Transcript 02_fluid properties

Chapter 2: Properties of Fluids
Introduction
Any characteristic of a system is called a property.
Familiar: pressure P, temperature T, volume V, and mass m.
Less familiar: viscosity, thermal conductivity, modulus of
elasticity, thermal expansion coefficient, vapor pressure, surface
tension.
Intensive properties are independent of the mass of the
system. Examples: temperature, pressure, and density.
Extensive properties are those whose value depends on
the size of the system. Examples: Total mass, total
volume, and total momentum.
Extensive properties per unit mass are called specific
properties. Examples include specific volume v = V/m
and specific total energy e=E/m.
Chapter 2: Properties of Fluids
Continuum
Atoms are widely spaced in the
gas phase.
However, we can disregard the
atomic nature of a substance.
View it as a continuous,
homogeneous matter with no
holes, that is, a continuum.
This allows us to treat properties
as smoothly varying quantities.
Continuum is valid as long as
size of the system is large in
comparison to distance between
molecules.
Chapter 2: Properties of Fluids
Density and Specific Gravity
Density is defined as the mass per unit volume r = m/V.
Density has units of kg/m3
Specific volume is defined as v = 1/r = V/m.
For a gas, density depends on temperature and
pressure.
Specific gravity, or relative density is defined as the
ratio of the density of a substance to the density of some
standard substance at a specified temperature (usually
water at 4°C), i.e., SG=r/rH20. SG is a dimensionless
quantity.
The specific weight is defined as the weight per unit
volume, i.e., gs = rg where g is the gravitational
acceleration. gs has units of N/m3.
Chapter 2: Properties of Fluids
Density of Ideal Gases
Equation of State: equation for the relationship
between pressure, temperature, and density.
The simplest and best-known equation of state
is the ideal-gas equation.
Pv=RT
or
P=rRT
Ideal-gas equation holds for most gases.
However, dense gases such as water vapor and
refrigerant vapor should not be treated as ideal
gases. Tables should be consulted for their
properties, e.g., Tables A-3E through A-6E in
textbook.
Chapter 2: Properties of Fluids
Vapor Pressure and Cavitation
Vapor Pressure Pv is defined
as the pressure exerted by its
vapor in phase equilibrium
with its liquid at a given
temperature
If P drops below Pv, liquid is
locally vaporized, creating
cavities of vapor.
Vapor cavities collapse when
local P rises above Pv.
Collapse of cavities is a
violent process which can
damage machinery.
Cavitation is noisy, and can
cause structural vibrations.
Chapter 2: Properties of Fluids
Energy and Specific Heats
Total energy E is comprised of numerous forms:
thermal, mechanical, kinetic, potential, electrical,
magnetic, chemical, and nuclear.
Units of energy are joule (J) or British thermal unit (BTU).
Microscopic energy
Internal energy u is for a non-flowing fluid and is due to
molecular activity.
Enthalpy h=u+Pv is for a flowing fluid and includes flow energy
(Pv).
Macroscopic energy
Kinetic energy ke=V2/2
Potential energy pe=gz
In the absence of electrical, magnetic, chemical, and
nuclear energy, the total energy is eflowing=h+V2/2+gz.
Chapter 2: Properties of Fluids
Coefficient of Compressibility
How does fluid volume change with P and T?
Fluids expand as T ↑ or P ↓
Fluids contract as T ↓ or P ↑
Need fluid properties that relate volume changes to changes in P and
T.
Coefficient of compressibility (also called as Bulk Modulus of Elasticity)
 P 
 P 


r



 v T
 r T
Coefficient of volume expansion
  v 
P
V
V
1  v 
1  r 
      
v  T  P
r  T  P
Combined effects of P and T can be written as
 v 
 v 
dv  
dT


  dP
 T  P
 P T
Chapter 2: Properties of Fluids
Values for Bulk Modulus for selected liquids
at atmospheric pressure and 20 deg C
LIQUID
BULK MODULUS (MPa)
Ethyl Alcohol
896
Benzene
1 062
Machine Oil
1 303
Water
2 179
Glycerine
4 509
Mercury
24 750
Chapter 2: Properties of Fluids
Example:
1.
Compute the change in pressure that must be applied to water to
change its volume by 1.0 per cent.
Solution:
The 1.0 % volume change means ∆V/V = - 0.01
From Table : Bulk Modulus of Water is κ = 2 179 MPa
Therefore, from the definition of the Bulk Modulus,
i.e., κ = - ∆P/(∆V/V), we get
∆P = - κ * ∆V/V = 2 179*0.01 = 21. 79 MPa. ANS.
2.
Compute the pressure change required to cause a decrease in
the volume of mercury by 1.0 per cent. Express the result in MPa.
[Answer: 247.5 MPa]
Chapter 2: Properties of Fluids
Isothermal Compressibility is the inverse of the coefficient of compressibility. It
represents the fractional change in volume or density corresponding to a unit
change in pressure and is denoted by α. Thus,
 
1


1 V
1 r
(
)T 
(
)T
V P
r P
The combined effects of pressure and temperature changes on the volume change
of a fluid can be determined by taking the specific volume to be function of T and P.
Differentiating V = V(T,P) and using the definitions of the compression and
expansion coefficients α and β give
V
r

  T   P
V
r
Chapter 2: Properties of Fluids
Example: Variation of Density with Temperature and Pressure
Consider water initially at 20 deg C and 1 atm. Determine the final density of
water (a) if it is heated to 50 deg C at a constant pressure of 1 atm, and
(b) if it is compressed to 100-atm pressure at a constant temperature of
20 deg C. Take the isothermal compressibility of water to be α = 4.80x105 atm-1.
Soln.
The density of water at 20 deg C and 1 atm is ρ1=998.0 kg/m3.
The average temperature is (20+50)/2 = 35 0C
The value of β (for water) at this average temperature is = 0.337x10-3 K-1
(From Table)
The value of α = 4.80x10-5 atm-1 (Given).
(a) The change in density due to change of temperature from 20 0C to 50 0C
at constant pressure is
∆ρ = -βρ∆T = -(0.337x10-3)(998)(50-20) = -10.0 kg/m3
Noting that ∆ρ = ρ2 – ρ1, the density of water at 50 0C and 1 atm is:
ρ2 = ρ1 + ∆ρ = 998.0 + ( -10.0) = 988.0 kg/m3 ANSWER
Chapter 2: Properties of Fluids
(b) The change in density due to change of pressure from 1 atm to 100
atm at constant temperature is given by
∆ρ = αρ∆P = (4.80x10-5)(998.0)(100 – 1)
= 4.7 kg/m3
Then the density of water at 100 atm and 20 0C becomes
ρ2 = ρ1 + ∆ρ = 998.0 + 4.7 = 1002.7 kg/m3 ANSWER
Discussion: The density of water decreases while being heated and
increases while being compressed, as expected.
Chapter 2: Properties of Fluids
Viscosity
Viscosity is a
property that
represents the
internal resistance of
a fluid to motion.
The force a flowing
fluid exerts on a body
in the flow direction is
called the drag force,
and the magnitude of
this force depends, in
part, on viscosity.
Chapter 2: Properties of Fluids
Viscosity
To obtain a relation for viscosity,
consider a fluid layer between
two very large parallel plates
separated by a distance ℓ
Definition of shear stress is t =
F/A.
Using the no-slip condition,
u(0) = 0 and u(ℓ) = V, the velocity
profile and gradient are u(y)=
Vy/ℓ and du/dy=V/ℓ
Shear stress for Newtonian fluid:
t = mdu/dy
m is the dynamic viscosity and
has units of kg/m·s, Pa·s, or
poise.
Chapter 2: Properties of Fluids
Values of Dynamic viscosity
Liquid
Temperature
deg C
Dynamic Viscosity N-s/m2
Water
20
1.0x10-3
Gasoline
20
3.1x10-4
SAE 30 Oil
20
3.5x10-1
SAE 30 Oil
80
1.9x10-2
Chapter 2: Properties of Fluids
Viscometry
How is viscosity measured? A rotating
viscometer.
Two concentric cylinders with a fluid in
the small gap ℓ.
Inner cylinder is rotating, outer one is
fixed.
Use definition of shear force:
du
F t A  mA
dy
If ℓ/R << 1, then cylinders can be
modeled as flat plates.
Torque T = FR, and tangential velocity
V=wR
Wetted surface area A=2pRL.
Measure T and w to compute m
Chapter 2: Properties of Fluids
MEASUREMENT OF VISCOSITY
T  FR
-----(1)
du
F t A  mA
dy
V
F  mA
A  2pRL
l
V  wR
w  2pN
2
3
4p R NL
T m
l
Chapter 2: Properties of Fluids
Example:
The viscosity of a fluid is to be measured by a viscometer constructed
of two 40-cm-long concentric cylinders. The outer diameter of the
inner cylinder is 12 cm, and the gap between the two cylinders is
0.15 cm. The inner cylinder is rotated at 300 rev/min, and the torque
is measured to be 1.8 N-m. Determine the viscosity of the fluid.
Soln:
The velocity profile is linear only when the curvature effects are
negligible, and the profile can be approximated as being linear in
this case since l/R << 1. Therefore,
μ = Tl/(4 π2R3NL), gives
μ = 0.158 N-s/m2
ANS.
Chapter 2: Properties of Fluids
Problem
The viscosity of a fluid is to be measured by a viscometer constructed of two
75-cm-long concentric cylinders. The outer diameter of the inner cylinder is
15 cm, and the gap between the two cylinders is 0.12 cm. The inner cylinder
is rotated at 200 rev/min, and the torque is measured to be 0.8 N.m.
Determine the viscosity of the fluid.
Chapter 2: Properties of Fluids
Problem: In regions far from the entrance, fluid flow through a circular pipe is ondimensional, and the velocity profile for laminar flow is given by u(r) = umax(1-r2/R2),
where R is the radius of the pipe, r is the radial distance from the center of the pipe, and
umax is the maximum flow velocity, which occurs at the center.
Obtain (a) a relation for the drag force applied by the fluid on a section of the pipe of
length L and (b) the value of the drag force for water flow at 20 0C with R = 0.08 m, L =
15 m, umax = 3 m/s and μ = 0.0010 kg/m.s
Chapter 2: Properties of Fluids
Prob: A flat plate 0.1 m2 area is pulled at 30 cm/s relative to another plate
located at a distance of 0.01 cm from it, the fluid separating them being
water with dynamic viscosity of 0.001 Pa-s. Find the force and power
required to maintain the velocity.
Prob: Determine the torque and power required to turn a 10 cm long, 5 cm
diameter shaft at 500 rev/min in a 5.1 cm diameter concentric bearing
flooded with a lubricating oil of viscosity 100 centipoise.
Note that: I poise = 0.1 Pa-s.
Prob: A skater weighing 800 N skates at 54 km/hr on ice at 0 deg C. The
average skating area supporting him is 10 cm2 and the effective dynamic
coefficient of friction between the skates and the ice is 0.02. If there is
actually a thin film of water between the skates and the ice, determine the
average thickness.
Note that the friction force is given by friction coefficient*Weight of the
skater. Dynamic viscosity is taken to be 0.001 Pa-s.
Chapter 2: Properties of Fluids
Surface Tension
Liquid droplets behave like small
spherical balloons filled with
liquid, and the surface of the
liquid acts like a stretched elastic
membrane under tension.
The pulling force that causes this
is
due to the attractive forces
between molecules
called surface tension ss.
Attractive force on surface
molecule is not symmetric.
Repulsive forces from interior
molecules causes the liquid to
minimize its surface area and
attain a spherical shape.
Chapter 2: Properties of Fluids
Capillary Effect
Capillary effect is the rise
or fall of a liquid in a smalldiameter tube.
The curved free surface in
the tube is call the
meniscus.
Water meniscus curves up
because water is a wetting
fluid.
Mercury meniscus curves
down because mercury is a
nonwetting fluid.
Force balance can
describe magnitude of
capillary rise.
Chapter 2: Properties of Fluids