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

Question 7 of 7: Wetted-Wall Stripping of TCE from Water

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

Notes on this paper

National Exams — May 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. Property data (diffusivities, vapour pressures, psychrometric enthalpies) are stated in each Given block; the psychrometric appendix supplied with the exam furnishes the saturated-air data used in Q6.

Reference texts: Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular diffusion, convective mass transfer, absorption and humidification; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — gas absorption, wetted-wall and cooling-tower design; Welty, Wicks, Wilson & Rorrer, Fundamentals of Momentum, Heat and Mass Transfer — boundary-layer and falling-sphere correlations; supporting property/psychrometric data from Perry's Chemical Engineers' Handbook (9th ed.).

Question 7: Wetted-Wall Stripping of TCE from Water (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 dilute wetted-wall stripping column with two film resistances in series; a very volatile solute (large Henry constant) shifts nearly all resistance to one phase.

QuantityValue
Column$D=4.0$ cm, height 2 m
Flowsair 2000 cm³/s, water 50 cm³/s
Liquid$\rho_L=998.2$, $\mu_L=9.93\times10^{-4}$, $D_L=8.90\times10^{-10}$ m²/s
Gas$\rho_G=1.19$, $\mu_G=1.84\times10^{-5}$, $D_G=8.08\times10^{-6}$ m²/s
Henry constant $H$$3.05\times10^{4}$ atm·m³/kmol

Find. (a) $k_L$; (b) $k_G$; (c) $K_L$ and the controlling phase.

wetted-wall column (D = 4 cm, 2 m)fallingwater filmair up (TCE stripped out)TCE transfers fromwater film to air
Figure 8 — Wetted-wall column: water films down the inside wall while air rises through the core; dissolved TCE transfers from the liquid film, across the interface, into the air. Both film resistances act in series.

Approach. Evaluate each film with $Sh=0.023\,Re^{0.83}Sc^{1/3}$ (the specified liquid $Re$, and a bulk-gas $Re$ from the air velocity), then combine the two resistances in series into $K_L$, converting the gas coefficient to compatible units through $H$.

  1. (a) Liquid film. Water mass rate $w=\rho_L Q_L = 998.2(5\times10^{-5}) = 0.0499$ kg/s, so $Re_L=\dfrac{4w}{\pi D\mu_L}=1600$ and $Sc_L=\dfrac{\mu_L}{\rho_L D_L}=1118.$ Then $Sh_L=0.023\,Re_L^{0.83}Sc_L^{1/3}=109$, giving $$k_L=\frac{Sh_L D_L}{D}=\frac{109(8.90\times10^{-10})}{0.04}=\boxed{2.4\times10^{-6}\ \text{m/s}}.$$
  2. (b) Gas film. Air velocity $v_G=Q_G/(\pi D^2/4)=1.59$ m/s, so $Re_G=\dfrac{\rho_G v_G D}{\mu_G}=4117$ and $Sc_G=\dfrac{\mu_G}{\rho_G D_G}=1.91.$ Then $Sh_G=0.023\,Re_G^{0.83}Sc_G^{1/3}=28.6$, giving $$k_G=\frac{Sh_G D_G}{D}=\frac{28.6(8.08\times10^{-6})}{0.04}=\boxed{5.8\times10^{-3}\ \text{m/s}}.$$
  3. (c) Combine the resistances. Put the gas coefficient on a partial-pressure basis, $k_{G,p}=k_G/RT=2.4\times10^{-4}$ kmol/m²·s·atm, so $H k_{G,p}=3.05\times10^{4}(2.4\times10^{-4})=7.3$ m/s. Then $$\frac{1}{K_L}=\frac{1}{k_L}+\frac{1}{Hk_G}=\frac{1}{2.4\times10^{-6}}+\frac{1}{7.3}=4.1\times10^{5}+0.14\ \text{s/m}.$$
  4. Overall coefficient and controlling phase. The gas term (0.14) is negligible beside the liquid term ($4.1\times10^{5}$), so $$K_L\approx k_L=\boxed{2.4\times10^{-6}\ \text{m/s}},$$ i.e. the transfer is essentially liquid-phase controlled (the liquid film carries >99.99% of the resistance).
QuantityResult
(a) $k_L$$2.4\times10^{-6}$ m/s
(b) $k_G$$5.8\times10^{-3}$ m/s
(c) $K_L$$2.4\times10^{-6}$ m/s
Controlling phaseliquid-phase controlled
Check
The huge Henry constant means TCE strongly favours the gas, so an equal gas-film velocity removes solute from the interface far faster than the liquid film can supply it; the gas resistance $1/(Hk_G)$ is therefore negligible. This conclusion is robust to the exact liquid correlation because $k_L\sim10^{-6}$ m/s is six orders of magnitude below $Hk_G\sim1$ m/s.
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