The first step in deriving the UH from the observed flow is:

Subtract the 10 m^3/s baseflow to get the DRH

The effective rainfall depth is found from:

Volume of DRH divided by basin area

2078 Bhadra · 8 marks

Derive 4-hr UH from observed streamflow

Short answer

Subtract baseflow (10 m³/s) to get DRH; effective rainfall depth is 2.124 cm (volume 12.744×10⁶ m³ over 600 km²). Divide DRH ordinates by 2.124 to obtain 4-h UH; peak 103.6 m³/s at 8 h.

Observed streamflow in a catchment (Area = 600 km $^2$ ) from a storm event of 4 hours' duration tabulated hereunder. If baseflow is 10 m $^3$ /s, derive 4 hr unit hydrograph.

Time (hr)	0	4	8	12	16	20	24	28	32	36	40	44	48
Observed flow (m $^3$ /s)	10	100	230	190	130	100	70	60	50	30	20	15	10

Method: subtract the baseflow (10 m $^3$ /s) to get the direct runoff hydrograph (DRH), find the volume and the effective rainfall depth, and divide the DRH ordinates by this depth to get the 4-h UH (ordinates per 1 cm).

t (h)	Observed Q (m $^3$ /s)	DRH = Q - 10	UH ordinate = DRH / 2.124
0	10	0	0.00
4	100	90	42.37
8	230	220	103.58
12	190	180	84.75
16	130	120	56.50
20	100	90	42.37
24	70	60	28.25
28	60	50	23.54
32	50	40	18.83
36	30	20	9.42
40	20	10	4.71
44	15	5	2.35
48	10	0	0.00
Sum		885

Direct runoff volume

V=\sum DRH\times\Delta t = 885\times4\times3600 = 12,744,000\ \text{m}^3 = 12.744\times10^6\ \text{m}^3

Effective rainfall depth over the catchment

d=\frac{V}{A}=\frac{12,744,000}{600\times10^6}=2.124\ \text{cm}

Unit hydrograph

U(t)=\frac{DRH(t)}{2.124}

Check: $\sum U\times\Delta t = 416.7\times14400 = 6,000,000$ m $^3$ = 1 cm over 600 km $^2$ (6,000,000 m $^3$ ).

Answer: The 4-h UH has a peak of 103.6 m $^3$ /s at 8 h and ordinates as shown in the last column (0, 42.4, 103.6, 84.8, 56.5, 42.4, 28.2, 23.5, 18.8, 9.4, 4.7, 2.4, 0 m $^3$ /s at 0, 4, ..., 48 h).

✓ Answer checked by Claude Sonnet 5.5

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