Question

A seaside community, with a population of 250,000 people, decides to build a seawater desalination plant to supply all of their water needs (approximately 70 liters of water per day per person). The proposed design models seawater entering the desalination plant as a 3.5% salt solution by mass, with a density of 1.029 kg per liter and an average temperature of 15°C. A concentrated brine (18.5% salt solution by mass) is discharged to the ocean as waste and the treated water (essentially salt free and allowed to cool for 6.5 hours) is sent out to the city for consumption. What is the required volume of seawater into the desalination plant that will be needed to supply the estimated city water demand for one day?

Answer 1

Basis:

1. Per person water consumption = 70 lt/day
2. For 250,000 people total consumption = 70 lt/day * 250,000 = 17500000
3. Considering Density if treated salt free water to be 1 kg/lt

We consider the basis of 70 lt/day per person consumption water for calculation. Then we would multiply the answer by 250,000.

Water consumption per person = 70 lt / day

W = 70 lt/day * Density of water consumed

W = 70 lt/day * 1 kg / lt

W = 70 kg/lt

Overall Mass Balance :

S = W + B

S - B = 70 ..... Equation 1

Material Balance of Salt

0.035 * S = 0* W + 0.185 B

S = 5.28*B .... Equation 2

Solving Eqaution 1 and 2 simultaneously

B = 16.33 kg/day

S = 86.24 kg/day

Sea water input for each person = 86.24 kg/day / density of sea water kg/lt

Sea water input for each person = 86.24 / 1.029

Sea water input for each person = 83.80 lt/day per person

Sea water input for 250000 people = 250000 * 83.80 lt/day per person

Sea water input for 250000 people = 2095*10⁴ lt/day