Question

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1) is desired to remove from liquid water in a single operating isletstream team of 300 kg when the molt stream 50 km and i t O of the by with The The is est od the Fouwing Toolsportowe out (1 ) Moll fraction is in exiting or stream (1 Tot exiting flow me d. Moltre of in exiting (0 Henry's the equilibr i ) ) X ) ! 2) By means of places, served from its with airby volatile b ol These co mode perto the estering is een free of the one in the vaporwa r d ed the exiting liquid contains 10 wele e Thevar inte strewe is 100 meles. The p r opisy (Seraphi Determine the use of ideal per repaired for the separation propically (1 ) Find ) d. Find (1 Find Vou ( ) Lo ) VI

Answer 1

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page2

Xx, Lal 2 , / ₂ Yz G- only air flow rate Ls - Onlywater an X = 0.02 = 0.0204 Gi= 200 kmolshor 4 = 450 kmollhr Y = 0, 1 = (1-0.9 Kg = 0.02 Y = 0 . ! = Bill! 12 1-0.02 - V = (1+0.7) Y, = 0.370. X -mole ratio = 0.033 ammoniá in lig Water inlig Gs=2008.0.9 = 180 kmilftor. air in gas Ls = 450 x0,98 = 441 kmol/bor solute balance to (X₂ – X) = +Gs (Y ₂ -t) 441 (0,0204 - xs) = 180 (0.033 -0.11) Xi = 0.05182 mole natio of ammone Y ammonia in gas 0,0204 I water at exit ! X = 0.049 = 4.9%

gas inlet - 200 kmolshor ( 180 + 20 ) air NH Yo = 1+0.033 0.033 – 0.03194 1 +0.033 G₂ ( 1.- H₂) = 1.80 s. air is not getting absorbed, so at exit G - 180 . I also ais will be 180 (1-0.03194) gas flowrate at exit = 185.94 kmithra Bi mole fraction of ammónia at exit gas stream de Y2 = 0.03194 c. water tig inlet = 450 kmolster ( 441 +9) X = 0.02 X; = 0,05182 (calculated from solate balanc) d) at exit lig mole fraction x = 0.049 water ammount L = at exit . 41 (1-0.049) = 441 (won't change, glig flowrate 4 = 463.72 kmolster

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