Residence-Time Distribution from Pulse-Tracer Data: 23-Chem-A4 Chemical Reactor Engineering, December 2018, Q5

Question 5: Residence-Time Distribution from Pulse-Tracer Data (25 marks: a 10, b 15)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Impulse (pulse) tracer response $C(t)$ at the reactor exit. The RTD is $E(t) = C(t)/\int_0^\infty C\,dt$; the mean residence time and variance follow from the first and second moments of $E(t)$.

Find. (a) mean residence time $\bar t$; (b) the RTD curve compared with an ideal CSTR.

Figure 5. Measured RTD $E(t)$ (red) versus the ideal mixed-flow response $E(t)=\frac{1}{\bar t}e^{-t/\bar t}$ (blue dashed). The two nearly coincide, and the tanks-in-series index $N = \bar t^2/\sigma^2 \approx 1.2$ confirms the vessel behaves like a single well-mixed tank.

Approach. Compute the tracer-curve area (normalisation), then the first moment for $\bar t$ and the second central moment for the variance; convert the variance to a tanks-in-series index and overlay the ideal exponential RTD.

(a) Normalising area. By the trapezoidal rule over the eight points, $\int_0^{30} C\,dt = 1.034\ \text{g}\cdot\text{min/cm}^3$. This is the pulse area used to normalise $E(t)$.
(a) Mean residence time (first moment). $$\bar t = \frac{\int_0^\infty t\,C\,dt}{\int_0^\infty C\,dt} = \frac{4.85}{1.034} = \boxed{4.69\ \text{min}.}$$
(b) Variance (second central moment). $\sigma^2 = \dfrac{\int t^2 C\,dt}{\int C\,dt} - \bar t^2 = 40.75 - 22.02 = 18.7\ \text{min}^2$, so the spread is broad relative to $\bar t$.
(b) Tanks-in-series comparison. The dimensionless variance gives $$N = \frac{\bar t^{\,2}}{\sigma^2} = \frac{4.69^{2}}{18.7} = 1.18 \approx 1\ \text{tank}.$$ An $N$ of about one is the signature of a single ideal CSTR.
(b) Overlay the ideal MFR curve. For an ideal mixed-flow reactor $E(t) = \frac{1}{\bar t}e^{-t/\bar t}$, which starts at $E(0) = 1/\bar t = 0.213$ min$^{-1}$. The measured curve starts at $E(0) = C(0)/\text{area} = 0.242$ min$^{-1}$ — within 14% — and decays with a similar tail (Figure 5). The slight early excess and finite rise indicate a small amount of short-circuiting/stagnancy, but overall the reactor is very close to perfectly mixed.

Quantity	Value
Tracer-curve area	1.034 g·min/cm³
Mean residence time $\bar t$	4.69 min
Variance $\sigma^2$	18.7 min²
Tanks-in-series index $N$	1.18 (≈ ideal CSTR)
$E(0)$ measured vs. ideal MFR	0.242 vs. 0.213 min$^{-1}$

Concept & Theory

Topic. Residence-time distribution analysis and non-ideal-flow diagnosis by the tanks-in-series model.

Key relations. $E(t)=C(t)/\int_0^\infty C\,dt$; $\bar t=\int_0^\infty tE\,dt$; $\sigma^2=\int_0^\infty t^2E\,dt-\bar t^2$; tanks-in-series $N=\bar t^2/\sigma^2$; ideal CSTR $E(t)=\frac{1}{\bar t}e^{-t/\bar t}$.

Why this works. A perfect impulse marks a packet of fluid; the exit tracer curve is therefore the fraction of fluid leaving at each age — exactly $E(t)$. The moments compress that curve into a mean age and a spread. A single perfectly mixed tank has the maximum possible spread for a single vessel ($\sigma^2=\bar t^2$, $N=1$), so an experimental $N\approx1$ certifies near-ideal mixing.

Common pitfall. Failing to normalise by the pulse area before taking moments; treating the non-uniform time grid with a fixed $\Delta t$ instead of trapezoidal weights; and confusing $E(t)$ (exit-age) with the internal-age distribution.

Source. Levenspiel, 3rd ed., Ch. 11–14 (non-ideal flow); Fogler, 5th ed., Ch. 16–17.

Question 5 of 5: Residence-Time Distribution from Pulse-Tracer Data

Question 5: Residence-Time Distribution from Pulse-Tracer Data (25 marks: a 10, b 15)