2075 Baisakh · 10 marks

Discharge from 0.2d and 0.8d velocities (left bank)

Short answer

Average velocity = (V_0.2d + V_0.8d)/2. With 2 m segment widths, mid-section discharge sum = 2 * Σ(depth * mean velocity) = 8.56 m³/s.

Compute the discharge of a stream with the following data.

Distance from left bank (m)	0	2	4	6	8	10	12
Depth (m)	-	0.9	2.4	2.2	1.0	0.6	-
Velocity at 0.2d	-	0.6	0.9	0.7	0.6	0.4	-
Velocity at 0.8d	-	0.4	0.6	0.5	0.4	0.3	-

The two-point method is used: mean velocity in each vertical $\bar{V} = (V_{0.2d} + V_{0.8d})/2$ . Discharge is by the mid-section method; verticals are 2 m apart, so each interior vertical represents a 2 m wide strip. Depth and velocity are zero at the banks (0 and 12 m).

Computation

Distance (m)	Width (m)	Depth (m)	$\bar{V}$ (m/s)	Area (m²)	ΔQ (m³/s)
2	2	0.9	(0.6+0.4)/2 = 0.50	1.8	0.90
4	2	2.4	(0.9+0.6)/2 = 0.75	4.8	3.60
6	2	2.2	(0.7+0.5)/2 = 0.60	4.4	2.64
8	2	1.0	(0.6+0.4)/2 = 0.50	2.0	1.00
10	2	0.6	(0.4+0.3)/2 = 0.35	1.2	0.42
Total				14.2	8.56

 0    2    4    6    8    10   12  (m)
 ~~~~~|~~~~|~~~~|~~~~|~~~~|~~~~~
  \   |    |    |    |    |   /
   \__|    |    |    |____|__/
      |____|____|____|

Q = \sum w_i\,d_i\,\bar{V}_i = 8.56\ \text{m}^3/\text{s}

Answer: Discharge of the stream ≈ 8.56 m³/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Discharge from 0.2d and 0.8d velocities (left bank)

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