How is base flow chosen from the hydrograph?

Constant 40 m3/s, the discharge at start and end, subtracted from all ordinates

How are UH ordinates derived from the DRH?

Divide DRH by the runoff depth V/A = 5.52 cm

What is the combined equation and result?

Q = 8U(t) + 11U(t-6) + 40, peak about 1206 m3/s at 30 h

Q = 9.5U(t) + 12.5U(t-6) + 40, peak at 24 h

Q = 8U(t) + 11U(t) + 40 with no lag

2067 Mangsir (old course) · 16 marks

6-h UH and flood from two storms, 700 km2

Short answer

Baseflow 40 m³/s subtracted; direct runoff volume gives effective depth 5.52 cm. Scaling DRH yields 6-h UH (peak 65.2 m³/s per cm). For two successive storms 9.5 and 12.5 cm with 1.5 cm loss each, effective depths 8.0 and 11.0 cm; superposition with baseflow gives flood peak 1205.9 m³/s at 30 h.

The following are the ordinates of the hydrograph of flow from a catchment area of 700 km $^2$ due to a 6-h rainfall.

Time (Hour)	0	6	9	12	18	24	30	33	36	42	48	54	60	66	72
Discharge (m $^3$ /s)	40	65	140	215	360	400	350	330	270	205	145	100	70	50	40

a) Derive the ordinates of 6-h unit hydrograph. b) Calculate the flood hydrograph for two successive storms of 9.5 and 12.5 cm of 6 hours duration rainfall and an average storm loss of 0.25 cm/hr.

(a) Ordinates of the 6-h unit hydrograph

Base flow: the discharge is 40 m³/s at the start and at the end (t = 0 and 72 h), so a constant base flow of 40 m³/s is assumed and subtracted. Ordinates at 6-h intervals are used (the 9 h and 33 h readings are not on the 6-h grid).

Volume of direct runoff = (sum of DRH ordinates) × 6 h × 3600:

\begin{aligned} \sum DRH &= 1790 \text{ m}^3/\text{s} \\ V &= 1790 \times 6\times 3600 = 38,664,000 \text{ m}^3 \\ \text{depth} &= \frac{V}{A} = \frac{38,664,000}{700\times 10^6} = 0.0552 \text{ m} = 5.52 \text{ cm} \end{aligned}

UH ordinate = DRH ordinate / 5.52 (m³/s per cm):

t (h)	0	6	12	18	24	30	36	42	48	54	60	66	72
Observed Q	40	65	215	360	400	350	270	205	145	100	70	50	40
DRH = Q − 40	0	25	175	320	360	310	230	165	105	60	30	10	0
UH (m³/s per cm)	0.0	4.5	31.7	57.9	65.2	56.1	41.6	29.9	19.0	10.9	5.4	1.8	0.0

6-h UH ordinates (m³/s per cm), t = 0 to 72 h: 0.0, 4.5, 31.7, 57.9, 65.2, 56.1, 41.6, 29.9, 19.0, 10.9, 5.4, 1.8, 0.0. Check: sum × 6 × 3600 / 700 km² = 1.00 cm ✓.

(b) Flood hydrograph for two successive storms

Loss = 0.25 cm/h × 6 h = 1.5 cm in each period.

Storm 1: 9.5 − 1.5 = 8.0 cm
Storm 2: 12.5 − 1.5 = 11.0 cm

Q(t) = 8.0\,U(t) + 11.0\,U(t-6) + 40

t (h)	UH (m³/s)	8 cm × UH	11 cm × UH	Base	Q (m³/s)
0	0.00	0.0	0.0	40.0	40.0
6	4.53	36.2	0.0	40.0	76.2
12	31.68	253.5	49.8	40.0	343.3
18	57.94	463.5	348.5	40.0	852.0
24	65.18	521.4	637.3	40.0	1198.7
30	56.12	449.0	716.9	40.0	1205.9
36	41.64	333.1	617.4	40.0	990.5
42	29.87	239.0	458.0	40.0	737.0
48	19.01	152.1	328.6	40.0	520.7
54	10.86	86.9	209.1	40.0	336.0
60	5.43	43.5	119.5	40.0	202.9
66	1.81	14.5	59.7	40.0	114.2
72	0.00	0.0	19.9	40.0	59.9
78	0.00	0.0	0.0	40.0	40.0

Answer: peak flood ≈ 1206 m³/s at t = 30 h (ordinates of the flood hydrograph are in the last column).

✓ Answer checked by Claude Sonnet 5.5

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6-h UH and flood from two storms, 700 km2

(a) Ordinates of the 6-h unit hydrograph

(b) Flood hydrograph for two successive storms

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