Question 6 of 6: Part B: Minimum Liquid Rate and Diameter of a Packed Absorber
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: open-book, three-hour exam; Part A (Heat Transfer, Q1–Q3) and Part B (Mass Transfer, Q4–Q6); all problems 25 points, at least two attempted per part. Every problem is worked in full below.
Reference texts: J. M. Coulson & J. F. Richardson, Chemical Engineering Vols. 1&2 (Coulson & Richardson) — boiling curves, tube-bank convection, gas absorption and column sizing; F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — free-convection correlations and the gas-property tables reproduced in the paper; J. R. Welty, C. E. Wicks & R. E. Wilson, Fundamentals of Momentum, Heat and Mass Transfer — the laminar vertical-plate local Nusselt relation; R. E. Treybal, Mass-Transfer Operations (McGraw-Hill) — stagnant-film diffusion, differential (Rayleigh) distillation and packed-tower flooding; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Question 6 — Part B: Minimum Liquid Rate and Diameter of a Packed Absorber (25 points)
Given. Gas: $1$ m³/s at 293 K, $1.013\times10^{5}$ Pa; $y_A=0.10$, $M_A=64$, $M_{inert}=30$; remove 95 % of A. Solvent: water ($M=18$). $\rho_L=1020$ kg/m³, $\mu'_L=0.902$ cP, $g_c=9.807$. Equilibrium (mole ratios $X$ = A/water, $Y$ = A/inert):
$X$
0.001
0.002
0.003
0.004
0.005
0.006
$Y$
0.024
0.055
0.090
0.129
0.170
0.212
Find. (a) minimum water rate $L_{min}$ (kg/s); (b) tower diameter $D$ at $L=1.2L_{min}$ and $u=0.6\,u_{flood}$.
Figure 6 — Countercurrent packed absorber. Rich gas enters the bottom ($Y_1$), scrubbed gas leaves the top ($Y_2$); water enters clean at the top and leaves loaded at the bottom ($X_1$). Minimum liquid corresponds to the bottom liquid in equilibrium with the entering gas.
Approach. Work in solute-free mole ratios: fix the inert gas flow, set $Y_1$ and $Y_2$ from the 95 % removal, take the exit liquid at its equilibrium maximum ($X_1^{*}$ with $Y_1$) for $L_{min}$; then for (b) build the actual mass flows, solve the flooding correlation for $G'_{flood}$, take 60 % of it, and size the area from the gas mass flow.
Gas molar flow and inert rate. Ideal gas: $n=PV/RT=(1.013\times10^{5})(1)/[(8.314)(293)]=41.6$ mol/s. Then A in $=4.16$ mol/s, inert $G_s=37.4$ mol/s.
Terminal gas ratios. Bottom (entering gas): $Y_1=A/\text{inert}=4.16/37.4=0.1111$. Removing 95 % of A leaves $Y_2=0.05\,Y_1=0.00556$.
Minimum liquid (part a). At $L_{min}$ the operating line touches equilibrium at the bottom, so the leaving liquid is in equilibrium with the entering gas. Interpolating the table at $Y=0.1111$ (between $Y=0.090\!\to\!X=0.003$ and $Y=0.129\!\to\!X=0.004$):
$$X_1^{*}=0.003+\frac{0.1111-0.090}{0.129-0.090}(0.001)=0.003541\ \text{mol A/mol water}.$$
Solute A absorbed $=G_s(Y_1-Y_2)=37.4(0.1111-0.00556)=3.95$ mol/s, so
$$W_{s,min}=\frac{G_s(Y_1-Y_2)}{X_1^{*}}=\frac{3.95}{0.003541}=1115\ \text{mol water/s},$$
$$L_{min}=W_{s,min}\,M_{H_2O}=(1115)(0.018)=\boxed{20.1\ \text{kg/s}.}$$
Actual terminal mass flows (part b). Gas mean $M=0.1(64)+0.9(30)=33.4$, so the bottom gas mass flow $G_{mass}=4.16(0.064)+37.4(0.030)=1.39$ kg/s and $\rho_G=PM/RT=1.39$ kg/m³. With $L=1.2L_{min}$: $L_{mass}=1.2(20.1)+\text{(absorbed A)}=24.4$ kg/s, giving $L'/G'=24.4/1.39=17.5$.
Flooding mass velocity. Solve the given correlation for $G'$ (SI: $\rho$ in kg/m³, $\mu'_L$ in cP, $g_c=9.807$):
$$G'^{2}=\frac{0.052\,(L'/G')\,(\rho_G/\rho_L)^{0.5}\,g_c\,\rho_G\,\rho_L}{538\,(\mu'_L)^{0.2}}=0.887\;\Rightarrow\;G'_{flood}=0.942\ \text{kg/m}^2\text{s}.$$
Operating velocity, area and diameter. Operating at 60 % of flooding, $G'_{op}=0.6(0.942)=0.565$ kg/m²s. The cross-section carries the gas mass flow:
$$A=\frac{G_{mass}}{G'_{op}}=\frac{1.39}{0.565}=2.46\ \text{m}^2,\qquad D=\sqrt{\frac{4A}{\pi}}=\boxed{1.77\ \text{m}.}$$
