- 2068 Chaitra · 14 marks
Flood hydrograph from 6-h UH, three storms
Short answer
Subtract φ-index loss1.2 cm per6-h period gives excesses1.8,3.8,2.8 cm. Applying superposition with the6-h UH and adding30 m³/s base flow yields a flood hydrograph peaking at6010 m³/s at24–30 h and returning to30 m³/s at78 h.
The ordinates of a 6-h unit hydrograph are given below. A storm had three successive 6-h intervals of rainfall magnitude of 3.0, 5.0 and 4.0 cm respectively. Assuming an index of 0.20 cm/h and a base flow of 30 m/s, determine and plot the resulting hydrograph of flow.
| Time (hr) | 0 | 3 | 6 | 9 | 12 | 18 | 24 | 30 | 36 | 42 | 48 | 54 | 60 | 66 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 6-h UH ordinate (m/s) | 0 | 150 | 250 | 450 | 600 | 800 | 700 | 600 | 450 | 320 | 200 | 100 | 50 | 0 |
Step 1: Rainfall excess. Each 6-h period loses cm.
| Period | Rain (cm) | Loss (cm) | Excess (cm) |
|---|---|---|---|
| 0 – 6 h | 3.0 | 1.2 | 1.8 |
| 6 – 12 h | 5.0 | 1.2 | 3.8 |
| 12 – 18 h | 4.0 | 1.2 | 2.8 |
(Intensities are 0.5, 0.83 and 0.67 cm/h, all above 0.20 cm/h, so every period produces runoff.)
Step 2: UH ordinates at 6-h interval (the given 3 h and 9 h values are not needed): t = 0, 6, 12, …, 66 h → 0, 250, 600, 800, 700, 600, 450, 320, 200, 100, 50, 0 m³/s.
Step 3: Superposition. The three excess depths are applied to the 6-h UH, each lagged by 6 h, and added with the base flow of 30 m³/s:
| t (h) | UH (m³/s) | 1.8 cm × UH | 3.8 cm × UH | 2.8 cm × UH | Base | Q (m³/s) |
|---|---|---|---|---|---|---|
| 0 | 0.00 | 0.0 | 0.0 | 0.0 | 30.0 | 30.0 |
| 6 | 250.00 | 450.0 | 0.0 | 0.0 | 30.0 | 480.0 |
| 12 | 600.00 | 1080.0 | 950.0 | 0.0 | 30.0 | 2060.0 |
| 18 | 800.00 | 1440.0 | 2280.0 | 700.0 | 30.0 | 4450.0 |
| 24 | 700.00 | 1260.0 | 3040.0 | 1680.0 | 30.0 | 6010.0 |
| 30 | 600.00 | 1080.0 | 2660.0 | 2240.0 | 30.0 | 6010.0 |
| 36 | 450.00 | 810.0 | 2280.0 | 1960.0 | 30.0 | 5080.0 |
| 42 | 320.00 | 576.0 | 1710.0 | 1680.0 | 30.0 | 3996.0 |
| 48 | 200.00 | 360.0 | 1216.0 | 1260.0 | 30.0 | 2866.0 |
| 54 | 100.00 | 180.0 | 760.0 | 896.0 | 30.0 | 1866.0 |
| 60 | 50.00 | 90.0 | 380.0 | 560.0 | 30.0 | 1060.0 |
| 66 | 0.00 | 0.0 | 190.0 | 280.0 | 30.0 | 500.0 |
| 72 | 0.00 | 0.0 | 0.0 | 140.0 | 30.0 | 170.0 |
| 78 | 0.00 | 0.0 | 0.0 | 0.0 | 30.0 | 30.0 |
Step 4: Plot. Plot Q (m³/s) against t (h). The hydrograph rises from 30 m³/s at t = 0 to the peak, and falls back to the base flow of 30 m³/s at t = 78 h.
t(h) Q (m3/s)
0 | 30.0
6 |### 480.0
12 |############ 2060.0
18 |######################### 4450.0
24 |################################## 6010.0
30 |################################## 6010.0
36 |############################# 5080.0
42 |####################### 3996.0
48 |################ 2866.0
54 |########### 1866.0
60 |###### 1060.0
66 |### 500.0
72 |# 170.0
78 | 30.0
Answer: peak flood = 6010 m³/s at t = 24 h to 30 h (equal ordinates at both); base flow 30 m³/s is restored at 78 h.
✓ Answer checked by Claude Sonnet 5.5
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