Question

Think of the below system like the Devils Punch Bowl on the Oregon Coast. When the waves approach the rocks they drive a pressure gradient under the rocks that cause the water to shoot up into the air. The system is very similar to the diagram below with a large reservoir of water and a free jet. What can we assume about the pressure in the free jet? What about the reservoir? What can we assume about the velocity of the water in the 'reservoir, compared to that of the jet? What is the velocity of water in the free jet? From the movie we watched in class it looked like the flow moving through the Devil's Punch Bowl was 50 m3/s just before a large jet Knowing all the above, what type of head (velocity, pressure, elevation) drives these jet events? And how much would be needed to drive a jet event of 3 m? A. B. C. D. 8 m 1 m

Answer 1

(A) The pressure in the jet is zero(gage), and equals to the atmospheric pressure(absolute.) = 101.3 kPa

In the reservoir, the pressure at the top water surface is equal to the atmospheric pressure, while it increases with the depth of the water in the reservoir.

(B) The velocity of the water in the reservoir can be assumed to be zero when compared to the velocity of the jet.

(C) Given that the jet has a rise of 3m, the velocity head at the outlet must be 3m, so velocity = root(2.g.h) = root(2*9.81*3) = 7.67 m/sec

(D) It can be seen that the top water surface in the tank is below the rise of the free jet, so either the air is the air in the empty space above the water surface in the reservoir must be pressurized, or there must be a pump installed in between the outlet pipe to provide the necessary head.

At the bottom of the jet, the head is purely a velocity head, but as it reaches towards to the top or maximum height, it converts into elevation head.