomes

00

1 01

2( , ) sin( ) ( )sin( )

( )

h

n

q n n r n z T r z z dz I

n b k h h hn I

h

=

=


4/11

J. McKelliget, 2002

4/4

Homogeneous Boundary Condit ions in the r-dir ection

Consider a cylindrical problem where the homogeneous direction is r. A cylinder haszero temperature at the outer surface, zero temperature at the lower surface and specified

temperature at the top surface.

r

0=

bT=0

T=Th

h

Assuming azimuthal symmetry, T=T(r,z),the steady-state heat conduction equation incylindrical polar coordinates becomes

2 2

2 2

10

T T T

rr r z

+ + =

Postulate that

( , ) ( ) ( )T r z R r Z z =

The heat conduction equation becomes

2 2

2 2

1 1 1d R dR d Z C

R r Zd r d r d z

+ = =

or,2

21 0d R dR CRrd r d r

+ =

2

20

d ZCZ

d z+ =

where C is the separation constant.


5/11

J. McKelliget, 2002

5/5

What makes this problem different is that the homogeneous boundary conditions are inthe rdirection. They are

( )0

0 ; , 0r

TT b z

r =

= =

and separate to

( )0

0 ; 0r

dRR b

dr == =

The radial equation is

2

2

10

d R dRCR

rd r d r

+ =

Case I C=0. The equation becomes

2

2

10

d R dR

rd r d r + =

or

10

d dRr

r d r d r

=

This is integrated once to give

dR A

d r r=

The symmetry axis boundary condition0

0r

dR

dr == immediately requires that A=0 and

0dR

R Bd r

= =

The boundary condition ( ) 0R b = requires thatB=0and we are left with the trivialsolution. Thus the separation constant cannot be zero if we are looking for non trivial

solutions.


6/11


7/11

J. McKelliget, 2002

7/7

0 0( ) ( )R EJ r DY r = +

Since 0Y as 0r we must setD=0to get

0 ( )R EJ r=

The boundary condition ( ) 0R b = requires

00 ( )EJ b=

If you refer back to the plot of 0J given in the section on Bessel Functions you will see

that it is an oscillatory function similar to a decaying sine wave. This means that we cansatisfy the boundary condition (without settingE=0) as long as the separation constant

satisfies the equation

00 ( )nJ b=

The values of n that satisfy this equation are called the eigenvalues of the problem.

Although we cannot give a simple solution to this equation, the eigenvalues are a discreteinfinite set of numbers that can be calculated or looked up in tables.

The axial equation becomes

2

22 0n n nd Z

Zd z

=

which has the general solution

cosh( ) sinh( )n n n n nZ D z E z = +

The boundary condition

( ,0) 0 ( ) (0)n nT r R x Z = =

is separable and gives Zn(0)=0. This can be satisfied if we setDn=0. The solution for the

temperature becomes

0( )sinh( )n n n nT F J r z =


8/11

J. McKelliget, 2002

8/8

These functions satisfy the heat conduction equation and the three homogeneousboundary conditions. Each one, by itself, is incapable of satisfying the the specified

temperature boundary condition atz=h. Previous experience, however, suggests that alinear combination of these functions might be able to satisfy this boundary condition.

Consider

01

( , ) ( )sinh( )n n nn

T r z F J r z

==

Putting in the boundary condition

01

( , ) ( )sinh( )h n n nn

T r h T F J r h

== =

How to determine the constantsFn? When we solved a similar problem in Cartesiancoordinates we used the properties of the Fourier series and function orthogonality. We

could do the same here if the Bessel Functions were orthogonal. We will state thefollowing orthogonality relationship that we will prove later.

0 0

0

( ) ( ) 0

b

n mrJ r J r dr if n m =

where n are the solutions of the equation 00 ( )nJ b=

Multiplying the equation

01

( )sinh( )h n n nn

T F J r h

==

by 0( )mrJ r and then integrating from 0to bgives

0 0 010 0

( ) ( ) ( ) sinh( )

b b

h m n m n nn

rT J r dr F rJ r J r dr h

=

=

The integral in square brackets is only non-zero when n=mand the above expressionbecomes

0 0 0

0 0

( ) ( ) ( ) sinh( )

b b

h m m m m mrT J r dr F rJ r J r dr h =


9/11

J. McKelliget, 2002

9/9

This gives

0

0

0 0

0

( )

( ) ( ) sinh( )

b

h m

m b

m m m

rT J r dr

F

rJ r J r dr h

=

In summary, the analytic solution becomes

01

( , ) ( )sinh( )n n nn

T r z F J r z

==

where

0

0

0 0

0

( )

( ) ( ) sinh( )

bh n

n b

n n n

rT J r dr

F

rJ r J r dr h

=

and n are the solutions of the equation 00 ( )nJ b=


10/11

J. McKelliget, 2002

10/10

Proof of Orthogonality Relationship

Prove that

0 0

0( ) ( ) 0

b

n mrJ r J r dr if n m =

and n are the solutions of the equation 00 ( )nJ b=

Use the notation

( ) ( );m o m n o nJ r J r = =

These functions satisfy the Bessel Equation

2

2

0

0

mm m

nn n

ddr r

dr dr

ddr r

dr dr

+ = + =

Multiply the first equation by n , the second by m , subtract the equations, and then

integrate from 0to b

2 2

0

0

b

m nn m m n m n m nd dd d

r r r r dr dr dr dr dr

+ =

Rearrange as

( )2 20 0 0

0

b b bm n

n m m n m nd dd d

r dr r dr r dr dr dr dr dr

+ =

The first two terms can be integrated by parts to give

( )2 20 00 0 0

0

b b bb bm m n n n m

n m m n m nd d d d d d

r r dr r r dr r dr dr dr dr dr dr dr

+ + =

In this equation the first two integrals cancel to give


11/11

J. McKelliget, 2002

11/11

( )2 20 0 0

0

bb bm n

n m m n m nd d

r r r dr dr dr

+ =

Both n and m are zero atx=b as a result of the eigenvalue equation 00 ( )nJ b= .

The derivatives of the Bessel functions are zero atx=0, (as is r), so both terms in squarebrackets are zero. The expression becomes

( )2 20

0

b

m n m nr dr =

If m n the integral must be zero and orthogonality is proved. This proof was verysimilar to the proof for the orthogonality of sines and cosines. It is the differentialequations and the boundary conditions that are producing orthogonal functions, it is not

just something peculiar to sines cosines and Bessel functions.

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