Question 4 of 6: Part B — Separation Processes

Question B1: Height and Diameter of a Packed Absorption Tower (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.

Find. (a) the packed height $Z=H_{oy}N_{oy}$; (b) the tower diameter operating at 50% of the flooding velocity.

(a) Packed height

Approach. Convert the streams to solute-free molar flows, fix the minimum (then operating) water rate at the bottom pinch, get $N_{oy}$ from the Colburn dilute expression, and multiply by the height per transfer unit $H_{oy}$.

1. Molar flows and terminal compositions. The entering gas has $\bar M=0.02(58.08)+0.98(28.97)=29.55\ \text{kg/kmol}$, so $G=450/29.55=15.23\ \text{kmol/hr}$; the inert-air carrier is $G_s=0.98(15.23)=14.92\ \text{kmol/hr}$ and acetone in $=0.3046\ \text{kmol/hr}$. Removing 96% leaves $0.0122\ \text{kmol/hr}$, giving mole ratios $$Y_1=\frac{0.3046}{14.92}=0.02041,\qquad Y_2=\frac{0.0122}{14.92}=8.16\times10^{-4},\qquad y_2=8.16\times10^{-4}.$$
2. Minimum and operating liquid rate. The minimum solvent leaves the bottom in equilibrium with the entering gas: $x_1^{*}=y_1/m=0.02/2.5=0.008$, i.e. $X_1^{*}=0.008/(1-0.008)=0.008065$. A solute balance on the carriers (pure water, $X_2=0$) gives $$L_{s,\min}=\frac{G_s(Y_1-Y_2)}{X_1^{*}}=\frac{14.92(0.02041-0.000816)}{0.008065}=\boxed{36.3\ \text{kmol/hr}},$$ and the operating rate is $L_s=1.20\,L_{s,\min}=\boxed{43.5\ \text{kmol/hr}}$ water.
3. Number of transfer units (Colburn). With the absorption factor $A=\dfrac{L_s}{mG_s}=\dfrac{43.5}{2.5(14.92)}=1.166$ (so $1/A=0.857$) and $x_2=0$, $$N_{oy}=\frac{\ln\!\left[\left(1-\tfrac1A\right)\dfrac{y_1}{y_2}+\tfrac1A\right]}{1-\tfrac1A}=\frac{\ln\!\left[0.1426(24.5)+0.857\right]}{0.1426}=\boxed{10.3}.$$
4. Height of a transfer unit and packed height. The paper's own relation, with $mG/L=1/A=0.857$: $$H_{oy}=H_y+\frac{mG}{L}H_x=0.54+0.857(0.32)=\boxed{0.814\ \text{m}},$$ hence $$\boxed{Z=H_{oy}N_{oy}=0.814(10.3)=8.4\ \text{m}.}$$ The large $N_{oy}\approx10$ is expected — 96% removal with $A$ only modestly above 1 (a near-pinched line) demands many transfer units.

(b) Tower diameter at 50% flooding

Approach. Enter the Sherwood/Eckert flooding chart on the flow-parameter abscissa using the bottom mass rates, read the flooding ordinate, solve for the flooding gas flux (with consistent SI units), halve it, and size the area from the actual gas mass flow.

1. Flow-parameter abscissa. Using the bottom (largest) mass rates — gas $G'=450\ \text{kg/hr}$, liquid $L'=L_sM_w+\text{(acetone absorbed)}=43.5(18.02)+0.29(58.08)\approx801\ \text{kg/hr}$: $$\frac{L'}{G'}\sqrt{\frac{\rho_G}{\rho_L}}=\frac{801}{450}\sqrt{\frac{1.181}{998.4}}=1.78(0.0344)=0.0612.$$
2. Read the flooding ordinate and solve the capacity group. On the supplied chart the flooding line at abscissa $0.061$ lies between the 0.10 and 0.20 gridlines and reads $\approx0.17$. The ordinate is $\dfrac{(G'_f)^2 F\Phi\mu_L^{0.1}}{\rho_G\rho_L g}$, which is dimensionless only when $F$ carries reciprocal length in the same units as $g$. The paper's $F=160$ is the tabulated value for 2.54-cm ceramic Raschig rings in ft$^{-1}$, so in SI $F=160/0.3048=525\ \text{m}^{-1}$. With $\Phi=1.0$ and $\mu_L^{0.1}=0.86^{0.1}=0.985$: $$G'_f=\sqrt{\frac{0.17\,\rho_G\rho_L g}{F\Phi\mu_L^{0.1}}}=\sqrt{\frac{0.17(1.181)(998.4)(9.81)}{525(1.0)(0.985)}}=\boxed{1.95\ \text{kg/m}^2\!\cdot\!\text{s}}.$$
3. Design flux and diameter. At 50% of flooding the design gas flux is $G'_\text{op}=0.5(1.95)=0.975\ \text{kg/m}^2\!\cdot\!\text{s}$. With the actual gas mass flow $450\ \text{kg/hr}=0.125\ \text{kg/s}$, $$A=\frac{0.125}{0.975}=0.128\ \text{m}^2\ \Longrightarrow\ D=\sqrt{\frac{4A}{\pi}}=\boxed{0.40\ \text{m}}.$$ The tower is therefore $\approx0.40$ m (400 mm) in diameter and 8.4 m tall — the slender, tall geometry typical of a dilute-gas absorber at high recovery.