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 






0
dx
Pressure, p
yL
y
yR
y
x

For a deferential fluid element, the body force, d F B , is



d F B  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


d F B  ρ g dx dy dz
By use of the Taylor series representation, the pressure at the left face of the differential
element is
pL  p 
p
 y L  y   p  p   dy   p  p dy
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
1
pR  p 
p
 yR  y   p  p dy
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 


d FS   p 
dydz i    p 
dydz  i 
x 2 
x 2 


front
rear


p dy 
p dy 
dxdz j    p 
dxdz j 
  p 
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

 p
p
p 
d F S    i 
j  k dx dy dz
y
z 
 x
or,

 p
p
p 
d F S   i 
j  k dx dy dz
y
z 
 x
(3.1a)
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
y
z   x
y
z 
 x
Using the gradient designation, Eq. 3.1a can be written as

d F S   grad p dx dy dz  p dx dy dz
(3.1b)
From Eq. 3.1b,

d FS
grad p  p  
dx dy dz
Total force act on a fluid element,





d F  d F S  d F B    grad p   g


dx dy dz

or on a unit volume basis



dF
dF

  grad p   g
d dx dy dz
(3.2)
2



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


dF
a0
d
Substituting for

 grad p   g  0
Let us review briefly our derivation of this equation. The physical significance of each
term is

 grad p
pressurefo rce

per unit volum
at a point

+


e


+
g
body force per 


unit volum e 
at a point



=0
=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
p

 ρ gx  0
x
x direction
=0
p

 ρ gy  0
y
y direction
p
 ρ gz  0
z
z direction

(3.4)
under this condition, the component equations become
p
0
x
p
0
y
(3.5)
p
  ρ gz
z
p
  ρ g z  
z
(3.6)
3
3.1.1 Pressure Variation in a Static Fluid
a. Incompressible Fluid
For incompressible fluid, ρ = ρo = constant. Then for constant gravity,
dp
  ρo g  constant
dz
If the pressure at the reference level, zo, is designated as po, then pressure, p, at location z
is found by integration

p
po
z
dp    ρo g dz
zo
p  po   o g z  zo   o g zo  z 
or
With h measured positive downward, then
zo  z  h
p  po  o g h
and
(3.7)
z
zo

g

po
h
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.
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FIND:
Determine the pressure difference, pA – pB, in units of lbf/in2.
SOLUTION:
Basic equations:
p
  ρ g z  
z
SG 
dp  z



H O
2
p2
p1

H O
2
z2
dp     dz
z1
For γ = constant
p2  p1  γ z  z o 
Beginning at point A and applying the equation between successive point around the
manometer gives
pA
p C  p A  γ H 2O d 1
p D  pC  γ Hg d 2
 γ H 2O d 1
p E  p D  γ H Oil d 3
pF  pE  γHg d 4
p B  p F  γ H 2O d 5
C
p A  p B   p A  pC    p C  p D    p D  p E    p E  p F    p F  p B 
  H 2O d1   Hg d 2   Oil d 3   Hg d 4   H 2O d 5
Substituting   SG. H 2O
p A  p B   H 2O d1  13.6  H 2O d 2  0.8 H 2O d 3  13.6 H 2O d 4   H 2O d 5
  H 2O  d1  13.6 d 2  0.8d 3  13.6d 4  d 5 
  H 2 O  10  40.8  3.2  68  8 in.
  H 2O  103.6 in.
5
 62.4
p A  p B  3.74
lbf
ft
ft 2

103.6
in.


12 in. 144 in. 2
ft 3
lbf
in 2
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
p2
SOLUTION:
d
p1
2
h
D
H
Equilibrium
liquid level
z
1
Oil, SG = 0.827
Basic equations:
dp
  g
dz
dp   ρ g dz
For ρ = constant
SG 

H O
2
and

p2
p1
z2
dp    gdz
z1
p  γ Δz 
p1  p2  g z 2  z1 
or
p1  p2    g z 2  z1    oil g h  H 
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.
6

4
D2H 

4
2
d 2h
d
H   h
D
or
Substituting gives
p1  p 2   oil
  d 2 
g h 1    
  D  
This equation can be simplified by expressing the applied pressure differential as an
equivalent water column
p1  p 2   H 2O g h
and noting that  oil  SGoil  H 2O . Then
  d 2 
g h 1    
  D  
 H O g h  SGoil  H O
2
2
or
h
1

h 0.827 1  d / D 2
Evaluating
h
1

 1.09
h 0.827 1  10 / 302




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,
Ev 
dp
dp /  
(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
gas equation of state
p   RT
(3.9)
where: R = the gas constant
7
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:
ρ = 24.8 in. Hg
z = 10,600 ft
o
dT
F
 0.003
dz
ft
T = 80o F
FIND:
a) Atmosphere pressure at Vail Pass.
b) Percent engine at Vail Pass compared to Denver.
SOLUTION:
Basic equations:
dp
  g
dz
p   RT   
p
RT
Assumptions: 1) Static fluid
2) Air behaves as an ideal gas
By substituting into the basic pressure-height relation,
dp
p

g
dz
RT
or
dp
g dz

p
RT
But temperature varies linearly with elevation, dT/dz = – m, so T = To – m(z– zo)
dp
g dz

p
R To  mz  zo 
8

g   md z  zo   g To   md z  zo  


mRTo  mz  zo  mR  To To  mz  zo 

g  To  md z  zo  
g  1 To  md z  zo 
=
 

 

mR  To To  mz  zo 
mR  To To  mz  zo  
By integrating from po in Denver to p at Vail,
 p
 T  mz  z o 
g
g T 
ln   
ln  o
ln  

To
 po  mR 
 mR  To 
or
p T

p o  To



g / mR
Evaluating gives
g
32.2 ft
ft
lbm R
slug
lbf . sec 2





 6.25
mR
0.003 F 53.3 ft lbf 32.2 lbm slug . ft
sec 2
and

T  0.003o F
1
 1 
 10,600  5,280 ft 
 0.970
To 
ft
460  80o R 
Note that To must be expressed as an absolute temperature because it came from the ideal
gas equation.
Thus
p T 
 
p o  To 
and
g / mR
 0.970
6.25
 0.827
p  0.827 po  0.827  24.8 in. Hg  20.5 in Hg
The percentage change in power is equal to the change in density, so that
P     o




1
Po
o
o
o

p
RT
o 
po
RTo
By substituting from the ideal gas equation,
P
p To
 1 

 1  0.827 
  1  0.145
Po
po T
 0.970 
or
P
  14.5 percent
Po
9
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
Temperature
Pressure
Density
Specific weight
Viscosity
Symbol
SI
English
T
p
ρ
γ
μ
288 oK
101.3 k Pa (abs)
1.225 kg/m3
1.781 x 10-5 kg/m sec
59 oF
14.696 psia
0.002377 slug/ft3
0.7651 lbf/ft3
3.719 x 10-7 lbf/ft2
3.3 ABSOLUTE AND GAGE PRESSURES
Pressure level
Pabsolute
Atmospheric Pressure
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
Pabsolute  Pgage  Patmosphere
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.
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