Question 5.12: Water is flowing with an average velocity of 6 m/s when it c......
Water is flowing with an average velocity of 6 m/s when it comes down the spillway of a dam as shown in Fig. 5–26. Within a short distance a hydraulic jump occurs, which causes the water to transition from a depth of 0.8 m to 2.06 m. Determine the head loss caused by the turbulence within the jump. The spillway has a constant width of 2 m.
Fluid Behavior. Steady flow occurs before and after the jump. The water is assumed to be incompressible.
Control Volume. Here we will consider a fixed control volume that contains the water within the jump and a short distance on either side of it, Fig. 5–26. Steady flow passes through the open control surfaces because these surfaces are removed from the regions within the jump
where the flow is not well defined.
Continuity Equation. Since the cross sections of AB and DE are known, we can determine the average velocity out of DE using the continuity equation.
0 – (1000 kg/m^3) (6 m/s) (0.8 m)(2 m) + (1000 kg/m^3)V_{out}(2.06 m)(2 m) = 0
V_{out} = 2.3301 m/s
Energy Equation. We will place the datum at the bottom of the control volume, Fig. 5–26. From this we can determine the hydraulic head (p/γ + z) at each open control surface. For example, if we select points A and D, then since p_{in} = p_{out} = 0, we get
\frac{p_{in}}{\gamma} + z_{in} = 0 + 0.8 m = 0.8 m\frac{p_{out}}{\gamma} + z_{out} = 0 + 2.06 m = 2.06 m
If instead we take points B and E, then since p = \gamma h, we again have
\frac{p_{in}}{\gamma} + z_{in} = \frac{\gamma (0.8 m)}{\gamma} + 0 = 0.8 m\frac{p_{out}}{\gamma} + z_{out} = \frac{\gamma (2.06 m)}{\gamma} + 0 = 2.06 m
Finally, if we use intermediate points C and F, then once again
\frac{p_{in}}{\gamma} + z_{in} = \frac{\gamma (0.5 m)}{\gamma} + 0.3 m = 0.8 m\frac{p_{out}}{\gamma} + z_{out} = \frac{\gamma (1 m)}{\gamma} + 1.06 m = 2.06 m
In all cases we obtain the same results, regardless of which pair of points on the control surface we choose. For application, we will choose points A and D. Since no shaft work is done, we have
\frac{p_{in}}{\gamma} + \frac{V_{in}^2}{2g} + z_{in} + h_{pump} = \frac{p_{out}}{\gamma} + \frac{V_{out}^2}{2g} + z_{out} + h_{turbine} + h_L0 + \frac{(6 m/s)^2}{2(9.81 m/s^2)} + 0.8 m + 0 = 0 + \frac{(2.3301 m/s)^2}{2(9.81 m/s^2)} + 2.06 m + 0 + h_L
h_L = 0.298 m
This head loss is due to turbulence and frictional heating within the jump.