--- title: "Discharge from multi-point current meter data" subject: "Engineering Hydrology" chapter: 4 key: "semester_5/engineering_hydrology:current-meter-two-point-0-3n" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/current-meter-two-point-0-3n site: BCT Engineering --- # Discharge from multi-point current meter data Engineering Hydrology · Chapter 4: Surface Runoff - Asked in: 2071 Shrawan, 6 marks - Video: [How to Measure Discharge of Water||Water Current Meter|| By Civil Guruji](https://www.youtube.com/watch?v=sXFGUjjTmdM) — Civil Engineering Delhi - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/current-meter-two-point-0-3n - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-4 ## Question The following data were collected for a stream at a gauging station. Compute the discharge. Rating equation of current meter: $V = 0.3N + 0.05$. | Distance from one end of water surface (m) | Depth, d (m) | at 0.6d Rev. | at 0.6d Sec. | at 0.2d Rev. | at 0.2d Sec. | at 0.8d Rev. | at 0.8d Sec. | |---|---|---|---|---|---|---|---| | 3 | 1.4 | 12 | 50 | | | | | | 6 | 3.3 | | | 38 | 52 | 23 | 55 | | 9 | 5.0 | | | 40 | 58 | 30 | 54 | | 12 | 9.0 | | | 48 | 60 | 34 | 58 | | 15 | 5.4 | | | 34 | 52 | 30 | 50 | | 18 | 3.8 | | | 35 | 52 | 30 | 54 | | 21 | 1.8 | 18 | 50 | | | | | ## Short answer Mid-section method with v = 0.3N+0.05, mean velocity from two-point (0.2d,0.8d) or single-point (0.6d) readings, total discharge is 20.61 m³/s. Q ≈ 20.6 m³/s. ## Answer **Method:** mid-section method. Velocity from $v = 0.3N + 0.05$ ($N$ in rev/s). Where the depth is small (1.4 m, 1.8 m) one reading at 0.6 d is the mean velocity; at other verticals two readings are used: $v_m=\dfrac{v_{0.2}+v_{0.8}}{2}$. The verticals are 3 m apart, so each strip width is 3 m. The banks are assumed at 0 m and 24 m (zero depth). | Dist. (m) | Depth (m) | Revolutions / time | $v_m$ (m/s) | Width (m) | Area (m$^2$) | $q$ (m$^3$/s) | |---|---|---|---|---|---|---| | 3 | 1.4 | 0.6d: 12/50 = 0.240 | 0.1220 | 3 | 4.20 | 0.512 | | 6 | 3.3 | 0.2d: 0.731, 0.8d: 0.418 rps → 0.269, 0.175 | 0.2223 | 3 | 9.90 | 2.201 | | 9 | 5.0 | 0.2d: 0.690, 0.8d: 0.556 rps → 0.257, 0.217 | 0.2368 | 3 | 15.00 | 3.552 | | 12 | 9.0 | 0.2d: 0.800, 0.8d: 0.586 rps → 0.290, 0.226 | 0.2579 | 3 | 27.00 | 6.964 | | 15 | 5.4 | 0.2d: 0.654, 0.8d: 0.600 rps → 0.246, 0.230 | 0.2381 | 3 | 16.20 | 3.857 | | 18 | 3.8 | 0.2d: 0.673, 0.8d: 0.556 rps → 0.252, 0.217 | 0.2343 | 3 | 11.40 | 2.671 | | 21 | 1.8 | 0.6d: 18/50 = 0.360 | 0.1580 | 3 | 5.40 | 0.853 | | **Total** | | | | | **89.10** | **20.610** | Mean velocity $= 0.231$ m/s. **Answer: Discharge $Q \approx 20.6\ \text{m}^3/\text{s}$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At verticals with two readings, mean velocity is A. v0.2 x v0.8 B. v0.2 only C. v0.8 only D. (v0.2 + v0.8)/2 Answer: D. Mean of the 0.2 d and 0.8 d velocities. (Key point: Two-point mean) 2. At 1.4 m and 1.8 m depths, which reading is used? A. One reading at 0.6 d B. None C. Two readings D. Five readings Answer: A. Shallow depth uses a single 0.6 d reading as mean velocity. (Key point: Shallow verticals) 3. At 3 m: 12 revolutions in 50 s, V = 0.3N + 0.05 gives A. 0.24 m/s B. 0.122 m/s C. 3.65 m/s D. 0.35 m/s Answer: B. N = 0.24 rps, V = 0.3 x 0.24 + 0.05 = 0.122 m/s. 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