Civil Engineering - Waste Water Engineering - Discussion
Discussion Forum : Waste Water Engineering - Section 1 (Q.No. 4)
4.
If 2% solution of a sewage sample is incubated for 5 days at 20°C and depletion of oxygen was found to be 5 ppm, B.O.D. of the sewage is
Discussion:
16 comments Page 2 of 2.
Asaithambi said:
9 years ago
Considering total quantity of sewage =100%, solution of sewage = 2%(sample), depletion of oxygen = 5ppm.
Dilution factor = 100/2 = 50.
B.O.D = 50 * 5 = 250ppm.
Dilution factor = 100/2 = 50.
B.O.D = 50 * 5 = 250ppm.
Roshan said:
1 decade ago
P = Dilution factor = Fraction of waste water in the sample of 300 ml BOD bottle = 2/100 = 0.02.
BOD = Oxygen consumed/P = 5/0.02 = 250 ppm.
BOD = Oxygen consumed/P = 5/0.02 = 250 ppm.
Dinesh Sharma said:
1 decade ago
BOD = Oxygen consumed x dilution factor. So, BOD = 5x100/2 = 250ppm.
Nilima said:
1 decade ago
Dilution ratio = 100/% of solution.
= 100/2.
= 50.
B.O.D = 5 x 50 = 250 ppm.
= 100/2.
= 50.
B.O.D = 5 x 50 = 250 ppm.
Satya said:
1 decade ago
2% solution bod 5 ppm.
100% solution 100/2*5 = 250 ppm.
100% solution 100/2*5 = 250 ppm.
Md seraj said:
1 decade ago
D.O consume = 5 ppm.
Dilution factor = 100/2 = 50.
B.O.D of the sewage = 5*50 = 250 ppm.
Dilution factor = 100/2 = 50.
B.O.D of the sewage = 5*50 = 250 ppm.
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