Why is the 0.6 depth reading used directly as the vertical's velocity?

It approximates the mean velocity of the vertical

How is N obtained at the 2 m vertical?

N = 80/180 = 0.444 rps, then v = 0.32(0.444) + 0.032 = 0.174 m/s

N = 0.444 rps and v = 0.444 m/s

2067 Shrawan (old course) · 10 marks

Discharge from current meter at 0.6 depth

Short answer

Using mid-section method with v=0.32N+0.032 m/s, discharge per segment as v·depth·width. Total Q ≈ 11.82 m^3/s.

The following are the data obtained from a stream gauging station. A current meter with a calibration equation $V = (0.32N + 0.032)$ m/s where N = revolutions per second, was used to measure the velocity at 0.6 depth. Calculate the discharge in the stream.

Distance from right bank (m)	2	4	6	9	12	15	18	20	22	23
Depth (m)	0.5	1.1	1.95	2.25	1.85	1.75	1.65	1.50	1.25	0.75
No. of revolutions	80	83	131	139	121	114	109	92	85	70
Time (sec)	180	120	120	120	120	120	120	120	120	150

Mid-section method; velocity at 0.6 d is the mean velocity: $v = 0.32N + 0.032$ with $N$ = rev/time (rps).

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)
2	0.5	80/180 = 0.444	0.174	2	1.000	0.174
4	1.1	83/120 = 0.692	0.253	2	2.200	0.557
6	1.95	131/120 = 1.092	0.381	2.5	4.875	1.859
9	2.25	139/120 = 1.158	0.403	3	6.750	2.718
12	1.85	121/120 = 1.008	0.355	3	5.550	1.968
15	1.75	114/120 = 0.950	0.336	3	5.250	1.764
18	1.65	109/120 = 0.908	0.323	2.5	4.125	1.331
20	1.5	92/120 = 0.767	0.277	2	3.000	0.832
22	1.25	85/120 = 0.708	0.259	1.5	1.875	0.485
23	0.75	70/150 = 0.467	0.181	1	0.750	0.136
Total					35.375	11.825

Total area $= 35.38$ m $^2$ ; mean velocity $= 0.334$ m/s.

Answer: Discharge $Q \approx 11.82\ \text{m}^3/\text{s}$ .

✓ Answer checked by Claude Sonnet 5.5

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