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 - - index is 0-4 cm-h and base flow is 15 m3-sIf there is rainfall of 5-4 cm in 1 hour the ordinate of the flood hydrograph at 15th hour is

Rainfall in 1 h = 5.4 cm Effective rainfall in 1 h nbsp; = 5.4 ϕ = 5.4 0.4 = 5 cm Ordinate of flood hydrograph = 5 [ordinate of UH at 15th hour] + base ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Other

Hydrology & Irrigation

Introduction and Basic Assumptions of Unit Hydrograph

1 h triangula...

Question

1 h triangular unit hydrograph of a watershed has the peak discharge of $60 m^{3} / s$ at 10 h and time base of 30h. The $ϕ - i n d e x$ is 0.4 cm/h and base flow is $15 m^{3} / s$

If there is rainfall of 5.4 cm in 1 hour the ordinate of the flood hydrograph at 15th hour is

225 m^{3} / s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

240 m^{3} / s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

249 m^{3} / s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

258 m^{3} / s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $240 m^{3} / s$

Rainfall in 1 h = 5.4 cm
Effective rainfall in 1 h
$= 5.4 - ϕ = 5.4 - 0.4 = 5 c m$

Ordinate of flood hydrograph = 5 [ordinate of UH at 15th hour] + base flow
$= 5 \times 45 + 15 = 240 m^{3} / s$

Suggest Corrections

Similar questions

Q. 1 h triangular unit hydrograph of a watershed has the peak discharge of 60

m^{3} / s

at 10 h and time base of 30h. The

ϕ

- index is 0.4 cm/h and base flow is 15

m^{3} / s

The catchment area of the watershed is

Q. The drainage area of a watershed is

50 k m^{2}

. The ϕ - index is 0.5 cm/hour and the base flow at the outlet is

10 m^{3} / 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

m^{3} / s

) of the hydrograph is

Q. The drainage area of a watershed is

50 k m^{2}

. The

ϕ - i n d e x

is 0.5 cm/hour and the base flow at the outlet is

10 m^{3} / 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

m^{3} / s

) of the unit hydrograph is

Q. An effective rainfall of 2 hour duration produced a flood hydrograph peak of

200 m^{3} / s .

The flood hydrograph has a base flow of

20 m^{3} / 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

m^{3} / s

up to one decimal place) is _____

Q. The peak of a flood hydrograph due to a 6-h storm is 470

m^{3} / 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 m^{3} / s

, then the peak discharge of 6-h unit hydrograph for this catchment is

Join BYJU'S Learning Program

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

The correct option is B 240 m3/s Rainfall in 1 h = 5.4 cm Effective rainfall in 1 h =5.4−ϕ=5.4−0.4=5 cm Ordinate of flood hydrograph = 5 [ordinate of UH at 15th hour] + base flow =5×45+15=240 m3/s

1 h triangular unit hydrograph of a watershed has the peak discharge of $60 m^{3} / s$ at 10 h and time base of 30h. The $ϕ - i n d e x$ is 0.4 cm/h and base flow is $15 m^{3} / s$

If there is rainfall of 5.4 cm in 1 hour the ordinate of the flood hydrograph at 15th hour is

The correct option is B $240 m^{3} / s$

Rainfall in 1 h = 5.4 cm
Effective rainfall in 1 h
$= 5.4 - ϕ = 5.4 - 0.4 = 5 c m$

Ordinate of flood hydrograph = 5 [ordinate of UH at 15th hour] + base flow
$= 5 \times 45 + 15 = 240 m^{3} / s$