23-Chem-A4 Chemical Reactor Engineering · December 2013
Question 1 of 5: Two CSTRs in Series (Bimolecular Liquid Reaction)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: Five questions, each 20 marks; any four constitute a complete paper (answer all five here). Open book, 3 hours, non-programmable calculator. Examiner asks that the origin of significant formulas be cited (e.g. Fogler).
Reference texts: H. S. Fogler, Elements of Chemical Reaction Engineering (5th ed., Prentice Hall) — CSTR/PFR design equations, multiple reactors, gas-phase variable-volume kinetics, adiabatic energy balance; O. Levenspiel, Chemical Reaction Engineering (3rd ed., Wiley) — reactor sizing and combinations; supporting data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Question 1: Two CSTRs in Series (Bimolecular Liquid Reaction) (20 marks)
Given. Two feed streams (A and B) each 0.10 mol/L at 1.80 L/s combine at the inlet of reactor 1. Because A and B enter in equal molar amounts and are consumed 1:1, $C_A = C_B$ everywhere, so $(-r_A)=kC_A^2$. The liquid density is constant, so the volumetric flow is constant at $q = 3.60$ L/s throughout.
Quantity
Value
Combined feed flow, $q$
3.60 L/s
Inlet concentration, $C_{A0}=C_{B0}$
0.050 mol/L
Reactor volumes $V_1,\ V_2$
250 L, 2000 L
Rate constant, $k$
0.65 L/(mol·s)
Find. (a) the conversion of A leaving reactor 1 and reactor 2; (b) whether reversing the reactor order raises or lowers the overall conversion, with justification.
Figure 1 — Two CSTRs in series. Equal streams of A and B mix to $C_{A0}=0.05$ M; the small tank precedes the large tank in the base case.
Approach. The inlet after mixing is 0.05 M (equal streams halve each concentration). Apply the CSTR mass balance $q(C_{in}-C_{out}) = V\,k\,C_{out}^2$ to each tank in turn, then compare the two orderings.
Inlet concentration after mixing. Two equal 0.10 M streams combine, so each species is halved:$$C_{A0}=\frac{(0.10)(1.80)}{3.60}=0.050\ \text{mol/L},\qquad C_{A0}=C_{B0}\ \Rightarrow\ (-r_A)=kC_A^2.$$
CSTR 1 balance (equimolar 2nd order). With $\tau_1=V_1/q=250/3.60=69.4$ s, the steady balance $C_{A0}-C_{A1}=\tau_1 k\,C_{A1}^2$ gives a quadratic $45.14\,C_{A1}^2+C_{A1}-0.050=0$:$$C_{A1}=\frac{-1+\sqrt{1+4(45.14)(0.050)}}{2(45.14)}=0.0240\ \text{mol/L}\;\Rightarrow\;X_1=1-\frac{0.0240}{0.050}=\boxed{0.52}.$$
CSTR 2 balance. Reactor 2 receives $C_{A1}=0.0240$ M with $\tau_2=2000/3.60=555.6$ s, so $361.1\,C_{A2}^2+C_{A2}-0.0240=0$:$$C_{A2}=\frac{-1+\sqrt{1+4(361.1)(0.0240)}}{2(361.1)}=0.00689\ \text{mol/L}.$$
Overall conversion. Referenced to the reactor-1 inlet:$$X_2=1-\frac{C_{A2}}{C_{A0}}=1-\frac{0.00689}{0.050}=\boxed{0.862\ (86.2\%)}.$$That is part (a): $X_1=52\%$ out of the small tank, $X_2=86\%$ out of the train.
Part (b) — reverse the order (2000 L first, then 250 L). Repeating the two balances with the large tank first gives $C_A=0.01046$ M after tank 1 and $C_A=0.00775$ M after tank 2, i.e.$$X_{2,\text{rev}}=1-\frac{0.00775}{0.050}=0.845\ (84.5\%)\;\Rightarrow\;\boxed{\text{about 1.7 points LOWER}}.$$The reaction is second order ($n>1$), so its rate falls steeply as concentration drops. A CSTR runs entirely at its low exit concentration; putting the small tank first converts a useful amount while $C_A$ is still high (rate is high, so little volume is needed), leaving the large tank to grind down the now-dilute stream. Reversing wastes the big tank’s volume on an early, easy concentration drop and starves the small tank at low $C_A$ — hence the lower overall conversion.
