Jump and chute block with hump control

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:

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

For the full width: $\gamma b y_1^2/2 = 5960$ N and $\rho QV_1 = 91\,288$ N, so the left side is $5960 + 91\,288 - 185 = 97\,063$ 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).