--- title: "Lined canal capacity and irrigable area" subject: "Irrigation and Drainage Engineering" chapter: 4 key: "semester_6/irrigation_and_drainage_engineering:lined-canal-trapezoid-5m-bed-capacity" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/lined-canal-trapezoid-5m-bed-capacity site: BCT Engineering --- # Lined canal capacity and irrigable area Irrigation and Drainage Engineering · Chapter 4: Design of Canals - Asked in: 2081 Chaitra, 8 marks; 2080 Chaitra, 7 marks (asked 2 times) - 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/lined-canal-trapezoid-5m-bed-capacity - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4 ## Question A lined irrigation canal with trapezoidal cross-section has 5 m bed width, 25 m depth [as printed] and 1.5:1 (θ = 0.588 radian) side slope. The longitudinal bed slope of the canal is 1:1000 and Manning's n = 0.16 [as printed]. What is the maximum carrying capacity of the canal? What area of the land in hectares the canal can irrigate if the crop has 150 mm field irrigation requirement in a Kor period of 10 days. ## Short answer Using corrected depth 2.5 m, n=0.016, side slope 1.5H:1V: A=25.55 m², P=15.44 m, R=1.655 m, V=2.77 m/s, Q=70.6 m³/s. In 10 days volume=6.1×10⁷ m³; with 150 mm requirement, irrigable area≈40,700 ha. ## Answer The printed data look like misprints, so these are assumed: depth $D = 2.5$ m (not 25 m) and $n = 0.016$ (not 0.16; typical for concrete lining). $B = 5$ m, $S = 1/1000$, side slope 1.5H:1V, so $\cot\theta = 1.5$ and $\theta = \tan^{-1}(1/1.5) = 0.588$ rad. #### Carrying capacity (Manning's formula) Giving $\theta$ in radians shows that the standard **lined trapezoidal section with rounded bottom corners** (radius = $D$) is intended, for which $$ A = BD + D^2(\theta + \cot\theta),\qquad P = B + 2D(\theta + \cot\theta) $$ Here $\theta + \cot\theta = 0.588 + 1.5 = 2.088$. $$ \begin{aligned} A &= 5\times 2.5 + 2.5^2\times 2.088 = 12.50 + 13.05 = 25.55\ \text{m}^2 \\ P &= 5 + 2\times 2.5\times 2.088 = 15.44\ \text{m} \\ R &= A/P = 1.655\ \text{m} \\ V &= \frac{1}{n}R^{2/3}S^{1/2} = \frac{1}{0.016}\times 1.655^{2/3}\times 0.001^{1/2} = 2.765\ \text{m/s} \\ Q &= AV = 25.55\times 2.765 = 70.65\ \text{m}^3/\text{s} \end{aligned} $$ **Maximum carrying capacity $Q \approx 70.6$ m³/s** (full supply depth 2.5 m). #### Irrigable area Water delivered during one kor period of 10 days: $$ \text{Volume} = Q\,t = 70.65\times 10\times 86400 = 6.104\times10^{7}\ \text{m}^3 $$ Field irrigation requirement (depth) $= 150$ mm $= 0.15$ m. Neglecting losses beyond the stated field requirement: $$ \text{Area} = \frac{\text{Volume}}{\text{Depth}} = \frac{6.104\times10^{7}}{0.15} = 4.069\times10^{8}\ \text{m}^2 \approx 40,690\ \text{ha} $$ **Answer:** $Q \approx 70.6$ m³/s; irrigable area $\approx 40,700$ ha. (Velocity 2.77 m/s is within the safe limit for concrete lining.) If the section is treated as a plain trapezoid with sharp corners, $A = 21.875$ m², $P = 14.01$ m, $V = 2.66$ m/s, $Q \approx 58.2$ m³/s and the area $\approx 33,510$ ha. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does the answer treat the printed depth 25 m and n = 0.16? A. As misprints: D = 2.5 m and n = 0.016 for concrete B. Uses 25 m and 0.16 as given C. Uses D = 25 m and n = 0.016 D. Uses D = 2.5 m and n = 0.16 Answer: A. Those values are unrealistic, so 2.5 m and 0.016 (concrete lining) are assumed. (Key point: Assumed data) 2. Given theta = 0.588 rad (tan = 1/1.5), what is the side slope z? A. 1.5V:1H, z = 0.667 B. 1.5H:1V, z = 1.5 C. 0.588H:1V D. 1H:1.5V Answer: B. tan(theta) = 1/1.5 for angle with horizontal, so horizontal run is 1.5 per unit rise. (Key point: Side slope) 3. Using Manning with R = 1.561 m, S = 0.001, n = 0.016, the discharge is about A. 2.66 m^3/s B. 21.9 m^3/s C. 58.2 m^3/s D. 5.8 m^3/s Answer: C. V = 2.66 m/s and A = 21.875 m^2, so Q = 58.2 m^3/s. (Key point: Capacity) 4. How is the area irrigated in a Kor period found? A. Q divided by depth only B. Volume multiplied by depth C. Q x 10 days divided by 150 m D. Volume Q x t divided by field irrigation depth, about 33,510 ha Answer: D. Volume = 58.18 x 10 x 86400 m^3, divided by 0.15 m gives 3.35e8 m^2 = 33,510 ha. 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