M p: =8; q: =2;
Diffusivity constant
M k: =1+0. 1* q;
Length of the rod
M L : = 100+10* p;
Initial temperature
M u0 : = 100;
p :=8
q :=2
(1)
k:=1.2
(2)
L :=180
(3)
u0 :=100
(4)
B o t h e n d s o f t h e r o d a r e i n a n i c e b a t h ( u ( 0 ,t ) =u ( L ,t ) =0 )
According to the analysis of Section 9.5 (separation of variables), the temperature at time t at the
position x on the bar is given by the Fourier series (here we compute only the partial sum up to N,
N=50 is good enough to get a good approximation)
M u : = ( x, t , N) -> 4* u0/Pi * sum( 1/( 2* j -1) * exp( -( 2* j -1) ^2*
Pi ^2* k* t /L^2) * si n( ( 2* j -1) * Pi * x/L) , j =1. . N) ;
22
I 2jI 12 q kt
2jI 1 qx
L
N e
sin
L
4 u0 <
2jI 1
j=1
u := x, t, N (1.1)
q
The spatial distribution of the temperature on the rod after 30s is
M pl ot ( u( x, 30, 50) , x=0. . L) ;
M
100
80
60
40
20
0
0
20
40
60
80
100
120
140
x
We can find the time at which the rod midpoint reaches 50 degrees:
M pl ot ( { u( L/2, t , L) , 50} , t =1. . 5000) ;
160
180
100
90
80
70
60
50
40
30
1 ,0 0 0
2 ,0 0 0
3 ,0 0 0
4 ,0 0 0
5 ,0 0 0
t
We could estimate the time graphically or just use the buit-in numerical methods in Maple
M T50 : = f sol ve( u( L/2, t , L) = 50 , t =0. . 5000) ;
T50:=2556.547907
(1.2)
The time in minutes and seconds is then:
M pr i nt f ( " %d: %. 2f \n" , f l oor ( T50/60) , f r ac( T50/60) * 60) ;
42: 36. 55
E n d x =L i s i n su l a t e d w h i l e x =0 i s i n i c e b a t h
Here we use the anaylsis from problem 9.5.24 to find that the temperature on the rod is given by
M c: = n-> ( 2/L) * i nt ( 100* si n( n* Pi * x/2/L) , x=0. . L) ;
L
1 nqx
2
100 sin
dx
2 L
0
c :=n(2.1)
L
M ubi s : = ( x, t , N) -> sum( c( 2* j +1) * exp( -( ( 2* j +1) * Pi /2/L) ^2* k* t ) *
si n( ( 2* j +1) * Pi * x/2/L) , j =0. . N) ;
(2.2)
2jA 1 2q2kt
I 1
4
1
L2
ubis := x, t, N - < c 2 j A 1 e
sin
2
j=0
N
2jA 1 qx
L
(2.2)
M
Here is the typical temperature distribution after 3000s
M pl ot ( ubi s( x, 3000, 50) , x=0. . L) ;
90
80
70
60
50
40
30
20
10
0
0
20
40
60
80
100
120
140
160
180
x
We can see that the maximum temperature of the rod is at the insulated end. We can find the time at
which the rod midpoint reaches 50 degrees:
M pl ot ( { ubi s( L, t , L) , 50} , t =1. . 30000) ;
100
90
80
70
60
50
40
30
20
10
0
1 0 ,0 0 0
2 0 ,0 0 0
3 0 ,0 0 0
t
We could estimate the time graphically or just use the buit-in numerical methods in Maple
M T50 : = f sol ve( ubi s( L, t , L) = 50 , t =0. . 30000) ;
T50:=10226.19162
The time in minutes and seconds is then:
M pr i nt f ( " %d: %. 2f \n" , f l oor ( T50/60) , f r ac( T50/60) * 60) ;
170: 26. 19
M
(2.3)