Engineering Thermodynamics Lecture Notes Chapter 1 (Draft )

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Engineering Thermodynamics
Lecture Notes
Chapter 1
(Draft )
Wayne Hacker
Copyright ©Wayne Hacker 2009.All rights reserved.
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.1
Contents
1 Introductory concepts,denitions,and principles 2
1.1 What is thermodynamics?..........................2
1.1.1 Articial subdivisions of thermodynamics..............3
1.1.2 Engineering Thermodynamics....................4
1.2 What is Energy?...............................5
1.3 Dening thermodynamic systems......................6
1.3.1 What is a system?..........................6
1.3.2 The continuum hypothesis......................9
1.3.3 Property,state,and process.....................9
1.3.4 Extensive and intensive properties..................11
1.3.5 Equilibrium and quasi-equilibrium states..............12
1.4 Physical quantities,dimensions,and units.................13
1.4.1 Dimensional consistency.......................14
1.4.2 Units..................................16
1.5 Density,specic volume,and specic weight................19
1.6 Pressure....................................20
1.7 Temperature..................................20
1.7.1 Conversion formulas.........................21
1.8 Static uids..................................22
1.9 Introduction to the 4 laws of thermodynamics...............22
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.2
1 Introductory concepts,denitions,and principles
About the Notes
The purpose of the exercises
No set of class notes would be complete without a set of exercises to accompany it.I view
the exercises as mental exercises in thermodynamics.You would not dare set foot on a
football eld,wrestling mat,or track for a competitive event without doing exercises on
your own time.You would then go to every practice (class lecture) before the big game
(the exams).While this analogy is not perfect,you get my point.
Disclaimer
This set of notes are not nished numerous topics are not fully developed.I am still new
at teaching thermodynamics and need a summer to think out my notes more carefully.
Unfortunately,my home department is physics and the needs of the physics department
take precedence over this class.That is to say,with my other class teaching obligations,
I simply don't have enough spare time in my life at this point to make this set of notes
into something good.Think of these notes as a random set of collected ramblings,a
screed.
1.1 What is thermodynamics?
Crudely,thermodynamics is the study of interaction between matter and energy.But this
is almost a vacuous statement since it hard to imagine any event in nature that does not
deal with the interaction between energy and matter.Today thermodynamics is a very
broad eld with applications in every area of science from biology to physics.However,it
wasn't always this way.The eld had humble beginnings as inventors struggled to gure
out how to turn heat energy into a form of energy that can do mechanical work.
Thermodynamics did not emerge as a science until the late 1600 hundreds (and some
would say much later) in the study of the rst steam engines.The name comes from
the Greek words Therme (heat) and dynamis (dynamite/power),and was rst used by
Lord Kelvin (formerly William Thomson) in an 1849 publication.The rst textbook was
written in 1859 by William Rankine,for whom the Rankine temperature scale is named
after.
Although most people associate thermodynamics as the study of heat,it has a long and
interesting history with the study of both heat and lack of heat (cold).One of the main
areas of research in modern thermodynamics is the study of cold.In fact,it could be
argued that Michael Faraday,the famous 19
th
century scientist,was one of the original
founders of man-made refrigerators.It was his experiments (and failures) with ammonia
that lead him to realize that by compressing and releasing gas in the right order,that
you could cool the surroundings.He even went so far as to suggest ways that compressed
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.3
gas could be used to cool food and other perishables and how developing this technology
would have great potential industrial applications.Fortunately,he was far too interested
in science to waist time on such a mundane endeavor.
From a practical engineering point of view,thermodynamics is the study of the transfor-
mation process of converting work into heat and heat back into work,and the eciency
with which these two processes can occur.
1.1.1 Articial subdivisions of thermodynamics
There are two basic areas of thermodynamics:microscopic and macroscopic.
• Microscopic study:This area focuses on microscopic issues.Its main focus is the
structure of matter and the interaction of molecules in collisions.It is known as
statistical physics,or statistical thermodynamics
1
.Topics studied in this discipline
include,but are not limited to:lasers,plasmas,high-speed gases,raried gases
(such as in the outer atmosphere),turbulence in uids.
• Macroscopic study:This area focuses on macroscopic (bulk) energy ow.It is typ-
ically referred to as Classical Thermodynamics,or Engineering Thermodynamics.
This eld focuses on
{ the transfer of energy
{ the transformation of energy
{ the storage of energy
We will not study the details of molecular motion in this course;however,we will not
ignore them either.While it is true that thermodynamics has its historic roots in exper-
imentation and observation (the macroscopic world),it is also a fact that modern-day
thermodynamics owes much of its success to the study of the behavior of the molecules
composing the matter under scrutiny.That said,the macroscopic study is theoretically
justied by the continuum hypothesis,an assumption used to avoid direct dealings with
molecular interactions.Mathematically,the continuumhypothesis says that if we started
with a xed amount of matter and kept cutting it in half,then the parts would look just
like the whole.We know that this is not the case.There is a limit where if we go below
it,we will be at the molecular level with lots of empty space.It is at these small length
scales that we need the methods of statistical physics to justify the continuumhypothesis.
We will try to balance the these two approaches by avoiding the technical mathematics
associated with statistical physics,but at the same time not forgetting that all matter
is composed of atoms and molecules,and action of these molecules necessarily aects
the behavior of the substance.Moreover,without studying the molecular behavior we
cannot understand the concepts behind internal energy.We will therefore be forced to
walk a proverbial tightrope between the macroscopic and microscopic worlds.Be careful,
don't fall o!
1
A common sub-division of statistical physics used by engineers is known as kinetic theory
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.4
1.1.2 Engineering Thermodynamics
Engineering thermodynamics,as a discipline,can be subdivided into three broad areas
of study:
• the transformation of energy from one form into another
(e.g.,a car skids to a stop )kinetic energy
work
!heat)
• the transfer of energy across boundaries
(heat + piston-cylinder container of water
work
!move piston)
• the storage of energy in molecules (heat,internal energy)
Of course,in practice these studies overlap.
The engineer's objective in studying thermodynamics is the analysis and design of large-
scale systems such as power plants,solar farms,air-conditioning systems,etc.Some
examples of areas of interest involving thermodynamics for engineers are listed below
• large-scale energy storage
• how energy is transferred through heat and work
• how energy transforms from one form of energy into another
(e.g.,heat!mechanical work)
• the economic impact of various materials used for heat insulators and conductors
Example 1.Which of the following situations are related to the main focus of the study
of thermodynamics.
(a) storing the energy made by a windmill
(b) converting the energy stored in coal into work via a steam engine
(c) determining the energy loss due to friction in the moving parts of an engine
(d) how energy is bought and sold on the open market
Solution:Statements (a)-(c) are activities that engineers participate in.
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.5
1.2 What is Energy?
Since thermodynamics involves the study of energy we should start with a little intro-
duction to energy.Energy comes in many forms.It can be thought of as\the ability to
cause change".Here are a few common forms of energy:
• kinetic energy [energy due to motion]
• potential energy [energy due to position,or the conguration of a system]
• internal energy [energy stored in the molecules{both potential and kinetic{associated
with temperature]
• chemical energy [energy due to chemical composition (i.e.,energy due to the ar-
rangement of molecules with respect to other molecules in the system)]
• nuclear energy [subatomic energy (i.e,energy stored in the nucleus of an atom)]
In order for energy to change from one form to another it must go through some sort of
transition.The transition process is a mechanism for converting one form of energy to
another.Here are two conversion mechanisms:
• work [mechanical energy in transit]
• heat [molecular energy in transit]
Think of these mechanisms as energy in transit.Do not confuse these concepts with the
concept of energy.Conversion mechanisms diers fromenergy in a fundamental way.The
energy stored in a system is independent of the way it goes into the object;whereas,the
work done on an object (i.e.,the amount of energy added or subtracted from a system)
depends on the process (the path) used to transfer the energy into or out of the system,
except when the net force doing the work is a conservative force.Mathematically,energy
in a system can be dened in terms of an exact dierential operator;whereas work and
heat can only be described as 1-forms (inexact dierential operators).
The main forms of energy that we will be interested in are:internal energy,chemical
(phase transitions),potential,and kinetic.
Example 2.If heat is added to a system and the temperature of a system increases,
without knowing anything else,which form of energy will be denitely increase?
(a) The kinetic energy of the system
(b) The potential energy of the system
(c) The work done by the system
(d) The internal energy (i.e.,the molecular energy) of the system
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.6
1.3 Dening thermodynamic systems
1.3.1 What is a system?
A rst step in any analysis is to describe precisely what is to be studied.If you can't do
this,then you have no hope of solving\the problem".The subject under investigation
is known as the system.In a broad sense,a system is whatever we choose to study.
That said,a more precise denition of system is needed if we are to conduct any kind of
quantitative rigorous analysis of a system.
For pedagogical reasons,it is advantageous to start our study of systems by giving a
general all-encompassing denition of a system together with denitions of the system's
surroundings and boundary,and then to make a further renement that distinguishes
between two basic types of systems:closed and open.We will nd that we gain insight
into the concept of a system by making this further decomposition of the concept of a
system into systems based on observed mass,and systems based observed space.
Denition:A thermodynamic system is dened as a quantity of matter or a region in
space chosen for study.
For brevity,we will use the verbiage system in place of thermodynamic system from here
forward.
Denition:Everything external to the system is the dened to be the system's sur-
roundings.
Denition:The area separating the system from its surrounding is the boundary of the
system.The boundary may be at rest or in motion.
The boundary is the collection of points that is in contact with both the system and its
surroundings.It is a surface,and since a surface is a two-dimensional object,it has zero
volume.
Denition:A closed system is a denite quantity of matter contained within some
closed surface.A closed system is sometimes referred to as a control mass because the
matter composing the system is assumed known for all time.
Thus,a closed system consists of a xed amount of mass.That is,no mass can ever
enter or leave the system.This means no mass can ever cross the boundary of a closed
system.
A closed system is appropriate for systems in some sort of enclosure,such as a gas being
compressed by a piston in a closed cylinder.If we choose the gas as our system,then the
gas cannot escape,although energy could escape through the piston walls.
Although mass cannot leave a closed system,energy can escape in the form of heat or
work.A special type of a closed system that does not allow the escape of energy is
known as an isolated system.An isolated system is necessarily a closed system (since
mass is energy,and if energy cannot cross the boundary,then neither can mass),but
a closed system need not be isolated since energy can be transferred across a boundary
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.7
in the form of heat without any mass transfer.This is related to the idea that a wave
can occur in a medium without any net mass transfer.For example,waves in the ocean
can transfer energy long distances,but the particles composing the wave at any given
moment in time are only displaced over a very short distance.
Denition:A system is said to be isolated if no energy is transferred across the bound-
aries.
Denition:An open system is a denite xed location in space.The system is called
open because mass may ow in or out of the system.An open system is sometimes
referred to as a control volume because the location composing the system is assumed
known for all time.
The surface surrounding the control volume is sometimes known as a control surface.
The control surface can be along a real surface of the system or it can be an imaginary
surface chosen for convenience.
Choosing a system boundary in practice is somewhat of an art.In general,choosing a
system's boundary depends on two factors:the objective of the analysis,and what is
known about the system at various locations,such as at an intake valve where we would
have explicit information about heat ow at the boundaries.
As a general rule of thumb,a control volume is used when studying things like turbines,
pumps,hot water heaters,or any device where mass ows through a piece of machinery.
Although we could in principle keep track of the mass of a uid as it ows through a
pump,in practice this method would become unwieldily.It turns out that when one
wishes to analyze a piece of machinery like a turbine,it is just easier to x a location in
space around the turbine and monitor the ow in and out of it.Notice that the turbine
could be on a jet and be moving through space.In this case we would just choose a
frame of reference moving with the system.
In summary,in a closed system the same matter (mass) is being studied.It follows that
there can be no transfer of matter across the boundary of the system;whereas with an
open system there must be mass ow across the boundary,otherwise it would be a closed
system (by denition).In general,a control mass can change shape and volume,but a
control volume cannot.Notice that in both types of systems (open and closed) there can
be a ow of energy in or out of the system.In fact,in an open system,just as there must
be a mass ow,there must also be a ow of energy in and/or out of the system.
Comments:
• We will use a movable piston-cylinder assembly as our standard closed-system with
movable boundaries example.
• If you have a ow through a device,you have an open system.Use a control volume
approach.
• In practice,most control volumes have xed boundaries because the investigator
gets to choose the boundary.
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.8
Example 3.(Dening Systems)
Which statement best describes the concept of a system?
(a) the subject of the analysis
(b) an object with xed set of molecules
(c) a uid-solid mixture
(d) an object that radiates heat into its environment
Solution:Only (a) describes the concept of a system.The other answers are particular
systems.
Example 4.(System's Surroundings)
Consider a refrigerator in a kitchen.Take the refrigerator and everything in it to be our
system.Which best describes the system's surroundings?
(a) All of the air in the kitchen.
(b) Any one standing in the kitchen.
(c) The air inside the refrigerator.
(d) Everything in the universe external to the system.
Solution:By denition of surroundings,everything that is not in the system is outside
it.The answer is (d).
Example 5.(System's Boundary)
Consider a refrigerator in a kitchen.Take the refrigerator and everything in it to be our
system.Which best describes the system's boundary?
(a) All of the air in the kitchen.
(b) The thin region separating the system from everything else.
(c) The air inside the refrigerator.
(d) The outer walls of the refrigerator.
Solution:In theory (b),in practice the outer wall (d)
Example 6.(Closed systems)
Which of the following statements are true?
(a) A closed system is necessarily an isolated system
(b) An isolated system is necessarily a closed system
(c) A system cannot be both closed and isolated.
(d) The concepts of closed and isolated in regards to a system are independent concepts.
Solution:Only statement (b) is true.
Example 7.(Control volume)
Circle all of the true statements:
(a) closed system = control mass
(b) closed system = control volume
(c) open system = control mass
(d) open system = control volume
Solution:Only statements (a) and (d) are true.
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.9
Example 8.(Control volume)
Which of the following situations would be best suited for using a control volume in the
thermodynamic analysis of the system?Do not worry about the details of the analysis.
(a) compression of air in a cylinder
(b) expansion of gases in a cylinder after a combustion
(c) the air in a balloon
(d) lling a bike tire with air from a compressor
Solution:Systems (a)-(c) are closed systems,so you'd want to use control mass for
them.Only system (d) is an open system with a xed boundary.
1.3.2 The continuum hypothesis
The key to the success of the macroscopic approach is the continuum hypothesis.It is
a well-established fact that all matter is composed of atoms and molecules.Thus air,
water,honey,and metal are necessarily discrete even though they feel continuous to the
touch.In fact,there is more space in a full glass of water than there is matter!
Fortunately we don't need to track each molecule in order to make predictions about
the behavior of various properties of a system.For example,if we are interested in the
pressure in a tire,we don't have to worry about how each molecule of air is interacting
with its neighbors and with the walls of the tire;all we need is a pressure gage (i.e.,a
tire gage).
For deniteness,we will demonstrate how this hypothesis is applied to one of the prop-
erties of a gas.In particular,we'll use density for its straight-forward graphical interpre-
tation.
Example 9 (continuum).Start with a xed collection of gas in a closed chamber.
FINISH
1.3.3 Property,state,and process
Matter may exist in one or more of several phases:solid,liquid,gas,or plasma.We will
not be interested in plasmas in this course,and we will do very little with the solid phase
of materials.
Denition:For a given quantity of matter,a phase is a state where all of the matter
has the same chemical composition throughout.Matter that is in the same phase is
homogeneous.
Example 10 (phase).Water can exist as a solid (ice),liquid,or gas.It can also exist
in mixed phases,such as an ice cube in water.
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.10
Denition:A property is any quantity which serves to describe a system.For our
purposes,a property is any macroscopic characteristic of a system that can be assigned a
numerical value at a given time without knowledge of the previous history of the behavior
of the system.
Clearly,work cannot be a property of a system,since it is something that is done on or
by a system,and not something that the system possess.
List of Examples 1.(properties) Here is a partial list of some typical system prop-
erties:
• mass
• volume
• pressure
• density
• temperature
• energy
• viscosity
• modulus of elasticity
• thermal expansion coecient
• velocity
• elevation
• electrical resistivity
Denition:The state of a system is the condition of the system as determined by its
properties.
Denition:A process is a transformation from one state to another.Whenever any
property of a system undergoes a change (e.g,a change in pressure),then by denition
the state changes,and the system is said to have undergone a process (also called a
transformation in older physics books).
Denition:If none of a system's properties change with time,then the system is said
to be in steady state.
Example 11.True or False:A systemis in steady state if its properties are independent
of time.
Solution:True.
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.11
1.3.4 Extensive and intensive properties
Thermodynamic properties can be placed into two general classes:extensive and inten-
sive.
Denition:A property of a system is called extensive if its value for the overall system
is the sum of the values of the parts to which the system has been divided into.
Some examples of extensive properties are:mass,volume,and energy
Extensive properties,as the name suggests,depend on the extent (size) of the system.
Denition:A property of a system is called intensive if its value is independent of
the extent (size) of the system,and may vary from place to place and from moment to
moment.
Some examples of intensive properties are:density,specic volume,pressure,and tem-
perature.
An easy test for whether a property is extensive or intensive is to imagine a xed amount
of matter and ask if you cut the matter into two pieces would the property in question
remain unchanged.
Example 12.(extensive-intensive test)
Consider a xed amount of gas in a closed insulated container.Assume the temperature,
pressure,and density of the gas were uniform throughout the volume.If we cut the
container in half by magically inserting an insulated impermeable wall that did not
disturb the gas,then the temperature of the two halves would not change,the pressure
would not change,and the density would not change;but the volume would be half that
of the original volume,the total mass would also be cut in half,and the total energy
would be cut in half as well.
Intensive properties may be functions of position and time;whereas extensive properties
can only be functions of time.
Example 13.(extensive properties)
Which of the following is not an extensive property?
(a) Kinetic Energy (b) Momentum
(c) Mass (d) Density
(e) None of these
Solution:Density is an intensive property.Answer (d).
Example 14.(intensive properties)
Which of the following is not an intensive property?
(a) Velocity (b) Volume
(c) Pressure (d) Temperature
(e) None of these
Solution:Volume is an extensive property.Answer (b).
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.12
1.3.5 Equilibrium and quasi-equilibrium states
A system is in equilibrium if at each moment in time its properties are approximately
independent of space.A systemundergoing a process (i:e:;a transformation) is said to be
in quasi-equilibrium if it is approximately in equilibrium at each point in time.Without
going into detail at this point in time,a quasi-equilibrium process is one where at each
moment in time the system is arbitrarily close to being in equilibrium.This means that
the system's properties are almost homogeneous through out the system.
For example,if a system such as a piston-cylinder container is lled with an explosive
gas and the gas is ignited,then that process cannot be a quasi-equilibrium process since
the pressure for example could not possibly be the same throughout the cylinder since
there would be shock waves everywhere!On the other hand,if the piston cylinder device
is a syringe that is slowly being compressed,then this process could be described as a
quasi-equilibrium process since the gas inside the cylinder would have time to adjust to
the increasing pressure and every point inside the syringe would approximately be at the
same pressure.It's as though the air inside the syringe were acting like a memory foam
that was uniformly being compressed by the piston.
You have undoubtedly see the classic case of a stable system as being a ball in the
bottom of a bowl,or two blocks that are touching one another that are both at the
same temperature.In this case we would say that there exists a temperature equilibrium
between two bodies.
Example 15.Two metal blocks,one at 50

,and the other at 0

are set next to each
other in a perfectly-insulted box at time t = 0.At this instant are the blocks in thermal
equilibrium?Now suppose you wait a long time.Are they in equilibrium now?
Solution:The blocks are not initially in thermal equilibrium.The hotter block will cool
and the colder block will warm.Eventually,the two blocks will come into equilibrium
and be at the same temperature.We cannot say what this temperature will be without
more detail about the blocks.
Example 16.In a quasi-equilibrium process,the pressure in a system
(a) is held constant throughout the entire process
(b) is spatially independent (uniform throughout the system) at each moment in time
(c) increases if volume increases
(d) always varies with temperature
Solution:Answer (b).
Example 17.Which of the following process is a quasi-equilibrium process?
(a) the stirring and mixing of cold creamer in hot coee
(b) a balloon bursting
(c) combustion
(d) the slow compression of air in a cylinder
Solution:Answer (d).
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.13
1.4 Physical quantities,dimensions,and units
Denition:A dimension is a name given to any measurable physical quantity.
Thus any physical quantity that you can think of can be characterized by dimensions.
Some examples of common physical quantities are:mass,length,time,velocity,temper-
ature,and pressure.
The set of all measurable physical quantities (dimensions) are divided into two categories:
fundamental and derived,which are sometimes referred to as primary and secondary.Just
what constitutes a fundamental dimension and a derived one is a matter of convenience,
tradition,and personal preference.There are no divine dimensions!One could build
a physical system with only one fundamental dimension,time.It would make working
with such a system dicult,but it could be done.
Once a set of fundamental quantities is chosen,all other quantities are described in
terms of these quantities using denitions and mathematical representations of physical
laws.Thus derived quantities are just that,they can be expressed entirely in terms of
the fundamental physical quantities.Of the set of all physical dimensions,the derived
dimensions are merely the ones that didn't make it into the collection of fundamental
quantities,they are the leftovers.In the language of mathematics:if S
dim
is the set of
all dimensions and S
fund
the set of fundamental dimensions,then the complement of the
fundamental set S
c
fund
= S
dim
S
fund
is the set of derived dimensions.
Most modern systems of dimensions and corresponding units take mass,length,and
time (MLT) as the fundamental dimensions.Although,the old-english system took
force,mass,length,and time as (FMLT) the fundamental dimensions.If this isn't
confusing enough,there is a system referred to as the absolute english system that takes
force,length,and time (FLT) as the fundamental/primary dimensions and treats mass
as a secondary unit.
We now list the fundamental dimensions that were chosen for the SI system and their
abbreviations:
The Fundamental Dimensions:
• Mass!M
• Length!L
• Time!T
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.14
1.4.1 Dimensional consistency
A formula that is not dimensionally homogenous,is not correct.A formula that is
dimensionally homogenous,is not necessarily correct.Thus,
(
formula dimensionally inconsistent (inhomogeneous) )formula is wrong
formula dimensionally consistent;formula is correct
In order to check physical equations for dimensional consistency we need a mathematical
tool (operator) known as the dimensionator.
Denition:The Dimensionator [ ] is the\take the fundamental dimensions of"op-
erator.
Example 18.Apply the dimensionator to nd the dimensions of the following funda-
mental quantities of Mass M,Length L,and Time T:
• d = distance;[d] = L
• x = distance;[x] = L
• R = radius;[R] = L
• h = height;[h] = L
• m = mass;[m] = M
• t = time;[t] = T
Example 19.Apply the dimensionator to nd the dimensions of the following derived
quantities in terms of the fundamental quantities:
• v = velocity;[v] = L=T
• r = rate (speed);[r] = L=T
• A = area;[A] = L
2
• V = volume;[V ] = L
3
In the following examples determine which of the formulas could not be correct because it
violates the consistency-of-dimensions principle discussed above.Do not worry about the
origin or application of the formulas:just focus on the issue of consistency of units.That
is,based solely on consistency of units in an equation,determine if the given formula
could be correct.
Warning:Do not let the subscripts confuse you.The subscripts are only used to label
variables,they are dimensionless numbers and letters and do not aect the outcome of
dimensionator one bit.
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.15
Example 20.d = rt
Solution:[d] = [rt] = [r][t] ) L =
?
=

L
T

T = L )dimensionally consistent
Example 21.v = x +t
Solution:[v] = [x +t] = [x] +[t] )
L
T
=
?
= L +T )dimensionally inconsistent
Example 22.Convert the expression v
2
,where  is density = mass/volume,into the
fundamental dimensions:L,T,and M.
Solution:Start by writing each component  and v in terms of the fundamental dimen-
sions. = density =
mass
volume
=
mass
length
3
and v = velocity =
length
time
.Using bracket notation:
[]  [v
2
] =
M
L
3

L
2
T
2
=
M
LT
2
.
Example 23.(The three fundamental kinematic equations) Verify that the three
fundamental equations are dimensionally consistent.
The general fundamental kinematic equations are:
Fundamental Equation 1:v
f
= v
i
+a(t
f
t
i
)
Fundamental Equation 2:v
2
f
= v
2
i
+2a(x
f
x
i
)
Fundamental Equation 3:x
f
= x
i
+v
i
(t
f
t
i
) +
1
2
a(t
f
t
i
)
2
where a is acceleration,v is velocity,x is position,t is time,and the subscript i and f
denotes initial and nal values,respectively.Typically,we take the initial time t
i
= 0
since,in practice,when you time an experiment you always start with the stopwatch set
to zero.
(a) Verify that the 1
st
fundamental equation is dimensionally consistent.
Solution:Start by rewriting the equation as v = at.Taking the units of the
right-hand side:[at] = (L/T
2
)T]= L/T,the units for velocity.
(b) Verify that the 2
nd
fundamental equation is dimensionally consistent.
Solution:Start by rewriting the equation as v
2
f
v
2
i
= 2ax.Taking the units of the
right and left-hand side yields [v
2
f
v
2
i
] = [2ax]],which leads to (L/T)
2
= (L/T
2
)L.
(c) Verify that the 3
rd
fundamental equation is dimensionally consistent.
Solution:Start by rewriting the equation as
x = v
i
t +
1
2
a(t)
2
t
!
x
t
= v
i
+
1
2
at
As we've already seen,at has units of velocity.Thus the equation is dimensionally self
consistent.
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.16
1.4.2 Units
In order to compare two similar quantities,such as length,we need to introduce a stan-
dard.The arbitrary,but xed,standardized dimensions of physical quantities are known
as units.For example,one way of measuring the length of an object is to compare it
to the standardized length of the foot,or the meter.Both the meter and the foot are
arbitrary standards,but they are xed standards,and that's the important part.
The units corresponding to the fundamental dimensions are known as Base Units.The
units corresponding to the fundamental dimensions must be\independent"of one an-
other,and they must be able to\span"the other physical quantities of interest in the
following sense:the units of all other quantities of interest must be able to be expressed
in terms of these proposed fundamental units.Such a complete set is known as a basis
2
and the collection is typically referred to as the Fundamental Physical Quantities,or the
Fundamental Dimensions;however,the corresponding fundamental units are referred to
as the Base Units.In the same spirit as the dimensionator operator will have need for
an operator that can express a physical quantity in terms of the base units.
Denition:The Uninator [ ] is the\take the Base Units of"operator.
The General Conference of Weights and Measures produced the SI system based on six
fundamental physical quantities (dimensions) and their corresponding units,which are
listed below.
The Base SI Units:
• For Length the meter,denoted by m.
• For Mass the kilogram,denoted by kg.
• For Time the second,denoted by s.
• For Temperature the kelvin,denoted by K (this is an absolute temperature).
• For Electric Current the ampere,denoted by A.
• For Luminous Intensity the candela,denoted by cd.
They also set down the following rules for abbreviations:
• The degree symbol was to be dropped from the absolute temperature scale.This
only applies to SI units,although some people apply the rule to the Rankine tem-
perature scale in the English system.
2
The name is derived fromthe concept of a basis in linear algebra,although it is a bit more complicated
in this case since the combination of units is nonlinear.However,by taking the natural log of the physical
quantities one can reduce all of the physical quantities to a linear system.For more details,look up the
Buckingham Pi Theorem.
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.17
• All unit names are to be written without capitalization,even if they were derived
from proper names.However,the abbreviation was to be capitalized if the unit
was derived from a proper name.
• The full name of a unit may be pluralized,but the abbreviation may not.
For example,the SI unit of temperature is the kelvin.It was named after Lord Kelvin,
so its abbreviation K is capitalized.They probably didn't want the unit capitalized so
that there would be a clear distinction between Kelvin the scientist,and kelvin the unit.
All other units are referred to as derived units.For example,the SI unit for force,the
newton (N),can be expressed in terms of the base units using Newton's second law:
F = ma
[ ]
![F] = [m][a] ) 1N = 1kg
m
s
2
:
Such a technique of dening derived units in terms of base units using a physical equation
to connect the units is known as aliasing.
The English System or United States Customary System of units
• The unit for length is the foot
• The unit for mass is the slug
• The unit for time is the second
The unit for mass in the old British system was the pound-mass,denoted lbm.The unit
for force,which was considered to be one of the fundamental units,was the pound-force,
denoted lbf.The reason the pound was a fundamental unit was mostly historical.The
pound lb is an abbreviation for libra,a unit used for weight by the Romans.Today,
we mostly just refer to the pound,but in thermodynamics there is a dierence.The
pound-mass requires the conversion factor:g
c
= 32.174 lbm = 1 slug.
The British/English system is no longer used in the UK.In fact,this system hasn't been
the standard for a long time.This system's last strong hold is the US.However,with
more and more outsourcing the British system,like the US industrial base,is becoming
obsolete.In this course we will give little attention to the United States Customary
System.
Example 24.Which of the following is not an acceptable\extended"SI unit?Recall:
The SI system is the MKS system (Meter,Kilogram,Second),but we'll allow the ex-
tended SI system to include the cgs system (centimeter,gram,second),but you can't
mix these two systems.
(a) distance measured in centimeters
(b) pressure measured in newtons per square meter
(c) volume measured in cubic meters
(d) density measured in grams per cubic meter
Solution:Statement (d) is not correct,because you can't mix meters and grams.Either
grams must be converted to kg,or meters to centimeters.
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.18
Converting between dierent sets of units
We can convert between dierent sets of units by using conversion factors.
Example 25.You have a measuring stick that has centimeters and inches on it.You
notice that 1.00 in = 2.54 cm.Using only this information together with the relationship
1 mile = 5,280 ft.,nd a relationship between miles and kilometers.That is,1 mile =
how many kilometers?
Solution:We will use a total of 5 conversion factors.That is,we will apply the one-
factor 5 times (via multiplication).
1 mile = 1 mile  1  1  1  1  1
= 1 mile 

5280ft
1mile

12in
1ft

2:54cm
1in

1m
100cm

1km
1000m

=
5280  12  2:54
10
5
km
= 1:61 km
where the second equality follows from the one-factor.Thus,1 mile = 1:61 km.Crudely,
1 mile 
3
2
km.
Example 26.An engine burns fuel at a rate of 11.2 g/s.What is the consumption rate
in kg/hr?Round your answer to the nearest 0.1 kg/hr.
Solution:Use the conversion factors:1000 g/kg and 3600 s/hr.To make the units
work,multiply by the second factor and the reciprocal of the rst:
g
s

s
hr


g
kg

1
=
g
s

s
hr

kg
g
=
kg
hr
The engine consumes fuel at a rate of
11:2
g
s


3600
s
hr


1 kg
1000 g

= 40:3
kg
hr
Example 27.The pressure at the bottom of a swimming pool is 18 lb/in
2
.What is the
pressure in pascals (Pa = N/m
2
)?Round your answer to 2 signicant gures.
Solution:We need to convert pounds to newtons,and inches to meters.Use the
conversion factors:4.45 N/lb and 2.54 cm/in.To make the units work,multiply by the
rst conversion factor,and divide by the square of the second,and lastly we'll need to
convert centimeters to meters.The pressure is
18
lb
in
2


4:45
N
lb

1 in
2.54 cm

2

100cm
1m

2
= 12 10
4
N
m
2
= 12 10
4
Pa
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.19
1.5 Density,specic volume,and specic weight
Denition:The density  of a homogeneous substance is dened as its mass mdivided
by its volume V.It is denoted by the Greek lower-case letter rho.In symbols,
 
m
V
Denition:The specic volume v of a homogeneous substance is dened as its volume
V divided by its mass m.It is denoted by the lower-case letter v.In symbols,
v 
V
m
=
1

Denition:The specic weight w of a homogeneous substance is dened as its weight
W divided by the volume occupied by the substance.It is denoted by the lower-case
letter w.In symbols,starting from the equation for weight and dividing by volume gives
W = mg
V
!w 
W
V
=
mg
V
=
m
V
g = g
Example 28.The mass of an unknown gas mixture in a room that is 4 m  3 m  5
m is known to be 600 kg.What is the density and specic volume of the gas?
Solution:The density is found using the equation
 =
m
V
=
600 kg
60 m
3
= 10 kg=m
3
:
The specic volume is found using the equation
v =
V
m
=
60 m
3
600 kg
=
1
10
kg=m
3
=
1

:
Example 29.Given that the density of water is approximately 1:0 10
3
kg/m
3
,what
is the specic weight of water?
Solution:Using the formula for specic weight
w = g  10
3
kg
m
3
 10
m
s
2
= 10
4
kg
m
2
s
2
Example 30.Determine the fundamental dimensions for density = mass/volume,de-
noted by .
Solution:Start by writing  = density =
mass
volume
=
m
V
,where m is mass and V is
volume.Using the dimensionator:[] =
[m]
[V]
=
M
L
3
.
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.20
Example 31.The mass of an unknown gas mixture in a room that is 3 m  4 m  5
m is known to be 500 kg.What is the density of the gas ( =
m
V
)?
Solution:From the denition of density and the volume of a rectangular room:
 =
M
LWH
=
500 kg
(5 m)(4 m)(3 m)
=
25
3
kg
m
3
= 8:3
3
kg
m
3
;
where M = mass,L = length,W = width,and H = height.
1.6 Pressure
Denition:The pressure P of a uid is dened as the normal force F
n
per unit area
A on a real or imaginary surface of the uid.
P 
F
n
A
:
Example 32.Express pressure,denoted by P,in terms of the fundamental dimensions.
For now,just take pressure to be force/area.Note:Using the mathematical expression
for Newton's second law we've seen that the dimensions of force are [F] = M
L
T
2
.
Solution:Start by writing P = pressure =
force
area
=
F
A
,where F is force and A is area.
Using the dimensionator:[P] =
[F]
[A]
=
[F]
L
2
= M
L
T
2

1
L
2
Example 33.Determine the expression for the unit of pressure (the pascal Pa) in terms
of the SI units for pressure and area from the equation dening pressure P =
F
A
,where
F is force and A is area.
Solution:Using the uninator on the mathematical expression for pressure gives
[P] =

F
A

=
[F]
[A]
=
N
m
2
) 1 Pa =
N
m
2
Example 34.Which one of the following expressions can be converted to the unit of a
joule (J = N m)?
(a) Pa  m
2
*(b) Pa  m
3
(c) Pa=m
2
(d) N=kg
(e) None of these
Solution:Answer (b).
1.7 Temperature
Derive formulas from temperature scales.
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.21
1.7.1 Conversion formulas
Between relative temperature scales:
• T
F
=
9
5
T
C
+32 (Convert from degrees Fahrenheit to degrees Celsius)
• T
C
=
5
9
(T
F
32) (Convert from degrees Fahrenheit to degrees Celcius)
Between absolute temperature scales:
• T
K
= T
C
+273:15 (Convert from degrees Celsius to Kelvin )
• T
R
= T
F
+459:67 (Convert from degrees Fahrenheit to degrees Rankine)
Example 35.Convert 98

F to

C.
(a) 32

C (b) 208

C
(c) 20

C *(d) 37

C
(e) None of these
Solution:T
C
=
5
9
(T
F
32) =
5
9
(9832)

C =
5
9
 66

C =
5
3
 22

C = (35+
5
3
)

C  37

C
Example 36.Convert 20

C to K.
(a) 253 K *(b) 293 K
(c) 68 K (d) 0 K
(e) None of these
Solution:T
K
= T
C
+273 K = (20 +273) K = 293 K.
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.22
1.8 Static uids
Pascal's law:Pressure applied to an enclosed uid is transmitted undiminished to every
portion of the uid and to the walls of the container.
Archimedes's principle:A body immersed in a uid is buoyed up with a force equal
to the weight of the uid displaced by the body.
Equations for static pressure:
(
P = P
atm
+gh (Total Pressure)
P
gage
= P P
atm
= gh (Gage Pressure)
1.9 Introduction to the 4 laws of thermodynamics
Of course no introductory chapter for thermodynamics would be complete without giving
an overview of the four fundamental laws on thermodynamics.
Much of engineering thermodynamics is based on experimental observations that have
been summarized and are expressed in mathematical statements known as physical laws.
For completeness of this introductory discussion,these laws are listed below after giving
and discussing the denition of the zeroth law.For technical reasons will not get to the
rst and second law until later in the course;furthermore,will never discuss the third
law,which involves topics far beyond the scope of this course.
The zeroth law is dened in terms of the transitive property.Recall from algebra class:
The Transitive property:If a = b and b = c,then a = c.
The zeroth law of thermodynamics:If two systems A and C are each in ther-
modynamic equilibrium with a third system B,then A and C are in thermodynamic
equilibrium with each other.
Note:This law may seem obvious,but it is not.It allows us to classify it from other
physical laws that do not satisfy a transitive law.
Example 37.Magnetic attraction does not satisfy a transitive law.Let A and C be
iron nails and B be a magnet.Then B attracts A and B attracts C,but A and C do
not attract each other.
Thermodynamics 232 Lecture notes Copyright ©Wayne Hacker 2009.All rights reserved.23
The Four Fundamental Laws of Thermodynamics:
• Zeroth Law (temperature satises a transitive property,which turns out to justify
the use of thermometers)
• First Law (conservation of energy)
• Second Law (asserts energy has a quality as well as quantity)
• Third Law (says you can't reach absolute zero in nitely many steps)
A brief history of zeroth law:
• The rst and second laws emerged in the middle of the 19
th
century out of the
works of Rudolph Clausius,William Rankine,and Lord Kelvin.
• In 1931 Robert Fowler realized that the fundamental physical principle that was to
become the zeroth law could not be derived from the rst or second laws,and was
even more fundamental than the rst and second law.
• Oh s**t,this posed a real problem for the scientic community!
• One option would be to call the new law the rst law,and rename the old rst law
as the new second law,and so on,but this would only lead to confusion for future
scientist that would be reading old papers.
• The only real option then was to name the new law the zeroth law.
About the 3rd Law:
• The 3
rd
law comes from the way that various gases were supercooled into a liquid
form on the race to absolute zero.
• A liquid was cooled using a special apparatus,then that liquid was used to cool
a second liquid,which was used to cool a third,and so on until the desired gas
liqueed,or the lab equipment exploded (the latter was usually the case).
• The 3
rd
law states that if one used this procedure to reach absolute zero,then it
would take innitely many steps.
• Absolute zero does not mean that the atoms cease to move.It means that they are
in a state of minimum energy.That is,they can only accept energy,they cannot
transmit any energy.