Question

The residence-time distribution (RTD) function of a reactor (in min⁻¹) is:

\[ E(t) = \begin{cases} 1 - 2t, & \text{if } t \leq 0.5\ \text{min} \\ 0, & \text{if } t > 0.5\ \text{min} \end{cases} \]

The mean residence time of the reactor is _____ min (rounded off to 2 decimal places).

Hint:

When calculating integrals for RTD functions, focus on the effective range of the RTD curve where $E(t)$ is non-zero to simplify the computation.

Answer 1

Step 1: Understand the RTD Function.
The RTD function is defined piecewise, with a linear decrease from 1 to 0 as time increases from 0 to 0.5 minutes and 0 thereafter. 
Step 2: Calculate the Mean Residence Time.
The mean residence time $\tau$ can be calculated using the integral of $t \cdot E(t)$ over the effective time range: \[ \tau = \int_0^\infty t \cdot E(t) \, dt = \int_0^{0.5} t \cdot (1 - 2t) \, dt \] \[ \tau = \int_0^{0.5} (t - 2t^2) \, dt = \left[ \frac{t^2}{2} - \frac{2t^3}{3} \right]_0^{0.5} \] \[ \tau = \left( \frac{0.5^2}{2} - \frac{2(0.5)^3}{3} \right) = \frac{0.25}{2} - \frac{0.25}{3} \] \[ \tau = \frac{0.125 - 0.0833}{1} = 0.0417 { min} \] % Conclusion The mean residence time of the reactor is $0.042$ min, which is approximately $2.5$ seconds, rounding off to two decimal places.