Question

Two reactors with average residence times \( t_1 \) and \( t_2 \) are placed in series. Reactor 1 has zero dispersion and reactor 2 has infinite dispersion. The E curve of the system is given by

• A. 0 for \( t \leq t_1 \), \( \frac{1}{t_2} \exp\left(\frac{t - t_1}{t_2}\right) \) for \( t>t_1 \)
• B. 0 for \( t \leq t_2 \), \( \frac{1}{t_1} \exp\left(\frac{t - t_2}{t_1}\right) \) for \( t>t_2 \)
• C. 0 for \( t \leq t_2 \), \( \frac{1}{t_1} \exp\left(\frac{t_1 - t_2}{t_2}\right) \) for \( t>t_2 \)
• D. 0 for \( t \leq t_1 \), \( \frac{1}{t_2} \exp\left(\frac{t_1 - t_2}{t_1}\right) \) for \( t>t_1 \)

Hint:

When analyzing reactor systems in series with different dispersion properties, the E curve is determined by the residence times and dispersion in each reactor.

Answer 1

The Correct Option is A

Step 1: Understanding the system.
The system consists of two reactors in series: Reactor 1 with zero dispersion and Reactor 2 with infinite dispersion. The E curve (exit age distribution curve) describes how the reactant concentration changes with time, which is influenced by the residence time and dispersion in each reactor.
Step 2: Analyzing the options.
(1) 0 for \( t \leq t_1 \), \( \frac{1}{t_2} \exp\left(\frac{t - t_1}{t_2}\right) \) for \( t>t_1 \): This is the correct answer. For Reactor 1 with zero dispersion, the reactant immediately exits at time \( t_1 \). For Reactor 2 with infinite dispersion, the E curve follows an exponential decay starting from \( t_1 \).
(2) 0 for \( t \leq t_2 \), \( \frac{1}{t_1} \exp\left(\frac{t - t_2}{t_1}\right) \) for \( t>t_2 \): This is incorrect because the dispersion and residence times are reversed.
(3) 0 for \( t \leq t_2 \), \( \frac{1}{t_1} \exp\left(\frac{t_1 - t_2}{t_2}\right) \) for \( t>t_2 \): This is incorrect as it incorrectly combines the times and dispersion effects.
(4) 0 for \( t \leq t_1 \), \( \frac{1}{t_2} \exp\left(\frac{t_1 - t_2}{t_1}\right) \) for \( t>t_1 \): This is incorrect as the exponential function does not properly represent the dispersion behavior.
Step 3: Conclusion.
The correct answer is option (1), which correctly describes the E curve behavior for the system of two reactors in series with the given dispersion characteristics.

Answer 2

The exit-age distribution (E curve) of reactors in series is found by combining the individual behaviours of each reactor. Reactor 1 has zero dispersion, meaning every element of fluid spends exactly the same time \( t_1 \) inside it (pure plug flow, a fixed time delay); Reactor 2 has infinite dispersion, meaning it behaves like an ideal well-mixed (CSTR) vessel with a decaying exponential exit-age distribution based on its own mean residence time \( t_2 \). Combining a pure time delay of \( t_1 \) with a CSTR's exponential response of mean \( t_2 \) simply shifts that exponential response forward by \( t_1 \). Checking each option against this combination:

1. 0 for \( t \le t_1 \), \( \frac{1}{t_2}\exp\left(\frac{t-t_1}{t_2}\right) \) for \( t>t_1 \): This correctly shows zero response before the plug-flow delay \( t_1 \) has elapsed, after which the response follows the CSTR-type exponential decay based on \( t_2 \), shifted to start at \( t_1 \) -- exactly the combination of a time delay with a CSTR response.
2. 0 for \( t \le t_2 \), \( \frac{1}{t_1}\exp\left(\frac{t-t_2}{t_1}\right) \) for \( t>t_2 \): This swaps the roles of the two reactors, using \( t_2 \) as the delay and \( t_1 \) as the exponential's own time constant, reversing which reactor contributes the delay and which contributes the mixing behaviour.
3. 0 for \( t \le t_2 \), \( \frac{1}{t_1}\exp\left(\frac{t_1-t_2}{t_2}\right) \) for \( t>t_2 \): Here the exponent no longer depends on the running time \( t \) at all, making the expression a constant rather than a decaying function of time, which cannot represent an exit-age distribution.
4. 0 for \( t \le t_1 \), \( \frac{1}{t_2}\exp\left(\frac{t_1-t_2}{t_1}\right) \) for \( t>t_1 \): The exponent here uses only the fixed values \( t_1 \) and \( t_2 \) with no dependence on the running variable \( t \), so this expression cannot decay over time as an exit-age distribution must.

Only the first option keeps the correct reactor assigned to the plug-flow delay, the correct reactor assigned to the exponential decay, and retains the running time variable \( t \) inside the exponential as required for a genuine time-dependent distribution.

Therefore, the correct answer is 0 for \( t \le t_1 \), \( \dfrac{1}{t_2}\exp\left(\dfrac{t-t_1}{t_2}\right) \) for \( t>t_1 \).