Fluid properties 24.pdf of engineering and technology

Fluid Properties
Density and Specific Weight
Fluid density is defined as mass per unit volume. The units of density are Kg/m3
or slug/ft3.
r = m/V
A fluid property directly related to density is the specific weight. Specific weight
is defined as the weight per unit volume.
g = W/V = mg/V = (m/V)g = rg
Where g is the local gravitational acceleration. The units of specific weight are
N/m3 or lb/ft3.
M. M. Razzaque, Department of ME, BUET 1

Specific gravity
The specific gravity is used to determine the specific weight or density of a fluid
(usually a liquid). It is defined as the ratio of the density of a substance to that of
water at a reference temperature of 4oC.
For example, the specific gravity of mercury is 13.6, a dimensionless number;
means the mass of mercury is 13.6 times that of water for the same volume.
The density, specific weight and specific gravity of air and water at standard
conditions are given in the following Table.

Stress Field
In fluid mechanics, we need to understand what kinds of forces act on fluid
particles. Each fluid particle can experience: surface forces (pressure, friction)
that are generated by contact with other particles or a solid surface; and body
forces (such as gravity and electromagnetic) that are experienced throughout the
particle.
The gravitational body force acting on an element of volume, is given by
where ρ is the density (mass per unit volume) and is the local
gravitational acceleration. Thus the gravitational body force per unit volume is
ρg and the gravitational body force per unit mass is g .
Surface forces on a fluid particle lead to stresses. The concept of stress is useful
for describing how forces acting on the boundaries of a medium (fluid or solid)
are transmitted throughout the medium.

When one stands on a diving board, stresses are generated within the board. On
the other hand, when a body moves through a fluid, stresses are developed within
the fluid. The difference between a fluid and a solid is that stresses in a fluid are
mostly generated by motion rather than by deflection.
Imagine the surface of a fluid particle in contact with other fluid particles, and
consider the contact force being generated between the particles as shown in Fig.
2.6. The force may be resolved into two components, one normal to and the other
tangent to the area. A normal stress and a shear stress are then defined as

Consider the stress on the element δAx, whose outwardly drawn normal is in the x
direction. The force, δF has been resolved into components along each of the
coordinate directions. Dividing the magnitude of each force component by the
area, δAx, and taking the limit as δAx approaches zero, we define the three stress
components shown in Fig. 2.7b:
The first subscript indicates the plane on which the stress acts (in this case, a
surface perpendicular to the x axis). The second subscript indicates the direction
in which the stress acts.

Viscosity
For a solid, stresses develop when the material is elastically deformed or
strained; for a fluid, shear stresses arise due to viscous flow. Hence we say solids
are elastic, and fluids are viscous. Many biological tissues are viscoelastic,
meaning they combine features of a solid and a fluid.
Each fluid can be categorized by examining the relation between the applied
shear stresses and the flow (i.e. the rate of deformation) of the fluid. Consider the
behavior of a fluid element between the two infinite plates shown in Fig. 2.9a.

The rectangular fluid element is initially at rest at time t. Let us now suppose a
constant rightward force δFx is applied to the upper plate so that it is dragged
across the fluid at constant velocity δu. The relative shearing action of the
infinite plates produces a shear stress, τyx, which acts on the fluid element and is
given by
where δAy is the area of contact of the fluid element with the plate and δFx is the
force exerted by the plate on that element.
Snapshots of the fluid element, shown in Figs. 2.9a–c, illustrate the deformation
of the fluid element from position MNOP at time t, to M'NOP' at time t +δt, to
M"NOP" at time t +2δt, due to the imposed shear stress. Focusing on the time
interval δt (Fig. 2.9b), the deformation of the fluid is given by

The distance, δl, between the points M and M' is given by δl = δuδt.
Alternatively, for small angles, δl = δyδα. Equating these two expressions for δl
gives
Taking the limits of both sides of the equality, we obtain
Thus, the fluid element of Fig. 2.9, when subjected to shear stress τyx,
experiences a rate of deformation (shear rate) given by du/dy.
Any fluid that experiences a shear stress will flow (it will have a shear rate).
Fluids in which shear stress is directly proportional to rate of deformation are
Newtonian fluids. The term non-Newtonian is used to classify all fluids in
which shear stress is not directly proportional to shear rate.

Newtonian Fluid
Most common fluids such as water, air, and gasoline are Newtonian under
normal conditions. If the fluid of Fig. 2.9 is Newtonian, then
We are familiar with the fact that some fluids resist motion more than others. For
example, a container of lube oil is much harder to stir than one of water. Hence
lub oil is much more viscous—it has a higher viscosity. (Note that a container of
mercury is also harder to stir, but for a different reason!)
The constant of proportionality in Eq. 2.14 is the absolute (or dynamic) viscosity,
μ. Thus in terms of the coordinates of Fig. 2.9, Newton’s law of viscosity is
given for one-dimensional flow by

Viscosity is the most important fluid property in the study of fluid flows.
-It can be thought of as the internal stickiness of a fluid.
-It is one of the properties that controls the fluid flow rate in a pipeline.
-It accounts for the energy losses associated with the transport of fluids in ducts,
channels, and pipes.
-It plays a primary role in the generation of turbulence.

The rate of deformation of a fluid is directly linked to the viscosity of the fluid.
For a given stress, a highly viscous fluid deforms at a slower rate than the fluid
with a low viscosity.
Consider a flow in which the fluid particles move in the x-direction at different
speeds, so that particle velocities u vary with the y-coordinate. The Figure shows
two particle positions at different times.
For such a simple flow field, in which u = u(y), we can define the viscosity m of
the fluid by the relationship, t = m du/dy, where t is the shear stress and u is the
velocity in the x-direction. The units of t are N/m2 or Pa, and of m are N.s/m2. The
quantity du/dy is a velocity gradient and can be interpreted as a strain rate.
Newton’s
Law of Viscosity

Since the dimensions of τ are [F/L2] and the dimensions of du/dy are [1/t], μ has
dimensions [Ft/L2].
The units of absolute (or dynamic) viscosity, m are thus N.s/m2 or Pa-s. The CGS
unit for m is the poise (P), named after Jean Leonard Marie Poiseuille. It is more
commonly expressed, as centipoise (cP).
Water at 20 °C has a viscosity of 1.0020 cP.
In fluid mechanics the ratio of absolute viscosity, μ, to density, ρ, often arises.
This ratio is given the name kinematic viscosity and is represented by the symbol
ν. Since density has dimensions [M/L3], the dimensions of ν are [L2/t].
The SI unit of kinematic viscosity is m2/s. The CGS physical unit for kinematic
viscosity is the stokes (St), named after George Gabriel Stokes.
M. M. Razzaque, Department of ME, BUET
13

It is sometimes expressed in terms of centistokes (cSt). 1 stoke ≡ 1 cm2/s. Water
at 20 °C has a kinematic viscosity of about 1 cSt.
1 P = 0.1 Pa·s, 1 cP = 1 mPa·s = 0.001 Pa·s = 0.001 N·s/m2.
1 St = 1 cm2·s−1 = 10−4 m2·s−1. 1 cSt = 1 mm2·s−1 = 10−6 m2·s−1.
Note that for gases, viscosity increases with temperature, whereas for liquids,
viscosity decreases with increasing temperature.
Viscosity can be thought of as the internal stickiness of a fluid and is the most
important fluid property in the study of fluid flows.
It controls the fluid flow rate in a pipeline and accounts for the energy losses
associated with the transport of fluids in ducts, channels, and pipes.
An important effect of viscosity is to cause the fluid to adhere to the surface; this
is known as the no-slip condition.

Rheology
Fluids which follow the linear pattern of the Newton‘s law of viscosity are called
newtonian fluids. There are many non-newtonian fluids and they are treated in
rheology.
An ex ample of a yielding fluid is toothpaste, which will not flow out of the tube
until a finite stress is applied by squeezing.
Figure: Rheological behavior
of various materials
A dilatant (or shear-thickening) fluid increases
resistance with increasing applied stress.
A pseudoplastic (or shear-thinning) fluid
decreases resistance with increasing stress.
If the thinning effect is very strong the fluid is
termed plastic. Bingham plastic is the limiting
case of a plastic which requires a finite yield
stress before it begins to flow. The flow behavior
after yield may be linera or nonlinear.

A further complication of non-newtonian behavior is the transient effect shown
in the following figure.
Some fluids require a gradually increasing shear stress to maintain a constant
strain rate and are called rheopectic.
The opposite case of a fluid which thins out with time and requires decreasing
stress is termed thixotropic.
Figure: Rheological behavior of
various materials
Effect of time on applied stress

EXAMPLE:
A 60-cm-wide belt moves as shown. Calculate the horsepower requirement
assuming a linear velocity profile in the 10oC water.
Solution: Newton‘s Law of Viscosity
du/dy = 10*1000/2 = 5000 s-1
m for 10oC water = 1.308 x 10-3 N.s/m2
t = m. du/dy = 5000*1.308 x 10-3 N/m2
F = t.A = 5000*1.308 x 10-3 *4*0.6 N
Power = F.U = 5000*1.308 x 10-3 *4*0.6*10 Nm/s
= 0.21 Hp

EXAMPLE:
A block of weight W slides down An inclined plane while lubricated by a thin film of oil as
shown. The film contact area is A and the thickness is h. Assuming a linear velocity
distribution in the film, derive an expression for the ―terminal‖ (zero acceleration) velocity
V of the block. Find the terminal velocity of the block if the block mass is 6 kg, A = 3 cm2,
q = 15o, and the film is 1 mm thick SAE 30 oil at 20oC.
W
F
Solution: Using the Newton‘s Law of Viscosity
Viscous shear stress, t = m.V/h
Viscous force along the sliding surface, F = t.A = m.VA/h
At the terminal velocity, the viscous force will be balanced by the component of
weight along the sliding surface.
W sin q = F = m.VA/h
Solving we get, V = Wh sin q/m.A
m for 20oC SAE 30 oil = 0.29 N.s/m2 , A = 0.0003 m2, W = 6 kg = 58.86 N, h =
0.001 m. Putting these values in the above expression, V = ??

EXAMPLE:
Consider a fluid within the small gap between two concentric cylinders. What
will be the torque to rotate the inner cylinder at constant speed while the outer
cylinder remains stationary?
Solution: This resistance to the rotation of the cylinder is due to viscosity. The
shear tress that resists the applied torque for this simple flow depends directly on
the velocity gradient in the fluid film in the gap between the cylinders, i.e.
For a small gap h<<R, this gradient can be approximated by assuming a linear
velocity distribution in the gap.

Thus using the Newton‘s Law of viscosity, the shear stress on the surface of the
inner cylinder may be written as
We can then relate the applied torque T to the viscosity and other parameters by
the equation, T = stress x area x moment arm = t x 2pRL x R
T
Here the shearing stress acting on the ends of the cylinder is neglected; L
represents the length of the rotating cylinder. Note that the torque depends
directly on the viscosity, thus the cylinders could be used as a viscometer, a
device that measures the viscosity of a fluid.
Example:
A 1.2 m long, 2 cm diameter shaft rotates inside an equally long cylinder that is
2.06 cm in diameter. Calculate the torque required to rotate the inner shaft at 2000
rpm if SAE-30 oil at 20oC fills the gap. Also, calculate the horsepower required.
Assume symmetric motion.

Solution:
N = 2000 rpm
w = 2*3.142*N/60 = 209.5 rad/s
du = wr - 0 = 209.5 * 0.01 = 2.095 m/s
dr = 0.06/2 = 0.03 cm = 0.03 x 10-2 m
du/dr = 6982.22 s-1
m for SAE-30 oil at 20oC = 0.4 N.s/m2
t = m. du/dr = 0.4* 6982.22 = 2792.89 N/m2
F = t.A = 2792.89 *3.142*2 x 10-2 *1.2 = 210.61 N
T = r . F = 0.01* 210.61 = 2.1061 Nm
Power = T. w = 2.1061 * 209.5 Nm/s = 441.15 watt = 0.6 Hp

EXAMPLE
A 25-cm-diameter horizontal disk rotates a distance of 2 mm above a solid
surface. Water at 10oC fills the gap. Estimate the torque required to rotate the
disk at 400 rpm.
Solution
N = 400 rpm, h = 0.002 m
w = 2*3.142*N/60 = 41.9 rad/s
m for 10oC water = 1.308 x 10-3 N.s/m2
du = wr - 0 = wr; du/dz = wr /h
t = m. du/dz = mwr /h
dF = t.dA = (mwr /h)*2prdr = 2pmwr2dr/h
dT = r*dF = 2pmwr3dr/h
= 3.142*1.308 x 10-3 *41.9/(2*2 x 10-3)* 0.1254 = 0.0105 Nm
Power = T. w = 0.44 watt
  4
R
0
4 R
0
3
R
h
2
4
r
h
2
dr
r
h
2
T
pm

pm

pm

dA
w
r

PROBLEM:
2.54 Calculate the approximate viscosity of the oil used to lubricate the sliding
surface shown below.
2.55 Calculate the approximate power lost in friction in this ship propeller shaft
bearing shown below.

Problem: Consider a fluid flow between two parallel fixed
plates 5 cm apart, as shown below. The velocity distribution
for the flow is given by u(y) = 120(0.05y - y2) m/s where y is
in meters. The fluid is water at 10°C. Calculate the magnitude
of the shear stress acting on each of the plates.
Problem: A solid cone of angle 2q, base ro, and density rc is
rotating with initial angular velocity wo inside a conical seat,
as shown in the following figure. The clearance h is filled with
oil of viscosity m. Neglecting air drag, derive an analytical
expression for the cone’s angular velocity w(t) if there is no
applied torque.
Problem: Calculate the torque needed to rotate the cone
shown in the Fig. at 2000 rpm if SAE-30 oil of 0.1 N.s/m2
viscosity fills the gap. Assume a linear velocity profile
between the cone and the fixed wall.
Problem: For the fluid in the annular gap between two 0.2-m-long rotating
concentric cylinders, the velocity distribution is given by u(r) = 0.4/r - 1000r m/s.
The diameters of the cylinders are 2 cm and 3 cm, respectively. Calculate the fluid
viscosity if the torque on the inner cylinder is measured to be 0.0026 N.m.

Compressibility
In the preceding section we discussed the deformation of fluids that results from
shear stresses. In this section, we discuss the deformation that results from
pressure changes.
All fluids compress if the pressure increases, resulting in an increase in density. A
common way to describe the compressibility of a fluid is by the following
definition of the bulk modulus of elasticity B:
In words, the bulk modulus is defined as the ratio of the change in pressure (Dp)
to relative change in density (Dr/r) while the temperature remains constant. The
bulk modulus obviously has the same units as pressure. The bulk modulus for
water at standard conditions is approximately 2100 MPa (310,000 psi), or 21 000
times the atmospheric pressure. For air at standard conditions, B is equal to 1
atm. In general, B for a gas is equal to the pressure of the gas.

26
To cause a 1% change in the density of water a pressure of 21 MPa (210 atm) is
required. This is an extremely large pressure needed to cause such a small
change; thus liquids are often assumed to be incompressible. For gases, if
significant changes in density occur, say 4%, they should be considered as
compressible; for small density changes they may also be treated as
incompressible.
Small density changes in liquids can be very significant when large pressure
changes are present. For example, they account for "water hammer," which can
be heard shortly after the sudden closing of a valve in a pipeline. When the valve
is closed an internal pressure wave propagates down the pipe, producing a
hammering sound due to pipe motion when the wave reflects from the closed
valve.
The bulk modulus can also be used to calculate the speed of sound in a liquid; it
is given by
This yields approximately 1450 m/s (4800 ft/sec) for the speed of sound in water
at standard conditions.

Vapor Pressure
When a small quantity of liquid is placed in a closed container, vaporization
continues until equilibrium is reached between the liquid and gaseous states of
the substance in the container - in other words, when the number of molecules
escaping from the water surface is equal to the number of incoming molecules.
The pressure resulting from molecules in the gaseous state is the vapor pressure.
The vapor pressure of water at standard conditions (15oC, 101.3 kPa) is 1.70 kPa
absolute and for ammonia it is 33.8 kPa absolute. The vapor pressure is highly
dependent on pressure and temperature; it increases significantly when the
temperature increases.
In general, a transition from the liquid state to the gaseous state occurs if the
local absolute pressure is less than the vapor pressure of the liquid.
In liquid flows, conditions can be created that lead to a pressure below the vapor
pressure of the liquid. When this happens, bubbles are formed locally. This
phenomenon is called cavitation.
Cavitation in a flow can be very damaging when bubbles are transported by the
flow to high pressure regions and collapse. It has the potential of damaging a
pipe wall or a ship‗s propeller.

Surface Tension
Suface tension is a property that results from the attractive forces between
molecules. As such, it manifests itself only in liquids. The forces between
molecules in the bulk of a liquid are equal in all directions, and as a result, no net
force is exerted on the molecules.
However, at the surface, the molecules exert a force that has a resultant in the
surface layer. This force holds a drop of water suspended on a rod and limits the
size of the drop that may be held. It also causes the small drops from a sprayer or
atomizer to assume spherical shapes.
Surface tension has units of force per unit length, N/m (lb/ft). The force due to
surface tension results from a length multiplied by the surface tension; the length
to use is the length of fluid in contact with a solid, or the circumference in the
case of a bubble.
If the interface is curved, a mechanical balance shows that there is
a pressure difference across the interface, the pressure being higher
on the concave side, as illustrated in the Fig. The pressure increase
in the interior of a liquid cylinder is balanced by two surface-
tension forces: 2RLDp = 2sL and thus, Dp = s/R.
sL
sL

29
Again consider the free body diagrams of half a droplet and half a bubble as
shown in Fig. 1.11. The droplet has one surface and the bubble is composed of a
thin film of liquid with an inside surface and an outside surface. The pressure
inside the droplet and bubble can now be calculated as follows.
The pressure force ppR2 in the droplet balances the surface tension force around
the circumference. Hence
Similarly, the pressure force in the bubble is balanced by the surface tension
forces on the two circumferences. Therefore,
So, we can conclude that the internal pressure in a bubble is twice as large as
that in a droplet of the same size.

Figure 1.12 shows the rise of a liquid in a clean glass capillary tube due to
surface tension.
The liquid makes a contact angle b with the
glass tube. Experiments have shown that this
angle for water and most liquids is zero.
There are also cases for which this angle is
greater than 90o (e.g. mercury); such liquids
have a capillary drop instead of capillary rise.
If h is the capillary rise, D the diameter, and r
the density, s can be determined from equating
the surface tension force to the weight of the
liquid column.
EXAMPLE 5
A 2-mm-diameter clean glass tube is inserted, as shown, in water at l5oC.
Determine the height that the water will climb up the tube. The water makes a
contact angle of 0o with the clean glass.

Solution
A free-body diagram of the water shows that the upward surface-tension force is
equal and opposite to the weight. Writing the surface-tension force as surface
tension times distance, we have
Solving for h, we get,
The numerical values for s and r were obtained from Table of water properties.
Surface tension of water at l5oC is 0.0741 N/m. It decreases with temperature.
Note that the nominal value used for the density of water is 1000 kg/m3.
The capillary rise h decreases as pipe diameter D increases. So, we see
significant amount of capillary rise in capillary tubes not in large diameter pipes.

Contact Angle
Another important surface effect is the contact angle b which appears when a
liquid interface intersects with a solid surface, as in the Fig. The force balance
would then involve both s and b. If the contact angle is less than 90o, the liquid
is said to wet the solid; if b > 90o, the liquid is termed nonwetting.
For example, water wets soap but does not wet wax. Water is extremely wetting
to a clean glass surface, with b = 0o. Like s, the contact angle b is sensitive to the
actual physico-chemical conditions of the solid-liquid interface. For a clean
mercury-air-glass interface, b = 130o.

33
P1 Determine the maximum diameter, in milliliters, of a solid aluminum ball,
density, 2700 kg/m3, which will float on a clean water-air surface at 20oC.
Ans: 4.1 mm
P2 A solid cylindrical needle of diameter d, length L, and density r may float in
liquid of surface tension s. Neglect buoyancy and assume a contact angle of 0o.
Derive a formula for the maximum diameter dmax able to float in the liquid.
Calculate dmax for a steel needle (SG 7.84) in water at 20oC.
P3 Derive an expression for the capillary height change h for a fluid of surface
tension s and contact angle q between two vertical parallel plates a distance W
apart, as in the Fig. What will h be for water at 20oC if W = 0.5 mm?
Ans: h = 2s.cosq/rgW

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