Fluid Statics 4p6b1y

Fluid Statics

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

force N P  2  Pa area m

Direction of fluid pressure on boundaries

Furnace duct

Pipe or tube Heat exchanger

Pressure is a Normal Force (acts perpendicular to surfaces) It is also called a Surface Force Dam

Absolute and Gauge Pressure • Absolute pressure: The pressure of a fluid is expressed relative to that of vacuum (=0) • Gauge pressure: Pressure expressed as the difference between the pressure of the fluid and that of the surrounding atmosphere.  Usual pressure guages record guage pressure. To calculate absolute pressure:

Pabs  Patm  Pgauge

Units for Pressure Unit 1 Pascal (Pa)

Definition or Relationship 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

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 gravity The sum of the forces acting on the fluid must equal zero

What are the z-direction forces? PS z  z Let Pz and Pz+z denote the pressures at the base and top of the cube, where the elevations are z and z+z respectively.

z

y x

mg  Szg PS z

Pressure distribution for a fluid at rest A force balance in the z direction gives:

F

z

 0  PS z  PS z z  Szg Pz  z  Pz   g z

For an infinitesimal element (z0)

dP   g dz

Incompressible fluid Liquids are incompressible i.e. their density is assumed to be constant:

P2  P1  g(z 2  z1 ) When we have a liquid with a free surface the pressure P at any depth below the free surface is:

P  gh  Po

Po is the pressure at the free surface (Po=Patm)

By using gauge pressures we can simply write:

P  gh

Static pressure is a concern !!!!!

Static pressure - 10 minute problem: Rainwater collects behind the concrete retaining wall shown below. If the water-saturated soil (specific gravity = 2.2) acts as a fluid, determine the force and center of pressure on a one-meter width of wall.

Water

1m

Soil

3m

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 floats The upper surface of the body is subjected to a smaller force than the lower surface

h1

 A net force is acting upwards

Free liquid surface F1

H h 2

F2

Buoyancy The net force due to pressure in the vertical direction is: FB = F2- F1 = (Pbottom - Ptop) (xy) The pressure difference is: Pbottom – Ptop =  g (h2-h1) =  g H Combining: FB =  g H (xy) Thus the buoyant force is: FB =  g V

Measurement of Pressure 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 Apply the basic equation of static fluids to both legs of manometer, realizing that P2=P3.

P2  Pa  b g (Z m  Rm ) P3  Pb  b g ( Z m )   a gRm

Pa  Pb  gRm (  a  b )

Inclined Manometer • To measure small pressure differences need to magnify Rm some way.

Pa  Pb  gR1 (  a  b ) sin 

Manometer - 10 minute problem

A simple U-tube manometer is installed across an orifice plate. The manometer is filled with mercury (specific gravity = 13.6) and the liquid above the mercury is water. If the pressure difference across the orifice is 24 psi, what is the height difference (reading) on the manometer in inches of mercury ?

Compressible Flow: Tall Mountains

Natural gas well

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:

dP   g dz

f or T  To  const :  g M ( z 2  z1 )  P2  P1 exp    RTo  

Compressible Linear Temperature Gradient

T  T0   ( z  z 0 ) p

dp gM   p R p0

z

dz  T   ( z  z0 ) z0 0

 T   ( z  z0 )  p( z )  p0  0  T0  

gM

R

Atmospheric Equations • Assume constant p ( z )  p0 e

 g M ( z  z0 )

RT0

• Assume linear  T   ( z  z0 )  p( z )  p0  0  T 0  

gM

R

Temperature variation with altitude for the U.S. standard atmosphere

Compressible Isentropic P





 

 constant 

P1

T P    T1  P1 



1



 1 y

Cv

    1  gMz    P2  P1 1       RT1 

  1

    1  gMz    T2  T1 1       RT1 

Compressible Fluid – 10 minute problem

The temperature of the earth’s surface drops about 5 C for every 1000 m of elevation above the earth’s surface. If the air temperature at ground level is 15 C and the pressure is 760 mm Hg, what is the air pressure on top of Mt. Everest at 8847 m ? Assume air behaves as an ideal gas.