--- title: "Discharge of triangular river from surface velocities" subject: "Engineering Hydrology" chapter: 4 key: "semester_5/engineering_hydrology:triangular-river-discharge-3m" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/triangular-river-discharge-3m site: BCT Engineering --- # Discharge of triangular river from surface velocities Engineering Hydrology · Chapter 4: Surface Runoff - Asked in: 2066 Magh (old course), 8 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/triangular-river-discharge-3m - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-4 ## Question Depth of a triangular shaped river is 3 m. Its top width is 10 m. The maximum depth of the river is at 4 m from the one side. The top velocity of the river measured at 3.0 and 7.0 m from the same side are 1.5 m/s and 1.8 m/s respectively. Calculate the discharge of the river. ## Short answer Depths at 3 m and 7 m were 2.25 m and 1.50 m; mean velocities (0.85 × surface) were 1.275 and 1.53 m/s. Dividing section at 5 m gave Q = 8.75×1.275 + 6.25×1.53 ≈ 20.7 m^3/s. ## Answer **Data:** triangular section, top width $T = 10$ m, maximum depth $D=3$ m at 4 m from one side (the left side). Surface velocities: 1.5 m/s at 3.0 m and 1.8 m/s at 7.0 m from the same side. Mean velocity in a vertical $\approx 0.85\times$ surface velocity (usual surface-to-mean coefficient for floats and surface readings). ```text 0 3 4 5 7 10 m +-----v---v---v---v---------+ water surface \ | | | | / \ | | | | / \| | | | / \ |3 m| | / \_ V __|_/ deepest point at x = 4 m ``` #### Depths Left side (0 to 4 m): depth $= 3\times\dfrac{x}{4}$. Right side (4 to 10 m): depth $= 3\times\dfrac{10-x}{6}$. - At $x=3$ m: $d_1 = 3\times3/4 = 2.25$ m. - At $x=7$ m: $d_2 = 3\times(10-7)/6 = 1.50$ m. #### Areas The total area is $\tfrac12\times10\times3=15$ m$^2$. The section is divided at $x = 5$ m (midway between the two verticals), where depth is $3\times5/6 = 2.5$ m. - Left part (vertical at 3 m): $A_1 = \tfrac12(4)(3) + \tfrac12(3+2.5)(1) = 6 + 2.75 = 8.75$ m$^2$. - Right part (vertical at 7 m): $A_2 = 15 - 8.75 = 6.25$ m$^2$. #### Mean velocities $$ v_1 = 0.85\times1.5 = 1.275\ \text{m/s},\qquad v_2 = 0.85\times1.8 = 1.53\ \text{m/s} $$ #### Discharge $$ Q = v_1A_1 + v_2A_2 = 1.275\times8.75 + 1.53\times6.25 = 11.156 + 9.563 = 20.72\ \text{m}^3/\text{s} $$ **Answer: Discharge $\approx 20.7$ m$^3$/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 from surface velocity? A. Use the surface velocity directly B. Multiply by 2 C. Multiply by about 0.85 D. Divide by 0.85 Answer: C. Surface readings are reduced by the surface-to-mean coefficient 0.85. (Key point: Mean velocity) 2. What is the depth at 7 m from the side? A. 2.25 m B. 3 m C. 2.1 m D. 1.5 m, since the right side depth is 3(10-7)/6 Answer: D. On the right side (4 to 10 m) depth = 3(10-x)/6 = 1.5 m. (Key point: Depth at vertical) 3. How is the section divided and what is Q? A. At x = 5 m: A1 = 8.75, A2 = 6.25 m^2, Q about 20.7 m^3/s B. Equal areas of 7.5 m^2 each, Q about 20.3 m^3/s C. At x = 4 m, Q about 15 m^3/s D. Whole area 15 m^2 times 1.8 m/s = 27 m^3/s Answer: A. The division is midway between the verticals; Q = 1.275 x 8.75 + 1.53 x 6.25 = 20.72 m^3/s. 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