2-D Steady Heat Equation

2 2 22 2 2

2 2 2

2 22

2 2

Heat Equation

( )

Steady state, 2-D: 0, 0

Bounday Conditions: ( 0, ) 0, ( , ) 0

( , 0) 0, ( , ) ( ) as shown below

p p

T k T T T kT c T

t c x y z c

T TT

x y

T x y T x a y

T x y T x y b f x

x

y

y=b

y=0x=0

x=a

T(x,b)=f(x)

2

2 2

2

Use separation of variables: ( , ) ( ) ( )

It can be shown that

Then 0 and 0

( ) cos( ) sin( )

(0) 0

( ) 0 sin( ), , 1,2,..

0

( ) cosh

n

T x y X x Y y

X Yp

X Y

X p X Y p Y

X x A px B px

X A

nX a B pa p n

a

nY Y

a

n yY y C

a

sinhn y

Da

1 1

1

(0) 0

sinh

( , ) sin sinh

( , ) ( , ) sin sinh

Apply the last boundary condition:

( , ) ( ) sin sinh

n n

n nn n

nn

Y C

n yY D

a

n x n yT x y C

a a

n x n yT x y T x y C

a a

n x n bT x y b f x C

a a

Example

1

1 1

If 3, 6

( ) 100(3 )sin( ) between 0 3

( ,6) ( ) 100(3 )sin( ) sin sinh 23

100(3 )sin( ) sinh sin sin

where sinh i

nn

n nn n

n n

a b

f x x x x

n xT x f x x x C n

n b n x n xx x C A

a a a

n bA C

a

22 2 2 20

s the Fourier sine coefficients

of the function 100(3- )sin( )

1 ( 1) cos( )200(3 )sin( )sin 400

( )

na

n

x x

n an xA x x dx a

a a a n

2

2 2 2 2

22 2

1

22 2

1 ( 1) cos( )400

sinh(2 ) sinh(2 ) ( )

3600 1 ( 1) cos(3)

sinh(2 ) 9

( , ) sin sinh

3600 1 ( 1) cos(3)sin

sinh(2 ) 9

n

nn

n

nn

n

n aA aC

n n a n

n

n n

n x n yT x y C

a a

n

n n

1

sinh3 3n

n x n y

Temperature Distribution in x at different y stations

f(x)=100(3-x)sin(x)

0 1 2 30

50

100

150

200169.322

0

T x 6( )

T x 5.75( )

T x 5.5( )

T x 5( )

T x 4( )

30 x

TT

Constant temperature contour plotRed: max temperaturePurple: min temperature

Each line corresponds to a constant temperature, therefore, the denser the line distribution, the steeper the temperature gradient and vice versa.

The direction of the local heat transfer is normal to the local constant temperature line; and its magnitude is inversely proportional to the local spacing between the two neighboring constant temperature lines.

Superposition of Two Solutions

T1(x,y) T=TA

T=TB

T2(x,y)

0

00

0

0

0

T=TA

T=TB T(x,y)=T1(x,y)+T2(x,y)

0

0