Two Adiabatic CSTRs in Series — Annual Production of B: 23-Chem-A4 Chemical Reactor Engineering, May 2018, Q4

Question 4: Two Adiabatic CSTRs in Series — Annual Production of B (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.

Given. Liquid-phase first-order reaction in two equal adiabatic CSTRs in series; the exothermic heat release raises the temperature, which in turn accelerates the reaction.

Quantity	Value
Annual feed of A	$2.1\times10^6$ lb/yr over 7000 hr/yr
Reactor volume (each)	1000 gal $\times$ 2 (series)
Feed temperature $T_0$	20 °C $=293.15$ K
Density $\rho_A$	7.5 lb/gal
Heat of reaction $\Delta H_{rxn}$	$-83$ cal/g
Heat capacity $C_p$	0.5 cal·g$^{-1}$·°C$^{-1}$

Find. The mass of B produced per year.

Figure 4 — Two equal adiabatic CSTRs in series. Cold feed enters reactor 1; because no heat is removed, the temperature climbs with conversion, so reactor 2 runs hotter and finishes the conversion.

Approach. The adiabatic energy balance ties temperature to conversion ($T=T_0+\Delta T_{ad}X$); solve each CSTR mole balance together with that relation, taking the ignited (upper) steady state as the operating point.

Feed rate and residence time. $\dot m=\dfrac{2.1\times10^6}{7000}=300\ \text{lb/hr};$ $v_0=300/7.5=40\ \text{gal/hr},$ so each reactor has $\tau=\dfrac{1000}{40}=25\ \text{hr}.$
Adiabatic temperature rise. Per unit mass, $\Delta T_{ad}=\dfrac{-\Delta H_{rxn}}{C_p}=\dfrac{83}{0.5}=166\ \text{°C}$ at complete conversion, so along the reactors $T=293.15+166\,X\ \text{(K)}.$
Reactor 1 (mole + energy balance). The first-order CSTR gives $k\tau=\dfrac{X_1}{1-X_1}$ with $k$ evaluated at $T_1=293.15+166X_1.$ This nonlinear pair has three roots (see the note below); the reactor operates on the upper, ignited branch: $$\boxed{X_1=0.990,\quad T_1=457\ \text{K}\ (184\ \text{°C}).}$$ The long residence time ($\tau=25$ hr) means that once hot, $k\tau\approx96$, so a single tank already reaches 99% conversion.
Reactor 2 (polishing). $\tau=\dfrac{X_2-X_1}{k(T_2)(1-X_2)}$ with $T_2=293.15+166X_2$ carries the conversion to $$\boxed{X_2=0.9999,\quad T_2=459\ \text{K}\ (186\ \text{°C}).}$$ The very high boiling points quoted confirm both streams stay liquid at 186 °C.
Production of B. Since $A\rightarrow B$ conserves mass, the B produced equals the A converted: $$\dot m_B=X_2\times2.1\times10^6=\boxed{2.10\times10^{6}\ \text{lb/yr}\ (\approx300\ \text{lb/hr}).}$$

Check — multiplicity of steady states

The reactor-1 balance has three steady states: a quenched one ($X\approx0$, reactor near the 20 °C feed, reaction negligibly slow), an unstable middle one ($X\approx0.70$, which no reactor can hold), and the ignited stable one ($X\approx0.99$) reported here. Bracketing between the middle and quenched roots would wrongly return $X\approx0.70$; the operated design sits on the upper branch (reached by start-up pre-heating), which the high boiling points make thermally feasible.

Stage	Conversion	Temperature
Reactor 1 exit	$X_1=0.990$	184 °C
Reactor 2 exit	$X_2=0.9999$	186 °C
B produced		$2.10\times10^{6}$ lb/yr

Concept & Theory

Topic: Adiabatic CSTRs in series with feedback between conversion and temperature (steady-state multiplicity).

Key relations: CSTR design $k\tau=X/(1-X)$ (first order); adiabatic energy balance $T=T_0+\dfrac{(-\Delta H_{rxn})}{C_p}X$; Arrhenius $k(T)$ (Fogler Ch. 12; the van Heerden ignition analysis).

Why this works: With no heat removed, every unit of conversion deposits a fixed temperature rise, so the operating temperature is a linear function of conversion. Substituting that into the strongly temperature-dependent $k$ makes the mole balance a single nonlinear equation; its upper root is the hot, high-conversion operating point that the series achieves.

Common pitfall: Converging on the unstable middle root ($X\approx0.70$) instead of the ignited upper root — always bracket the solver near $X\to1$; also, evaluating $k$ at the feed temperature (orders of magnitude too small). In an adiabatic series, reactor 2’s temperature is measured from the original feed (total conversion), not from its own inlet — both give $T=T_0+166X$.

Source: Fogler, Ch. 12 (steady-state nonisothermal reactor design); Levenspiel, Ch. 9 (temperature and pressure effects).

Question 4 of 5: Two Adiabatic CSTRs in Series — Annual Production of B

Question 4: Two Adiabatic CSTRs in Series — Annual Production of B (25 marks)