1
MECH 221 FLUID MECHANICS(Fall 06/07)Tutorial 3
2
Outline
1. Absolute and gage pressure2. Forces on Immersed surface
1. Plane surface2. Curved surface
3. 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
Atmospheric pressure
Gage pressure
Absolute 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)AOR F = (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:
A
A
A
A
dA
ydAy
dA
xdAx
9
2.1 Forces on Immersed Surfaces(plane surface)
Centre of pressure:
....
.. gcgc
xcpc y
Ay
Iy
10
2.1 Forces on Immersed Surfaces(plane surface)
What is Ixc (or Iyc).? By definition:
Ayc
Axc
dAyI
dAxI
2
2
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
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
20
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
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 FB is called Buoyant Force
23
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?
24
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
25
3 Example (Answer)
By ideal gas law: pV = mRT
γ = pg/RT
For [email protected],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
26
3 Example (Answer)
For [email protected],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
27
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
28
The End