Question 5 of 6: Hydraulic jump in a rectangular channel

Question 5: Hydraulic jump in a rectangular channel (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A rectangular channel carrying a fixed discharge with a known shallow supercritical approach depth.

Find. (a) the momentum (specific-force) equation of the jump; (b) the conjugate downstream depth $z_2$; (c) why momentum, not energy, governs.

(a) Momentum equation. Apply the momentum principle to the control volume spanning the jump in a horizontal, rectangular channel, neglecting boundary friction over the short length. The net hydrostatic pressure force equals the change in momentum flux. Per unit width (with $q=Q/B$): $$\underbrace{\tfrac{1}{2}\gamma z_1^2 + \dfrac{\gamma q^2}{g\,z_1}}_{\text{section 1}} = \underbrace{\tfrac{1}{2}\gamma z_2^2 + \dfrac{\gamma q^2}{g\,z_2}}_{\text{section 2}},\qquad\text{i.e. the specific force } M = \dfrac{q^2}{g\,z}+\dfrac{z^2}{2}\ \text{is conserved.}$$ Dividing through and rearranging yields the Bélanger conjugate-depth relation $$\dfrac{z_2}{z_1} = \tfrac{1}{2}\left(\sqrt{1+8\,\mathrm{Fr}_1^{\,2}}-1\right).$$

(b) Downstream depth. A short calculation gives the approach Froude number and then $z_2$:

1. Approach velocity and Froude number. $$q = \dfrac{20}{6.5}=3.077\ \text{m}^2/\text{s},\quad V_1 = \dfrac{q}{z_1}=\dfrac{3.077}{0.2}=15.38\ \text{m/s},\quad \mathrm{Fr}_1 = \dfrac{V_1}{\sqrt{g\,z_1}}=\dfrac{15.38}{\sqrt{9.81\cdot0.2}}=10.98.$$ The flow is strongly supercritical (a strong jump).
2. Conjugate depth. $$z_2 = \dfrac{z_1}{2}\left(\sqrt{1+8\,\mathrm{Fr}_1^{\,2}}-1\right)=\dfrac{0.2}{2}\left(\sqrt{1+8(10.98)^2}-1\right)=\boxed{3.01\ \text{m}}.$$ Then $V_2 = q/z_2 = 1.02$ m/s and $\mathrm{Fr}_2 = 0.19$ (subcritical), confirming a valid jump. The energy dissipated is $\Delta E = \left(z_1+\tfrac{V_1^2}{2g}\right)-\left(z_2+\tfrac{V_2^2}{2g}\right)=12.26-3.06=9.20$ m.

(c) Why momentum, not energy. A hydraulic jump is a highly turbulent, roller-dominated transition that dissipates a large but a priori unknown amount of mechanical energy (here 9.2 m of head, converted to turbulence and heat). The energy equation cannot be applied across the jump because the loss term is exactly the unknown we lack. The momentum equation, by contrast, involves only the external forces on the control volume — the hydrostatic pressure forces and (negligible) boundary friction — and the momentum flux, none of which require knowledge of the internal dissipation. It therefore closes with the given depths and discharge alone, which is why the conjugate-depth relation is derived from momentum.