23-Chem-A4 Chemical Reactor Engineering · May 2014
Question 1 of 5: Ethyl-Acetate Saponification in a CSTR (and Two in Series)
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 1: Ethyl-Acetate Saponification in a CSTR (and Two in Series) (20 marks: a 10, b 10)
Given. Two feeds blend before the reactor: ethyl acetate (A) 0.1 M at 70 L/min and NaOH (B) 0.2 M at 50 L/min, so the combined volumetric feed is $q=120$ L/min. Second-order irreversible saponification $-r_A=kC_AC_B$. Rate constants $k(0^\circ\text{C})=23.5$ and $k(20^\circ\text{C})=92.4$ L/(gmol·min). (a) $V=200$ L, $T=30^\circ$C. (b) two equal CSTRs matching the same overall conversion.
Quantity
Value
Ethyl acetate (A) after mixing $C_{A0}=0.1(70)/120$
0.0583 gmol/L
NaOH (B) after mixing $C_{B0}=0.2(50)/120$
0.0833 gmol/L
Combined feed $q$
120 L/min
$k$ at 0 °C / 20 °C
23.5 / 92.4 L/gmol·min
Reactor volume / temperature (a)
200 L / 30 °C
Find. (a) The outlet composition (conversion and all four concentrations) of the single 200-L CSTR; (b) the volume of each of two equal CSTRs in series that reproduces the same overall conversion.
Figure 1 — The two reactant streams blend to a single 120 L/min feed; part (b) replaces the one 200 L tank with two equal CSTRs in series delivering the same overall conversion.
Approach. Blend the feeds (A is limiting), fix $k$ at 30 °C from the two-temperature Arrhenius ratio, solve the single-CSTR design equation for conversion, then solve the two coupled series balances for the equal tank volume.
Blend the feeds and identify the limiting reactant. Mixing dilutes each species to $C_{A0}=0.1\times70/120=0.0583$ M and $C_{B0}=0.2\times50/120=0.0833$ M. Ethyl acetate (A) is present in smaller amount, so A is limiting and conversion $X$ is referenced to A: $C_A=C_{A0}(1-X)$, $C_B=C_{B0}-C_{A0}X$.
Rate constant at 30 °C from the two data points. The Arrhenius ratio removes the pre-exponential: $E_a/R=\dfrac{\ln(k_2/k_1)}{1/T_1-1/T_2}=\dfrac{\ln(92.4/23.5)}{1/273.15-1/293.15}=5482$ K ($E_a\approx10.9$ kcal/mol). Extrapolating to 303.15 K:$$k_{30}=k_2\exp\!\Big[-\tfrac{E_a}{R}\big(\tfrac{1}{303.15}-\tfrac{1}{293.15}\big)\Big]=\boxed{171\ \text{L/(gmol}\cdot\text{min)}}.$$
Single-CSTR design equation (part a). The mole balance $C_{A0}X=\tau(-r_A)=\tau kC_AC_B$ with $\tau=V/q=200/120=1.667$ min ($k\tau=285$) becomes, dividing by $C_{A0}$,$$X=k\tau\,(1-X)\,(C_{B0}-C_{A0}X).$$Solving this quadratic gives $\boxed{X=0.898}$.
Outlet composition (part a). With $X=0.898$: unreacted ethyl acetate $C_A=0.0583(1-0.898)=0.0059$ M; residual NaOH $C_B=0.0833-0.0583(0.898)=0.0309$ M; and each product (sodium acetate, ethanol) $=C_{A0}X=0.0524$ M.
Two equal CSTRs in series (part b). Each tank of space time $\tau_t=V_t/q$ obeys its own balance. Tank 1: $X_1=k\tau_t(1-X_1)(C_{B0}-C_{A0}X_1)$; tank 2: $X_2-X_1=k\tau_t(1-X_2)(C_{B0}-C_{A0}X_2)$. Requiring the overall $X_2=0.898$ and solving the two equations simultaneously for $\tau_t$ gives $\tau_t=0.346$ min, i.e.$$V_t=\tau_t\,q=\boxed{41.5\ \text{L each}}\quad(\text{total }82.9\ \text{L, with }X_1=0.712).$$The two staged tanks reach the same conversion in only ~41% of the single-tank volume — staging is far more volume-efficient because each tank runs at a higher average concentration.
