--- title: "Mid-section method discharge from current meter data" subject: "Engineering Hydrology" chapter: 4 key: "semester_5/engineering_hydrology:midsection-0-3n-0-004" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/midsection-0-3n-0-004 site: BCT Engineering --- # Mid-section method discharge from current meter data Engineering Hydrology · Chapter 4: Surface Runoff - Asked in: 2075 Chaitra, 8 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/midsection-0-3n-0-004 - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-4 ## Question 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) | 0 | 2 | 4 | 6 | 9 | 12 | 15 | 18 | 20 | |---|---|---|---|---|---|---|---|---|---| | Depth (m) | 0 | 0.50 | 1.10 | 1.90 | 2.2 | 1.8 | 1.1 | 0.7 | 0 | | Number of revolutions | 0 | 80 | 83 | 130 | 121 | 116 | 100 | 90 | 0 | | Time (s) | 0 | 170 | 110 | 100 | 100 | 100 | 100 | 90 | 0 | ## 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. ## Answer **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 $Q \approx 8.38\ \text{m}^3/\text{s}$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the mid-section method, what is the width of the strip assigned to vertical i? A. The full distance between the two banks divided by the number of verticals B. The distance b(i) - b(i-1) C. The distance b(i+1) - b(i) D. Half the distance between the verticals on either side, (b(i+1) - b(i-1))/2 Answer: D. Each vertical represents a strip from the mid-point of the previous interval to the mid-point of the next, so w = (b(i+1) - b(i-1))/2. (Key point: Strip width in mid-section method) 2. At 2 m: 80 revolutions in 170 s. Which velocity is correct for V = 0.3N + 0.004? A. N = 0.471 rps, V = 0.145 m/s B. N = 2.125, V = 0.64 m/s C. 0.3 x 80 + 0.004 = 24 m/s D. N = 0.471, V = 0.471 m/s Answer: A. N = 80/170 = 0.471 rps, V = 0.3 x 0.471 + 0.004 = 0.145 m/s. (Key point: Rev per second) 3. How are the end verticals at 0 m and 20 m treated? A. Ignored completely including widths B. Zero depth, so they carry no flow but fix the end strip widths C. Given velocity of neighbour D. Given depth equal to neighbour Answer: B. Banks have zero depth; they only define the last strip widths, e.g. (20 - 15)/2 = 2.5 m for the vertical at 18 m. 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