CHAPTER 2 Fluid Statics and Its Applications Nature of fluids Hydrostatic Equilibrium Applications of fluid statics
CHAPTER 2 ❖Fluid 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
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 liquid and gas or slurry
Fluids include ❖liquid , ❖gas ❖and solid particles suspended in liquid and gas or slurry
Fluids also can be divided as Incompressible-the density changes 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)
Fluids also can be divided as ❖Incompressible——the density changes 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 acting on this volume: (1)the force from pressure p acting in an upward direction which is os; (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 Figure2.1 Hydrostatic equilibrium gpsdz
Hydrostatic Equilibrium ❖ 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 Figure2.1 Hydrostatic equilibrium p p +dp g
Then +pS-(p+dp)S-gpSdZ=0 (2.1) After simplification and division by S,Eq.(2.1) becomes dp+gpdz-0 (2.2) Integration of Eq.(2.2)on the assumption that density is constant gives +gZ=const (2.3) p
❖ 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) + − + − = pS p dp S g SdZ ( ) 0 dp g dZ + = 0 p gZ const + =
Between the two definite heights Za and Zp shown in Fig.2.1, 卫P6-=g(Z。-Z) 00 (2.4) Equation(2.3)expresses mathematically the condition of hydrostatic equilibrium
❖ Between the two definite heights Za and Zb shown in Fig.2.1, (2.4) Equation (2.3) expresses mathematically the condition of hydrostatic equilibrium. ( ) b b a b p p g 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 absolute pressure P(vacuum)=P(atmosphere)-P(absolute) Or P(vacuum)=-P(gauge)
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 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 H20.If the atmospheric pressure is 1 atm,what happens to the above cases in absolute pressure?
❖ 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 pM 0 (2.5) RT Substitution from Eq.(2.5)intoEq.(2.2)gives ppNM中+84dz=0 (2.6) RT P RT
❖ 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 gM dZ p RT + =