The one-dimensional heat
equation is
∂T /∂t =c 2 ∂2 T /∂x 2
where c 2 =k /(C p ρ )
where k is thermal conductivity, C p
is specific heat at constant pressure, and ρ is the
mass density. (You may find a derivation
here . ) The general solution of this equation is
T (x,t )=exp(-c 2 s 2 t )[A sin(sx )+B cos(sx )]+Cx+D
where A , B , C ,
D , and s are constants to be determined for
the specific problem. (For a derivation, go
here . Go to Section 2.2)
The bar is of length L . Suppose
the boundary conditions are that (1) the temperature of the bar
at one end is held at temperature T 0 , (2)
the other end of the bar (and the sides) are insulated, and that
(3) the temperature at t =0 decreases linearly from
T 0 :
T (0,t )=T 0
∂T (x ,t )/∂x|x= L =0
T (x ,0)=T 0 (1-Fx /L )
These boundary conditions lead to the
following:
B exp(-c 2 s 2 t)+D=T 0
B =0, D =T 0
exp(-c 2 s 2 t )[sA cos(sL )]+C= 0
C =s (exp(-c 2 s 2 t) [A cos(sL )])
s ≠0 so C =0A n cos(s n L )=0⇒s n = ½nπ /L,
n odd
∫T 0 (1-Fx /L )sin(s m x )dx =-Σ∫{A n sin(s n x )sin(s m x )}dx =(L /2)A n δmn
=- 4T 0 ( An =- 4T 0 (1-2(-1)(n+1/2) F /(nπ ))/(nπ )
So, finally,
T (x,t )=T 0 +Σ{exp(-c 2 s n 2 t )A n sin(s n x )}
c 2 =k /(C p ρ ),
An =- 2T 0 (1-2F /(nπ ))/(nπ )
sn = ½nπ/L n odd
We should tabulate the constants to be
used:
T (x,t )=100[1-Σ{(2[1-1.8/(nπ ))])(e(-0.00206n2 t )sin(nπx /40)/(nπ )}
(t is in hours here.)
The plot shows the calculation for only
n=1. It can be seen that even with only one term, the linear
temperature across the ends of the bar is quite well
represented. Calculations for the higher n values show that they
are very small and totally negligible above a couple hundred
degrees. The conclusions are the same as for the boundary
condition T (x,0)=T 1 that several
hundred hours are required to get near to the equilibrium
condition.