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Consider an evaporation pond. A Feed water containing solute A enters the open p

ID: 528200 • Letter: C

Question

Consider an evaporation pond. A Feed water containing solute A enters the open pond, and exits at the same rate. Solute A transfers from the feed water in the open pond to the surrounding air which is essentially free of the solute. At these conditions, the individual liquid-film mass-transfer coefficient, k_L, is 5 times 10^-4 kgmol/(m^2 s (kgmol/m^3)) the individual gas-film mass-transfer coefficient, k_G is 0.01 kg mol/(m^2 s atm). The concentrations are in the Henry's law region where p_Ai = H_cAi with H = 10 atm/(kgmol/m^3). (a) Using two-film theory, calculate the the overall coefficient in the liquid phase, K_L? (b) What is the percent resistance to mass transfer in the liquid film?

Explanation / Answer

Given, kL = 5 * 10-4 kgmole/(m2 s (kgmole/m3))

kG = 0.01 kgmole/(m2 s atm)

m = Henry's constant = 10 atm/(kgmole/m3)

(a) Overall coefficient in the liquid phase 1/KL = 1/kL + 1/(m.kG)

= 1/(5 * 10-4) + 1/(10 * 0.01)

= 2000 + 10

= 2010

Therefore, KL = 1/2010

= 4.975 * 10-4 kgmole/(m2 s (kgmole/m3))

(b) Percent resistance to mass transfer in the liquid film = (Resistance in liquid film / Total resistance in both the films) * 100 %

= (1/kL / 1/KL) * 100 %

= 1/(5 * 10-4) / 1/(4.975 * 10-4) * 100 %

= 2000 / 2010.05 * 100 %

= 99.5 %