Andrew Foose
HW#6 – MEAE6330
Conduction Heat Transfer
March 2, 2000
1.)
a. To show that the solution is T(r)=g(b2-r2)/4k it is a simple ordinary differential
equation (after T/t=0). Substituting Ar2+Br+C for T(r) and using the boundary
conditions you can easy show that A=-g/4k, B=0, and C=gb2/4k.
b. The following equations are derived to solve the problem for a control volume.
T2  T1 
1 2g
x
2
k
Ti 1  2T2  T1  x 2
g
k
Tm 1  0
Using a spreadshed and the goal seek function to find T1 that makes Tm-1=0 will yield the
solution. The following shows the results of different numbers of nodes.
r
0
0.05
0.1
exact m=3 error m=11 error m=21 error m=41 error m=101 error
10
15 50%
11 10% 10.5
5% 10.25 2.5% 10.1 1.00%
7.5
10 33%
8
7% 7.75
3% 7.625 1.7% 7.55 0.67%
0
0
0
0
0
0
c. The above table shows that 101 nodes are needed to get 1% error
2)
a. To solve this problem explicitly a very similar method to above is used. T1 for each
time step will be determined using goal seek to make Tm-1 = 0. The following
equations are developed from the transient heat equation.
x 2
(T1k 1  T1k )
2t
x 2 k 1
 2Ti k 1  Ti k11 
(Ti  Ti k )
t
k 1
Tm1  0
T2k 1  T1k 1 
Ti k11
This will result for a mesh spacing of 3 nodes in x with 5 time steps with the following
results.
x
T
0 105.2361
0.5 65.13597
0.1 -2.8E-14
b. Based on the Heisler charts, this answer is accurate to within 1-5%. However, varying
dx and dt will greatly affect the accuracy in this answer.
c. Again the equations for the explicit method are found in a similar manner, but goal
seek is not necessary for these answers. The equations are as follows.
2t k
(T2  T1k )
2
x
t
 Ti k  2 (Ti k1  2Ti k  Ti k1 )
x
Tmk11  0
T2k 1  T1k 
Ti k 1
The results of different time steps are shown in the following table
T0
n
5
7
9
11
16
21
26
31
36
41
156
3156
882
125
105
108
109.5
110.2
110.6
110.9
111.1
112.2
d. For m=3; n=11 will give an accurate answer. As n increases above n=11 the error
begins to add up.