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23-Chem-A3 Heat and Mass Transfer · December 2013

Question 4 of 6: Overall Mass-Transfer Coefficients for a Gas Absorber

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

Notes on this paper

National Exams — December 2013 — 04-Chem-A3 Mass Transfer Operations. Three-hour, open-book exam; any non-communicating calculator permitted. Format: six questions in two parts — answer two of Q1–Q3 (Part A) and two of Q4–Q6 (Part B); four questions of equal value constitute a complete paper. All six 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 enthalpies used in Q5.

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

Question 4: Overall Mass-Transfer Coefficients for a Gas Absorber (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 counter-current absorber (subscript 1 = bottom/gas-rich end, 2 = top). A linear equilibrium line $y^*=12x$ lets us use log-mean driving forces.

QuantityValue
Equilibrium$y^*=12.0\,x$
$y_{A1}$ (bottom) / $y_{A2}$ (top)0.04 / 0.0065
$x_{A2}$ (top, inlet liquid)0.0002
$G_s$ / $L_s$25 / 398 kmol/m²·h
Packed height $Z$2.0 m

Find. $K_ya$ and $K_xa$ (kmol/m³·h).

liquid mole fraction x_Agas mole fraction y_Ay* = 12.0xbottom (1)top (2)operating line
Figure 4 — Operating line between the two column ends lies above the equilibrium line $y^*=12x$; the vertical gap is the gas-phase driving force $(y-y^*)$.

Approach. Close the solute balance for the unknown bottom liquid composition, form the log-mean driving forces on each phase, get the numbers of transfer units, and back out $K_ya=G\,N_{OG}/Z$ and $K_xa=L\,N_{OL}/Z$.

  1. Outlet liquid composition. A solute balance (dilute, $G\approx G_s$, $L\approx L_s$) gives $x_{A1}=x_{A2}+\dfrac{G_s}{L_s}(y_{A1}-y_{A2}) = 0.0002+\dfrac{25}{398}(0.0335)=\boxed{0.00230}.$
  2. Gas-phase driving forces. $y_1^*=12(0.00230)=0.0276$ and $y_2^*=12(0.0002)=0.0024$, so $\Delta y_1=0.04-0.0276=0.01235$ and $\Delta y_2=0.0065-0.0024=0.0041$. The log mean is $(\Delta y)_{lm}=\dfrac{0.01235-0.0041}{\ln(0.01235/0.0041)}=0.00748.$
  3. Gas-phase transfer units and $K_ya$. $N_{OG}=\dfrac{y_{A1}-y_{A2}}{(\Delta y)_{lm}}=\dfrac{0.0335}{0.00748}=4.48$, hence $$K_ya=\frac{G\,N_{OG}}{Z}=\frac{25\times4.48}{2.0}=\boxed{56.0\ \text{kmol/m}^3\text{h}}.$$
  4. Liquid-phase driving forces. $x_1^*=y_{A1}/12=0.00333$, $x_2^*=y_{A2}/12=0.000542$, so $\Delta x_1=0.00333-0.00230=0.00103$, $\Delta x_2=0.000542-0.0002=0.000342$, and $(\Delta x)_{lm}=\dfrac{0.00103-0.000342}{\ln(0.00103/0.000342)}=0.000623.$
  5. Liquid-phase transfer units and $K_xa$. $N_{OL}=\dfrac{x_{A1}-x_{A2}}{(\Delta x)_{lm}}=\dfrac{0.00210}{0.000623}=3.38$, so $$K_xa=\frac{L\,N_{OL}}{Z}=\frac{398\times3.38}{2.0}=\boxed{672\ \text{kmol/m}^3\text{h}}.$$ As a check the two coefficients carry the same interfacial load: $K_ya(\Delta y)_{lm}=K_xa(\Delta x)_{lm}=0.419$ kmol/m³·h.
QuantityResult
Outlet liquid $x_{A1}$0.00230
$N_{OG}$ / $N_{OL}$4.48 / 3.38
$K_ya$≈ 56.0 kmol/m³·h
$K_xa$≈ 672 kmol/m³·h
Check
The dilute approximation ($G\approx G_s$, $L\approx L_s$, constant flows) is used because the largest mole fraction is only 0.04. Retaining the true inert (solute-free) fluxes would shift $K_ya$ and $K_xa$ by under 3%.