Treated water flows to a reverse osmosis membrane process Th

Treated water flows to a reverse osmosis membrane process. The water contains 900 mg/l TDS. If it is 97.5% rejected at a recovery of 70%, what is the brine concentration and flow?

Solution

Feed flow= Permeate flow + Brine flow---------(1)

Also Q_f X C_f = ( Q_p X C_p ) + (Q_b X C_b )----------(2)

Where

Qf- feed flow

Qp- Permeate flow

Qb- Brine flow

Cf- Feed salt concentration=900 mg/l

Cp-Permeate salt concentration

Cb-Brine salt concentration

Let Qf be one litre. Qf= 1 litre

Then Qp= 1 X 70/100 = 0.7 litre       (with 70% recovery)

Qb= 1-0.7= 0.3 litre    {using (1)}

Salt passage %= 100 X Cp/Cf

Salt rejetion % = {1-(Cp/Cf)} 100= 97.5 (given)

Salt passage= Cp/Cf= 0.025

Cp=0.025 Cf

Substitituting these values in eqn (2)

1X900 = 0.7 X (0.025X900) + (0.3 X Cb)

Solving, Brine Salt concentration Cb= 2947.5 mg/l

Brine flow will be 0.3 times the volume of feed flow