Non-uniform rapidly varied flow (RVF)

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 $q$:

   y ^           /   45 deg line
     |          /
  y2 |- - - - -*---.  (subcritical branch)
     |        /|   .
  yc |-------*-+---.--- minimum E
     |        \|   .
  y1 |- - - - -*<--|
     |          E1  E2
     +--------------------> E
         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: $E_2 < E_1$. The horizontal distance between them 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 |- -\- - - -/- -   same F
     |     \    /
  yc |------ * ------   minimum F
     |      /  \
  y1 |- - -/- - -\- -   same F
     +---------------> F

$y_1$ and $y_2$ have the same specific force (the momentum is conserved across the jump, since friction on the short length is negligible and the bed is horizontal), so they are on the same vertical line.

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:

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

Given: horizontal channel, $y_1 = 1$ m (upstream, subcritical), $y_3 = 0.3$ m (supercritical, after the bump), then a hydraulic jump to $y_4$. Friction is neglected. Section 1 and section 3 are at the same bed level.

$v_1$ and $v_3$

Continuity: $v_1y_1 = v_3y_3$, so $v_3 = v_1(1/0.3) = 3.333\,v_1$.

Energy (smooth flow over the bump, no loss between 1 and 3):

$$y_1 + \frac{v_1^2}{2g} = y_3 + \frac{v_3^2}{2g}$$ $$1 - 0.3 = \frac{v_1^2}{2g}\left[\left(\frac{1}{0.3}\right)^2 - 1\right] \;\Rightarrow\; v_1^2 = \frac{2(9.81)(0.7)}{10.111} = 1.358$$ $$v_1 = 1.165\ \text{m/s},\qquad v_3 = 3.885\ \text{m/s}$$

$y_4$ (after the jump)

$$Fr_3 = \frac{3.885}{\sqrt{9.81 \times 0.3}} = 2.265$$ $$y_4 = \frac{0.3}{2}\left(\sqrt{1 + 8(2.265)^2} - 1\right) = 0.822\ \text{m}$$

Bump height $h$

The flow changes from subcritical to supercritical smoothly over the bump, so it passes through critical depth at the crest (section 2). With $q = v_1y_1 = 1.165\ \text{m}^2/\text{s}$:

$$y_c = \left(\frac{1.165^2}{9.81}\right)^{1/3} = 0.517\ \text{m},\qquad E_c = 1.5\,y_c = 0.776\ \text{m}$$ $$E_1 = 1 + \frac{1.165^2}{2g} = 1.069\ \text{m}$$ $$h = E_1 - E_c = 1.069 - 0.776 = 0.293\ \text{m}$$

Answer: $v_1 = 1.17$ m/s; $v_3 = 3.89$ m/s; $y_4 = 0.82$ m; $h \approx 0.29$ m (critical flow assumed at the crest).

Given: horizontal channel, $y_1 = 1$ m (upstream, subcritical), $y_2 = 0.3$ m (supercritical, after the bump). The jump takes the flow from section 3 (start of jump, still at $y_2 = 0.3$ m because friction is neglected) to section 4. Friction is neglected, and sections 1 and 2 are at the same bed level.

$v_1$ and $v_2$

Continuity: $v_1y_1 = v_2y_2$, so $v_2 = v_1/0.3$.

Energy (no loss between 1 and 2):

$$y_1 + \frac{v_1^2}{2g} = y_2 + \frac{v_2^2}{2g}$$ $$1 - 0.3 = \frac{v_1^2}{2g}\left[\left(\frac{1}{0.3}\right)^2 - 1\right] \;\Rightarrow\; v_1^2 = \frac{2(9.81)(0.7)}{10.111} = 1.358$$ $$v_1 = 1.165\ \text{m/s},\qquad v_2 = 3.885\ \text{m/s}$$

$y_4$ (depth after the jump)

The supercritical depth at the jump toe is $y_3 = y_2 = 0.3$ m (the same flow, no losses before the jump):

$$Fr_3 = \frac{3.885}{\sqrt{9.81 \times 0.3}} = 2.265$$ $$y_4 = \frac{y_3}{2}\left(\sqrt{1 + 8Fr_3^2} - 1\right) = 0.15\left(\sqrt{1 + 41.04} - 1\right) = 0.822\ \text{m}$$

Check by specific force (per unit width): $\dfrac{q^2}{g\,y_3} + \dfrac{y_3^2}{2} = \dfrac{1.358}{9.81 \times 0.3} + 0.045 = 0.506\ \text{m}^2$ and $\dfrac{1.358}{9.81 \times 0.822} + \dfrac{0.822^2}{2} = 0.168 + 0.338 = 0.506\ \text{m}^2$. The two are equal.

Answer: $v_1 = 1.17$ m/s; $v_2 = 3.89$ m/s; $y_4 = 0.82$ m.

Given: $b = 5$ m, $n = 0.011$, $S_0 = 0.0004$. At the gate $y_1 = 4$ m upstream and $y_2 = 0.5$ m at the vena contracta. The sequent depth after the jump equals the normal depth $y_3 = y_0$.

Discharge (energy equation across the gate, no loss):

$$Q = b\,y_1y_2\sqrt{\frac{2g}{y_1 + y_2}} = 5(4)(0.5)\sqrt{\frac{19.62}{4.5}} = 20.88\ \text{m}^3/\text{s}$$ $$q = 4.176\ \text{m}^2/\text{s},\quad V_2 = \frac{20.88}{5 \times 0.5} = 8.352\ \text{m/s},\quad Fr_2 = 3.77$$

Normal depth by Manning's equation (trial, $R = A/P$): $y_0 = y_3 = 2.102$ m, $V_3 = 20.88/(5 \times 2.102) = 1.986$ m/s.

(Without blocks, the sequent depth of 0.5 m would be $y_2' = 2.428$ m. The blocks reduce the required depth to 2.102 m.)

i) Head loss in the jump

$$E_2 = 0.5 + \frac{8.352^2}{2g} = 0.5 + 3.556 = 4.056\ \text{m}$$ $$E_3 = 2.102 + \frac{1.986^2}{2g} = 2.102 + 0.201 = 2.303\ \text{m}$$ $$\Delta E = E_2 - E_3 = 4.056 - 2.303 = 1.75\ \text{m}$$

ii) Force on the blocks

Momentum equation between sections 2 and 3 for the control volume (the force of the blocks on the water, $F_b$, acts upstream; bed friction and the gravity component are neglected):

$$\frac{\gamma b y_2^2}{2} + \rho Q V_2 - F_b = \frac{\gamma b y_3^2}{2} + \rho Q V_3$$ $$F_b = 6131 + 174\,400 - 108\,400 - 41\,480 \approx 30\,700\ \text{N}$$

The force on the blocks by the water is equal and acts downstream.

Answer: (i) $\Delta E = 1.75$ m; (ii) $F_b \approx 30.7$ kN.

Given: $b = 10$ m, $Q = 150\ \text{m}^3/\text{s}$ ($q = 15\ \text{m}^2/\text{s}$), tailwater $y_2 = 4$ m. Without blocks the jump needs a sequent depth of $4 + 1 = 5$ m.

i) Initial depth, energy loss and area of blocks

Initial depth $y_1$ (from the sequent depth of 5 m without blocks):

$$y_1 = \frac{y_2'}{2}\left(\sqrt{1 + 8Fr_2'^2} - 1\right)\ \text{or}\ \frac{q^2}{g} = \frac{y_1y_2'(y_1 + y_2')}{2}$$

Solving $\dfrac{225}{9.81} = 22.94 = \dfrac{y_1(5)(y_1 + 5)}{2}$ gives $y_1 = 1.427$ m.

$$V_1 = \frac{15}{1.427} = 10.51\ \text{m/s},\quad E_1 = 1.427 + \frac{10.51^2}{2g} = 7.056\ \text{m}$$

Downstream (tailwater): $V_2 = 15/4 = 3.75$ m/s, $E_2 = 4 + \dfrac{3.75^2}{2g} = 4.717$ m.

Energy loss with blocks:

$$\Delta E = E_1 - E_2 = 7.056 - 4.717 = 2.34\ \text{m}$$

Force on blocks (momentum equation, horizontal bed, no friction):

$$F_b = \frac{\gamma b}{2}(y_1^2 - y_2^2) + \rho Q(V_1 - V_2)$$ $$F_b = 99\,935 + 1\,576\,315 - 784\,800 - 562\,500 = 328\,950\ \text{N} \approx 329\ \text{kN}$$

Frontal area from $F_b = 2\gamma A E_1$:

$$A = \frac{328\,950}{2 \times 9810 \times 7.056} = 2.376\ \text{m}^2$$

Answer (i): $A \approx 2.38\ \text{m}^2$, $\Delta E \approx 2.34$ m.

ii) Without blocks

On the specific force curve, $F = \dfrac{q^2}{gy} + \dfrac{y^2}{2}$ (per unit width, in $\text{m}^2$), the initial depth $y_1 = 1.427$ m and the sequent depth $y_2' = 5$ m have the same specific force:

$$F(1.427) = \frac{225}{9.81 \times 1.427} + \frac{1.427^2}{2} = 16.07 + 1.02 = 17.09\ \text{m}^2$$ $$F(5) = \frac{225}{9.81 \times 5} + \frac{5^2}{2} = 4.59 + 12.5 = 17.09\ \text{m}^2$$

Both points lie on the vertical line $F = 17.09$.

iii) With blocks

The tailwater depth 4 m has a specific force $F_2 = 15^2/(9.81 \times 4) + 4^2/2 = 5.73 + 8 = 13.73\ \text{m}^2$, which is less than $F_1 = 17.09\ \text{m}^2$. The difference is supplied by the blocks:

$$\frac{F_b}{\gamma b} = F_1 - F_2 = 17.09 - 13.73 = 3.36\ \text{m}^2\ \Rightarrow\ F_b = 9810 \times 10 \times 3.36 \approx 330\ \text{kN}\ (\text{agrees})$$

   y ^
     |                          * y2'=5   (no blocks)
   5 |- - - - - - - - - - - - -/- -
   4 |- - - - - - - - - - -  *-/ - - y2 = 4 (with blocks)
     |                     /
  yc |------------------- *  minimum F
     |                    \
 1.43|- - - - - - - - - - -*- - y1
     +-------+-------+-------> F
            13.73   17.09
        (F2)   <-Fb/gb->  F1

• Without blocks: $y_1 = 1.427$ m and $y_2' = 5$ m are at $F = 17.09$.
• With blocks: $y_1$ is at $F_1 = 17.09$ and the tailwater $y_2 = 4$ m is at $F_2 = 13.73$; the gap equals $F_b/(\gamma b) = 3.36\ \text{m}^2$.

iv) Effect of 10% taller blocks

Frontal area increases by 10% (width unchanged), $A' = 1.1 \times 2.376 = 2.614\ \text{m}^2$. With the tailwater fixed at 4 m, the momentum equation and $F_b = 2\gamma A'E_1$ are solved together for the new initial depth:

$$y_1' = 1.375\ \text{m},\quad E_1' = 7.438\ \text{m},\quad F_b' = 381\ \text{kN}$$ $$\Delta E' = 7.438 - 4.717 = 2.72\ \text{m}$$ $$\frac{2.72 - 2.34}{2.34} \times 100 = +16.3\%$$

Answer (iv): the energy loss increases by about 16%.

Hydraulic jumps are classified by the upstream Froude number $Fr_1$ (USBR classification):

The percentage of energy lost, $\Delta E/E_1$, rises with $Fr_1$. Stilling basins are designed to produce a steady jump ($Fr_1 = 4.5$ to $9$); oscillating jumps are avoided.

Given: $b = 5$ m, $Q = 15\ \text{m}^3/\text{s}$ ($q = 3\ \text{m}^2/\text{s}$), hump $\Delta z = 0.4$ m with critical flow at section 3. Losses are neglected, except in the jump.

Critical conditions on the hump:

$$y_c = \left(\frac{3^2}{9.81}\right)^{1/3} = 0.972\ \text{m},\qquad E_c = 1.5\,y_c = 1.458\ \text{m}$$

i) Depths $y_2$ and $y_1$

Energy equation between section 2 (upstream of the hump, after the jump) and section 3:

$$E_2 = E_c + \Delta z = 1.458 + 0.4 = 1.858\ \text{m}$$

Section 2 is subcritical:

$$y_2 + \frac{3^2}{2g\,y_2^2} = 1.858 \;\Rightarrow\; y_2 + \frac{0.4587}{y_2^2} = 1.858 \;\Rightarrow\; y_2 = 1.699\ \text{m}$$

Section 1 is the supercritical depth conjugate to $y_2$:

$$\frac{q^2}{g} = \frac{y_1y_2(y_1 + y_2)}{2} \;\Rightarrow\; 0.9174 = \frac{1.699\,y_1(y_1 + 1.699)}{2} \;\Rightarrow\; y_1 = 0.493\ \text{m}$$

Check: $V_1 = 6.09$ m/s, $Fr_1 = 2.77$, and $\dfrac{y_1}{2}\left(\sqrt{1 + 8(2.77)^2} - 1\right) = 1.699$ m.

Answer (i): $y_2 = 1.70$ m, $y_1 = 0.49$ m.

ii) With a chute block

Assumed: the block's frontal area is height $\times$ width $= 0.1 \times 1 = 0.1\ \text{m}^2$ (the 1 m dimension is across the channel). The drag force on the block (acting on the water in the upstream direction), using the velocity at section 1:

$$F_D = C_D\,\rho A\frac{V_1^2}{2} = 0.1 \times 1000 \times 0.1 \times \frac{6.086^2}{2} = 185\ \text{N}$$

Momentum equation between sections 1 and 2:

$$\frac{\gamma b y_1^2}{2} + \rho Q V_1 - F_D = \frac{\gamma b y_2^2}{2} + \rho Q V_2$$

Per metre: $\gamma b y_1^2/2 = 5960$ N, $\rho QV_1 = 91\,288$ N, so the left side is $97\,063$ N. The right side is $24\,525\,y_2^2 + 45\,000/y_2$. Solving:

$$24\,525\,y_2^2 + \frac{45\,000}{y_2} = 97\,063 \;\Rightarrow\; y_2 = 1.696\ \text{m}$$

Answer (ii): $y_2 \approx 1.70$ m (a reduction of about 3 mm, since the block force is very small compared with the momentum flux of the jump).

For a 90° triangular channel the side slopes are 1H : 1V, so $z = 1$.

Derivation

Geometry at depth $y$: $A = y^2$, $T = 2y$, and the centroid of the area is at $\bar y = y/3$ below the surface.

Specific force:

$$F = \frac{Q^2}{gA} + A\bar y = \frac{Q^2}{gy^2} + \frac{y^3}{3}$$

For a jump in a horizontal channel, $F_1 = F_2$:

$$\frac{Q^2}{g}\left(\frac{1}{y_1^2} - \frac{1}{y_2^2}\right) = \frac{y_2^3 - y_1^3}{3}$$ $$\frac{Q^2}{g}\cdot\frac{y_2^2 - y_1^2}{y_1^2y_2^2} = \frac{y_2^3 - y_1^3}{3}$$ $$\boxed{Q^2 = \frac{g}{3}\cdot\frac{y_1^2y_2^2\,(y_2^3 - y_1^3)}{y_2^2 - y_1^2}}$$

Flow rate

With $y_1 = 0.5$ m and $y_2 = 1.0$ m:

$$Q^2 = \frac{9.81}{3}\cdot\frac{(0.25)(1)(1 - 0.125)}{1 - 0.25} = 3.27 \times 0.29167 = 0.9538$$ $$Q = 0.977\ \text{m}^3/\text{s}$$

Froude numbers

The hydraulic depth is $D = A/T = y/2$.

$$V_1 = \frac{Q}{y_1^2} = \frac{0.9766}{0.25} = 3.906\ \text{m/s},\qquad Fr_1 = \frac{3.906}{\sqrt{9.81 \times 0.25}} = 2.49$$ $$V_2 = \frac{0.9766}{1.0} = 0.977\ \text{m/s},\qquad Fr_2 = \frac{0.977}{\sqrt{9.81 \times 0.5}} = 0.44$$

Answer: $Q \approx 0.98\ \text{m}^3/\text{s}$; $Fr_1 \approx 2.49$ (supercritical) and $Fr_2 \approx 0.44$ (subcritical).

Geometry (triangular, side slope $z$): $A = zy^2$, $T = 2zy$, centroid depth $\bar y = y/3$.

Specific force (momentum function):

$$F = \frac{Q^2}{gA} + A\bar y = \frac{Q^2}{gzy^2} + \frac{zy^3}{3}$$

Momentum equation for a horizontal channel, $F_1 = F_2$:

$$\frac{Q^2}{gz}\left(\frac{1}{y_1^2} - \frac{1}{y_2^2}\right) = \frac{z}{3}\left(y_2^3 - y_1^3\right)$$ $$\frac{Q^2}{gz}\cdot\frac{y_2^2 - y_1^2}{y_1^2y_2^2} = \frac{z}{3}\left(y_2^3 - y_1^3\right)$$

Put $y_2 = ry_1$:

$$\frac{Q^2}{gz}\cdot\frac{y_1^2(r^2 - 1)}{y_1^4r^2} = \frac{z\,y_1^3}{3}(r^3 - 1)$$ $$\frac{Q^2}{gz^2y_1^5} = \frac{1}{3}\cdot\frac{r^2(r^3 - 1)}{r^2 - 1}$$

Froude number at section 1. For a triangular channel, $V_1 = Q/(zy_1^2)$ and the hydraulic depth is $D_1 = A/T = y_1/2$. The problem defines $Fr_1^2 = V_1^2/(gy_1)$, so

$$Fr_1^2 = \frac{V_1^2}{gy_1} = \frac{Q^2}{gz^2y_1^5}$$

which is exactly the left side of the equation above. Hence

$$\boxed{3Fr_1^2 = \frac{r^2(r^3 - 1)}{r^2 - 1}}$$

Check: for $y_1 = 0.5$, $y_2 = 1.0$ ($r = 2$), the right side is $\dfrac{4 \times 7}{3} = 9.33$, so $Fr_1^2 = 3.11$ and $Fr_1 = 1.76$ (with $Fr_1 = V_1/\sqrt{gy_1}$, which is $\sqrt2$ times smaller than the usual $V_1/\sqrt{gD_1} = 2.49$).

Given: $b = 6$ m, upstream depth $y_1 = 6$ m, depth of the stream just after the gate $y_2 = 1.2$ m, depth after the jump $y_3 = 3.1$ m.

Flow rate

Energy equation across the gate (no loss):

$$y_1 + \frac{V_1^2}{2g} = y_2 + \frac{V_2^2}{2g} \;\Rightarrow\; Q = b\,y_1y_2\sqrt{\frac{2g}{y_1 + y_2}}$$ $$Q = 6(6)(1.2)\sqrt{\frac{19.62}{7.2}} = 43.2 \times 1.651 = 71.3\ \text{m}^3/\text{s}$$

Force on the blocks

$$V_2 = \frac{71.31}{6 \times 1.2} = 9.905\ \text{m/s},\qquad V_3 = \frac{71.31}{6 \times 3.1} = 3.834\ \text{m/s}$$

Without blocks the sequent depth of 1.2 m would be $\dfrac{1.2}{2}\left(\sqrt{1 + 8(2.887)^2} - 1\right) = 4.336$ m. Since the actual depth after the jump (3.1 m) is smaller, the blocks are needed to hold the jump in place.

Momentum equation between sections 2 and 3 (horizontal bed, friction neglected). The blocks push on the water in the upstream direction with force $F_b$:

$$\frac{\gamma b y_2^2}{2} + \rho Q V_2 - F_b = \frac{\gamma b y_3^2}{2} + \rho Q V_3$$ $$F_b = \frac{\gamma b}{2}(y_2^2 - y_3^2) + \rho Q(V_2 - V_3)$$ $$F_b \approx 192\,500\ \text{N} \approx 192\ \text{kN}$$

The water pushes on the blocks with an equal force in the downstream direction.

Answer: $Q = 71.3\ \text{m}^3/\text{s}$; force on the blocks $\approx 192\ \text{kN}$.

Given: $b = 4$ m, $Q = 100\ \text{m}^3/\text{s}$ ($q = 25\ \text{m}^2/\text{s}$), upstream depth $y_1 = 20$ m. Loss through the gate $= 0.05\,V_2^2/2g$. The jump forms immediately downstream of the gate.

Flow through the gate ($V_1 = 25/20 = 1.25$ m/s):

$$y_1 + \frac{V_1^2}{2g} = y_2 + 1.05\frac{V_2^2}{2g},\qquad V_2 = \frac{25}{y_2}$$ $$20.080 = y_2 + \frac{1.05 \times 25^2}{2g\,y_2^2} \;\Rightarrow\; y_2 = 1.336\ \text{m}$$ $$V_2 = 18.71\ \text{m/s},\qquad Fr_2 = \frac{18.71}{\sqrt{9.81 \times 1.336}} = 5.17$$

i) Depth downstream of the jump

$$y_3 = \frac{y_2}{2}\left(\sqrt{1 + 8Fr_2^2} - 1\right) = \frac{1.336}{2}\left(\sqrt{1 + 8(26.73)} - 1\right) = 9.12\ \text{m}$$ $$V_3 = \frac{25}{9.12} = 2.74\ \text{m/s}$$

ii) Thrust on the gate

Momentum equation between section 1 (upstream) and section 2 (just downstream of the gate). $F$ is the force of the gate on the water, and the thrust of the water on the gate is equal and opposite:

$$\frac{\gamma b y_1^2}{2} - \frac{\gamma b y_2^2}{2} - F = \rho Q(V_2 - V_1)$$ $$F = 7\,848\,000 - 35\,012 - 1\,746\,473 \approx 6.07 \times 10^6\ \text{N}$$

iii) Energy loss in the jump

$$\Delta E = \frac{(y_3 - y_2)^3}{4y_2y_3} = \frac{(9.12 - 1.336)^3}{4(1.336)(9.12)} = 9.68\ \text{m}$$

Check: $E_2 = 1.336 + \dfrac{18.71^2}{2g} = 19.19$ m and $E_3 = 9.12 + \dfrac{2.74^2}{2g} = 9.50$ m, so $E_2 - E_3 = 9.68$ m.

Answer: (i) $y_3 = 9.12$ m; (ii) thrust on the gate $\approx 6.07$ MN; (iii) energy loss in the jump $\approx 9.68$ m.

Given: $b = 4$ m, $Q = 16\ \text{m}^3/\text{s}$ ($q = 4\ \text{m}^2/\text{s}$), $n = 0.0165$, downstream slope $S_{02} = 0.0024$, upstream uniform depth $Y_{01} = 0.4$ m (steep reach).

$$y_c = \left(\frac{4^2}{9.81}\right)^{1/3} = 1.177\ \text{m}$$

a) Depth in the downstream channel

For the jump to end in uniform flow, the depth after the jump must equal the normal depth of the downstream channel. From Manning's equation with $R = A/P$ (trial):

$$Y_{02} = y_n = 1.495\ \text{m}\quad (> y_c,\ \text{so the slope is mild; } V = 2.68\ \text{m/s})$$

For comparison, the sequent depth of $Y_{01} = 0.4$ m is:

$$V_{01} = 10\ \text{m/s},\ Fr = 5.05,\quad y_2 = \frac{0.4}{2}\left(\sqrt{1 + 8(5.05)^2} - 1\right) = 2.66\ \text{m}$$

Since $2.66 > 1.495$ m, the tailwater is too low for the jump to form at the break. The supercritical flow continues on the mild slope as an M3 curve until its depth $y_j$ has a sequent depth equal to 1.495 m:

$$y_j:\ \frac{y_j}{2}\left(\sqrt{1 + 8\frac{q^2}{g\,y_j^3}} - 1\right) = 1.495 \;\Rightarrow\; y_j = 0.908\ \text{m}$$

b) Length of the M3 profile up to the jump (3 steps)

Depth from 0.40 m to 0.908 m, with $\Delta y = 0.169$ m:

$$E = y + \frac{V^2}{2g},\quad S_f = \frac{n^2V^2}{R^{4/3}},\quad \Delta x = \frac{E_2 - E_1}{S_{02} - \bar S_f}$$ $$L = 31.6 + 32.1 + 28.9 \approx 92.5\ \text{m}$$

The jump itself then raises the depth from 0.908 m to 1.495 m over a length of about $6(y_2 - y_1) = 6(1.495 - 0.908) \approx 3.5$ m.

Answer: (a) $Y_{02} = 1.495$ m; (b) the toe of the jump is about 92 m downstream of the slope change (M3 profile), and the jump is about 3.5 m long.

Given: $b = 10$ m, $Q = 7\ \text{m}^3/\text{s}$, upstream depth $y_1 = 2.5$ m, downstream depth $y_2 = 0.25$ m, $\beta = 1.15$.

Velocities

$$V_1 = \frac{7}{10 \times 2.5} = 0.28\ \text{m/s},\qquad V_2 = \frac{7}{10 \times 0.25} = 2.80\ \text{m/s}$$

Momentum equation between sections 1 and 2 (horizontal bed, friction neglected). $F$ is the force of the gate on the water, so the force of the water on the gate is also $F$ (downstream):

$$\frac{\gamma b y_1^2}{2} - \frac{\gamma b y_2^2}{2} - F = \rho Q\,\beta\,(V_2 - V_1)$$ $$F = \frac{\gamma b}{2}(y_1^2 - y_2^2) - \beta\rho Q(V_2 - V_1)$$ $$F = 303\,497 - 20\,286 = 283\,211\ \text{N}$$

Answer: Force of water on the gate $\approx 283\ \text{kN}$ (about 286 kN if $\beta = 1$).

Hydraulic jump

A hydraulic jump is the sudden, turbulent rise of the water surface that occurs when a supercritical flow ($Fr > 1$) changes into subcritical flow ($Fr < 1$) in an open channel. It has a violent roller, strong mixing and large energy loss.

Why the energy principle is not used

The jump contains intense turbulence and eddies, and the energy dissipated (converted to heat) is unknown and cannot be written as a simple term. Hence the energy equation cannot be solved for the unknown depth. Instead the momentum equation is used, because the forces on the short control volume are known (the hydrostatic forces at the ends), while friction on the short, horizontal bed is negligible.

Numerical part

Given: $b = 5$ m, $Fr_1 = 3.5$, $y_1 = 1.2$ m.

Sequent depth

$$y_2 = \frac{y_1}{2}\left(\sqrt{1 + 8Fr_1^2} - 1\right) = 0.6\left(\sqrt{1 + 8(12.25)} - 1\right) = 5.37\ \text{m}$$

Velocities

$$V_1 = Fr_1\sqrt{g\,y_1} = 3.5\sqrt{9.81 \times 1.2} = 12.01\ \text{m/s}$$ $$V_2 = V_1\frac{y_1}{y_2} = 12.01 \times \frac{1.2}{5.37} = 2.68\ \text{m/s}$$

Froude number after the jump

$$Fr_2 = \frac{V_2}{\sqrt{g\,y_2}} = \frac{2.68}{\sqrt{9.81 \times 5.37}} = 0.37$$

Answer: $Fr_2 \approx 0.37$ (subcritical; $y_2 = 5.37$ m).

Given: $b = 1.1$ m, $Q = 7.2\ \text{m}^3/\text{s}$ ($q = 6.545\ \text{m}^2/\text{s}$), $n = 0.025$, $S_{01} = 0.065$ changing to $S_{02} = 0.0085$.

$$y_c = \left(\frac{6.545^2}{9.81}\right)^{1/3} = 1.635\ \text{m}$$

Normal depths (Manning's equation with $R = A/P$, by trial):

Flow on the upstream reach is supercritical ($y_{n1} < y_c$) and on the downstream reach it is subcritical ($y_{n2} > y_c$). The change from supercritical to subcritical flow can occur only through a hydraulic jump, so a jump occurs.

Location

Sequent depth of $y_{n1} = 1.224$ m:

$$y_2 = \frac{1.224}{2}\left(\sqrt{1 + 8(1.54)^2} - 1\right) = 2.128\ \text{m}$$

Because $y_2 = 2.128\ \text{m} < y_{n2} = 2.962$ m, the downstream depth is higher than needed. The jump is pushed upstream of the break, onto the steep slope. After the jump the depth rises from 2.128 m to 2.962 m along an S1 curve ending at the break. The jump forms where the S1 depth equals 2.128 m.

Direct step on the S1 curve (from 2.962 m at the break to 2.128 m, 4 steps), with $S_0 = 0.065$:

$$L = 3.04 + 2.95 + 2.81 + 2.58 \approx 11.4\ \text{m}$$

Answer: A hydraulic jump occurs on the steep reach about 11.4 m upstream of the slope change (depths 1.224 m to 2.128 m), followed by an S1 curve up to the break.

Given: $b = 5$ m, $n = 0.02$, $S_0 = 0.04$, uniform depth $y_n = 0.5$ m; a low dam raises the depth to 2 m at the dam.

Discharge

$$A = 2.5\ \text{m}^2,\ P = 6\ \text{m},\ R = 0.4167\ \text{m}$$ $$Q = \frac{1}{n}AR^{2/3}S_0^{1/2} = \frac{1}{0.02}(2.5)(0.4167)^{2/3}(0.2) = 13.95\ \text{m}^3/\text{s}$$ $$q = 2.789\ \text{m}^2/\text{s},\quad V = 5.58\ \text{m/s},\quad Fr = 2.52,\quad y_c = \left(\frac{2.789^2}{9.81}\right)^{1/3} = 0.926\ \text{m}$$

Uniform flow is supercritical ($y_n < y_c$; steep slope). Upstream of the dam the water is deep and subcritical, so a jump is possible.

Does a jump occur?

Sequent depth of 0.5 m:

$$y_2 = \frac{0.5}{2}\left(\sqrt{1 + 8(2.52)^2} - 1\right) = 1.549\ \text{m}$$

The depth just upstream of the dam (2 m) is greater than 1.549 m, so the dam backwater is deeper than the jump needs. A hydraulic jump does occur, forming upstream of the dam where the S1 backwater curve has depth 1.549 m.

Distance upstream of the dam

Direct step method along the S1 curve from $y = 2.0$ m (dam) to 1.549 m (5 steps, $S_0 = 0.04$):

$$L = 2.05 + 2.02 + 1.98 + 1.93 + 1.87 \approx 9.9\ \text{m}$$

The steep bed slope ($S_0 = 0.04$) dominates the friction slope, so the S1 curve is short.

Answer: A jump occurs (0.5 m to 1.55 m). Its downstream end is about 10 m upstream of the dam; the jump itself (length about $6(y_2 - y_1) \approx 6.3$ m) lies just upstream of this point.