3.1 THE BASIC EQUIATION OF FLUID STATICS - FLUID MECH Lecture 3

  LECTURE 3 FLUID STATICS

  By definition, a fluid must deform continuously when a shearing stress of any magnitude is applied.

3.1 THE BASIC EQUIATION OF FLUID STATICS

  dy z _ dz

   p y      p  dxdz j

   

      y

  2     

    _ p y

        p dxdz j

        y

  2        dx

  Pressure, p y y L y

  R y x

   _B

  For a deferential fluid element, the body force, d F , is

     _B d F  g dm  g ρ d 

   Where g is the local gravity vector, ρ the density, and d  is the volume of the element.

  In Cartesian coordinates, d   dx dy dz , so

    _B d F  ρ g dx dy dz

  By use of the Taylor series representation, the pressure at the left face of the differential element is

  p p dy p dy

  

 p  p  y  y  p     p  _{L  L   } y y

  2 y

  2

  

 

  (Terms of higher order omitted because in the limit they vanish.) The pressure on the right face of the deferential element is

   p  p dy

  p  p   y  y   p  _{R R}

   y  y

  2 Stress and forces on the other faces of element are obtained in the same way. Combining all such forces gives the surface force acting on the element. Thus

   p dx p dx

    _S     d F  p  dydz i  p  dydz  i

  

         

 x 2  x

  2     front rear

    p dy    p dy   p   dxdz   j  p   dxdz    j

       y 2  y

  2    

  left right  

   p dz  p dz       

 p   dxdy  k  p   dxdy    k

           z 2  z

  2         lower upper

  Collecting and canceling terms, we obtain

   _S   p  p  p  d F i j k dx dy dz

       

   x  y  z  

   _S   p  p  p  or, d F i j k dx dy dz (3.1a)

       

   x  y  z  

  The term in parentheses is called the gradient of the pressure or simply the gradient, and is designated as grad p or  p . In rectangular coordinates    

   p  p  p   

  

grad p   p  i  j  k  i  j  k p

       x  y  z  x  y  z

      Using the gradient designation, Eq. 3.1a can be written as

   _S

d F   grad p  dx dy dz    p dx dy dz (3.1b)

  From Eq. 3.1b,

   d F _S grad p   p   dx dy dz

  Total force act on a fluid element,     _{S B}

 

d F  d F  d F   grad p   g dx dy dz

  

 

 

  or on a unit volume basis

     d F d F grad p g (3.2)

       d  dx dy dz

    

  For a fluid particle, Newton’s second law gives d F  a dm  a ρ d  . For a static fluid,

    a = 0. Thus d F / d from Eq. 3.2, becomes

     d F

   a 

   d

   Substituting for

  

   grad p   g  Let us review briefly our derivation of this equation. The physical significance of each term is

     g = 0 + grad p

   pressurefo rce body force per        per unit volum e unit volum e = 0    

•     at a point at a point

      This is a vector equation , which means that it really consists of three component equations that must be satisfied individually. Expanding into components, we find _{= 0}

   p ρ g x direction

     _x  x _{= 0}

   p   ρ g  y direction (3.4) _y

  y

  

   p   ρ g  z direction _z z

  

  under this condition, the component equations become

   p   x

   p 

  (3.5)  y

   p ρ g

    _z  z  p

  ρ g    (3.6)

_z z

  

3.1.1 Pressure Variation in a Static Fluid

a. Incompressible Fluid

  For incompressible fluid, ρ = ρ o = constant. Then for constant gravity,

  dp   ρ g  constant _o dz

  If the pressure at the reference level, z o , is designated as p o , then pressure, p , at location z is found by integration _{p z p z} dp   ρ g dz _o

    _{o o}

  or p  p    g z  z   g z  z _{o o  o  o  o } With h measured positive downward, then

  z  z  h _o

  and p  p   g h _{o o} (3.7) z p o

  

  z o

  

  h

  g

  z p y x

  

Fig. Coordinate for determination of pressure variation in a static liquid.

  Example 3.1

  Water flows through pipes A and B. oil, with specific gravity 0.8, is in the upper portion of the inverted U. Mercury (specific gravity 13.6) is in the bottom of the manometer bends.

• – p

  13 8 . 6 .

  _{O H Hg Oil Hg O H} d d d d d           

  Substituting _{O H}

  SG ₂ .

     _{5 4}

₃

_{2 1 2 2 2 2 2} 6 .

  13 8 . 6 .

  13 d d d d d p p _{O H O H O H O H O H B A}

             

    _{5 4 3 2 1} 6 .

  13 ₂ d d d d d _{O H}         

            _{B F F E E D D C C A B A} p p p p p p p p p p p p

  8

  68 2 .

  3 8 .

  40

  10 ₂       _{O H}  in.

  103 6 . ₂   _{O H}  in.

  C p A

  1 ₂ γ d _{O H}

             _{5 4 3 2 1 2 2}

  γ d p p _{O H F B}  

  FIND: Determine the pressure difference, p

  

  A

  B,

  in units of lbf/in

  2 .

  SOLUTION: Basic equations:

         _z

  ρ g z p _{O H O H}

  SG _{2 2}   

   

  γ d p p _{Hg E F}   _{5 2}

  z dp    dz dp ^z _z ^p _p ² ₁ ² ₁

      

  For γ = constant

    o γ z z p p    _{1 2} Beginning at point A and applying the equation between successive point around the

  manometer gives _{1 2}

  γ d p p _{O H A C}   ₂

  γ d p p _{Hg C D}   ₃

  γ d p p _{Oil H D E}   ₄

  

  2

  lbf ft ft  62 . 4  103.6 in.   _{3 2} ft

  12 in. 144 in.

  lbf p  p  _{A B} 3 .

  74 ₂

  in Example 3.2

  A reservoir manometer is built with a tube diameter of 10 mm and a reservoir diameter of 30 mm. The manometer liquid is Meriam red Oil with SG = 0.827 . Determine the manometer deflection in millimeters per millimeter of applied pressure deferential.

  FIND

  Liquid deflection, h, in millimeter per millimeter of water applied pressure p

  2 SOLUTION: d

  p

  1

  2 h D H

  Equilibrium liquid level

  z

  1 Oil, SG = 0.827

  dp 

  Basic equations:    g SG 

  

dz 

_{p 2 z H O 2 2} dp   ρ g dz and dp    gdz _{p 1 z 1}  

  For ρ = constant  p   γ   Δz

  p p g z z ₁        _{2 2 1} or p p g z z g h H ₁             _{2 2 1 oil}

  To eliminate H, note that volume of manometer liquid must remain constant. Thus the volume displaced from the reservoir must be the same as that which rises into the tube.

  2   d _{2 2  }

  

D H  d h or H  h

 

  4

  4 D  

  Substituting gives ₂  

  d

   

  p  p  g h

  1  _{1 2 oil  }    D  

     

  This equation can be simplified by expressing the applied pressure differential as an equivalent water column

  p  p   g  h _{1 2 H O 2}

  and noting that   SG  . Then _{oil oil H O 2 2}  

  d

   

  g h SG g h

  1

       _{H O oil H O     2}

₂

D

       

  h

  1

  or  ₂

   h . 827 1   d / D 

    h

1 Evaluating  

  ₂ 1 .

  09 h . 827 1   10 /

30 

   

  This problem illustrates the effect of manometer design and choice of gage liquid on sensitivity.

b. Compressible Fluid

  Pressure variation in any static fluid is described by the basic pressure-height relation

  dp    g dz

  For many liquids, density is only a weak function of temperature. Pressure and density of liquids are related by the bulk compressibility modulus , or modulus of elasticity,

  dp

  (3.8)

  E  _v  dp /   If the bulk modulus is assumed constant, then density is only a function of pressure.

  The density of gases generally depends on pressure and temperature. The ideal

  gas equation of state p   R T

  (3.9) where: R = the gas constant T = the absolute temperature

  Example 3.3

  The maximum power output capability of an internal combustion engine decreases with altitude (sea level) because the air density and hence the mass flow rate of fuel and air decrease. A truck leaves Denver (jenenge kutho ing monconegoro) (elevation 5,280 ft ).

  o

  Determine the local temperature and barometric pressure are

  80 F and 24.8 in. of

  mercury, respectively. It travels through Vail Pass (jenenge kutho ing monconegoro)

  o

  (elevation 10,600 ft ). The temperature decreases at the rate of 3 F/1000 ft of elevation change. Determine the local barometric pressure at Vail Pass and the percent decrease in maximum power available, compared to that at Denver. GIVEN: Truck travels from Denver to Vail Pass. Engine power output is directly proportion to air density.

  Denver : z = 5,280 ft Vail Pass : z = 10,600 ft _o dT F

  ρ = 24.8 in. Hg   . 003

  dz ft o

  T = 80 F

  FIND: a) Atmosphere pressure at Vail Pass.

  b) Percent engine at Vail Pass compared to Denver.

  :

  SOLUTION

dp p

  Basic equations:   g p  R T  

  

  

RT dz

  Assumptions: 1) Static fluid 2) Air behaves as an ideal gas

  By substituting into the basic pressure-height relation,

  dp p dp g dz   g or   dz RT p RT

  But temperature varies linearly with elevation, dT/dz = – m, so T = T o – m(z– z o )

  dp g dz   p R T  m z  z

      _{o o}

             

       

  and

  T T p p

     ^{mR g} _{o o}

    . 970 827 . ^{25 . 6 /}      

  Note that T o must be expressed as an absolute temperature because it came from the ideal gas equation. Thus

  T T _o ^o _o

  R ft ft F

    

  20 . 8 . . 24 827 827 .   

    

  1 

  10 003 .

  1 280 , , 5 600

  970 . 80 460

     

  and

    Hg in Hg in p p _o 5 .

  The percentage change in power is equal to the change in density, so that  1 

  ft slug lbf lbm slug ft lbf R lbm

  1 970 .

  or percent 5 .

  P P _{o o o}

  T T p p

     

    

    

  1 827 . 1   

    145 .

   

  By substituting from the ideal gas equation,

  RT p  

  RT p   _o ^o _o

      

  P P  

   _{o o o o o}

   

  F ft ft mR g

  32 ² ₂      

    

    

    

    

     

  =

  T z z m z z md T T mR g

    _{o o} ^o _o ^o

     

    

    _{o o} ^{o o} _o

     

  T

  T z z m z z md T T

mR

g

z z m mR z z md g _o

     _{o o} ^o _o ^o _o ^o

     

     

    

     

  T z z m T z z md T mR g

  32 53.3 003 . sec 2 .

  T T mR g

  6 . . sec 2 .

  Evaluating gives 25 .

    

     

  T T p p ^/

  ln ln ln or ^{mR g} _{o o}

  T T z z m mR g p p

     _{o o o o o}

  1 By integrating from p

     

      

     

    

    

   

  in Denver to p at Vail,

  o

   14   _o P P

  3.2 THE STANDARD ATMOSPHERE

  Several International Congresses for Aeronautics have been held so that aviation experts around the world might better be able to communicate.

Table 3.1 Sea Level Condition of the U.S. Standard Atmosphere

  

Property Symbol SI English

o o

  Temperature T 288 K

  59 F Pressure p 101.3 k Pa (abs) 14.696 psia

  3

  3 Density ρ 1.225 kg/m 0.002377 slug/ft

  3 Specific weight - γ 0.7651 lbf/ft

• 5 -7

  2 Viscosity μ 1.781 x 10 kg/m sec 3.719 x 10 lbf/ft

  3.3 ABSOLUTE AND GAGE PRESSURES

  Pressure level

  Atmospheric Pressure

  P

  absolute

  101.3 kPa (14.696 psia) at standard sea level conditions

  Vacuum Absolute pressures must be used in all calculations with the ideal gas or other equations of state. Thus

  P P P   absolute gage atmosphere

  BIBLIOGRAPHY: nd

  1. Fox & Mc Donald, Introduction to fluid mechanics, 2 edition, John Wiley & Sons, Canada.

  2. Irving H. Shames, Mechanics of Fluids, Fourth Edition, Mc Graw Hill, Singapore.