23-Chem-A4 Chemical Reactor Engineering · May 2017
Question 3 of 5: Rate Equation from Plug-Flow Catalytic Data (Variable Density)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. Chemical Engineering — 23-Chem-A4 Chemical Reactor Engineering, May 2017. Open-book, 3 hours. Five questions; answering any four constitutes a complete paper (each 25 marks). All five are solved below.
Reference texts. O. Levenspiel, Chemical Reaction Engineering, 3rd ed. (Wiley); H. S. Fogler, Elements of Chemical Reaction Engineering, 5th ed. (Pearson); Perry's Chemical Engineers' Handbook, 9th ed.
Question 3: Rate Equation from Plug-Flow Catalytic Data (Variable Density) (25 marks)
Given. Catalytic plug-flow reactor, gas-phase, pure A feed, $A\rightarrow4R$ (strong mole increase).
Quantity
Value
Pressure / temperature
3.2 bar, 115 °C (388.15 K)
Volumetric feed $v_0$
20 L/hr pure A
Catalyst $W$ (kg)
0.020, 0.040, 0.080, 0.120, 0.160
Effluent $C_A$ (mol/L)
0.074, 0.060, 0.044, 0.035, 0.029
Find. A rate expression $-r_A^{'}=f(C_A)$ per unit mass of catalyst (order and rate constant).
Figure 3.1 — Log–log plot of the differential rate $-r_A^{'}$ against $C_A$. The points fall on a line of slope $\approx1$, so the reaction is first order.
Approach. Because a mole of A makes four moles of R, the gas expands and $C_A$ is not proportional to $(1-X)$; convert each datum to a true conversion with the expansion factor, differentiate the plug-flow performance equation to obtain the point rate, then read the order and constant from a log–log plot.
Feed concentration and molar feed. For pure A, $C_{A0}=\dfrac{P}{RT}=\dfrac{3.2}{(0.08314)(388.15)}=0.0992\ \text{mol/L},$ and $F_{A0}=C_{A0}v_0=0.0992\times20=1.98\ \text{mol/hr}.$
Expansion factor. Pure A with $\delta=(4-1)/1=3$ gives $\varepsilon_A=y_{A0}\delta=1\times3=3.$ Hence $$X=\frac{1-C_A/C_{A0}}{1+\varepsilon_A\,C_A/C_{A0}}.$$ Evaluating at the five catalyst loadings: $X=0.078,\ 0.140,\ 0.239,\ 0.314,\ 0.377.$
Point rate by the differential method. Along the packed bed $-r_A^{'}=\dfrac{dF_A}{d(-W)}=F_{A0}\dfrac{dX}{dW}.$ Finite differences of $F_A=F_{A0}(1-X)$ between successive points give the rate at the interval mid-concentrations: $$-r_A^{'}\approx 6.1,\ 4.9,\ 3.8,\ 3.1\ \tfrac{\text{mol}}{\text{kg}\cdot\text{hr}}\ \text{at}\ C_A\approx0.067,\ 0.052,\ 0.040,\ 0.032\ \text{mol/L}.$$
Determine the order. A log–log fit of $-r_A^{'}$ versus $C_A$ (Figure 3.1) has slope $$n=\frac{d\ln(-r_A^{'})}{d\ln C_A}\approx 0.93\approx 1,$$ so the reaction is first order in $C_A$.
Determine the rate constant. With $n=1$, $k=-r_A^{'}/C_A$ at each point averages to $$k\approx\frac{1}{4}\big(91.7+93.8+94.9+97.0\big)\approx\boxed{94\ \text{L}/(\text{kg}\cdot\text{hr}).}$$ The rate equation is therefore $$\boxed{-r_A^{'}=94\,C_A\quad\left[\tfrac{\text{mol}}{\text{kg cat}\cdot\text{hr}},\ C_A\ \text{in mol/L}\right].}$$
Check: the integral check confirms first order — substituting $-r_A^{'}=kC_A$ into the plug-flow design equation gives $\dfrac{kW}{v_0}=(1+\varepsilon_A)\ln\dfrac{1}{1-X}-\varepsilon_A X$, and back-solving $k$ from every $(W,X)$ pair yields 91–95 L/(kg·hr), constant to within 4%.
