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MECH 221 FLUID MECHANICS (Fall 06/07) Tutorial 3

MECH 221 FLUID MECHANICS (Fall 06/07) Tutorial 3. Outline. Absolute and gage pressure Forces on Immersed surface Plane surface Curved surface Buoyant force. 1. Absolute and Gage pressure. Absolute pressure: Measured from absolute zero Gage pressure: Measured from atmospheric pressure

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MECH 221 FLUID MECHANICS (Fall 06/07) Tutorial 3

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  1. MECH 221 FLUID MECHANICS(Fall 06/07)Tutorial 3

  2. Outline • Absolute and gage pressure • Forces on Immersed surface • Plane surface • Curved surface • Buoyant force

  3. 1. Absolute and Gage pressure • Absolute pressure: • Measured from absolute zero • Gage pressure: • Measured from atmospheric pressure • If negative, it is called vacuum pressure • Pabs = Patm + Pgage

  4. 1. Absolute and Gage pressure Gage pressure Absolute pressure Atmospheric pressure

  5. 1. Example • A scuba diver practicing in a swimming pool takes enough air from his tank to fully expand his lungs before abandoning the tank at depth L and swimming to the surface. When he reaches the surface, the different between the external pressure on him and the air pressure in his lung is 9.3kPa. From what depth does he start?

  6. 1. Example (Answer) • When the diver fills his lungs at depth L, the external pressure on him (and thus the air pressure within his lungs) is,P = P0+ρgL • When he reaches the surface, the pressure difference between his lung and surrounding is,ΔP = P–P0 = ρgLL = ΔP/ρg = 9300/(1000x9.81) = 0.948m

  7. 2.1 Forces on Immersed Surfaces (plane surface) • For plane surface: F = (Patm + ghc.g)AORF = (Patm + γhc.g)A hc.g.=vertical distance from the fluid surface to the centroid of the area

  8. 2.1 Forces on Immersed Surfaces (plane surface) • Where is the centroid.? • By definition:

  9. 2.1 Forces on Immersed Surfaces (plane surface) • Centre of pressure:

  10. 2.1 Forces on Immersed Surfaces (plane surface) • What is Ixc (or Iyc).? • By definition:

  11. 2.1 Example • The rectangular gate CD shown in the figure is 1.8m wide and 2.0 long. Assuming the material of the gate to be homogeneous and neglecting friction at the hinge C, determine the weight of the gate necessary to keep it shut until the water level rises to 2.0m above the hinge.

  12. 2.1 Example (Answer) • Procedure: • Magnitude of the resultant force: • FR = ρghc.g.A → hc.g. = ? • Centre of pressure yc.p.: • yc.p.= (Ixc/yc.g.A) + yc.g.→ yc.g. =? ; Ixc = ? • Moment balance at hinge C • ΣM = 0

  13. 2.1 Example (Answer) • hc.g.=2+0.5(4/5)(2)=2.8m • FR=(9.81)(1000)(2.8)(2)(1.8)=98.885kN • yc.p.= (Ixc/yc.g.A) + yc.g. • yc.g.=2.8(5/4)=3.5m • Ixc=(1/12)(1.8)(2)3=1.2m4 • yc.p.=[1.2/(3.5x2x1.8)]+3.5=3.595m • Moment equilibrium • Resultant force: MF=FR(yc.p.-2(5/4)) =108.279kNm • Weight of the gate: Mg=W(0.5)(2)(3/5)=0.6W • Since MF=Mg→0.6W=108.279; W=180.465kN

  14. 2.2 Forces on Immersed Surfaces (curved surface) • For curved surface: • Horizontal force: horizontal force on a curved surface equals the force on the plane area formed by the projection of the curved surface onto a vertical plane

  15. 2.2 Forces on Immersed Surfaces (curved surface) • For curved surface: • Vertical force: • Similar to the previous approach, FaV = Fa cos = Pa Aacos  • Aacos is the horizontal projection of 'a', but this is only at a point! • Notice that if one looks at the entire plate, the pressures on the horizontal projection are not equal to the pressures on the plate • Consequently, one needs to integrate along the curved plate

  16. 2.2 Example • The concrete seawall has a curved surface and restrains seawall at a depth of 24ft. The trace of the surface is a parabola as illustrated. Determine the moment of the fluid force (per unit length) with respect to an axis through the toe (point A).

  17. 2.2 Example (Answer) • Procedure: • Magnitude of the horizontal force: • FH = γhc.g.A → hc.g. = ? • Magnitude of the vertical force: • FV = γV • Volume? Location of centroid? • Moment at hinge A

  18. 2.2 Example (Answer) • Horizontal force and pressure centre: • hc.g.=y1=24/2 = 12ft • FH=F1= γhc.g.A =(64)(12)(24) = 18432lb/ft • y1=24/3=8ft

  19. 2.2 Example (Answer) • Volume of the seawater: • Given the function of the surface: • y=0.2x2 • When y=24ft, x0=√120

  20. 2.2 Example (Answer) • Location of the centroid: • Given the function of the surface: • y=0.2x2, x0=√120, A=175.271ft2

  21. 2.2 Example (Answer) • Moment at point A: • MH=FHy1=(18432)(8)=147456lb·ft/ft (CW) • MV=W(15-xc)=(64)(175.271)(15-4.108)=122179.311lb·ft/ft (CCW) • MA=MH-MV=147456-122179.311=24276.689lb·ft/ft (CW)(moment per unit length)

  22. 3. Buoyant force • FB=g(vol. a-b-c-d) • This force FBis called Buoyant Force

  23. 3. Example • A hot-air balloon weights 500lb. The air outside the balloon has a temperature of 80F, and the heated air inside the balloon has a temperature of 150F. Assume the inside and outside air to be at standard atmospheric pressure of 14.7psi. Determine the required volume of the balloon to support the weight. If the balloon had a spherical shape, what would be the required diameter?

  24. Wair, heated Wloading 3 Example (Answer) • Procedure: • Buoyant force of air: • FB = γair, outsideV • Total weight of the balloon: • W = Wloading + Wair, inside • ΣFvert = 0

  25. Wair, heated Wloading 3 Example (Answer) • By ideal gas law: • pV = mRTγ = pg/RT • For air@14.7psi,80F • γair, outside= pg/RT = (14.7)(144)(32.2)/(1716)(80+460)= 0.07356lb/ft3 • Buoyant force of air: • FB = γair, outsideV = 0.07356V

  26. Wair, heated Wloading 3 Example (Answer) • For air@14.7psi,150F • γair, inside= pg/RT = (14.7)(144)(32.2)/(1716)(150+460)= 0.06512lb/ft3 • Total weight of the balloon: • W = Wloading + Wair, inside • W = 500 + γair, insideV = 500 + 0.06512V

  27. Wair, heated Wloading 3 Example (Answer) • By force equilibrium, • FB = W • 0.07356V = 500 + 0.06512V • V = 59241.706ft3 • Also, V = (π/6)D3 • D = 48.366ft

  28. The End

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