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pPREFACE
SUBJECT: FLUID MECHANICS
EVALUATION METHOD:
PRESENT
= 10 %
TASK
= 20 %
Mid Test
= 30 %
Final Test
= 40%
TOTAL
= 100%
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LECTURE 1
INTRODUCTION
In beginning the study of any subject, a number of questions come to mind immediately.
Among those that a student in the first course in fluid mechanics may ask are the
following:
What is fluid mechanics all about?
Why do I have to study it?
Why should I want to study it?
How does it relate to subject areas with which I am already familiar?
1.1
DEFINITION OF A FLUID
A fluid is a substance that deforms continuously under the application of a shearing (i.e.
tangential) stress no matter how small the shearing stress.
F
F
t o t1
(a)
t2
(b)
Fig.1.1 Behavior of (a) solid and (b) fluid under the action of a constant shear force.
A solid is a substance that deforms when a shear stress is applied, but it does not continue
to deform.
Fig (a) Deformation of the solid is directly proportional to applied shear stress (τ), where

F
and A is the area of the surface in contact with the plate.
A
Fig (b) upon application of the force (F), to upper plate, we notice that the fluid element
continues to deform as long as the force is applied. The shape of the fluid element,
at successive instants of time, t2 > t1 > t0 is shown by dashed lines.
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1.2
BASIC EQUITION
An analysis of any problem in fluid mechanics necessarily begins, either directly or
indirectly, with statements of the basic laws governing the fluid motion. These laws,
which are independent of the nature of the particular fluid, are:
1. Conservation of mass.
2. Newton’s second law of motion.
3. Moment of momentum.
4. The first law of thermodynamics.
5. The second law of thermodynamics.
Clearly, not all of these laws are always required in the solution of any one problem. In
some problems, it is necessary to bring into the analysis additional relations, in the form
of constitutive equations describing the behavior of physical properties of fluids under
given conditions.
1.3
METHODS OF ANALYSIS
In basic mechanics, extensive use was made of free body diagram. In thermodynamics
you referred to the system under analysis as either a closed system or open system. We
shall employ the terms system and control volume.
System boundary
Piston
GAS
Fig. 1.2 Piston-cylinder assembly.

A system is defined as a fixed, identifiable quantity of mass.

The system boundaries separate the system from the surroundings. The boundaries
of the system may be fixed or movable, but no mass crosses them.
Example 1.1
A radiator of steam engine heating system has a volume of 0.7 ft3. When the radiator is
filled with dry saturated steam at a pressure of 20 psia, all valve to the radiator are closed.
How much heat will have been transferred to the room when the pressure of the steam is
10 psia?
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GIVEN:
Steam radiator with volume,  = 0.7 ft3.
=
System
0.7 ft3
Dry saturated steam pressure p = 20 psia.
Steam
Q
FIND:
Amount of heat transferred when pressure reaches 10 psia.
SOLUTION:
We are dealing with a system.
At: State 1
 = 0.7 ft3
p1 = 20 psia
State 2
 = 0.7 ft3
p2 = 10 psia
dry saturated steam
W12  0
Basic equation: First law for system: Q12  W12  E 2  E1
Assumptions: W12  0 , since no shaft work is done and  = constant
E = U, since the system is stationary
Under these assumptions, the first law statement reduces to
Q12  U 2  U1  m u 2  u1 
Thus, to determine Q12, we must determine m, u2, and u1.
At state 1,
p1 = 20 psia
from steam tables:
dry saturated steam
Since v 
u1 = 1081.9 Btu/lbm (internal energy)
v1 = 20.089 ft3/lbm
(specific volume)

, then
m
m1 
 0.7 ft 3
lbm

x
 0.0348 lbm
v1
20.089 ft 3
For steam
p
Vg
Liquid
Vf
x
Vfg
Steam
T = constant
v
4
Consequently, when p is reduced at constant v, then state 2 must lie in the wet region.
Thus at state 2
v 2  v1  v 2 f  x v 2 fg
and
x
v1  v 2 f
v 2 fg
v 2fg
38.40 ft 3

lbm
From the steam tables at p2 = 10 psia.
v 2f
0.017 ft 3

lbm
x
v1  v 2 f 20.089  0.017

 0.523
v 2 fg
38.40
Then
u 2  u 2f  x u 2 fg  161.14  0.523(911.1)  638 Btu/lbm
Finaly,
Q12  mu 2  u1   0.0348 lbm  638  1081.9
Btu
 15.4 Btu
lbm
The minus sign
indicates heat transfer
out of the system
The purpose of this problem was to review the use of:
i.
The first law of thermodynamics for a system, and
ii.
The steam tables
1.4
METHODS OF DISCRIPTION
Mechanics and Thermodynamics deal almost exclusively with systems. Thus, you
have made extensive use of the basic equations applied to a fixed, identifiable quantity of
mass.
Consider for example the application of Newton’s second law to a particle system
of mass, m, in one of the following forms:


F  ma

or

dV
d  
 F  m dt  dt  mV 

or
(1.1a)

d2 r d2
F

m
 2

dt 2
dt
 
m r 


(1.1b)
(1.1c)
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
where:
 F  the sum of all external forces acting on the system.

a  the acceleration of the center of mass of the system.

V  the velocity of the center of mass of the system.

r  the position vector of the center of mass of the system relative to a fixed
coordinate system.
Thus, in describing the motion of a particle in a rectangular coordinate system:




F  i Fx  j Fy  k Fz



(1.2a)

a  i ax  j ay  k az



(1.2b)

V  i ux  j vy  k wz



(1.2c)

r  i x jykz
(1.2d)

where: Fx, Fy, and Fz are the components of F in the x, y, and z directions, respectively.

ax, ay, and az are the components of a in the x, y, and z directions, respectively.

u, v, and w are the components of V in the x, y, and z directions, respectively.
x, y, and z are the coordinates of the particle.

Since particle moves under the action of the force ( F ), we recognize (mengenali) that its



position ( r ), velocity ( V ), and acceleration ( a ) are in functions of time. That is,


r  r (t)

x  x(t) ,
y  y(t)
z  z(t)
and, hence,
u  u(t) ,
v  v(t)
w  w(t)
and, hence,
a x  a x (t) ,
a y  a y (t)
a z  a z (t)

V  V(t)

and, hence,

a  a (t)
From elementary mechanics we know further (as has already been implied in Eq 1.1) that


dV
a
dt

and

du
,
ax 
dt


dv
,
ay 
dt


dw
az 
dt
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

dr
V
dt
ux 
and
dx
,
dt
v
dy
,
dt
w
dz
dt
Example 1.2
A ball is thrown vertically upward with an initial speed of 30 m/sec. Neglecting air
resistance, determine the maximum height to which it will rise, and the time required for
it to reach the maximum height.

V
GIVEN:
x
A ball is thrown vertically upward.
At

V  u o i  30i m/sec
t = 0, x = 0

g
Neglect air resistance.
t=0, x=0
FIND:
y
a) The maximum height to which the ball will rise.
b) The time required to reach maximum height.
SOLUTION:

Basic equations:
F
x

F  ma
ax 
 max
d 2x
dt 2
ux 
dx
dt
Drawing a free body diagram, we obtain



F  ma

Vui


F  W
x

d2x
 mg
dt 2

d2x
 g
dt 2
u = dx/dt = 0

W  mg i
d  dx 
   g
dt  dt 
m
t=?
 dx 
r 0 d dt     g dt
Xmax =?
r
Integrating respect to time between limits of 0 and t, we have
dx  dx  = uo

 gt
dt  r dt  r 0
or
dx
 u0  g t
dt
The maximum height is reached when u = dx/dt = 0. At this condition
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dx
 0  u0  g t
dt
g t  uo
Thus the time required to reach the maximum height is given by
t
u o 30 m sec 2

x
 3.06 sec
g
sec 9.81 m
The maximum height is obtained from
x  uot 
 x max
1.4
1 2
gt
2
t
with
u
1 u
 u o o  g  o
g 2  g
uo
g
2

1 u o2 1 30 m 2
sec 2
 
 x
x
 45.9 m
2 g
2
9.81 m
sec 2

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SYSTEM OF DIMENTIONS
There are three basic systems of dimensions, corresponding to the three basic ways of
specifying the primary dimensions. These systems use primary dimension of:
1. Mass M  , length L  , time t  , temperature T  .
2. Force F , length L  , time t  , temperature T  .
3. Force F , Mass M  , length L  , time t  , temperature T  .


We recognize that Newton’s second law ( F m a ) relates the four dimensions, F, M, L,


and t. The dimension of Newton’s second law must in fact be F  ML / t 2 .
1.5
SYSTEM OF UNITS
We shall present only the most common engineering system of units for each of the basic
systems of dimensions.
a. MLtT
In the SI system of units:

the unit of mass is the kilogram (kg).

the unit length is the meter (m).

the unit of time is the second (sec).

the unit temperature is the Kelvin (oK).
Force is a secondary dimension, and its unit, the Newton (N), is defined from
Newton’s second law as
1N 
1 kg m
sec 2
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In the Absolute Metric system of units:

the unit of mass is the gram (g).

the unit length is the centimeter (cm).

the unit of time is the second (sec).

the unit temperature is the Kelvin (oK).
Force is a secondary dimension, the unit of force, the dyne, is defined in terms of
Newton’s second law as
1 dyne 
1 g cm
sec 2
b. FLtT
In the British Gravitational system of units:

the unit of force is the pound (lbf).

the unit length is the foot (ft).

the unit of time is the second (sec).

the unit temperature is the Rankine (oR).
Mass is a secondary dimension, the unit mass, the slug, is defined from Newton’s
second law as
1 lbf sec 2
1 slug 
ft
c. FMLtT
In the English Engineering system of units:

the unit of force is the pound force (lbf).

the unit of mass is the pound mass (lbm).

the unit length is the foot (ft).

the unit of time is the second (sec).

the unit temperature is the Rankine (R).
Since both force and mass are chosen as primary dimensions, Newton’s second law is
written as

F

ma
gc
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A force of one pound (1 lbf) is the force that gives a pound mass (1 lbm) an
acceleration equal to the standard acceleration of gravity on Earth (=32.17 ft/sec2).
From Newton’s second law we see that
1 lbm x 32.2 ft /sec 2
1 lbf 
gC
gC 
32.2 ft . lbm
lbf . sec 2
Since a force of 1 lbf accelerates 1 lbm at 32.2 ft/sec2. it would accelerate 32.2 lbm at 1
ft/sec2. A slug is also accelerated at 1 ft/sec2 by a force 1 lbf.
Therefore,
1 slug  32.2 lbm
Example 1.3
The density of mercury is given as 26.3 slug/ft3. Calculate the specific weight in lbf/ft3 on
Earth and on the moon (acceleration of gravity on the moon is 5.47 ft/sec2). Calculate the
specific volume in m3/kg and the specific gravity of the mercury.
GIVEN:
Density of mercury, ρ = 26.3 slug/ft3.
Acceleration of gravity on the moon is 5.47 ft/sec2.
FIND:
a) Specific weight γ of the mercury, in lbf/ft3, on Earth and on the moon.
b) Specific volume, v, in m3/kg.
c) Specific gravity, SG.
SOLUTION:
Definitions of specific weight, specific volume, and specific gravity are
Specific weight:
γ
weight
 ρg
volume
Specific volume:
v
1



m
10
SG 
Specific gravity:

H O
2


H O
2
g Earth = 32.2 ft/sec2,
Properties are
g moon = 5.47 ft/sec2
 H O  1.94 slug/ft3 (see Appendix A)
2
(a) Specific weight:
γ Earth  ρ g Earth 
26.3 slug 32.2 ft lbf . sec 2
lbf
x
x
 847 3
3
2
slug . ft
ft
sec
ft
(Recall from Newton’s second law that 1 lbf = 1 slug x 1 ft/sec2)
γ moon  ρ g moon 
26.3 slug 5.47 ft lbf . sec 2
lbf
x
x
 144 3
3
2
slug . ft
ft
sec
ft
(b) Specific volume:
3

1
ft 3
0.3048 m 
slug
lbm
5 m
v 
x
x
x
 7.37x10
ρ 26.3 slug
32.2 lbm 0.4536 kg
kg
ft 3
3
(c) Specific gravity:
SG 
ρ
ρ H 2O

26.3 slug
ft 3
x
 13.6
1.94 slug
ft 3
REFERENCE:
1. Fox & Mc Donald, Introduction to fluid mechanics, 2nd edition, John Wiley &
Sons, Canada.
2. Irving H. Shames, Mechanics of Fluids, Fourth Edition, Mc Graw Hill, Singapore.
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Capek ...dehh..!!
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