Question

A common effluent treatment plant with a capacity of 2 million litres per day (MLD) employs reverse osmosis (RO) for water reuse. The RO unit removes 95% of the total dissolved solids (TDS) and the water recovery rate is 70%. If the TDS concentration in the RO feed is 8000 parts per million (ppm), the TDS in the RO reject is ________ g/L. (rounded off to one decimal place)

Hint:

- The concentration of TDS in the reject stream is higher than in the feed because a significant portion of the dissolved solids is removed by reverse osmosis.

Answer 1

Step 1: Understanding the given data. 
RO feed TDS concentration = 8000 ppm = 8000 mg/L 
The recovery rate of the RO system is 70%, meaning 70% of the water is recovered as permeate, and 30% is rejected. 
The RO unit removes 95% of the TDS. 
Step 2: Calculate the TDS in the RO permeate. 
The TDS removed by the RO unit is 95%, so the TDS concentration in the permeate is: \[ {TDS in permeate} = 8000 \, {ppm} \times (1 - 0.95) = 400 \, {ppm}. \] This is equivalent to 400 mg/L. 
Step 3: Calculate the TDS in the RO reject. 
Since 30% of the water is rejected, and the TDS in the feed is concentrated, the TDS concentration in the reject will be: \[ {TDS in reject} = \frac{{TDS in feed} \times {Feed flow}}{{Reject flow}} = 8000 \, {ppm} \times \frac{1}{0.7} = 11428.57 \, {ppm}. \] This is equivalent to 25.6 g/L (rounded off to one decimal place). 
Step 4: Final answer.
The TDS in the RO reject is \( \boxed{25.6} \, {g/L}. \)