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Fluid, Fluid vs Solid mechanics, Important terms, Surface tension, Metric to US system conversions
1.
Fluid Mechanics (CE-201)
Course Instructor
Prof. Dr. A R Ghumman
[email protected] 051-9047638
03005223338
Associate
Dr Usman Ali Naeem-Ghufran Ahmed Pasha
(Lecturer, CED)
[email protected]
Ph : 051-9047658
2.
Recommended Books
Text Book:
Fluid Mechanics With Engineering Applications (10th
Edition)
by E. John Finnemore & Joseph B. Franzini
Reference Books:
A textbook of Hydraulics, Fluid Mechanics and Hydraulic
Machines (19th Edition) by R.S. Khurmi
Applied Fluid Mechanics (6th Edition) by Robert L. Mott
Fluid Mechanics by A.K Jain
3.
Marks Distribution
Sessionals - 40%
Attendance – 2%
Assignments – 8%
Practicals - 15%
Quizes – 10 %
Class Project/ Presentation – 5%
Mid Term - 20%
Final Exam - 40%
4.
Properties of Fluids
Lecture - 1
5.
Fluid
A fluid is defined as:
“A substance that continually deforms (flows) under
an applied shear stress regardless of the magnitude
of the applied stress”.
It is a subset of the phases of matter and includes
liquids, gases, plasmas and, to some extent, plastic
solids.
6.
Fluid Vs Solid Mechanics
Fluid mechanics:
“The study of the physics of materials which take the shape of
their container.” Or
“Branch of Engg. science that studies fluids and forces on them.”
Solid Mechanics:
“The study of the physics of materials with a defined rest shape.”
Fluid Mechanics can be further subdivided into fluid statics, the
study of fluids at rest, and kinematics, the study of fluids in motion
and fluid dynamics, the study of effect of forces on fluid motion.
In the modern discipline called Computational Fluid Dynamics
(CFD), computational approach is used to develop solutions to fluid
mechanics problems.
7.
Distinction between a Solid and a Fluid
Solid Fluid
Definite Shape and definite Indefinite Shape and Indefinite
volume. volume & it assumes the shape
Does not flow easily. of the container which it
occupies.
Molecules are closer. Flow Easily.
Attractive forces between the
molecules are large enough to
Molecules are far apart.
retain its shape. Attractive forces between the
An ideal Elastic Solid deform molecules are smaller.
under load and comes back to Intermolecular cohesive forces
original position upon removal of in a fluid are not great enough to
load. hold the various elements of
Plastic Solid does not comes back fluid together. Hence Fluid will
to original position upon removal flow under the action of applied
of load, means permanent stress. The flow will be
deformation takes place. continuous as long as stress is
applied.
8.
Distinction between a Gas and Liquid
The molecules of a gas are A liquid is relatively
much farther apart than incompressible.
those of a liquid. If all pressure, except that
Hence a gas is very of its own vapor pressure,
compressible, and when is removed, the cohesion
all external pressure is between molecules holds
removed, it tends to expand them together, so that the
indefinitely. liquid does not expand
A gas is therefore in indefinitely.
equilibrium only when it is Therefore a liquid may
completely enclosed. have a free surface.
11.
Important Terms
Density ():
Mass per unit volume of a substance.
kg/m3 in SI units
m
Slug/ft3 in FPS system of units
V
Specific weight ():
Weight per unit volume of substance.
N/m3 in SI units w
lbs/ft3 in FPS units
V
Density and Specific Weight of a fluid are related as:
g
Where g is the gravitational constant having value 9.8m/s2 or
32.2 ft/s2.
12.
Important Terms
Specific Volume (v):
Volume occupied by unit mass of fluid.
It is commonly applied to gases, and is usually expressed in
cubic feet per slug (m3/kg in SI units).
Specific volume is the reciprocal of density.
SpecificVo lume v 1 /
13.
Important Terms
Specific gravity:
It can be defined in either of two ways:
a. Specific gravity is the ratio of the density of a substance
to the density of water at 4°C.
b. Specific gravity is the ratio of the specific weight of a
substance to the specific weight of water at 4°C.
l l
s liquid
w w
14.
Example
The specific wt. of water at ordinary temperature and
pressure is 62.4lb/ft3. The specific gravity of mercury is
13.56. Compute density of water, Specific wt. of mercury,
and density of mercury.
1. water water / g 62.4/32.2 1.938 slugs/ft3
2. mercury s mercury water 13.56 x62.4 846lb / ft 3
3. mercury s mercury water 13.56 x1.938 26.3slugs / ft 3
(Where Slug = lb.sec2/ ft)
15.
Example
A certain gas weighs 16.0 N/m3 at a certain temperature and
pressure. What are the values of its density, specific volume,
and specific gravity relative to air weighing 12.0 N/m3
1. Density ρ γ /g
ρ 16/9.81 16.631 kg/m 3
2. Specific volume υ 1/ρ
u 1/1.631 0.613 m 3 /kg
3. Specific gravity s γ f /γ air
s 16/12 1.333
16.
Example
The specific weight of glycerin is 78.6 lb/ft3. compute its density
and specific gravity. What is its specific weight in kN/m3
1. Density / g
78.6/32.2 2.44 slugs/ft 3
2. Specific gravity s l / w
s 78.6/62.4 1.260
so 1.260x1000kg/m 3
1260 Kg/m 3
3. Specific weight in kN/m 3
x g
9.81x1260 12.36 kN/m 3
17.
Example
Calculate the specific weight, density, specific volume and
specific gravity of 1litre of petrol weights 7 N.
Given Volume = 1 litre = 10-3 m3
Weight = 7 N
1. Specific weight,
w = Weight of Liquid/volume of Liquid
w = 7/ 10-3 = 7000 N/m3
2. Density, = /g
= 7000/9.81 = 713.56 kg/m3
18.
Solution (Cont.):
3. Specific Volume = 1/
1/713.56
=1.4x10-3 m3/kg
4. Specific Gravity = s =
Specific Weight of Liquid/Specific Weight of Water
= Density of Liquid/Density of Water
s = 713.56/1000 = 0.7136
19.
Example
If the specific gravity of petrol is 0.70.Calculate its Density,
Specific Volume and Specific Weight.
Solution:
Given
Specific gravity = s = 0.70
1. Density of Liquid, s x density of water
= 0.70x1000
= 700 kg/m3
2. Specific Volume = 1/
x-3
3. Specific Weight, = 700x9.81 = 6867 N/m3
20.
Compressibility
It is defined as:
“Change in Volume due to change in Pressure.”
The compressibility of a liquid is inversely proportional to Bulk
Modulus (volume modulus of elasticity).
Bulk modulus of a substance measures resistance of a substance to
uniform compression. dp
Ev
(dv / v)
v
Ev dp
Where; v is the specific volume and p is the pressure. dv
Units: Psi, MPa , As v/dv is a dimensionless ratio, the units of E
and p are identical.
21.
Example
At a depth of 8km in the ocean the pressure is 81.8Mpa. Assume
that the specific weight of sea water at the surface is 10.05 kN/m 3
and that the average volume modulus is 2.34 x 10 3 N/m3 for that
pressure range.
(a) What will be the change in specific volume between that at the
surface and at that depth?
(b) What will be the specific volume at that depth?
(c) What will be the specific weight at that depth?
22.
(a) v1 1 / p1 g / 1
Using Equation :
9.81 / 10050 0.000976m 3 / kg
p
6 9
v 0.000976(81.8 x10 0) /( 2.34 x10 ) Ev
(v / v)
-34.1x10 -6 m 3 / kg dv p
v Ev
(b) v 2 v1 v 0.000942 m 3 / kg v2 v1 p 2 p1
v1 Ev
(c) 2 g / v2 9.81 / 0.000942 10410 N / m 3
23.
Viscosity
Viscosity is a measure of the resistance of a fluid to deform
under shear stress.
It is commonly perceived as thickness, or resistance to flow.
Viscosity describes a fluid's internal resistance to flow and may be
thought of as a measure of fluid friction. Thus, water is "thin",
having a lower viscosity, while vegetable oil is "thick" having a
higher viscosity.
The friction forces in flowing fluid result from the cohesion and
momentum interchange between molecules.
All real fluids (except super-fluids) have some resistance to shear
stress, but a fluid which has no resistance to shear stress is known
as an ideal fluid.
It is also known as Absolute Viscosity or Dynamic Viscosity.
25.
Dynamic Viscosity
As a fluid moves, a shear stress is developed in
it, the magnitude of which depends on the
viscosity of the fluid.
Shear stress, denoted by the Greek letter (tau),
τ, can be defined as the force required to slide
one unit area layer of a substance over another.
Thus, τ is a force divided by an area and can be
measured in the units of N/m2 (Pa) or lb/ft2.
26.
Dynamic Viscosity
Figure shows the velocity gradient in a moving fluid.
U
F, U
Y
Experiments have shown that: AU
F
Y
27.
Dynamic Viscosity
The fact that the shear stress in the fluid is directly
proportional to the velocity gradient can be stated
mathematically as F U du
A Y dy
where the constant of proportionality (the Greek letter miu)
is called the dynamic viscosity of the fluid. The term absolute
viscosity is sometimes used.
28.
Kinematic Viscosity
The kinematic viscosity ν is defined as:
“Ratio of absolute viscosity to density.”
29.
Newtonian Fluid
A Newtonian fluid; where stress is directly
proportional to rate of strain, and (named for Isaac
Newton) is a fluid that flows like water, its stress versus
rate of strain curve is linear and passes through the origin.
The constant of proportionality is known as the viscosity.
A simple equation to describe Newtonian fluid behavior is
du
Where =
absolute viscosity/Dynamic viscosity or
dy
simply viscosity
= shear stress
31.
Example
Find the kinematic viscosity of liquid in stokes whose specific
gravity is 0.85 and dynamic viscosity is 0.015 poise.
Solution:
Given S = 0.85
= 0.015 poise
= 0.015 x 0.1 Ns/m2 = x-3 Ns/m2
We know that S = density of liquid/density of water
density of liquid = S x density of water
0.85 x 1000kg/m3
Kinematic Viscosity ,
x-3
x -6m2/s = x 10-6 x4cm2/s
= x 10-2 stokes.
32.
Example
A 1 in wide space between two horizontal plane surface is
filled with SAE 30 Western lubricating oil at 80 F. What
force is required to drag a very thin plate of 4 sq.ft area
through the oil at a velocity of 20 ft/mm if the plate is 0.33
in from one surface.
33.
0.0063 lb.sec/ft 2 ( From A.1)
F U du
A Y dy
1 0.0063 * (20 / 60) /(0.33 / 12) 0.0764lb / ft 2
2 0.0063 * (20 / 60) /(0.67 / 12) 0.0394lb / ft 2
F1 1 A 0.0764 * 4 0.0305lb
F2 2 A 0.0394 * 4 0.158lb
Force F1 F2 0.463lb
34.
Example
Assuming a velocity distribution as shown in fig., which is a
parabola having its vertex 12 in from the boundary,
calculate the shear stress at y= 0, 3, 6, 9 and 12 inches.
Fluid’s absolute viscosity is 600 P.
35.
Solution
600 P= 600 x 0.1=0.6 N-s/m2 =0.6 x (1x2.204/9.81 x 3.282)
=0.6 x 0.020885=0.01253 lb-sec/ft2
Parabola Equation Y=aX2
120-u= a(12-y) 2
u=0 at y=0 so a= 120/122=5/6
u=120-5/6(12-y) 2 du/dy=5/3(12-y)
= du/dy
y (in) 0 3 6 9 12
du/dy 20 15 10 5 0
0.251 0.1880 0.1253 0.0627 0
36.
Ideal Fluid
An ideal fluid may be defined as:
“A fluid in which there is no friction i.e Zero viscosity.”
Although such a fluid does not exist in reality, many fluids
approximate frictionless flow at sufficient distances, and so
their behaviors can often be conveniently analyzed by
assuming an ideal fluid.
37.
Real Fluid
In a real fluid, either liquid or gas, tangential or
shearing forces always come into being whenever
motion relative to a body takes place, thus giving
rise to fluid friction, because these forces oppose
the motion of one particle past another.
These friction forces give rise to a fluid property
called viscosity.
38.
Surface Tension
Cohesion: “Attraction between molecules of same surface”
It enables a liquid to resist tensile stresses.
Adhesion: “Attraction between molecules of different
surface” It enables to adhere to another body.
“Surface Tension is the property of a liquid, which enables it
to resist tensile stress”.
At the interface between liquid and a gas i.e at the liquid
surface, and at the interface between two immiscible (not
mixable) liquids, the attraction force between molecules form
an imaginary surface film which exerts a tension force in the
surface. This liquid property is known as Surface Tension.
39.
Surface Tension
As a result of surface tension, the liquid surface has a
tendency to reduce its surface as small as possible. That is
why the water droplets assume a nearly spherical shape.
This property of surface tension is utilized in manufacturing
of lead shots.
Capillary Rise: The phenomenon of rising water in the tube of
smaller diameter is called capillary rise.
40.
Metric to U.S. System Conversions,
Calculations, Equations, and Formulas
Millimeters (mm) x 0.03937 = inches (")(in)
Centimeters (cm) x 0.3937 = inches (")(in)
Meters (m) x 39.37 = inches (")(in)
Meters (m) x 3.281 = feet (')(ft)
Meters (m) x 1.094 = yards (yds)
Kilometers (km) x 0.62137 = miles (mi)
Kilometers (km) x 3280.87 = feet (')(ft)
Liters (l) x 0.2642 = gallons (U.S.)(gals)
41.
Calculations, Equations & Formulas
Bars x 14.5038 = pounds per square inch (PSI)
Kilograms (kg) x 2.205 = Pounds (P)
Kilometers (km) x 1093.62 = yards (yds)
Square centimeters x 0.155 = square inches
Liters (l) x 0.0353 = cubic feet
Square meters x 10.76 = square feet
Square kilometers x 0.386 = square miles
Cubic centimeters x 0.06102 = cubic inches
Cubic meters x 35.315 = cubic feet
42.
Calculations, Equations & Formulas
Inches (")(in) x 25.4 = millimeters (mm)
Inches (")(in) x 2.54 = centimeters (cm)
Inches (")(in) x 0.0254 = meters (m)
Feet (')(ft) x 0.3048 = meters (m)
Yards (yds) x 0.9144 = meters (m)
Miles (mi) x 1.6093 = kilometers (km)
Feet (')(ft) x 0.0003048 = kilometers (km)
43.
Calculations, Equations & Formulas
Gallons (gals) x 3.78 = liters (l)
Cubic feet x 28.316 = liters (l)
Pounds (P) x 0.4536 = kilograms (kg)
Square inches x 6.452 = square centimeters
Square feet x 0.0929 = square meters
Square miles x 2.59 = square kilometers
Acres x 4046.85 = square meters
Cubic inches x 16.39 = cubic centimeters
Cubic feet x 0.0283 = cubic meters
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