The 3-h UH is converted to a 1-h UH because:

Storm durations 4 h and 3 h are not both multiples of 3 h

2073 Shrawan · 10 marks

Peak flow from 3-h UH, two successive storms

Short answer

The 3-h UH was converted to a 1-h UH by the S-curve method, then convolved with rainfall excess: 2 cm/h for the first 4 h and 1 cm/h for the next 3 h. Peak flow is approximately 76.5 m³/s, occurring 7 h after commencement.

The 3 h unit hydrograph of a basin with an area of 20 km $^2$ at one hour interval are as given below: 0, 0.41, 1.38, 4, 7.72, 10.06, 9.24, 6.62, 4.57, 3.86, 2.76, 2.07, 1.38, 0.83, 0.41, 0 (m $^3$ /s). If rainfall excess with intensity of 2.0 cm/h for a period of 4 h followed immediately by another 3 h storm with an intensity of 1 cm/h occurs on the basin, what is the peak flow produced by this rainfall and at what time after the commencement of rainfall would this peak flow occur? Assume baseflow is negligible.

Check of the UH: $\sum U\times3600=55.31\times3600=199\,116$ m $^3$ , which is 0.996 cm over 20 km $^2$ , so it is a 1 cm, 3-h UH.

The storm consists of 4 h at 2 cm/h followed by 3 h at 1 cm/h. These durations are not multiples of 3 h, so the 3-h UH is first converted to a 1-h UH by the S-curve method.

Step 1: S-curve

$S(t)=U_3(t)+S(t-3)$ (response to a continuous 1 cm per 3 h = 0.333 cm/h).

Step 2: 1-h UH

U_1(t)=\frac{3}{1}\left[S(t)-S(t-1)\right]

Step 3: Hourly effective rainfall

Hours 1 to 4: 2 cm each; hours 5 to 7: 1 cm each.

Q(t)=\sum_{j=0}^{6} P_j\,U_1(t-j),\qquad P=(2,2,2,2,1,1,1)\ \text{cm}

t (h)	$U_3$	S(t)	$U_1$	Q (m $^3$ /s)
0	0	0.00	0.00	0.00
1	0.41	0.41	1.23	2.46
2	1.38	1.38	2.91	8.28
3	4	4.00	7.86	24.00
4	7.72	8.13	12.39	48.78
5	10.06	11.44	9.93	67.41
6	9.24	13.24	5.40	75.30
7	6.62	14.75	4.53	76.50
8	4.57	16.01	3.78	70.44
9	3.86	17.10	3.27	64.14
10	2.76	17.51	1.23	53.34
11	2.07	18.08	1.71	39.84
12	1.38	18.48	1.20	28.53
13	0.83	18.34	0.00	19.86
14	0.41	18.49	0.00	14.10
15	0	18.48	0.00	8.61
16	0	18.34	0.00	4.14
17	0	18.49	0.00	2.91
18	0	18.48	0.00	1.20
19	0	18.34	0.00	0.00

The $U_1$ ordinates after 12 h are rounding noise and are taken as zero. The total ER is $2\times4+1\times3=11$ cm.

Answer: Peak flow $\approx 76.5$ m $^3$ /s, occurring 7 h after the commencement of rainfall (at the end of the 7-h storm).

✓ Answer checked by Claude Sonnet 5.5

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Peak flow from 3-h UH, two successive storms

Step 1: S-curve

Step 2: 1-h UH

Step 3: Hourly effective rainfall

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