Civil Engineering - GATE Exam Questions - Discussion
Discussion Forum : GATE Exam Questions - Section 4 (Q.No. 5)
5.
In a BOD test, 5 ml of waste is added to 295 ml of aerated pure water. Initial dissolved oxygen (D.O) content of the diluted sample is 7.8 mg/l. After 5 days of incubation at 20°C, the D.O. content of the sample is reduced to 4.4 mg/l. The BOD of the waste water is :
Discussion:
14 comments Page 1 of 2.
Lencho Edeto said:
3 years ago
BOD = [DO( initial) -DO final)] * Dilution factor.
= (7.8-4.4)mg/lit * (295/5).
= 200.6 mg/lit.
= (7.8-4.4)mg/lit * (295/5).
= 200.6 mg/lit.
Sharath Manohar said:
4 years ago
295/5 * (7.8-4.4) = 200 mg/l is the actual right answer.
Abhishek thakur said:
4 years ago
The volume of dilution sample over the undiluted raw sample is a formula not aerated water over an undiluted raw sample.
204 is the right answer.
204 is the right answer.
(1)
Abhishek thakur said:
4 years ago
The volume of dilution sample over the undiluted raw sample is a formula not aerated water over an undiluted raw sample.
204 is the right answer.
204 is the right answer.
(1)
Abhishek said:
4 years ago
Df = 295÷5
Bod = (intial Do-final DO) * D.F,
= 200.6.
Bod = (intial Do-final DO) * D.F,
= 200.6.
Sachin s Naik said:
6 years ago
The correct answer is 200.6mg/l.
Manisha nimje said:
6 years ago
The correct answer is 204 mg/l.
Keshav Kaushik said:
7 years ago
Dilution Factor = volume of wastewater sample/(volume of wastewater sample + dilution sample).
B.O.D = Initial D.O - Final D.O / dilution factor = 204mg/l.
B.O.D = Initial D.O - Final D.O / dilution factor = 204mg/l.
Himanshu said:
8 years ago
BOD = (initial D.O - Final D.O.)/ Dilution factor.
= (7.8 - 4.4)/(5/300).
= 204 mg/l.
= (7.8 - 4.4)/(5/300).
= 204 mg/l.
(2)
Erick said:
8 years ago
The first initial DO of the mixture is required.
DO = 7.8 x (295/300) = 7.67,
BOD = (7.67-4.4)/(5/300) = 196.2 mg/L.
DO = 7.8 x (295/300) = 7.67,
BOD = (7.67-4.4)/(5/300) = 196.2 mg/L.
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