Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — December 2015 — 04-Chem-A4 Chemical Reactor Engineering. Three-hour open-book exam; the designated Fogler textbook (any edition), unit-conversion / mathematical tables (CRC Handbook) and a non-communicating programmable calculator are permitted. Five questions are printed and any four constitute a complete paper (each worth 25 marks); all five are solved below for completeness. No credit is given for re-deriving standard rate expressions, so the batch / CSTR / PFR / packed-bed design equations are quoted and applied. Property look-ups not printed on the paper (the gas constant, molar volumes) are stated explicitly in each Given block as permitted open-book references.
Reference texts: H. S. Fogler, Elements of Chemical Reaction Engineering (4th/5th ed., Prentice Hall) — the designated open-book text: batch/CSTR/PFR design equations, the stoichiometric table with expansion factor $\varepsilon$ for gas-phase reactions with a change in moles, parallel-reaction selectivity/yield, the semibatch mole balances, and the adiabatic energy balance; O. Levenspiel, Chemical Reaction Engineering (3rd ed., Wiley) — reactor sequencing for >1-order kinetics and packed-bed (catalyst-weight) design; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Question 5: Recycle (Gradientless) Reactor → Packed-Bed Catalyst Sizing (25 marks: a 10, b 12, c 3)
Given. A very large recycle ratio makes the recycle reactor gradientless (differential / mixed): composition is uniform at the exit value, so the rate can be read directly. Catalytic second-order rate $-r_A'=k'C_A^2$ (per gram of catalyst).
Quantity
Recycle test
Part (b) PBR
Feed conc. $C_{A0}$
2 mol/L
1 mol/L
Feed rate $v_0$
1 L/hr
1000 L/hr
Exit / conversion
$C_A=0.5$ mol/L
$X=0.80$
Catalyst $W$
3 g (given)
to be found
Find. (a) $k'$ with units; (b) catalyst weight $W$ for 80% conversion of 1000 L/hr; (c) $W$ and total bed mass with 1:4 catalyst:inert dilution.
Figure 5 — Part (b)/(c) run the kinetics from the recycle test in a once-through packed-bed reactor (plug flow in catalyst weight $W$); inert dilution changes the bed volume but not the catalyst mass.
Approach. Treat the high-recycle unit as gradientless to extract $-r_A'$ directly and back out $k'$; then integrate the packed-bed design equation $\frac{dX}{dW}=\frac{-r_A'}{F_{A0}}$ for a second-order rate to size the once-through bed; finally reason about inert dilution.
(a) Read the rate from the gradientless reactor. Conversion $X=\dfrac{C_{A0}-C_A}{C_{A0}}=\dfrac{2-0.5}{2}=0.75$; $F_{A0}=C_{A0}v_0=2(1)=2$ mol/hr. Because the reactor is uniform at exit, $-r_A'=\dfrac{F_{A0}X}{W}=\dfrac{2(0.75)}{3}=0.5$ mol·g⁻¹·hr⁻¹.
(a) Rate constant. With $-r_A'=k'C_A^2$ evaluated at the uniform exit $C_A=0.5$ mol/L, $$k'=\frac{-r_A'}{C_A^2}=\frac{0.5}{0.5^2}=\boxed{2.0\ \text{L}^2\,\text{mol}^{-1}\text{g}^{-1}\text{hr}^{-1}.}$$
(b) Size the once-through packed bed. Plug flow in catalyst weight: $\dfrac{dX}{dW}=\dfrac{-r_A'}{F_{A0}}$ with $-r_A'=k'C_{A0}^2(1-X)^2$ (liquid, constant density). Integrating, $$W=\frac{F_{A0}}{k'C_{A0}^2}\Big[\frac{1}{1-X}-1\Big].$$ Now $F_{A0}=C_{A0}v_0=1(1000)=1000$ mol/hr, $C_{A0}=1$ mol/L, $X=0.80$: $$W=\frac{1000}{2(1)^2}\Big[\frac{1}{0.2}-1\Big]=500(4)=\boxed{2000\ \text{g}=2.0\ \text{kg}.}$$
(c) Effect of 1:4 catalyst-to-inert dilution. Inert solid carries no kinetics, so the conversion depends only on the mass of active catalyst. The required catalyst mass is therefore unchanged: $$W_\text{cat}=\boxed{2.0\ \text{kg}}.$$ With 1 part catalyst to 4 parts inert, the total packed mass is $5\times$ the catalyst, i.e. $W_\text{cat}+4W_\text{cat}=10$ kg of solids (2 kg catalyst + 8 kg inert). Dilution buys better isothermality and fewer hot spots at the cost of a larger bed, not more catalyst.
