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23-Chem-A4 Chemical Reactor Engineering · May 2018

Question 1 of 5: Ethane Cracking — PFR Volume for 60% Conversion

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National Exams / EGBC — May 2018 — 16-Chem-A4 Chemical Reactor Engineering. Three-hour open-book exam; one textbook of the candidate’s choice (Fogler or Levenspiel), 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 25 marks); all five are solved below for completeness. No credit is given for re-deriving standard rate expressions, so the batch / CSTR / PFR / packed-bed design equations are quoted (with their source) and applied. Property look-ups not printed on the paper (the gas constant, molar volumes, unit conversions) 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) — batch/CSTR/PFR design equations, the stoichiometric table with expansion factor $\varepsilon$ for gas-phase reactions with a change in moles, packed-bed (catalyst-weight) mole balances, and the adiabatic CSTR energy balance and multiplicity of steady states; O. Levenspiel, Chemical Reaction Engineering (3rd ed., Wiley) — reactors in series for $>$1-order kinetics and batch turnaround; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).

Question 1: Ethane Cracking — PFR Volume for 60% Conversion (25 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.

Check — kinetic model used
The equation is written reversibly, but the data supply only a single forward first-order constant ($k=12.8\ \text{s}^{-1}$) and no equilibrium constant $K_e$. We therefore size the reactor for the stated 60% conversion as an irreversible homogeneous first-order reaction $-r_A=kC_A$, as the given rate data dictate (exam instruction #1).

Given. Isothermal gas-phase PFR; ethane fed with steam diluent; first-order kinetics; the reaction increases the mole count ($1\rightarrow2$).

QuantityValue
Ethane feed, $\dot m_{A0}$$2\times10^4$ kg/hr ($M=30$)
Steam dilution0.5 mol steam / mol ethane
Temperature, $T$900 °C $=1173.15$ K
Pressure, $P$$1.4\times10^5$ Pa
Rate constant, $k$$12.8\ \text{s}^{-1}$
Target conversion, $X$0.60
Gas constant, $R$$8.314\ \text{J}\,\text{mol}^{-1}\text{K}^{-1}$

Find. The plug-flow reactor volume $V$ needed for 60% conversion of ethane.

MixerIsothermal PFR900 C, 1.4 barEthane2x10^4 kg/hSteam0.5 mol/molfeedC2H4 + H2 + C2H6X = 0.60
Figure 1 — Ethane and steam are mixed and fed to an isothermal, isobaric plug-flow reactor at 900 °C. Because cracking splits one mole into two, the gas expands along the tube, which the expansion factor $\varepsilon$ accounts for.

Approach. Build the stoichiometric expansion factor $\varepsilon=y_{A0}\delta$, write the first-order gas-phase PFR design equation with volume change, and integrate to 60% conversion.

  1. Molar feed rates and inlet mole fraction. $\dot F_{A0}=\dfrac{2\times10^4\ \text{kg/hr}}{30\ \text{kg/kmol}}=666.7\ \text{kmol/hr}=185.2\ \text{mol/s}.$ Steam $=0.5\times666.7=333.3$ kmol/hr, so the total feed is $1000$ kmol/hr and $y_{A0}=\dfrac{666.7}{1000}=0.667.$
  2. Inlet concentration. With the ideal-gas law $C_{tot}=\dfrac{P}{RT}=\dfrac{1.4\times10^5}{(8.314)(1173.15)}=14.35\ \text{mol/m}^3,$ so $C_{A0}=y_{A0}C_{tot}=9.57\ \text{mol/m}^3.$ The inlet volumetric flow is $v_0=\dot F_{A0}/C_{A0}=185.2/9.57=19.35\ \text{m}^3/\text{s}.$
  3. Expansion factor. For $\text{C}_2\text{H}_6\rightarrow\text{C}_2\text{H}_4+\text{H}_2$, $\delta=\tfrac{1+1-1}{1}=+1,$ hence $\varepsilon=y_{A0}\delta=0.667.$ The concentration profile is $C_A=C_{A0}\dfrac{1-X}{1+\varepsilon X}.$
  4. Integrate the PFR design equation. For $-r_A=kC_A$, $V=\dfrac{\dot F_{A0}}{kC_{A0}}\displaystyle\int_0^{X}\frac{1+\varepsilon X}{1-X}\,dX=\dfrac{v_0}{k}\Big[(1+\varepsilon)\ln\tfrac{1}{1-X}-\varepsilon X\Big].$ The bracket $=(1.667)\ln(2.5)-(0.667)(0.6)=1.527-0.400=1.127,$ so $$V=\frac{19.35}{12.8}(1.127)=\boxed{1.70\ \text{m}^3\ (\approx1700\ \text{L}).}$$
QuantityResult
Inlet concentration $C_{A0}$9.57 mol/m$^3$
Expansion factor $\varepsilon$0.667
Inlet volumetric flow $v_0$19.35 m$^3$/s
Reactor volume $V$1.70 m$^3$ (1704 L)
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