23-Chem-A4 Chemical Reactor Engineering · May 2015
Question 4 of 5: Gas-Phase PFR with Mole Change — Third-Order Kinetics
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — May 2015 — 04-Chem-A4 Chemical Reactor Engineering. Three-hour open-book exam; the designated Fogler textbook (any edition), unit-conversion / mathematical tables (CRC Handbook) 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 batch / MFR / PFR design equations are quoted and applied. Property look-ups not printed on the paper (molar masses, the gas constant) are stated explicitly in each Given block as permitted open-book references.
Reference texts: H. S. Fogler, Elements of Chemical Reaction Engineering (4th/5th ed., Prentice Hall) — the designated open-book text: batch/CSTR/PFR design equations, the stoichiometric table with expansion factor $\varepsilon$ for gas-phase reactions with a change in moles, and reversible-reaction kinetics; O. Levenspiel, Chemical Reaction Engineering (3rd ed., Wiley) — non-ideal flow (dead-zone / bypass models), the dispersion and tanks-in-series RTD models, and rate-equation determination from a differential (mixed) catalytic reactor; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Given. Gas-phase $2A+B\rightarrow2C$, third-order rate $-r_A=kC_AC_B^2$ (first order in A, second order in B, as printed) with $k=10^5$ L²/(mol²·s) at $T=500\,{}^{\circ}\text{C}=773.15$ K and $P=1$ atm. Feed 6 L/s of 25% A / 25% B / 50% inert; isothermal, isobaric PFR to $X_A=0.90$.
Find. The PFR volume $V$ (L) that converts 90% of the feed A.
Figure 4 — Isothermal, isobaric plug-flow reactor: the equimolar A/B feed (25/25/50) contracts as $2A+B\rightarrow2C$ reduces the mole count, so the volumetric flow falls along the reactor (expansion factor $\varepsilon=-0.125$).
Approach. Build the stoichiometric table on a per-mole-A basis to get the expansion factor, write $C_A$ and $C_B$ as functions of $X$ for the isobaric gas, then integrate the PFR design equation numerically.
Feed concentrations. At 773.15 K, 1 atm, $C_{T0}=P/RT=1/(0.082057\times773.15)=0.01576$ mol/L. With $y_{A0}=y_{B0}=0.25$: $C_{A0}=0.003941$ mol/L and $F_{A0}=C_{A0}v_0=0.003941\times6=0.02364$ mol/s; $\theta_B=1$.
Expansion factor. On a per-mole-A basis divide the stoichiometry by 2: $A+\tfrac12B\rightarrow C$, so $\delta=1-\tfrac12-1=-\tfrac12$ mol change per mol A. Thus $\varepsilon=y_{A0}\delta=0.25(-0.5)=-0.125$ — the gas contracts as it reacts.
Concentrations vs. conversion (isobaric gas). $C_A=C_{A0}\dfrac{1-X}{1+\varepsilon X}$ and $C_B=C_{A0}\dfrac{\theta_B-\tfrac12X}{1+\varepsilon X}=C_{A0}\dfrac{1-\tfrac12X}{1+\varepsilon X}.$ Substituting into $-r_A=kC_AC_B^2$ collects the composition dependence into one dimensionless integral.
PFR design equation. $V=F_{A0}\displaystyle\int_0^{0.9}\frac{dX}{-r_A}=\frac{F_{A0}}{kC_{A0}^3}\int_0^{0.9}\frac{(1+\varepsilon X)^3}{(1-X)\,(1-\tfrac12X)^2}\,dX.$ The prefactor is $F_{A0}/(kC_{A0}^3)=0.02364/[10^5(0.003941)^3]=3.86$ L, and Simpson quadrature gives the integral $=3.974.$
Reactor volume. Combining,$$V=3.86\times3.974=\boxed{15.4\ \text{L}}.$$The $(1-X)^{-1}$ factor makes the integrand climb steeply as $X\to1$: the last ten points of conversion (0.80 to 0.90) account for 38% of the integral. The contraction also helps — with $\varepsilon=0$ the same integral would give 20.0 L, so the falling volumetric flow trims the reactor by about a quarter.
