Question

the BOD of a sewage incubated for one day at 300c has been found to be 100mg/l what will be the 5days 200c BOD assume k=0.12(base10)at 200c

Answer 1

The answer is i don't know

Answer 2

Answer: 197.5 mg/l

Explanation:

BOD at 30oC = 100 mg/L K20 = 0.12

K30 = K20 θ(T-20)   K 30 = 0.12 (1.056)30-20 = 0.207 per day

BODt = Lo (1 – 10-kt) 100 = Lo (1 – 10-0.207 x 1) Lo = 263.8 mg/L

Now BOD5 at 20oC can be calculated as: BOD5 at 20oC  = Lo (1 – 10-kt) = 263.8 (1 – 10- 0.12 x 5) = 197.5 mg/L