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Uniform flow in rectangular channel (Chezy)

Short answer

Using Chezy's formula: A=16.875 m², P=12 m, R=1.406 m. Velocity V=55√(1.406×1/1000)=2.063 m/s, discharge Q=34.8 m³/s.

Find the rate of flow for a rectangular channel 7.5 m wide for uniform flow at a depth of 2.25 m. The channel is having bed slope as 1 in 1000. Take Chezy's constant C = 55.

Uniform flow in an open channel is given by Chezy's formula Q=ACRSQ = AC\sqrt{RS}.

Section properties

A=by=7.5×2.25=16.875 m2P=b+2y=7.5+2×2.25=12.0 mR=AP=16.87512=1.406 m\begin{aligned} A &= by = 7.5 \times 2.25 = 16.875\ \text{m}^2 \\ P &= b + 2y = 7.5 + 2 \times 2.25 = 12.0\ \text{m} \\ R &= \frac{A}{P} = \frac{16.875}{12} = 1.406\ \text{m} \end{aligned}

Velocity and discharge

V=CRS=551.406×11000=55×0.0375=2.063 m/sQ=AV=16.875×2.063=34.80 m3/s\begin{aligned} V &= C\sqrt{RS} = 55\sqrt{1.406 \times \frac{1}{1000}} = 55 \times 0.0375 = 2.063\ \text{m/s} \\ Q &= AV = 16.875 \times 2.063 = 34.80\ \text{m}^3/\text{s} \end{aligned}

Answer: Rate of flow Q≈34.8Q \approx 34.8 m³/s (velocity ≈ 2.06 m/s).

✓ Answer checked by Claude Opus 5.5 (high effort)

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