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A five day B.O.D. at 15°C of the sewage of a town is 100 kg/day. If the 5 day B.O.D. per head at 15°C for standard sewage is 0.1 kg/day, the population equivalent is

A. 100

B. 1000

C. 5000

D. 10000

Answer: Option D


This Question Belongs to Civil Engineering >> Waste Water Engineering

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Comments (1)

  1. Omar Rahat
    Omar Rahat:
    7 months ago

    Bod of 1 day= 0.1 kg but waste per day is 100 so bod = 100*0.1
    Waste per day is 100 so 5 days waste = 5*100
    And 5 days bod = 5 * 0.1( 0.1= 1 days bod).
    Total population = (bod per waste * waste per 5 day)/ 5 days bod.
    = (100*0.1 * 5 *100)/( 0.1* 5)
    = 10,000.

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