Hydraulic jump: derivation and numerical

(a) Hydraulic jump

A hydraulic jump is the sudden rise in water surface when a rapid (supercritical, $Fr > 1$) flow changes to a slow (subcritical, $Fr < 1$) flow. It is a turbulent phenomenon with large energy dissipation.

Derivation (horizontal channel, unit width, neglect friction on the bed): apply the momentum equation to a control volume enclosing the jump.

Hydrostatic forces: $F_1 = \tfrac12\rho g y_1^2$ and $F_2 = \tfrac12\rho g y_2^2$. Discharge per unit width: $q = V_1y_1 = V_2y_2$.

$$\tfrac12\rho g y_1^2 - \tfrac12\rho g y_2^2 = \rho q (V_2 - V_1)$$

Using $V_1 = q/y_1$, $V_2 = q/y_2$:

$$\frac{g}{2}(y_1^2 - y_2^2) = q^2\left(\frac{1}{y_2} - \frac{1}{y_1}\right) = \frac{q^2 (y_1 - y_2)}{y_1y_2}$$

Dividing by $(y_1 - y_2)$ and rearranging:

$$\frac{q^2}{g} = \frac{y_1y_2(y_1 + y_2)}{2}$$

With $q = V_1y_1$ and $Fr_1 = V_1/\sqrt{gy_1}$, this gives the quadratic in $y_2/y_1$:

$$\left(\frac{y_2}{y_1}\right)^2 + \frac{y_2}{y_1} - 2Fr_1^2 = 0 \quad \Rightarrow \quad \frac{y_2}{y_1} = \frac12\left(\sqrt{1 + 8Fr_1^2} - 1\right)$$

Head loss: $h_L = E_1 - E_2$, where $E = y + \dfrac{V^2}{2g}$. Substituting from above:

$$h_L = \frac{(y_2 - y_1)^3}{4y_1y_2}$$

(b) Numerical

Given $y_1 = 0.3\ \text{m}$, $V_1 = 6\ \text{m/s}$.

$$\begin{aligned} q &= 0.3 \times 6 = 1.8\ \text{m}^2/\text{s} \\ Fr_1 &= \frac{6}{\sqrt{9.81 \times 0.3}} = 3.497 \\ y_2 &= \frac{0.3}{2}\left(\sqrt{1 + 8(3.497)^2} - 1\right) = 1.341\ \text{m} \\ V_2 &= \frac{q}{y_2} = \frac{1.8}{1.341} = 1.342\ \text{m/s} \\ Fr_2 &= \frac{1.342}{\sqrt{9.81 \times 1.341}} = 0.370 \end{aligned}$$

Head loss and power:

$$\begin{aligned} h_L &= \frac{(1.341 - 0.3)^3}{4 \times 0.3 \times 1.341} = 0.702\ \text{m} \\ P &= \rho g q h_L = 1000 \times 9.81 \times 1.8 \times 0.702 = 12\,390\ \text{W/m} = 12.4\ \text{kW per m width} \end{aligned}$$

Since $Fr_2 < 1$, the downstream flow is subcritical, as expected.

Answer: $y_2 = 1.34\ \text{m}$, $V_2 = 1.34\ \text{m/s}$, $Fr_2 = 0.37$, $h_L = 0.70\ \text{m}$, $P = 12.4\ \text{kW per metre width}$