1. Pratical Lecture 4 Mecânica de Fluidos Ambiental 2015/2016

3. Resolution

7. With a new iteration the calculation can be improved

11. Resolution Exit Surface

12. Momentum flux at the entrance The fluid leaving through the area of lower momentum entered through an surface shorter than 2L. Using the mass conservation one gets: The Momentum entering is:

13. And the force and Cd

14. Let us consider a large number of plates, as in a real turbine.

15. Resolution Maximum power is obtained when the derivative in relation to the angular velocity is null: In this case there is movement of the plate. The linear speed is: The relative velocity has the same modulus and opposite sign. The discharge is the jet discharge, if we assume many plates.

16. In this case Whole jet the kinetic energy would be extracted. Efficiency would be 100%.

17. Mass balance Momentum Balance Energy Balance

18. Resolution

20. Bernoulli Mass conservation

21. Resolution considering This equation states that the force must balance the hydrostatic pressure plus the momentum exiting from under the sluice gate. The force tends to zero when h 2 goes to h 1. The force would grow linearly to be maximum for h 2 =0 If the momentum entering into the system was relevant one had to subtract it from the force:

23. Description of the problem Vj is the jet velocity relative to the boat and Vin is the inlet velocity also relative to the boat. Qj is the jet discharge (equal to the inlet discharge). The boat “sees” the water entering into the control volume at its own velocity and leaving it at the relative velocity (Vj). V Vj Vin V The boat velocity depends on the jet velocity, on the discharge and on friction. If there was no friction there would be no force and the maximum boat speed would be the jet speed.

24. If we had considered a control volume without the green area, we should consider Vin instead of V, but in that case we should have computed the pressure force in front of the boat, computed as: V Vj Vin This is the force applied over the fluid inside the green volume. The pressure reduction at the interface between both is the symmetrical of that force. This force must appear in the application to the blue control volume: Which is the same result as before.