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,denitions,and principles 2

1.1 What is thermodynamics?..........................2

1.1.1 Articial subdivisions of thermodynamics..............3

1.1.2 Engineering Thermodynamics....................4

1.2 What is Energy?...............................5

1.3 Dening 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,specic volume,and specic 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,denitions,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 eciency

with which these two processes can occur.

1.1.1 Articial 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,raried 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

justied 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 aects

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 conguration 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 diers 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 dened in terms of an exact dierential operator;whereas work and

heat can only be described as 1-forms (inexact dierential 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 denitely 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 Dening 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 denition 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 denition of a system together with denitions of the system's

surroundings and boundary,and then to make a further renement 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.

Denition:A thermodynamic system is dened 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.

Denition:Everything external to the system is the dened to be the system's sur-

roundings.

Denition: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.

Denition:A closed system is a denite 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.

Denition:A system is said to be isolated if no energy is transferred across the bound-

aries.

Denition:An open system is a denite 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 denition).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.(Dening 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 denition 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 deniteness,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.

Denition: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

Denition: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 coecient

• velocity

• elevation

• electrical resistivity

Denition:The state of a system is the condition of the system as determined by its

properties.

Denition: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 denition

the state changes,and the system is said to have undergone a process (also called a

transformation in older physics books).

Denition: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.

Denition: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.

Denition: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,specic 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 coee

(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

Denition: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 dicult,but it could be done.

Once a set of fundamental quantities is chosen,all other quantities are described in

terms of these quantities using denitions 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.

Denition: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 aect 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 = at.Taking the units of the

right-hand side:[at] = (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

= 2ax.Taking the units of the

right and left-hand side yields [v

2

f

v

2

i

] = [2ax]],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

at

As we've already seen,at 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.

Denition: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 dening 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 dierence.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 dierent sets of units

We can convert between dierent 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 signicant 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,specic volume,and specic weight

Denition:The density of a homogeneous substance is dened as its mass mdivided

by its volume V.It is denoted by the Greek lower-case letter rho.In symbols,

m

V

Denition:The specic volume v of a homogeneous substance is dened as its volume

V divided by its mass m.It is denoted by the lower-case letter v.In symbols,

v

V

m

=

1

Denition:The specic weight w of a homogeneous substance is dened 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 specic volume of the gas?

Solution:The density is found using the equation

=

m

V

=

600 kg

60 m

3

= 10 kg=m

3

:

The specic 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 specic weight of water?

Solution:Using the formula for specic 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 denition 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

Denition:The pressure P of a uid is dened 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 dening 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

(9832)

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 denition 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 dened 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 satises 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 scientic 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

liqueed,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 innitely 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.

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