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
A. 100
B. 1000
C. 5000
D. 10000
Answer: Option D
The effluent of a septic tank is
A. Fit for discharge into any open drain
B. Foul and contains dissolved and suspended solids
C. As good as that from a complete treatment
D. None of these
The bottom of the sewage inlet chamber of septic tanks, is provided an outward slope
A. 1 in 5
B. 1 in 10
C. 1 in 15
D. 1 in 20
Drop manholes are the manholes
A. Without entry ladders
B. Without manhole covers
C. With depths more than 3.5 m
D. Having drains at different levels
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.