Design a circular sewage sedimentation tank for town with a population of 50,000. The seaverage water demand is 170 liters/capacity/day. Assume that 75% water reaches at the treatment unit and maximum demand is 3 times the average demand.

Given data:-

Population of town=50,000

Average water demand = 170 litre/capacity/day

75% water reaches at the treatment unit

Maximum demand = 3*average demand

1. The avg. water requirement=170*50,000

$=8.5*10^6 liter/day=8.5*10^3m^3/day$

∴ Average quantity of sewage=$\frac{75*8.5*10^3}{100}$

$=6.375*10^3$

∴maximum quantity of sewage=3*average demand

$=3*6.37*10^3$

$=19.125*10^3 m^3/day$

$=\frac{19.125*10^3}{24}$

$=796.875m^3/h$

∴maximum quntity of sewage …