--- title: "Lacey design: Q 30, f 1.1" subject: "Irrigation and Drainage Engineering" chapter: 4 key: "semester_6/irrigation_and_drainage_engineering:lacey-q30-f11" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/lacey-q30-f11 site: BCT Engineering --- # Lacey design: Q 30, f 1.1 Irrigation and Drainage Engineering · Chapter 4: Design of Canals - Asked in: 2076 Bhadra, 6 marks - Video: [Lacey's sum | Type 1 when Silt factor is given & Side slope is 1/2 : 1 | Lacey's Design of canal](https://www.youtube.com/watch?v=U4SwFYekkLg) — Professor Nidhi Trivedi - 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/lacey-q30-f11 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4 ## Question Design a regime channel (using Lacey's equation) for the following data: Discharge Q = 30 cumecs, Silt factor f = 1.1, Side slope = 0.5:1. Find also the longitudinal slope. ## Short answer Lacey's equations yield V = 0.80 m/s, A = 37.6 m², R = 1.45 m, P = 26.0 m, slope S = 0.000199 (1/5023). For 0.5:1 side slopes, depth D = 1.62 m and bed width B = 22.4 m. ## Answer #### Lacey's regime values Given: $Q = 30\ \text{m}^3/\text{s}$, $f = 1.100$, side slope 0.5H:1V. $$ \begin{aligned} V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{30\times 1.100^2}{140}\right)^{1/6} = 0.799\ \text{m/s} \\ A &= \frac{Q}{V} = \frac{30}{0.799} = 37.57\ \text{m}^2 \\ R &= \frac{5V^2}{2f} = \frac{5\times 0.799^2}{2\times 1.100} = 1.449\ \text{m} \\ P &= 4.75\sqrt{Q} = 4.75\sqrt{30} = 26.02\ \text{m} \\ S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{1.100^{5/3}}{3340\times 30^{1/6}} = 0.000199 = \frac{1}{5023} \end{aligned} $$ #### Trapezoidal section For side slope $z = 0.5$: $A = (B + zD)D$ and $P = B + 2D\sqrt{1+z^2} = B + 2.2361D$. Eliminating $B$: $$ \begin{aligned} 1.7361\,D^2 - 26.02\,D + 37.57 &= 0 \\ D &= \frac{26.02 - \sqrt{26.02^2 - 4\times 1.7361\times 37.57}}{2\times 1.7361} = 1.619\ \text{m} \\ B &= P - 2.2361D = 26.02 - 2.2361\times 1.619 = 22.40\ \text{m} \end{aligned} $$ (The other root of the quadratic gives a very deep, narrow section, so the smaller root is taken.) Check: $A = (22.40 + 0.5\times 1.619)\times 1.619 = 37.57\ \text{m}^2$ against required 37.57 $\text{m}^2$. **Answer:** Bed width $B \approx 22.4$ m, depth of water $D \approx 1.62$ m, side slope 0.5H:1V, bed slope $S = 1$ in 5023 (0.000199), velocity $V = 0.80$ m/s, hydraulic mean radius $R = 1.45$ m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the Lacey design procedure, which quantity is found first from Q and f? A. Velocity V = (Qf^2/140)^(1/6) B. Slope S from Manning's formula C. Depth D by trial and error D. Bed width B from Kutter's formula Answer: A. V is computed directly from Q and f, then A = Q/V, R = 5V^2/2f, P = 4.75 sqrt(Q) and S follow. (Key point: Order of Lacey design) 2. For Q = 30 m^3/s, what wetted perimeter does Lacey's P = 4.75 sqrt(Q) give? A. 13.01 m B. 26.02 m C. 39.03 m D. 36.02 m Answer: B. P = 4.75 x sqrt(30) = 26.02 m, independent of silt factor. (Key point: Wetted perimeter) 3. Using S = f^(5/3)/(3340 Q^(1/6)) with f = 1.1, Q = 30, the bed slope is about A. 1 in 502 B. 1 in 50230 C. 1 in 5023 D. 1 in 2511 Answer: C. 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 = 1.619 m, B = 22.40 m)? A. Assume D and iterate until V = V0 B. Take D = R directly and find B from A C. Take the larger root, giving a deep narrow section D. Eliminate B using A = (B + 0.5D)D and P = B + 2.236D, solve the quadratic and take the smaller root Answer: D. 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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