Transcript Chapter 1 FLUID PROPERTIES

Chapter 1
FLUID PROPERTIES
1.1 DEFINITION OF A FLUID
Fluid - substance that deforms continuously when
subjected to a shear stress, no matter how small
that shear stress may be.
Shear stress = Tangential (Shear )Force / Area
Normal Stress (pressure) = Normal Force /Area
 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 over the area.
 Shear stress at a point is the limiting value of shear force
to area as the area is reduced to the point.
Figure 1.1
Deformation
resulting from
application of
constant shear
force.
A substance is placed between two closely spaced parallel plates
so large that conditions at their edges may be neglected. The
lower plate is fixed, and a force F is applied to the upper plate,
which exerts a shear stress F/A on any substance between the
plates. A is the area of the upper plate. When the force F causes
the upper plate to move with a steady (nonzero) velocity, no
matter how small the magnitude of F, one may conclude that the
substance between the two plates is a fluid.
 The fluid in immediate contact with a solid boundary has the same
velocity as the boundary (no slip at the boundary)
 Fluid in the area abcd flows to the new position ab'c'd, each fluid
particle moving parallel to the plate and the velocity u varying
uniformly from zero at the stationary plate to U at the upper plate.
 Experiments:
other quantities being held constant, F is directly
proportional to A and to U and is inversely proportional to
thickness t.
In equation form
μ is the proportionality factor and includes the effect of the
particular fluid
 If τ = F/A for the shear stress,
 The ratio U/t: angular velocity of line ab, or it is the rate of
angular deformation of the fluid (rate of decrease of angle
bad)
 The angular velocity may also be written du/dy – more
general
 The velocity gradient du/dy may also be visualized as the
rate at which one layer moves relative to an adjacent layer,
in differential form,
- Newton's law of viscosity
- proportionality factor μ: viscosity of the fluid
 Materials other than fluids cannot satisfy the definition
of a fluid
– A plastic substance will deform a certain amount
proportional to the force, but not continuously when
the stress applied is below its yield shear stress.
– A complete vacuum between the plates would cause
deformation at an ever-increasing rate.
– If sand were placed between the two plates,
Coulomb friction would require a finite force to
cause a continuous motion.
–  plastics and solids are excluded from the
classification of fluids.
 Fluids:
– Newtonian
– non-Newtonian
 Newtonian fluid:
linear relation between the magnitude
of applied shear stress and the resulting
rate of deformation
[μ constant in
Eq. (1.1.1)]
Figure 1.2
Theological diagram.
 Non-Newtonian fluid: nonlinear relation between the
magnitude of applied shear stress and the rate of angular
deformation
- An ideal plastic has a definite yield stress
and a constant linear relation of τ to du/dy.
- A thixotropic substance, such as
printer's ink, has a viscosity that is
dependent upon the immediately prior
angular deformation of the substance and
has a tendency to take a set when at rest.
- Gases and thin liquids tend to be
Newtonian fluids, while thick, longchained hydrocarbons may be nonNewtonian.
Figure 1.2 Theological diagram.
 For purposes of analysis, the assumption is frequently made
that a fluid is nonviscous
 With zero viscosity the shear stress is always zero, regardless
of the motion of the fluid.
 If the fluid is also considered to be incompressible, it is then
called an ideal fluid and plots as the ordinate in Fig. 1.2.
1.2 FORCE, MASS, LENGTH, AND TIME UNITS
 Force, mass, length, and time: consistent units
– greatly simplify problem solutions in mechanics
– derivations may be carried out without reference to any
particular consistent system
 A system of mechanics units: consistent when unit force causes
unit mass to undergo unit acceleration
 The International System (SI)
– newton (N) as unit or force
– kilogram (kg) as unit of mass
– metre (m) as unit of length
– the second (s) as unit of time
 With the kilogram, metre, and second as defined units, the
newton is derived to exactly satisfy Newton's second law of
motion
 The force exerted on a body by gravitation is called the force
of gravity or the gravity force. The mass m of a body does not
change with location; the force of gravity of a body is
determined by the product of the mass and the local
acceleration of gravity g:
 For example, where g = 9.876 m/s2, a body with gravity force
of 10 N has a mass m = 10/9.806 kg. At the location where g =
9.7 m/s2, the force of gravity is
 Standard gravity is 9.806 m/s2. Fluid properties are often
quoted at standard conditions of 4oC and 760 mm Hg.
Table 1.1 Selected prefixes for powers of 10 in SI units
1.3 VISCOSITY
 Viscosity requires the greatest consideration in the study of
fluid flow.
 Viscosity is that property of a fluid by virtue of which it
offers resistance to shear
 Newton's law of viscosity [Eq. (1.1.1)] states that for a given
rate of angular deformation of fluid the shear stress is
directly proportional to the viscosity.
 Molasses and tar are examples of highly viscous liquids;
water and air have very small viscosities.
 The viscosity of a gas increases with temperature, but the
viscosity of a liquid decreases with temperature – it can
be explained by examining the causes of viscosity.
– The resistance of a fluid to shear depends upon its
cohesion and upon its rate of transfer of molecular
momentum.
– A liquid, with molecules much more closely spaced
than a gas, has cohesive forces much larger than a gas.
Cohesion - predominant cause of viscosity in a liquid;
and since cohesion decreases with temperature, the
viscosity does likewise.
– A gas, on the other hand, has very small cohesive
forces. Most of its resistance to shear stress is the
result of the transfer of molecular momentum.
 Fig.1.3: rough model of the way in which momentum
transfer gives rise to an apparent shear stress, considering
two idealized railroad cars loaded with sponges and on
parallel tracks
 Assume each car has a water tank and pump so arranged
that the water is directed by nozzles at right angles to the
track. First, consider A stationary and B moving to the right,
with the water from its nozzles striking A and being
absorbed by the sponges. Car A will be set in motion owing
to the component of the momentum of the jets which is
parallel to the tracks, giving rise to an apparent shear stress
between A and B. Now if A is pumping water back into B at
the same rate, its action tends to slow down B and equal and
opposite apparent shear forces result. When both A and B
are stationary or have the same velocity, the pumping does
not exert an apparent shear stress on either car.
Figure 1.3
Model illustrating
transfer of momentum.
 Within fluid there is always a transfer of molecules back and
forth across any fictitious surface drawn in it. When one layer
moves relative to an adjacent layer, the molecular transfer of
momentum brings momentum from one side to the other so
that an apparent shear stress is set up that resists the relative
motion and tends to equalize the velocities of adjacent layers
in a manner analogous to that of Fig. 1.3. The measure of the
motion of one layer relative to an adjacent layer is du/dy.
 Molecular activity gives rise to an apparent shear stress in
gases which is more important than the cohesive forces, and
since molecular activity increases with temperature, the
viscosity of a gas also increases with temperature.
 For ordinary pressures viscosity is independent of pressure
and depends upon temperature only. For very great pressures,
gases and most liquids have shown erratic variations of
viscosity with pressure.
 A fluid at rest or in motion so that no layer moves relative to
an adjacent layer will not have apparent shear forces set up,
regardless of the viscosity, because du/dy is zero throughout
the fluid
–  fluid statistics - no shear forces considered, and the
only stresses remaining are normal stresses, or pressures
 greatly simplifies the study of fluid statics, since any
free body of fluid can have only gravity forces and normal
surface forces acting on it
 Dimensions of viscosity: from Newton's law of viscosity –
solving for the viscosity μ
and inserting dimensions F, L, T for force, length, and time,
shows that μ has the dimensions FL-2T.
 With the force dimension expressed in terms of mass by
use of Newton's second law of motion, F = MLT-2,
the dimensions of viscosity may be expressed as ML-1T-1.
 The SI unit of viscosity which is the pascal second
(symbol Pa·s) has no name.
Kinematic Viscosity
 μ - absolute viscosity or the dynamic viscosity
 ν - kinematic viscosity (the ratio of viscosity to mass density):
- occurs in many applications (e.g., in the dimensionless
Reynolds number for motion of a body through a fluid, Vl/ν, in
which V is the body velocity and l is a representative linear
measure or the body size)
 The dimensions of ν are L2T-1.
 SI unit: 1 m2/s, has no name.
 Viscosity is practically independent of pressure and depends
upon temperature only.
– The kinematic viscosity of liquids, and of gases at a given
pressure, is substantially a function of temperature.
Example 1.1
A liquid has a viscosity or 0.005 Pa·s and a density or 850
kg/m3. Calculate the kinematic viscosity:
Example 1.2
In Fig. 1.4 the rod slides inside a concentric sleeve with a
reciprocating motion due to the uniform motion of the crank.
The clearance is δ and the viscosity μ. Write a program in
BASIC to determine the average energy loss per unit time in
the sleeve. D = 0.8 in, L = 8.0 in, δ = 0.001 in, R = 2 ft, r =
0.5 ft, μ = 0.0001 lb s/ft2, and the rotation speed is 1200 rpm.
The energy loss in the sleeve in one rotation is the product of
resisting viscous (shear) force times displacement integrated
over the period of the motion. The period T is 2π/w, where w =
dθ/dt. The sleeve force depends upon the velocity. The force Fi
and position xi are found for 2n equal increments of the period.
Then by the trapezoidal rule the work done over the half period
is found
Using the law of sines to eliminate φ, we get
Figure 1.5 lists the program, in which the variable RR
represents the crank radius r.
Figure 1.4 Notation for sleeve motion
Figure 1.5 BASIC program to determine loss in sleeve motion
1.4 CONTINUUM
 In dealing with fluid-flow relations on a mathematical or
analytical basis: consider that the actual molecular structure is
replaced by a hypothetical continuous medium - continuum.
 Example: velocity at a point in space is indefinite in a
molecular medium, as it would be zero at all times except
when a molecule occupied this exact point, and then it would
be the velocity of the molecule and not the mean mass
velocity of the particles in the neighborhood.
 This is avoided if consider velocity at a point to be the
average or mass velocity of all molecules surrounding the
point. With n molecules per cubic centimetre, the mean
distance between molecules is of the order n-1/3 cm.
 Molecular theory, however, must be used to calculate fluid
properties (e.g., viscosity) which are associated with
molecular motions, but continuum equations can be employed
with the results of molecular calculations.
 Rarefied gases (the atmosphere at 80 km above sea level): the
ratio of the mean free path [the mean free path is the average
distance a molecule travels between collisions] of the gas to a
characteristic length for a body or conduit is used to
distinguish the type of flow.
 The flow regime is called gas dynamics for very small values
of the ratio; the next regime is called slip flow; and for large
values of the ratio it is free molecular flow.
 In this text only the gas-dynamics regime is studied.
1.5 DENSITY, SPECIFIC VOLUME, UNIT
GRAVITY FORCE, RELATIVE DENSITY,
PRESSURE
 The density ρ: its mass per unit volume.
– Density at a point: the mass Δm of fluid in a small volume
ΔV surrounding the point
– For water at standard pressure (760 mm Hg) and 4oC, ρ =
1000 kg/m3.
 The specific volume vs: the volume occupied by unit mass of
fluid
 The unit gravity force, γ: the force of gravity per unit volume. It
changes with location depending upon gravity
– Water: γ = 9806 N/m3 at 5oC, at sea level.
 The relative density S of a substance: the ratio of its mass to the
mass of an equal volume of water at standard conditions. (may
also be expressed as a ratio or its density to that of water).
 The average pressure : the normal force pushing against a plane
area divided by the area.
– The pressure at a point is the ratio of normal force to area as
the area approaches a small value enclosing the point.
– If a fluid exerts a pressure against the walls or a container,
the container will exert a reaction on the fluid which will be
compressive.
– Units force per area, which is newtons per square metre,
called pascals (Pa).
– Absolute pressure : P, gage pressures : p.
1.6 PERFECT GAS
 This treatment: thermodynamic relations and compressiblefluid-flow cases are limited generally to perfect gases
(defined in this section)
 The perfect gas : substance that satisfies the perfect-gas-law
and that has constant specific heats. P is the absolute pressure;
vs is the specific volume; R is the gas constant; T is the
absolute temperature.
 The perfect gas must be carefully distinguished from the
ideal fluid. An ideal fluid frictionless and incompressible.
The perfect gas has viscosity and can therefore develop shear
stresses, and it is compressible according to Eq. (1.6.1).
 Eq.(1.6.1): the equation of state for a perfect gas; may be
written
 The units of R can be determined from the equation
 Real gases below critical pressure and above the critical
temperature tend to obey the perfect-gas law. As the pressure
increases, the discrepancy increases and becomes serious near
the critical point.
 The perfect-gas law encompasses both Charles' law and
Boyle's law.
– Charles' law: for constant pressure the volume of a given
mass of gas varies as its absolute temperature.
– Boyle's law (isothermal law): for constant temperature the
density varies directly as the absolute pressure.
 The volume v of m mass units of gas is mvs 
 With
being the volume per mole
 If n is the number of moles of the gas in volume ϑ
 The product MR, called the universal gas constant, has a
value depending only upon the units employed
 The gas constant R can then be determined from
 knowledge of relative molecular mass leads to the value
of R
 The specific heat cv of a gas : number of units of heat added
per unit mass to raise the temperature of the gas one degree
when the volume is held constant.
 The specific heat cp : the number of heat units added per unit
mass to raise the temperature one degree when the pressure is
held constant.
 The specific heat ratio k = cp/cv.
 The intrinsic energy u (dependent upon P, ρ and T) : the
energy per unit mass due to molecular spacing and forces.
 The enthalpy h : important property of a gas given by
h=u+P/ρ.
 cv and cp : units joule per kilogram per kelvin (J/kg·K)
– 4187 J of heat added raises the temperature of one
kilogram of water one degree Celsius at standard
conditions
 R is related to cv and cp by
Example 1.2
A gas with relative molecular mass of 44 is at a pressure or 0.9
MPa and a temperature of 20oC. Determine its density.
From Eq.(1.6.7),
Then, from Eq.(1.6.2)
1.7 BULK MODULUS OF ELASTICITY
 For most purposes a liquid may be considered as
incompressible, but for situations involving either sudden or
great changes in pressure, its compressibility becomes
important; also when temperature changes are involved, e.g.,
free convection.
 The compressibility of a liquid is expressed by its bulk
modulus of elasticity.
 If the pressure of a unit volume of liquid is increased by dp ,
it will cause a volume decrease -dV ; the ratio dp/dV is the
bulk modulus of elasticity K
 For any volume V
 Expressed in units of p. For water at 20oC K = 2.2. GPa.
Example 1.3
A liquid compressed in a cylinder has a volume of 1 L
(1000cm3) at 1 MN/m2 and volume of 995 cm3 at 2 MN/m2.
What is its bulk modulus of elasticity?
1.8 VAPOR PRESSURE
 Liquids evaporate because or molecules escaping from the
liquid surface : vapor molecules exert partial pressure in the
space - vapor pressure.
 If the space above the liquid is confined, after a sufficient
time the number of vapor molecules striking the liquid
surface and condensing is just equal to the number escaping
in any interval of time, and equilibrium exists.
 Depends upon temperature and increases with it. Boiling:
when the pressure above a liquid equals the vapor pressure of
the liquid
 20oC water: 2.447 kPa, mercury: 0.173 Pa
 When very low pressures are produced at certain locations in
the system, pressures may be equal to or less than the vapor
pressure  the liquid flashes into vapor - cavitation.
1.9 SURFACE TENSION
Capillarity
 At the interface between a liquid and a gas, or two
immiscible liquids, a film or special layer seems to form
on the liquid, apparently owing to attraction of liquid
molecules below the surface
 The formation or this film may be visualized on the basis
of surface energy or work per unit area required to bring
the molecules to the surface. The surface tension is then
the stretching force required to form the film, obtained by
dividing the surface-energy term by unit length of the film
in equilibrium.
 The surface tension of water varies from about 0.074 N/m
at 20oC to 0.059 N/m at 100oC (Table 1.2)
Table 1.2 Approximate properties of common liquids at 20oC and
standard atmospheric pressure
 The action of surface tension is to increase the pressure
within a droplet of liquid or within a small liquid jet.
 For a small spherical droplet of radius r the internal pressure
p necessary to balance the tensile force due to the surface
tension σ calculated in terms of the forces which act on a
hemispherical free body (see Sec. 2.6),
 For the cylindrical liquid jet of radius r, the pipe-tension
equation applies:
 Both equations: the pressure becomes large for a very small
radius of droplet or cylinder
 Capillary attraction is caused by surface tension and by the
relative value of adhesion between liquld and solid to
cohesion of the liquid.
 A liquid that wets the solid has a greater adhesion than
cohesion. Surface tension in this case: causes the liquid to
rise within a small vertical tube that is partially immersed
in it
 For liquids that do not wet the solid, surface tension tends
to depress the meniscus in a small vertical tube. When the
contact angle between liquid and solid is known, the
capillary rise can be computed for an assumed shape of the
meniscus.
 Figure 1.4: the capillary rise for water and mercury in
circular glass tubes in air
Figure 1.5 Capillarity in circular glass tubes