Pulse-Tracer RTD — Tracer Amount, Reactor Volume, and Variance: 23-Chem-A4 Chemical Reactor Engineering, December 2016, Q5

Question 5: Pulse-Tracer RTD — Tracer Amount, Reactor Volume, and Variance (25 marks: a 9, b 5, c 11)

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. A pulse-tracer experiment at $v=0.1$ L/min with the $C(t)$ table above, sampled every $\Delta t = 1$ min with zero concentration at both ends.

Find. (a) injected tracer $M$; (b) reactor volume $V$; (c) variance $\sigma^2$ of the residence-time distribution.

Pulse-response curve $C(t)$. Area under the curve gives the injected tracer; its centroid is the mean residence time $\bar t = 8.94$ min (dashed line); its spread about that centroid is the variance.

Approach. Evaluate the three moments of the pulse response by the trapezoidal rule (which reduces to simple sums here because the endpoints are zero and $\Delta t=1$), then convert them to tracer mass, volume, and variance.

Zeroth moment → injected tracer (a). The area under the curve is $\int_0^\infty C\,dt = \Delta t\sum C_i = 152$ mmol·min/L. A tracer mass balance gives $M = v\!\int C\,dt = 0.1(152) = \boxed{15.2\ \text{mmol}}.$
First moment → mean residence time. $\bar t = \dfrac{\int t\,C\,dt}{\int C\,dt} = \dfrac{\sum t_i C_i}{\sum C_i} = \dfrac{1359}{152} = 8.94$ min.
Reactor volume (b). For an incompressible flow the mean residence time is the space time, so $$V = v\,\bar t = 0.1(8.94) = \boxed{0.894\ \text{L}}.$$
Second moment → variance (c). $\sigma^2 = \dfrac{\int t^2 C\,dt}{\int C\,dt} - \bar t^{\,2} = \dfrac{13313}{152} - (8.94)^2 = 87.6 - 79.9 = \boxed{7.65\ \text{min}^2}.$ The spread ($\sigma=2.77$ min) against a mean of 8.94 min ($\sigma^2/\bar t^{\,2}=0.096$) indicates a fairly narrow distribution — behaviour between plug flow and a single stirred tank, consistent with a few tanks-in-series.

Quantity	Result
(a) Tracer injected	15.2 mmol
Mean residence time $\bar t$	8.94 min
(b) Reactor volume	0.894 L
(c) Variance $\sigma^2$	7.65 min$^2$ ($\sigma = 2.77$ min)

Concept & Theory

Topic. Residence-time distribution from a pulse tracer: moments of the exit-age curve $C(t)$.

Key relations. $E(t)=C(t)/\int C\,dt$; $M=v\int C\,dt$; $\bar t=\int tE\,dt=\int tC\,dt/\int C\,dt$; $V=v\bar t$; $\sigma^2=\int t^2 E\,dt-\bar t^{\,2}$ (Fogler, Ch. 16; Levenspiel, Ch. 11).

Why this works. A pulse deposits a known area of tracer whose subsequent washout is the exit-age distribution. Its zeroth moment recovers how much was injected, its first moment (centroid) is the mean residence time — equal to $V/v$ for a closed vessel — and its second central moment measures how broadly molecules spread in age, a direct diagnostic of mixing/dispersion.

Common pitfall. Forgetting the flow rate when converting the concentration–time area to tracer mass (the area is in mmol·min/L, not mmol), and computing variance as $\int t^2 C\,dt/\int C\,dt$ without subtracting $\bar t^{\,2}$. Trapezoidal integration is exact enough here because the endpoints are zero.

Source. Fogler, Elements of Chemical Reaction Engineering, 5th ed., Ch. 16–17; Levenspiel, Chemical Reaction Engineering, 3rd ed., Ch. 11–13.

Question 5 of 5: Pulse-Tracer RTD — Tracer Amount, Reactor Volume, and Variance

Question 5: Pulse-Tracer RTD — Tracer Amount, Reactor Volume, and Variance (25 marks: a 9, b 5, c 11)