Concepts, Definitions, and Basic Principles

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Concepts, Definitions, and Basic Principles


Thermodynamics

is a science in which the storage, the
transformation, and the transfer of energy

are studied. Energy is
stored
as internal energy (associated with temperature), kinetic energy (due
to

motion), p
otential energy (due to elevation) and chemical energy
(due to chemical composition); it is

transformed
from one of these
forms to another; and it is
transferred
across a boundary as either

heat
or work. In thermodynamics we will develop mathematical equat
ions
that relate the transformations

and transfers of energy to material
properties such as temperature, pressure, or enthalpy.


Substances and their properties thus become an important secondary
theme. Much of our work will be

based on experimental observ
ations
that have been organized into mathematical statements, or
laws;

the
first and second laws of thermodynamics are the most widely used.

The engineer’s objective in studying thermodynamics is most often
the analysis or design of a

large
-
scale system
-
an
ything from an air
-
conditioner to a nuclear power plant. Such a system may be

regarded
as a continuum in which the activity of the constituent molecules is
averaged into

measurable quantities such as pressure, temperature, and
velocity. This outline, then,

will be

restricted to
macroscopic
or
engineering thermodynamics.
If the behavior of individual molecules
is

important, a text in
statistical thermodynamics
must be consulted.















THERMODYNAMIC SYSTEMS AND CONTROL VOLUMES


A

thermodynamic
system
is a definite quantity of matter most often
contained within some closed

surface. The surface is usually an
obvious one like that enclosing the gas in the cylinder
;

however, it
may be an imagined boundary like the deforming boundary of a
certain amount of

mass as

it flows through a pump.
T
he system is the
compressed gas, the
working fluid,
and the

system boundary is shown
by the dotted line.


All matter external to a system is collectively called its
surroundings.
Thermodynamics is concerned

with the inter
actions of a system and
its
surroundings, or one system interacting with another.


A system interacts with its surroundings by transferring energy across
its boundary. No material

crosses the boundary of a given system. If
the system does not exchange ener
gy with the surroundings,

it is an
isolated
system.


In many cases, an analysis is simplified if attention is focused on a
volume in space into which,

and or from which, a substance flows.
Such a volume is a
control volume.
A pump, a turbine, an

inflating
balloon, are examples of control volumes. The surface that completely
surrounds the control

volume is called a
control
s u
rf

a c e .
We thus
must choose, in a particular problem, whether a system is to be
considered or whether a

control volume is more use
ful. If there is
mass
flux across a boundary of the region, then a control

volume is
required; otherwise, a system is identified. We will present the
analysis of a system first and

follow that with a study using the control
volume.







UNITS



EXAMPLE 1.
2
Newton’s second law,
F
= ma, relates a net force
acting on a body to its mass and acceleration.

Thus, a force of one
N
ewton accelerates a mass of one kilogram at one m/s
2
; or, a force of
one lbf accelerates

32.2 lbm (1 slug) at a rate of one ft/sec
2
. Hen
ce, the
units are related as

1 N = 1 kg
-

m/s
2

or 1
l
bf = 32.2
l
bm
-
ft/sec
2



EXAMPLE 1.3
Weight is the ‘force of gravity; by Newton’s second
law,
L(
W
)

=
mg
.
As mass remains constant, the

variation of
L(
W
)

with elevation is due to changes in the acceleration

of gravity g (from
about 9.77 m/s2 on the

highest mountain to 9.83 m/s2 in the deepest
ocean trench). We will use the standard value 9.81 m/s2 (32.2

ft/sec2),
unless otherwise stated.



EXAMPLE 1.4
To express the energy unit J (joule) in terms of
SI
base
units, recall that energy or work is force

times distance. Hence,
by Example 1.2,

1
J
=
(1 N)(1 m)
=
(1 kg
-

m/s2)(1 m)
=
1 kg
*
m2/s2

In the English system both the lbf and the lbm are base units. As
indicated in Table 1
-
1, the primary energy

unit is the f
t
-
lbf. By
Example 1.2,

1 ft
-
lbf
=
32.2 lbm
-
ft2/sec2
=
1 slug
-
ft2/sec2

analogous to the SI relation found abov
e












DENSITY, SPECIFIC VOLUME, SPECIFIC WEIGHT

D
ensity is mass per unit volume; specific volume is volume per unit
mass.

Therefore,

ρ

= 1/v

Associated with (mass) density is
weight density
or
spec
ific

weight


EXAMPLE 1.5
The mass of air in a room 3
X
5
X
20 m is known to
be 350 kg. Determine the density, specific

volume, and specific

w
eight.



EXAMPLE 1.6
Express a pressure gage reading of 3
5 psi in absolute
pascals.

First we convert the pressure reading into
P
ascals. We have

(35
lbf/inch
2
) (144
inch2
-
ft2) (0.04788kPa
/
lbf
-
ft
2

1
=
241 kPa gage

kPa

To find the absolute pressure we simply add the atmospheric pressure
to the above value. Using
Patm
=
100 kPa,

we obtain

P
=
241
+
100
=
341 kPa



EXAMPLE 1.7
The manometer shown in Fig. 1
-
10 is used to
measure the pressure in the water pipe. Determine

the water pressure if the manometer reading is 0.6 m. Mercury is 13.6
times heavier than water.

Fig. 1
-
10

To solve the manometer problem we use the fact that
P,
=
P b .
The
pressure
P,
is simply the pressure
P
in

the water pipe plus the pressure due to the 0.6 m of water; the pressure
Pb
is the pressure due to 0.6 m of

mercury. Thus,

P
+
(0.6 m)(9810 N/m3)
=
(0.6 m)(13.6)(9810 N/m3)

This gives
P
=
74 200 Pa or 74.2 kPa gage.

EXAMPLE 1.8
Calculate the force due to the pressure acting on the
1
-
m
-
diameter horizontal hatch of a

submarine submerged 600 m below the surface.

The pressure acting on the hatch at a de
pth of 600 m is found from
(1.11)as

P
=
pgh
=
(1000 kg/m3)(9.81 m/s2)(600 m)
=
5.89 MPa

The pressure is constant over the area; hence, the force due to the
pressure is given by




THE IDEAL
-
GAS EQUATION OF STATE


When the vapor of a substance has relativel
y low density, the
pressure, specific volume, and

temperature are related by the simple
equation

p
v
=
RT

where
R
is a constant for a particular gas and
is

called the
gas
constant.
This equation is an
equation
of
state

in that it relates the
state properti
es
p
,

v
,
and
T ;
any gas for which this equation is valid
is

called an
ideal gas
or a
perfect gas.
Note that when using the ideal
-
gas equation the pressure and

temperature must be expressed as
absolute quantities.


The gas constant
R
is related to a
unive
rsal gas constant
,
which has
the same value for all gases,

by the relationship
-

In the SI system it is

c
onvenient to use instead the kilomole (kmol),
which amounts to
x
kilograms of a substance of

molecular weight
x,
For instance, 1 kmol of carbon is a ma
ss of 12 kg
(exactly); 1 kmol of molecular

oxygen is 32 kg (very nearly). Stated
otherwise,
M

=
12 kg/ kmol
for C,
and
M
=
32
kg/ kmol for 0
2
,.

In the English s
ystem one uses the pound
-
mole (l
bmol); for 0
2
,
M
=
32
lbm/lbmol.

The value of is

=
8.314
kJ/( km
ol
-

K)
=
1545
ft
-
lbf/( lbmol
-

OR)
(
2
-
81

For air
M

is 28.97 kg/kmol(28.97 Ibm/lbmol), so that for air
R
is
0.287 kJ/kg
-

K (53.3 ft
-
lbf/lbm
-

OR),

a value used extensively in calculations involving air.

Other forms of the ideal
-
gas equation are

p
V
=
mRT


p

=
ρ
RT

p
V
=
nRT
(
2.9)

where
n
is the number of moles.

Care must be taken in using this simple convenient equation of state.
A
low
-
density
ρ

can be

experienced by either having a low pressure or a high temperature. For
air the ideal
-
gas equation is

surprising
ly accurate for a wide range of temperatures and pressures;
less than 1 percent error is

encountered for pressures as high as 3000 kPa at room temperature, or
for temperatures as low as

-

130°C
at atmospheric pressure.

The
compressibility factor
y

helps us

in determining whether or not
the ideal
-
gas equation should

be used. It is defined as

z=
RP
-
LT’

and is displayed in Fig. 2
-
6 for nitrogen. Since air is composed mainly
of nitrogen, this figure is

acceptable for air also. If
Z
=
1, or very nearly 1, the ideal
-
gas
equation can be used.









EXAMPLE 2.5
An

automobile tire with a volume of 0.6 m3

is
i
nflated to a gage pressure of 200 kPa.
Calculate
the mass of air in the
tire if the temperature is 20°C.

Air is assumed to be an id
eal gas at the
conditions of this example. In the ideal
-
gas equation,
p
V

=
mRT, we

use absolute pressure and absolute temperature. Thus, using
P
at
=
100
kPa,

P
=
200
+
100
=
300 kPa and T
=
20
+
273
=
293 K

The mass is then calculated to be

m
=
2.14 kg

The
units in the above equation should be checked.