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

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

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Outline

1. Absolute and gage pressure2. Forces on Immersed surface

1. Plane surface2. Curved surface

3. Buoyant force

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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

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1. Absolute and Gage pressure

Atmospheric pressure

Gage pressure

Absolute pressure

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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?

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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

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2.1 Forces on Immersed Surfaces(plane surface)

For plane surface:

F = (Patm + ghc.g)AOR F = (Patm + γhc.g)A

hc.g.=vertical distance from the fluid surface to the centroid of the area

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2.1 Forces on Immersed Surfaces(plane surface)

Where is the centroid.? By definition:

A

A

A

A

dA

ydAy

dA

xdAx

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2.1 Forces on Immersed Surfaces(plane surface)

Centre of pressure:

....

.. gcgc

xcpc y

Ay

Iy

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2.1 Forces on Immersed Surfaces(plane surface)

What is Ixc (or Iyc).? By definition:

Ayc

Axc

dAyI

dAxI

2

2

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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.

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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

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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

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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

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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

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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).

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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

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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

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2.2 Example (Answer)

Volume of the seawater: Given the function of the surface:

y=0.2x2

When y=24ft, x0=√120

length)unit per (volume /271.175,

271.175

120

3

2.024)2.024(

3

2

0

0

3

0

2

0

24

2.0

00

0

2

ftftVAlso

ftA

x

xxdxxA

dxdyA

xx

x

x

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2.2 Example (Answer)

Location of the centroid: Given the function of the surface:

y=0.2x2, x0=√120, A=175.271ft2

ftx

ftAx

A

xx

A

dxxx

x

A

dxxx

x

c

xx

c

x

c

108.4

271.175,120

42.0

122.024

2.024(

20

0

42

0

3

0

2

00

0

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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)

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3. Buoyant force

FB=g(vol. a-b-c-d)

This force FB is called Buoyant Force

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3. Example

A hot-air balloon weights 500lb. The air outside the balloon has a temperature of 80F, and the heated air inside the balloon has a temperature of 150F. 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?

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3 Example (Answer)

Procedure: Buoyant force of air:

FB = γair, outsideV

Total weight of the balloon: W = Wloading + Wair, inside

ΣFvert = 0

Wair, heated

Wloading

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3 Example (Answer)

By ideal gas law: pV = mRT

γ = pg/RT

For air@14.7psi,80F γair, outside= pg/RT

= (14.7)(144)(32.2)/(1716)(80+460)= 0.07356lb/ft3

Buoyant force of air: FB = γair, outsideV = 0.07356V

Wair, heated

Wloading

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3 Example (Answer)

For air@14.7psi,150F 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

Wair, heated

Wloading

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3 Example (Answer)

By force equilibrium, FB = W 0.07356V = 500 + 0.06512V V = 59241.706ft3

Also, V = (π/6)D3 D = 48.366ft

Wair, heated

Wloading

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The End