2.
(a) A bar 100 cm long, with insulated sides, has its ends kept at 0° Celsius and 100° Celsius until
steady state condition prevails. The two ends are then suddenly insulated and kept so. Find
the temperature distribution.
Answers
Answer:
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Answer:
u (x,t)={400\over π}\sum_ {oddn}{1\over n}e^{−({nπα\over 100})^2t }sin {nπx\over 100}u(x,t)=
π
400
∑
oddn
n
1
e
−(
100
nπα
)
2
t
sin
100
nπx
Step-by-step explanation:
The heat flow flow equation is
\nabla ^2u(x,t)={1\over \alpha ^2}{\partial u\over \partial t}...................(i)∇
2
u(x,t)=
α
2
1
∂t
∂u
...................(i)
where u(x, t)u(x,t) is the temperature. Because the sides of the bar are insulated, the heat flows only in the xx direction; the same happens for a slab of finite thickness but infinitely large. The initial condition is u(x, 0) = 100u(x,0)=100 and the boundary condition for t > 0t>0 is u(0, t) = u(100, t) = 0u(0,t)=u(100,t)=0 . We search for a solution of the form u(x, t) = F(x)T(t)u(x,t)=F(x)T(t) ; the differential equation (i)(i) gives
{∂ ^2F\over ∂x^2} + k ^2F = 0 , \implies F = A cos kx + B sin kx.............(ii)
∂x
2
∂
2
F
+k
2
F=0,⟹F=Acoskx+Bsinkx.............(ii)
{∂T\over ∂t} = −k^2α^2T , \implies T = e^{−k^2α^2t}....................(iii)
∂t
∂T
=−k
2
α
2
T,⟹T=e
−k
2
α
2
t
....................(iii)
Because of the boundary condition in x = 0x=0 , we have A = 0A=0 . To satisfy u(100, t) = 0u(100,t)=0 , we find discrete values of k:k:
sin 100k = 0 ,\implies k_n ={nπ\over100},sin100k=0,⟹k
n
=
100
nπ
, .........(iv).........(iv) ,for n = 1, 2, 3, . . .n=1,2,3,...
Finally, we must satisfy the initial condition
u(x, 0) =\displaystyle\sum_{n=1}^∞b_n sin{nπx\over 100}= 100 ...............(v)u(x,0)=
n=1
∑
∞
b
n
sin
100
nπx
=100...............(v) ,for 0 ≤ x ≤ 10.0≤x≤10.
Now,solving for b_nb
n
we have
b_n ={2\over 100} \int_0^{100 }sin{nπx\over 100}dx = −{200\over nπ}cos{nπx\over 100}\mid_0^{100}b
n
=
100
2
∫
0
100
sin
100
nπx
dx=−
nπ
200
cos
100
nπx
∣
0
100
={200\over nπ}[1 − (−1)^n]=
nπ
200
[1−(−1)
n
]
= \begin{cases} {400\over \pi n}&\text{for }{odd} {n} \\ 0 &\text{for } {even} n \end{cases}={
πn
400
0
for oddn
for evenn
..............(vii)..............(vii)
after using cos (nπ )= (−1)^ncos(nπ)=(−1)
n
for integer nn . The solution to the differential equation (i)(i) with the given boundary conditions is then
u (x,t)={400\over π}\sum_ {oddn}{1\over n}e^{−({nπα\over 100})^2t }sin {nπx\over 100}u(x,t)=
π
400
∑
oddn
n
1
e
−(
100
nπα
)
2
t
sin
100
nπx
where “odd n” means a sum over n = 1, 3, 5, . . ..