- 2076 Asoj · 10 marks
S-curve: 4-hr UH to 3-hr UH
Short answer
Construct S-curve by adding 4-h UH lagged every4 h, lag S-curve by3 h, multiply difference by4/3. The3-h UH peaks at113.3 m³/s at4 h; ordinates at1-h intervals:0,8.0,48.0,88.0,113.3,101.3,84.0,72.0,62.7,54.7,44.0,36.0,30.7,25.3,20.0,13.3,12.0,6.7,5.3,0,2.0 m³/s.
The ordinates of a 4 hr UH of a basin area of 300 km measured at 1-hr intervals are 6, 36, 66, 91, 106, 93, 79, 68, 58, 49, 41, 34, 27, 23, 17, 13, 9, 6, 3 and 1.5 m/s respectively. Obtain the ordinates of a 3 hr UH using the S-curve technique.
The 4-h UH is given at 1-h intervals (the ordinate is 0 at t = 0 and 6 m/s at t = 1 h). Check: the sum of the ordinates is 826.5 m/s, so the volume under the UH is m. This is 0.99 cm over 300 km, so the UH is a 1 cm UH.
Method
- Build the S-curve: (add the 4-h UH repeatedly, each lagged by 4 h).
- Lag the S-curve by the new duration of 3 h: .
- The difference is the response to 3 h of rain at the 4-h rate (1 cm per 4 h). Scale it by :
| t (h) | (m/s) | ||||
|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0.00 |
| 1 | 6 | 6 | 0 | 6 | 8.00 |
| 2 | 36 | 36 | 0 | 36 | 48.00 |
| 3 | 66 | 66 | 0 | 66 | 88.00 |
| 4 | 91 | 91 | 6 | 85 | 113.33 |
| 5 | 106 | 112 | 36 | 76 | 101.33 |
| 6 | 93 | 129 | 66 | 63 | 84.00 |
| 7 | 79 | 145 | 91 | 54 | 72.00 |
| 8 | 68 | 159 | 112 | 47 | 62.67 |
| 9 | 58 | 170 | 129 | 41 | 54.67 |
| 10 | 49 | 178 | 145 | 33 | 44.00 |
| 11 | 41 | 186 | 159 | 27 | 36.00 |
| 12 | 34 | 193 | 170 | 23 | 30.67 |
| 13 | 27 | 197 | 178 | 19 | 25.33 |
| 14 | 23 | 201 | 186 | 15 | 20.00 |
| 15 | 17 | 203 | 193 | 10 | 13.33 |
| 16 | 13 | 206 | 197 | 9 | 12.00 |
| 17 | 9 | 206 | 201 | 5 | 6.67 |
| 18 | 6 | 207 | 203 | 4 | 5.33 |
| 19 | 3 | 206 | 206 | 0 | 0.00 |
| 20 | 1.5 | 207.5 | 206 | 1.5 | 2.00 |
The S-curve reaches an equilibrium value of about 207 m/s, in agreement with m/s.
Answer: The 3-h UH ordinates at 1-h intervals are 0, 8.0, 48.0, 88.0, 113.3, 101.3, 84.0, 72.0, 62.7, 54.7, 44.0, 36.0, 30.7, 25.3, 20.0, 13.3, 12.0, 6.7, 5.3, 0, 2.0 m/s (t = 0 to 20 h), with a peak of 113.3 m/s at 4 h. The small ripples at the tail come from rounding of the given ordinates.
✓ Answer checked by Claude Sonnet 5.5
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