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Solve this problem from the picture from Advanced Water Resource / Hydrology topic

Solve this problem from the picture from Advanced Water Resource / Hydrology topic-example-1
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User Shautieh
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The estimated catchment area for a 5-year return period is 101520 m². Actual time of concentration is 5.05 min, yielding a peak runoff of 2036504 m³/s. Assuming te = 10 min introduces a 49.5% error in peak runoff.

(a) Estimate Catchment Area:

Given:

te = 10 min

T = 5 yr

S = 0.002

Calculate i from the IDF curve:

i = 7836 * (48.67 * 0.11 + (0.5895 + 5^-0.67))

i ≈ 25.23 mm/h

Now, substitute into the catchment area formula:

A = (10 * 25.23) / (0.0125 * 0.002)

A ≈ 101520 m²

(b) Actual Time of Concentration (tc) and Peak Runoff:

Use the catchment area obtained in part (a) and the calculated i to find tc:

tc = (0.0125 * 0.002 * 101520) / 25.23

tc ≈ 5.05 min

Now, calculate the peak runoff:

Q = 0.8 * 101520 * 25.23

Q ≈ 2036504 m³/s

(c) Percentage Error in Peak Runoff:

Assuming te = 10 min, find the estimated tc and Q:

tc_assumed = 10 min

Q_assumed = 0.8 * 101520 * 25.23

Calculate the percentage error:

Percentage Error = |(5.05 - 10) / 10| * 100

Percentage Error ≈ 49.5%

Considering a 49.5% error, it may be significant depending on the project's sensitivity to runoff variations. Further evaluation is needed based on project requirements.