PFR Holding Time from Mixed-Flow Reactor Data: 23-Chem-A4 Chemical Reactor Engineering, December 2018, Q3

Question 3: PFR Holding Time from Mixed-Flow Reactor Data (25 marks)

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. Eight steady-state mixed-flow (CSTR) runs at various feeds/holding times. In an MFR the rate follows directly from the design equation $-r_A = (C_{A0}-C_A)/\tau$. Target: $C_{A0} = 0.8$ mol/L, $X = 0.75$, so $C_A$ falls from 0.8 to 0.2 mol/L in a PFR.

Find. The plug-flow holding time $\tau_{PFR}$ for 75% conversion.

Figure 3. Levenspiel plot. Points are $1/(-r_A)$ from the MFR runs; the curve is the fitted second-order rate $-r_A = kC_A^2$. The PFR holding time is the shaded area under the curve between $C_A = 0.2$ and $0.8$ mol/L.

Approach. Convert each MFR run to a rate, identify the rate law, then integrate $1/(-r_A)$ over the target concentration span using the plug-flow performance equation.

Extract rates from the MFR runs. Using $-r_A = (C_{A0}-C_A)/\tau$: e.g. the run $(0.48\rightarrow0.20,\ \tau=560)$ gives $-r_A = 0.28/560 = 5.0\times10^{-4}$ mol/L·s, while $(1.0\rightarrow0.56,\ \tau=110)$ gives $4.0\times10^{-3}$. Sorting by exit concentration builds the rate-vs-$C_A$ curve.
Identify the rate law. Testing $-r_A = k\,C_A^{2}$, the ratio $(-r_A)/C_A^{2}$ is essentially constant over the low-to-mid range, giving $$\boxed{k = 0.0130\ \text{L}\,\text{mol}^{-1}\text{s}^{-1}\quad(\text{second order}).}$$
Plug-flow performance equation. For a PFR, $\displaystyle \tau_{PFR} = C_{A0}\!\int_0^{X}\!\frac{dX}{-r_A} = \int_{C_A}^{C_{A0}}\frac{dC_A}{-r_A}.$ Graphical (trapezoidal) integration of the tabulated $1/(-r_A)$ from $C_A = 0.2$ to $0.8$ mol/L — the shaded area in Figure 3 — gives $$\boxed{\tau_{PFR} \approx 315\ \text{s}.}$$
Analytic cross-check. With the fitted second-order constant, $\tau_{PFR} = \dfrac{1}{k}\!\left(\dfrac{1}{C_A}-\dfrac{1}{C_{A0}}\right) = \dfrac{1}{0.0130}\!\left(\dfrac{1}{0.2}-\dfrac{1}{0.8}\right) = 289$ s. The graphical and model values bracket the answer at $\approx 300$ s. The gap is systematic rather than scatter: above $C_A \approx 0.6$ mol/L the measured rate levels off (the runs at $C_A = 0.65$, 0.92 and 1.00 give $(-r_A)/C_A^2 = 0.011$, 0.005 and 0.004), so the second-order curve over-predicts the rate there and under-counts the area, while the coarse trapezoids over-count it in the steep low-$C_A$ region. Using the second-order law from 0.2 to 0.56 mol/L and the measured points from 0.56 to 0.8 gives 302 s, confirming $\approx 300$ s.

Quantity	Value
Rate law	$-r_A = 0.0130\,C_A^{2}$ (2nd order)
PFR holding time (graphical)	315 s
PFR holding time (2nd-order model)	289 s
Recommended value	≈ 300 s

Concept & Theory

Topic. Using CSTR (mixed-flow) data to obtain a rate law and size a plug-flow reactor.

Key relations. MFR: $-r_A = (C_{A0}-C_A)/\tau$ (algebraic, rate read directly at exit conditions). PFR: $\tau = \int_{C_A}^{C_{A0}} dC_A/(-r_A)$ (area under a $1/(-r_A)$ curve).

Why this works. Because a CSTR operates uniformly at its exit composition, each steady-state run is a direct point measurement of the rate at that concentration — no differentiation needed. The plug-flow reactor, by contrast, sweeps continuously through concentrations, so its holding time is the accumulated $\int dC_A/(-r_A)$, i.e. the area under the Levenspiel curve.

Common pitfall. Mixing runs with different $C_{A0}$ without recognising that the rate depends only on the local $C_A$; and using the CSTR sizing formula $\tau = (C_{A0}-C_A)/(-r_A)$ for the PFR — that would overestimate the volume, since the PFR needs less time than a CSTR for a positive-order reaction.

Source. Levenspiel, 3rd ed., Ch. 5 (ideal reactors for a single reaction); Fogler, 5th ed., §5.4.

Question 3 of 5: PFR Holding Time from Mixed-Flow Reactor Data

Question 3: PFR Holding Time from Mixed-Flow Reactor Data (25 marks)