2074 Bhadra · 10 marks

Stream flow from 0.2d and 0.8d measurements

Short answer

The mid-section method is applied with 0.6 m wide strips. Mean velocity in each vertical is the average of velocities at 0.2d and 0.8d. Computing Σ(0.6 × depth × mean velocity) yields stream flow = 4.83 m³/s.

Compute the stream flow for the measurement data given below.

Distance (m)	0.6	1.2	1.8	2.4	3.0	3.6	4.2	4.8	5.4	6.0
Depth (m)	0.3	1.29	2.16	2.55	2.22	1.68	1.41	1.05	0.63	0.42
Velocity (m/s) at 0.2d	0.42	0.57	0.78	0.87	0.81	0.75	0.69	0.63	0.54	0.45
Velocity (m/s) at 0.8d	0.21	0.36	0.54	0.60	0.30	0.51	0.45	0.39	0.33	0.30

Mean velocity in each vertical is taken as the average of the velocities at 0.2d and 0.8d. The mid-section method is used with equal spacing of 0.6 m, so each interior vertical represents a strip 0.6 m wide; the bank points (0 and 6.6 m) have zero depth.

Computation

Dist. (m)	Depth (m)	$\bar{V}$ (m/s)	Area = 0.6d (m²)	ΔQ (m³/s)
0.6	0.30	0.315	0.180	0.0567
1.2	1.29	0.465	0.774	0.3599
1.8	2.16	0.660	1.296	0.8554
2.4	2.55	0.735	1.530	1.1246
3.0	2.22	0.555	1.332	0.7393
3.6	1.68	0.630	1.008	0.6350
4.2	1.41	0.570	0.846	0.4822
4.8	1.05	0.510	0.630	0.3213
5.4	0.63	0.435	0.378	0.1644
6.0	0.42	0.375	0.252	0.0945
Total			8.226	4.833

Sample: at 2.4 m, $\bar{V} = (0.87 + 0.60)/2 = 0.735$ m/s, $\Delta Q = 0.6 \times 2.55 \times 0.735 = 1.1246$ m³/s.

Q = \sum 0.6\,d_i\,\bar{V}_i = 4.83\ \text{m}^3/\text{s}

Answer: Stream flow ≈ 4.83 m³/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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