Author: Prof.Parimal Pal, PhD, FRSC Ref: Membrane-based Technologies for Environmental Pollution Control, Parimal Pal, Elsevier Science, New York, 2020 Case Studies on water treatment by membrane technology • Case 1: Calculation of osmotic pressure of salt solution • Calculate the osmotic pressure of a solution containing 0.10 gmol NaCl/1000 g H2O at 250C. • Solution: • For dilute water solutions, • • where, • n = the number of kmol of solute • Vm = the volume of pure solvent water in m3 associated with n kmol of solute • R = the gas law constant = 82.057 × 10 –3 (m3.atm/kmol.K) • T = temperature, K If a solute exists as two or more ions in solution, n represents the total number of ions. • NaCl gives two ions (Na+ and Cl–). • Quantity of pure solvent water = 1000g = 1 kg • Density of water at 25 Deg C = 997 kg/m3 • From the above equation, The osmotic pressure of the solution containing 0.1 gmol NaCl/1 kg water at 250C = 4.87 atm. Problem 2: Experimental determination of membrane permeability • Experiments at 250C were performed to determine the permeability of a cellulose acetate membrane. The laboratory test section shown in figure has membrane area A = 2.00×10 –3 m 2 . The inlet feed solution concentration of NaCl is c1 = 10.0 kg NaCl/m 3 solution (10.0 g NaCl /L, ρ 1 = 1004 kg solution /m 3 ). The water recovery is assumed low so that the concentration c1 in the entering feed solution flowing past the membrane and the concentration of the exit feed solutions are essentially equal. The product solution contains c2 = 0.39 kg NaCl/m 3 solution ( ρ 2 = 997 kg solution /m 3 ) and its measured flow rate is 1.92 1.92 ×10 –8 m 3 solution/s. A pressure differential of 5514 kPa (54.42 atm) is used. Calculate the permeability constants of the membrane and the solute rejection R. Solution: The Flux through the membrane: • Ns = the solute (salt) flux, kg solute/m 2 .s • Ds = the diffusivity of solute in membrane, m 2 /s • As = the solute permeability constant, m/s • c1 = the solute concentration in upstream or feed (concentrate) Solution, kg solute/m 3 • c2 = the solute concentration in downstream or product (permeate) solution, kg solute/m 3 At steady state, the solute diffusing through the membrane = the amount of solute leaving in the downstream or product (permeate) solution. Where, • Nw = the solvent (water) flux, kg/m 2 .s • cw2 = the concentration of solvent in stream 2 (permeate), kg solvent/m 3 • Since the concentration c1 in the entering feed solution flowing past the membrane and the concentration of the exit feed solutions are essentially equal, assuming c2 is very low (dilute solution). Therefore, the value of cw2 can be assumed as the density of solvent (here water). Therefore, cw2 = 997 kg water/ m 3 Calculation of water (solvent) flux: • The membrane area, A = 2.00×10 –3 m 2 • The product flow rate = 1.92 ×10 –8 m 3 solution/s (given) • The product flow rate ≈ water flow rate (volumetric) Calculation of solute (NaCl) flux: • Since c 2 = 0.39 kg NaCl/ m 3 , from equation (2) From equation (1), • The solute permeability constant of the membrane (As) = 3.895×10 -7 m/s • Calculation of the solvent permeability constant of the membrane (Aw): Design parameters : • N w = the solvent (water) flux, kg/m 2 .s • P w = the solvent membrane permeability, kg solvent/m.atm.s • L m = the membrane thickness, m • A w = the solvent permeability constant, kg solvent/ m 2 .atm.s • ∆ P = hydrostatic pressure difference, atm = P 1 - P 2 • P 1 = pressure on feed side, atm • P 2 = pressure on product side, atm • ∆ π = osmotic pressure of feed solution – osmotic pressure of product solution, atm Calculation of osmotic pressure of feed solution ( π 1 ): • Basis: 1 m 3 solution • We know, • n 1 = the number of kmol of solute (NaCl) • V m1 = the volume of pure solvent water in m 3 associated with n kmol of solute • Molecular weight of NaCl = 58.5 • Since c 1 = 10.0 kg NaCl/m 3 solution As NaCl gives two ions (Na + and Cl – ) Since ρ 1 = 1004 kg solution /m 3 Pure water in 1 m 3 solution = 1004 – 10 = 994 kg Density of pure water at 25 0 C = 997 kg water / m 3 Since ρ 1 = 1004 kg solution /m 3 Pure water in 1 m 3 solution = 1004 – 10 = 994 kg Density of pure water at 25 0 C = 997 kg water / m 3 • R = the gas law constant = 82.057 × 10 –3 (m 3 .atm/kmol.K) • T = temperature, K = 289 K • From literature (obtained by linear interpolation), • • Therefore, the calculated value of π 1 is higher than the experimental value.