Question 12.15: The sluice gate in Fig. 12–32 is partially opened in a 2-m-w......
The sluice gate in Fig. 12–32 is partially opened in a 2-m-wide channel, and water passing under the gate forms a hydraulic jump. At the low level, just before the jump, the water depth is 0.2 m, and the flow is 1.30 m³/s. Determine the depth farther downstream and also the head loss across the jump.
Fluid Description. We will assume the water is incompressible and the level in the reservoir is maintained, so steady flow occurs past the sluice gate.
Analysis. Just before the jump, the Froude number is
Thus the flow is supercritical, so indeed a jump can occur. Applying Eq. 12–29 to determine the water height after the jump, we have
\frac{y_2}{y_1} = \frac{1}{2} (\sqrt{1 + 8Fr_1^2} – 1); \frac{y_2}{0.2 m} = \frac{1}{2} (\sqrt{1 + 8(2.320)^2} – 1)
y_2 = 0.5638 m = 0.564 m
At this depth,
Fr_2 = \frac{Q/A_2}{\sqrt{gy_2}} = \frac{(1.30 m^3/s)/[(2 m)(0.5638 m)]}{\sqrt{(9.81 m/s^2)(0.5638 m)}} = 0.4902 < 1The flow is subcritical, as expected. The head loss can be determined from Eq. 12–30,
h_L = \frac{(y_2 – y_1)^3}{4y_1y_2} = \frac{(0.5638 m – 0.2 m)^3}{4 (0.2 m)(0.5638 m)} = 0.1068 m = 0.107 mNote that since the original specific energy of the flow is
E_1 = \frac{q^2}{2gy_1^2} + y_1 = \frac{[(1.30 m^3/s) / (2 m)]^2}{2(9.81 m/s^2)(0.2 m)^2} + 0.2 m = 0.7384 mthen, after the jump, the specific energy of the flow becomes
E_2 = E_1 – h_L = 0.7384 m – 0.1068 m = 0.6316 m
This produces the following percent of energy lost within the jump,