In the mid-section method, what is the width of the strip assigned to vertical i?

Half the distance between the verticals on either side, (b(i+1) - b(i-1))/2

The full distance between the two banks divided by the number of verticals

How are the end verticals at 0 m and 20 m treated?

Zero depth, so they carry no flow but fix the end strip widths

Ignored completely including widths

2075 Chaitra · 8 marks

Mid-section method discharge from current meter data

Short answer

Mid-section method with velocity v=0.3N+0.004 (N rev/s); strip width halfway between adjacent verticals. Summing partial discharges gives Q≈8.38 m³/s.

Calculate the discharge in a stream by using mid-section method from provided data. A current meter is used to measure velocity at 0.6 depth and calibrated as $V = 0.3N + 0.004$ .

Distance from right bank (m)	2	4	6	9	12	15	18
Depth (m)	0.50	1.10	1.90	2.2	1.8	1.1	0.7
Number of revolutions	80	83	130	121	116	100	90
Time (s)	170	110	100	100	100	100	90

Mid-section method: the cross-section is divided into strips, each centred on a measuring vertical, the strip extending from the mid-point of the preceding interval to the mid-point of the next.

q_i = v_i\,d_i\,\frac{b_{i+1}-b_{i-1}}{2},\qquad Q=\sum q_i

Velocity at 0.6 depth is the mean velocity of the vertical: $v = 0.3N + 0.004$ with $N$ = revolutions/second.

Distance (m)	Depth d (m)	N = rev/time (rps)	v (m/s)	Width w (m)	Area $w\,d$ (m $^2$ )	$q = v\,w\,d$ (m $^3$ /s)
2	0.5	80/170 = 0.471	0.145	2	1.000	0.145
4	1.1	83/110 = 0.755	0.230	2	2.200	0.507
6	1.9	130/100 = 1.300	0.394	2.5	4.750	1.872
9	2.2	121/100 = 1.210	0.367	3	6.600	2.422
12	1.8	116/100 = 1.160	0.352	3	5.400	1.901
15	1.1	100/100 = 1.000	0.304	3	3.300	1.003
18	0.7	90/90 = 1.000	0.304	2.5	1.750	0.532
Total					25.000	8.382

Total area $= 25.00$ m $^2$ ; mean velocity $= 0.335$ m/s.

Answer: Discharge $Q \approx 8.38\ \text{m}^3/\text{s}$ .

✓ Answer checked by Claude Sonnet 5.5

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