--- title: "Discharge from current meter at 0.6 depth" subject: "Engineering Hydrology" chapter: 4 key: "semester_5/engineering_hydrology:current-meter-0-32n-0-032" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/current-meter-0-32n-0-032 site: BCT Engineering --- # Discharge from current meter at 0.6 depth Engineering Hydrology · Chapter 4: Surface Runoff - Asked in: 2067 Shrawan (old course), 10 marks - Video: [Discharge (Stream flow) Measurement Methods: [ Mid-Section and Mean-Section Methods ]](https://www.youtube.com/watch?v=QoeX_fyuDCE) — Hydraulics and Hydrology Tutorials - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/current-meter-0-32n-0-032 - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-4 ## Question 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) | 0 | 2 | 4 | 6 | 9 | 12 | 15 | 18 | 20 | 22 | 23 | 24 | |---|---|---|---|---|---|---|---|---|---|---|---|---| | Depth (m) | 0 | 0.5 | 1.1 | 1.95 | 2.25 | 1.85 | 1.75 | 1.65 | 1.50 | 1.25 | 0.75 | 0 | | No. of revolutions | 0 | 80 | 83 | 131 | 139 | 121 | 114 | 109 | 92 | 85 | 70 | 0 | | Time (sec) | 0 | 180 | 120 | 120 | 120 | 120 | 120 | 120 | 120 | 120 | 150 | 0 | ## 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. ## Answer 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 $Q \approx 11.82\ \text{m}^3/\text{s}$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the 0.6 depth reading used directly as the vertical's velocity? A. It is the surface velocity B. It is the maximum velocity C. It approximates the mean velocity of the vertical D. It is the bed velocity Answer: C. The velocity at 0.6 d from the surface is close to the mean in the vertical. (Key point: Velocity at 0.6 depth) 2. How is N obtained at the 2 m vertical? A. N = 80 rps, v = 25.6 m/s B. N = 180/80 = 2.25 rps C. N = 0.444 rps and v = 0.444 m/s D. N = 80/180 = 0.444 rps, then v = 0.32(0.444) + 0.032 = 0.174 m/s Answer: D. Revolutions are divided by time in seconds before applying the equation. (Key point: Computing N) 3. What is the width for the vertical at 9 m (neighbours at 6 and 12 m)? A. 3 m B. 6 m C. 1.5 m D. 9 m Answer: A. (12-6)/2 = 3 m. (Key point: Strip width) 4. What is the discharge? A. About 35.4 m^3/s B. About 11.82 m^3/s C. About 0.334 m^3/s D. About 13.17 m^3/s Answer: B. The sum of v x w x d over the verticals is 11.825 m^3/s, with area 35.38 m^2. 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