Principle Concept Of Fluid Mechanic

1. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/1
PRINCIPLE CONCEPTS OF FLUID MECHANICS
OBJECTIVES
General Objective : To know, understand and apply the measurement of temperature,
pressure and physical properties of fluid.
Specific Objectives : At the end of the unit you should be able to :
 list the temperature scales.
 convert the temperature to Celsius (ºC), Fahrenheit (ºF),
Rankine (R), Kelvin (K) scales.
 define Pressure ( P ) , Atmospheric Pressure (Patm) , Gauge
Pressure ( PG ) , Absolute Pressure ( PA ) and Vacuum ( Pv )
 calculate pressure gauge using the formula given.
 differentiate mass density ( ρ ), specific weight ( ω ), specific
gravity ( s ) , specific volume ( v ) and viscosity.
 calculate fluid properties using the formula given.
UNIT 1

2. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/2
INTRODUCTION TO FLUID MECHANICS
Fluid Mechanics - Introduction
Fluid Mechanics is a section of applied mechanics, concerned with the static and
dynamics of liquids and gases.
Knowledge of fluid mechanics is essential for the chemical engineer, because the
majority of chemical processing operations are conducted either partially or totally in
the fluid phase.
The handling of liquids is much simpler, cheaper, and less troublesome than handling
solids.
Even in many operations a solid is handled in a finely divided state so that it stays in
suspension in a fluid.
Fluids and their Properties
Fluids
In everyday life, we recognize three states of matter: solid, liquid and gas. Although
different in many respects, liquids and gases have a common characteristic in which
they differ from solids. Both are fluids, but lack the ability of solids to offer a
permanent resistance to a deforming force.
A fluid is a substance which deforms continuously under the action of shearing forces,
however small they may be. Conversely, if a fluid is at rest, there can be no shearing
forces acting and, therefore, all forces in the fluid must be perpendicular to the planes
upon which they act.

3. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/3
1.1 TEMPERATURE UNITS
Temperature scales are defined by the numerical value assigned to a standard fixed
point. By international agreement the standard fixed point is the easily reproducible
triple point of water. These are represented by the state of equilibrium between steam,
ice and liquid water.
In this unit we learn how to convert temperatures into Celsius, Fahrenheit Kelvin and
Rankine scales.
The Celsius temperature scale uses the unit degree Celsius (ºC), which has the same
magnitude as the Kelvin. Thus the temperature differences are identical on both scales.
However, the zero point on the Celsius scale is shifted to 273K, as shown by the
following relationship between the Celsius temperature and the Kelvin temperature:
( ) ( ) 273−=° KTCT …(1)
By definition, the Rankine scale, the unit of which is the degree Rankine (R) is
proportional to the Kelvin temperature according to
( ) ( )KTRT 8.1= …(2)
A degree of the same size as that on the Rankine scale is used in the Fahrenheit scale,
but the zero point is shifted according to the relation
( ) ( ) 460−=° RTFT …(3)
INPUTINPUT

4. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/4
substituting Eqs. (1) and (2) into Eq. (3), it follows that
( ) ( ) 328.1 +°=° CTFT
Example 1.1
Convert 200ºC to K.
Solution to Example 1.1
273+°= CK
273200 +=
K473=
Example 1.2
Convert 250 ºC to ºF
Solution to Example 1.2
CF °+=° 8.132
( )2508.132 +=
F°= 482
Example 1.3
Convert 365 ºF to R
Solution to Example 1.3
FR °+= 460
365460 +=
R825=
Example 1.4

5. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/5
Convert 200 ºF to R
Solution to Example 1.4
FR °+= 460
200460 +=
R660=
Example 1.5
Convert 450 R to K
Solution to Example 1.5
8.1
R
K =
8.1
450
=
K250=
Example 1.6
Convert 410 K to ºF
Solution to Example 1.6
( ) 322738.1 +−=° KF
( ) 322734108.1 +−=
F°= 6.278
ACTIVITY 1A

6. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/6
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT
INPUT…!
1.1 Write the formula to convert the following temperature scales.
1.2 Solve the problems below:
i) Air entering a wet scrubber is at 153 ºC. What is the temperature expressed in
degree Rankine?
ii) The gas stream temperature entering a fabric filter is 410 ºF. What is the
temperature expressed in degree Kelvin?
FEEDBACK ON ACTIVITY 1A
ºC to K
ºF to R
R to K

7. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/7
1.1
1.2
i) Convert temperature from Celsius to Fahrenheit.
ºF = 32 + 1.8ºC
ºF = 32 + 1.8(153)
= 307 ºF
Convert temperature from Fahrenheit to Rankine.
R = ºF + 460
R = 307 + 460
= 767 R
ii) Convert temperature from Fahrenheit to Rankine.
R = ºF + 460
R = 410 + 460
= 870 R
Convert temperature from Rankine to Kelvin.
ºC to K
ºF to R
R to K
K = ºC + 273
R = 460 + ºF
K =

8. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/8
8.1
R
K =
8.1
870
=
K483=
1.2 DEFINITION OF PRESSURE
INPUTINPUT

9. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/9
Fluid will exert a normal force on any boundary it is in contact with. Since these
boundaries may be large and the force may differ from place to place it is convenient
to work in terms of pressure, p, which is the force per unit area.
Units : Newton’s per square metre, Nm-2
,kgm-1
s-2
.
(The same unit is also known as Pascal, Pa i.e 1 Pa = 1Nm-2
)
Also frequently used is the alternative SI unit the bar,
where 1 bar = 105
Nm-2
1.3 MEASUREMENT OF PRESSURE GAUGES
Absolute Pressure, Gauge Pressure, and Vacuum
In a region such as outer space, which is virtually void of gases, the pressure is
essentially zero. Such a condition can be approached very nearly in a laboratory when
a vacuum pump is used to evacuate a bottle. The pressure in a vacuum is called
absolute zero, and all pressures referenced with respect to this zero pressure are
termed absolute pressures.
Many pressure-measuring devices measure not absolute pressure but only
difference in pressure. For example, a Bourdon-tube gage indicates only the difference
appliedisforcethewhichoverArea
Force
pressure =

10. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/10
between the pressure in the fluid to which it is tapped and the pressure in the
atmosphere. In this case, then, the reference pressure is actually the atmospheric
pressure. This type of pressure reading is called gauge pressure. For example, if a
pressure of 50 kPa is measured with a gauge referenced to the atmosphere and the
atmospheric pressure is 100 kPa, then the pressure can be expressed as either p = 50
kPa gauge or p = 150 kPa absolute.
Whenever atmospheric pressure is used as a reference, the possibility exists
that the pressure thus measured can be either positive or negative. Negative gauge
pressure is also termed as vacuum pressure. Hence, if a gauge tapped into a tank
indicates a vacuum pressure of 31 kPa, this can also be stated as 70 kPa absolute, or
-31 kPa gauge, assuming that the atmospheric pressure is 101 kPa .
1.3.1 Atmospheric Pressure, patm
- The earth is surrounded by an atmosphere many miles high. The
pressure due to this atmosphere at the surface of the earth depends upon the
head of the air above the surface.
- The air is compressible, therefore the density is different at different
height.
- Due to the weight of atmosphere or air above the surface of earth, it
is difficult to calculate the atmospheric pressure. So, atmospheric pressure
is measured by the height of column of liquid that it can support.
- Atmospheric pressure at sea level is about 101.325 kN/m2
, which is
equivalent to a head of 10.35 m of water or 760 mm of mercury
approximately, and it decreases with altitude.
1.3.2 Gauge Pressure, pG
- It is the pressure, measured with the help of a pressure measuring
instrument, in which the atmospheric pressure is taken as datum; in other
words the atmospheric pressure at the gauge scale is marked zero.
- The gauge pressure can be either positive or negative depending on
whether the pressure is above atmospheric pressure (a positive value) or
below atmospheric pressure (a negative value).
1.3.3 Absolute Pressure, pA
- It is the pressure equal to the algebraic sum of the atmospheric
and gauge pressures.

11. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/11
pressurecAtmospheripressureGaugepressureAbsolute +=
atmGA ppp +=
1.3.4 Vacuum, pv
- In a perfect vacuum which is a completely empty space, the pressure
is zero.
Example 1.7
Define the following terms :
i. Pressure (p )
ii. Atmospheric Pressure ( patm)
iii. Gauge Pressure ( pG )
iv. Absolute Pressure ( pA )
v. Vacuum ( pv )
Solution to Example 1.7
Local
atmospheric
pressure
reference
absolute pressure
gauge pressure +ve
pressure –ve
absolute pressure
Figure 2.1 atmospheric, gauge, absolute and vacuum pressures
Pgauge
Figure 1.2 Pressure Gauges

12. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/12
i. Pressure ( p ):
Pressure is force ( F ) per unit area ( A ).
ii. Atmospheric Pressure ( patm) :
The pressure due to atmosphere at the surface of the earth depends
upon the head of the air above the surface.
iii. Gauge Pressure ( pG ):
It is the pressure, measured with the help of a pressure measuring
instrument, in which the atmospheric pressure is taken as datum.
iv. Absolute Pressure ( pA )
It is the pressure that equals to the algebraic sum of the atmospheric
and gauge pressures.
v. Vacuum ( pv )
A completely empty space where the pressure is zero.
Example 1.8
What is the pressure gauge of air in the cylinder if the atmospheric gauge is
101.3 kN/m2
and absolute pressure is 460 kN/m2
.
Solution to Example 1.8
pA = 460 kN/m2
patm = 101.3 kN/m2
pG = ?
With reference to the formula below :

13. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/13
Absolute pressure, pA = Gauge pressure, pG + atmospheric pressure, patm
Therefore ,
pG = pA – patm
= 460 – 101.3
=
2
/7.358 mkN
ACTIVITY 1B
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT
INPUT…!
1.3 Label the diagram below :

14. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/14
1.4 A Bourdon pressure gauge attached to a boiler located at sea level shows a reading
pressure of 7 bar. If atmospheric pressure is 1.013 bar, what is the absolute pressure in
that boiler (in kN/m2
) ?
FEEDBACK ON ACTIVITY 1B
1.3
Local
atmospheric
pressure
reference
a
b
c
d

15. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/15
1.4
pA = ?
patm = 1.013 bar
pG = 7 bar
With reference to the formula below :
Absolute pressure,pA = Gauge pressure,pG + atmospheric pressure,patm
Therefore ,
pA = pG + patm
Local
atmospheric
pressure
reference
a = absolute pressure
b = gauge pressure +ve
d = pressure –ve
d = absolute pressure

16. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/16
= 7 x 105
+ 1.013 x 105
= 801300 N/m2
=
2
/3.801 mkN
1.4 PHYSICAL PROPERTIES OF FLUID
INPUTINPUT

17. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/17
Fluid properties are intimately related to fluid behaviour. It is obvious that different
fluids can have grossly different characteristics. For example, gases are light and
compressible, whereas liquids are heavy and relatively incompressible.
To quantify the fluid behaviour differences certain fluid properties are used. The fluid
properties are mass density, specific weight, specific gravity, specific volume and
viscosity.
1.4.1 Mass density, ρ is defined as the mass per unit volume.
( SI units, kg/m3
)
Vvolume
mmass
,
,
=ρ
1.4.2 Specific weight, ω is defined as the weight per unit volume.
( SI units, N/m3
)
Vvolume
Wweight
,
,
=ω
V
mg
=ω
gρω = (where g = 9.81m/sec2
)
In SI units the specific weight of water is 9.81 x 1000 = 9810 N/m3
1.4.3 Specific gravity or relative density, s is the ratio of the weight of the substance
to the weight of an equal volume of water at 4 ºC.
water
cesubs
s
ω
ω tan
=
water
cesubs
s
ρ
ρ tan
=
1.4.4 Specific volume, v is defined as the reciprocal of mass density. It is used to
mean volume per unit mass. (SI units, m3
/kg ).
ρ
ν
1
=

18. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/18
mmass
Vvolume
,
,
=ν
1.4.5 Viscosity
A fluid at rest cannot resist shearing forces but once it is in motion, shearing
forces are set up between layers of fluid moving at different velocities. The
viscosity of the fluid determines the ability of the fluid in resisting these
shearing stresses.
Example 1.9
What is the mass density, ρ of fluid (in kg/m3
) if mass is 450 g and the volume is 9
cm3
.
Solution to Example 1.9
V
m
=ρ
6
3
109
10450
−
−
×
×
=
33
/1050 mkg×=
Example 1.10
What is the specific weight, ω of fluid (in kN/m3
) if the weight of fluid is 10N and the
volume is 500 cm2
.
Solution to Example 1.10
V
W
=ω
6
3
10500
1010
−
−
×
×
=
= 20 000 N/m3
3
/20 mkN=
Example 1.11
What is the specific gravity of fluid in Example 1.10.

19. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/19
Solution to Example 1.11
water
cesubs
s
ω
ω tan
=
( )81.91000
1020 3
×
=
039.2=
Example 1.12
What is the specific volume, v of fluid in Example 1.9.
Solution to Example 1.12
m
V
v =
3
6
10450
109
−
−
×
×
=
kgm /102 35−
×=
ACTIVITY 1C
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT
INPUT…!
1.5 Match the following
ρ
ω
s
v
Vvolume
Wweight
,
,
mmass
Vvolume
,
,
Vvolume
mmass
,
,
forwater
cesubsfor
ω
ω tan
m3
/kg
kg/m3
N/m3
Specific
gravity
Specific
volume
Mass
density
Specific
weight

20. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/20
FEEDBACK ON ACTIVITY 1C
1.5
ρ
ω
s
v
Vvolume
Wweight
,
,
mmass
Vvolume
,
,
Vvolume
mmass
,
,
forwater
cesubsfor
ω
ω tan
m3
/kg
kg/m3
N/m3
Specific
gravity
Specific
volume
Mass
density
Specific
weight

21. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/21
SELF-ASSESSMENT
You are approaching success. Try all the questions in this self-assessment section
and check your answers with those given in the Feedback on Self-Assessment. If you
face any problems, discuss it with your lecturer. Good luck.
1.1 Convert the temperatures below according to the specified scales:
a) 175 ºC to ºF, K and R.
b) 518 ºR to ºC, ºF and K.
c) 214 ºF to ºC, R and K.

22. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/22
d) 300 K to ºC, R and ºF.
1.2 Assume the density of water to be 1000 kg/m3
at atmospheric pressure 101
kN/m2
. What will be:
a) the gauge pressure
b) the absolute pressure of water at a depth of 2000 m below the free surface?
1.3 Determine in Newton per square metre, the increase in pressure intensity per
metre depth in fresh water. The mass density of fresh water is 1000 kg/m3
.
1.4 Given specific weight of fluid is 6.54 kN/m3
and its mass is 8.3 kg, calculate
the following:
a) volume of fluid
b) specific volume of fluid
c) density of fluid
1.5 Given oil specific gravity is 0.89, find :
a) density of oil
b) specific weight of oil
c) specific volume of oil
FEEDBACK ON SELF-ASSESSMENT
Answers :
1.1 a) 347 F,807 R,448.3 K ,
b) 58 °F,287.7 K,14.7 C ,
c) 674 °C,374.4 R,101.4 K
d) 27 °C,80.6 °F,540.6 R
1.2 a) 117.72 kN/m2
,

23. PRINCIPLE CONCEPTS OF FLUID MECHANICS J3008/1/23
b) 218.72 kN/m2
1.3 9.81 x 103
N/m2
1.4 a) 0.072 m3
b) 0.0015 m3
/kg
c) 691.67 kg/m3
1.5 a) 0.89 x 103
kg/m3
b) 8730.9 N/m3
c) 0.00112 m3
/kg