Civil Engineering - Waste Water Engineering - Discussion
Discussion Forum : Waste Water Engineering - Section 4 (Q.No. 6)
6.
3.0 ml of raw sewage is diluted to 300 ml. The D.O. concentration of the diluted sample at the beginning of the test was 8 mg/l. After 5 day-incubation at 20°C, the DO. concentration was 5 mg/l. The BOD of raw sewerage is
Discussion:
12 comments Page 1 of 2.
Jahangir Abbas Sahil said:
4 years ago
DO = Dissolved Oxygen.
DO I= Initial = 8 mg/l,
DOF = Final = 5 mg/l.
DF= Dilution Factor.
Formula of BOD:
BOD = (DO I - DO F) * DF ---> (1)
DF = Total Vol/Vol of Sample
DF = 300/3
DF = 100
So, eq(1) becomes:
BOD = (8-5)*100.
BOD = 300 mg/l.
DO I= Initial = 8 mg/l,
DOF = Final = 5 mg/l.
DF= Dilution Factor.
Formula of BOD:
BOD = (DO I - DO F) * DF ---> (1)
DF = Total Vol/Vol of Sample
DF = 300/3
DF = 100
So, eq(1) becomes:
BOD = (8-5)*100.
BOD = 300 mg/l.
(7)
Bhabani said:
4 years ago
Thanks @Pravin @Riju.
(1)
Praveen kumar said:
1 decade ago
At first.
Bod = 8*(300/3) = 800.
2nd:
Bod = 5*(300/3) = 500.
Now 800-500 = 300mg/l.
Bod = 8*(300/3) = 800.
2nd:
Bod = 5*(300/3) = 500.
Now 800-500 = 300mg/l.
Mahi said:
8 years ago
Thank you @Praveen.
Sapna said:
7 years ago
Thanks @Praveen.
Raj said:
7 years ago
Thanks @Praveen.
Sumit Narayan Bose said:
7 years ago
Thanks @Praveen.
Kasturi said:
7 years ago
Thank you @Praveen.
Atanu said:
6 years ago
Thank you for explaining the answer @Praveen.
Shree said:
6 years ago
Thanks for explaining @Praveen.
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