Transcript Document 7215030

MECH 221 FLUID MECHANICS
(Fall 06/07)
Tutorial 3
1
Outline
1.
2.
Absolute and gage pressure
Forces on Immersed surface
1.
2.
3.
Plane surface
Curved surface
Buoyant force
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1. Absolute and Gage pressure

Absolute pressure:


Gage pressure:


Measured from absolute zero
Measured from atmospheric pressure
If negative, it is called vacuum pressure

Pabs = Patm + Pgage
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1. Absolute and Gage pressure
Gage
pressure
Atmospheric
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 = ρgL
L = 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)A
OR
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:
xdA

x
 dA
ydA

y
 dA
A
A
A
A
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2.1 Forces on Immersed Surfaces
(plane surface)

Centre of pressure:
yc . p . 
I xc
yc . g . A
 yc . g .
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2.1 Forces on Immersed Surfaces
(plane surface)


What is Ixc (or Iyc).?
By definition:
I xc   x dA
2
A
I yc   y dA
2
A
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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:


→ hc.g. = ?
Centre of pressure yc.p.:


FR = ρghc.g.A
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:


→ hc.g. = ?
Magnitude of the vertical force:



FH = γhc.g.A
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
 24 
A     dy dx


0  0 .2 x 2 
x0
x0

0. 2 x 
A   (24  0.2 x 2 )dx  24 x 

3

0
0
x0
3
 x0  120
 A  175.271 ft 2
Also , V  175.271 ft 3 / ft (volume per unit length)
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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
 x(24  0.2 x dx
x0
2
xc 
0
A
x0
 2 0.2 x 
0 24 x  0.2 x dx 12 x  4  0
xc 

A
A
 x0  120 , A  175.271 ft 2
x0

 xc  4.108 ft
3

4
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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 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?
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3 Example (Answer)

Procedure:

Buoyant force of air:


FB = γair, outsideV
Total weight of the balloon:


Wair, heated
Wloading
W = Wloading + Wair, inside
ΣFvert = 0
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3 Example (Answer)
Wair, heated

By ideal gas law:


Wloading
For [email protected],80F


pV = mRT
γ = pg/RT
γair, outside= pg/RT
= (14.7)(144)(32.2)/(1716)(80+460)
= 0.07356lb/ft3
Buoyant force of air:

FB = γair, outsideV = 0.07356V
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3 Example (Answer)
Wair, heated

For [email protected],150F


γair, inside= pg/RT
= (14.7)(144)(32.2)/(1716)(150+460)
= 0.06512lb/ft3
Wloading
Total weight of the balloon:


W = Wloading + Wair, inside
W = 500 + γair, insideV = 500 + 0.06512V
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3 Example (Answer)
Wair, heated

By force equilibrium,
 FB = W
 0.07356V = 500 + 0.06512V
 V = 59241.706ft3
 Also, V = (π/6)D3
 D = 48.366ft
Wloading
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The End
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