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.
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.
Dinesh Sharma said:
1 decade ago
BOD = Oxygen consumed x dilution factor. So, BOD = 5x100/2 = 250ppm.
Nahid Farjana said:
8 years ago
DF= 2%= 0.02.
DO= 5ppm,
BOD= consumed DO/DF.
= 5/0.02= 250.
DO= 5ppm,
BOD= consumed DO/DF.
= 5/0.02= 250.
(1)
Satya said:
1 decade ago
2% solution bod 5 ppm.
100% solution 100/2*5 = 250 ppm.
100% solution 100/2*5 = 250 ppm.
Bruno said:
6 years ago
(Initial do - final do )*dilution factor.
5*50 = 250.
5*50 = 250.
(2)
Raaj said:
7 years ago
Thank you all for explaining it.
(1)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers