Adiabatic Mixed-Flow Reactor Volume for 90% Conversion: 23-Chem-A4 Chemical Reactor Engineering, December 2016, Q2

Question 2: Adiabatic Mixed-Flow Reactor Volume 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.

Given. A liquid-phase (constant-density) adiabatic CSTR; the reaction is elementary and first order, $-r_R = k C_R$.

Quantity	Value
Volumetric feed $v_0$	56.64 L/min
Molar feed $F_{R0}$, $F_{I0}$	0.67, 0.33 mol/min
Feed / reference $T_0$, $T_R$	300 K, 298 K
$C_{pR},C_{pS},C_{pT},C_{pI}$	7, 4, 4, 8 cal/(mol·°C)
$k(298\,\text{K})$, $E_a$	0.12 hr$^{-1}$, 25 kcal/mol
$\Delta H_r$, target $X$	$-333$ cal/mol, 0.90
Gas constant $R$ (open-book)	1.987 cal/(mol·K)

Find. The CSTR volume $V$ for $X = 0.90$, accounting for the adiabatic temperature rise that accelerates the reaction.

Approach. Couple the adiabatic energy balance (which fixes the exit temperature at $X=0.9$) with the Arrhenius law (which gives $k$ at that temperature) and finally the mixed-flow design equation $V = F_{R0}X/(-r_R)$.

Feed ratios and mixture heat capacity. Per mole of $R$, the inert ratio is $\Theta_I = F_{I0}/F_{R0} = 0.33/0.67 = 0.4925$. The reactant-plus-inert heat capacity carried into the balance is $\sum \Theta_i C_{pi} = C_{pR} + \Theta_I C_{pI} = 7 + 0.4925(8) = 10.94$ cal/(mol·K), and the reaction heat-capacity change is $\Delta C_p = C_{pS}+C_{pT}-C_{pR} = 4+4-7 = 1$ cal/(mol·K).
Adiabatic energy balance → exit temperature. With no work or heat exchange, $X\,[-\Delta H_r - \Delta C_p (T-T_R)] = \sum\Theta_i C_{pi}\,(T-T_0)$. Solving this linear equation for $T$ at $X=0.9$: $$T = \frac{-X\Delta H_r + X\Delta C_p T_R + (\sum\Theta_i C_{pi})T_0}{\sum\Theta_i C_{pi} + X\Delta C_p} = \boxed{325.2\ \text{K}\ (52.0\,{}^{\circ}\text{C})}.$$ The exothermic reaction self-heats the adiabatic vessel by about 25 K.
Arrhenius correction of the rate constant. $k(T) = k(298)\exp\!\big[\tfrac{E_a}{R}(\tfrac{1}{298}-\tfrac{1}{T})\big] = 0.12\,\exp\!\big[\tfrac{25000}{1.987}(\tfrac{1}{298}-\tfrac{1}{325.2})\big]$. Evaluating the exponent ($\approx 3.53$) gives $$\boxed{k(325.2\,\text{K}) = 4.08\ \text{hr}^{-1} = 0.0680\ \text{min}^{-1}}.$$ The 25 K rise multiplies the rate constant roughly 34-fold — the payoff of running hot.
Mixed-flow design equation. The liquid feed concentration is $C_{R0} = F_{R0}/v_0 = 0.67/56.64 = 0.01183$ mol/L, so the exit concentration is $C_R = C_{R0}(1-X) = 0.001183$ mol/L and the exit rate $-r_R = kC_R = 0.0680(0.001183) = 8.05\times10^{-5}$ mol/(L·min). Then $$V = \frac{F_{R0}\,X}{-r_R} = \frac{0.67(0.90)}{8.05\times10^{-5}} = \boxed{7.49\times10^{3}\ \text{L} \approx 7.5\ \text{m}^3}.$$

Quantity	Result
Adiabatic exit temperature	325.2 K (52.0 °C)
Rate constant at exit $T$	4.08 hr$^{-1}$
Feed concentration $C_{R0}$	0.0118 mol/L
Reactor volume for $X=0.90$	7.49 × 10$^3$ L (7.5 m$^3$)

Check: $\Delta H_r$ and $k$ are both quoted near 298 K, so the heat of reaction is referenced to $T_R = 298$ K in the energy balance; the small $\Delta C_p$ correction ($\pm1$ cal/mol·K) shifts $T$ by under 0.2 K and $V$ by $<1\%$.

Concept & Theory

Topic. Simultaneous mass and energy balance on an adiabatic CSTR — the design temperature is unknown a priori and is fixed by conversion.

Key relations. Adiabatic EB: $X(-\Delta H_r) = \sum\Theta_i C_{pi}(T-T_0)$ (with $\Delta C_p$ correction); Arrhenius: $k(T)=k_0 e^{-E_a/RT}$; CSTR design: $V=F_{A0}X/(-r_A)$; liquid phase $C_A=C_{A0}(1-X)$ (Fogler, Ch. 11–12).

Why this works. In an adiabatic reactor the enthalpy released by reaction has nowhere to go but into sensible heat of the stream, so temperature and conversion move together along a single "adiabatic line." Fixing $X=0.9$ therefore fixes $T$, which fixes $k$, which closes the otherwise-underdetermined design equation. The three relations must be solved as a set, not in isolation.

Common pitfall. Using $k$ at the feed temperature (298–300 K). That understates the rate 30-fold and oversizes the reactor by the same factor. Always evaluate $k$ at the exit temperature of a CSTR, since the whole vessel sits at exit conditions.

Source. Fogler, Elements of Chemical Reaction Engineering, 5th ed., Ch. 11–12 (steady-state nonisothermal design).

Question 2 of 5: Adiabatic Mixed-Flow Reactor Volume for 90% Conversion

Question 2: Adiabatic Mixed-Flow Reactor Volume for 90% Conversion (25 marks)