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The return period for the annual maximum flood of a given magnitude is 8 years. The probability that this flood
magnitude will be exceeded at least once during the next 5 years is :
The following steps are involved in arriving at a unit hydrograph :
(1) Estimating the surface runoff in depth
(2) Estimating the surface runoff in volume
(3) Separation of base flow
(4) Dividing the surface runoff ordinates by the depth of runoff. The correct sequence of the steps is :
The flood hydrograph due to rainfall of intensity 1 cm/h is given by
Q = 2  (1 + t) e^{2t}
(t is in hours and Q is in m^{3}/s).
Find the area of catchment ?
The number of revolution of a current meter is 50 seconds were found to be 12 and 30 corresponding to the velocities of 0.25 m/s and 0.46 m/s respectively. What velocity (in m/s) would be indicated by 50 revolutions of that current meter in 1 minute ?
(Important  Enter only the numerical value in the answer)
The probability that a 10 year return period flood will occur atleast once in the next 4 year is :
If the base period of a 6 hr unit hydrograph (UH) of a basin is 84 hours, then a 12 hr unit hydrograph derived from this 6 hr unit hydrograph will have a base period of :
To obtain 12 hr UH, we lag 6hr UH by another 6 hours hence, base period of 12 hr UH is 84 + 6 = 90 hr.
A 6 hr storm has 6 cm of rainfall and the resulting runoff of 3 cm. If φ index remains at the same value,
which one of the following is the runoff due to 12 cm of rainfall in 9 hr in the catchment ?
Linked Questions 09 and 10
The drainage area of a watershed is 50 Km^{2}. The φindex is 0.5 cm/hr and the base flow at the outlet is 10m^{3}/s. One hour unit hydrograph of a watershed is triangular in slope with the time base of 20 hr. The
peak ordinate occurs at 10 hours. 9.
Q. The peak ordinate (in m^{3}/s) of the unit hydrograph is :
Linked Questions 09 and 10
The drainage area of a watershed is 50 Km^{2}. The φindex is 0.5 cm/hr and the base flow at the outlet is 10m^{3}/s. One hour unit hydrograph of a watershed is triangular in slope with the time base of 20 hr. The peak ordinate occurs at 10 hours.
Q. For the storm of depth 6.5 cm and duration 1 hr, the peak ordinate (in m^{3}/s) of the flood hydrograph is :
A 3 hour storm over a water shed had an average depth of 27mm. The resulting flood hydrograph was found to have a peak flow of 200 m^{3}/s and a base flow of 20 m^{3}/s. If the loss rate could be estimated as 0.3 cm/h, a 3h unit hydrograph for this water shed will
The 3 hr unit hydrograph U_{1} of a catchment of area 250 km^{2} is in the form of a triangle with peak discharge of 40 m^{3}/s. Another 3 hour unit hydrograph U_{2} is also triangular in shape and has the same base width as U_{1} but with a peak flow of 80m^{3}/s. The catchment which U_{2} refers to has an area of :
The storm hydrograph was due to 3h of effective rainfall. It contained 6cm of direct runoff. The ordinates of DRH of this storm :
The return period that a designs must use in the estimation of a flood for a hydraulic structure, if he is willing to accept 20% risk that the flood of that or higher magnitude will occur in the next 10 year, is
A reservoir with surface area of 250 hectares has saturation vapour pressure at water surface = 17.54 mm of Hg and actual vapour pressure of air = 7.02mm of Hg. Wind velocity at 1m above the ground surface = 16 km/hr. Estimate the average daily evaporation from the lake using Meysis formula……….mm/days.
(Take K_{M} = 0.36)
How is the average velocity along the vertical in a wide stream obtained?
Average velocity in a wide stream can be calculated by measuring the velocity at half the depth.
Average velocity in a deep stream can be calculated by averaging the velocity at 0.2 and 0.8 times depth from surface.
The intensity of rainfall time interval of a typical storm are :
The maximum intensity of rainfall for 20 minute duration of storm is:
The coefficient of variation of the rainfall for six rain gauge stations in a catchment was found to be 29.54% The optimum number of stations in the catchment for an admissible 10% error in the estimation of mean rainfall will be:
A 4hr storm had 4 cm of rainfall and the resulting direct runoff was 2 cm. If the φindex remain at the same value, the runoff due to 10 cm of rainfall in 8 hr in the catchment is ?
Consider the following chemical emulsions :
1. Methyl Alcohol
2. Cetyl Alcohol
3. Stearye Alcohol
4. Kerosene
Which of the above chemical emulsions is/are used to minimize the loss of water through the process of evaporation ?
The average surface area of a reservoir in the month of June is 20 km^{2}. In the same month, the average rate of inflow is 10 m^{3}/s, outflow rate is 15 m^{3}/s, monthly rainfall is 10 cm, monthly seepage loss is 1.8 cm and the storage change is 16 million m^{3}. The evaporation (in cm) in that month is:
Concept:
Water budget method:
By applying mass balance around water body and by writing water budget equation, evaporation is estimated.
Water budget equation around water body is
Σinflow  Σoutflow = ± ΔS
Σinflow includes rainfall and runoff
Σoutflow includes water used, evaporation and seepage
ΔS = change in storage
ΔS is positive when water level rises.
ΔS is negative when water level falls.
Calculation:
Let ‘x’ cm evaporation takes place in month of June.
Total inflow in cm
=10×30×24×60×6020×106×100+10
= 139.6 cm
Total outflow in cm
=15×30×24×60×6020×106×100+1.8+x
= (196.2 + x) cm
As total outflow is more than total inflow, therefore depression in storage takes place.
Depression in storage (ΔS)
ΔS=−16×10620×106×100=−80cm
Σinflow  Σoutflow = ± ΔS
139.6  (196.2 + x) =  80
– x =  80 + 56.6
∴ Evaporation in month of June = x = 23.4 cm
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