CivilHydraulicsmediumCourseworkPE / Advanced

Critical Depth in a Rectangular Channel

A rectangular channel carries a unit discharge (discharge per unit width) of $q = 3.0\ \text{m}^2/\text{s}$ .

Take $g = 9.81\ \text{m/s}^2$ .

Find the critical depth $y_c$ .

Given

3 m^2/s

Unit discharge

Hint 1

Critical flow is where the Froude number equals 1 and specific energy is minimised.

Hint 2

For a rectangular channel there is a closed-form result in terms of unit discharge.

Hint 3

y_c = (q²/g)^(1/3) — a cube root, not a square root.

Worked solution — try the problem first

For a rectangular channel, critical depth follows from setting the Froude number to 1:

$y_c = \left(\frac{q^2}{g}\right)^{1/3}$

$= \left(\frac{3.0^2}{9.81}\right)^{1/3} = \left(\frac{9.0}{9.81}\right)^{1/3} = (0.9174)^{1/3}$

$= 0.972\ \text{m}$

$\boxed{y_c = 0.972\ \text{m}}$

What critical depth means. It is the depth at which specific energy is minimised for a given discharge, and where $Fr = 1$ . Flow deeper than $y_c$ is subcritical (tranquil, controlled from downstream); shallower is supercritical (rapid, controlled from upstream). A hydraulic jump is the abrupt transition from supercritical back to subcritical.

Sanity check. Critical velocity is $V_c = q/y_c = 3.0/0.972 = 3.09$ m/s, and $\sqrt{gy_c} = \sqrt{9.81 \times 0.972} = 3.09$ m/s. Equal, so $Fr = 1$ ✓

Concepts:Critical depthFroude numberSpecific energyOpen channel flow

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