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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 ppm


Option: 2

1330 ppm


Option: 3

1500 ppm


Option: 4

1750 ppm


Answers (1)

best_answer

 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).

Posted by

Pankaj Sanodiya

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