Question

Problem: The following data are given for an osmosis process. The universal gas constant R = 8.314 k Pa m/kg mole K, Temperature of the RO process T= 350 K, E X; = 700 kg mole K ,X = 7000 PPM, X = 10000 PPM, Mp=50 kg, Mr = 200 kg. Find the osmotic pressure, Salt rejection and permeate recovery

Answer 1

Ans: 1 Given values are. universal gas constant R = 8.314 k Patm?) kg moke k Temperature of the Ro Poocess T: 350k EXi = 700 kg mole k Xp = 7000 PPM xf = 10000 PPM Mp = sokg Не : Эoo K9 Now, we have to find i) osmotic pressuse 2.) Salt rejection 3:) Deo meate Becovery T = iMRT 1.) osmotic pressure = EX;X RXT = 700*8.314 x 350 - 2036930 kipa osmatic pressure TT - 2036930 K.Pas

9) Soll Rejection SR: 1- xp 1 - 7000 10000 = 1- to 1-07 : 0:31. :: Salt Rejection sR 0.34 3) Permeate Recovery = / Peomeat flow L. X I flow 100 feed PR - Cho) 2100 ) *100 X100 = 0.25 x 100 Permeate Recovery = 951 i u osmatic Poessure (TT) - 2036930 k pa g) Salt Rejection (SR) - 03/ 3) Peomerte Recovery (PR) :25)