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

For a deferential fluid element, the body force, dFB



, is 



 



ρ d g dm g F d B

Where



gis the local gravity vector, ρ the density, and dis the volume of the element. In Cartesian coordinates, ddx dy dz, so

dx dy dz ρ g

F d B

 



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





2 2

dy y p p dy y p p y y y p p

p_{L L}

                

  

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

x

y

z

dx dy

dz

  

     

  



 

 

 y dxdz j y

p p

2

 

 

 

 

   

  



 

 

 y dxdz j y

p p

2

0

Pressure, p

y





2 dy y p p y y y p p

p_{R R}

        

Stress andforces 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



 

dx



dydz

 

i

x p p i dydz dx x p p F

d S  

                    2 2



 

dy



dxdz

 

j

y p p j dxdz dy y p

p _ 

                   2 2









                               

  dz dxdy k

z p p k dxdy dz z p p 2 2

Collecting and canceling terms, we obtain

dx dy dz k z p j y p i x p F

d S 

               

or, k dx dy dz

z p j y p i x p F

d S 

                (3.1a)

The term in parentheses is called thegradient of the pressure or simply the gradient, and is designated asgradp or p. In rectangular coordinates

p z k y j x i z p k y p j x p i p

grad p _

                              

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



dx dy dz



p dx dy dz grad p

F

d S  



(3.1b) From Eq. 3.1b,

dx dy dz F d p

grad p S



   

Total force act on a fluid element,

dx dy dz g grad p F d F d F

d S B 

    _{ }        

or on a unit volume basis

      

For a fluid particle, Newton’s second law gives   

 



ρ d a dm a F

d . For a static fluid,



a= 0. Thus 



d F

d / from Eq. 3.2, becomes

0   

 

a d

F d



Substituting for

0   grad p g

Let us review briefly our derivation of this equation. The physical significance of each term is

grad p

 +



g

 = 0

    

    

point a at

e unit volum per

rce pressurefo

+

    

    

point a at

e unit volum

per force body

= 0

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

   

 ρ g_x

x p

x direction

0    

 ρ g_y

y p

y direction (3.4)

0

   

 ρ g_z

z p

z direction under this condition, the component equations become

0

  

x p

0    y p

(3.5)

z ρ g z

p

   



     

z ρ g z

p

(3.6)

= 0

3.1.1 Pressure Variation in a Static Fluid

a. Incompressible Fluid

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

constant

   ρ g dz

dp o

If the pressure at the reference level,zo, is designated aspo, then pressure,p, at location z

is found by integration

dz g ρ dp z

z o p

po



o





or p p_o _og



zz_o



_og



z_o z



Withh measured positive downward, then

h z z_o  

and p p_o _og h (3.7)

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.

x

y h





g z

zo

z

po

FIND:

Determine the pressure difference, pA– pB, in units of lbf/in2.

SOLUTION:

Basic equations:  

 

z ρ g z

p

O H O H SG

2 2 

 

  

z

dp dp z dz

z p

p

2

1 2

1





 

For γ = constant



z zo



γ p

p₂  ₁ 

Beginning at point A and applying the equation between successive point around the manometer gives

1

2 d γ p

p_C  _A  _HO

2

d γ p p_D  _C  _Hg

3

d γ p p

Oil H D E  

4

d γ p p_F  _E  _Hg

5

2 d γ p

p_B  _F  _HO



A C

 

C D

 

D E

 

E F

 

F B



B

A p p p p p p p p p p p

p           

5 4

3 2

1 2

2Od Hgd Oild Hgd HOd

H    

    

 

Substituting SG _HO 2

.  

5 4

3 2

1 2 2 2 2

2 d 13.6 d 0.8 d 13.6 d d p

p_A  _B _HO  _HO  _HO  _HO _HO



1 13.6 2 0.8 3 13.6 4 5



2O d d d d d

H     





10 40.8 3.2 68 8



2     

_HO in.

6 . 103

2 

_HO in.

C pA

1

2 d

γ_HO

2 2

3 _144in.

ft in. 12

ft in. 103.6 ft

lbf 4 .

62   



2

74 . 3

in lbf p

p_A  _B 

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

SOLUTION:

Basic equations: g

dz dp



 

O H SG

2 

 

ρ g dz

dp and







2

1 2

1

z z p

p dp gdz For ρ = constant pγ

 

Δz



2 1



2

1 p g z z

p   

or p₁ p₂  g



z₂ z₁



 _oil g



hH



To eliminate H, note that volume of manometer liquid must remain constant. Thus the D

d

z p2

p1

h 2

1

Oil, SG = 0.827

H

Equilibrium

h d H

D2 2

4 4

 

 or h

D d H

2

      

Substituting gives

    

  

       



2

2

1 h 1

D d g

p

p _oil

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

h

2

2

1 p  g

p _HO

and noting that _oil SG_{oil HO} 2



  . Then

    

  

       



2

O

H h 1

h

2 2

D d g

SG

g _oil

O

H 



or







2



/ 1 827 . 0

1

D d h

h

 



Evaluating









1.09

30 / 10 1 827 . 0

1

2 

 

h h

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 g

dz dp



 

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/



dp

E_v  (3.8)

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 gasequation of state

T R

p (3.9)

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). Determine the local temperature and barometric pressure are 80oF and 24.8 in. of mercury, respectively. It travels through Vail Pass (jenenge kutho ing monconegoro) (elevation 10,600 ft). The temperature decreases at the rate of 3 oF/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

ρ = 24.8 in. Hg

ft F dz

dT o

003 . 0  

T = 80o F FIND:

a) Atmosphere pressure at Vail Pass.

b) Percent engine at Vail Pass compared to Denver.

SOLUTION:

Basic equations: g

dz dp





 p RT

RT p





Assumptions: 1) Static fluid

2) Air behaves as an ideal gas

By substituting into the basic pressure-height relation, g

RT p dz

dp



 or

RT dz g p dp

 

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

dz g dp

 













_ 











_          o o o o o o o z z m T z z md T T mR g z z m mR z z md g o T









_          o o o o o z z m T z z md T T mR g =









_          o o o o

o T m z z z z md T T mR g 1

By integrating frompo in Denver top at Vail,





                      o o o o o T T mR g T z z m T mR g p p ln ln ln or mR g o o T T p p /        Evaluating gives 25 . 6 . sec . 2 . 32 53.3 003 . 0 sec 2 . 32 2

2     

 ft slug lbf lbm slug ft lbf R lbm F ft ft mR g and







460 80



0.970

1 280 , 5 600 , 10 003 . 0

1 _

           R ft ft F T T o o o

Note that To must be expressed as an absolute temperature because it came from the ideal

gas equation. Thus



0.970



6.25 0.827

/          mR g o o T T p p

and p0.827 p_o 



0.827



24.8in.Hg 20.5inHg

The percentage change in power is equal to the change in density, so that 1        o o o o o P P        RT p   o o o RT p  

By substituting from the ideal gas equation,





1 0.145

970 . 0 1 827 . 0

1  

         T T p p P P _o o o

or  14.5 percent

o 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

Temperature T 288oK 59oF

Pressure p 101.3 k Pa (abs) 14.696 psia

Density ρ 1.225 kg/m3 0.002377 slug/ft3

Specific weight γ - 0.7651 lbf/ft3

Viscosity μ 1.781 x 10-5 kg/m sec 3.719 x 10-7 lbf/ft2

3.3 ABSOLUTE AND GAGE PRESSURES

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

BIBLIOGRAPHY:

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

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

Pressure level

Atmospheric Pressure

101.3 kPa (14.696 psia) at standard sea level conditions

Vacuum

atmosphere gage

absolute

P

P