Question

Mean Residence Time Calculation for a given system: \[ t_m = \int_0^{\infty} t E(t) \, dt \] Where \( E(t) \) is the distribution function. For the given system, the distribution function is: \[ E(t) = 1 - 2t, \quad t \leq 0.5 \] \[ E(t) = 0, \quad t > 0.5 \]

Hint:

In residence time calculations, remember to carefully consider the boundaries of the distribution function and break the integral into appropriate intervals.

Answer 1

To calculate the mean residence time, we integrate the expression for \( t_m \) using the given distribution \( E(t) \): \[ t_m = \int_0^{0.5} t (1 - 2t) \, dt + \int_{0.5}^{\infty} t (0) \, dt \] The second integral is zero because \( E(t) = 0 \) for \( t > 0.5 \). Now, solving the first integral: \[ t_m = \int_0^{0.5} t (1 - 2t) \, dt \] \[ t_m = \int_0^{0.5} (t - 2t^2) \, dt \] \[ t_m = \left[ \frac{t^2}{2} - \frac{2t^3}{3} \right]_0^{0.5} \] Substituting the limits: \[ t_m = \left( \frac{(0.5)^2}{2} - \frac{2(0.5)^3}{3} \right) - (0) \] \[ t_m = \left( \frac{0.25}{2} - \frac{2 \times 0.125}{3} \right) \] \[ t_m = 0.125 - 0.0833 = 0.0417 \, \text{units of time} \] Thus, the mean residence time is \( t_m = 0.0417 \) units of time.