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
 Fy  Py x z  Ps x s sin 
Y:  
 Fz  Pz x y  Pz x s cos  
 x y z
2
 x y  z
az
2
ay
Z:  
 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   )
What happen at a pt. ?
p y  ps
p z  ps
z
2
 x,  y, z  0
 p y  p z  ps
 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)
1
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: y : Fy  ( p 
Fy  
p  y
p  y
) x z  ( p 
) x z
y 2
y 2
p
 x y z
y
Similarly, in z and x directions: Fx  



Fs  Fx i  Fy j  Fz k  (

p
 x y z
x
Fz  
p
 x y z
z
p  p  p 
i 
j  k ) x y z  (p) x y z
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
2
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
 
dz
(Eq. 2.4)
2.3.1 Incompressible
  g  const
p2
p
1
dp   z dz  p1  p2   ( z2  z1 )  h
z1
2
Hydrostatic Distribution p1  h  p2
*see Fig. 2.2 h 
pressure head
10 psi  p1  p2  h  23.1 ft or 518mmHg
Ex:
(  62.4 lb 2 ) (  133 KN 3 )
ft
m
p  h  p0
Pressure in a homogeneous, incompressible fluid at rest
~ reference level, X size or shape of the container.
The required equality of pressures at equal elevations
Throughout a system.  F2 
A2
F1
A1
( Fig. 2.5)
Transmission of fluid pressure
3
p1  p2

2.3.2 Compressible Fluid
perfect gas:
dp
gp
    g  
dz
RT
p  RT
p dp
p2
g Z dz
p p  ln p   R Z T
1
2
1
2
1
Assume
g , R const. (z1  z 2 )
T  T0 over z1 , z2  isothermal conditions
 g ( z2  z1 ) 
p2  p1 exp 
RT0 

2.4
Troposphere:
4
T  Ta  z
Ta @ z  0
0.0065  K
  lapose rate
m
0.00357  R
p  pa (1 
ft
z
Ta
g
)
R
2.5 Measurement of Pressure
See Fig. 2.7 explain the gage and absolute pressure
patm  h  pvapor
(Mercury barometer) Example 2.3 pa 
N
( pascal)
m2
2.6 Manometry
1. Piezometer Tube:
1. p  pa
2. h 1 is reasonable p  pa 不大 2. U-Tube Manometer:
3. liquid, not a gas
p A   2 h2   1h1
3. Inclined-tube manometer see examples
*explain Fig. 2.11 Differential U-tube manometer
p A  pB   2 h2   3h3   1h1
Example 2.5
Ex. 2.5
5
u , p , p  p A  pB
Q(the volume rate of the flow)  k p A  pB
p A   1h1   2 h2   1 (h1  h2 )  pB
p A  pB  h2 ( 2   1 )
2.6.3 Fig. 2.12 Inclined tube manometer
p A  pB   2l2 sin 
p  pB
l2  A
 2 sin 
Small difference in gas pressure
If pipes A & B c o n t a ai ng a s
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
6
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 elementmeas. pressure are static or
only changing slowly (quasistatic).
relatively mass of Bourdon tube
<diaphragm>
*strain-gage pressure transducer *
Fig. 2.15 (arterial blood pressure)
piezo-electric crystal. (Refs. 3, 4, 5)
1Hz
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.
7
. Pressure varies linearly with depth if incompressible
dp
    g p  h for open tank, Fig. 2.16
dz
The resultant force acts through the centroid of the area
dθ
Ri
τ
R0
* Exercise 1.66
d  RidA
torque
shearing stress
dA  ( Ri d )l
d  Ri ld
2
2
  Ri2l 0 d  2Ril
Assume velocity distribution in the gap is linear   
2Ri3lw
 
R0  Ri
8
Ri w
R0  Ri
dF  hdA
FR  A hdA  A y sindA
if  ,  are constants.
FR   sin A ydA
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 yR  A ydF  A  siny 2 dA
 FR  A C sin
 yR


A
y 2 dA
yc A
I x  A y 2 dA  second moment of the area (moment of inertia)
yR 
Ix
; I x  I xc  Ayc2
yc A
9
yR 
I xc
 yc
ycA
yR  yc
xR 
I xyc
 xc
ycA
I xc , I xyc ect see Fig. 2.18
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. FR ; ( xR , yR )
b. M ( moment )
10
a.
FR  Eq. 2.18
FR  1.23  106 N
xR  Eq. 2.19, 2.20 xR  0
yR  11.6m
yR
b.
M
c
0
(shaft ; water)
M  FR ( yR  yc )  1.01  105 N  m
2.9 Pressure Prismthe
pressure varies linearly with depth. See Fig. 2.19
11
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.
dp
First, draw the pressure prism out.
 
dz
p  z  p0
Example 2.8
F1  (h1  ps ) A  2.44  10 4 N
h2  h1
) A  0.954  103 N
2
FR  F1  F2  25.4 KN
F2   (
FR y0  F1 (0.3m)  F2 (0.2m)
y0  0.296 m
12
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
See Fig. 2.18
13
F1  hc A
lb 3
 ft  3  1 ft 2 
3
ft
2
 281lb
 62.4
lb   32 2
  ∀ g∀ 62.4 3 
ft  1 ft  441lb at C.G (Centroid;
ft
4
center of pressure, CP; center of gravity)
1
4

3
I C 3
yR  yC 
 ft  12
ft  2 ft
3 2
yc A 2
3
2
4R 4  3
Similarly xR ≈1.27 ft

 1.27 ft
3 3  
∴ F1  FH  281lb; FV    441lb; F2  0
∴ FR  FH2  FV2  523lb
tan  
FH
F
⇒  tan -1 H  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 equilibriumExplain Fig. 2.25; 26; 27; 28
14