CHAPTER 2Fluid Statics and Its Applications

Nature of fluids

Hydrostatic Equilibrium

Applications of fluid statics

Nature of fluids 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

Fluids include

liquid ,gas and solid particles suspended in li

quid and gas or slurry

Fluids also can be divided asIncompressible——the density change

s only slightly with moderate changes in temperature and pressure

Compressible——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)

Hydrostatic Equilibrium There is a vertical column

of fluid shown in Fig.2.1 Three vertical forces are a

cting 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 Figure2.1 Hydrostatic equilibrium

p

p +dp

g

Then

(2.1)

After simplification and division by S,Eq.(2.1) becomes

(2.2)

Integration of Eq.(2.2) on the assumption that

density is constant gives

(2.3)

( ) 0pS p dp S g SdZ

0dp g dZ

pgZ const

Between the two definite heights Za and Zb show

n in Fig.2.1,

(2.4)

Equation (2.3) expresses mathematically the condition of hydrostatic equilibrium.

( )b ba b

p pg Z Z

Gauge pressure, absolute pressure and vacuum

The relationship between gauge pressure and absolute pressure

P(gauge)=P(absolute)-P(atmosphere) The relationship between vacuum and absol

ute 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?

Barometric equation

For an ideal gas , the density and pressure are related by the equation

(2.5)

Substitution from Eq.(2.5)intoEq.(2.2)gives

(2.6)

or

pM=

RT

pM= 0

RT

dp gMdZ

p RT

(2.7)

Equation(2.7)is known as the barometric equation.

Integration of Eq.(2.6)between levels and ,

on the assumption that T is constant,gives

a b

ln ( )bb a

a

p gMZ Z

p RT

exp ( )bb a

a

p gMZ Z

p RT

or

Hydrostatic equilibrium in a centrifugal field

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.b

r1r2

r

dr

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 dr at a

radium r.

If ρ is the density of the liquid and b the breadth of the ring.

2dF rdm

2dm rbdr

Eliminating dm gives

The change in pressure over the element is the force exerted by the element of liquid, divided by the area of the ring.

2 22dF b r dr

2

2

dFdp rdr

rb

The pressure drop over the entire ring is

Assuming the density is constant and integration gives

(2.8)

2

1

22 1

r

r

p p rdr

2 2 22 1

2 1

( )

2

r rp p

Applications of fluid statics

Manometer (pressure gauge) The manometer is an important device for

measuring pressure differences. U tube manometer (or reverse U tube) Inclined manometer Differential manometer

U tube manometer It is the simplest form of manometer.A pressure pa is exerted in one

arm of U tube and a pressure pb

in the other.As a result of the difference in pressure, the meniscus in one branch of the tube is higher than

that in the other.Vertical distance between the

two meniscuses Rm may be

used to measure the difference

in pressure.

pa pb

Rm1

3

2

zm4

ρB

ρA

The pressure at the point 1 is

The pressure at the point 2 is

p1 is equal to p2 for the continuous fluid at the same level, thus

1 ( )a m m Bp p g z R

2 b m B m Ap p gz gR

( )a m m B b m B m Ap g z R p gz gR

Simplification of this equation gives

Note that this relationship is independent of the distance zm, 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, ρB is usually negligible compared to ρA and may be omitted from Eq. (2.10)

a b m A Bp p gR （ ） (2.10)

Inclined manometer

Used for measuring small differences in pressure.

By making α small, the magnitude of Rm is mu

ltiplied into a long distance R1, and large reading becomes equivalent to a small pressure difference

(2.11)

1 sina b A Bp p g R

Differential manometer

B

C

P1 P2

R

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

pah

R

A'A

1. 11图