Thermodynamics Module 2.3 Key Concepts

Module 2
Licence Category
B1 and B2
Physics
2.3 Thermodynamics
For Training Purposes Only

3-2 Module 2.3 Thermodynamics
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Module 2.3 Thermodynamics
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Knowledge Levels — Category A, B1, B2, B3 and C Aircraft Maintenance
Licence
Basic knowledge for categories A, B1, B2 and B3 are indicated by the allocation of knowledge levels indicators (1,
2 or 3) against each applicable subject. Category C applicants must meet either the category B1 or the category B2
basic knowledge levels.
The knowledge level indicators are defined as follows:
LEVEL 1
• A familiarisation with the principal elements of the subject.
Objectives:
• The applicant should be familiar with the basic elements of the subject.
• The applicant should be able to give a simple description of the whole subject, using common words and
examples.
• The applicant should be able to use typical terms.
LEVEL 2
• A general knowledge of the theoretical and practical aspects of the subject.
• An ability to apply that knowledge.
Objectives:
• The applicant should be able to understand the theoretical fundamentals of the subject.
• The applicant should be able to give a general description of the subject using, as appropriate, typical
examples.
• The applicant should be able to use mathematical formulae in conjunction with physical laws describing the
subject.
• The applicant should be able to read and understand sketches, drawings and schematics describing the
subject.
• The applicant should be able to apply his knowledge in a practical manner using detailed procedures.
LEVEL 3
• A detailed knowledge of the theoretical and practical aspects of the subject.
• A capacity to combine and apply the separate elements of knowledge in a logical and comprehensive
manner.
Objectives:
• The applicant should know the theory of the subject and interrelationships with other subjects.
• The applicant should be able to give a detailed description of the subject using theoretical fundamentals
and specific examples.
• The applicant should understand and be able to use mathematical formulae related to the subject.
• The applicant should be able to read, understand and prepare sketches, simple drawings and schematics
describing the subject.
• The applicant should be able to apply his knowledge in a practical manner using manufacturer's
instructions.
• The applicant should be able to interpret results from various sources and measurements and apply
corrective action where appropriate.

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Table of Contents
Module 2.3 Thermodynamics__________________________________________________9
Temperature ____________________________________________________________9
The Gas Laws __________________________________________________________15
Thermal Expansion ______________________________________________________27
Heat __________________________________________________________________37
Refrigeration and Heat Pumps______________________________________________53
Thermodynamics and the 1st and 2nd Laws ___________________________________61

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Module 2.3 Enabling Objectives and Certification Statement
Certification Statement
These Study Notes comply with the syllabus of Singapore Airworthiness Requirements Part 66 -
Aircraft Maintenance Licensing:
Objective
SAR-66
Reference
Licence Category
B1 B2
Thermodynamics 2.3 2 2
(a)
Temperature: thermometers and temperature
scales: Celsius, Fahrenheit and Kelvin; Heat
definition
(b) 2 2
Heat capacity, specific heat
Heat transfer: convection, radiation and
conduction;
Volumetric expansion
First and second law of thermodynamics
Gases: ideal gases laws; specific heat at
constant volume and constant pressure, work
done by expanding gas
Isothermal, adiabatic expansion and
compression, engine cycles, constant volume
and constant pressure, refrigerators and heat
pumps
Latent heats of fusion and evaporation, thermal
energy, heat of combustion

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Temperature
Temperature Scales
Our common notion of hot and cold has its precise expression in the concept of temperature. As
objects are heated their molecules move faster. In a solid the molecules vibrate more rapidly. In
liquids and gases the molecules move all over in the container at a faster rate of speed. These
variations in speed of the molecules cause objects to expand when they are heated.
This expansion can be used to construct instruments called thermometers. The ordinary
mercury thermometer uses the expansion of a volume of mercury contained in a bulb to indicate
temperature.
A number of temperature scales are currently in use. The Fahrenheit scale is the one we have
used most extensively. On this scale the freezing point of water is 32°and its boiling point is
212°
. The metric scale is the Celsius or centigrade scale. On this scale the freezing point of
water is zero and the boiling point is 100°
.
In theory, if we cool any substance enough, we can cause all molecular motion to cease. We
call this lowest possible temperature “absolute zero”. Ordinary gases like air would be rock solid
at this temperature. Low temperature physicists have never been able to reach this extremely
low temperature in their laboratories. However, they have come close—down to a fraction of a
centigrade degree. Absolute zero is a limiting temperature which can never be reached. Two
other temperature scales are used by engineers and experimental scientists. In both of these
scales the zero of the scale is placed at absolute zero, the coldest possible temperature. These
scales are the metric Kelvin scale and the English Rankin scale.
In Table 3-3, the four temperature scales are compared.
There are formulas that enable us to change from a centigrade reading to a Fahrenheit reading
and vice versa. These formulas are:
C =
9
5
(F - 32) and F =
5
9
C + 32
Note that there are parentheses in the first formula but not in the second formula. Be careful!
There are also formulas that change from a centigrade reading to a Kelvin reading and from a
Fahrenheit reading to a Rankin reading. These formulas are very important to us at this time
since we will have to use absolute temperatures in the gas laws.
These formulas are:

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K = C + 273
Boiling
Point of
Water
Centigrade
Kelvin
Fahrenheit
Rankin
Table 3.1: Comparisons of boiling points, freezing points and
absolute zero in different units
Notes:
• Kelvin has no o
sign in-
• The accurate conversion factor for
Figure 3.1: Temperature scale comparison
K = C + 273 and R = F + 460
Boiling
Point of
Water
Freezing
Point of
Water
Absolute
Zero
100° 0° -273°
373 273 0
212o
32° -460°
672° 492° 0°
: Comparisons of boiling points, freezing points and
absolute zero in different units
-front of the K.
The accurate conversion factor for 0
C to K is +273.15
Figure 3.1: Temperature scale comparison

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Problems
1. Change 20°
C to degrees F.
2. Change -15 °
C to degrees F.
3. Change 86°
F to degrees C.
4. Change -4°
F to degrees C.
5. Change 100°
F to degrees R.
6. Change 450°
R to degrees F.
7. Change 100°
C to degrees K.
8. Change 383 K to degrees C.
9. Gas turbine engine performance is very sensitive to variations in the temperature
of the air. All engines are rated with the air at a standard temperature of 59°
F.
What is the equivalent Centigrade temperature?
10. On some large commercial turbojet engines, the temperature at the front end of
the combustion section is approximately 400°
C. What is this temperature on the
Fahrenheit scale?
11. As air enters the combustion chamber of a turbojet fuel is added and the
temperature is raised to about 3,500°
F in the hotte st part of the flame. What is this
temperature on the Centigrade scale?

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Answers
1. 68°
F
2. 5°
F
3. 30°
C
4. -20°
C
5. 560°
R
6. -10°
F
7. 373 K
8. 110°
C
9. 15°
C
10. 752°
F
11. 1,930°
C

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The Gas Laws
We will next discuss the volume and density of gases under varying conditions of temperature
and pressure. Three gas laws, named after the scientists that discovered them, will be
considered.
Boyle’s Law
A cylinder containing gas is fitted with a light piston. This cylinder contains a certain mass of gas
and therefore a certain number of molecules of gas. The gas has a definite absolute
temperature. This temperature is a measure of the average speed of the gas molecules in the
sample. Some of the molecules are moving faster and some are moving slower. The average
speed determines the temperature.
If the temperature of the gas remains constant and the volume of the gas sample is decreased,
the molecules, still moving with the same average speed, are “squashed” into a smaller space
(see figure 3.2).
The result is that the sides of the container experience more collisions per unit time. This results
in an increase in the absolute pressure the molecules exert on the walls of the container.
Note that a decrease in volume produces an increase in absolute pressure. This is
characteristic of an inverse proportion. We write the equation as:
1
2
2
1
V
V
P
P
=
If we cross multiply in the above equation we
reach the form in which Boyle’s Law is usually
written:
P1V1 = P2V2
Here P1 and P2 are the absolute pressures
corresponding to the volumes V1 and V2
respectively. In working with Boyle’s Law, it
must always be remembered to use absolute
pressures.
Figure 3.2: Boyle’s Law

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EXAMPLE:
A cylinder fitted with a piston contains gas at a pressure of 35.5 lbs./in2
as indicated by a
gauge mounted to the outside of the cylinder. The atmospheric pressure is 14.5 lbs./in2
if the
piston is forced down reducing the volume in the cylinder to one fourth of its original volume
while holding the temperature of the gas constant, determine the new reading on the pressure
gauge.
P1 = (35.5 + 14.5) lbs./in2
P1 = 50 ibs./in2
V2 = ¼ V1
P1V1 = P2V2
(50 lbs./in2
) (V1) = P2 ( ¼ V1)
Solving for P2 gives,
P2 = 200 ibs./in2
absolute
We still must express this new pressure as a gauge pressure since the problem asked for the
new reading on the pressure gauge. Our final answer is:
P2 = (200 - 14.5) lbs./in2
= 186 lbs./in2
Charles’ Law
Toward the end of the 18th century, investigations carried out by French physicists, Jacques
Alexandre Charles and Joseph Louis Gay-Lussac led to the discovery of a relation between the
volume and absolute temperature of gases under conditions of constant pressure.
Let us again consider a sample of gas containing a definite number of molecules. We stipulate
that the pressure on this sample of gas will remain constant. If the pressure is to remain
constant, an increase in absolute temperature must be accompanied by a corresponding
increase in volume (see figure 3.3).

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We say that the volume is directly proportional to
the absolute temperature. provided that the
pressure remains constant. We write the
equation as:
2
2
1
1
T
V
T
V
=
The absolute temperatures must be either
Kelvin, or Rankin degrees.
EXAMPLE:
A quantity of air occupies a volume of one cubic foot on a day when the temperature is 15°
F.
What will be the volume of this quantity of air when the temperature increases to 85°
F, and
the pressure stays the same?
R
545
V
R
475
ft
1
o
2
o
3
=
Note that we have changed the temperatures from degrees Fahrenheit to degrees Rankin,
because we must express the temperatures in absolute units. Cross multiplying, we obtain:
3
o
o
3
2 ft
15
.
1
R
475
R
545
ft
1
V =
×
=
Failure to convert to absolute temperatures will always lead to incorrect answers when
working with the gas laws!
Figure 3.3: Charles’ Law

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Gay-Lussac’s Law
This third gas law relates the absolute pressure
to the absolute temperature of a gas when its
volume is held constant.
Again we consider a certain number of
molecules of gas in a closed container where the
volume of the gas is held constant. If we
increase the absolute temperature of the gas,
the average speed of the molecules increases.
As these molecules strike the walls of the
container they exert a greater pressure since
they are moving faster (see figure 3.4).
Using absolute pressures and temperatures the
following simple relationship is obtained:
2
2
1
1
T
P
T
P
=
This equation is referred to as Gay-Lussac’s Law.
EXAMPLE:
The tyre of a bicycle is filled with air to a gauge pressure of 50.0 lbs./in. at 58°
F. What is the
gauge pressure in the tyre on a day when the temperature rises to 86°
F? Assume that the
volume of the tyre does not change and the atmospheric pressure is 14.7 lbs.fin.2
We must first convert to absolute temperatures and pressures.
P1 = 50.0 lbs./in2
+ 14.7 lbs./in2
= 64.7 ibs./in2
T1 = 460 + 58°
F = 518°
R
T2 = 460 + 86°
F = 546°
R
Substituting these values into Gay-Lussac’s Law gives:
R
546
P
R
518
in
/
lbs
7
.
64
o
2
o
2
=
Solving for P2, we obtain P2 = 68.2 ibs./in2
. Finally, the new gauge pressure is obtained by
subtracting the atmospheric pressure from P2.
68.2 lbs./in2
- 14.7 lbs./in2
= 53.5 lbs./in2
Figure 3.4: Gay Lussac’s Law

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The General (Ideal) Gas Law
The three properties, pressure, temperature, and volume are interrelated for a fixed mass
(number of molecules) of gas in such a way that if two of them change in value the third can
immediately be determined. Combining the three gas laws the following general gas law can be
written:
2
2
2
1
1
1
T
V
P
T
V
P
=
Note that this equation gives us the three gas laws that we have studied.
If the temperature of the gas remains constant, we can cancel the temperatures in the
denominators and obtain:
P1V1 = P2V2 Boyle’s Law
If the pressure remains constant, we can cancel the pressures in the numerators and obtain:
2
2
1
1
T
V
T
V
= Charles’ Law
If the volume remains constant, we can cancel the volumes in the numerators and obtain:
2
2
1
1
T
P
T
P
= Gay-Lussac’s Law
EXAMPLE:
A tank of helium gas has a gauge pressure of 50.2 lbs/in2
and a temperature of 45°
F. A piston
decreases the volume of the gas to 68% of its original volume and the temperature drops to
10°
F. What is the new gauge pressure? Assume normal atmospheric pressure.
We must change both temperatures to absolute units. We must change the original gauge
pressure to absolute pressure. We remember that when the final pressure is obtained it will be
in absolute units. We also note that V2 = 0.68 V1.
2
2
2
1
1
1
T
V
P
T
V
P
=

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We transfer V2 from the numerator on the right to the denominator on the left. We also
transfer T2 from the denominator on the right to the numerator on the left. In this way, we
solve our formula for P2.
2
2
1
2
1
1
P
V
T
T
V
P
=
Next we substitute our known values:
)
V
68
.
0
)(
R
505
(
)
R
470
)(
V
)(
in
/
lbs
9
.
64
(
P
1
o
o
1
2
2 =
P2 = 88.8 lbs/in2
Absolute
P2 = 74.1 lbs/in2
(New Gauge Pressure)
Alternate Form of the General (Ideal) Gas Law
The general gas law tells us that for a fixed quantity of gas, the expression PV/T is constant.
Since PV/T is a constant for a fixed mass of gas, we can set this expression equal to the
product of the mass (m) of the gas and what is referred to as a gas constant (R). This gas
constant (R) varies according to the type of gas. Table 3.2 gives values of R for various gases.
We can write:
mR
T
PV
=
PV = mRT
If we divide both sides of this equation by V. we obtain:
V
mRT
P =
We remember that the density of any substance is given by:
V
m
=
ρ
Values of the Gas Constant, R,
for Some Common Gases
Pa m3
/kg K ft.lbs/slug o
R
Air 287 1,710
Carbon Dioxide 189 1,130
Helium 2,077 12,380
Nitrogen 297 1,770
Oxygen 260 1,550
Water Vapour 462 2,760
Table 3.2: Gas Constants, R

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Therefore we can write:
P = ρRT
The most important application of this formula enables us to obtain the density of any particular
kind of gas if we know its absolute pressure and absolute temperature.
We write the equation in the form:
RT
P
=
ρ
Note: When comparing the density of one type of gas to another, we need to use equal
temperatures and pressures for each gas (since, as the above equation shows, density
changes with pressure and temperature changes). The temperature and pressure we use for
this is known as Standard Temperature and Pressure. These are 0o
C and 1 atmosphere
(273.15 K and 760 mmHg).
EXAMPLE:
Find the density of air if the temperature is 80°
F and the absolute pressure is 2,150 lbs./ft2
)
R
540
)(
R
slug
/
lbs
.
ft
1710
(
ft
/
lbs
2150
RT
P
o
o
2
=
=
ρ
= 0.00233 slug/ft3
Application of the General Gas Law to Compressors
We can apply the general gas law to the flow of air through the compressor of a turbojet engine.
The function of the compressor is to provide a large quantity of high pressure air to the limited
space of the combustion chamber. The reason for this is that the energy released in the
combustion chamber is proportional to the mass of air consumed. The pressure of the air when
it leaves the compressor is called the compressor discharge pressure (CDP) and the ratio of
this to the compressor inlet pressure (CIP) is the compression ratio. That is,
Compression Ratio =
CIP
CDP
Note that the compression ratio can also be expressed as:
Compression Ratio =
1
2
P
P
where the 1’s refer to the inlet pressure and the 2’s to the discharge pressure.

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Air entering a compressor having a compression ratio of 12.5:1 at a pressure of 14.7 PSIA will
leave with a pressure of:
(12.5)(14.7) = 184 PSIA
If however, the temperature of the air is increased too much in the compression process the
volume of a quantity of air entering the combustion chamber will not be reduced significantly
and the compressor efficiency will be low.
EXAMPLE:
A quantity of air occupying 1 cu.ft. at pressure of 14.7 PSIA and a temperature of 59°
F enters
the compressor of a turbojet engine having a compression ratio of 12.5:1 and is discharged at
a temperature of 2,000°
F. With what volume will thi s quantity of air enter the combustion
chamber?
Solving our general equation:
2
2
2
1
1
1
T
V
P
T
V
P
= for V2 yields
















=
=
1
2
1
2
1
2
1
2
1
1
2
P
/
P
1
T
T
V
P
T
T
V
P
V
Therefore, we can substitute our given values:












+
+
=
5
.
12
1
460
59
460
2000
)
ft
1
(
V 3
2
V2 = 0.379 ft.3
EXAMPLE:
With what volume would the quantity of air of the previous problem enter the combustion
chamber if the discharge temperature of the compressor were 750°
F instead of 2,000°
F?












+
+
=
5
.
12
1
460
59
460
750
)
ft
1
(
V 3
2
V2 = 0.187 ft.3
We see that the volume of the original cubic foot of air is less (0.187 ft.3
) when the
temperature is 750°
F than it is (0.379 ft. 3
) when the temperature is 2,000°
F.

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Problems
1. A quantity of gas is contained in a cylinder fitted with a piston. The absolute pressure of
the gas is 240 kPa when the volume is 0.15 m3
. What will the volume be when the
absolute pressure of the gas is changed to 80 kPa while the temperature is held
constant?
2. A quantity of gas is contained in a cylinder fitted with a piston. The gauge pressure of the
gas in the cylinder is 335 lbs/in2
when the volume occupied by the gas is 72in3
What is
the gauge pressure when the volume is decreased to 60 in3
? Assume atmospheric
pressure to be 15 lbs/in2
, and assume that the temperature is held constant.
3. A sample of nitrogen is held at an absolute pressure of 1.50 atmospheres and a volume
of 7.80 m3
. A piston gradually reduces the volume to 6.30 m3
. The temperature does not
change. What is the new absolute pressure in atmospheres?
4. A volume of 1.35 m3
of air at 17°
C is heated to 427°
C while its pressu re is held constant.
What is the volume of the gas at this elevated temperature.
5. A tank of carbon dioxide is maintained at an absolute pressure of 5,000 lbs/ft2
. The
temperature is 190°
F. What is the density of this c arbon dioxide?
6. The air pressure and density at a point on the wing of a Boeing 747 flying at altitude are
70 kPa, and 0.9 kg/m3
respectively. What is the temperature at this point on the wing in
degrees Centigrade?
7. The Goodyear non-rigid airship, the Mayflower, has a volume of 4000 m3
and is filled with
helium to an absolute pressure of 100 kPa. The temperature is 27°
C. Find the density
and total mass of the helium in the ship.
8. At an altitude of 8,000 ft. the absolute temperature of air is 500°
R and the absolute
pressure is 1600 lbs/ft2
. What is the density of air at this altitude?
9. A tank of carbon dioxide is maintained at an absolute pressure of 5,830 lbs/ft2
and a
temperature of 70°
F. What is the density of this ca rbon dioxide?
10. A quantity of air occupying 0.9 ft3
at a pressure of 15 PSIA and a temperature of 40°
F
enters the compressor of a turbojet engine having a compression ratio of 13:1 and is
discharged at a temperature of 1,540°
F. With what v olume will this quantity of air enter
the combustion chamber?

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Answers
1. 0.45m3
2. 405 lb./in2
3. 1.86 atmosphere
4. 3.26 m3
5. 0.007 slug/ft3
6. - 2°
C
7. 0.16 kg/m3
, 640 kg
8. 0.00 188 slug/ft3
9. 0.01 slug/ft3
10. 0.3 ft3

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Thermal Expansion
The temperature of a body is a measure of the average kinetic energy of the molecules of that
body. It follows that molecules of warm liquids and gases move around faster in their containers
than molecules of cool liquids and gases. As a solid is heated its molecules vibrate faster about
their equilibrium positions. As a result of this increased motion of molecules as they are heated,
solids and liquids expand as the temperature is raised.
We’re going to return to the idea of how temperature is related to molecular motion later, but
first let’s look at what happens to materials when they change temperature. Let’s say that
you’ve got a jar lid that’s stuck and you want to get it off. How can you do this? One common
way is by running the jar under hot water so that the jar lid expands and can come off the jar.
Thinking back to our model of solids, if temperature is a measure of how fast things are moving,
when a solid heats up, the molecules vibrate about their normal positions. At higher
temperature, they vibrate more and the material actually grows in size. When a material is
cooled, the molecules don’t move as much and the material shrinks.
If we look at a long strip of metal, with length Lo, we might want to find out what its change in
length is under certain conditions. This is important, for instance, in building roads that must
undergo temperature extremes. Experimentally, we find that the change in length is directly
related to the change in temperature and to the initial length of the bar. The dependence on the
initial length of the bar comes about because there are that many more molecules moving, so
the change in length will be greater than that of a shorter bar.
But let’s think back to the jar. When you heat the lid, you’re also heating the glass, too. Doesn’t
the glass also expand? The answer is that it does, but it expands less than the material from
which the lid is made. This means that we somehow have to account for the fact that different
materials expand or contract by different amounts under the same temperature change.
Let’s try to arrange these materials on a scale. The way we account for the different rates of
different materials in out equation is via the coefficient of linear expansion, α. α has units of
/°
C (pronounced ‘per degree Celsius’) .

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Linear Expansion
A rod of a substance will increase its length for a given temperature change. The increase in
length depends on the original length of the rod, the tem
the rod. The increase in size of the object comes about by the fact that an increase in
temperature results in a n increase in kinetic energy of the molecules or atoms which make up
the material. Increasing the movement of the molecules forces it to occupy more space.
We define alpha (α), the coefficient of linear expansion. Tables of values of alpha for various
substances are found in handbooks of physics.
The formula is:
∆
∆
∆
∆L = α
α
α
α Lo ∆
∆
∆
∆T
In this formula,
Lo = the original length of the rod
α = the coefficient of linear expansion
∆L = the change in length of the rod
∆T = the change in temperature
Area Expansion
Two-dimensional solid bodies also experience ther
follows:
∆
∆
∆
∆A = 2α
α
α
α Ao
In this formula,
Ao = the original area of the body
α = the coefficient of linear
expansion
∆A = the change in area of the
body
∆T = the change in
temperature
length of the rod, the temperature change, and the material of
ial. Increasing the movement of the molecules forces it to occupy more space.
), the coefficient of linear expansion. Tables of values of alpha for various
stances are found in handbooks of physics.
the original length of the rod
the coefficient of linear expansion
the change in length of the rod
the change in temperature
dimensional solid bodies also experience thermal area expansion. The formula is as
o ∆
∆
∆
∆T
the original area of the body
the coefficient of linear
the change in area of the
Table 3.3: Coefficients of Linear Expansion
perature change, and the material of
ial. Increasing the movement of the molecules forces it to occupy more space.
), the coefficient of linear expansion. Tables of values of alpha for various
expansion. The formula is as
Table 3.3: Coefficients of Linear Expansion

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Volume Expansion
Three-dimensional solid bodies experience volume expansion.
∆
∆
∆
∆V = 3α
α
α
α Vo ∆
∆
∆
∆T
In this formula:
Vo = the original volume of the body
α = the coefficient of linear expansion
∆V = the change in volume of the body
∆T = the change in temperature
Expansion of Liquids and Gases
We have a minor problem with our expression for the thermal expansion of solids, which is that
it only works for solids. Neither liquids nor gases have a fixed shape when left on their own. The
expression also fails if you have to consider the expansion of a solid in all directions.
β is called the coefficient of volume expansion. For solids, β is approximately equal to 3α. This
is true only when the change in volume is small compared to the original volume. The problem
is that for liquids and gases, β is very large and this formula sometimes won’t work.
Liquids also experience thermal
expansion. We introduce beta
(β), the coefficient of volume
expansion. There are also tables
of the coefficients of volume
expansion.
∆V = β Vo ∆T
Generally, liquids expand more than solids, and gases much more than liquids, for any given
change in temperature. This is because the molecules of liquids are not tied to each other and
have more room and freedom to vibrate than do the molecules or atoms in solids. The
molecules of gases of course, are completely free to move, and thus will move much more
vigorously when heated than either solids or liquids
Table 3.4: Coefficients of Volume Expansion

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The Interesting Case of Water
Most materials expand when heated and contract when cooled. Water is an exception. Between
0°
C and 4°
C, water actually expands when cooled. Ab ove this range, it behaves normally.
Water therefore has its greatest density at 4°
C. Th is turns out to be quite important for things
that live underwater. In the winter, you notice that the top of a pond always freezes first. As the
temperature decreases, there is a temperature gradient in the water. The top will be cooler than
the bottom because it is in contact with the cold air. When the water on the top of the lake
reaches 4°
C, it becomes denser and sinks to the bot tom of the lake, being replaced by warmer
water from the bottom. The water that is now on top cools to 4°
C, and so on, until the whole
lake is at 4°
C. The surface water cools even more, but now it is less dense than the water below
it, so it stays on the top of the lake and turns to ice (which is even less dense than cold water). If
the ice sank instead of floating, the lake would freeze all the way through and pretty much
everything inside would die. The layer of ice additionally acts as an insulator, keeping the rest of
the water away from the surface and the colder environment.
EXAMPLE:
A steel rail of length 140 ft. Is laid down when the temperature is 20°
F. What is the increase in
length of this rail when the temperature is 95°
F?
∆L = α Lo ∆T
∆L = (11 x 10-6
/o
F) (140 ft.) (75 o
F)
∆L = 0.116 ft.
EXAMPLE:
An aluminium tank has volume 35 ft.3
What is the increase in volume of this tank when the
temperature increases from 30°
F to 90°
F?
It should be noted that a solid block of a substance increases in volume as the body is heated.
Also, a container has a bigger volume as the temperature of the container increases.
∆V = 3α Vo ∆T
∆V = 3 (13 x 10-6
/o
F) (35 ft.3
) (60 o
F)
∆V = 0.0819 ft.3

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EXAMPLE:
The manager of an airport accepts delivery of 1,000 gallons of avgas on a cool evening when
the temperature is 35°
F. This avgas completely fill s a 1,000 gallon aluminium tank. A warm
front moves in the next morning and the temperature rises to 95°
F. How much avgas will
overflow?
If the avgas costs £1.25/gal. what is the loss to the airport?
For the gasoline:
∆V = β Vo ∆T
∆V = (0.58 x 1 0-3
/o
F) (1,000 gal.) (60 o
F)
∆V = 34.8 gal.
For the tank:
∆V = β Vo ∆T
∆V = 3(13 x 10-6
/o
F) (1,000 gal.) (60 o
F)
∆V = 2.3 gal.
The new volume of the avgas is 1,034.8 gal. and the new volume of the tank is 1,002.3 gal.
We note that 32.5 gallons of avgas will overflow!
Loss £1.25/gal. x 32.5 gal. = £40.63.
EXAMPLE:
A motorist puts 20.1 gallons of petrol in his gas tank on a hot summer day when the
temperature is 95°
F. He uses 0.1 gal. in driving ho me. The temperature falls to 45°
F that
evening after a cool front has moved into the area. How many gallons are in his tank the next
morning when he leaves for work?
∆V = β Vo ∆T
∆V = (0.58 x 10-6
/o
F) (20 gal.) (50 o
F)
∆V = 0.58 gal.
There are 19.42 gallons of petrol in his tank the next morning!

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Problems
1. A 90 ft. aluminium rail is put in place on a hot summer day when the temperature is 85°
F.
What is the decrease in length of this rail when the temperature is 35 °
F?
2. A 150 ft. steel rail is put in place when the temperature is 35°
F. What is the increase in
length of this rail when the temperature is 95°
F?
3. A concrete bridge is laid down in sections with some space between sections to allow for
expansion. The length of one section is 250 ft. The lowest recorded temperature in the
area is - 45 °
F and the highest recorded temperatu re is 115 °
F. How much space should
the builders leave between each section?
4. The volume of an aluminium tank is 200 gallons on a day when the temperature is 30 °
F.
It is completely filled with gasoline from a supply truck. The temperature rises to 70°
F
when a warm front moves in. How many gallons of gasoline overflow?

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Intentionally Blank

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Answers
1. 0.0585 ft
2. 0.099 ft.
3. 0.20 ft
4. 3.7 gallons

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Intentionally Blank

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Heat
We recall that temperature is a measure of the average kinetic energy of molecules or atoms in
a gas, and therefore the average velocity, of the molecules of the substance whose temperature
is being measured.
Heat is a measure of the total energy of molecular motion. The more molecules that are moving,
the greater is the heat energy. Let us compare a teaspoon of water at 100°
F with a cup of water
at 50°
F. The molecules of water in the teaspoon are moving faster than the molecules of water
in the cup. However, since we have so many more molecules in the cup, the heat energy in the
cup is greater than the heat energy in the teaspoon. If the teaspoon of water is placed on a
large block of ice and the cup of water is also placed on this block of ice, the cup of water at
50°
F would melt more ice than the teaspoon of water at 100°
F.
There are definite units for measuring heat energy. The units are the Btu (British thermal unit)
and the metric units, the large Calorie (written with a capital “C”) and the small calorie.
The definitions are:
1 British thermal unit (Btu) = the amount of heat needed to raise the temperature of 1
lb of water 1°
F
1 Calorie = the amount of heat needed to raise the temperature of 1 kilogram of
water 1°
C
(Note: 1 Calorie = 4186 J, 1 Btu = 0.252 Cal.)
1 calorie = the amount of heat needed to raise the temperature of 1 gram of water
1°
C
1 Celsius Heat Unit (CHU) = the amount of heat needed to raise the temperature of 1
lb of water 1°
C
(Note: The CHU is a mix of English and Metric units and is rarely used).
When we talk about the heat content of fuel (which must be burned to be released) –
commonly called the heat of combustion, we talk about Calories per lb. of fuel, or Btu per lb. of
fuel, or Joules per kg of fuel. Since 1 Btu = 252 calories, and 1 calorie = 4.186 Joules, there are
1055 Joules in 1 Btu. And since 1 lb. = 2.2 kg, 1 Btu/lb. = 2326 J/kg.
We note that the Calorie is the famous dietary Calorie. The body stores excess food as fat and
we measure the Calories in a certain foodstuff by burning these foodstuffs and measuring the
heat produced!
In the solution of heat problems, we will limit our discussion to the English system, since this is
the system that is most often encountered in our society.

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As heat is added to a body its temperature increases. However, the same amount of heat
added to a piece of aluminium and a piece of copper will not produce the same temperature
change. Aluminium and copper have different “specific heats”.
The important equation is the following:
Q = wC∆
∆
∆
∆T (when using English units)
In this equation:
Q = heat gained or lost (Btu)
w = weight of the body (lb.)
C = the specific heat of the substance Btu/lb.o
F
∆T = the temperature change (o
F or o
R)
Q = mC∆
∆
∆
∆T (when using Metric units)
In this equation:
Q = heat gained or lost (J)
m = mass of the body (kg)
C = the specific heat of the substance (J/kgo
C)
∆T = the temperature change (o
C or K)
It is important to note that this equation deals with substances that are not changing their states
of matter. Another equation will deal with heat added or lost as a body changes from one state
(solid, liquid, or gas) to another.
Since there are two equations, (depending on whether you are using English or Metric units)
there are also two sets of Specific Heat Capacity constants.
Table 3.5 shows various specific heats of substances in English and Metric units.

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SPECIFIC HEAT CAPACITIES
Liquids J/kg K Btu/lbo
F
Acetic acid 2130 0.51
Alcohol 2930 0.70
Ammonia 470 0.11
Paraffin 2140 0.51
Petroleum 2090 0.50
Turpentine 1980 0.33
Water, fresh 4190 1.00
Water, sea 4o
C 3940 0.93
Metals
Aluminium 912 0.212
Antimony 214 0.051
Copper 389 0.093
Gold 130 0.031
Iron 460 0.110
Lead 130 0.031
Mercury 138 0.033
Nickel 452 0.108
Platinum 134 0.032
Silver 234 0.056
Tin 230 0.055
Zinc 393 0.094
Solids
Asbestos 84 0.20
Ashes 84 0.20
Asphalt 80 0.19
Brick 92 0.22
Carbon 71 0.17
Coal 1310 0.314
Coke 850 0.203
Concrete 1130 0.27
Cork 2030 0.485
Glass 840 0.20
Granite 750 0.18
Graphite 710 0.17
Ice 2110 0.504
Wood 2300 - 2700 0.55 - 0.65
Table 3.5: Specific Heat Capacities of some
common substances

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Gases have a different specific heat capacity depending upon whether they are held at constant
volume or constant pressure. Therefore all gases, as well as having a specific heat capacity,
also has a value of its ratio of CP to Cv.
V
P
C
C
=
γ
For example:
Specific Heat Capacity of Dry Air:
at Constant Pressure,
at Constant Volume,
CP
Cv
1004 J / K kg
717 J / K kg
Therefore
V
P
air
C
C
=
γ = 1.4
Also CP – Cv = R (the ideal gas constant we saw in Chapter 4).
Since both R and γ are always positive values, and greater than 1, CP is always greater than
Cv.
EXAMPLE:
How much heat must be supplied to raise the temperature of a 32 lb. aluminium fitting from
60°
F to 90°
F?
Q = wC∆T (imperial)
)
F
30
.)(
lbs
32
(
.
ft
.
lb
Btu
212
.
0
Q o






=
Q = 204 Btu.
EXAMPLE:
How much heat is given up as 100 lbs. of sea water cools from 90°
F to 50°
F?
Q = wC∆T (imperial)
)
F
40
.)(
lbs
100
(
.
ft
.
lb
Btu
93
.
0
Q o






=
Q = 3720 Btu.

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Heat Exchange
When hot bodies and cool bodies are mixed heat exchange occurs. The heat lost by the hot
body equals the heat gained by the cold body:
Heat Lost = Heat Gained
On each side of this equation there is a wC∆T term. In writing an expression for ∆T, we always
express this change as the larger temperature minus the smaller temperature.
EXAMPLE:
If 5,000 lbs. of sea water at 100°
F are mixed with 7,000 lbs. of ordinary water at 40°
F, what is
the final temperature of the mixture?
We note that, if the final temperature is T, the temperature 100°is more than T and the
temperature 40°is less than T. Therefore the tempe rature change of the sea water is (100 -
T) and the temperature change of the ordinary water is (T - 40).
Heat Lost = Heat Gained
In setting up the wC∆T left and right members of the above equation, we will not include the
units. However we will note that the weights must be in lbs. and the temperature changes in
Fahrenheit degrees.
(0.93) (5,000) (100 - T) = (1.00) (7,000) (T - 40)
465,000 – 4,650 T = 7,000 T – 280,000
745,000 = 11,650 T
T = 63.9 o
F

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Cooling and Heating Curves
When a substance changes phas
the energy, it requires energy to do so. The potential energy stored in the inter
between molecules needs to be overcome by the kinetic energy of the motion of the particles
before the substance can change phase.
If we measure the temperature of the substance which is initially solid as we heat it we produce
a graph like Figure 3.5.
Figure 3.5
changes are indicated by flat regions
energy used to overcome attractive forces between
molecules
Starting a point A, the substance is in its solid phase, heating it brings the temperature up to its
melting point but the material is still a solid at point B. As it is heated further, the energy from the
heat source goes into breaking the bonds holding the atoms in place. This takes place from B to
C. At point C all of the solid phase has been transformed into the liquid phase. Once again, as
energy is added the energy goes into the kinetic energy of the particles ra
(C to D). At point D the temperature has reached its boiling point but it is still in the liquid phase.
From points D to E thermal energy is overcoming the bonds and the particles have enough
kinetic energy to escape from the liquid.
further heating under pressure can raise the temperature still further is how a pressure cooker
works.
Note the temperature stays constant during the state changes of melting and boiling.
When a substance changes phase, that is it goes from either a solid to a liquid or liquid to gas,
the energy, it requires energy to do so. The potential energy stored in the inter
before the substance can change phase.
3.5: Temperature change with time. Phase
changes are indicated by flat regions where heat
molecules
goes into breaking the bonds holding the atoms in place. This takes place from B to
energy is added the energy goes into the kinetic energy of the particles ra
kinetic energy to escape from the liquid. The substance is entering the gas phase. Beyond E,
e, that is it goes from either a solid to a liquid or liquid to gas,
the energy, it requires energy to do so. The potential energy stored in the inter-atomic forces
: Temperature change with time. Phase
where heat
goes into breaking the bonds holding the atoms in place. This takes place from B to
energy is added the energy goes into the kinetic energy of the particles raising the temperature,
The substance is entering the gas phase. Beyond E,

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Note: Since ‘fusion’ (meaning ‘to melt’) is the opposite of ‘solidification’, the Latent Heat of
Fusion is the same as the Latent Heat of Solidification. Also, since ‘vaporisation’ is the opposite
of ‘condensation’, the Latent Heat of Vaporisation is the same as the Latent Heat of
Condensation.
A heating curve summarises the changes:
solid ⇒
⇒
⇒
⇒ liquid ⇒
⇒
⇒
⇒ gas
The principle of latent heat (especially of vaporization) is what is behind the operation of the
fridge and air conditioning system, water injection of gas turbine engines, and the cooling effect
you feel when you perspire.
That principle is that if you make a fluid vaporize, it extracts heat (latent heat) to cause it to
vaporize, but the fluid does not change temperature.
Latent Heat of Fusion and Vaporisation
The energy required to change the phase of a substance is known as a latent heat. The word
latent means hidden. When the phase change is from solid to liquid we must use the latent heat
of fusion, and when the phase change is from liquid to a gas, we must use the latent heat of
vaporisation.
The latent heat energy required is given by the formula:
Q= m L
where m is the mass of the substance and L is the specific latent heat of fusion or vaporisation
which measures the heat energy to change 1 kg of a solid into a liquid.
Some values of Specific Latent Heats of Fusion and Vaporisation are shown in table 3.6.
Substance
Specific
latent heat of
fusion kJ/kg
Freezing
Temperature
°
C
Specific
latent heat
of
vaporisation
kJ/kg
Boiling
Temperature
°
C
Water 334 0 2258 100
Ethanol 109 -114 838 78
Chloroform 74 -64 254 62
Mercury 11 -39 294 357
Sulphur 54 115 1406 445
Hydrogen 60 -259 449 -253
Oxygen 14 -219 213 -183
Nitrogen 25 -210 199 -196
Table 3.6: Latent heats, freezing points and boiling points of some common
substances

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Further Discussion on Latent Heat
Each time water changes physical state, energy is involved.
In the vapour state, the water molecules are very energetic. The
each other, but move around as single molecules. Water vapour is invisible to us, but we can
feel its effect to some extent, and water vapour in the atmosphere is a very important factor in
weather and climate.
In the liquid state, the individual molecules have less energy, and some bonds form, break, then
re-form. At the surface of liquid water, molecules are continually moving back and forth from the
liquid state to the vapour state. At a given temperature, there will be an e
number of molecules leaving the liquid, and
In solid water--ice--the molecules are locked together in a crystal structure:
a framework. They are not moving around, and they contain less energy.
How do you make water evaporate? Here is a bowl of water.
water evaporate. Go ahead.
How did you make the water evaporate?
might have put it out in the sun, or possibly put it over a fire.
water evaporate, you put energy into it. The individual molecules in the water absorb that
energy, and get so energetic that they break the hydrogen bonds connecting them to other
water molecules. They become molecules of water vapour. Evaporation is the change of state
from liquid to vapour. In the process of evaporation, the molecule absorbs energy. This energy
is latent heat. Latent means hidden, so latent heat is "hidden" in the water molecule
feel it, but it is there. Wherever that individual molecule of
latent heat with it. To get the molecule of water vapour to become liquid again, we have to take
the energy away, that is, we have
change from the vapour state to
it releases latent heat.
Now, how do you make ice melt?
state. Make it melt. Go ahead.
Again, you probably melted the ice by adding energy.
energy was absorbed by the individual molecules of water, which became
so energetic that they broke some of the hydrogen bonds holding the ice crystal together, and
became liquid (that is, the ice melted).
liquid water is holding that latent heat.
that latent heat away, or in other words, cool the water.
Water could change directly from the frozen state to the vapour state without passing through
the liquid state first. This process is called sublimation.
vapour state to the frozen state without passing through the liquid state.
deposition, and is what you see when frost forms on grass or wind
(Sometimes the term sublimation is used
from solid to vapour, or vapour to solid).
Further Discussion on Latent Heat
Each time water changes physical state, energy is involved.
In the vapour state, the water molecules are very energetic. The molecules are not bonded with
tate, the individual molecules have less energy, and some bonds form, break, then
form. At the surface of liquid water, molecules are continually moving back and forth from the
liquid state to the vapour state. At a given temperature, there will be an equilibrium between the
number of molecules leaving the liquid, and the number of molecules returning.
the molecules are locked together in a crystal structure:
a framework. They are not moving around, and they contain less energy.
w do you make water evaporate? Here is a bowl of water. Make the
How did you make the water evaporate? Probably you added heat. You
might have put it out in the sun, or possibly put it over a fire. To make
you put energy into it. The individual molecules in the water absorb that
They become molecules of water vapour. Evaporation is the change of state
In the process of evaporation, the molecule absorbs energy. This energy
Latent means hidden, so latent heat is "hidden" in the water molecule
Wherever that individual molecule of water vapour goes, it takes that
To get the molecule of water vapour to become liquid again, we have to take
the energy away, that is, we have to cool it down so that it condenses (condensation is the
the liquid state). When water condenses,
Now, how do you make ice melt? Here is a block of ice, water in the solid
Again, you probably melted the ice by adding energy. The additional
was absorbed by the individual molecules of water, which became
became liquid (that is, the ice melted). This energy is also latent heat, and each molecule of th
liquid water is holding that latent heat. To change the liquid water back to ice, you have to take
that latent heat away, or in other words, cool the water.
This process is called sublimation. Water can also change from the
vapour state to the frozen state without passing through the liquid state. This is usually called
deposition, and is what you see when frost forms on grass or windows on a cold night.
(Sometimes the term sublimation is used when water changes state in either direction, that is,
from solid to vapour, or vapour to solid).
molecules are not bonded with
tate, the individual molecules have less energy, and some bonds form, break, then
form. At the surface of liquid water, molecules are continually moving back and forth from the
quilibrium between the
the number of molecules returning.
the molecules are locked together in a crystal structure:
You
you put energy into it. The individual molecules in the water absorb that
They become molecules of water vapour. Evaporation is the change of state
In the process of evaporation, the molecule absorbs energy. This energy
Latent means hidden, so latent heat is "hidden" in the water molecule--we can't
water vapour goes, it takes that
To get the molecule of water vapour to become liquid again, we have to take
to cool it down so that it condenses (condensation is the
When water condenses,
Here is a block of ice, water in the solid
was absorbed by the individual molecules of water, which became
This energy is also latent heat, and each molecule of the
To change the liquid water back to ice, you have to take
Water can also change from the
This is usually called
ows on a cold night.
when water changes state in either direction, that is,

The really important thing to remember is that each time water changes state, energy is
absorbed or released. This energy is latent heat.
released when a substance changes its physical state. Latent heat is absorbed upon
evaporation, and released upon condensation to liquid (as in clouds). Latent heat is also
absorbed when water melts, and released when it freezes.
How much heat does it take to get water to change state?
100 degrees C (that is, the boiling point, or 212 degrees F) it takes an additional 540 calories of
heat to convert one gram of water from the liquid state to the vapour state.
converts to the liquid state, 540 calories of energy will be released per gram of water. If you are
converting solid water (ice) to liquid water at 0 degrees C, it w
heat to melt one gram of ice, and the 80 calories will be released when the liquid water is frozen
to the solid state.
Water does not have to be at the boiling point to evaporate.
of water out in the sun and watch it slowly disappear.
but it is evaporating it. in a given amount of water at a
water will have more energy than others, so some molecules will be able to evaporate, while
others remain in the liquid state.
required for evaporation. if the water is liquid at a temperature of 0 degrees C, the latent heat of
vaporization is 597 cal/g, compared to 540 cal/g at 100 degrees C.
C, an input of 569 cal/g would be required for evaporation.
It will take a total of about 720 calories per gram to
from ice at 0 degrees C, to vapour at 100 degrees C: this includes 80 calories from latent heat
of fusion (melting) + 100 calories to raise the temperature of the water 100 degrees C + 540
calories to make the liquid water evaporate (latent heat of vaporization).
Figure 3.6: Ice, water and water
ed or released. This energy is latent heat. Latent heat is the energy absorbed or
absorbed when water melts, and released when it freezes.
How much heat does it take to get water to change state? if the water is at a temperature of
t to convert one gram of water from the liquid state to the vapour state.
converting solid water (ice) to liquid water at 0 degrees C, it will require about 80 calories of
Water does not have to be at the boiling point to evaporate. if you don't believe this, set a pan
of water out in the sun and watch it slowly disappear. The sun's heat is not boiling the water,
in a given amount of water at a given temperature, some molecules of
The lower the temperature of the water, the more energy is
e water is liquid at a temperature of 0 degrees C, the latent heat of
vaporization is 597 cal/g, compared to 540 cal/g at 100 degrees C. in between, at 50 degrees
C, an input of 569 cal/g would be required for evaporation.
0 calories per gram to sublimate water, that is change it directly
calories to make the liquid water evaporate (latent heat of vaporization). Similarly, about 720
Figure 3.6: Ice, water and water-vapour
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Latent heat is the energy absorbed or
if the water is at a temperature of
t to convert one gram of water from the liquid state to the vapour state. When the vapour
ill require about 80 calories of
if you don't believe this, set a pan
The sun's heat is not boiling the water,
given temperature, some molecules of
The lower the temperature of the water, the more energy is
e water is liquid at a temperature of 0 degrees C, the latent heat of
in between, at 50 degrees
water, that is change it directly
Similarly, about 720

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© Copyright 2012
calories per gram will be released when water is changed directly from vapour to ice, the
process called deposition.
Methods of Heat Transfer
Heat can be transferred from one place to another by one or more of the following processes:
Convection is the transfer of heat by the actual
movement of the warmed matter. Heat leaves a coffee
cup as the currents of steam and air rise. Convection
is the transfer of heat energy in a gas or liquid by
movement of currents. Think of air and water currents!
(it can also happen is some solids, like sand.) The
heat moves with the fluid. Consider this: convection is
responsible for making macaroni rise and fall in a pot
of heated water. The warmer portions of the water are
less dense and therefore, they rise. Meanwhile, the
cooler portions of the water fall because they are
denser.
Conduction is the transfer of energy
through matter from particle to particle. It is
the transfer and distribution of heat energy
from atom to atom within a substance. For
example, a spoon in a cup of hot soup
becomes warmer because the heat from the
soup is conducted along the spoon.
Conduction is most effective in solids-but it
can happen in fluids. Have you ever noticed
that metals tend to feel cold? Believe it or
not, they are not colder! They only feel
colder because they conduct heat away from
your hand. You perceive the heat that is
leaving your hand as cold.
Radiation: Electromagnetic waves that directly transport ENERGY through space. Sunlight is a
form of radiation that is radiated through space to our planet at the speed of light without the aid
of fluids or solids. The energy travels through nothingness! Just think of it! The sun transfers
heat through 93 million miles of space. Because there are no solids (like a huge spoon)
touching the sun and our planet, conduction is not responsible for bringing heat to Earth. Since
there are no fluids (like air and water) in space, convection is not responsible for transferring the
heat. Thus, radiation brings heat to our planet.
Figure 3.7: Heat - convection
Figure 3.8: Heat - conduction

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Heat Transfer
We know that heat flows through insulating materials from the warm side to the cool side. It is
possible to predict how many Btu will flow through a given insulator in a given amount of time.
The equation is:
L
T
kA
t
Q ∆
=
The equation is less difficult than it
seems at first. We will carefully
define each symbol.
Q = heat flow in Btu
t = time in hours
A = the surface area of the
insulation in square feet
∆T = the temperature difference
in o
F
L = the thickness of the
insulation in inches
k = the thermal conductivity of the
material from which the insulation is made
EXAMPLE:
An outside wall of a house has total cross-sectional area of 2,000 ft.2
The thickness of the
fibreboard insulation is 3 inches. The inside temperature is 70°
F and the outside
temperature is 20°
F. What is the heat loss per hour through this outside wall?
L
T
kA
t
Q ∆
=
in
3
)
F
50
)(
ft
000
,
2
)(
F
hr
ft
/
.
in
Btu
42
.
0
(
t
Q o
2
=
hr
/
Btu
000
,
14
t
Q
=
Table 3.7: Thermal conductivities of some
common materials

3-49
© Copyright 2012
Problems
1. How much heat must be supplied to raise the temperature of 67 lbs. of ethyl alcohol
from 32°
F to 76 o
F?
2. How much heat is given up as 780 lbs. of steel cool from 90°
F to 45°
F?
3. If 1 lb. of vodka (alcohol) at 90°
F is mixed wit h 0.2 lb. of water at 40°
F what is the final
temperature?
4. If 3 lbs. of hot water at 200 °
F are poured into a 1.5 lbs. aluminium container at 40 °
F,
what is the final temperature?
5. A house has an outside wall area of 3,000 ft2
These walls are insulated with corkboard
4 in. thick. The inside temperature is 75°
F and the outside temperature is 15°
F. What is
the heat loss per hour through these outside walls?

3-53
© Copyright 2012
Refrigeration and Heat Pumps
First of all, did you know that there is no such thing as cold? You can describe something as
cold and everyone will know what you mean, but cold really only means that something contains
less heat than something else. All there really is, is greater and lesser amounts of heat. The
definition of refrigeration is The Removal and Relocation of Heat. So if something is to be
refrigerated, it is to have heat removed from it. If you have a warm can of pop at say 80 degrees
Fahrenheit and you would prefer to drink it at 40 degrees, you could place it in your fridge for a
while, heat would somehow be removed from it, and you could eventually enjoy a less warm
pop. (oh, all right, a cold pop.) But lets say you placed that 40 degree pop in the freezer for a
while and when you removed it, it was at 35 degrees. Even "cold" objects have heat content that
can be reduced to a state of "less heat content". The limit to this process would be to remove all
heat from an object. This would occur if an object was cooled to Absolute Zero which is -273º C
or -460º F. They come close to creating this temperature under laboratory conditions and
strange things like electrical superconductivity occur
How do things get colder?
Things get cold because they lose heat by one or more of the following methods:
• Radiation
• Conduction
• Convection
The latter two are used extensively in the design of refrigeration equipment. If you place two
objects together so that they remain touching, and one is hot and one is cold, heat will flow from
the hot object into the cold object. This is called conduction. This is an easy concept to grasp
and is rather like gravitational potential, where a ball will try to roll down an inclined plane. If you
were to fan a hot plate of food it would cool somewhat. Some of the heat from the food would be
carried away by the air molecules. When heat is transferred by a substance in the gaseous
state the process is called convection. And if you kicked a glowing hot ember away from a
bonfire, and you watched it glowing dimmer and dimmer, it is cooling itself by radiating heat
away. Note that an object doesn’t have to be glowing in order to radiate heat, all things use
combinations of these methods to come to equilibrium with their surroundings. So you can see
that in order to refrigerate something, we must find a way to expose our object to something that
is colder than itself and nature will take over from there. We are getting closer to talking about
the actual mechanics of a refrigerating system, but there are some other important concepts to
discuss first.
The States of Matter
Figure 3.9: Summary of radiation, conduction and convection

3-54
ST Aerospace Ltd
© Copyright 2012
They are of course; solid, liquid and gas. It is important to note that heat must be added to a
substance to make it change state from solid to liquid and from liquid to a gas. It is just as
important to note that heat must be removed from a sub
gas to a liquid and from a liquid to a solid.
The Magic of Latent Heat
Long ago it was found that we needed a way to quantify heat.
Something more precise than "less hea
deal of heat" was required. This was a fairly easy task to accomplish.
They took 1 Lb. of water and heated it 1 degree Fahrenheit. The
amount of heat that was required to do this was called 1 BTU
Thermal Unit).
definition. You can for example purchase a 6000 BTU/H window air
conditioner. This would be a unit that is capable of relocating 6000
BTU's of heat per hour. A larger unit capable of 12,000 BTU/H could
also be called a One Ton unit. There are 12,000 BTU's in 1 Ton.
To raise the temperature of 1 LB of water from 40 degrees to 41 degrees would take 1 BTU. To
raise the temperature of 1 LB of water from 177 degrees to 178 degrees would also take 1 BTU.
However, if you tried raising the temperature of water from 212 degrees to 213 degrees you
would not be able to do it. Water boils at 212 degrees and would prefer to change into a gas
rather than let you get it any hotter. Something of utmost importance occurs at
of a substance. If you did a little experiment and added 1 BTU of heat at a time to 1 LB of water,
you would notice that the water temperature would increase by 1 degree each time. That is until
you reached 212 degrees. Then something ch
water would not get any hotter! it would change state into a gas and it would take 970 BTU's to
vaporize that pound of water. This is called the
water it is 970 BTU's per pound.
So what! you say. When are you going to tell me how the refrigeration effect works? Well hang
in there, you have just learned about 3/4 of what you need to know to understand the process.
What keeps that beaker of water from boiling when it i
because it is not hot enough, sorry but you are wrong. The only thing that keeps it from boiling
is the pressure of the air molecules pressing down on the surface of the water. When you heat
that water to 212 degrees and then continue to add heat, what you are doing is supplying
sufficient energy to the water molecules to overcome the pressure of the air and allow them to
escape from the liquid state. If you took that beaker of water to outer space where there is no ai
pressure the water would flash into a vapour. If you took that beaker of water to the top of Mt.
Everest where there is much less air pressure, you would find that much less heat would be
needed to boil the water. (it would boil at a lower temperature th
at 212 degrees at normal atmospheric pressure. Lower the pressure and you lower the boiling
point. Therefore we should be able to place that beaker of water under a bell jar and have a
vacuum pump extract the air from within
room temperature. This is indeed the case!
Figure 3.10: Raising
1lb of by water 1o
F
important to note that heat must be removed from a substance to make it change state from a
gas to a liquid and from a liquid to a solid.
Something more precise than "less heat" or "more heat" or "a great
amount of heat that was required to do this was called 1 BTU
Thermal Unit). The refrigeration industry has long since utilized this
be called a One Ton unit. There are 12,000 BTU's in 1 Ton.
, if you tried raising the temperature of water from 212 degrees to 213 degrees you
rather than let you get it any hotter. Something of utmost importance occurs at
you reached 212 degrees. Then something changes. You would keep adding BTU's, but the
vaporize that pound of water. This is called the Latent Heat of Vaporization
s per pound.
What keeps that beaker of water from boiling when it is at room temperature? if you say it's
and then continue to add heat, what you are doing is supplying
escape from the liquid state. If you took that beaker of water to outer space where there is no ai
needed to boil the water. (it would boil at a lower temperature than 212 degrees). So water boils
vacuum pump extract the air from within the bell jar and watch the water come to a boil even at
room temperature. This is indeed the case!
stance to make it change state from a
t" or "more heat" or "a great
amount of heat that was required to do this was called 1 BTU (British
frigeration industry has long since utilized this
be called a One Ton unit. There are 12,000 BTU's in 1 Ton.
, if you tried raising the temperature of water from 212 degrees to 213 degrees you
rather than let you get it any hotter. Something of utmost importance occurs at the boiling point
anges. You would keep adding BTU's, but the
Latent Heat of Vaporization and in the case of
s at room temperature? if you say it's
and then continue to add heat, what you are doing is supplying
escape from the liquid state. If you took that beaker of water to outer space where there is no air
an 212 degrees). So water boils
the bell jar and watch the water come to a boil even at

A liquid requires heat to be added to it in order for it to overcome the air pressure pressing down
on its' surface if it is to evaporate into a gas. We ju
liquids surface is reduced it will evaporate easier. We could look at it from a slightly different
angle and say that when a liquid evaporates it absorbs heat from the surrounding area. So,
finding some fluid that evaporates at a handier boiling point than water (i.e. lower) was one of
the first steps required for the development of mechanical refrigeration.
Chemical Engineers spent years experimenting before they came up with the perfect chemicals
for the job. They developed a family of hydroflourocarbon refrigerants which had extremely low
boiling points. These chemicals would boil at temperatures below 0 degrees Fahrenheit at
atmospheric pressure. So finally, we can begin to describe the mechanical refrigeration
process.
There are 4 main components in a mechanical refrigeration system. Any components beyond
these basic 4 are called accessories. The compressor is a vapour compression pump which
uses pistons or some other method to compress the refrigerant gas and send it on it's way to
the condenser. The condenser is a heat exchanger which removes heat from the hot
compressed gas and allows it to condense into a liquid. The liquid refrigerant is then routed to
the metering device. This device restricts the flow by forcing the refrigerant to go through a
small hole which causes a pressure drop. And what did we say happens to a l
pressure drops? if you said it lowers the boiling point and makes it easier to evaporate, then you
are correct. And what happens when a liquid evaporates? Didn't we agree that the liquid will
absorb heat from the surrounding area? This is in
refrigeration works. This component where the evaporation takes place is called the evaporator.
The refrigerant is then routed back to the compressor to complete the cycle. The refrigerant is
used over and over again absorbing heat from one area and relocating it to another. Remember
the definition of refrigeration? (The removal and relocation of heat
(Accepts heat)
(Low Pressure)
Figure 3.11: Main components of a refrigeration system
on its' surface if it is to evaporate into a gas. We just learned that if the pressure above the
evaporates at a handier boiling point than water (i.e. lower) was one of
the first steps required for the development of mechanical refrigeration.
y developed a family of hydroflourocarbon refrigerants which had extremely low
called accessories. The compressor is a vapour compression pump which
gas and allows it to condense into a liquid. The liquid refrigerant is then routed to
small hole which causes a pressure drop. And what did we say happens to a l
absorb heat from the surrounding area? This is indeed the case and you now know how
rbing heat from one area and relocating it to another. Remember
he removal and relocation of heat.)
(Rejects heat)
(High Pressure)
3-55
ST Aerospace Ltd
© Copyright 2012
st learned that if the pressure above the
evaporates at a handier boiling point than water (i.e. lower) was one of
y developed a family of hydroflourocarbon refrigerants which had extremely low
called accessories. The compressor is a vapour compression pump which
gas and allows it to condense into a liquid. The liquid refrigerant is then routed to
small hole which causes a pressure drop. And what did we say happens to a liquid when the
deed the case and you now know how
rbing heat from one area and relocating it to another. Remember
(Rejects heat)
(High Pressure)

ST Aerospace Ltd
© Copyright 2012
So what is a refrigerant?
Remember that a refrigerant, in order to cool the space, must evaporate at the temperature of
the space. So in the case of an air conditioning unit, we need a chemical which will evaporate
(boil) at the temperature of the room you are trying to cool. In the case of a fridge, you need a
chemical which will boil at the temperature of the inside of the fridge. There are hundreds of
chemicals which will do this. Several of the commonest are as follows:
Refrigerant Formula
Boiling
temperature
o
C
Properties Applications
Ammonia NH3 -33
Penetrating odour,
soluble in water.
harmless in concentration
up to 1/30%, non
flammable, explosive
Large industrial plants
R12
C Cl2F2
Chlorodi-
fluoromethane
-29.8
Little odour, colourless as
gas or liquid, non
flammable, non corrosive
of ordinary metals, stable
Small plants with
reciprocating
compressors.
R11 C Cl3F 8.9
Non flammable, non
corrosive non toxic,
stable
Commercial plants with
centrifugal compressors.
R22 CH Cl F2 -40.8
Little odour, colourless as
gas or liquid, non toxic,
non irritating, non
flammable, non
corrosive, stable
Packaged air
conditioning units where
size of equipment and
economy are important.
R500
C Cl2 F2
(73.8%)
CH3CH F2
(26.2%)
-33 Similar to R12
Offers approx. 20%
more refrigeration
capacity than R12 for
same compressor.
R502
C Cl F2 (48.8%)
C Cl
F2 - CF3 (51.2%)
-45.6
Non flammable, non
toxic, non corrosive,
stable
Capacity comparable to
R22.
R134a
F3CCH2F
Tetrafluoro-
ethane
-26.6
Non flammable, non
toxic, non corrosive,
stable
Totally replaces other
Freon types for auto and
aircraft applications
Table 3.8: Some common refrigerants

3-57
© Copyright 2012
The range of refrigerants beginning with the ‘R’ prefix (R12, R22 etc.) are complex compounds
of fluorocarbons or hydroflourocarbons collectively known as ‘Freon’.
Other chemicals could be used and have been used in the past. Methyl bromide is a refrigerant
but has almost completely been phased out for safety and environmental reasons. Carbon
dioxide (dry ice) has also been used as a refrigerant (infact at one time Carbon Dioxide and
Ammonia were the only two refrigerants in use) but is no longer used because other complex
formulas are much more efficient.
So is water a refrigerant?
Remember that a refrigerant, in order to cool the space, must evaporate at the temperature of
the space. Unless the room you are trying to cool, or the inside of your fridge, is 100oC or
more, water will not work as a refrigerant at normal atmospheric pressure. You would have to
reduce the pressure to almost nothing before water will boil at 0o
C (0.089 PSI to be exact). This
is impractical. However, if you want to cool something which is very hot (i.e. over 100oC) like
the rods of a nuclear reactor, then pouring water on it will do the job. In turning to steam
(evaporating) the rods are cooled. This is how cooling towers work – water is poured onto the
hot object at the base of the cooling tower. The water turns to steam and rises (because water
vapour is lighter than air) and expands, which forces it to cool (Charles’ Law). The cooled
steam condenses into water which runs down the inside of the cooling tower and back onto the
hot object where the process repeats. So you have water, as a refrigerant, cooling an object in
much the same way as Freon or Ammonia cools the interior of your car, office or fridge.
But also, water can vaporise at temperatures below 100o
C, thus missing out the boiling process.
Hence water which exudes from the pores of your skin when you are hot (sweat) evaporates,
and in changing state absorbs the heat from your skin, and cools you. Again water is a
refrigerant. However, to be efficient, it usually helps to have a breeze over the skin to help the
evaporation process.
Heat Pumps
A heat pump is a machine or device that moves heat from one location (the 'source') to another
location (the 'sink' or 'heat sink'), using work. Most heat pump technology moves heat from a
low temperature heat source to a higher temperature heat sink. Common examples are food
refrigerators and freezers and air conditioners and reversible-cycle heat pumps for providing
thermal comfort.
Heat pumps can be thought of as a heat engine which is operating in reverse. One common
type of heat pump works by exploiting the physical properties of an evaporating and condensing
a refrigerant. In heating, ventilation, and cooling (HVAC) applications, a heat pump normally
refers to a vapour-compression refrigeration device that includes a reversing valve and
optimized heat exchangers so that the direction of heat flow may be reversed. Most commonly,
heat pumps draw heat from the air or from the ground. Air-source heat pumps do not work well
when temperatures fall below around −5°
C (23°
F).
According to the second law of thermodynamics heat cannot spontaneously flow from a colder
location to a hotter area; work is required to achieve this. Heat pumps differ in how they apply
this work to move heat, but they can essentially be thought of as heat engines operating in
reverse. A heat engine allows energy to flow from a hot 'source' to a cold heat 'sink', extracting a

3-58
ST Aerospace Ltd
© Copyright 2012
fraction of it as work in the process. Conversely, a heat pump requires work to move thermal
energy from a cold source to a warmer heat sink.
Since the heat pump uses a certain amount of work to move the heat, the amount of energy
deposited at the hot side is greater than the energy taken from the cold side by an amount
equal to the work required. Conversely, for a heat engine, the amount of energy taken from the
hot side is greater than the amount of energy deposited in the cold heat sin
heat has been converted to work.
Figure 3.12. A heat pump's vapo
cycle: 1) condenser, 2)
4) compressor.
The working fluid, in its gaseous state, is pressurized and circula
compressor. On the discharge side of the compressor, the now hot and highly pressurized gas
is cooled in a heat exchanger called a condenser until it condenses into a high pressure,
moderate temperature liquid. The condensed ref
lowering device like an expansion valve, capillary tube, or possibly a work
such as a turbine. This device then passes the low pressure, barely liquid (saturated vapo
refrigerant to another heat exchanger, the evaporator where the refrigerant evaporates into a
gas via heat absorption. The refrigerant then returns to the compressor and the cycle is
repeated.
In such a system it is essential that the refrigerant reaches a sufficiently high temperat
compressed, since the second law of thermodynamics prevents heat from flowing from a cold
fluid to a hot heat sink. Similarly, the fluid must reach a sufficiently low temperature when
allowed to expand, or heat cannot flow from the cold region int
pressure difference must be great enough for the fluid to condense at the hot side and still
evaporate in the lower pressure region at the cold side. The greater the temperature difference,
the greater the required pressure
compress the fluid. Thus as with all heat pumps, the energy efficiency (amount of heat moved
per unit of input work required) decreases with increasing temperature difference. Thus a
ground-source heat pump, which has a very small temperature differential, is relatively efficient.
(Figures of 75% and above are quoted.)
energy from a cold source to a warmer heat sink.
ted at the hot side is greater than the energy taken from the cold side by an amount
hot side is greater than the amount of energy deposited in the cold heat sin
heat has been converted to work.
A heat pump's vapour-compression refrigeration
condenser, 2) expansion valve, 3) evaporator,
The working fluid, in its gaseous state, is pressurized and circulated through the system by a
moderate temperature liquid. The condensed refrigerant then passes through a pressure
lowering device like an expansion valve, capillary tube, or possibly a work
such as a turbine. This device then passes the low pressure, barely liquid (saturated vapo
exchanger, the evaporator where the refrigerant evaporates into a
In such a system it is essential that the refrigerant reaches a sufficiently high temperat
allowed to expand, or heat cannot flow from the cold region into the fluid. In particular, the
e greater the required pressure difference, and consequently more energy is needed to
pump, which has a very small temperature differential, is relatively efficient.
(Figures of 75% and above are quoted.)
ted at the hot side is greater than the energy taken from the cold side by an amount
hot side is greater than the amount of energy deposited in the cold heat sink since some of the
compression refrigeration
ted through the system by a
rigerant then passes through a pressure-
lowering device like an expansion valve, capillary tube, or possibly a work-extracting device
such as a turbine. This device then passes the low pressure, barely liquid (saturated vapour)
exchanger, the evaporator where the refrigerant evaporates into a
In such a system it is essential that the refrigerant reaches a sufficiently high temperature when
o the fluid. In particular, the
difference, and consequently more energy is needed to
pump, which has a very small temperature differential, is relatively efficient.

Thermodynamics Module 2.3 Key Concepts

Thermodynamics Module 2.3 Key Concepts

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