Problem. If 2% solution is used from sewage of 5 days at 20°C & dissolved oxygen depletion was 5 ppm, determine BOD5?
We Know that,
BOD5= {(D1─ D2)/P}
Here, D1
=Initial sample dissolved oxygen
D2 =Final sample dissolved oxygen
So, (D1─ D2)= 5
P = Volumetric sample of fraction used
= 2%
=0.02
BOD5 = ( 5/0.02)
=250 ppm
=250mg/l
Answer: 250mg/l
No comments:
Post a Comment