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  • A reverse osmosis system is designed to produce of pe

A reverse osmosis system is designed to produce 100 \ \text{m}^3/\text{day} of permeate water

from seawater containing 40,000 ppm of dissolved salts. The membrane has a salt rejection

coefficient of 0.98, and the applied pressure is 25 bar. If the feed solution is maintained at a

temperature of 25 \ ^\circ\text{C}, what is the osmotic pressure of the solution?

Option: 1

23.6 bar


Option: 2

25.3 bar


Option: 3

27.5 bar


Option: 4

29.2 bar


Answers (1)

best_answer

The osmotic pressure of the solution can be calculated using the Van 't Hoff equation: \pi = iMRT where i is the van't Hoff factor (1 for salts), M is the molar concentration, R is the gas constant, and T is the temperature.
Given, feed flow rate = 100 \ \text{m}^3/\text{day},

salt concentration in feed = 40,000 ppm,

rejection coefficient = 0.98, and

applied pressure = 25\ bar.

The effective pressure applied for reverse osmosis can be calculated as: P_\text{eff} = P_\text{applied} - \pi

where \pi  is the osmotic pressure.

Rearranging the above equation,

we get:\pi = P_\text{applied} - P_\text{eff}.

To calculate P_\text{eff}, we can use the following equation: P_\text{eff} = RT \times \text{ln} \left( \frac{1}{1 - C_\text{feed}/C_\text{permeate}} \right)

where C_\text{feed}\ and\ C_\text{permeate} are the concentrations of solute in the feed and permeate, respectively.

Given,T = 25 \ ^\circ\text{C},

P_\text{applied} = 25 \ \text{bar}, C_\text{feed} = 40,000 \ \text{ppm}, and \ C_\text{permeate} = 400 \ \text{ppm} (since the rejection coefficient is 0.98).

Substituting the values in the above equation, we get:

P_\text{eff} = 25 \times 10^5 \ \text{Pa} \pi = P_\text{applied} - P_\text{eff} = 0.25 \ \text{bar} \approx 0.24 \ \text{atm} \approx 23.6 \ \text{bar}

Therefore, the correct option is (a).

Posted by

avinash.dongre

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