--- title: "Lacey and Kennedy bed width, Q 40" subject: "Irrigation and Drainage Engineering" chapter: 4 key: "semester_6/irrigation_and_drainage_engineering:lacey-kennedy-q40-comparison" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/lacey-kennedy-q40-comparison site: BCT Engineering --- # Lacey and Kennedy bed width, Q 40 Irrigation and Drainage Engineering · Chapter 4: Design of Canals - Asked in: 2070 Bhadra, 4+6 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/lacey-kennedy-q40-comparison - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4 ## Question A stable channel is to be designed for a discharge of 40 m³/s and the silt factor of unity. Calculate the dimensions of the channel using Lacey's regime equations. What would be the bed-width of this channel if it were to be designed on the basis of Kennedy's method with critical velocity ratio equal to unity and the ratio of bed-width to depth of flow the same as obtained from Lacey's method. ## Short answer Lacey: B=25.9 m, D=1.84 m, S=1/6177. Kennedy with m=1 and same B/D yields identical dimensions (B=25.9 m, D=1.84 m) because Kennedy's critical velocity equals Lacey's velocity. ## Answer Side slope is not stated, so 0.5H:1V is assumed. #### Lacey's design Given: $Q = 40\ \text{m}^3/\text{s}$, $f = 1.000$, side slope 0.5H:1V. $$ \begin{aligned} V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{40\times 1.000^2}{140}\right)^{1/6} = 0.812\ \text{m/s} \\ A &= \frac{Q}{V} = \frac{40}{0.812} = 49.29\ \text{m}^2 \\ R &= \frac{5V^2}{2f} = \frac{5\times 0.812^2}{2\times 1.000} = 1.647\ \text{m} \\ P &= 4.75\sqrt{Q} = 4.75\sqrt{40} = 30.04\ \text{m} \\ S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{1.000^{5/3}}{3340\times 40^{1/6}} = 0.000162 = \frac{1}{6177} \end{aligned} $$ #### Trapezoidal section For side slope $z = 0.5$: $A = (B + zD)D$ and $P = B + 2D\sqrt{1+z^2} = B + 2.2361D$. Eliminating $B$: $$ \begin{aligned} 1.7361\,D^2 - 30.04\,D + 49.29 &= 0 \\ D &= \frac{30.04 - \sqrt{30.04^2 - 4\times 1.7361\times 49.29}}{2\times 1.7361} = 1.835\ \text{m} \\ B &= P - 2.2361D = 30.04 - 2.2361\times 1.835 = 25.94\ \text{m} \end{aligned} $$ (The other root of the quadratic gives a very deep, narrow section, so the smaller root is taken.) Check: $A = (25.94 + 0.5\times 1.835)\times 1.835 = 49.29\ \text{m}^2$ against required 49.29 $\text{m}^2$. **Answer:** Bed width $B \approx 25.9$ m, depth of water $D \approx 1.84$ m, side slope 0.5H:1V, bed slope $S = 1$ in 6177 (0.000162), velocity $V = 0.81$ m/s, hydraulic mean radius $R = 1.65$ m. #### Kennedy's method with $m = 1$ and the same $B/D$ Lacey's $B/D = 25.94/1.835 = 14.13$. Let $B = 14.13D$, so $A = (B + 0.5D)D = 14.63D^2$. Kennedy's critical velocity: $V_0 = 0.55\,m\,D^{0.64}$ with $m = 1$, and $Q = V_0A$: $$ \begin{aligned} 40 &= 0.55D^{0.64}\times 14.63D^2 = 8.048\,D^{2.64} \\ D^{2.64} &= 4.970 \\ D &= 1.836\ \text{m} \\ B &= 14.13\times 1.836 = 25.94\ \text{m} \\ V_0 &= 0.55\times 1.836^{0.64} = 0.811\ \text{m/s} \end{aligned} $$ **Answer:** Lacey: $B \approx 25.9$ m, $D \approx 1.84$ m, $V = 0.81$ m/s, $S$ = 1 in 6177. Kennedy (with $m = 1$, same $B/D$): bed width $\approx 25.9$ m, $D \approx 1.84$ m, essentially the same as Lacey's because $V_0 = 0.811$ m/s is almost equal to Lacey's $V = 0.812$ m/s. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For Q = 40 m^3/s and f = 1, which relation gives Lacey's regime velocity? A. V = 0.55 m D^0.64 B. V = 4.75 sqrt(Q) C. V = (Q f^2/140)^(1/6) D. V = (Q/140 f)^(1/2) Answer: C. Lacey's velocity is V = (Q f^2/140)^(1/6) = 0.812 m/s here. 0.55 m D^0.64 is Kennedy's critical velocity. (Key point: Lacey velocity) 2. How is Kennedy's design set up so that it uses the same B/D as Lacey's (14.13)? A. Take Kennedy's D equal to Lacey's R and find B from P B. Use Lacey's V as V0 and then choose D arbitrarily C. Assume B = D and scale the discharge afterwards D. Put B = 14.13 D, so A = (B + 0.5D)D = 14.63 D^2, and solve Q = V0 A with V0 = 0.55 m D^0.64 Answer: D. Substituting B = 14.13 D into the area and using Q = V0 A with m = 1 gives 40 = 8.048 D^2.64, so D = 1.836 m. (Key point: Same B/D in Kennedy) 3. Why does Kennedy's design give almost the same bed width (about 25.9 m) as Lacey's? A. Kennedy's V0 = 0.811 m/s is almost equal to Lacey's V = 0.812 m/s B. Kennedy's method ignores the side slope just like Lacey's C. Both methods use the same silt factor f = 1 D. Kennedy's depth was fixed equal to Lacey's depth Answer: A. With the same B/D and nearly the same velocity, the area and hence D and B come out nearly equal. 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