For a catchment area of 120 km2, the equilibrium discharge in m3/hour of an S-curve obtained by the summation of 6 hour unit hydrograph is
A. 0.2 x 106
B. 0.6 x 106
C. 2.4 x 106
D. 7.2 x 106
Answer: Option A
Solution (By Examveda Team)
The correct answer is Option A: 0.2 x 106The question involves understanding S-curve hydrographs and their relationship to unit hydrographs.
An S-curve hydrograph is created by summing an infinite series of unit hydrographs, each lagged by a duration equal to the unit hydrograph duration.
In this case, we have a 6-hour unit hydrograph.
The equilibrium discharge of the S-curve represents the constant rate of runoff when the rainfall continues indefinitely at a rate of 1 cm (or 1 unit) over the basin for the duration of the unit hydrograph.
The equilibrium discharge (Qeq) is calculated as:
Qeq = (Area of catchment x Rainfall depth) / Duration of Unit Hydrograph
Where, Area = 120 km2 = 120 x 106 m2
Rainfall depth = 1 cm = 0.01 m
Duration = 6 hours
Qeq = (120 x 106 m2 x 0.01 m) / 6 hours
Qeq = (1.2 x 106 m3) / 6 hours
Qeq = 0.2 x 106 m3/hour
Therefore, the equilibrium discharge of the S-curve is 0.2 x 106 m3/hour.
This Question Belongs to Civil Engineering >> Irrigation Engineering
A=120 km²=120 *10^6 m²
t=6 hr
h=1 cm= 0.01 m (unit hydrograph)
Q=V/t
=(120*10^6*0.01)/6
=0.2*10^6 m³/hr
Jsk
Q= A/D *10 to the power 4 bcz m3/hr
Where D =duration of time
A= Area of catchment
Q= 120/6 *10 4= 0.2*10 to the power 6
Explain please