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 If negative, it is called vacuum pressure Pabs = Patm + Pgage

1. Absolute and Gage pressure Absolute pressure Atmospheric pressure

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?

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 = ρgL L = ΔP/ρg = 9300/(1000x9.81) = 0.948m

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

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

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

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

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.

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

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

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

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

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).

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

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

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

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

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)

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

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?

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

3 Example (Answer) By ideal gas law: For air@14.7psi,80F Wair, heated Wloading By ideal gas law: pV = mRT γ = 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

3 Example (Answer) For air@14.7psi,150F Total weight of the balloon: Wair, heated Wloading 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

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

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