--- title: "Surface drainage design, 40 ha Terai field" subject: "Irrigation and Drainage Engineering" chapter: 9 key: "semester_6/irrigation_and_drainage_engineering:surface-drainage-40ha-terai" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/surface-drainage-40ha-terai site: BCT Engineering --- # Surface drainage design, 40 ha Terai field Irrigation and Drainage Engineering · Chapter 9: Water Logging and Drainage - Asked in: 2069 Bhadra, 10 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/surface-drainage-40ha-terai - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-9 ## Question Design a surface drainage for a field of 40 ha area in Terai with following data. Design maximum yearly precipitation for three consecutive days = 50 mm, longitudinal slope of channel 1:400, Manning roughness coefficient 0.025, Maximum water level is 300 mm which may persist for up to one day and depth in excess of 200 mm may persist for up to 3 days. Assume other suitable values if necessary. ## Short answer Drainage rate D = 50 mm / 3 days = 16.7 mm/day (1.93 l/s/ha). Design discharge Q = 1.93 × 40 = 77.2 l/s. Trapezoidal drain: bed width 0.5 m, flow depth 0.21 m, side slope 1:1, longitudinal slope 1:400, total depth 0.40 m. ## Answer Surface drainage of the 40 ha Terai field is designed from the design rainfall, the permissible ponding of the rice crop, and Manning's formula for the drain. #### Step 1: Design rainfall and drainage rate Design 3-day rainfall $R$ = 50 mm (16.7 mm/day). Assume the initial water depth in the bunded field $h_0$ = 50 mm (not given). The water level at the end of the storm without drainage = 50 + 50 = 100 mm. This is below the permissible 300 mm and also below 200 mm, so the crop limits are met even without drainage. To avoid accumulation over later storms, the 3-day rainfall is removed within the same 3 days: $$ D = \frac{R}{3} = \frac{50}{3} = 16.7\ \text{mm/day} = 16.7\times\frac{10}{86.4} = 1.93\ \text{l/s/ha} $$ #### Step 2: Design discharge $$ Q = D\,A = 1.93\times40 = 77.2\ \text{l/s} = 0.0772\ \text{m}^3/\text{s} $$ #### Step 3: Section of the drain Assume an earthen trapezoidal drain, side slope 1:1, $n$ = 0.025 (given), slope $S$ = 1/400 (given), bed width $b$ = 0.5 m. $$ Q=\frac{1}{n}AR^{2/3}S^{1/2} $$ By trial: depth $d$ = 0.208 m (say 0.21 m), $A=(b+d)d$ = 0.147 m², $P=b+2\sqrt{2}d$ = 1.088 m, $R$ = 0.135 m; check $Q$ = (1/0.025) × 0.147 × 0.135^(2/3) × (1/400)^0.5 = 0.0776 m³/s ≈ 0.0772 m³/s. $V$ = 0.0772/0.147 = 0.53 m/s, which is non-silting (> 0.3 m/s) and non-scouring (< 0.6 m/s for earth). Provide a free board of 0.15 m: total depth = 0.21 + 0.15 = 0.36 m; adopt 0.40 m. #### Step 4: Layout Field drains (laterals) run along the slope into one collector drain along the lower edge of the 40 ha block; the collector discharges into the natural drain through an outlet structure with its sill 0.3 m below the lowest field level. The drain is made deep enough for the field outlets to discharge freely, with the drain water level at least 0.3 m below the field outlets. **Answer: D = 16.7 mm/day (1.93 l/s/ha), Q = 77.2 l/s; trapezoidal drain with bed width 0.5 m, flow depth 0.21 m (total 0.40 m), side slopes 1:1, slope 1 in 400.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can the drainage rate be taken as R/3 here? A. Even with no drainage the level (50 + 50 = 100 mm) is under both the 200 mm and 300 mm limits, so only accumulation for later storms is avoided B. Because the level would reach 300 mm without drainage C. Because the Terai needs no drainage at all D. Because the 200 mm limit is exceeded for 3 days Answer: A. The crop limits are already met, so the 3-day rainfall is simply removed within the 3 days to prevent build-up. (Key point: Check without drainage) 2. What are D in mm/day and l/s/ha? A. 50 mm/day = 5.8 l/s/ha B. 16.7 mm/day = 1.93 l/s/ha C. 16.7 mm/day = 0.19 l/s/ha D. 150 mm/day = 17.4 l/s/ha Answer: B. D = 50/3 = 16.7 mm/day; multiplied by 10/86.4 gives 1.93 l/s/ha. (Key point: Drainage rate) 3. What is the design discharge for the 40 ha block? A. 7.72 l/s B. 1.93 l/s C. 77.2 l/s D. 772 l/s Answer: C. Q = D x A = 1.93 x 40 = 77.2 l/s = 0.0772 m3/s. (Key point: Design discharge) 4. Which range should the drain velocity satisfy? A. Above 2 m/s to keep it clean B. Below 0.1 m/s to avoid erosion C. Any value, since velocity is not checked D. Above 0.3 m/s (non-silting) and below about 0.6 m/s (non-scouring in earth) Answer: D. The calculated V = 0.53 m/s lies between the non-silting and non-scouring limits. 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