ECH5842 CHEMICAL ENGINEERING ANALYSIS R.Chella

HOMEWORK # 4 Fall'97

1(a) . Show that a ``separation of variables'' approach to the unsteady-state, one-dimensional heat conduction equation:

$$\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} \qquad \ 0 \leq x \leq l$$

with initial conditions:

u(t=0,x) = u₀(x)

and boundary conditions:

$$u(t,0) = 0, \qquad \cos (\beta)u(t,l) + \ \sin(\beta)\frac{\partial u(t,l)}{\partial x} = 0$$

leads to the eigenvalue problem[$u(x,t) = \phi(x)\tau(t)$]:

$$\phi'' + \lambda\phi = 0 \qquad \ \hbox{[$'$ denotes differentiation with respect to x]}$$

$$\phi(0) = 0, \qquad \cos(\beta)\phi(l) + \sin(\beta)\phi'(l) = 0$$

(b) Discuss the various physical situations corresponding to values of $\beta$ between 0 and $\pi/2$ (e.g. $\beta=0$ corresponds to the relative temperature at the end of the rod at x=l being fixed at zero).

(c) Solve for the eigenvalues and eigenvectors of $\phi$ for $\lambda \geq 0$.

2. Solve by separation of variables:

$$u_{t} - u_{xx} = x \sin t, \qquad 0 \leq x \leq 1 , t \geq 0$$

u(x,0) = x(1-x)

$$u(0,t) = u_{x}(1,t) = 0 \qquad t \gt 0$$

to show that:

$$u(x,t) = \sum_{n=0}^{\infty}A_{n}(t) \sin \lambda_n \, x$$

$$A_{n}(t) = A_{n}(0)e^{-\lambda_{n}^{2}t} + \frac{b_{n}}{\lambda_n^4 + 1}$$

$$A_n(0) = \frac{32 - 4\pi^2(2n+1)^2}{\pi^3(2n+1)^3}$$

3. Show using the separation of variables technique, that the solution to:
$$\begin{align*} u_t &= \kappa u_{xx}\qquad 0 < x < l \quad t\gt 0 \ \ u(x,0) &=... ... < x < l \ \ u(0,t) = f_0(t) &\quad u(l,t) = f_1(t) \qquad t \gt 0\end{align*}$$
with f₀(0) = f₁(0) = 0, is:
$$\begin{align*} u(x,t) &=\left( 1 - \frac{x}{l} \right) f_0(t) + \frac{x}{l} f_1(... ... x/l) \ c_n(t) &= \frac{2}{n \pi} [(\cos n \pi) f_1'(t) - f_0'(t) ]\end{align*}$$

4. Solve problem 3 using Laplace transforms instead, to show:

$$u(x,t) = \int_0^t M_x(x,t-\tau) f_0(\tau) \, d\tau + \int_0^t M_x(l-x,t-\tau) f_1(\tau) \, d\tau$$

$$M(x,t) = - \sqrt{\frac{\kappa}{\pi t}} \sum_{n= -\infty}^{n= \infty} \ e^{-(2nl -x)^2/4\kappa t}$$

Comparing the solutions obtained in problems 3 and 4, which one would converge more rapidly for small values of t, and which one for large values of t.

Note:

$${\cal{L}}^{-1} \left\{ \frac{\sinh a \sqrt{s}}{\sinh b \sqrt... ...fty} \frac{(2n+1)b +a}{\sqrt{4\pi t^3}} \ e^{-[(2n+1)b+a]^2/4t}$$