written 5.7 years ago by
teamques10
★ 64k
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• modified 3.9 years ago |
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.
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written 5.7 years ago by
teamques10
★ 64k
|
• modified 5.7 years ago |
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
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written 5.7 years ago by
teamques10
★ 64k
|
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
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