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ÓBUDA UNIVERSITY

ÓBUDA UNIVERSITY. Heat and Flow Technology II. Use only inside. Dr. Ferenc Szlivka Professor. Momentum equation and it ’s applications Chapter 7. Momentum equation. Integral form. With single ( loads ) forces. Force acting on a flat plate.

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ÓBUDA UNIVERSITY

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  1. ÓBUDA UNIVERSITY Heat and Flow Technology II. Use only inside Dr. Ferenc Szlivka Professor Dr. Szlivka: Heat and Flow Technology II_7

  2. Momentum equation and it’s applicationsChapter 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. 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. StepWrite 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. dI2 I1 Control surface Moving together with the fat plate dI3 IV. Step The closed surface momentum integral can be calculated sum of part integrals. 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. dI2 I1 Control surface Moving together with the fat plate dI3 V. Step "x" direction:

  11. Pelton-turbine

  12. Pelton-turbine

  13. Pressure tube Turbine casing Járókerék Quick Jet regulator Nose Valve Regulator 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 I2’’ Control surface w2’’ X component of the momentum law I1’ w1 w2’ I2’

  15. I2’’ w2’’ I1’ w1 w2’ I2’ a./ Force acting on one blade 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. 1000 900 800 700 600 500 400 300 200 100 0 0 20 40 60 80 100 120 c./ The function of power respect to „u” (circumferential speed) Pe[kW] u [m/s]

  21. c./ The power with the given data (Pe) and the maximum power (Pemax) The maximum power 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. 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 65-90 %-a, so the power is 600 kW.

  38. Windmill in Kulcs in Hungary http://www.winfo.hu

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