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Best hydraulic rectangular canal design

Short answer

For the best hydraulic rectangular section, width b = 2y. Solving Manning's equation with Q=15 m³/s, n=0.02, S=0.001 gives depth 2.13 m, width 4.26 m, and velocity 1.65 m/s.

Design the best hydraulic rectangular cross-section for a canal conveying 15 m³/s discharge with Manning's n = 0.02 and a bed slope of 0.001.

For the best hydraulic rectangular section, width is twice the depth: b=2yb = 2y, so A=2y2A = 2y^2 and R=y/2R = y/2.

Manning's equation

Q=1nAR2/3S1/215=10.02(2y2)(y2)2/3(0.001)1/215=50×2×0.63×0.03162 y8/3=1.992 y8/3y8/3=7.529  ⇒  y=7.5293/8=2.132 m\begin{aligned} Q &= \frac{1}{n}AR^{2/3}S^{1/2} \\ 15 &= \frac{1}{0.02}(2y^2)\left(\frac{y}{2}\right)^{2/3}(0.001)^{1/2} \\ 15 &= 50 \times 2 \times 0.63 \times 0.03162\, y^{8/3} = 1.992\,y^{8/3} \\ y^{8/3} &= 7.529 \;\Rightarrow\; y = 7.529^{3/8} = 2.132\ \text{m} \end{aligned}

Section

b=2×2.132=4.264 mA=2×2.1322=9.091 m2,R=1.066 mV=15/9.091=1.65 m/s\begin{aligned} b &= 2 \times 2.132 = 4.264\ \text{m} \\ A &= 2 \times 2.132^2 = 9.091\ \text{m}^2, \quad R = 1.066\ \text{m} \\ V &= 15/9.091 = 1.65\ \text{m/s} \end{aligned}

Check: 50×9.091×1.0662/3×0.03162=15.050 \times 9.091 \times 1.066^{2/3} \times 0.03162 = 15.0 m³/s.

   |<---- b = 4.26 m ---->|
   |~~~~~~~~~~~~~~~~~~~~~~|
   |                      |  y = 2.13 m
   |______________________|

Answer: Width ≈ 4.26 m, depth ≈ 2.13 m (plus freeboard), velocity ≈ 1.65 m/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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