How is the velocity at each vertical found?

Compute N = revolutions/time, then v = 0.045 + 0.76N

v = 0.76 + 0.045N using total revolutions

How is the strip width of a vertical found in the mid-section method?

Half the distance between the previous and next verticals, (b(i+1)-b(i-1))/2

The distance to the left bank only

The average depth of the section

What is the total discharge?

About 6.23 m^3/s, the sum of q = v x w x d over the verticals

Asked 2/27
2067 Mangsir (old course) · 8 marks

Stream flow from current meter, V=0.045+0.76N

Short answer

Using the mid-section method with v = 0.045+0.76 N and segment widths, the computed stream flow is 6.23 m³/s.

Compute the stream flow from the following data. The calibrated equation of current meter is: $V = 0.045 + 0.76N$ , where V is in m/sec and N is revolution/sec.

Distance from bank (m)	0.6	1.5	2.5	3.5	5.0	6.0	7.0
Depth (m)	0.3	0.75	1.2	1.7	1.3	0.7	0.3
No. of revolutions	15	95	110	120	110	80	20
Time (sec)	45	85	95	90	100	70	40

Mid-section method with $v = 0.045 + 0.76N$ , $N$ = revolutions per second. The meter reading is taken as the mean velocity of each vertical.

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

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)
0.6	0.3	15/45 = 0.333	0.298	0.75	0.225	0.067
1.5	0.75	95/85 = 1.118	0.894	0.95	0.712	0.637
2.5	1.2	110/95 = 1.158	0.925	1	1.200	1.110
3.5	1.7	120/90 = 1.333	1.058	1.25	2.125	2.249
5	1.3	110/100 = 1.100	0.881	1.25	1.625	1.432
6	0.7	80/70 = 1.143	0.914	1	0.700	0.639
7	0.3	20/40 = 0.500	0.425	0.75	0.225	0.096
Total					6.812	6.230

Total area $= 6.81$ m $^2$ ; mean velocity $= 0.915$ m/s.

Answer: Stream flow $Q \approx 6.23\ \text{m}^3/\text{s}$ .

✓ Answer checked by Claude Sonnet 5.5

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