Download Ch2 Fluid Statics

Ch2 Fluid Statics
• Fluid either at rest or moving in a
manner that there is no relative motion
between adjacent particles.
• No shearing stress in the fluid
• Only pressure (force that develop on
the surfaces of the particles)
2.1 Pressure at a point N/m2 (Force/Area)


F  ma
Y:
 Fy  p yxz  Psxs sin 
F
z
 p z xy  p z xs cos   
xyz
2
az

Z:
xyz
2

ay
xyz
2
az
y  s cos  ; z  s sin 
y : p y  p s  a y
y
2
z : p z  p s  ( a z   )
z
2
What happen at a pt. ?
p y  ps
p z  ps

p y  pz  ps
x, y, z  0
θ is arbitrarily chosen
Pressure at a pt. in a fluid at rest, or in motion, is
independent of direction as long as there are no shearing
stresses present.
(Pascal’s law)
2.2 Basic equation for Pressure Field
How does the pressure in a fluid which there are no shearing stresses vary
from pt. to pt.?
Surface & body forces acting on small fluid element
pressure
weight
Surface forces:
p y
p y
y : Fy  ( p 
)xz  ( p 
)xz
y 2
y 2
Fy  
p
xyz
y
Similarly, in z and x directions:
p
Fx   xyz
x
p
Fz   xyz
z



p  p  p 
Fs  Fx i  Fy j  Fz k  ( i 
j  k )xyz
x
y
z
 (p)xyz
     
 i 
j k
x
y
z
Newton’s second law




F  ma  Fs  W  pxyz  xyz

  xyz a


  p  k  a
General equation of motion for a fluid in which there
are no shearing stresses.
2.3 Pressure variation in a fluid at rest


a  0  p  k  0
p
 
z
dp
 
(Eq. 2.4)
dz
2.3.1 Incompressible
γ  ρ g  const

p2
p1
dp  γ

z1
z2
dz  p1  p2  γ ( z2  z1 )  γ h
Hydrostatic Distribution
h
p1  p 2
p1  γ h  p2
pressure head

Ex: 10 psi  p1  p 2  h  23.1 ft or 518mmHg
(  62.4 lb
ft
*see Fig. 2.2
2
) (  133 KN
m
3
p  γ h  p0
)
Pressure in a homogeneous, incompressible fluid at rest: ~ reference level,
indep. of size or shape of the container.
The required equality of pressures at equal elevations
Throughout a system.  F2  A2 F1
( Fig . 2.5)
A1
Transmission of fluid pressure
2.3.2 Compressible Fluid
dp
gp
    g  
dz
RT
p2
g Z 2 dz
p 2 dp

ln


 p1
Z1
p
p1
R
T
Assume
perfect gas: p  ρ RT
g , R const. (z1  z 2 )
T  T0 overz1 , z 2  isothermal conditions
 g ( z 2  z1 ) 
p 2  p1 exp 

RT
0


2.4 Standard Atmosphere
Troposphere: T  Ta  β z
0.0065  K
m
0.00357  R
ft
Ta @ z  0
 β lapose rate
β z g Rβ
p  pa ( 1 
)
Ta
2.5 Measurement of Pressure
See Fig. 2.7
Absolute &
Gage pressure
patm  γ h  pvapor
(Mercury barometer)
Example 2.3
N
pa  2 ( pascal )
m
2.6 Manometry
1. Piezometer Tube: 1. p  p 2. h is reasonable
2. U-Tube Manometer: p A  γ 2 h2  γ 1h1
3. Inclined-tube manometer
a
1
p  pa不大 3. liquid, not a gas
see examples
*explain Fig. 2.11 Differential U-tube manometer
p A  pB  γ 2 h2  γ 3h3  γ 1h1
Example 2.5
Ex. 2.5
Δ u  , Δ p  , Δ p  p A  pB
Q( the volume rate of the flow )  k p A  pB
p A  γ 1h1  γ 2 h2  γ 1 ( h1  h2 )  pB
p A  pB  h2 ( γ 2  γ 1 )
2.6.3 Fig. 2.12 Inclined tube manometer
p A  pB  γ 2l2 sin θ
l2 
p A  pB
γ 2 sin θ
Small difference in gas pressure
If pipes A & B contain a gas
2.7 Mechanical and Electronic Pressure Measuring Device
. Bourdon pressure gage (elastic structure)
Bourdon Tube
p , curved tube  straight
deformation  dial
. A zero reading on the gage indicates that the measured
pressure
. Aneroid barometer－measure atmospheric pressure
(absolute pressure)
. Pressure transducer－pressure V.S. time
Bourdon tube is connected to a linear variable
differential transformer(LVDT), Fig. 2.14
coil; voltage
This voltage is linear function of the pressure, and could
be recorded on an oscillograph, or digitized for storage
or processing on computer.
Disadvantage-elastic sensing element
meas. pressure are static or only changing
slowly(quasistatic).
relatively mass of Bourdon tube
<diaphragm>
1Hz
*strain-gage pressure transducer *
Fig. 2.15 (arterial blood pressure)
piezo-electric crystal. (Refs. 3, 4, 5 )
2.8 Hydrostatic Force on a Plane Surface
Fig. 2.16 Pressure and resultants hydrostatic force
developed on the bottom of an open tank.
FR  pA
Storage tanks, ships
. For fluid at rest we know that the force must be
perpendicular to the surface, since there are no shearing
stress present.
. Pressure varies linearly with depth if incompressible
dp
    g
dz
p  h for open tank, Fig. 2.16
The resultant force acts through the centroid of the area
dθ
* Exercise 1.66
d  RidA
torque shearing stress
dA  ( Ri d )l
d  Ri 2ld
Ri
τ
R0
  Ri2l 02 d  2Ri l
Assume velocity distribution in the gap is linear
2Ri3lw
 
R0  Ri
 
Ri w
R0  Ri
dF  hdA
FR   hdA   y sin dA
A
A
if  ,  are constants.
FR   sin   ydA
A
first moment of the area
∫ydA = y A
A
c
 FR  AyC sin   hc A
Indep. Of 
The moment of the resultant force must equal the moment of the
Distributed pressure force
FR y R   A ydF   A  sin y 2 dA
 FR  A C sin 
 yR
2
 A y dA

yc A
I x   A y dA  second moment of the area (moment of inertia)
2
Ix
yR 
; I x  I xc  Ayc2
yc A
I xc
yR 
 yc
y R  yc
ycA
xR 
I xyc
 xc
I xc , I xyc ect see Fig. 2.18
ycA
Note: Ixy-the product of inertia wrt the x& y area.
Ixyc-the product of inertia wrt to an orthogonal
coord. system passing through the centroid
of the area.
If the submerged area is symmetrical wrt an axes passing
through the centroid and parallel to either the x or y axes,
the resultant force must lie along the line x=xc,
since Ixyc= 0.
Center of pressure (Resultant force acts points)
Example 2.6 求a. F ; ( x , y )
R
R
R
b. M (moment)
a.
FR  Eq. 2.18
FR  1.23 106 N
xR  Eq. 2.19, 2.20 xR  0
yR
b.  M c  0
y R  11.6m
(shaft ; water)
M  FR ( y R  yc )  1.01105 N  m
2.9 Pressure Prism
the pressure varies linearly with depth. See Fig. 2.19
h
FR  PAve A   ( ) A
2
FR  volume of pressure prism
1
h
 (h)(bh)  A
2
2
No matter what the shape of the pressure prism is, the resultant
force is still equal in magnitude to the volume of the pressure
Prism, and it passes through the centroid of the volume.
First, draw the pressure prism out.
p  z  p0
dp
 
dz
Example 2.8
F1  (h1  ps ) A  2.44 10 4 N
h2  h1
F2   (
) A  0.954 103 N
2
FR  F1  F2  25.4 KN
FR y0  F1 (0.3m)  F2 (0.2m)
y0  0.296m
2.10 Hydrostatic Force on a Curved Surface
.
Eqs. Developed before only apply to the plane surfaces
magnitude and location of FR
. Integration:
tedious process/ no simple, general formulas can be
developed.
.
Fig. 2.23
F1; F2 → plane surface
W  xV ; through C.G(center of gravity )
FH , FV  The compoments of force that the tank
exerts on the fluid.
For equilibrium, FH  F2 ; collinear. through pt
FV  F1  W
Example 2.9 排水管受力情形
F1 = γh c A
lb 3
= 62.4 3 × ft × (3 ×1ft 2 )
ft 2
= 281lb
lb π × 3 2
ω = γ∀= ρg∀= 62.4 3 ×
ft ×1ft
ft
4
= 441lb at C.G
2
See Fig. 2.18
(Centroid; center of pressure, CP; center of gravity)
1
4
×
3
I×C
3
yR = yC +
=
ft + 12
ft
3
yc A
2
× 32
2
= 2ft
Similarly
x R ≈1.27ft
4R
4×3
=
= 1.27 ft
3π
3× π
∴ F1 = FH = 281lb ; FV = ω = 441lb ; F2 = 0
∴ FR =
FH2 + FV2 = 523lb
FH
-1 FH
tan θ =
⇒θ = tan
= 32.5°
FV
FV
2.11 Buoyancy, Flotation, and Stability
2.11.1 阿基米德原理
請看圖2.24, 來分析其受力情形
FB  V
任意形狀的物體之體積
2.11.2 Stability
stable equilibrium
stable
Light
Heavy
Stable
neutral
Heavy
Light
unstable
Explain Fig. 2.25; 26; 27; 28
unstable
L
H
H
L