--- title: "River discharge by mid-section method" subject: "Hydro Power" chapter: 2 key: "semester_6/hydro_power:mid-section-discharge" source: https://www.bctengineer.com/q/semester_6/hydro_power/mid-section-discharge site: BCT Engineering --- # River discharge by mid-section method Hydro Power · Chapter 2: Introduction to Hydrology - Asked in: 2077 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 Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/hydro_power/mid-section-discharge - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-2 ## Question Compute the river discharge by mid-section with the help of following data (y is the depth of flow): | Distance from the left bank (m) | Depth (m) | Velocity at 0.6y (m/s) | Velocity at 0.2y (m/s) | Velocity at 0.8y (m/s) | |---|---|---|---|---| | 0 | 0 | - | - | - | | 1.2 | 0.7 | 0.4 | - | - | | 2.4 | 1.7 | - | 0.7 | 0.5 | | 3.6 | 2.5 | - | 0.9 | 0.6 | | 4.8 | 1.3 | - | 0.6 | 0.4 | | 6.0 | 0.5 | 0.35 | - | - | | 7.2 | 0 | - | - | - | ## Short answer Using the mid-section method, each vertical strip is 1.2 m wide. Mean velocity is V0.6 for depth ≤0.7 m, else (V0.2+V0.8)/2. Summing width × depth × mean velocity gives discharge = 4.80 m³/s. ## Answer In the **mid-section method**, each vertical represents a strip extending halfway to its neighbours; here all verticals are 1.2 m apart, so each strip is 1.2 m wide. Mean velocity in a vertical: - Shallow verticals (one reading): $\bar{V} = V_{0.6y}$ - Deeper verticals (two readings): $\bar{V} = (V_{0.2y} + V_{0.8y})/2$ #### Computation | Distance (m) | Width (m) | Depth (m) | $\bar{V}$ (m/s) | Area (m²) | ΔQ (m³/s) | |---|---|---|---|---|---| | 1.2 | 1.2 | 0.7 | 0.40 | 0.84 | 0.336 | | 2.4 | 1.2 | 1.7 | (0.7+0.5)/2 = 0.60 | 2.04 | 1.224 | | 3.6 | 1.2 | 2.5 | (0.9+0.6)/2 = 0.75 | 3.00 | 2.250 | | 4.8 | 1.2 | 1.3 | (0.6+0.4)/2 = 0.50 | 1.56 | 0.780 | | 6.0 | 1.2 | 0.5 | 0.35 | 0.60 | 0.210 | | **Total** | | | | **8.04** | **4.800** | (The bank points at 0 and 7.2 m have zero depth and give no flow.) $$ Q = \sum w_i\,y_i\,\bar{V}_i = 4.80\ \text{m}^3/\text{s} $$ **Answer:** River discharge ≈ 4.80 m³/s. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the mean velocity found at a shallow vertical with a single reading at 0.6y? A. It is halved B. It is taken directly as the mean velocity C. It is doubled D. It is averaged with the neighbouring vertical Answer: B. Shallow verticals use V mean = V0.6y; deeper ones average V0.2y and V0.8y. (Key point: Mean velocity by depth) 2. What is dQ at 3.6 m (depth 2.5 m, V0.2 = 0.9, V0.8 = 0.6, strip width 1.2 m)? A. 1.80 m3/s B. 2.70 m3/s C. 2.25 m3/s D. 3.00 m3/s Answer: C. V mean = 0.75 m/s, so dQ = 1.2 x 2.5 x 0.75 = 2.25 m3/s. (Key point: Segment discharge) 3. Why do the points at 0 and 7.2 m contribute no discharge? A. They have zero depth B. They lie outside the river C. Their velocity is infinite D. They are used only for width Answer: A. Depth is zero at the banks, so w*y*V = 0. (Key point: Bank points) 4. What is the total river discharge? A. 8.04 m3/s B. 6.00 m3/s C. 3.40 m3/s D. 4.80 m3/s Answer: D. 0.336 + 1.224 + 2.250 + 0.780 + 0.210 = 4.80 m3/s. 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