Question 4 of 6: Diameter and Pressure Drop of a Packed Absorption Tower

Question B1: Diameter and Pressure Drop of a Packed Absorption Tower (25 points)

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) tower diameter at 70% of flooding; (b) pressure drop per foot of packing.

Approach. Evaluate the gas and liquid mass rates at the bottom, form the Sherwood flow-parameter abscissa, read the flooding-line ordinate, solve for the flooding gas mass flux, de-rate to 70% and size the area/diameter; the operating pressure drop follows from the $(G')^2$-scaled ordinate on the generalised chart.

1. Bottom mass rates. Entering gas: 2 lbmol/hr NH₃ + 38 lbmol/hr air $\Rightarrow G=38(29)+2(17)=\boxed{1136\ \text{lbm/hr}}$. Absorbing 90% of 2 lbmol NH₃ adds $1.8(17)=30.6$ lbm/hr to the liquid: $L=3200+30.6=3230.6\ \text{lbm/hr}$.
2. Gas density at 20 °C, then the flow-parameter abscissa. The printed 0.0791 lbm/ft³ is exactly the 0 °C value for this mixture ($M=0.95(29)+0.05(17)=28.4$, and $28.4/359=0.0791$). The tower runs at 20 °C, so $\rho_G=0.0791(273.15/293.15)=0.0737\ \text{lbm/ft}^3$. Then $$X=\frac{L}{G}\sqrt{\frac{\rho_G}{\rho_L}}=\frac{3230.6}{1136}\sqrt{\frac{0.0737}{62.3}}=\boxed{0.098}.$$
3. Read flooding, solve for gas flux. On the supplied chart the vertical gridline just right of the one labelled 0.1 is the true $X=0.1$ line; check it against the 0.01 and 1.0 gridlines. At $X=0.098$ the flooding line reads $Y_\text{flood}\approx0.134$, whereas 0.19 is its value near $X\approx0.045$. With $Y=\dfrac{(G')^2F\Phi\mu_L^{0.1}}{\rho_G\rho_L g_c}$ and $\mu_L^{0.1}=1$: $$G'_\text{flood}=\sqrt{\frac{0.134(0.0737)(62.3)(32.174)}{155(1)(1)}}=0.357\ \text{lbm/ft}^2\text{s}.$$
4. (a) Area and diameter at 70% of flooding. $G'_\text{op}=0.70(0.357)=0.250\ \text{lbm/ft}^2\text{s}$; with $G=1136/3600=0.3156\ \text{lbm/s}$: $$A=\frac{G}{G'_\text{op}}=1.26\ \text{ft}^2\ \Rightarrow\ D=\sqrt{\tfrac{4A}{\pi}}=\boxed{1.27\ \text{ft}\ (0.386\ \text{m})}.$$ Using the printed 0 °C density without correction gives $D\approx1.25$ ft, so the temperature correction moves the answer by under 2%.
5. (b) Pressure drop. The ordinate scales with $(G')^2$, so $Y_\text{op}=(0.70)^2(0.134)=0.066$. At $X=0.098$ the 0.50-in curve sits at $Y\approx0.050$ and the 1.00-in curve at $Y\approx0.074$. The operating point falls between them, about 70% of the way up on the log scale: $$\boxed{\Delta P/Z\approx0.8\ \text{in H}_2\text{O per ft}\ (\approx0.66\ \text{kPa/m})}.$$ That is above the 0.25–0.5 in/ft usually preferred for absorbers, which is typical of a design at 70% of flooding. A 60% design would bring it down to about 0.5 in/ft.