Civil Engineering - GATE Exam Questions - Discussion
Discussion Forum : GATE Exam Questions - Section 5 (Q.No. 4)
4.
An outlet irrigates an area of 20 ha. The discharge (l/s) required at this outlet to meet the evapotranspiration requirement of 20 mm occurring uniformly in 20 days neglecting other field losses is
Discussion:
4 comments Page 1 of 1.
Prashanth said:
3 years ago
Q=V/t =A*D/t.
Q=20*10^4X0.02/20
Q=200m^3/day.
But the question asked is discharge (L/s) so we should convert to L/s from m^3/day..
Q=200*1000/86400,
Q=2.31
Q=20*10^4*10^3*0.02/20*86400.
Q=2.31L/s.
Where,
Q-discharge.
V-volume.
Q=20*10^4X0.02/20
Q=200m^3/day.
But the question asked is discharge (L/s) so we should convert to L/s from m^3/day..
Q=200*1000/86400,
Q=2.31
Q=20*10^4*10^3*0.02/20*86400.
Q=2.31L/s.
Where,
Q-discharge.
V-volume.
RAKESH KUMAR said:
6 years ago
Can anyone explain this properly?
Snehal Wankhede said:
7 years ago
Q = A * depth/days = 2.31.
Elvis Presley said:
7 years ago
D = 8.64 *20/.02
= 8640 ha/cumec.
Q = A/D.
= 20/8640.
= 2.31* 10^-3 cumec.
= 8640 ha/cumec.
Q = A/D.
= 20/8640.
= 2.31* 10^-3 cumec.
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