1 h triangular unit hydrograph of a watershed has the peak discharge of 60m3-s at 10 h and time base of 30h- The - - index is 0-4 cm-h and base flow is 15 m3-sThe catchment area of the watershed is

For UH, depth of effective rainfall = 1 cm nbsp; Volume of runoff = A × 10^ 2m^3 Volume of runoff = Area of UH nbsp; A × 10^ 2=12× 60× (30× 3600) A = ...

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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 $ϕ$ - index is 0.4 cm/h and base flow is 15 $m^{3} / s$

The catchment area of the watershed is

3.24 k m^{2}

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32.4 k m^{2}

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324 k m^{2}

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3240 k m^{2}

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Solution

The correct option is C $324 k m^{2}$
For UH, depth of effective rainfall = 1 cm

Volume of runoff = $A \times 10^{- 2} m^{3}$

Volume of runoff = Area of UH

$A \times 10^{- 2} = \frac{1}{2} \times 60 \times (30 \times 3600)$

$A = 324 \times 10^{6} m^{2} = 324 k m^{2}$

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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 correct option is C 324 km2For UH, depth of effective rainfall = 1 cm Volume of runoff = A×10−2m3 Volume of runoff = Area of UH A×10−2=12×60×(30×3600) A=324×106 m2=324 km2

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

The correct option is C $324 k m^{2}$
For UH, depth of effective rainfall = 1 cm

Volume of runoff = $A \times 10^{- 2} m^{3}$

Volume of runoff = Area of UH

$A \times 10^{- 2} = \frac{1}{2} \times 60 \times (30 \times 3600)$

$A = 324 \times 10^{6} m^{2} = 324 k m^{2}$