#### A reverse osmosis system is designed to produce $50 \ \text{m}^3/\text{day}$ of permeate water from seawater containing 35,000 ppm of dissolved salts. The system has a water recovery rate of 70% and the applied pressure is 20 bar. If the membrane has a salt rejection coefficient of 0.97, what is the concentration of salts in the permeate water?Option: 1 1050 ppmOption: 2 1330 ppmOption: 3 1500 ppmOption: 4 1750 ppm

The water recovery rate can be defined as the ratio of the permeate flow rate to the feed flow rate:

Water recovery rate = $\frac{\text{Permeate flow rate}}{\text{Feed flow rate}} \times 100%$ .

Given,

$\\\text{feed flow rate} = 50 \text{m}^3/\text{day}, \\ \text{salt concentration in feed = 35,000 ppm,} \\ \text{water recovery rate} = 70%, \\\ \text{rejection coefficient} = 0.97, \\ \text{and applied pressure = 20 bar.}$

We can calculate the permeate flow rate as:

$\text{Permeate flow rate }= \text{Feed flow rate} \times \text{Water recovery rate} = 35 \ \text{m}^3/\text{day}$

Using the salt rejection coefficient, we can calculate the concentration of salts in the permeate as:

$C_\text{permeate} = C_\text{feed} \times (1 - \text{Rejection coefficient})$

$= 35,000 \times (1 - 0.97) = 1,050 \ \text{ppm}$

Therefore, the correct option is (1).