Download ME 345 Heat Transfer

ME 345 Heat Transfer
Lecture 9
Lecture Outline:
1) Conduction in a wall with internal generation
2) Conduction in a cylinder or pipe with internal generation
Conduction in a wall with internal generation: (i.e.
q  0 )
For starters, let’s assume that the heat transfer is 1-d, steady-state, with k and q constant.
Then the heat equation becomes:
d2T
dx
2

q
0
k
d2T
or
dx
2

q
k
Separate and integrate twice to get the temperature distribution T(x): (BE SURE that you know how to
do this…)
T( x)  
q 2
x  C1 x  C 2 (the result will look like a parabola!)
2k
So we need two boundary conditions again to solve for C1 and C2. Let’s assume that we know the
temperatures on each side of the wall:
1) T = Ts,1 at x = -L
and
2) T = Ts,2 at x = L
Ts,2
Ts,1
If you use these BCs in the equation for T(x), you get:
T(x) 
Ts,2  Ts,1 x Ts,1  Ts,2
 2
qL
x2
(1  2 ) 

2k
2
L
2
L
-L
0
L
If needed, you can now take the derivative of this to get q”.
IF Ts,2 = Ts,1 = Ts, then we have SYMMETRY about the center line. Essentially the two sides of the wall
are mirror images.
The equation for T(x) becomes:
T(x) 
 2
qL
x2
(1  2 )  Ts
2k
L
If we wanted to find the point where the maximum temperature occurs, we will need to take a
derivative…
dT
.
dx
REMEMBER: whenever you need a minimum or a maximum, you need to take the derivative, and then
the second derivative to determine whether it is a minimum or a maximum.
Take the derivative of the T(x) equation on the previous page:
  2  2 x

qx
dT qL

( 2 )  
dx
2 k L
k
Take the second derivative:
this will equal zero only at x = 0!!!!
d 2T
dx
2

q
is negative, so the temperature is MAX at x = 0.
k
AND, since dT/dx = 0 at x = 0, this means that there is NO heat flux at x = 0! All the heat generated on
the right side of the wall goes out the right side, and all the heat generation on the left side goes out the
left side…NONE of it crosses the middle!
IMPORTANT!!!! Whenever you have symmetry of a problem, you can use the boundary condition that
q” = -k dT/dx at that line of symmetry.
I.E. it is as though the line of symmetry were INSULATED!!!
So, if the temperatures on each side of the wall are the same, we could also draw the problem this way:
then we could have actually solved the
problem using the following two boundary
conditions:
insulation
1) dT/dx = 0 at x = 0
Ts
2) T = Ts at x = L
We would get the exact same answer as we
got before.
0
L
ALSO NOTE that all of this IS NOT TRUE if Ts,1 Ts,2 because then we do not have symmetry!
Conduction in a cylinder or pipe with internal generation: (i.e.
Once again, let’s assume that heat transfer is 1-d, steady, with k and q constant.
The basic heat equation is then:
q
1 d dT
(r )   0
r dr dr
k
or
d dT
q
(r )   r
dr dr
k
or
C
q
dT

r  1
dr
2k
r
Again, integrate with respect to r:
r
q 2
dT

r  C1
dr
2k
q  0 )
If we integrate again:
T(r )  
q 2
r  C1 ln r  C 2
4k
Once again, we need two boundary conditions to get C1 and C2.
Now let’s practice…
GROUPS:
1) Form into groups and divvy up jobs as usual. Choose a group favorite color!
2) You are given the following as your boundary conditions for a copper pipe, with q = 100 W/m3.
a) T = 20C at r = 5 cm (the inside surface) and
b) T = 25C at r = 6 cm (the outside surface).
What is the temperature distribution T( r) through the pipe wall?
What is the heat flux at the surface of the pipe?
BE SURE to draw a picture of the pipe and show how the heat is flowing!!!
3) You now are given a copper cylinder in which there is q = 100 W/m3. The outer temperature of the
cylinder is 30C, and the radius of the cylinder is 2.0 mm. (You can think of this as being something
like an electrical cable that is carrying electricity.
What is the temperature distribution T ( r) through the cylinder
4) Reporters will tell how each group decided on a second boundary condition for the cylinder problem.