23-Chem-A4 Chemical Reactor Engineering · December 2015
Question 4 of 5: Nitration in a CSTR — Adiabatic vs Cooled Volume and Space-Time
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 4: Nitration in a CSTR — Adiabatic vs Cooled Volume and Space-Time (25 marks: a 12, b 13)
The factor $0.0033\ \text{K}^{-1}\approx1/303$ K, so $k=0.090$ L²·mol⁻²·min⁻¹ at the feed temperature 303 K; the exponential corrects $k$ to the actual operating temperature. We take $E_a=40$ kJ/mol as the $\Delta E_a$ in the correlation and use $1/303$ for 0.0033 (the difference is below 0.2% in $k$).
Given. First order in A and second order in B: $-r_A=kC_AC_B^{2}$ (consistent with the printed units of $k$, L²·mol⁻²·min⁻¹). $\theta_B=F_{B0}/F_{A0}=3$; strongly exothermic ($\Delta H_R=-370.1$ kJ/mol A). Feed molar rate $F_{A0}=10$ mol/min in both parts; $X=0.35$. Assumption: the printed $\Delta H_R$ applies at $T_R=T_0=303$ K, corrected with $\Delta C_p$ since all four heat capacities are given.
Quantity
Value
$\Delta H_R$ / $E_a$
$-370.1$ kJ/mol / 40 kJ/mol
$k(303\text{ K})$
0.090 L²·mol⁻²·min⁻¹
$F_{A0}$ / $F_{B0}$ ($\theta_B$)
10 / 30 mol/min (3)
$\sum F_{i0}C_{pi}=10(84.5)+30(137)$
4955 J·min⁻¹·K⁻¹
$\Delta C_p=2(170)+75-84.5-2(137)$
56.5 J·mol⁻¹·K⁻¹
Target conversion, $X$
0.35
Find. (a) Adiabatic exit temperature, then $V$ and $\tau$ for $X=0.35$. (b) $V$ and $\tau$ if the reactor is held at 323 K, plus the coolant temperature the $U\!A_H$ implies.
Figure 4 — Nitration CSTR. In (a) the released heat raises the exit temperature (adiabatic); in (b) cooling holds it at 323 K, so the two cases size very differently despite the same conversion.
Approach. The CSTR temperature is fixed independently of kinetics — by the adiabatic energy balance in (a) and by specification in (b). Evaluate $k$ at that temperature, form $-r_A=kC_AC_B^2$ from the exit concentrations, then size the CSTR from $V=F_{A0}X/(-r_A)$ and $\tau=V/v_0$; finally close a heat balance for the coolant in (b).
(a) Adiabatic exit temperature. The steady-state CSTR energy balance (Fogler) with $\Delta H_R(T)=\Delta H_R^\circ+\Delta C_p(T-T_R)$ is $$\sum F_{i0}C_{pi}(T-T_0)=F_{A0}X\big[-\Delta H_R^\circ-\Delta C_p(T-T_0)\big].$$ So $4955(T-303)=3.5\,[370{,}100-56.5(T-303)]$, giving $(4955+197.75)(T-303)=1{,}295{,}350$ and $$T=303+251.4=\boxed{554\ \text{K}}.$$
(a) Rate and volume. $k(554\text{ K})=0.090\exp\!\big[\tfrac{40000}{8.314}(\tfrac1{303}-\tfrac1{554.4})\big]=120.6$ L²·mol⁻²·min⁻¹. Exit concentrations: $C_A=C_{A0}(1-X)=0.0065$, $C_B=C_{A0}(\theta_B-2X)=0.01(2.3)=0.023$ mol/L. Then $-r_A=kC_AC_B^{2}=120.6(0.0065)(0.023)^2=4.15\times10^{-4}$ mol·L⁻¹·min⁻¹, so
$$V=\frac{F_{A0}X}{-r_A}=\frac{10(0.35)}{4.15\times10^{-4}}=8.44\times10^{3}\ \text{L}=\boxed{8.44\ \text{m}^3},\quad \tau=\frac{V}{v_0}=\frac{8440}{1000}=\boxed{8.44\ \text{min}}.$$ If $\Delta C_p$ were neglected, $T$ would be 564 K and $V=7.24$ m³, so the product heat capacities change the answer by about 15%.
(b) Cooled to 323 K — rate and volume. Now $v_0=100$ L/min, $C_{A0}=0.1$ mol/L (still $F_{A0}=10$ mol/min). $k(323\text{ K})=0.090\exp\!\big[\tfrac{40000}{8.314}(\tfrac1{303}-\tfrac1{323})\big]=0.241$. Exit $C_A=0.065$, $C_B=0.23$ mol/L $\Rightarrow -r_A=0.241(0.065)(0.23)^2=8.27\times10^{-4}$ mol·L⁻¹·min⁻¹. Thus $$V=\frac{10(0.35)}{8.27\times10^{-4}}=4.23\times10^{3}\ \text{L}=\boxed{4.23\ \text{m}^3},\quad \tau=\frac{4230}{100}=\boxed{42.3\ \text{min}}.$$ Although $k$ is 500 times smaller, the ten-fold higher concentrations raise $C_AC_B^{2}$ by $10^3$, so the cooled reactor is actually smaller.
(b) Coolant temperature check. Steady-state heat balance: heat generated minus sensible heating of the feed to 323 K must be removed. With $-\Delta H_R(323)=370{,}100-56.5(20)=368{,}970$ J/mol, $\dot Q_\text{rem}=3.5(368{,}970)-4955(20)=1.192\times10^{6}$ J/min. With $\dot Q=U\!A_H(T-T_a)$, $$T_a=T-\frac{\dot Q_\text{rem}}{U\!A_H}=323-\frac{1.192\times10^{6}}{9000}=190.5\ \text{K}.$$
Check — reading of “cooled to 323 K” in (b)
Read literally, the reactor contents are held at 323 K. That is the answer boxed above, but it needs a coolant at $T_a\approx190$ K ($-83\ ^\circ$C), which is impractical with $U\!A_H=9000$ J·min⁻¹·K⁻¹. If the examiner meant a coolant at 323 K, the energy balance $\sum F_{i0}C_{pi}(T-T_0)+U\!A_H(T-323)=F_{A0}X[-\Delta H_R(T)]$ gives $T=407$ K, $k=5.24$, $-r_A=1.80\times10^{-2}$ mol·L⁻¹·min⁻¹, $V=194$ L and $\tau=1.94$ min. State which reading you adopt, as exam instruction 1 asks.
