Calculate the population equivalent of city where average sewage from the city is 96*$10^6$ lit/day and avg BO$D_5$ is 300 mg/lit. Assume per capita BOD of sewage/day as 0.08kg/day/person. calculate the population equivalent.

Given:-avg BO$D_5$=300 mg/l

avg sewage($Q_s$)=95*$10^6$ lit/day

standard BO$D_5$=0.08 kg/day/person

Total BOD=Avg sewage * avg BO$D_5$

=$95*10^6$*300

=$2.85*10^3*10^7$ mg/day

=$2.85*10^4$ kg/day

Population Equivalent=$\frac{Industrial BOD_5}{Avg standard BOD_5}$

=$\frac{2.85*10^4}{0.08}$

=3,56,250

Given:-avg BO$D_5$=300 mg/s

avg sewage($Q_s$)=95*$10^6$ lit/day

standard BO$D_5$=0.08 kg/day/person

Total BOD=Avg sewage * avg BO$D_5$

=$95*10^6$*300

=$2.85*10^3*10^7$ mg/day

=$2.85*10^4$ kg/day

Population Equivalent=$\frac{Industrial BOD_5}{Avg standard BOD_5}$

=$\frac{2.85*10^4}{0.08}$

=3,56,250