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Flood hydrograph from 6-hr UH, two storms

Short answer

Separate constant baseflow40 m³/s, derive6-h UH from DRH (effective rainfall5.02 cm). For12-h storm with3.3 cm then5.5 cm effective rain, superposition gives flood hydrograph peaking at638 m³/s at30 h.

Following are the ordinates of hydrograph from a catchment area of 770 km2^2 due to 6-hr rainfall. Derive the ordinates of flood hydrograph due to 3.3 cm and 5.5 cm effective rainfall of duration 12-hr.

t (hr)061218243036424854606672
Discharge (m3^3/s)4065215360400350270205145100705040

Assumptions: the 6-h hydrograph is the response to one 6-h effective-rainfall pulse. A 12-h effective storm is two successive 6-h pulses, the first giving 3.3 cm and the second 5.5 cm. The base flow of 40 m3^3/s (the initial and final flow) is constant.

Step 1: Direct runoff and 6-h UH

DRH = Q - 40. Sum of DRH ordinates =1790= 1790 m3^3/s.

V=1790×6×3600=38.66×106 m3,d=VA=38.66×106770×106=5.02 cmV=1790\times6\times3600 = 38.66\times10^6\ \text{m}^3,\qquad d=\frac{V}{A}=\frac{38.66\times10^6}{770\times10^6}=5.02\ \text{cm}

So the observed storm gave about 5.02 cm of effective rainfall, and the 6-h UH is U=DRH/5.02U = DRH/5.02.

Step 2: Superposition

Q(t)=40+3.3 U(t)+5.5 U(t−6)Q(t)=40+3.3\,U(t)+5.5\,U(t-6)
t (h)Observed QDRHUH (m3^3/s per cm)3.3 U(t)5.5 U(t-6)Flood Q (m3^3/s)
04000.000.00.040.0
665254.9816.40.056.4
1221517534.85115.027.4182.4
1836032063.73210.3191.7442.0
2440036071.69236.6350.5627.1
3035031061.74203.7394.3638.1
3627023045.80151.2339.6530.7
4220516532.86108.4251.9400.4
4814510520.9169.0180.7289.7
541006011.9539.4115.0194.4
6070305.9719.765.7125.4
6650101.996.632.979.4
724000.000.011.051.0
78--0.000.00.040.0

Answer: The flood hydrograph peaks at about 638 m3^3/s at 30 h; ordinates are in the last column.

✓ Answer checked by Claude Sonnet 5.5

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