Standard seawater is led to a reverse osmosis module with a cellulose acetate me

Question

Standard seawater is led to a reverse osmosis module with a cellulose acetate membrane. It is desired to achieve a solute rejection R of 99.5 %. The solvent permeability constant is Q_W = 5.0 times 10^-4 kg water/m^2.s.atm and the solute permeability constant is Q_s = 4.0 times 10^-7 m/s. The density of the feed solution is 1021.4 kg solution/m^3 and you may consider the product solution to be dilute. Calculate the pressure drop required in atm to achieve this separation. (tip: use a basis of 1 kg of feed solution).

Explanation / Answer

R = [Qw (P - )] / (Qs x ) ----->1

where R = Solute rejection = 99.5% = 0.995 (in terms of fraction)

Qw = Solvent permeability constant = 5.0 x 10-4 kg/m2.s.atm

P = Pressure drop

= Osmotic pressure difference = Osmotic pressure of feed solution (Standard salt water) - Osmotic pressure of product solution (0.5% salt solution) = 25.02 atm - 3.63 atm (Values taken from literature) = 21.39 atm

Qs = Solute permeability constant = 4.0 x 10-7 m/s

= Density of feed solution = 1021.4 kg/m3

Rearranging equation 1, we get

R = [Qw (P - )] / (Qs x )

P - = (R x Qs x ) / Qw

P - 21.39 atm = (0.995 x 4.0 x 10-7 m/s x 1021.4 kg/m3) / (5.0 x 10-4 kg/m2.s.atm)

P - 21.39 atm = 0.813 atm

P = 0.813 atm + 21.39 atm = 22.203 atm

Pressure drop = 22.203 atm

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