Question

The concentration \( S(x, t) \) of pollutants in a one-dimensional reservoir at position \\(( x \text{ and time } t \text{ satisfies the diffusion equation}\) 

\[\frac{\partial s(x, t)}{\partial t} = D \frac{\partial^2 s(x, t)}{\partial x^2}\]

 on the domain \( 0 \le x \le L \), where \( D \) is the diffusion coefficient of the pollutants. The initial condition \( S(x, 0) \) is defined by the step-function shown in the figure. 

The boundary conditions of the problem are given by \( \frac{\partial s(x, t)}{\partial x} = 0 \) at the boundary points  \(x = 0 \text{ and }\) \(x = L \text{ at all times}\). Consider \(D = 0.1 \ \text{m}^2/ \text{s}, S_0 = 5 \ \mu \text{mol/m}\), \(\text{and} L = 10 \ \text{m}.\) The steady-state concentration} \(\hat{s} \left( \frac{L}{2} \right) = s \left( \frac{L}{2}, \infty \right)\) at the center \(x = \frac{L}{2} \)\( \text{ of the reservoir is}\) _________ (in \(\mu \text{mol/m}) \)(in integer).

Hint:

In diffusion problems with constant boundary conditions, the solution typically involves finding a steady-state concentration by applying the boundary conditions.

Answer 1

Correct Answer: 2

In steady-state, the time derivative of the concentration becomes zero (\( \frac{\partial s}{\partial t} = 0 \)), and the equation reduces to: \[ \frac{\partial^2 s(x)}{\partial x^2} = 0 \] This equation has the general solution: \[ s(x) = A x + B \] Using the boundary conditions \( \frac{\partial s(x)}{\partial x} = 0 \) at \( x = 0 \) and \( x = L \), we find that the concentration is constant across the reservoir. The concentration at the center \( x = \frac{L}{2} \) is: \[ \hat{s} \left( \frac{L}{2} \right) = \frac{S_0}{2} \] Substituting the given value of \( S_0 = 5 \ \mu \text{mol/m} \), we get: \[ \hat{s} \left( \frac{L}{2} \right) = \frac{5}{2} = 2.5 \ \mu \text{mol/m} \] Thus, the steady-state concentration at the center is: \[ \boxed{0.02 \ \mu \text{mol/m}} \]