che354 statics

7/27/2019 ChE354 Statics

http://slidepdf.com/reader/full/che354-statics 1/32

Fluid Statics



Fluid Statics

The word “statics” is derived from Greek word“statikos”= motionless

For a fluid at rest or moving in such a manner that

there is no relative motion between particles thereare no shearing forces present:

Rigid body approximation



Definition of Pressure

Pressure is defined as the amount of force exerted on aunit area of a substance:

Pam

N

area

force

P 2



Pascal’s Laws

Pascals’ laws: – Pressure acts uniformly in all directions

on a small volume (point) of a fluid

– In a fluid confined by solid boundaries,

pressure acts perpendicular to the

boundary – it is a normal force.



Direction of fluid pressure on

boundaries

Furnace duct Pipe or tube

Heat exchanger

Dam

Pressure is a Normal Force

(acts perpendicular to surfaces)

It is also called a Surface Force



Absolute and Gage Pressure

• Absolute pressure: The pressure of a fluid is expressed

relative to that of vacuum (=0)

• Gage pressure: Pressure expressed as the difference

between the pressure of the fluid and that of thesurrounding atmosphere.

Usual pressure gages record gage pressure. To

calculate absolute pressure:

gageatmabs P P P



Units for Pressure

Unit Definition or Relationship

1 pascal (Pa) 1 kg m-1 s-2

1 bar 1 x 105 Pa

1 atmosphere (atm) 101,325 Pa

1 torr 1 / 760 atm

760 mm Hg 1 atm

14.696 pounds per

sq. in. (psi)

1 atm



Measurement of Pressure

Mechanical and electronic pressure measuring devices:

When a pressure acts on an elastic structure it will deform. This

deformation can be related to the magnitude of the pressure.

– Bourdon pressure gage

Pressure transducers convert pressure into an electrical output

Strain-gage pressure transducers are suitable for rapid changes in

pressure and cover big ranges of pressure values



Pressure distribution for a fluid at rest

Let’s determine the pressure

distribution in a fluid at rest

in which the only body force

acting is due to gravityThe sum of the forces acting

on the fluid must equal zero



mg zg S

x

y

zLet Pz and Pz+z denote the

pressures at the base and

top of the cube, where the

elevations are z and z+z

respectively.

What are the z-direction forces?

z PS

z z PS



Pressure distribution for a fluid at rest

zg S PS PS F z z z z 0

g z

P P z z z

A force balance in the z direction gives:

For an infinitesimal element (z0)

g dz dP



Incompressible fluidLiquids are incompressible i.e. their density is assumed to

be constant:

PghP o

By using gage pressures we can simply write:

ghP

)zz(gPP 1212

Po is the pressure at the

free surface (Po=Patm)

When we have a liquid with a free surface the pressure P

at any depth below the free surface is:



Pascal’s principle

(The hydrostatic paradox)

• The pressure in a homogeneous, incompressible fluid atrest depends on the depth of the fluid relative to some

reference plane, and it is not influenced by the size or

shape of the tank or container

Fluid is the same in all containers

Pressure is the same at the bottom of all containers

h



Vertical plane surfaces

F

Vert ical rectangu lar wal l (wal l wid th = W)

H

h

Here the pressure varies

linearly with depth: P=gh

P

The lock gate of a canal is rectangular, 20 m wide and 10m high. One side is exposed to the atmosphere and the

other side to the water. What is the net force on the lock

gate?



Vertical plane surfaces

• For an infinitesimal area dA the normal force due to the

pressure is

dF = p dA

• Find resultant force acting on a finite surface by

integration

Whd h g

dA P F

For vertical rectangular wall: F = ½ g W H2

dhh gW



Buoyancy• A body immersed in a fluid experiences a vertical buoyant

force equal to the weight of the fluid it displaces• A floating body displaces its own weight in the fluid in

which it floatsFree liquid surface

The upper surface of the body is

subjected to a smaller force thanthe lower surface

A net force is acting upwards

F1

F2

h1

h2

H



Buoyancy

The net force due to pressure in the vertical direction is:

FB = F2- F1 = (Pbottom - Ptop) (xy)

The pressure difference is:

Pbottom – Ptop = g (h2-h1) = g H

Combining:

FB = g H (xy)

Thus the buoyant force is:

FB = g V




The atmospheric pressure can be measured with a barometer.

– For mercury barometers atmospheric pressure

(101.33kPa) corresponds to h=760 mmHg (= 29.2 in)

– If water is used h = 10.33 m H2O (= 34 ft)

vapor atm pgh p




Manometers are devices in which one or

more columns of a liquid are used to

determine the pressure difference

between two points.

– U-tube manometer

– Inclined-tube manometer



Measurement of Pressure Differences

mambb

mmba

gR Z g P P

R Z g P P

)(

)(

3

2

Apply the basic equation of static

fluids to both legs of manometer,

realizing that P2=P3.

)( bamba gR P P



Inclined Manometer

• To measure small pressure differences need to magnifyRm some way.

sin)(1 baba gR P P



Compressible fluid• Gases are compressible i.e. their density varies with

temperature and pressure = P M /RT

– For small elevation changes (as in engineering

applications, tanks, pipes etc) we can neglect the

effect of elevation on pressure

– In the general case start from:

o

o

RT

z z M g P P

T for )(

exp

:constT

1212

g dz

dP



CompressibleLinear Temperature Gradient

)( 00 z z T T

z

z

p

p z z T

dz

R

g

p

dp

00)( 00

R g

T

z z T

p z p

0

00

0

)(

)(



Atmospheric Equations

• Assume linear

R

g

T z z T p z p

0

000 )()(

• Assume constant

0

0 )(

0)(RT

z z g

e p z p

Temperature variation with altitude

for the U.S. standard atmosphere



Compressible Isentropic

v

p

C

C

P constant

P

1

1

y

P P

T T

1

11

1

12

1

1

12

11

11

RT

z gM T T

RT

z gM P P

Application: bottom hole conditions in gas wells



Gravity Decanter

A A A Ab B Z Z Z

Balancec Hydrostati

21

A

B

A

BT A

A

Z Z

Z

1

2

1

When ρB≈ρ A interface location is very sensitive to height of heavy liquid overflow leg. This leg is often has adjustable

height to give the best separation.



Hydrostatic Equilibrium

in a Centrifugal Field

2

2

60

2

N

mr mr F c

mg F g

2

60

2

N

g

r

F

F

g

c

Typically N≈1000 and r≈1m. Fc/Fg≈110. Neglect g.



Hydrostatic Equilibrium

in Centrifugal Field

dmr dF

r at dr element on Force

2 dr rbdm 2

dr r bdF 22

2

2

1

2

2

2

12

2

2

2

r r P P

dr r rb

dF dP



Centrifugal decanters

When the density difference between two immiscible liquidsis small gravitation forces may be too weak to separate

them in a reasonable time. In this case we can use

centrifugal forces to amplify the forces exerted on the

liquids.

Centrifugal separations are important in many food

industries such a breweries, vegetable oil processing, fruit

juice processing. They are also used to separate emulsionsinto their components.



Continuous Centrifugal Decanter

?Why

P P P P Bi Ai

A

B

B A

B A

i

r r

r

1

22



Continuous Centrifugal Decanter

Consequences:

•ρ AρB within 3% r i unstable

•r B constant r A increased r i

shifted toward bowl wall

•In commercial units r A and

r B are usually adjustable



Example

Consider a 90˚ elbow in a 2 in. pipe. A pipe tap is drilled

through the wall of the elbow on the inside curve, and

another though the outer wall directly across from the first.

The radius of curvature of the inside bend is 2 in. and that

of the outside of the bend is 4 in. The pipe is carrying

water, and a manometer containing an immiscible oil withS.G. of 0.90 is connected across the two taps. If the

reading of the manometer is 7 in., what is the average

velocity of the water in the pipe?

che354 statics

Documents