--- title: "Drainage canal for 5 ha paddy field" subject: "Irrigation and Drainage Engineering" chapter: 9 key: "semester_6/irrigation_and_drainage_engineering:drainage-canal-paddy-5ha-300mm" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/drainage-canal-paddy-5ha-300mm site: BCT Engineering --- # Drainage canal for 5 ha paddy field Irrigation and Drainage Engineering · Chapter 9: Water Logging and Drainage - Asked in: 2078 Baisakh, 7 marks - Video: [surface drainage design | water balance equation | Irrigation numerical](https://www.youtube.com/watch?v=YYmufytG5M8) — Sudip Dhakal - 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/drainage-canal-paddy-5ha-300mm - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-9 ## Question Design the drainage canal in paddy field for 5 ha of land to meet the following conditions. Maximum yearly precipitation for 3 consecutive days = 300 mm. The designed rainfall is to be taken as 10 years return periods. Initial water level in the field = 40 mm. Maximum water level is 300 mm, which may persist for one day and depth in excess of 200 mm may persist for up to 3 days. Take growth factor for 10 years return period of 1.5. Assume other suitable data if necessary. ## Short answer Design rainfallR=1.5×300=450 mm in3 days. Trial givesD=60.6 mm/day=7.01 l/s/ha, Q=35.0 l/s. Canal: trapezoidal, bed width0.5 m, flow depth0.17 m (total depth0.32 m), side slopes1:1, slope1:1000, Manningn=0.025, velocity0.30 m/s. ## Answer The drainage canal must remove water from the paddy field fast enough that the level stays within the permissible depth and duration. The design discharge is the drainage rate multiplied by the area, and the canal is then sized by Manning's formula. #### Step 1: Design rainfall Mean annual 3-day maximum = 300 mm; for the 10-year return period the growth factor is 1.5, so $R = 1.5 \times 300 = 450$ mm in 3 days, i.e. 150 mm/day. #### Step 2: Drainage rate Initial level $h_0$ = 40 mm. Permissible: 300 mm for 1 day, and above 200 mm for up to 3 days. With rain falling uniformly and drainage $D$ starting with the rain: $$ \begin{aligned} h_{peak} &= 40 + 450 - 3D \\ T_{200} &= \left(3-\frac{200-40}{150-D}\right)+\frac{h_{peak}-200}{D} \le 3 \end{aligned} $$ By trial, $T_{200}$ = 3.00 days at $D$ = 60.6 mm/day, giving $h_{peak}$ = 308 mm; the time above 300 mm is only 0.23 day (< 1 day). So $$ D = 60.6\ \text{mm/day} = 7.01\ \text{l/s/ha} $$ #### Step 3: Design discharge $$ Q = D\,A = 7.01 \times 5 = 35.0\ \text{l/s} = 0.0350\ \text{m}^3/\text{s} $$ #### Step 4: Section of the drain Assumptions: earthen canal, side slope 1:1, Manning's $n$ = 0.025, bed slope $S$ = 1 in 1000, bed width $b$ = 0.5 m. Solving $Q=\frac{1}{n}AR^{2/3}S^{1/2}$ by trial: - Depth of flow $d$ = 0.172 m (say 0.17 m) - $A=(b+d)d$ = 0.116 m², $P=b+2\sqrt{2}\,d$ = 0.986 m, $R$ = 0.117 m - $V$ = 0.0350/0.116 = 0.30 m/s (about 0.3 m/s: low but non-eroding; silting is controlled by periodic cleaning) Provide a free board of 0.15 m, so the total depth = 0.17 + 0.15 ≈ 0.32 m. **Answer: D = 60.6 mm/day (7.01 l/s/ha), Q = 35.0 l/s; canal with bed width 0.5 m, flow depth 0.17 m (total depth 0.32 m), side slopes 1:1, slope 1 in 1000.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the 10-year, 3-day design rainfall obtained? A. Mean annual 3-day maximum x growth factor = 1.5 x 300 = 450 mm B. 300 mm directly C. 300 / 1.5 = 200 mm D. 300 + 1.5 = 301.5 mm Answer: A. The return-period rainfall is the 300 mm mean maximum multiplied by the growth factor 1.5, giving 450 mm (150 mm/day). (Key point: Design rainfall) 2. Which limit fixes the drainage rate D? A. Water above 300 mm for 1 day, giving D = 150 mm/day B. Water above 200 mm must not persist more than 3 days, giving D = 60.6 mm/day C. The initial depth of 40 mm, giving D = 40 mm/day D. The canal velocity limit Answer: B. T200 = 3.00 days at D = 60.6 mm/day; the peak is then 308 mm with only 0.23 day above 300 mm. (Key point: Governing limit) 3. How is the design discharge of the canal found? A. Q = 60.6 mm/day x 5 = 303 l/s B. Q = rainfall 450 mm x area converted directly C. Q = drainage rate x area = 7.01 l/s/ha x 5 ha = 35 l/s D. Q = 7.01 / 5 = 1.4 l/s Answer: C. Discharge is the drainage rate per hectare multiplied by the area drained: 35.0 l/s = 0.035 m3/s. (Key point: Design discharge) 4. How is the drain section sized once Q is known? A. By Lacey's silt theory only B. By setting velocity = 3 m/s and dividing Q by it C. By using the rainfall intensity formula D. By trial using Manning's formula Q = (1/n) A R^(2/3) S^(1/2) with assumed b, side slope, n and S Answer: D. Assuming bed width, side slope, n and slope, depth is found by trial until Manning's Q equals 0.035 m3/s (d = 0.17 m), then free board is added. (Key point: Canal section) ## Close variants set in other exams - [Drainage rate: 300 mm in 3 days, growth factor 1.5](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Adrainage-rate-3-day-300mm-growth) — Irrigation and Drainage Engineering, chapter 9, asked 2× ## Related in Irrigation and Drainage Engineering - [Surface drainage design, 40 ha Terai field](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Asurface-drainage-40ha-terai) — Irrigation and Drainage Engineering, chapter 9 - [Drainage rate (mm/day), initial 55 mm](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Adrainage-rate-mmday-bunded-55mm) — Irrigation and Drainage Engineering, chapter 9, asked 3× - [Drainage rate (l/s/ha) for bunded rice fields](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Adrainage-rate-lsha-bunded-rice-70bh) — Irrigation and Drainage Engineering, chapter 9, asked 3× - [Days inundated above 200 mm at 3 l/s/ha](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Ainundation-days-drainage-3lsha) — Irrigation and Drainage Engineering, chapter 9 - [Cross drainage: canal 300, drainage 500](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Acd-q300-canal-n0016-drain500) — Irrigation and Drainage Engineering, chapter 8 - [Short note: Design of drainage of irrigated land](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Adrainage-irrigated-land-design-note) — Irrigation and Drainage Engineering, chapter 9 ## Related in other subjects - [24-hr maximum rainfall for 50-year return period](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-24hr-max-50yr-semilog) — Engineering Hydrology, chapter 6