--- title: "Canal in alluvium, Q 40, d 0.5 mm" subject: "Irrigation and Drainage Engineering" chapter: 4 key: "semester_6/irrigation_and_drainage_engineering:alluvial-canal-q40-d05mm" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/alluvial-canal-q40-d05mm site: BCT Engineering --- # Canal in alluvium, Q 40, d 0.5 mm Irrigation and Drainage Engineering · Chapter 4: Design of Canals - Asked in: 2065 Shrawan (old course), 8 marks - Video: [Design of Unlined Canal | avg shear stress | irrigation numerical | most asked problem](https://www.youtube.com/watch?v=Q3S46argeco) — 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/alluvial-canal-q40-d05mm - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4 ## Question An irrigation canal carrying a discharge of 40 m³/s has to be constructed in alluvial soil. If the mean diameter of the soil is 0.5 mm, design a suitable section and bed slope of such a canal. ## Short answer Lacey's design with f=1.245: bed width 26.3 m, depth 1.69 m, side slope 0.5H:1V, bed slope 1 in 4290, velocity 0.87 m/s. ## Answer No side slope is given, so 0.5H:1V is assumed. #### Silt factor $$ f = 1.76\sqrt{d_{mm}} = 1.76\sqrt{0.5} = 1.245 $$ #### Lacey's regime values Given: $Q = 40\ \text{m}^3/\text{s}$, $f = 1.245$, side slope 0.5H:1V. $$ \begin{aligned} V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{40\times 1.245^2}{140}\right)^{1/6} = 0.873\ \text{m/s} \\ A &= \frac{Q}{V} = \frac{40}{0.873} = 45.82\ \text{m}^2 \\ R &= \frac{5V^2}{2f} = \frac{5\times 0.873^2}{2\times 1.245} = 1.531\ \text{m} \\ P &= 4.75\sqrt{Q} = 4.75\sqrt{40} = 30.04\ \text{m} \\ S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{1.245^{5/3}}{3340\times 40^{1/6}} = 0.000233 = \frac{1}{4290} \end{aligned} $$ #### Trapezoidal section For side slope $z = 0.5$: $A = (B + zD)D$ and $P = B + 2D\sqrt{1+z^2} = B + 2.2361D$. Eliminating $B$: $$ \begin{aligned} 1.7361\,D^2 - 30.04\,D + 45.82 &= 0 \\ D &= \frac{30.04 - \sqrt{30.04^2 - 4\times 1.7361\times 45.82}}{2\times 1.7361} = 1.690\ \text{m} \\ B &= P - 2.2361D = 30.04 - 2.2361\times 1.690 = 26.26\ \text{m} \end{aligned} $$ (The other root of the quadratic gives a very deep, narrow section, so the smaller root is taken.) Check: $A = (26.26 + 0.5\times 1.690)\times 1.690 = 45.82\ \text{m}^2$ against required 45.82 $\text{m}^2$. **Answer:** Bed width $B \approx 26.3$ m, depth of water $D \approx 1.69$ m, side slope 0.5H:1V, bed slope $S = 1$ in 4290 (0.000233), velocity $V = 0.87$ m/s, hydraulic mean radius $R = 1.53$ m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For d = 0.5 mm, f is about A. 0.88 B. 1.76 C. 1.245 D. 2.49 Answer: C. f = 1.76 x sqrt(0.5) = 1.245. (Key point: Silt factor) 2. For Q = 40 m^3/s, what wetted perimeter does Lacey's P = 4.75 sqrt(Q) give? A. 30.04 m B. 15.02 m C. 45.06 m D. 40.04 m Answer: A. P = 4.75 x sqrt(40) = 30.04 m, independent of silt factor. (Key point: Wetted perimeter) 3. Using S = f^(5/3)/(3340 Q^(1/6)) with f = 1.245, Q = 40, the bed slope is about A. 1 in 429 B. 1 in 4290 C. 1 in 42900 D. 1 in 2145 Answer: B. Slope is computed directly from Lacey's slope equation; a finer silt (smaller f) gives a flatter slope. (Key point: Regime slope) 4. After finding A and P, how is the depth D of the 0.5H:1V trapezoid obtained, and which root is taken (result D = 1.690 m, B = 26.26 m)? A. Assume D and iterate until V = V0 B. Take D = R directly and find B from A C. Eliminate B using A = (B + 0.5D)D and P = B + 2.236D, solve the quadratic and take the smaller root D. Take the larger root, giving a deep narrow section Answer: C. The two equations in B and D give a quadratic; the smaller root is the sensible section, the larger gives a very deep narrow channel. 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