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Normal depth in triangular channel (1:1)

Short answer

For a triangular channel with 1:1 side slopes (m=1), longitudinal slope 0.0004, discharge 1 m³/s, and Manning's n=0.014, the normal depth is calculated using Manning's equation as y ≈ 1.13 m. The geometry gives A=y² and R=0.3536y.

Find the normal depth of flow in a triangular channel having longitudinal slope of 0.0004 and side slopes of 1:1 when it carries 1 m³/sec, manning roughness coefficient (n) = 0.014.

Normal depth is the uniform flow depth from Manning's equation.

Geometry (side slope 1:1, m=1m = 1)

A=my2=y2P=2y1+12=22 y=2.828yR=y22.828y=0.3536y\begin{aligned} A &= my^2 = y^2 \\ P &= 2y\sqrt{1+1^2} = 2\sqrt{2}\,y = 2.828y \\ R &= \frac{y^2}{2.828y} = 0.3536y \end{aligned}

Manning's equation

Q=1nAR2/3S1/21=10.014 y2 (0.3536y)2/3(0.0004)1/21=71.43×0.5×0.02 y8/3=0.7143 y8/3y8/3=1.400  ⇒  y=1.4003/8=1.134 m\begin{aligned} Q &= \frac{1}{n}AR^{2/3}S^{1/2} \\ 1 &= \frac{1}{0.014}\,y^2\,(0.3536y)^{2/3}(0.0004)^{1/2} \\ 1 &= 71.43 \times 0.5 \times 0.02\, y^{8/3} = 0.7143\, y^{8/3} \\ y^{8/3} &= 1.400 \;\Rightarrow\; y = 1.400^{3/8} = 1.134\ \text{m} \end{aligned}

(Here 0.35362/3=0.50.3536^{2/3} = 0.5.)

Check

A=1.1342=1.287A = 1.134^2 = 1.287 m², P=3.209P = 3.209 m, R=0.401R = 0.401 m, V=1/1.287=0.777V = 1/1.287 = 0.777 m/s, Q=71.43×1.287×0.4012/3×0.02=1.00Q = 71.43 \times 1.287 \times 0.401^{2/3} \times 0.02 = 1.00 m³/s.

Answer: Normal depth yn≈1.13y_n \approx 1.13 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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