1. Heat and Flow Technology I.
ÓBUDA UNIVERSITY
Heat and Flow Technology I.
Use only inside
Dr. Ferenc Szlivka
Professor
Dr. Szlivka: Heat and Flow Technology I_7

2. Momentum equation and it’s applications Chapter 7.

3. Momentum equation
Integral form
With single (loads) forces

4. Force acting on a flat plate
A horizontal water jet with "A" cross section area and "v" absolute velocity acting on a flat plate that is perpendicular to the jet. The flat plate moving with "u" horizontal velocity.
How big force acting on the flat plate from the water jet ?
Dr. Szlivka: Fluid Mechanics 7.

5. Force acting on a flat plate
Solution:
Investigate the case when the flat plate is moving with "u" velocity. The relative velocity of the yet to the flat plate is:

6. Solution: 2 w2
I. Step: Calculate the outgoing velocity magnitude with the help of Bernoulli ‘s equation in a relative coordinate system. z2 1 z3 w1 3 w3
Neglect the height difference among the points 1,2 and 3.

7. Moving together with the fat plate
Control surface
Control surface
Moving together with the fat plate
II. Step draw a control surface
-the flow should be steady (pl. the rigid body does not go out from the surface),
-if we are searching for the force acting on the rigid body, the body should be inside the surface
-where fluid is going in or coming out the surface should be perpendicular to the yet or parallel with it

8. III. Step Write the momentum equation
Control surface
Moving together with the fat plate
The left integral argument is not equal to zero where the fluid is crossing the control surface. Denote these parts of integral with
The first integral on the right hand side is the result of the pressure forces acting on the control surface. It can be calculated also a sum of parts integrals.

9. IV. Step
The closed surface momentum integral can be calculated sum of part integrals. I1
dI2
dI3
Control surface
Moving together with the fat plate
The result of the pressure forces is zero, because the pressure is everywhere constant, p0.
The only term is and w1 is opposite than dA so the direction of momentum vector is directed out from the surface. In usual case the momentum vector is directed out from the surface.

10. V. Step dI2 "x" direction: I1 dI3 Control surface
Moving together with the fat plate
V. Step
"x" direction:

11. Pelton-turbine

12. Pelton-turbine

13. Pressure tube
Járókerék
Turbine casing
Quick Jet regulator
Valve
Regulator
Nose
elv.
Downstream
data:
Questions:
a./ Calculate the force acting on one blade of the turbine!
b./ Calculate the average force acting on the wheel!
c./ Calculate the power of the turbine with the given data!
d./ Calculate the function of power respect to „u” (circumferential speed) !

14. Solution: a./ Force acting on one blade Using the momentum equation
Control surface
w2’’
X component of the momentum law
I1’ w1
w2’
I2’

15. a./ Force acting on one blade
w2’’
w2’ w1
I1’
I2’’
I2’
R1x

16. b./ Force acting on the weel
Using the momentum equation
Control surface
X component of the momentum law

17. b./ Force acting on the weel
Using the momentum equation
Control surface
X component of the
momentum law
v2x

18. b./ Force acting on the weel
The average circumferential force
acting on the weel:
Force acting on one blade :
The average force is bigger than the one blade force. The jet can act not only one blade but two or more blades too.

19. b./ a kerékre ható erőt nagysága számszerűen
A kerületi sebesség:

20. c./ The function of power respect to „u” (circumferential speed)
Pe [kW]
100
200
300
400
500
600
700
800
900
1000 20 40 60 80
120
u [m/s]

21. c./ The power with the given data (Pe) and the maximum power (Pemax)
Circumferential speed

22. d./ The circumferential force change
In the same time more than one blade is working. Sometimes "1",and sometimes "2" are the acting force. The working time of two blades is denoted by „t2",and the working time of one blade is „t1".

23. d./ The circumferential force change

24. Airscrew theory
The air screw make a pressure jump in the air. The incoming and outgoing flow is approximately ideal flow without losses.

25. Airscrew theory
The pressure changing around the airscrew.

26. Airscrew theory (with the momentum equation)

27. Airscrew theory (with the momentum equation)
Continuity equation

28. Airscrew theory (with the momentum equation)

29. Airscrew theory (with the momentum equation)

30. Propulsion efficiency
Usefull power
Total power

31. a./ Calculate the force acting on an aircrew, with cross section area „A l„. The aircraft is moving with „ v1" velocity. The air velocity after the aircrew is „ v2" in the coordinate system fixed to the aircraft!
b./ Calculate the ideal propulsion efficiency of the screw!
data:

32. Solution:

33. Windmill

34. Windmill
Inverse airscrew

35. Maximum power of a windmill
The question is the optimum velocity after the windmill „v2" if the „v1" is constant? How big is the maximum power of the windmill?

36. Maximum power of a windmill
The expression shows that the maximum power of the windmill is only the 16/27 (59,26 %) part of the power crossing through the Asz area!

37. The windmill at Kulcs in Hungary
A Windmill on a farm has 2 m diameter. Calculate the maximum ideal power of it at
wind velocity.
The windmill at Kulcs in Hungary
The height of the pile is 60 m, the diameter of the impeller is 50 m. The maximum power calculated with the upper formula is 689 kW. The effective power is approximately %-a, so the power is 600 kW.

38. Windmill in Kulcs in Hungary