Pyrolysis — PFR Length for 90% Conversion: 23-Chem-A4 Chemical Reactor Engineering, May 2018, Q3

Question 3: Pyrolysis — PFR Length for 90% 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 — cross-sectional area used

A 6-in-diameter circle has area $0.196\ \text{ft}^2$, which is inconsistent with the printed cross-section $3.88\times10^{-2}\ \text{ft}^2$. Reactor length follows from $L=V/A$, so we use the stated design area $A=3.88\times10^{-2}\ \text{ft}^2$, because the paper gives it explicitly as the flow area, and report $L$ on that basis (exam instruction #1). If the 6-in diameter were taken as governing instead, the same volume would give $L=5.75/0.196=29.3$ ft.

Given. Pure gaseous A pyrolyses first order with a mole increase ($1\rightarrow2$); isothermal isobaric PFR.

Quantity	Value
Molar mass $M$	146 lb/lbmol
Temperature $T$	500 °C $=773.15$ K
Pressure $P$	5 atm
Feed rate	500 lb/hr (pure A)
Cross-section $A$	$3.88\times10^{-2}\ \text{ft}^2$
Conversion $X$	0.90

Find. The reactor length $L$ for 90% conversion of A.

Figure 3 — Isothermal tubular pyrolysis reactor at 500 °C and 5 atm. Pure A splits into two moles ($B+C$), so $\varepsilon=1$; the length follows from the required volume divided by the fixed cross-section.

Approach. Evaluate the Arrhenius $k$ at 773 K, size the variable-volume first-order gas PFR ($\varepsilon=1$), then convert volume to length using the given cross-section.

Rate constant at 500 °C. $k=7.8\times10^{9}\,e^{-19220/773.15}=7.8\times10^{9}\,e^{-24.86}=0.1247\ \text{s}^{-1}.$
Molar feed and inlet concentration. $\dot F_{A0}=\dfrac{500}{146}=3.425\ \text{lbmol/hr}=0.4315\ \text{mol/s}.$ $C_{A0}=\dfrac{P}{RT}=\dfrac{5}{(0.08206)(773.15)}=0.0788\ \text{mol/L}=78.8\ \text{mol/m}^3,$ giving $v_0=\dot F_{A0}/C_{A0}=5.48\times10^{-3}\ \text{m}^3/\text{s}.$
Expansion factor. Pure feed, $\delta=+1\Rightarrow\varepsilon=y_{A0}\delta=1.$
Reactor volume. $V=\dfrac{v_0}{k}\big[(1+\varepsilon)\ln\tfrac{1}{1-X}-\varepsilon X\big]=\dfrac{5.48\times10^{-3}}{0.1247}\big[2\ln 10-0.9\big]=0.0439\times3.705=0.1627\ \text{m}^3=5.75\ \text{ft}^3.$
Reactor length. $L=\dfrac{V}{A}=\dfrac{5.75\ \text{ft}^3}{3.88\times10^{-2}\ \text{ft}^2}=\boxed{148\ \text{ft}.}$

Quantity	Result
Rate constant $k(773\ \text{K})$	$0.1247\ \text{s}^{-1}$
Reactor volume $V$	0.163 m$^3$ (5.75 ft$^3$)
Reactor length $L$	148 ft

Concept & Theory

Topic: Length of an isothermal gas-phase pilot PFR for a first-order pyrolysis with mole expansion.

Key relations: Arrhenius $k=Ae^{-E_a/RT}$; first-order variable-volume PFR volume $V=\tfrac{v_0}{k}[(1+\varepsilon)\ln\tfrac{1}{1-X}-\varepsilon X]$; geometry $L=V/A$ (Fogler Ch. 4–5).

Why this works: With negligible pressure drop the reactor is isobaric, so the stoichiometric table applies directly with $\varepsilon=1$; dividing the required volume by the constant cross-section returns the tube length. The large $L$ reflects both the strong dilution ($\varepsilon=1$ doubles the gas volume) and the modest rate at 500 °C.

Common pitfall: Neglecting $\varepsilon$ under-predicts the length by nearly 40% (bracket $\ln10=2.30$ instead of 3.705); also mixing lb/lbmol with SI when forming $\dot F_{A0}$ and $C_{A0}$ — keep $v_0=\dot F_{A0}/C_{A0}$ in one consistent system before converting the final volume to ft$^3$.

Source: Fogler, Ch. 4–5 (Arrhenius kinetics and isothermal PFR design).

Question 3 of 5: Pyrolysis — PFR Length for 90% Conversion

Question 3: Pyrolysis — PFR Length for 90% Conversion (25 marks)