Baffle blocks for jump, tailwater 4 m

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 ^
     |                             /
   5 |- - - - - - - - - - - - - - *   y2' = 5 m (no blocks)
     |                        .'
   4 |- - - - - - - - - - *   y2 = 4 m (with blocks)
     |                .'
  yc |- - - - - - -*   minimum F
     |               '.
     |                   '-.
1.43 |- - - - - - - - - - - - - - *   y1
     |                             \
     +--------------------+-------+---> F
                        13.73   17.09
                          |<----->|  Fb/(gamma b) = 3.36 m^2

• 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%.