
 Nature of fluids
 Hydrostatic Equilibrium
 Applications of fluid statics

 A fluid is a substance that does not permanently resist distortion.
 During the change in shape, shear stresses exist, the magnitudes of which depend upon the viscosity of the fluid and the rate of sliding






——the density changes only slightly with moderate changes in temperature and pressure

——the changes in density caused by temperature and pressure are significant
 (Pressure concept : the pressure at any point in the fluid is independent of direction)






There is a vertical column of fluid shown in Fig.2.1
Three vertical forces are acting on this volume:
(1)the force from pressure p acting in an upward direction , which is pS ;
(2) the force from pressure p+dp acting in a downward direction , which is ( p+dp)S ;
(3)the force of gravity acting downward, which is gρsdz p + dp p g
Figure2.1 Hydrostatic equilibrium

 Then
 pS

( p

)
  
0 (2.1)
 After simplification and division by S ,Eq.(2.1) becomes dp
  
0 (2.2)
 Integration of Eq.(2.2) on the assumption that density is constant gives p

 gZ
 const (2.3)

 Between the two definite heights Z a and Z b shown in Fig.2.1, p
 b
 p
 b

( Z a
 b
)
(2.4)
Equation (2.3) expresses mathematically the condition of hydrostatic equilibrium.

 The relationship between gauge pressure and absolute pressure
 P(gauge)=P(absolute)-P(atmosphere)
 The relationship between vacuum and absolute pressure
 P(vacuum)=P(atmosphere)-P(absolute)
 Or P(vacuum)=- P(gauge)

 The reading in the gauge is 1.5 kgf /cm2 =
=(?)N/m2, and the reading of the vacuum gauge is 736 mmHg = ( )m H2O .If the atmospheric pressure is 1 atm, what happens to the above cases in absolute pressure?

or
 Barometric equation
 For an ideal gas , the density and pressure are related by the equation


= pM
RT
(2.5)
 Substitution from Eq.(2.5)intoEq.(2.2)gives

= pM
RT dp
 gM p RT dZ

0 (2.6)

on the assumption that T is constant,gives ln p b p a
  gM
RT
( Z b

Z a
) or p b p a
 exp

 gM
RT
( Z b

Z a
)

 (2.7)
Equation(2.7)is known as the barometric equation.

In a rotating centrifuge a layer of liquid is thrown outward from the axis of rotation and is held against the wall by centrifugal force.
The free surface of the liquid takes the shape of a paraboloid of revolution.

The rotational speed is so high and the centrifugal force is so much greater than the force of gravity that the liquid surface is virtually cylindrical and coaxial with the rotation.

The situation in shown in Fig.
dr r
2 r
1 r

The entire mass of liquid indicated in Figure is rotating as a rigid body, with no sliding of layer of liquid over another.

 Under these conditions the pressure distribution in the liquid may be found from the principles of fluid static.
• The pressure drop over any ring of rotating liquid is calculated as follows.

The volume element of thickness d r at a radium r .
dF
 
2 rdm
If ρ is the density of the liquid and b the breadth of the ring.
dm
 rbdr

Eliminating dm gives dF

2
 
2 2 b r dr
The change in pressure over the element is the force exerted by the element of liquid, divided by the area of the ring.
dp
 dF
2
 rb
 
2 rdr

The pressure drop over the entire ring is p
2
 p
1
 r
2  r
1

2 rdr
Assuming the density is constant and integration gives p
2
 p
1

 
( r
2
2  r
1
2
)
2
(2.8)


The manometer is an important device for measuring pressure differences.
 U tube manometer (or reverse U tube)
 Inclined manometer
 Differential manometer

 It is the simplest form of manometer.
A pressure p a is exerted in one arm of U tube and a pressure p b in the other.
As a result of the difference in pressure, the meniscus in one branch of the tube is higher than p a p b that in the other
.
ρ
B z m
3
4
Vertical distance between the R m
1 2 two meniscuses R m may be used to measure the difference
ρ
A in pressure.

 The pressure at the point 1 is p
1
 p a
 g z m

R )
 m B
 The pressure at the point 2 is p
2
 p b
 gz
 m B
 gR
 m A
 p 1 is equal to p 2 for the continuous fluid at the same level, thus p a

( m

R )
 m B
 p b
 gz
 m B
 gR
 m A

Simplification of this equation gives p a
 p b
 gR （  m A
 
B
） (2.10)
Note that this relationship is independent of the distance z m
, and of the dimensions of the tube, provided that pressure pa and pb are measured in the same horizontal plane.
If fluid B is a gas, ρ
B compared to
Eq. (2.10)
ρ
A is usually negligible and may be omitted from

 Used for measuring small differences in pressure.

 By making α small, the magnitude of R m is multiplied into a long distance R
1
, and large reading becomes equivalent to a small pressure difference
 p a
 p b
 g
 
A
 
B

R
1 sin

(2.11)

P
2
P
1
B
C
A

 Example: H2O flows through the pipe as shown in Fig. A U-tube manometer is used to measure the pressure P in the pipe. If the atmosphere pressure pa is 1 atm, R and h of mercury and water columns are 0.1 and 0.5 m, respectively, what is pressure P in the pipe,
N/m2?
p h p a
R
A A'
图 1.11