2016 Case Afternoon Question 7

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It’s a question that a colleague wrote down; it might relate to the strengths and weaknesses of our workplace. I’m posting it here for everyone to take a look at. -------------------- 7. In a counter-current packed recovery tower, component A of the mixture is absorbed using clean water. The molar fraction of A in the gas entering the tower is 0.03, and the absorption efficiency for A is 99%; the molar fraction of A in the liquid exiting the tower is 0.013. The operating pressure of the absorption tower is 101.33 KPa, and the temperature is 27°C. The equilibrium equation under these operating conditions is Y = 2X (where X and Y are molar fractions). Given that the flow rate of the inert gas per unit area is 54 kmol/(m2·h), and the total gas-phase absorption coefficient is 0.95 kmol/(m3·s·KPa), what is the required height of the packing in the absorption tower? (A) 1300 (B) 1500 (C) 1731 (D) 1742-------------------Y1=0.03/(1-0.03)=0.03093; Y2= Y1*(1-η)= 0.03093*0.99=0.00031 X1=0.013/ (1-0.013)=0.01317; X2=0 V(Y1-Y2)=L(X1-X2)→V *(0.03093-0.00031)=L*(0.01317-0) →V / L=0.01317/(0.03093-0.00031)=0.43 S=m* V / L = 2*0.43=0.86 NOG=*ln ； Where Y2*=m* X2=0=*ln=19.26 V/Ω=54 kmol/(m2*h)= 0.015 kmol/(m2*s); KGα=0.95 kmol/(m3*s*KPa), KY*α= KG*α*P=0.95*101.33=96.235 kmol/(m3*s); HOG=V/(KY*α*Ω)= 0.015/96.235=0.0001558 m ； Z = HOG * NOG = 19.26 * 0.0001558 = 0.003 m--------------------------------------------In the part I’ve marked in red, I think there’s clearly an issue with the units. KGα is clearly too large, or V/Ω is too small. What do you all think? Is it a problem with the question, or is there an issue with my answer?