1.
FLUID PROPERTIES
Definitions
Fluid Mechanics - Behavior of fluids
Fluid Mechanics is the study of fluids either at rest or in motion and the
subsequent effects of the fluid on the boundaries
There are three branches:
1. Fluid statics - Fluids at rest
2. Kinematics - Moving fluids without consideration of associated forces
3. Hydrodynamics - Forces involved with fluid motion
A Fluid :
• Latin - fluidus = free to change form
• A fluid is a substance which can not resist shear
• A fluid will yield to shear stress
There are two types of fluids
1. Gases
- are compressible
- expand when external pressure is removed
- at equilibrium only when completely enclosed
- (A vapor) = gas with temperature and pressure near the liquid phase
2. Liquid
- incompressible
- exhibits cohesion between molecules
- can have a free surface
Rheology - study of deformation and flow
Definition of a fluid (after Websters Ninth New Collegiate Dictionary)
• Having particles that easily move and change their relative position without a
separation of the mass, and that easily yield to pressure.
• Capable of flowing
• Likely or tending to change or move
• A substance (a liquid or gas) tending to flow or conform to the outline of a
container
Some non "technical definitions are as follows
• characterized or employing a smooth easy style
• easily converted to cash
2.
• available for a different use
• A fluid is a substance that is deformed continuously when acted upon by a
shearing stress of any magnitude.
• A fluid is a substance that deforms continuously under the action of an applied
shear stress.
• A fluid is a substance that deforms continuously under the action of an applied
shear (tangential) stress no matter how small that shear stress may be.
• A fluid is a substance that deforms continuously when subjected to shear stress ; a
fluid is not capable of sustaining shear stress at rest
• A fluid is often compared to a solid to define its properties
• A solid can resist shear stress by static deformation. This deformation can be
either elastic or plastic in nature.
Comparison of a liquid and a solid
A solid can resist shear stress by static deformation. A fluid can not resist shear stress,
hence motion must occur. A shear force is the force component tangent to a surface, and
this force divided by the area of the surface is the average shear stress.
τ = Force / Area = Shear Stress
In the case of a fluid when a force F is applied to the upper plate and the lower surface is
fixed motion will be produced and this motion will continue as long as the force is
applied.
Moving Plate
!y
Fixed Plate
When the moving plate is pulled with force F the motion will continue and the plate will
never reach an a stationary position.
A solid with the same type of force applied, however will reach an equilibrium location
at some spatial coordinates. These conditions are true for both linear or angular forces.
3.
" # "
!y
!x !x
Size Speed Pressure
Figure 1.1 of text
4.
DIMENSIONS AND UNITS
In the study of fluid mechanics it is necessary to describe the characteristics of a
fluid in both a qualitative and quantitative sense. Further the descriptions must be
translated into reasonable mathematical functions so that engineering analysis and design
can be accomplished.
The qualitative description identifies the nature or type of the characteristic.
Examples of the primary qualitative measures are:
Length L
Time T
Mass M
Temperature θ
Force F
Examples of secondary qualitative measures are:
Area L2
Velocity L T-1
Density M L-3
These qualitative measures are the basic dimensions used in the study of fluid mechanics.
A Dimension is the measure by which a physical variable is expressed quantitatively.
The quantitative description requires both a number (measure) and a standard (system) by
which the quantities can be compared.
The Unit is a method of attaching a number (value) to the quantitative Dimension.
Typical examples are :
L = feet or meters
Velocity = ft / sec or m / sec.
Density = kg / m3 or lbs / ft3 or slugs / ft3
Table 1.1 provides a list of dimensions for many of the commonly used physical
quantities.
5.
The two major systems of Unit are:
British Gravitational (BG) System
Length foot (ft)
Time Second (s)
Force Pound (lb)
Mass Slug (slug)
Temperature Fahrenheit (° F)
Absolute Temperature Rankine (° R)
Power Horsepower (HP) = γ Q H / 550
γ = Specific Weight of the Fluid
Q = Flowrate
H = Hydraulic Head
International System (SI)
Length meter (m)
Time Second (s)
Force Newton (N)
Mass Kilogram (kg)
Temperature Celsius* (° C) note: the actual scale is ° Kelvin
Absolute Temperature Kelvin ( k )
Work Joule (J) = 1 N · m
Power Watt (W) = 1 J / s = 1 N · m / s
Water boils 373.15°K 100.00°C 212.00°F 671.67°R
Ice forms 273.15°K 0.00°C 32.00°F 491.67°R
KELVIN
SI ABSOLUTE FAHRENHEIT
CELSIUS RELATIVE
RELATIVE RANKINE
ABSOLUTE
Absolute zero 0.00°K -459.69°F 0.00°R
-273.15°C
6.
GASES
Air is a mixture of gases. Since the mixture remains nearly constant over a relatively
large temperature range (100 °K to 2200 °K) and is a high temperature and low pressure
relative to the critical point it can be assumed to follow the ideal gas law.
P νs = R T
Where;
P = Absolute pressure
= Specific volume
T = Absolute Temperature (°K)
R = Gas constant
7.
Since a gas has a viscosity and can develop shear stresses and is compressible;
P =ρRT And R =P/ρT
If one sets units to the gas law where;
= Kg / m3, P = Pascals ( N / m2 ), T = (°K) = ( °C + 273)
Then R = N 1 = m3 N = m N
m Kg / m3
2
°K 2
m Kg °K Kg °K
When;
= slugs / ft3, P = lb / ft2 , T = (°R) = ( °F + 459.6)
Then R = lb 1 = ft lb = ft lb = ft3
ft2 slugs / ft3 °R slug °R 2 2 4 2
ft lb sec / ft °R sec °R
When;
= lbm / ft3, P = lb / ft2 , T = (°R) = ( °F + 459.6)
Then R = lb 1 = ft lb The magnitude of R in slugs is 32
2
ft lbm / ft3 °R lbm °R times greater the value in lbmass
The perfect gas law encompasses both Charles Law and Boyles Law.
Charles Law states that for a constant pressure the pressure varies with temperature.
That is:
When P = Constant T increases, and V increases.
for Boyles Law (under isothermal conditions) and a constant temperature;
varies as Pabs.
ρ increases as Pabs. increases
Under these conditions the Ideal Gas Law can be expressed as;
P Vv = m R T
8.
This relationship can be simplified using a molar basis. A kilogram mole of gas is equal
to the number of kilograms mass of gas equal to the molecular weight. That is;
A kilogram mole of O2 is 2 (16) = 32 Kg
A kilogram mole of N2 is 2 (14) = 28 Kg
When Vs is the volume per mole, the Perfect Gas Law can be rewritten as,
P Vs = nM R T
Where M is the molecular weight. Then if n is the number of moles of gas in
volume ν.
P ν = nM R T,
and where nM = Mm (Mm is the molecular mass)
From Avogadro’s Law, equal volumes of gas at absolute temperature and pressure have
the same number of molecules. So then the gas masses are proportional to the molecular
weights. That is;
Mm R = P ν /n T = a constant.
as P ν /n T is the same for any perfect gas, then MmR is the universal gas constant.
Mm R = 8312 m N / Kg mole °K
or
R = 8312 m N / kg °K = 8312 m2 / sec2 °K
Mm Mm
Since the molecular mass of air is 28.98
R = 8312 m N / kg °K = 286.8 m N / kg °K (Ref. Table 1.8)
28.98
Since 1 Joule = 1 N · m the above units are also J / kg °K
Using BG Units
R = 49,709 ft lb / slug °R = 1715 ft lb / slug °R (Ref. Table 1.7)
28.98
Specific Heat (Cv) is the number of units of heat added per unit mass to raise the
temperature of a gas one degree when the volume is held constant.
Specific Heat (Cp) is the number of units of heat added per unit mass to raise the
temperature of a gas one degree when the pressure is held constant.
9.
The Specific Heat Ratio (k) Is the ratio of Cp to Cv
K = Cp / Cv
The units of Cp And Cv are :
SI Units Kcal / Kg °K
BG Units BTU / lbm °R
Try example 1.3 (Page 15, Text)
A compressed air tank has a volume 0f 0.84 ft3. When the tank is filled with air at a
gauge pressure of 50 psig, Determine the density of the air and the weight of air in the
tank. Assume that the temperature is 70 oF, and the atmospheric pressure is 14.7 psia
The density of the air can be calcaulated using the ideal gas law (Eq. 1.8)
ρ = p/RT
The pressure in the tank is: 50 psi + 14.7 psi = 64 7 psia
(64.7 lb/in 2 ) (144 in 2/ ft2 )
! = = 0.01020slugs/ft 3
(1716 ft lb/slug oR) [(70 + 460) oR]
Weight = γ x Volume = ( γ g) Volume
= (0.01020 slugs/ft3) (32.2 ft/s2) (0.84 ft3)
= 0.276 lb
Note 1 slug = 1.0 lb. s2/ft.
(lb.) ( s2) (ft) (ft3)
(ft) (ft3) (s2)
10.
Parameters That Describe a Fluid
1. Density = ρ = Mass / Unit Volume
ρwater = γw / g = 62.4 lbs /ft3 / 32.2 ft / sec2 = 1.94 lbs / sec2 ft4
ρwater = 1.94 slugs / ft3 or, 1 slug = 1 lb - sec2 / ft.
1000
990
Density, ! kg/m 3
980
@ 4°C, ! = 1000 kg/m3
970
960
950
0 20 40 60 80 100
Temperature °C
Density of Water as a Function of Water
2. Specific Weight γ = Gravitational force acting on a unit volume of the
weight per unit volume
γ = ρ • g
γwater = 62.4 lbs / ft3
γ water = ρ •g = 1.94 slugs / ft3 • 32.2 ft / sec2 = 62.4 lbs /ft3
3. Specific Volume = υ = The volume occupied by a unit weight of the fluid.
υ=1/ρ
4. Specific Gravity = S = SG = ρ liquid /ρ water
For pure water @ standard pressure and 20 ° C
ρ water = 1.0 grams / cm3 @ 4 ° C
11.
5. Viscosity (either µ or ν )
µ is called the dynamic viscosity, absolute viscosity, or simply the viscosity
The units are (N•s/m2) or lb •s/ft2)
ν is called the kinematic viscosity and includes the property density ρ.
ν = µ / ρ . The units are (m2 / sec) and (ft2 / sec)
For the given system of a moving plate sliding over a fixed plate with a fluid of thickness
b.
For a small ∆ time or ∆t , line AB will rotate through the angle δB.
Then:
tan δB ≈ δB ≈ δa / b
Since this represents the “deformation” of the fluid, the rate of shearing strain
(γ) ≈ δB / δt
(γ) = U / b = du / dy
As; τ = P /A, τ ≈ ( γ) ≈ du / dy with the measure of the viscosity
τ =µ du /dy (Equation 1.9)
12.
Compressibility of Fluids
Bulk Modulus
The bulk modulus is defined to describe the compressibility of a fluid. As the
pressure in the system shown below increases, the volume of fluid decreases.
p
p + dp
V V - dV
The bulk modulus is defined as
Ev = - dp/(dV/V) Equation 1.12
Since the decrease in volume of a given mass will result in an increase in density
m = ρV.
Ev = - dp/(dρ/ρ) Equation 1.13
Compression and Expansion of Gases
As gases are expanded or compressed the relationship between pressure (p) and
density (ρ) depends upon the nature of the process.
When the change takes place at a constant temperature the process is isothermal.
p/ρ = constant Equation 1.14
If the change is frictionless and without heat exchange to the surrounding
environment the process is isentropic.
p/ρk = constant Equation 1.15
so that
k = cp/cv
where k is the ratio of the specific heat at a constant pressure, cp, to the specific heat at
a constant volume, cv.
13.
Due to the fact that the two specific heats are related to the gas constant, R, using the
relationship
R = cp -cv and all pressures are expressed as absolute pressure
Then for an isothermal process
Ev = p Equation 1.16
And for a isentropic process
Ev = kp Equation 1.17
As a example for air at standard conditions
Ev = kp = 1.4 x 14.7 psia - 20.6 psia
For water
Ev = 312,000 psia
So air is approximately 15,000 times more compressible than water
Speed of sound
The speed of sound is related to the change of pressure and density of the fluid medium.
It is expressed as:
c = (dp/dρ)0.5 Equation 1.18
or in terms of the bulk modulus
c = (Ev/ρ)0.5 Equation 1.19
as the disturbance caused by sound is small and there is negligible heat transfer then the
process is considered isentropic. For this isentropic process Ev = kp , or
c = (kp/ρ)0.5
and using the ideal gas law
14.
c = (kRT)0.5 Equation 1.20
Vapor Pressure
Boiling which is the formation vapor bubbles in a fluid mass, is initiated when the
absolute pressure in the fluid reaches the vapor pressure (pv).
15.
Surface Tension (σ )
2πRσ = ∆pπR2
∆p = pi - pe = 2σ /R
MOLECULE AT THE
SURFACE ARE ACTED
UPON BY A NET INWARD
COHESIVE FORCE
INTERIOR MOLECULES ARE ACTED OPON BY MORE
OR LESS EQUAL FORCES IN ALL,DIRECTIONS
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