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 is $796.875 m^3/hour$

1. capacity of tank by assuming detention period of 1 hour

∴capacity of tank=(maximum quantity sewage*detention time)

$=797*1=797m^3$

1. Provide the effective depth of tank as 0.3 m

surface area of the tank=$\frac{max. quantity of sewage}{effective depth}$

$=\frac{797}{3} sqm$

D=diameter of the tank

=$\frac{797*4}{3*π}^\frac{1}{2}$

D=18.4m

surface area of tank=$\frac{π*D^2}{4}=\frac{π*18.4^2}{4}$

=$265.9m^2$

4.surface loading of tank$=\frac{797*24}{265.9}=71.93m^3/m^2/day$

$71.93m^3/m^2/day\gt50 m^3/m^2/day$ here unsafe

so depth of the tank has to be reduced and surface area has to be increased

1. effective depth as 2.0 m

surface area of the tank$=\frac{797}{2}$

diameter of the tank(D)=$\frac{797*4}{2*π}^\frac{1}{2}$

D=22.5m

surface area of the tank$=\frac{π*D^2}{4}==\frac{π*22.5^2}{4}=397.6m^2$

surface loading=$\frac{797*24}{397.6}=48.1m^3/m^2/day$

$48.1m^3/m^2/day\lt50m^3/m^2/day$ hence safe

so, provide circular sedimentation tank of 22.5 m diameter and 2 m as effective water depth with overflow weir around in circumference