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Flow depth in triangular channel

Short answer

Using Manning's equation for uniform flow in a triangular channel (m=1.5, S=1/2000, n=0.025), the normal depth is calculated as y ≈ 1.57 m. The geometry gives A=1.5y², R=0.416y, and solving Q=(1/n)AR^(2/3)S^(1/2) yields y=1.57 m.

Find the depth of flow in a triangular channel having a longitudinal slope of 1 in 2000 and side slopes of 1.5H:1V. The channel carries a discharge of 2.5 m³/s. Take Manning's roughness coefficient n = 0.025.

Uniform flow depth is found with Manning's equation Q=1nAR2/3S1/2Q = \dfrac{1}{n}AR^{2/3}S^{1/2}.

Geometry (side slope m=1.5m = 1.5 H : 1 V)

A=my2=1.5y2P=2y1+m2=2y3.25=3.606yR=AP=1.5y23.606y=0.416y\begin{aligned} A &= my^2 = 1.5y^2 \\ P &= 2y\sqrt{1+m^2} = 2y\sqrt{3.25} = 3.606y \\ R &= \frac{A}{P} = \frac{1.5y^2}{3.606y} = 0.416y \end{aligned}

Manning's equation

2.5=10.025(1.5y2)(0.416y)2/3(12000)1/22.5=40×1.5×0.5573×0.02236×y8/32.5=0.7477 y8/3y8/3=3.344  ⇒  y=3.3443/8=1.572 m\begin{aligned} 2.5 &= \frac{1}{0.025}(1.5y^2)(0.416y)^{2/3}\left(\frac{1}{2000}\right)^{1/2} \\ 2.5 &= 40 \times 1.5 \times 0.5573 \times 0.02236 \times y^{8/3} \\ 2.5 &= 0.7477\, y^{8/3} \\ y^{8/3} &= 3.344 \;\Rightarrow\; y = 3.344^{3/8} = 1.572\ \text{m} \end{aligned}

Check

A=1.5(1.572)2=3.709A = 1.5(1.572)^2 = 3.709 m², P=5.670P = 5.670 m, R=0.654R = 0.654 m, V=2.5/3.709=0.674V = 2.5/3.709 = 0.674 m/s, and 10.025(3.709)(0.654)2/3(0.02236)=2.50\frac{1}{0.025}(3.709)(0.654)^{2/3}(0.02236) = 2.50 m³/s. Correct.

Answer: Depth of flow y≈1.57y \approx 1.57 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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