Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 4 (Q.No. 28)
28.
The total observed runoff volume during a 4 hour storm with a uniform intensity of 2.8 cm/hr is 25.2 x 106 m3 from a basin of 280 km2 area. What is the average infiltration rate for the basin ?
Discussion:
4 comments Page 1 of 1.
Chhaya said:
8 years ago
Total volume water run off per hour=(25.2*10^6)/4 = 6300000 m3
Per hour volume of rain fall= 0.028*280*1000*1000 = 7840000
Total infiltration per hour= 7840000-6300000 = 1540000
Infiltration water per m2 area to per hour = 1540000/(280*1000*1000)
= 5.5*10^-3
= 5.5 mm/hr
So ans is D.
Per hour volume of rain fall= 0.028*280*1000*1000 = 7840000
Total infiltration per hour= 7840000-6300000 = 1540000
Infiltration water per m2 area to per hour = 1540000/(280*1000*1000)
= 5.5*10^-3
= 5.5 mm/hr
So ans is D.
Gaja said:
8 years ago
Option D is correct.
(1)
Khushwant rana said:
7 years ago
Precipitation depth , P = Intensity*duration.
= 2.8*4 = 11.2 cm,
= 11 2 mm.
Runoff dept, R, = Runoff volumeArea.
= 25.2*106280*106m,
= 90*10-3 m = 90 mm.
Avg. infiltration rate = 5.5 mm/hr.
= 2.8*4 = 11.2 cm,
= 11 2 mm.
Runoff dept, R, = Runoff volumeArea.
= 25.2*106280*106m,
= 90*10-3 m = 90 mm.
Avg. infiltration rate = 5.5 mm/hr.
(1)
Priyanka said:
5 years ago
I think D is the correct option.
(3)
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