Question 12.7: The flow in the 2-m-wide channel in Fig. 12–16a is controlle......
The flow in the 2-m-wide channel in Fig. 12–16a is controlled by the sluice gate, which is partially opened so that it causes the depth of the water near the gate to be 3 m and the mean velocity there to be 0.5 m/s. Determine the depth of the water far upstream, where it is essentially at rest, and also find its depth downstream from the gate where free outflow occurs.
Fluid Description. The water a distance far upstream from the sluice gate is assumed to have a constant depth so that the flow at points 1 and 2 will be steady. Also, the water is assumed to be an ideal fluid.
Analysis. If the Bernoulli equation is applied between points 0 and 1, located on a streamline at the water surface, we have
\frac{p_0}{\gamma} + \frac{V_0^2}{2g} + y_0 = \frac{p_1}{\gamma} + \frac{V_1^2}{2g} + y_10 + 0 + y_0 = 0 + \frac{(0.5 m/s)^2}{2(9.81 m/s^2)} + 3 m
y_0 = 3.01274 m = 3.0127 m
The Bernoulli equation can be applied between points 1 and 2, but we can also apply it between points 0 and 2. If we do this, we have
\frac{p_0}{\gamma} + \frac{V_0^2}{2g} + y_0 = \frac{p_2}{\gamma} + \frac{V_2^2}{2g} + y_2
0 + 0 3.01274 m = 0 + \frac{V_2^2}{2(9.81 m/s^2)} + y_2 (1)
Continuity requires the flow at 1 and 2 to be the same.
Q = V_1A_1 = V_2A_2
V_2y_2 = 1.5
Substituting V_2 = 1.5/y_2 into Eq. 1 yields
y_2^3 – 3.01274y_2^2 + 0.11468 = 0
Solving for the three roots, we get
y_1 = 3 m Subcritical (as before)
y_2 = 0.2020 m Supercritical
y_2 = -0.1892 m Not realistic
The first root indicates the depth y_2 = y_1 = 3 m, and the second root is the depth downstream from the gate. Thus,
y_2 = 0.202 m
The specific energy for the flow can be determined either from point 0, 1, or 2 since the bed elevation of the channel is constant and friction losses through the gate have been neglected (ideal fluid). Using point 1,
A plot of the specific energy is shown in Fig. 12–16b. Here
y_c = \frac{2}{3} y_0 = \frac{2}{3}(3.01274 m) = 2.0085 mE_{min} = \frac{q^2}{2gy_c^2} + y_c = \frac{[(0.5 m/s)(3 m)]^2}{2(9.81 m/s^2)(2.0085 m)^2} + 2.0085 m = 2.037 m
As noted, subcritical flow occurs upstream near the gate, and supercritical flow occurs downstream.