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23-Chem-A3 Heat and Mass Transfer · December 2014

Question 7 of 7: Wetted-wall absorption of ammonia — film coefficient and rate

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2014 — 04-Chem-A3 Mass Transfer Operations. Three-hour, open-book exam; any non-communicating calculator permitted. Format: seven questions in three parts — answer one of Q1–Q2 (Part A), one of Q3–Q4 (Part B) and two of Q5–Q7 (Part C); four questions of equal value constitute a complete paper. All seven are solved below for completeness. Every property datum (vapour pressures, diffusivities, Henry constants) is stated in the question, so each answer is self-contained.

Reference texts: Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular diffusion, convective mass transfer and gas absorption; Welty, Wicks, Wilson & Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (6th ed., Wiley) — boundary-layer mass transfer and diffusion with reaction; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (species diffusion, sphere in a stagnant medium); Treybal, Mass-Transfer Operations (3rd ed.) — packed-tower and wetted-wall design; supporting data from Perry's Chemical Engineers' Handbook (9th ed.).

Question 7: Wetted-wall absorption of ammonia — film coefficient and rate (Part C — equal value)

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. A two-film (two-resistance) gas absorption at a point in a wetted-wall column: the overall gas-phase coefficient is measured and the gas holds a known fraction of the resistance, which lets us extract the individual gas-film coefficient and the local flux.

QuantityValue
Overall coefficient $K_G$$2.75\times10^{-6}$ kmol/m²·s·kPa
Gas / liquid composition$y=0.080$ / $x=0.00115$
Temperature / pressure300 K / 1 atm $=101.325$ kPa
Gas-phase resistance fraction0.80
Equilibrium slope $m$1.64 ($y_A^*=1.64x_A$)

Find. (a) the gas-film coefficient $k_y$; (b) the local ammonia absorption rate $N_A$ (kmol/m²·s).

bulk gas y = 0.080 bulk liquid x = 0.00115 gas film liquid film interface (y_i, x_i) N_A (NH₃) 80% resistance 20% resistance
Figure 7 — Two-film picture of NH₃ absorption. The overall gas-phase coefficient $K_y$ lumps both films; since 80% of the resistance is in the gas film, the individual gas-film coefficient $k_y$ is larger than $K_y$ by that fraction.

Approach. Convert the pressure-basis overall coefficient to a mole-fraction basis ($K_y=K_GP$); split off the gas-film coefficient using the 80% resistance fraction; then compute the flux from the overall gas driving force $y-y^*$.

  1. Overall mole-fraction coefficient. Since $p=Py$, $$K_y=K_G\,P=(2.75\times10^{-6})(101.325)=2.79\times10^{-4}\ \text{kmol/m}^2\!\cdot\!\text{s}.$$
  2. (a) Gas-film coefficient. Resistances add in series, $1/K_y=1/k_y+m/k_x$; the gas holds 80% of the total, so $1/k_y=0.80/K_y$, giving $$k_y=\frac{K_y}{0.80}=\frac{2.79\times10^{-4}}{0.80}=\boxed{3.48\times10^{-4}\ \text{kmol/m}^2\!\cdot\!\text{s}}.$$
  3. Equilibrium gas composition. $y^*=1.64\,x=1.64(0.00115)=0.00189$; the overall gas driving force is $y-y^*=0.080-0.00189=0.0781$.
  4. (b) Absorption rate. $$N_A=K_y\,(y-y^*)=(2.79\times10^{-4})(0.0781)=\boxed{2.18\times10^{-5}\ \text{kmol/m}^2\!\cdot\!\text{s}}.$$ (Identically, $N_A=K_G(Py-Py^*)=2.75\times10^{-6}\times101.325\times0.0781$.)
QuantityResult
Overall coefficient $K_y=K_GP$$2.79\times10^{-4}$ kmol/m²·s
Gas-film coefficient $k_y$$3.48\times10^{-4}$ kmol/m²·s
Driving force $y-y^*$0.0781
Absorption rate $N_A$$2.18\times10^{-5}$ kmol/m²·s
Check
The overall gas driving force $y-y^*=0.0781$ is close to $y=0.080$ because the liquid is very dilute (small $y^*$). The gas holding 80% of the resistance makes $k_y$ only $1/0.8=1.25\times$ the overall $K_y$ — consistent with a soluble gas like ammonia, whose large liquid solubility ($m=1.64$, but with $k_x\gg k_y$) puts most resistance in the gas film.
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