Question

A bar 100 cm long, with insulated sides, has its ends kept at 0

Answer 1

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.​

Solution

The heat flow flow equation is

$\nabla ^2u(x,t)={1\over \alpha ^2}{\partial u\over \partial t}...................(i)$

where $u(x, t)$ is the temperature. Because the sides of the bar are insulated, the heat flows only in the $x$ direction; the same happens for a slab of finite thickness but infinitely large. The initial condition is $u(x, 0) = 100$ and the boundary condition for $t > 0$ is $u(0, t) = u(100, t) = 0$ . We search for a solution of the form $u(x, t) = F(x)T(t)$ ; the differential equation $(i)$ gives

${∂ ^2F\over ∂x^2} + k ^2F = 0 , \implies F = A cos kx + B sin kx.............(ii)$

${∂T\over ∂t} = −k^2α^2T , \implies T = e^{−k^2α^2t}....................(iii)$

Because of the boundary condition in $x = 0$ , we have $A = 0$ . To satisfy $u(100, t) = 0$ , we find discrete values of $k:$

$sin 100k = 0 ,\implies k_n ={nπ\over100},$ $.........(iv)$ ,for $n = 1, 2, 3, . . .$

Finally, we must satisfy the initial condition

$u(x, 0) =\displaystyle\sum_{n=1}^∞b_n sin{nπx\over 100}= 100 ...............(v)$ ,for $0 ≤ x ≤ 10.$

Now,solving for $b_n$ we have

$b_n ={2\over 100} \int_0^{100 }sin{nπx\over 100}dx = −{200\over nπ}cos{nπx\over 100}\mid_0^{100}$

$={200\over nπ}[1 − (−1)^n]$

$= \begin{cases} {400\over \pi n}&\text{for }{odd} {n} \\ 0 &\text{for } {even} n \end{cases}$ $..............(vii)$

after using $cos (nπ )= (−1)^n$ for integer $n$ . The solution to the differential equation $(i)$ with the given boundary conditions is then

$u (x,t)={400\over π}\sum_ {oddn}{1\over n}e^{−({nπα\over 100})^2t }sin {nπx\over 100}$

where “odd n” means a sum over n = 1, 3, 5, . . ..