Introduction and Basic Assumptions of Unit Hydrograph
Trending Questions
Q. A 6-hour Unit Hydrograph (UH) of a catchment is triangular in shape with a total time base of 36 hours and a peak discharge of 18 m3/s. The area of the catchment (in sq. km) is
- 233
- 117
- 1.2
- Sufficient information not available
Q.
What do you mean by integral value?
Q. A 4 h unit hydrograph of a catchment is triangular in shape with base of 80 h. The area of the catchment is 720 km2 . The base flow and ϕ index are 30 m3/s and 1 mm/h, respectively. A storm of 4 cm occurs uniformly in 4 h over the catchment.
The peak flood discharge due to the storm is
The peak flood discharge due to the storm is
- 210 m3/s
- 230 m3/s
- 260 m3/s
- 720 m3/s
Q. A 4-hour unit hydrograph of a basin can be approximated as a triangle with a base period of 48 hours and peak ordinate of 300 m3/s. What is the area of the catchment basin?
- 7776 km2
- 5184 km2
- 2592 km2
- 1294 km2
Q. If the base period of a 6-hour hydrograph of a basin is 84 hours, then a 12 hours unit hydrograph derived from this 6 hour unit hydrograph will have a base period of
- 72 hours
- 78 hours
- 84 hours
- 90 hours
Q. A direct-runoff hydrograph due to isolated storm was triangular in shape with a base of 80 h and peak of 200 m3/s. If the catchment area is 1440 km2, the effective rainfall of the storm is
- 20 cm
- 10 cm
- 5 cm
- 2 cm
Q. A 3 hr unit hydrograph U1 of a catchment of area 235 km2 is in the form of a triangle with peak discharge 30 m3/s. Another 3 hr unit hydrograph U2 is also triangular in shape and has the same base width as U1, but has a peak flow of 90 m3/s.
What is the catchment area of U2?
What is the catchment area of U2?
- 117.5 km2
- 235 km2
- 470 km2
- 705 km2
Q. If a 4-hour unit hydrograph of a certain basin has a peak ordinate of 80 m3/s, the peak ordinate of a 2-hour unit hydrograph for the same basin will be
- equal to 80 m3/s
- greater than 80 m3/s
- less than 80 m3/s
- between 40 m3/s to 80 m3/s
Q. 1 h triangular unit hydrograph of a watershed has the peak discharge of 60m3/s at 10 h and time base of 30h. The ϕ - index is 0.4 cm/h and base flow is 15 m3/s
The catchment area of the watershed is
The catchment area of the watershed is
- 3.24 km2
- 32.4 km2
- 324 km2
- 3240 km2
Q. A 252 km2 catchment area has a 6 hr UH which is a triangle with time base of 35 hours. What is the peak discharge of the DRH due to 5cm effective rainfall in 6hr from that catchment?
- 45 cumecs
- 115 cumecs
- 200 cumecs
- 256 cumecs
Q. The ordinates of a 2 h unit hydrograph at 1 h intervals starting from time = 0 are 0, 3, 8, 6, 3, 2, and 0 m3/s. Use trapezoidal rule for numerical integration, if required.
What is the catchment area represented by the unit hydrograph ?
What is the catchment area represented by the unit hydrograph ?
- 1.00 km2
- 2.00 km2
- 7.92 km2
- 8.64 km2
Q. 1 h triangular unit hydrograph of a watershed has the peak discharge of 60 m3/s at 10 h and time base of 30h. The ϕ−index is 0.4 cm/h and base flow is 15 m3/s
If there is rainfall of 5.4 cm in 1 hour the ordinate of the flood hydrograph at 15th hour is
If there is rainfall of 5.4 cm in 1 hour the ordinate of the flood hydrograph at 15th hour is
- 225 m3/s
- 240 m3/s
- 249 m3/s
- 258 m3/s
Q. The peak of a flood hydrograph due to a 6-h storm is 470 m3/s. The mean depth of rainfall is 8.0 cm. If average infiltration loss is 0.25 cm/h and a constant base flow of 15 m3/s, then the peak discharge of 6-h unit hydrograph for this catchment is
- 58.25 m3/s
- 60.65 m3/s
- 70 m3/s
- 72.3 m3/s
Q. The peak of a flood hydrograph due to 2 hour duration isolated storm in a catchment area is 135m3/s. The total depth of rainfall is 54 mm. Assume a constant base flow of 10m3/s and ϕ−index to be equal to 4 mm/hr, then peak of 2-hr UH for the catchment in m3/s is
- 25.0
- 29.23
- 23.15
- 27.17
Q. A 2-hour unit hydrograph can be approximated as trapezoidal as shown in figure. The unit hydrograph refers to catchment of area
- 138.24 km2
- 0.0384 km2
- 384 km2
- 3840 m2
Q. The peak magnitude of a flood hydrograph during 4 hr study duration over a catchment is 300 m3/s. The total depth of rainfall is 6 cm, and the infiltration loss during the said 4 hr period is 2 cm. A constant uniform base flow of 20 m3/s is premised throughout. The peak value of the corresponding 4 hr unit hydrograph is
- 75 m3/s
- 70 m3/s
- 50 m3/s
- 40 m3/s
Q. An effective rainfall of 2 hour duration produced a flood hydrograph peak of 200 m3/s. The flood hydrograph has a base flow of 20 m3/s. If the spatial average rainfall in the watershed for the duration of storm is 2 cm and the average loss rate is 0.4 cm/hour, the peak of 2-hour unit hydrograph (in m3/s up to one decimal place) is _____
- 150
Q. The peak ordinate of 4 h unit hydrograph of a basin is 60 m3/s. An isolated storm of 4 hour duration in the basin was recorded to have a total rainfall of 8 cm. If it is assumed that the base flow and the ϕ−index are 30 m3/s and 0.25 cm/hr respectively, the peak of the flood discharge due to the storm could be estimated as.
- 210 m3/s
- 420 m3/s
- 450 m3/s
- 240 m3/s
Q. Which of the following principles relate to a Unit Hydrograph?
- The hydrographs of direct runoff due to effective rainfall of equal duration have the same time base.
- Effective rainfall is not uniformly distributed within its duration.
- Effective rainfall is uniformly distributed throughout the whole area of drainage basin.
- Hydrograph of direct runoff from a basin due to a given period of effective rainfall reflects the combination of all the physical characteristics of the basin.
- 1, 2 and 3
- 1, 2 and 4
- 2, 3 and 4
- 1, 3 and 4
Q. A unit hydrograph for a watershed is triangular in shape with base period of 20 hours. The area of the watershed is 500 ha. What is the peak discharge in m3/hour?
- 7000
- 6000
- 5000
- 4000
Q. Ordinates of a 1-hour unit hydrograph at 1 hour intervals, starting from time t = 0 , are 0, 2, 6, 4, 2, 1 and 0 m3/s
Catchment area represented by this unit hydrograph is
Catchment area represented by this unit hydrograph is
- 1.0 km2
- 2.0 km2
- 4.0 km2
- 5.4 km2
Q. The following four hydrological features have to be estimated or taken as inputs before one can compute the flood hydrograph at any catchment outlet:
- Unit hydrograph
- Rainfall hydrograph
- Infiltration index
- Base flow
- 1, 2, 3, 4
- 2, 1, 4, 3
- 2, 3, 1, 4
- 4, 1, 3, 2
Q. The drainage area of a watershed is 50 km2. The ϕ - index is 0.5 cm/hour and the base flow at the outlet is 10 m3/s. One hour unit hydrograph (unit depth = 1 cm) of the watershed is triangular in shape with a time base of 15 hours. The peak ordinate occurs at 5 hours.
For a storm of depth of 5.5 cm and duration of 1 hour, the peak ordinate (in m3/s ) of the hydrograph is
For a storm of depth of 5.5 cm and duration of 1 hour, the peak ordinate (in m3/s ) of the hydrograph is
- 55.00
- 82.60
- 92.60
- 102.60
Q. A two-hour storm hydrograph has 5 units of direct runoff. The two-hour unit hydrograph for this storm can be obtained by dividing the ordinates of the storm hydrograph by
- 2
- 2/5
- 5
- 5/2
Q. A peak flow of a flood hydrograph due to a six-hour storm is 470 m3/s. The corresponding average depth of rainfall is 8 cm. Assume infiltration loss of 0.25 cm/hour and a constant base flow of 15 m3/s. What is the peak discharge of 6 hour unit hydrograph for this catchment?
- 60 m3/s
- 70 m3/s
- 80 m3/s
- 90 m3/s
Q. A rainfall with an intensity of 1.5 cm/h occurred over a catchment area in a 3 hour storm, resulting in a peak discharge of 210 m3/s in the river. The base flow and average storm loss ϕ−index were 25 m3/s and 8 mm/hour respectively. The peak of the 3-hour unit hydrograph is ______ m3/s. (Upto two decimal places).
- 88.095
Q. Viewing watershed as a system, which one of the following assumptions is made in the Unit Hydrograph theory?
- Non-linearity
- Both linearity and time variance
- Both time invariance and non-linearity
- Both linearity and time invariance
Q. The drainage area of a watershed is 50 km2. The ϕ−index is 0.5 cm/hour and the base flow at the outlet is 10m3/s. One hour unit hydrograph (unit depth = 1 cm) of the watershed is triangular in shape with a time base of 15 hours. The peak ordinate occurs at 5 hours.
The peak ordinate (in m3/s) of the unit hydrograph is
The peak ordinate (in m3/s) of the unit hydrograph is
- 10.00
- 18.52
- 37.03
- 185.20
Q. Consider the following steps which are involved in arriving at a unit hydrograph:
1. Separation of base flow
2. Estimating the surface runoff in volume
3. Estimating the surface runoff in depth
4. Dividing surface runoff ordinate by depth of runoff
Which is the correct sequence of these steps?
1. Separation of base flow
2. Estimating the surface runoff in volume
3. Estimating the surface runoff in depth
4. Dividing surface runoff ordinate by depth of runoff
Which is the correct sequence of these steps?
- 4, 3, 2 and 1
- 1, 2, 3 and 4
- 4, 2, 3 and 1
- 1, 3, 2 and 4
Q. A 4 h unit hydrograph of a catchment is triangular in shape with base of 80 h. The area of the catchment is 720km2 . The base flow and ϕ−index are 30 m3/s and 1 mm/h, respectively. A storm of 4 cm occurs uniformly in 4 h over the catchment.
The peak discharge of 4 h unit hydrograph is
The peak discharge of 4 h unit hydrograph is
- 40 m3/s
- 50 m3/s
- 60 m3/s
- 70 m3/s