Adiabatic CSTR — Operating Temperature and Heat Duty: 23-Chem-A4 Chemical Reactor Engineering, December 2013, Q3

Question 3: Adiabatic CSTR — Operating Temperature and Heat Duty (20 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. An adiabatic (no heat exchange) CSTR with an exothermic first-order reaction. Feed and reactor data are listed below; the conversion $X_A=0.9$ is specified.

Quantity	Value
Inlet conc. $C_A^\circ$ / conversion $X_A$	1.75 mol/L / 0.90
Heat capacity $C_p$ / density $\rho$	2.5 J/g·K / 500 g/L
$\Delta H_r$ / inlet $T^\circ$	−27 000 J/mol / 315 K
Feed $q^\circ$ / volume $V$	30 L/min / 28 L

Find. (a) units of the pre-exponential factor; (b) the steady operating temperature; (c) the rate constant prevailing in the reactor; (d) the rate of heat release.

Figure 3 — Adiabatic CSTR: the heat of reaction raises the exit stream from 315 K to the operating temperature; no heat leaves through the walls.

Approach. (a) dimensional analysis of a first-order rate law; (b) an adiabatic energy balance (all reaction heat becomes sensible heat of the stream); (c) the given Arrhenius rate law evaluated at the operating temperature, cross-checked against the CSTR mass balance; (d) heat released = (moles reacted)$\times(-\Delta H_r)$.

Part (a) — units of $1.8\times10^{13}$. For first order $(-r_A)=kC_A$ with $(-r_A)$ in mol/(L·s) and $C_A$ in mol/L, $k$ has units s⁻¹. The exponential is dimensionless, so$$\boxed{[\,1.8\times10^{13}\,]=\text{s}^{-1}}.$$
Part (b) — adiabatic energy balance. With no external heat, the reaction heat raises the throughput’s temperature. Per unit time, $\dot m C_p (T-T^\circ)=(-\Delta H_r)\,q^\circ C_A^\circ X$, i.e. the adiabatic rise is$$\Delta T_{ad}=\frac{(-\Delta H_r)C_A^\circ X}{\rho C_p}=\frac{(27000)(1.75)(0.9)}{(500)(2.5)}=34.0\ \text{K}\;\Rightarrow\;\boxed{T=315+34=349\ \text{K}}.$$
Part (c) — rate constant at reactor conditions. A CSTR is uniform, so everything inside it is at the operating temperature from part (b), $T=349.0$ K. Evaluating the given rate law there:$$k=1.8\times10^{13}\exp\!\left(\frac{-84000}{(8.314)(349.0)}\right)=1.8\times10^{13}\,e^{-28.95}=\boxed{4.82\ \text{s}^{-1}}.$$Cross-check (mass balance). The first-order CSTR design equation with $\tau=V/q^\circ=28/30=0.933$ min gives $k=\dfrac{X}{\tau(1-X)}=\dfrac{0.9}{(0.933)(0.1)}=9.64\ \text{min}^{-1}=0.161\ \text{s}^{-1}$, about 30 times smaller. The given data do not agree with each other; see the Verify note.
Part (d) — rate of heat evolved. The molar rate of A reacting is $q^\circ C_A^\circ X=30\times1.75\times0.9=47.25$ mol/min, so$$\dot Q=(-\Delta H_r)\,q^\circ C_A^\circ X=(27000)(47.25)=1.276\times10^{6}\ \text{J/min}=\boxed{2.13\times10^{4}\ \text{J/s}\ (21.3\ \text{kW})}.$$This exactly equals the sensible-heat load $\dot m C_p\Delta T_{ad}$, as the adiabatic balance requires.

Check

Assumption (boxed per exam Note 1): the data are over-specified. The energy balance fixes $T=349$ K using only $X$, so part (b) does not depend on the conflict. Part (c) asks for the rate constant "for conditions within the CSTR". Here that means the given rate law, whose units part (a) establishes, evaluated at that temperature: $k=4.82$ s⁻¹. The mass balance with the stated $V$ and $X$ implies $k=0.161$ s⁻¹ instead. Reproducing that value from the rate law would need $T\approx312$ K, which is below the 315 K feed and impossible for an adiabatic exothermic reactor. With $k=4.82$ s⁻¹ the 28 L vessel would actually reach $X\approx0.996$. So the Arrhenius constants, $V$ and $X$ cannot all be exact. The heat duty in (d) comes from the stated conversion, $(-\Delta H_r)F_{A0}X$, which is consistent with the energy balance. Using $(-r_A)V$ with $k=4.82$ s⁻¹ would give $6.4\times10^{5}$ J/s, which contradicts the 34 K temperature rise.

Quantity	Result
(a) Units of $1.8\times10^{13}$	s⁻¹
(b) Operating temperature	349 K (76 °C)
(c) Rate constant in reactor	4.82 s⁻¹ (rate law at 349 K); mass balance implies 0.161 s⁻¹, so the data are inconsistent
(d) Heat evolved	$2.13\times10^{4}$ J/s (21.3 kW)

Concept & Theory

Topic. Coupled mass and energy balances for an adiabatic, exothermic CSTR.

Key relations. Adiabatic rise $\Delta T_{ad}=\dfrac{(-\Delta H_r)C_{A0}X}{\rho C_p}$; Arrhenius $k=A\,e^{-E/RT}$ at the reactor temperature; first-order CSTR $\tau=\dfrac{X}{k(1-X)}$ (consistency check); heat release $\dot Q=(-\Delta H_r)F_{A0}X$.

Why this works. With no wall heat transfer, an energy balance around the reactor equates the enthalpy of reaction to the sensible enthalpy change of the flowing stream — a relation independent of kinetics, so $T$ follows from $X$ alone.

Common pitfall. Mixing per-mass ($C_p$ in J/g·K, $\rho$ in g/L) and per-mole quantities — keep the group $\rho C_p$ on a volumetric basis so it matches $C_{A0}$ in mol/L. Also, "heat evolved" is the reaction heat, $(-\Delta H_r)F_{A0}X$, not $(-r_A)V$ unless the two are consistent.

Source. Fogler §11–12 (steady-state nonisothermal design; adiabatic operation); Levenspiel Ch. 9.

Question 3 of 5: Adiabatic CSTR — Operating Temperature and Heat Duty

Question 3: Adiabatic CSTR — Operating Temperature and Heat Duty (20 marks)