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.
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