Hydraulic jump on energy and force diagrams

Hydraulic jump profile

A hydraulic jump is the abrupt rise in water surface when supercritical flow (depth $y_1$, $Fr_1 > 1$) changes to subcritical flow (depth $y_2$, $Fr_2 < 1$). The two depths are conjugate (sequent) depths. A turbulent roller forms and much energy is dissipated.

                      roller
   y1 ______       ,-~~~~-.______ y2
  (supercritical)  |<-- Lj -->|  (subcritical)
          V1 ====>     turbulence     ====> V2
  -----------------------------------------  bed

On the specific energy diagram

Specific energy $E = y + q^2/(2gy^2)$ for constant $q$ (the curve has its minimum at $y_c$ and approaches the line $E = y$ at large depths):

   y ^
     |                         .
  y2 |- - - - - - - - - - - - *   subcritical branch
     |                     .'
     |                 .-'
  yc |- - - - - - - - *   minimum E (critical depth)
     |                 '-.
     |                     '-._
  y1 |- - - - - - - - - - - - - - -*   supercritical branch
     +------------------------+----+---> E
                              E2   E1
                              |<-->|  energy loss dE = E1 - E2

• $y_1$ lies on the supercritical (lower) branch at energy $E_1$.
• $y_2$ lies on the subcritical (upper) branch at energy $E_2$.
• $y_1$ and $y_2$ are not alternate depths, because $E_2 < E_1$. The horizontal distance between the two points is the energy loss $\Delta E = E_1 - E_2$.

On the specific force diagram

Specific force $F = \dfrac{q^2}{gy} + \dfrac{y^2}{2}$ (per unit width):

   y ^
     |                         /
  y2 |- - - - - - - - - - - - *   subcritical branch
     |                    .'
     |                 .'
  yc |- - - - - - - - *   minimum F (critical depth)
     |                 '.
     |                    '.
  y1 |- - - - - - - - - - - - *   supercritical branch
     |                         \
     +------------------------+-----> F
                           F1 = F2

$y_1$ and $y_2$ have the same specific force (momentum is conserved across the jump, since friction over its short length is negligible and the bed is horizontal), so they lie on the same vertical line. The minimum specific force occurs at $y_c$.

Momentum equation for a rectangular channel

Take a horizontal rectangular channel and a control volume from section 1 to section 2. Per unit width, with $q = V_1y_1 = V_2y_2$:

Net force $=$ rate of change of momentum:

$$\frac{\gamma y_1^2}{2} - \frac{\gamma y_2^2}{2} = \rho q\,(V_2 - V_1)$$

Substituting $V = q/y$ and $\gamma = \rho g$:

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

Since $y_1^2 - y_2^2 = (y_1 - y_2)(y_1 + y_2)$, dividing by $(y_1 - y_2)$ gives:

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

Using $Fr_1^2 = q^2/(gy_1^3)$ and $r = y_2/y_1$:

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

This is the Belanger equation; it gives the sequent depth $y_2$ for a given $y_1$ and $Fr_1$.

Energy loss:

$$\Delta E = E_1 - E_2 = \frac{(y_2 - y_1)^3}{4y_1y_2}$$