Question 12 on the second morning of 2016: The inner diameter of the packing absorption tower is 900 mm
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As shown in the figure, my calculated answer seems to differ a bit!
I calculated that Kya = 156.7874 kmol/m3·h, and 48.5 kg of propylene was recovered. Please have Haiyou take a look.
Kya=156.7874 kmol/m3.h. How was this value calculated? When I asked later, the value given was 268
I’ll share my approximate steps: the molar fraction of propane in the mixture is 0.049; thus, Y1 = 0.049/0.951 = 0.0515. The molar fraction of propane in the exhaust gas is 2.62×10^(-3), so Y2 = 0.00262/(1–0.00262) = 0.00263. The absorption by fresh water results in X2 = 0. There are 60 g of propane in the liquid at the bottom of the tower; the molar mass of propane is 58, so X1 = 60/58/((1000–60)/18) = 0.0198. Y* = 2X; therefore, Nog = 7.9622 and Hog = 6/Nog = V/(Kya×0.9). Here, V needs to be converted to standard conditions, and using the molar volume of gases at standard conditions of 22.4 L/mol, we get V = 106333.89 mol. Thus, Kya = 156787.36 (mol/m²·h). I hope someone can help me check for any mistakes. Thank you!
The last edit to this post was made by yeti on 2017-9-11 at 15:10. The amount of inert gas in HOG is (106.333/22.4) * (1-0.049) = 101.1 KMOL/h； And Hog=6/Nog=V/（Kya*0.785*0.9*0.9）
It should be that Hog=6 and Nog=V/(Kya*0.785*0.9^2). V=101.3*2600*(1-0.049)/8.314*298=101.1. Substituting this value into the formula gives Kya=211 (mol/m2.h). The amount of propylene is 101.1*（Y1-Y2）*58=286.56 kg
It should be that Hog=6 and Nog=V/(Kya*0.785*0.9^2). V=101.3*2600*(1-0.049)/8.314*298=101.1. Substituting this value into the formula gives Kya=211 (mol/m2.h). The amount of propylene is 101.1*（Y1-Y2）*58=286.56 kg
I’m too weak; I even didn’t convert the diameter to area.:) Thank you to the two fellow users above
This question is similar to question 5-29 from that association*, but the answers seem to differ a bit when calculated