23-Chem-A4 Chemical Reactor Engineering · May 2014
Question 4 of 5: Second-Order Reaction in Two Unequal CSTRs — Ordering and Parallel
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — May 2014 — 04-Chem-A4 Chemical Reactor Engineering. Three-hour open-book exam; the designated Fogler textbook, unit-conversion/mathematical tables and a non-communicating programmable calculator are permitted. Five questions are printed and any four constitute a complete paper (each worth 20 marks); all five are solved below for completeness. No credit is given for re-deriving standard rate expressions, so the design equations are quoted from Fogler and applied. Property data not printed on the paper (gas constant in imperial units, Rankine conversion) are stated explicitly in each Given block as open-book look-ups.
Reference texts: H. S. Fogler, Elements of Chemical Reaction Engineering (4th/5th ed., Prentice Hall) — the designated open-book text: CSTR/PFR/batch design equations, gas-phase reactions with change in moles, reversible reactions and optimum temperature progression; O. Levenspiel, Chemical Reaction Engineering (3rd ed., Wiley) — reactor-staging (CSTRs in series/parallel) and the optimum-temperature-progression arguments; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Question 4: Second-Order Reaction in Two Unequal CSTRs — Ordering and Parallel (20 marks: a 10, b 10)
Given. Second-order $-r_A=kC_A^2$, two CSTRs with $V_1=2V_2$ (space times $\tau_1=2\tau_2$), feed $C_{A0}$ with no product. Because only ratios matter, take dimensionless $k\tau_2 C_{A0}=1$ (small tank) and $k\tau_1 C_{A0}=2$ (large tank).
Find. (a) which series ordering (small-first vs large-first) maximises conversion; (b) whether a parallel arrangement can do better.
Figure 4 — The higher-conversion series arrangement places the smaller tank first, so the downstream large tank finishes the reaction at low concentration.
Approach. Solve the single-CSTR quadratic for the exit concentration, cascade it through both orderings, then compare with the best possible parallel split (which is equivalent to one CSTR of the combined volume).
Single second-order CSTR (building block). The balance $C_{in}-C_{out}=k\tau C_{out}^2$ is a quadratic whose physical root is$$C_{out}=\frac{-1+\sqrt{1+4k\tau C_{in}}}{2k\tau}.$$This is applied tank-by-tank, the exit of one feeding the next.
Series ordering (part a). Cascading (dimensionless $C_{A0}=1$): small tank first ($k\tau=1$ then $2$) gives exit $C_A=0.360$, i.e. $X=0.640$; large tank first ($k\tau=2$ then $1$) gives $C_A=0.366$, $X=0.634$. Therefore$$\boxed{\text{smaller tank first} \Rightarrow X=0.640 > 0.634.}$$Checking a range of Damköhler magnitudes ($k\tau_2C_{A0}=0.2$ to $10$) confirms small-first wins at every scale — the effect is real but modest here.
Why small-first wins. For order $n>1$ the rate falls steeply with concentration. Putting the small reactor first lets the mixture drop only part-way while still at high concentration (where a little volume buys a lot of conversion), leaving the large reactor to grind out the low-concentration tail where volume is used less effectively.
Parallel arrangement (part b). The best possible parallel operation splits the feed between the tanks in proportion to their volumes so both run at the same space time and exit concentration — which is identical to a single CSTR of the combined volume $3V_2$ ($k\tau=3$): $C_A=0.434$, $X=0.566$. Since $0.566<0.634$, parallel is strictly worse than either series ordering:$$\boxed{\text{No merit in parallel: }X_{\parallel}=0.566<X_{\text{series}}.}$$
Conclusion. Connect the tanks in series with the smaller ($V_2$) first; do not use parallel operation, which sacrifices the concentration-staging benefit that makes series superior for $n>1$ kinetics.
