Question

13.9-2 (From Geankpolis 4th edition) A cellulose acetate-membrane with an area of 4.0 x 10-3m^2 is used at 25 degC to determine the permeability constants for reverse osmosis of a feed salt solution containing 12.0 kg NaCl/m^3(density = 1005.5 kg/m^3). The product solution has a concentrationof 0.468 kg NaCl/m^3(density = 997.3 kg/m^3). The measured product flow rate is 3.84 x 10^-8 m^3/s and the pressure difference used is 56.0 atm. Calculate the permeability constants and the solute rejection R.

Answer 1

The total salt from across the membrane (N 3) N s = 0.468 times kg/m^3 times 3.84 times 19 ^- 8 m^3/s/4.0 times 10^- 3 m^2 = 4.4923 times kg/m^2 The rejection R R = C s1 - C s2/Cs1 = 12.0.468/12 = 0.961 The totlal water for N w = (3.84 times 10^- 8 m^3/s) 1 times 10^3 kg id 20/m^3/4.0 times 10^- 3 m^2 = 9.5 times 10^- 3 kg H 20 /m^2 pi 1 = c s1 RT = (12 g/2) (mal/58 g)(2 ions/ mol) (82.05 times 10^- 3 atm.L/mol cdot K) (298K) 10.1176 atm pi 2 = (0.498 g/2)(mol/58 g) (ions/mol)(32.05 times 10^- 3)(298) 0.42 atm K = HW/ sp - spi = 9.6 times 10 - 3/56 - 9.69 = 2.07 times 10^- 4 kg H 2 O/m^2 atm permeability coeffciant K = 2.07 times 10^- 4 kg H 2 O/m^2 cdot atm