GATE 2023 CE (CE2) – Question 63
The cross-section of a small river is sub-divided into seven segments of width 1.5 m each. The average depth, and velocity at different depths were measured during a field campaign at the middle of each segment width. The discharge computed by the velocity area method for the given data is ______ m$^3$/s (round off to one decimal place).
| Segment | Average depth D (m) | Velocity at 0.2D (m/s) | At 0.6D | At 0.8D |
|---|---|---|---|---|
| 1 | 0.40 | — | 0.40 | — |
| 2 | 0.70 | 0.76 | — | 0.70 |
| 3 | 1.20 | 1.19 | — | 1.13 |
| 4 | 1.40 | 1.25 | — | 1.10 |
| 5 | 1.10 | 1.13 | — | 1.09 |
| 6 | 0.80 | 0.69 | — | 0.65 |
| 7 | 0.45 | — | 0.42 | — |
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Show answer and explanation
Correct answer: 8.46 to 8.54
Explanation
**Method (velocity-area).** Discharge $Q=\sum(\text{width}\times\text{depth}\times\bar v)$. For each vertical, use the mean velocity:
- If only the velocity at 0.6D is given, use it as the mean.
- If the velocities at 0.2D and 0.8D are given, the mean is their average.
| Seg | D (m) | Mean velocity (m/s) | $D\bar v$ |
|---|---|---|---|
| 1 | 0.40 | 0.40 | 0.1600 |
| 2 | 0.70 | (0.76+0.70)/2 = 0.73 | 0.5110 |
| 3 | 1.20 | (1.19+1.13)/2 = 1.16 | 1.3920 |
| 4 | 1.40 | (1.25+1.10)/2 = 1.175 | 1.6450 |
| 5 | 1.10 | (1.13+1.09)/2 = 1.11 | 1.2210 |
| 6 | 0.80 | (0.69+0.65)/2 = 0.67 | 0.5360 |
| 7 | 0.45 | 0.42 | 0.1890 |
$$\sum D\bar v=5.654\text{ m}^2/\text{s}$$
$$Q=1.5\times5.654=8.481\approx\mathbf{8.5\ m^3/s}.$$