Non-Ideal Tank — Equivalent Ideal Mixed-Flow Reactor: 23-Chem-A4 Chemical Reactor Engineering, May 2015, Q2

Question 2: Non-Ideal Tank — Equivalent Ideal Mixed-Flow Reactor (20 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. Second-order reaction with $C_{A0}=C_{B0}$, so $-r_A=kC_A^2$. From the tracer flow model: total 6 m³, but only $V_a=2$ m³ is actively mixed, a 4 m³ dead zone is inert, and 20% of the feed ($0.2v$) bypasses the active tank while $0.8v$ passes through it. The measured overall conversion of the real unit is $X_\text{ov}=0.60$.

Quantity	Value
Total / active / dead volume	6 / 2 / 4 m³
Bypass fraction / through-flow fraction	0.20 / 0.80 of $v$
Measured overall conversion	0.60
Rate law ($C_{A0}=C_{B0}$)	$-r_A=kC_A^2$

Find. The volume $V_\text{ideal}$ of a single ideal mixed-flow (CSTR) reactor that, fed the whole stream $v$, would give the same 60% conversion.

Figure 2 — Tracer-derived flow model: only the 2 m³ stirred pocket reacts, fed by $0.8v$; the 4 m³ dead zone is stagnant and $0.2v$ bypasses unconverted, then remixes with the active outlet.

Approach. Back out the conversion actually achieved in the active pocket from the bypass mixing balance, use the second-order MFR design equation to fix the group $kC_{A0}/v$, then apply that same group to an ideal MFR fed the whole stream at the target conversion.

Bypass mixing balance. The bypass stream ($0.2v$) is unconverted, so blending it with the active-tank outlet ($0.8v$ at conversion $X_1$) gives the measured overall conversion: $X_\text{ov}=0.8X_1+0.2(0)$. Hence $X_1=X_\text{ov}/0.8=0.60/0.8=0.75.$
Second-order MFR design equation for the active pocket. With $-r_A=kC_A^2=kC_{A0}^2(1-X)^2$, the CSTR balance $kC_{A0}\tau_a=X_1/(1-X_1)^2$ gives $kC_{A0}\tau_a=0.75/(0.25)^2=12.$ Here $\tau_a=V_a/(0.8v)$ uses only the through-flow.
Extract the kinetic–throughput group. From $kC_{A0}\dfrac{V_a}{0.8v}=12$ with $V_a=2$ m³, $\dfrac{kC_{A0}}{v}=\dfrac{12\times0.8}{2}=4.8\ \text{m}^{-3}.$ This constant carries the (unknown) $k$ and $C_{A0}$ together and is all we need.
Ideal MFR at the same conversion. A single ideal CSTR fed the whole stream $v$ needs $kC_{A0}\tau_\text{id}=X_\text{ov}/(1-X_\text{ov})^2=0.60/(0.40)^2=3.75,$ with $\tau_\text{id}=V_\text{ideal}/v.$ Therefore$$V_\text{ideal}=\frac{3.75}{kC_{A0}/v}=\frac{3.75}{4.8}=\boxed{0.78\ \text{m}^3}.$$
Interpret. A properly mixed 0.78 m³ tank matches the real 6 m³ unit — the dead zone and bypass waste roughly seven-eighths of the installed volume, a dramatic illustration of how non-ideal flow destroys reactor performance.

Quantity	Result
Conversion in the active pocket $X_1$	0.75
Kinetic group $kC_{A0}/v$	4.8 m⁻³
Equivalent ideal MFR volume	0.78 m³

Concept & Theory

Topic. Non-ideal flow (dead volume + bypassing) and the equivalent ideal mixed-flow reactor for a second-order reaction.

Key relations. Second-order MFR design $kC_{A0}\tau=X/(1-X)^2$; bypass mixing $X_\text{ov}=fX_\text{active}$ with $f$ = active flow fraction; space time $\tau=V/v$.

Why this works. Only the actively mixed pocket reacts, and it sees only the through-flow $0.8v$, so its space time is $V_a/(0.8v)$. The unconverted bypass dilutes the exit, so the pocket must reach a higher conversion (0.75) than the measured 0.60. Once $kC_{A0}/v$ is known, an ideal tank of the same throughput is sized directly from the design equation.

Common pitfall. Using the full 6 m³ (or the total feed $v$) in the active-pocket balance, or forgetting that the bypass lowers the blended conversion — both give the wrong $kC_{A0}/v$.

Source. Levenspiel, Chemical Reaction Engineering §12–13 (compartment models: dead water and bypassing; conversion in non-ideal flow).

Question 2 of 5: Non-Ideal Tank — Equivalent Ideal Mixed-Flow Reactor

Question 2: Non-Ideal Tank — Equivalent Ideal Mixed-Flow Reactor (20 marks)