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.
Join The Discussion
Comments (4)
Related Questions on Irrigation Engineering
Finally formed berms in canals are provided for
A. Protection of banks erosion by the waves
B. Control of seepage losses
C. Strengthening of banks
D. All the above
Canals constructed for draining off water from water logged areas, are known
A. Drains
B. Inundation canals
C. Valley canals
D. Contour canals
A hydroelectric scheme operating under a head of 80 m will be classified as
A. low head scheme
B. medium head scheme
C. high head scheme
D. none of the above
A. Low sodium water
B. Medium sodium water
C. High sodium water
D. Very high sodium water

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