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23-Chem-A3 Heat and Mass Transfer · December 2016

Question 4 of 7: Overall Gas-Phase Coefficient for CO₂ Absorption in NaOH

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

Notes on this paper

Format: open-book, three-hour paper; seven questions in three parts (Part A Q1–2, Part B Q3–4, Part C Q5–7). A candidate answers one from A, one from B and two from C (four questions, equal value). All seven are solved here as a complete study resource.

Reference texts: R.E. Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — film theory, wetted-wall and flat-plate convective mass transfer, packed-tower absorption, distillation; Coulson & Richardson, Chemical Engineering Vol. 1 (6th ed.) and Vol. 2 (5th ed., Richardson, Harker & Backhurst) — diffusion, drying/evaporation, absorption with reaction, adsorption, membrane separations; C.J. Geankoplis, Transport Processes and Separation Process Principles (4th ed.) — Sherwood-number correlations, McCabe–Thiele with side streams; property data from Perry’s Chemical Engineers’ Handbook (9th ed.).

Question 4: Overall Gas-Phase Coefficient for CO₂ Absorption in NaOH (Part B — 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. Packed height $Z=3$ m, gas mass flux $G=0.34$ kg/m²s (essentially air), $P_T=101.3$ kPa. CO₂ mole fraction falls from $y_1=315$ ppm (bottom) to $y_2=31$ ppm (top). The fast reaction with NaOH makes the equilibrium partial pressure $p^\ast\approx0$.

Find. the volumetric overall gas-phase coefficient $K_Ga$ [kmol/(m³·s·kPa)].

3 m packed height19 mm Raschig, D = 250 mmgas in, CO₂ 315 ppmgas out, CO₂ 31 ppmNaOH 100kg/m³liquid out
Figure 4 — Dilute CO₂ is scrubbed from rising air by falling caustic; with $p^\ast=0$ the whole tower is a simple $N_{OG}=\ln(y_1/y_2)$ problem.

Approach. For a dilute gas with $p^\ast=0$, integrate the differential gas-phase balance over the height to relate $K_Ga$ to the number of transfer units.

  1. Molar gas flux. The gas is 99.97% air, so $G_m=\dfrac{G}{M_{air}}=\dfrac{0.34}{28.97}=1.174\times10^{-2}$ kmol/m²s.
  2. Differential balance. Over height $dZ$, $G_m\,dy=-K_Ga\,(yP_T-p^\ast)\,dZ$. With $p^\ast=0$ this separates to $\dfrac{dy}{y}=-\dfrac{K_Ga\,P_T}{G_m}dZ$.
  3. Integrate over the packing. $$\ln\!\frac{y_1}{y_2}=\frac{K_Ga\,P_T}{G_m}Z\;\Rightarrow\; N_{OG}=\ln\!\frac{315}{31}=2.32.$$
  4. Solve for the coefficient. $$K_Ga=\frac{G_m\ln(y_1/y_2)}{P_T\,Z}=\frac{1.174\times10^{-2}(2.32)}{101.3(3)}=\boxed{8.95\times10^{-5}\ \text{kmol/(m}^3\text{s}\cdot\text{kPa)}}.$$

Check (units): $K_Ga$ is a volumetric coefficient, so its units are kmol/(m³·s·kPa); the “m²” printed in the question is read as the per-unit-packed-volume “$a$” (interfacial area per m³) folded into $K_Ga$.

QuantityResult
Molar gas flux $G_m$$1.174\times10^{-2}$ kmol/m²·s
Transfer units $N_{OG}$2.32
Overall coefficient $K_Ga$$8.95\times10^{-5}$ kmol/(m³·s·kPa)