- 2071 Shrawan · 6 marks
Discharge from multi-point current meter data
Short answer
Mid-section method with v = 0.3N+0.05, mean velocity from two-point (0.2d,0.8d) or single-point (0.6d) readings, total discharge is 20.61 m³/s. Q ≈ 20.6 m³/s.
The following data were collected for a stream at a gauging station. Compute the discharge. Rating equation of current meter: .
| Distance from one end of water surface (m) | Depth, d (m) | at 0.6d Rev. | at 0.6d Sec. | at 0.2d Rev. | at 0.2d Sec. | at 0.8d Rev. | at 0.8d Sec. |
|---|---|---|---|---|---|---|---|
| 3 | 1.4 | 12 | 50 | ||||
| 6 | 3.3 | 38 | 52 | 23 | 55 | ||
| 9 | 5.0 | 40 | 58 | 30 | 54 | ||
| 12 | 9.0 | 48 | 60 | 34 | 58 | ||
| 15 | 5.4 | 34 | 52 | 30 | 50 | ||
| 18 | 3.8 | 35 | 52 | 30 | 54 | ||
| 21 | 1.8 | 18 | 50 |
Method: mid-section method. Velocity from ( in rev/s). Where the depth is small (1.4 m, 1.8 m) one reading at 0.6 d is the mean velocity; at other verticals two readings are used: .
The verticals are 3 m apart, so each strip width is 3 m. The banks are assumed at 0 m and 24 m (zero depth).
| Dist. (m) | Depth (m) | Revolutions / time | (m/s) | Width (m) | Area (m) | (m/s) |
|---|---|---|---|---|---|---|
| 3 | 1.4 | 0.6d: 12/50 = 0.240 | 0.1220 | 3 | 4.20 | 0.512 |
| 6 | 3.3 | 0.2d: 0.731, 0.8d: 0.418 rps → 0.269, 0.175 | 0.2223 | 3 | 9.90 | 2.201 |
| 9 | 5.0 | 0.2d: 0.690, 0.8d: 0.556 rps → 0.257, 0.217 | 0.2368 | 3 | 15.00 | 3.552 |
| 12 | 9.0 | 0.2d: 0.800, 0.8d: 0.586 rps → 0.290, 0.226 | 0.2579 | 3 | 27.00 | 6.964 |
| 15 | 5.4 | 0.2d: 0.654, 0.8d: 0.600 rps → 0.246, 0.230 | 0.2381 | 3 | 16.20 | 3.857 |
| 18 | 3.8 | 0.2d: 0.673, 0.8d: 0.556 rps → 0.252, 0.217 | 0.2343 | 3 | 11.40 | 2.671 |
| 21 | 1.8 | 0.6d: 18/50 = 0.360 | 0.1580 | 3 | 5.40 | 0.853 |
| Total | 89.10 | 20.610 |
Mean velocity m/s.
Answer: Discharge .
✓ Answer checked by Claude Sonnet 5.5
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