Question

The BOD5 of a waste has been measured 500 mg/l. If K= 0.26/day (base e), what is the

ultimate BOD, of waste? What proportion of BOD, remains unoxidised after 20 days?​

Answer 1

The answer is "687 mg/l"

Explanation:

BOD is the amount of oxygen required by the micro-organisms to degrade organic matter in the waste water

The formula to calculate BOD ultimate is

BOD ultimate = BOD5/(1-e^-kt)

Thus, it will become

BOD ultimate = 500/(1-e^-(0.26*5)

BOD ultimate = 687 mg/l

Now, we need to find BOD after 20 days. For that we will use

BOD20= BOD ultimate (1-e^-kt)

BOD20 = 687 (1-e^-0,26*20)

BOD20 = 683 mg/l

Answer 2

lo=687ppm

bod20=683.2ppm

Explanation:

bod5 =Lo(1-10^-kt); find Lo

bod20=Lo(1-10^-kt);find bod20