Civil Engineering - GATE Exam Questions - Discussion
Discussion Forum : GATE Exam Questions - Section 4 (Q.No. 12)
12.
A direct runoff hydrograph due to an isolated storm with an effective rainfall of 2 cm was trapezoidal in shape as shown in the below figure. The hydrograph corresponds to a catchment area (in sq. km) of
Discussion:
7 comments Page 1 of 1.
Yesika said:
6 years ago
Area of given DRH = volume of direct run off.
[(1/2)(70+4) * (60)*(60) * (90) ] m3=11988000 m3,
Effective rainfall = 2 cm,
Catchment area = Volume of DRH/Effective rainfall.
11988000 / (2 * 10-2) m2 = 599.4 m2.
[(1/2)(70+4) * (60)*(60) * (90) ] m3=11988000 m3,
Effective rainfall = 2 cm,
Catchment area = Volume of DRH/Effective rainfall.
11988000 / (2 * 10-2) m2 = 599.4 m2.
Keshav Kaushik said:
7 years ago
How come 900 and why 5 has been used?
Reshma said:
7 years ago
Why 900, is it 90 cumecs right?
(1)
Sarvesh said:
8 years ago
1/2*(70+4)*900*60*60*(10)^(-6) = 199.88 * 5 = 599.4.
Akhil said:
8 years ago
Area =1/2 * sum of parallel side *height.
= 1/2*(70+4) * 900 *60*60*(10)^(-6),
60*60 is conversion hour to secs 10^(-6) sqm to sqkm.
= 1/2*(70+4) * 900 *60*60*(10)^(-6),
60*60 is conversion hour to secs 10^(-6) sqm to sqkm.
RAKESH KUMAR said:
9 years ago
Can anyone explain even more clearly?
Pulkit said:
9 years ago
AREA = 90 * 1/2 * (70 + 4) * 60 * 60 * 10^(-6).
60 * 60 to convert hr to secs, 10^(-6) for sqm to sq km.
60 * 60 to convert hr to secs, 10^(-6) for sqm to sq km.
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