Andrew Foose
HW#3 – MEAE6330
Conduction Heat Transfer
February 10, 2000
3.1
Consider a hollow cylinder with a  r  b. The initial Temperature is F(r) and both
surfaces are insulated. Find T(r,t).
Reducing the 3-D cylindrical heat equation to 1-D results in the following equation.
 2T 1 T 1 T


r 2 r r  t
With the following B.C.’s and I.C.
T
 0  at  r  a  and  r  b
r
T0  F ( r )
After using separation of variables, applying the I.C. and solving for Cm, Equation 3-73
in the book provides us with T(r,t).

b
2
1
e mt R0 ( m , r )  r R0 (  m , r ) F ( r )dr 
a
m1 N (  m )
T ( r, t )  
Applying the B.C.’s gives us equations for R0 and 1/N.
R0 (  m , r )  J 0 (  m r )Y0(  m b)  J 0 (  m b)Y0 (  m r )
 m2 J 0 (  m a )
1
2

N (m )
2 J 0 2 (  m a )  J 02 (  m b)
2
Finally, the roots of the following transcendental eqn. provide the values of m.
J 0 (  m a )Y0(  m b)  J 0 (  m b)Y0(  m a )
12-5
The attached spreadsheet, prob1214.xls sheet 12-5, shows the results of the finite
difference approach using alg61.for from Burden and Faires, Numerical Analysis.
The steady state Temp at the center of the slab is 0.13.
12-14 The attached spreadsheet, prob1214.xls sheet 12-14, shows the results of a finite
difference numerical solution of this problem. The results show that at t=.1 the
relative error is 4.2% and at t=.2 the relative error is 8.3%. Essentially, the error is
doubled. This makes sense because the truncation error is compounded with each
time step. The reason it is not exactly double may be round off error.