What is the strip area and total discharge?

1.5 x d for each vertical; Q about 7.06 m^3/s

2080 Bhadra · 8 marks

Mid-section discharge with 0.2d, 0.6d, 0.8d velocities

Short answer

Mid-section method: for depths >0.75 m, mean velocity = (v_0.2 + v_0.8)/2; for shallower, use v_0.6. Multiplying each velocity by depth and 1.5 m width gives segment discharges, total Q ≈ 7.06 m³/s.

Calculate the flow discharge of a river from the following data using area velocity method using Mid sectional area.

Distance from bank (m)	0	1.5	3.0	4.5	6.0	7.5	9.0
Depth of flow (m)	0	0.75	2.50	3.50	2.30	0.80	0
Velocity 0.2 depth	-	-	0.65	0.90	0.50	-	-
Velocity 0.6 depth	0	0.35	-	-	-	0.30	0
Velocity 0.8 depth	-	-	0.25	0.40	0.20	-	-

Mid-section method. The spacing is 1.5 m, so every strip has width 1.5 m (the banks at 0 and 9 m have zero depth).

Mean velocity of each vertical:

Shallow verticals (0.75 m and 0.80 m deep) have one reading at 0.6 d: $v_m=v_{0.6}$ .
Deeper verticals have readings at 0.2 d and 0.8 d: $v_m=(v_{0.2}+v_{0.8})/2$ .

x (m)	d (m)	Mean velocity (m/s)	$v_m$	Area $1.5\,d$ (m $^2$ )	$q=v_m\times$ area
1.5	0.75	0.6d: 0.35	0.350	1.125	0.394
3.0	2.5	(0.65+0.25)/2	0.450	3.750	1.688
4.5	3.5	(0.90+0.40)/2	0.650	5.250	3.412
6.0	2.3	(0.50+0.20)/2	0.350	3.450	1.208
7.5	0.8	0.6d: 0.30	0.300	1.200	0.360
Total				14.775	7.061

Mean velocity $= 0.478$ m/s.

Answer: Discharge $Q \approx 7.06\ \text{m}^3/\text{s}$ .

✓ Answer checked by Claude Sonnet 5.5

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