Thermo Handouts 20110

8/12/2019 Thermo Handouts 20110

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School of Mechanical Engineering 2 of 36 26/02/2014

University of Birmingham

Fluid flow, Thermodynamics and Heat Transfer

1. Introduction

Thermodynamics:Science of relations between heat and other (mechanical, electrical etc.) forms of energy

(The Concise Oxford Dictionary). The word thermodynamicsstems from the Greek words therme(heat) and

dynamis(force).

Engineers seek to improve quality of life. Engineering activity requires transfer of energy from sources such as

fossil and nuclear fuels, solar energy in its various forms (direct solar radiation, wind, hydro), and other forms

regarded as renewable (geothermal, tidal) to heat, light, electricity etc.

Although this transfer brings to mind the picture of energy in motion, the classic methods of thermodynamics

(which we will study) deal only with systems in equilibrium and with changes of matter from one equilibrium

state to another (we could thus call it equilibrium thermodynamics). Dealing formally with the true rates of

everyday processes requires a much more complex approach. We can say, however, that real processes, which

occur within finite time, will always be less efficient than the theoretical processes. Theoretical processes

could be visualised as a series of very small steps, with the rate of progress so slow that at each step

equilibrium is attained within the whole system. Such theoretical processes would produce work, but theywould take infinitely long time to complete, thus power produced would be zero. Classical thermodynamics

also uses macroscopic approach and deals with average properties of very large numbers of molecules, as

opposed to statistical thermodynamics, which deals with properties of individual particles (molecules) and

statistical distribution of these properties. An example of such an approach is the kinetic theory of gases.

Despite all its limitations, the science of classical equilibrium thermodynamics is extremely important to an

engineer, as it describes the laws and limitations of the idealised world of energy conversion. Thus we can

learn the rules, develop the conceptual and mathematical tools of thermodynamic modelling and know the

limits of possibilities, even if in real life we will never be able to achieve these limits.

Thermodynamics isa physical science which endeavours to explain many aspects of processes occurring in

nature and which has great bearing on practically every aspect of human activity. It is based on two masterconcepts and two fundamental principles. The concepts are that of energyand entropy. The principles are

the first and second laws of thermodynamics.

The First Law states: the sum of all forms of energy remains constant in the physical universe(despite

the obvious flux of energy).

Not all forms of energy, however, are equally useful. A radiator full of hot water is more useful for heating the

bathroom than a bathtub full of tepid water, although they may both have the same total energy content.

Mechanical energy (such as shaft work and potential energy) is the most useful form of energy, we can convert

it into many other forms, in some cases with high efficiency, and we can completely convert it into heat.

Count B Rumford published this last observation in 1796, and James Joule (1818-89, English) proved formallyin 1846 the mechanical equivalence of heat, thus finally disproving the previously believed theory of a caloric

fluid which represented heat transferring from one body to another.

We cannot, however, convert all heat energy back into work. The limitations imposed by nature on this

conversion have been stated by Sadi Carnot (1796-1832, an officer in the French Army engineers) in the only

paper he published during his lifetime Reflections on the Motive Power of Heat. There he devised the use of

cycles for thermodynamic analysis and analysed the Carnot cycle. By stating the Carnot principle he laid the

foundations for the Second Law of Thermodynamics (he was then only 23 or 24!).

The measure of usefulness of energyto complete a task (for example to run a machine or to heat a room) is

called availability or exergy. Every flux and transformation of energy in real life is accompanied by a

penalty, which is the loss of the usefulness of energy. Entropy is a measure of the loss of usefulness of

energy.


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The Second Law states that the usefulness of energy decreases all the time or at least it cannot increase

(entropy of the universe increases, or at least cannot decrease). Another statement of the Second Law says that

heat can only spontaneously flow from a hot body to a cold one (Rudolph Clausius, 1822-88, German). Time

will come in the future when all the universe will reach uniform temperature and there will be no more fuels to

burn either. This is the heat death of the universe as envisioned by Clausius.

In this first year course we will concentrate on understanding the thermodynamic properties of working fluids,

the First Law and its applications to energy balances in closed and open systems. We will also analyse somebasic power cycles and in this analysis we will refer to the implications of the Second Law through the Carnot

principle.

Some recommended books: (Note possibly there are new book editions)

Cengel Y.A., Boles M.A. , Thermodynamics: An Engineering Approach,2nd Edition, McGraw-Hill, New

York 1994, ISBN 0-07-113249-X (lots of examples)

Moran M.J., Shapiro H.N., Fundamentals of Engineering Thermodynamics, 2nd Edition, SI version, J. Wiley ,

New York 1993, ISBN 0-471-59275-7 (as good and in some respects better than Cengel, more to the point)

Eastop T.D., McConkey A.,Applied Thermodynamics For Engineering Technologists(5th Edition), Longman

Scientific & Technical, Harlow, 1993, ISBN 0-582-09193-4 (a bit simpler, quite practical, more of a

Polyversity approach, but everything is there and it makes a faster reading)

Look, D.C., Sauer H.J.,Engineering Thermodynamics,SI Edition, Van Nostrandn Reinhold, 1988, ISBN 0 278

00052 5 (one of my favourites, lots of good examples)

Rogers, G.F.C., Mayhew, Y.R.Engineering Thermodynamics Work and Heat Transfer. Longman, London

and New York . (an evergreen British classic)

Russell L.D., Adebiyi G.A., Classical ThermodynamicsInternational Edition, Saunders College Publishing,Fort Worth 1993, ISBN 0-03-032417-3 (not as good as the first two, half of examples in imperial units)

Fundamentals of Engineering Thermodynamics,J.R Howell and R.O Buckius, Published by McGraw Hill.


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1.1 The first law of thermodynamics

The 1stLa of Thermo!ynamics tells "s that energy is neither create! nor !estroye!# th"s the energy of the

"niverse is a constant$ %oever# energy can certainly &e transferre! from one 'art of the "niverse to another$

To or( o"t thermo!ynamic 'ro&lems e ill nee! to isolate a certain 'ortion of the "niverse )the system*

from the remain!er of the "niverse )the s"rro"n!ings*$

The 1stLa of Thermo!ynamics ill &e m"ch more "sef"l if e can e+'ress it as an e,"ation$

E - .

E - The change energy of the system# . - The heat transferre! into )*/o"t )* of the system# - The or( !one &y )* /on )* the system$

This reform"lation of the 1stLa tells "s that once e !efine a system )remem&er e can !efine the system in

any ay that is convenient* the energy of the system ill remain constant "nless there heat a!!e! or ta(enaay from the system# or some or( ta(es 'lace$

1.2 Forms of energy

Energy can e+ist in vario"s forms s"ch as mechanical# 'otential# (inetic# thermal# electric# magnetic# chemical#

n"clear# an! their s"m constit"tes the total energy Eof a system$

The magnetic# electric an! s"rface tension effects are significant in some s'ecialie! cases only an! ill not &e

consi!ere! in the content of 1styear thermo!ynamics$ n the a&sent of these effects# the total energy of a

system consists of the (inetic# 'otential# flo an! internal energies$

Internal Energy U: sum of all microscopic forms of energy (molecular structure: chemical energy, nuclear

energy and degree of molecular activity: sensible and latent energy)

Kinetic Energy: The energy possessed by a moving body. A body with mass, m, moving with velocity, v, has

kinetic energy:

KE=1

2mv

2

Potential Energy:Energy possessed by a body at a height relative to a datum. A body of mass, m, at height, z,

above a datum has potential energy:

PE= mgz

Flow Energy or Flow work: Energy required to push element across boundary

FE = pv

TOTAL ENERGY:

1.3 Closed and open systems

thermodynamic systemis !efine! as a ,"antity of matter or a region in s'ace for st"!y$ The mass or region

o"tsi!e the system is calle! the surroundingan! the s"rface that se'arates the system from its s"rro"n!ings is

calle! the boundary$ System may &e consi!ere! to &e closedor open# !e'en!ing on hether a fi+e! mass or a

fi+e! vol"me in s'ace is chosen for st"!y$

Closed systemconsists of a fi+e! amo"nt of mass that can5t cross its &o"n!aries$ o mass can enter of leave a

close! system# &"t energy# in form of heat or or(# can cross the &o"n!ary$


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Closed system

Closed systemwill be considered to be stationary, hence total energy E will be identical to internal energyU, since kinetic, potential and flow work are equal to zero.

.12 12- U28 U1

First Law of Thermodynamics per unit mass is:

,12 12- "28 "1

Open Systems# or a control volume# is a 'ro'erly selecte! region in s'ace$ t is "s"ally encloses a !evice hich

involves mass flo s"ch as a com'ressor# t"r&ine# or nole$

pen !ystem

Hence the 1stlaw of thermodynamics can be written:

Q12+ W12= E2 E1

or, Q12+ W12=

1." #roperties of a system

9ontrol

:ol"me

9ontrol s"rface

9ol! ir

)Mass*

%ot ir

)Mass*

%eat

Energy $E!

Mass %Energy $E!


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A thermodynamic property is a property that is independent of the shape, position or velocity of a particular

system or quantity of substance, and does not require a reference frame to determine its value:

Mass, pressure, volume and temperature are all example of thermodynamic properties.

Pressure (p)

We define pressure as the normal component of force per unit area. Pressure is therefore an intensiveproperty.

Hence

p= F/A [N/m2]

The SI unit of pressure is N/m2, this is also known as the Pascal, i.e

1Pa = 1N/m2

Two other units of pressure are also common, these are the bar, and the standard atmosphere, where

1bar = 105Pa = 0.1MPa

1atm = 0.101325MPa

Volume (V)

The volume of a substance is an extensive property and has SI units of m 3. A unit commonly used for volume

is the litre:

1 litre = 10-3m3= 0.001m3

It is common in thermodynamics however to deal in terms of specific volume (v):

v =V/m [m3/kg]

By dividing the volume of a substance (extensive property) by the mass (extensive property) we obtain an

intensive property, and this is denoted by the term specific

Note:An intensive property is often denoted by using lower case; and extensive quantity by upper case.

Note:The density ;of a substance is simply the inverse of the specific volume, i.e.

;= m/V [kg/m3]

Temperature (T)

A number of temperatures scales have been devised, although only two are now in regular use: the Fahrenheitand SI Celsius scale.

Temperature (oF) = 1.8xTemperature (oC) + 32

Another temperature scale arises from the Second Law of Thermodynamics and is know as the absolute

temperature scale. This predicts an absolute zero of temperature which can then be adopted as zero on our

temperature scale. A common absolute temperature scale is the Kelvin scale; To convert to Celsius we add

273.15, hence

K= oC + 273.15

This may be thought of as thermodynamic temperature and temperature must always be expressed in degree

Kelvin when performing thermodynamic calculations.


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Properties of a system are considered to be either intensive or extensive

nintensi&e propertyis a 'ro'erty hich is in!e'en!ent of the sie of the system# i$e$ it !oesnot !e'en! on the amo"nt of the s"&stance 'resent$ )=ress"re# >ensity an! Tem'erat"re*

n e'tensi&e property!e'en!s "'on the sie of the system$ )Mass# :ol"me*

m

V

T

p

(

m/2 m/2

V/2 V/2

T T

p p

( (

Extensive

properties

Intensive

properties

)ntensi&e and e'tensi&e properties

Thermodynamic equilibrium

A system is said to be in equilibrium if no further changes will occur in it, when it is isolated from its

surroundings. This requires that all the properties (pressure, temperature etc.) must be uniform throughout the

system. It is only under equilibrium conditions that a single value of a property can be used to characterise the

system, and, hence, its state be determined. In this course the system will always be assumed to be in

equilibrium at the beginning and end of a process.

Zeroth law of thermodynamics

Consider a system A consisting of a gas in a closed vessel of fixed volume. If this system is placed in contactwith a second system B consisting of a block of metal and no change of pressure occurs in the gas, then

systems A and B must have the same temperature, and are in equilibrium with each other. If A is now placed in

contact with another system C, and again no change in pressure occurs, then A and C have the same

temperature and are in equilibrium. Therefore, as A is in equilibrium with both B and C, then B and C must

also have the same temperature and are in equilibrium with each other. This principle is known as the Zeroth

Law, and is the basis for temperature measurement.


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2$ Thermo!ynamic 'ro'erties an! 'rocesses

2.1 Thermodynamic #roperties of *aterials

Specific Heats

It is commonly observed that if heat is transferred into an object, its temperature will rise. Experimentation

shows that different substances react differently: some require a greater heat input to produce a specified

change in temperature than others. This leads to the idea that each substance has a property, called its specific

heatthat describes its reaction to heating.

Definition The specific heat, C, of a substance is the amount of heat that must be transferred into a unit mass

of that substance to raise its temperature by one degree.

Observe that, by referring to unit quantities, the effects of mass and temperature difference have been factored

out of the definition. Also note that the numerical value of C depends on the units that are used to measure

heat transfer, mass and temperature. We will always adhere to SI units and specify C as the number of kJ

required to raise 1 kg through 1 K so that C will always have the units kJ/(kg K).

For the general case of a mass m, the heat transfer Q required to bring about a temperature change T is givenby the equation:

Q = m C T

This completes the theory of specific heats for solids and liquids. Gases require more careful consideration

because temperature changes are often accompanied by significant volume changes, which have important

effects as discussed below.

The figures below show two ways in which heat, Q, may be transferred into a mass of gas. In the first figure,the gas is contained within a rigid box so that the volume remains constant throughout the process. In the

second figure, the gas is trapped under a piston that is free to rise or fall without friction.

Different Processes for Transferring Heat to a Gas

All the heat energy transferred into container (a) will be used to raise the temperature of the gas: there is

nowhere else for it to go. This is nottrue for case (b). As heat is transferred into the second container, thetemperature of the gas will begin to rise, but so will the internal pressure. However, the downward pressure on

the gas (due to the atmosphere and the weight of the piston) remains unchanged, so there will be a pressure

difference that will cause the piston to rise until the internal pressure regains its original value.

Q Q Q

(a) (b) (c)


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Movement of the piston implies that work is being done since a force (that on the inside face of the piston) is

being moved through a distance. Thus, onlypartof the heat transfer will be used to raise the temperature of

the gas; the remainder of the energy will be used to lift the piston (i.e. to do work). It follows that the tempera-

ture change for container (b) will be lessthan that for container (a). To obtain the same temperature increase,

more heat must be transferred into (b) and hence the specific heat will be higher since C = Q/(m T). Thespecific heats for these two cases are designated by the symbols C vand Cp, the subscripts referring to the fact

that one process occurs at constant volume, whilst the other occurs at constant pressure.

Example For air, Cv= 0.718 kJ/(kg K) and Cp= 1.005 kJ/(kg K), so Cpis larger than Cv, as expected from

the previous discussion. In fact, for the constant pressure case, 0.718/1.005 = 71.4% of the heat input is used

to raise the temperature whilst (1.005 0.718)/1.005 = 28.6% provides the work needed to expand the gass

volume.

Cases (a) and (b) are not the only ways in which heat may be transferred into a system. Case (c) shows the

situation where the piston is restrained by a spring. As the piston rises, the spring will become more

compressed and hence the downward force will increase. This means that the internal pressure, and hence the

work done, will vary in a way that is quite different to cases (a) and (b). In turn, this implies the need for a

different heat transfer and hence different value for the specific heat. Using this logic, it is quite easy to infer

that gases have an infinite numberof specific heats, one for each of the different ways in which heat transfer

may occur. (Just think of rerunning case (c) but with springs of different stiffness.) Although this observation

is true, only the values of Cvand Cpare of practical importance, as shown in later sections.

The Perfect Gas

Thermodynamics describes what happens when a substance undergoes a process that changes its energy

content. A complete description requires two kinds of information: a description of the process and a

description of the substanceon which the process acts. Only when we have both kinds of information is it

possible to derive equations that predict changes in temperature, pressure, energy, etc.

To understand what happens when a gas undergoes a process, we need equations that describe how the gasresponds to changes in its environment. Specifically, we require equations that express relationships between

the gass various properties: its temperature, pressure, volume, etc. This can lead to some fairly complex

theory if we try to model the gas to high precision. Instead, we will start with an approximate but very useful

model - the so-calledperfect gasmodel.

A perfect gas is definedby the two equations:

pv = RT

the equation of stateand

u = CvT

Because the above define a perfect gas, the equations are alwaysvalid, irrespective of what happens to the gas.

R and Cvare constants whose values depend upon the particular gas being considered. A third perfect gas

equation describes the relationship between the temperature, T, and the enthalpy, h. Enthalpy is the property

defined by the equation:

h = u + pv

Substituting, we obtain:

h = u + pv = CvT + RT = (Cv+ R)T = CpT

where the constant Cpis given by:

Cp= Cv+ R


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The following values apply for air.

Symbol Name Value for air in kJ/(kg K)

R Gas constant 0.287

Cv Specific heat at constant volume 0.718Cp Specific heat at constant pressure 1.005

A molis a unit that measures of the amount of substancein terms of the number of atoms or molecules that are

present. Thus, 1 mol of oxygen contains the same number of molecules as 1 mol of nitrogen, or 1 mol of

hydrogen or 1 mol of any other substance. 1kmol of substance contains 6.022 x 10 26atoms or molecules,

depending on how the substance exists in nature. The molar mass, M, is number of kilograms that contain 1

kmol of substance; it is numerically equal to the substances atomic or molecular mass. The gas constant, R,

for a given gas may be computed from the Universal Gas Constant, Ru, and the gass molar mass, M.

R = Ru/M

where

Ru= 8.314 kJ/(kmol K)

2.2 #hases of a substance

n this mo!"le e ill 'rimarily# &"t not e+cl"sively# consi!er the thermo!ynamic &ehavio"r of gases$

%oever s"&stances can of co"rse e+ist in !ifferent 'hases 8 gases# li,"i!s an! soli!s$ hich 'hase act"ally

e+ists !e'en!s 'rimarily "'on the 'ress"re# tem'erat"re an! s'ecific vol"me of the str"ct"re$ Aor some

com&ination of these 'arameters !ifferent 'hases can e+ist at the same time$ The fig"re &elo shos a ty'ical

'lot of the 'hase of a s"&stance against 'ress"re an! tem'erat"re$ long the lines in the !iagram to !ifferent

'hases can coe+ist$ t the 'oint mar(e! T# hich is (non as the tri'le 'oint# all 'hases can e+ist together$

#hase of substance

Aor many a''lications it is necessary to consi!er ho a s"&stance &ehaves if !ifferent 'hases e+ist together#

an! ho the energy content varies as a s"&stance changes from one 'hase to another$ These iss"es ill &e

consi!ere! to only a limite! e+tent in this mo!"le# &"t ill &e ta(en f"rther in some secon! an! thir! year

mo!"les$

2.3 +e&ersible and irre&ersible processes

What Is a Reversible Process?

Soli!

=ress"re

Tem'erat"re

Tas

Li,"i!


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The First Law of Thermodynamics for a closed system containing a unit mass of substance may be written:

q12 + w12 = u2 - u1

where subscripts 1 and 2 denote the start and end states, respectively. Frequently we wish to calculate separate

values for the specific heat transfer, q, and the specific work done, w. (Why? Think of calculations for an

engine, where q is related to the fuel consumption and w is directly related to the power output bothimportant but distinctparameters of engine performance.) In such situations, we have two unknowns but only

one equation (the First Law) and so a second equation is required.

The standard way of obtaining the second equation is to make a further assumption about the process 1 2;namely, that it is reversible. A reversible process may be thought of as one that occurs perfectly, with no

friction or other kind of losses. This simplicity allows us to derive a further equation (usually for the work, w)

as described in the following sections.

A Ball Rolling in a Track

The motion of a ball rolling in a U-shaped track provides a simple example of a reversible process and

provides some clues as to why these processes are useful in analysis.

If there is no rolling friction between the ball and track and no aerodynamic drag on the ball, then it will

oscillate backwards and forwards forever, once it is set in motion. Each oscillation will be the exact reverse of

the previous one, hence the term reversible process.

This is a very simple, very predictable situation. Since the motion is frictionless, there are no energy losses (or

gains) and hence the motion can be predicted using energy analysis: the ball will exchange potential energy

and kinetic energy in such a way that their sum remains constant. This can be described by the equation of

motion:

1/2V2 + gz = C, a constant

where C = gzmaxis the balls potential energy when it reaches its highest point, where its velocity is zero.

Using this equation, we can calculate the velocity at any other height.

Now consider the case where the effects of friction are significant and cannot be ignored. In this situation, the

motion of the ball will notbe reversible: the balls maximum height will decrease with each oscillation until it

finally comes to rest. The analysis will also be more complex. Instead of using simple energy arguments, we

must apply Newtons laws of motion to formulate an equation that includes terms for the aerodynamic and

rolling friction. This analysis results in a second order differential equation that is difficult to solve.

A third and final case is where there is a finite but small amount of friction; that is, where the ball and track arefairly smooth and the velocities are not too high. In this situation, we can approximatethe real motion by its

reversible equivalent. The benefits of doing this are obvious. We regain the simple analysis and all the


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predictions that follow from it. However, we must remember that our results are now only approximate, not

exact. We must also be careful not to over-extend the approximation: in the case of the ball, it might work for

one or two oscillations but it would probably fail if we tried to apply it to 20 or 30 oscillations where the

cumulative effects of friction begin to be significant again. Equally, we must not forget the assumptions on

which the approximation was based. For example, we could not use this type of approximation if the ball and

track were submerged in oil because there would be large viscous friction forces, which would invalidate the

original assumption.

When making first estimate thermodynamic analyses, this type of reversibility assumption turns out to be

very useful. There are many situations where the energy losses are not so great as to invalidate the assumption

and where approximate but reliable results are quite sufficient for design purposes.

The previous discussion was designed to provide a working introduction to the concept of reversibility and its

role in thermodynamics. It should be mentioned that the reversibility concept has a rigorous definition and that

it is fundamental to much thermodynamic theory.

Having set the scene, we now use the reversibility assumption to derive a veryimportant equation that enables

the work, w, to be calculated from the pressure and volume changes that occur during a process.

Relationship between Work, Pressure and Volue

Consider the figure below, which shows a mass of gas being compressed inside a piston/cylinder. It will be

assumed that the piston is free to move without friction or other losses i.e. that the motion is reversible. We

will analyse the pistons movement to find an equation that relates the work done to the applied force, F, and

thus to the internal pressure, p. This will lead to the desired relationship between work, pressure and volume.

When performing the analysis, we will follow a standard methodology for constructing theoretical models of

physical systems. Using this, we will consider only infinitesimal (i.e. verysmall) changes. This simplifies the

mathematics since the squares and higher powers of infinitesimals are so small that they can be ignored.

Having derived an equation involving infinitesimals, we will take limits so that a differential equation can be

obtained.

Work Done during Compression

In thermodynamic terminology, the gas trapped inside the piston/cylinder is the systemand the walls of the

piston/cylinder comprise the systems boundary. The fact that the externally applied force, F, is being moved

through a small displacement, x, means that work is being done on the system. Under the standard signconvention, the work is reckoned to bepositive since it is being done onthe system - i.e. it acts to increasethe

gass energy by raising its pressure and temperature.

The small displacement, x, will cause a small amount of work, W, to be done:

W = F x

Assuming that the small size of the displacement causes only a negligible change in pressure, p, the applied

force can be found from:

F = p A

Cross-sectional

area, A

Pressure, p Force, F

Displacement, x

System boundary


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where A is the pistons cross-sectional area. Combining these equations and recalling that the volume of a

cylinder is the product of the cross-sectional area and the height:

W = F x = ( p A ). x = p (A x) = p (-V) = - p V

where V is the change in the systems volume due to displacement, x. Observe the minus sign, which isrequired because the displacement causes a negative change in volume i.e. a reduction in the systems

volume. Taking limits, we obtain:

dW = - p dV

Integrating between the initial volume, V1, and the final volume, V2, gives an equation that relates the work

done to the pressure and volume:

Note that this equation applies to the total system. It may also be written in per unit mass form:

where w is the specific work done (kJ/kg), p is the pressure (kPa) and v is the specific volume (m3/kg).

When making the above analysis, it was assumed that the piston was free to move without friction or other

losses. Thus, if the applied force F was removed after compressing the gas, we would expect the piston to

move back down the cylinder and return to its original position. In a similar manner, all the other values (the

gas pressure, temperature, etc) would also return to their original values. In other words, the analysis and the

equation obtained from it depend on the assumption of reversibility.

Consider events if friction were present, as shown in the diagram below.

Irreversible Compression

In this case, part of the applied force, F, would be needed to overcome the friction force, f, between the piston

and cylinder. In energy terms, this implies that onlypartof the applied work (i.e. F x the distance piston is

moved) will be available to compress the gas, because some of the work will be used to overcome the friction.

Thus, in the irreversible compression case, the actual work, w, must be larger than -p dv because someadditional work will be needed to overcome friction. The reverse argument applies in the irreversible

expansion case. Here, the actual work output, w, will be less than suggested by the integral -p dv becausepart of the energy will be dissipated by friction.

To summarisethe above analysis is valid only for reversible processes.

Applied Force, FAriction force#

f

2

1

=V

V

dVpW

!v'

2

1

v

v

=


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3. pplication of first law of thermodynamics to closed systems with

ideal gases

3.1 How the First -aw is pplied

Having set-up the background to thermodynamics, we now apply the First Law to construct a theoreticaltoolkit that can be used to predict the performance of various kinds of machinery. This toolkit comprises sets

of equations, each set describing a specific type of process.

Five kinds of process will be considered. For each process, the following method will be used to derive

detailed equations that apply only to that specific case:

1) Assume a specific type of process and write down the equation describing it. For example, the

equation for a constant pressure process is p1 = p = p2 which states that the pressure at all stages

of the process remains the same: at the start, p1, anywhere in the middle, p, and at the end, p2.

2* Assume that the process is reversible. This allows us to use the equation w = - p dv. The pressure

(p) will be expressed as a function of specific volume (v) so that the integral can be evaluated toobtain specific work (w) in terms of properties at the start and end states.

3) Write down the First Law and insert the result of step 2,so as to obtain an equation for q in terms of

properties at the start and end states.

4) Make the additional assumption that the process involves a perfect gas . This will allow us to

combine the perfect gas equations with the results of steps 2 and 3 to obtain to a more detailed

description of the process.

Note that, as we make each assumption, the resulting theory onlyapplies when that assumption and those

that precede it are valid. Thus, the theory obtained in steps 2 and 3 onlyapply to reversible processes.

Similarly, the equations obtained in step 4 onlyapply when a perfect gas undergoes a reversible process.

The processes that will be considered are as follows.

Constant volume processes

Constant pressure processes

Constant temperature (isothermal) processes

Polytropic processes (with a defined relationship between pressure and volume)

Adiabatic and isentropic processes (an adiabatic process is one that takes place without heattransfer, and an isentropic process is an adiabatic process that is also reversible).


15/36


16/36




3.3 Constant #ressure #rocess

Process assumption. Since the process occurs at constant pressure:

p1 = p = p2

TT

Q(+)Q

(+)

W

(-)

W

(-)

V

p

V

p

p=Const

V1V1 V2V2

p1= p = p2p1= p = p2

State 1 State 2

1) 2)1) 2)

WW

QQ

T1

T2

P-V Diagram: Constant Pressure

Assume reversibility. Use the fact that p = constant to move p outside the integral for the work done:

First Law of Thermodynamics per unit mass is:

q12 + w12 = u2 - u1

Insert the expression for w into the First Law to find q in terms of quantities at the end states:

q12 = u2 - u1 + p( v2 - v1)

Use the constant pressure assumption to evaluate p at end states: p = p1 = p2

So: q12 = (u2+ p2v2) - (u1 + p1v1)

The combination u + pv occurs so frequently that it is treated as a property in its own right. It is known as

the specific enthalpyand given the symbol, h.

h = u + pv

So, finally:

q12 = h2 - h1

Assume a Perfect Gas: If the process involves a perfect gas, then pv = RT and h = CpT and so:

w12 = - p(v2 - v1) = - (p2v2 - p1v1) = - R(T2 - T1)

q12 = h2 - h1 = Cp(T2 - T1)

*)!v'!v 1212

2

1

2

1

vvppw

v

v

v

v ===


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3." Constant Temperature #rocess

Process assumption. Since the process occurs at constant temperature:

T1 = T = T2

V

p

V

p

T=Const

V1V1

State 1 State 2

W-W-

1)

2)

1)

2)

WWW

Q+Q+

QQ

p1p1

p2p2

p-v Diagram: Constant temperature

Assume reversibility. The reversibility assumption does not lead immediately to results of general utility.

We simply restate the fact that the following integral may be used if p can be expressed as a function of v:

Assume a Perfect Gas: If the process involves a perfect gas, then pv = RT. But T is a constant so:

pv = RT = C

where C = p1v1 = RT1 = RT2 = p2v2

It follows that:

pv = C

so

p = C / v

Inserting this in the w-p-v integral:

Substituting the various expressions for the constant, C:

By definition u = CvT for a perfect gas, so the First Law may be written:

q12 + w12 = u2 - u1 = Cv(T2 - T1) = 0 since T2 = T1

Thus q12 = - w12 where w12 may be calculated from any of the expressions given above.

!v'

2

1

v

v

=

v

vln9

v

vln

v

!v9!v

2

1

1

2

12

2

1

2

1

=

=== Cpw

v

v

v

v

v

vlnCT

v

vlnCT

v

vlnv'

v

vlnv'

2

1

2

2

1

1

2

1

22

2

1

1112

=

=

=

=w


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3./ #olytropic #rocess

Process assumption. Polytropic processes are observed in many practical situations where gases or vapours

undergo expansions or compressions. They are characterised by a curved p-v profile, as illustrated below:

Q+

State 2

State 1

W-

Q+Q+

State 2

State 1State 1

W-

Q-

State 2

State 1

W+

Q-

State 2

State 1

Q-Q-

State 2

State 1

W+

Process 1 2: Expansion

Process 2 1: Compression

T

V

p

2)

1)

W

p1

p2

V1 V2

TT

V

p

2)

1)

W

p1

p2

V1 V2V

p

V

p

2)

1)

2)

1)

WW

p1p1

p2p2

V1V1 V2V2

p-V Diagram: Polytropic Process

The process 1 2 is an expansion whilst the process 2 1 is a compression. Expressing the shape of the p-v curve mathematically results in the equation:

p1v1k = pvk = p2v2k = C a constant

where k is known as the index of expansionor index of compression, as appropriate. In practice, k often has

values around 1.3. An important sub-case, where the process is reversible, adiabatic and involves a perfectgas, will be described in detail below. For that case, k = , the ratio of the gass specific heats, which is 1.4 forair. We now return to the general case of a polytropic process.

Assume reversibility. Using this assumption, we may substitute p = C / vk in the w-p-v integral:

Recalling that for a polytropic process C v1-k = p1, etc, we finally obtain:

The heat transfer, q, may be obtained by rearranging the First Law:

q12 = (u2 - u1) - w12

where w may be calculated from the above equation

Assume a Perfect Gas: If the process involves a perfect gas, then pv = RT and the equation for w may be

rewritten in terms of start and end temperatures:

( ) ( )[ ]

1122

1((

(12

1(

9v

1(

9v9

v

!v9!v

2

1

2

1

2

1

2

1

vvvvdvpw kkv

v

v

v

v

v

v

v

+ =+

====

( )[ ]

1(

1

112212 vpvpw =

( )[ ] etc$#CTv'since

1(

C

2221212 == TTw


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3.0 diabatic and )sentropic #rocess

In thermodynamics, the word adiabaticis used to describe a process with zero heat transfer. An isentropic

process is one that is adiabatic and also occurs reversibly.

At first sight, isentropic processes appear to be a fairly restricted sub-case of the First Law. In fact, isentropic

approximations are frequently used because many real processes occur sufficiently quickly that there is little

time for significant heat transfer to take place. With the exception of combustion, heat transfer is usually a

slow process and hence a long period is required if a large amount of energy is to be transferred. As an

everyday example of this, consider how long it takes for heat to be conducted up the handle of a spoon when

stirring a cup of hot tea or coffee - minutes rather than seconds.

V

p

V

p

TT2)

1)

2)

1)

WW

State 2

State 1

W-

State 2

State 1

W-

State 2

State 1

W+

State 2

State 1

State 2

State 1

W+

Process 1 2: Expansion

Process 2 1: Compression

p1p1

p2p2

V1V1 V2V2

Q=0

p-V Diagram: Adiabatic and Isentropic Processes

Process assumption. By definition, an adiabatic process is one that occurs with zero heat transfer, so:

q12 = 0

First Law of Thermodynamics

w12 = u2 - u1

since q = 0.

The rest of the analysis is easier if we work with differentials. That is, if we work in terms of the changes that

occur in an infinitesimally small part of the process. Differentiating the First Law, we find that a small

amount of work, dw, is equivalent to a small change in specific internal energy, du:

dw = du

Assume reversibility. Then, as usual:

The equation may be differentiated to express it in terms of differentials:

dw = - p dv

Assume a Perfect Gas: For a perfect gas u = CvT which may also be expressed in terms of differentials:

du = CvdT

=2

1

!v12

v

v

pw


20/36




We thus obtain

-p dv = CvdT

Since the process involves a perfect gas, we may use the equation of state pv = RT to replace the dT term in

equation by p and v terms. The aim is to obtain an equation involving just p, v and some constants. Again,

we may take differentials to obtain:

d(pv) = p dv + v dp = R dT

Eliminating dT gives:

-p dv = (Cv/R) (p dv + v dp)

Gathering terms:

-(1 + Cv/R) p dv = (Cv/R) v dp

Recall that Cp = R + Cv so (1 + Cv/R) = Cp/R

Inserting this in the above, canceling R and using the definition = Cp/Cv gives

- p dv = v dp

Or

This may be integrated. Recognising the integrals as natural log terms and recalling that

c ln(x) = ln(xc)

gives:

Taking anti-logs finally gives the p-v relationship for an isentropic process involving a perfect gas:

p1v1 = pv = p2v2

Combining the above with the equation of state, pv = RT, and eliminating either v or p provides some other

important relationships that apply when a perfect gas undergoes an isentropic process:

The specific work done, w, for an isentropic process is given by:

Since we are dealing with a perfect gas RT = pv, we can also use:

'

!'

v

!v=

=

=

1

2

2

1

1

2

''ln

vvln

vvln

v

v

T

T

'

'

T

T

1

2

1

1

2

1

1

2

1

2

=

=

( )( )

( )T1

CT9""

1212v1212TTw

===

( )

( )

'' 1122

121

vvw =


21/36




". pplication of first law of thermodynamics to open systems with

ideal gases

A large number of engineering problems (e.g. turbines, compressor, radiators, heaters) involve mass flow in

and out of the system.

Mass and volume flow rates

The amount of mass flowing thorough a cross section per unit time is called the mass flow rate and is denoted

m . The mass flow rate thorough a differential area dA can be expressed as:

dm = dAVav$

For one dimensional flow (i.e. constant velocity across the entire cross section) the mass flow rate from the

integration of the above equation is:

AVm av= (kg/s)

Where = density (kg/m3), Vav. = average fluid velocity (m/s), A = cross-sectional area (m2).

The volume flow rate V and the mass flow rate m are related by:

VVm

==

Steady-flow process (no change with time)

A process during which a fluid flow through the system boundary and the fluid properties can change from

point to point (position) within the system, but at any fixed point they remain the same during the entire

process. In most practical problems including a large number of engineering devices such as turbines,

compressors and nozzles operate for long period of time under the same conditions, and they are classified as

steady-flowdevices.

As we discussed earlier the total energy of a simple compressible system consists of internal, kinetic, potential

and flow energy.

E = U + (mV2) + mgz + pV

However, the sum of the terms U and pV gives the enthalpy H = U + pV, or if both side divided the mass gives

the specific enthalpy h= u +pv.

Hence:

Q12+ W12=

or, q12+ w12=

pplications

!ir heater"


22/36




Potential energy PE12= mgz = 0, z1= z2= 0

Kinetic energy KE 12= (mV2) = 0, V1= V2= assume very small change

Work W= 0, there is not shaft or moving parts

Hence,

q12= h2 h1

Nozzle

Potential energy PE12= mgz = 0, z1= z2= 0

Work w= 0, there is not shaft or moving parts

Heat q=0, assume adiabatic

Hence,

0 = (h2 h1) + (V22

V12

)/2

Turbine

Potential energy PE12= mgz = 0, assume z1= z2= 0


23/36


24/36




5. Thermodynamic cycles for ideal gases and the second law of

thermodynamics

5.1 Introduction to Cycles

Many machines operate using a cycle of processesthat are repeated continually while the machine is running.

A common example is the petrol engine which works on the four stoke cycle shown below

The process comprising a petrol engines operating cycle are:

1 Induction Starting with the piston at the top of its stroke (the smallest volume, marked as point 1 on

the p-v chart), the inlet valves are opened and the piston begins to move down the cylinder, thus

drawing in a fresh charge of air and fuel vapour. This stoke corresponds to an increase in volume at a

pressure just less atmospheric, as shown by the bottom curve on the chart. The pressure in the

cylinder is explained by slight pressure drop required to draw air from the atmosphere through the

carburetor and the inlet manifold.

Operating Cycle of a Petrol Engine

2 Compression At the bottom of the stroke (i.e. the largest volume, point 2 on the chart), the inlet

valve is closed and the piston begins to rise, thus compressing the air/fuel mixture. Close to the top

of the stroke, the spark plug fires, thus initiating combustion and causing a rapid rise in gas pressure

and temperature. As shown on the chart, combustion may cause the pressure to climb even after the

piston as passed the top of its stroke (point 3) and the volume has begun to increase again.

3 Expansion The high pressure gases produced by combustion force the piston down the cylinder, thus

delivering work. This continues until the piston reaches the bottom of its stoke (point 4), when the

exhaust valves open to allow the waste products to escape.

4 Exhaust The cycle is completed as the piston moves back up the cylinder and the waste gases are

ejected through the exhaust open valves. This corresponds to a decrease in volume at a pressure just

above atmospheric, as shown by the second to bottom curve on the chart. The pressure level is set by

the need to push the waste gases through the exhaust system and out into the atmosphere.

It is possible to approximate the true cycle by a sequence of simple processes and use the equations developed

in the previous section to obtain first estimates of pressures, temperatures, power, etc. Before doing this, it is

helpful to develop some general theory about the thermodynamics of cycles.


25/36




/.2 Cycles efinition and Theory

A thermodynamic cycleis defined to be a closed sequence of processes that returns the system to the state from

which it began. This is shown below where a sequence of n processes begins and ends at state 1.

A Cycle of n Processes

If we apply the First Law to each of the n processes in the cycle and sum the results, we find:

Process 1 -> 2: q1,2 + w1,2 = u2 - u1

Process 2 -> 3: q2,3 + w2,3 = u3 - u2Process 3 -> 4: q3,4 + w3,4 = u4 - u3 ... etc ... ... etc ...

Process (n-1) -> n: q(n-1),n + w(n-1),n = un - u(n-1)Process n -> 1: qn,1 + wn,1 = u1 - un

Summing over cycle: q + w = 0 since the u terms cancel out

or: q = - w

It is common to refer to the two sums as the net heat transfer, qnet, and the net work, wnet. Thus, the previous

result shows that the net heat transfer is the negative of the net work done for a complete cycle. This providesa useful check when performing calculations: if your answers dont confirm this result, you have made an

error! On occasion, it can also provide a useful shortcut. There are some cases where direct calculation of

the net work is a lengthy and tedious procedure, and it is much easier to find the net heat transfer and infer the

work using the previous result.

Cycles lead to the concept of cycle efficiency, c, which provides a measure of the overall efficiency of amachine. The definition of cycle efficiency is:

c = -wnet/ qin

where:

wnet The sum of all the work terms, irrespective of their sign.

qin The sum of thepositiveheat transfers; i.e. the sum of all the heat inputsto the system.

The efficiency is usually quoted as a percentage rather than a fraction - e.g. a machine is said to be 30%

efficient rather than 0.3 efficient. The formal definition also agrees with common sense ideas about the

concept of efficiency. In laymans terms, it is the ratio:

What you want - i.e. the work output, which is related to an engines power

What you must pay - i.e. the heat input, which is related to the fuel consumption

Observe that the work term is wnet - that is, the sum of allthe works, not just the work outputs. Using this

value takes account of the fact that engines consumework in part of their cycle (e.g. during compression) and

well asproducingit in others parts. Recall that wnet is the overall work produced during the complete cycle.

5.3 The second law of thermodynamics

1

3

n

42

n-2n-1

Process 1 -> 2

Process n -> 1

Process 2 -> 3


26/36




TheSecond Law of Thermodynamicsstates that all engines reject heat. In other words, it is fundamentally

impossible to design an engine that converts all the supplied heat (i.e. all the fuel) into useful work. Some heat

is always wasted. As an example, observe that all cars have radiators and hot exhausts, both of which involve

the rejection of heat.

Let us define qinto be the sum of all the heat inputs (i.e. all the positive heat transfers) and q outto be the sum of

all the heat rejections (i.e. all the negative heat transfers). Thus, the net heat transfer is:

qnet= qin+ qout

Combining this with the definition of cycle efficiency and the fact that qnet= -wnetfor a cycle:

c = -wnet/ qin = qnet/ qin= (qin+ qout)/ qin = 1 + qout/ qin < 100%,

since qoutis a negative value. This leads to another expression of the Second Law: no engine can be 100%

efficient, no matter how well it is designed.

Subject to some additional assumptions, it is possible to relate the two heat transfers, q in and qout, to the

maximum and minimum temperatures found in a cycle. This enables a theoretical upper limit to be set on anengines cycle efficiency:

(c) max = 1 - Tmin/ Tmax

where both temperatures must be specified in Kelvin. This is the so-called Carnot efficiency, Carnot being the

Frenchman who first discovered the result. The equation serves two purposes: 1) As guide for designers the

way to get best efficiency is to burn fuel at the highest possible temperature, T max. The temperature at which

heat is rejected, Tmin, is generally set by the atmosphere to which all waste heat is eventually dumped; 2) More

prosaically, it acts as a quick and easy check on calculations if you obtain a c value that is larger than theCarnot value, (c) max , you have definitely made a mistake.

Finally let us consider the work done during a cycle. Recall that the work done during a reversible process is

given by the integral w = -p dv, which is represented graphically by the area under the p-v curve. Let usapply this observation to the figure below which show a cycle of two processes, a -> b -> a, that are performed

in the order shown by the arrows. Process a -> b is a compression (observe that the volume decreases) during

which the system is consumes work. Numerically, a positive amount of work is done during this process.

Process b > a (is an expansion where the system delivers work to the surroundings. This involves a negative

work term since the system is giving up some of its energy to do work on the surroundings.

(a) (b) (c)

Interpretation of the p-v Chart for a Cycle

Taking the two processes together, the positive work consumed during compression is cancelled by part of the

negative work delivered during expansion, leaving the area inside the loop, which represents the net work

delivered over the complete cycle.

a

v

'

&

v

'

a

&

v

'

a

&


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School of Mechanical Engineering 2< of 36 26/02/2014



5.4 The Carnot Cycle

In this section we consider and idealised heat engine, that is taking heat from a high constant temperature

reservoir and rejecting it to a low constant temperature reservoir. As a consequence the engine delivers some

work. We assume that the engine considered is of the piston and cylinder type, that is operating with an ideal

gas as the working fluid in a cycle with the properties at the start of the cycle the same as those at the end. Thepv curve for the Carnot cycle is shown in the figure below. It is assumed that heat is taken in between states 1

and 2 at constant temperature and rejected between states 3 and 4 also at a constant temperature (i.e. isothermal

processes). We also assume that the parts of the cycle between states 2 and 3 and 4 and 1 are adiabatic and

reversible (i.e. isentropic).

At point 1 the piston is at the inner limit of its stroke, and the gas then expands reversibly and isothermally

between 1 and 2, as the piston moves outwards, whilst in contact with the high temperature heat source.

Between 2 and 3 the gas is insulated from the heat source and expands reversibly and adiabatically as the

piston continues to move outwards. Between 3 and 4 the gas is in contact with the lower temperature source,

and, as the piston moves inwards, is compressed isothermally and reversibly. Between 3 and 4, the gas is

isolated from the source and compressed further adiabatically and reversibly.

For an isothermal process between states 1 and 2 we have from section 5.4

and also

q12= - w12

Similarly for the isothermal process between 3 and 4 we have

and alsoq34= - w34

v

vlnCT

2

1112

=w

v

vlnCT

4

3

334

=

'

3

42

1

v

%eat o"t )isothermal*

%eat in )isothermal*

!ia&atic# reversi&le

!ia&atic# reversi&le


28/36

School of Mechanical Engineering 2? of 36 26/02/2014



For the adiabatic, isentropic process between 2 and 3 we have from section 5.6

And similarly between 4 and 1 we have

The sum of the work done is

+

=+++=

4

3

3

2

1

141342312v

vnCT

v

vnCTD

The total heat input is q12which is given by

v

v

lnCT,2

1

112

=

Thus the efficiency is given by

2

3

21

43

2

3

12T

T1

*/vln)v

*/vln)v

T

T1

,

DE =

+=

=

The last step arises from the simple application of the ideal gas equation and the p v relation for isentropic and

adiabatic flows which shows that

4

3

1

2

v

v

v

v=

The above is the expression for the Carnot efficiency introduced in an earlier section.

5.5 The Otto cycle

The Otto cycle provides a zeroth order approximation of the performance of petrol engines. The aim is to

construct a theoretical model that will give rough, working estimates of power output, fuel consumption,

temperatures, pressures, etc., but with a minimum of calculation. The first step is to simplify the model by

ignoring those parts of the cycle that make only small contributions to the overall performance. This is done

using information taken from the engines p-v chart.

Now applying the concept of work done during a cycle to that for a petrol engine, it is clear that the work done

during the compression, combustion and expansion phases of the cycle (i.e. in the large loop in the upper part

of the figure) is significantly greater than that done during the exhaust and induction phases (the thin

horizontal loop at the bottom of the figure). Thus we may simplify calculations by ignoring the exhaust and

induction stokes, providing we are willing to incur a small error.

The final step is to approximate the remaining processes using equations taken from the theoretical toolkit

that was developed in section 5.

Inspection of the above figure shows that combustion occurs very rapidly near the top of the pistons stroke,

where there are only small changes in volume. Thus, we may idealise combustion by a constant volume

process.

v

p

1

2

3

4

Swept volumeClearance volume

Constant volume processes

Isentropic processes

( )( )TT

1

C 2323

=

( )( )

( )( )TT

1

CTT

1

C 324141

==


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School of Mechanical Engineering 2@ of 36 26/02/2014



The Otto Cycle

The compression and expansion stokes occur very quickly (a typical engine turns at 2,500 - 4,000 rpm, so both

processes are completed in milliseconds) and hence there is insufficient time for a significant amount of heattransfer through the cylinder walls, heat conduction being a relatively slow process. It follows that isentropic

processes are suitable for modelling these parts of the cycle.

The cycle is completed by a final process that simulates the exhaust of the hot waste gases and the induction of

a cool, fresh charge of fuel and air. Since these processes begin and end with the piston at the bottom of its

stroke (i.e. at maximum volume), they may be simulated by a constant volume cooling process. Applying

these approximations results in the Otto cycle, which is shown in the p-v chart below

The ratio of the maximum volume (vmax= v1 = v4), to the minimum volume (vmin= v2 = v3) is known as the

compression ratio, rv:

rv = vmax/ vmin

Sample Calculation

An engine works on the Otto cycle with a compression ratio of 10:1. The air/fuel mixture enters the engine at

a temperature and pressure of 17oC and 1 bar, respectively. If combustion is simulated by a specific heat

transfer of 195 kJ/kg, find the temperature and pressure at the end of each process in the cycle. Also calculate

the specific net work done and the cycle efficiency.

Solution

Refer to above figure for the sequence of processes that define the Otto cycle and the notation used to mark the

various states. Assume air is a perfect gas with constants: R = 0.287 kJ/kg K, CV= 0.718 kJ/kg K, CP= 1.005

kJ/kg K and = CP/CV = 1.4.

State 1 T1= 17oC => (273 + 17) = 290K Alwayswork in Kelvin, nor Celsius

Process 1 -> 2 This is an isentropic compression with compression ratio, rv= v1/v2= 10.

For isentropic processes involving a perfect gas, we have:

p1v1 = p2v2

p2 = p1 (v1/ v2) = p1 rv = 1 x 101.4 = 25.15 b

Also for an isentropic process:

K728.410x290rTv

vTT

v

v

T

T0.4

v112 ===

=

=

1

2

1

2

1

1

2

11


30/36




q1,2 = 0 and w1,2 = (u2 - u1) = Cv(T2 - T1) = 0.718 x (728.4 290) = 314.8 kJ/kg

Process 2 -> 3 Combustion is simulated by heat addition (q = 195 kJ/kg) at constant volume.

For a constant volume process involving a perfect gas

w2,3 = 0

q2,3 = (u3 - u2) = Cv(T3 - T2) T3 = T2 + q / Cv = 728.4 + 195/0.718 = 1000 K

For a constant volume process involving a perfect gas, the equation of state shows that:

p3/ p2 = T3/ T2 p3 = p2 (T3/ T2) = 25.15 x ( 1000/728.4) = 34.5 b

Process 3 -> 4 This is an isentropic expansion with expansion ratio, rv= v4/v3= 10.

For isentropic processes involving a perfect gas:

p3v3 = p4v4

p4 = p3 (v3/ v4) = p3/rv = 34.5 / 101.4 = 1.37 b

Again, q3,4 = 0 since the process is isentropic. The specific work done is:

w3,4 = (u4 - u3) = Cv(T4 - T3) = 0.718 x (398.1 - 1000) = -432.2 kJ/kg

Process 4 -> 1 Exhaust of burnt fuel and induction of a fresh charge of air/fuel is simulated by heat rejection

at constant volume. For such process, w4,1 = 0 and q4,1 = u. Treating air as perfect gas:

q4,1 = (u1 - u4) = Cv(T1 - T4) = 0.718 x (290 - 398.1) = -77.6 kJ/kg

Overall Results Recall that, for a cycle, the net heat transfer and the net work done should sum to zero. This

provides a useful check on the calculations. Adding results from each of the four processes:

Net value = Process 1 -> 2 + Process 2 -> 3 + Process 3 -> 4 + Process 4 -> 1

wnet = 314.8 + 0 + (-432.2) + 0 = -117.4 kJ/kg

qnet = 0 + 195 + 0 + (-77.6) = 117.4 kJ/kg

which confirms the expected result. Observe that net work done is negative, indicating that the overall cycle

produceswork. Similarly, the net heat transfer is positive showing that the cycle consumesheat (i.e. fuel).Also observe that the cycle rejects heat during process 4 -> 1. This is true of all cycles. No matter how well a

cycle is designed, it can convert only part of the heat input into work, the remaining heat being rejected.

The cycle efficiency, c, is defined to be the ratio of the net work done (i.e. what we want from an engine) tothe heat input (i.e. the fuel we must burn). Conventionally, signs are not shown when calculating cand theresult is expressed as a percentage:

c = wnet/ q+ve = 117.4 / 195 = 60.2%

5.6 Entropy

$ack#round

K398.110/1000/ rTv

vTT

v

v

T

T0.4

v334 ===

=

=

1

4

3

4

3

3

4

11


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The concept of reversibilityhas been discussed above It is shown that, for a reversible process, there exists a

relationship between work, pressure and volume:

It is useful to interpret the relationship in terms of its underlying physics. By definition, work is done when aforce moves through a distance. What causes the force? Answer: pressure, which acts to move the boundaries

of the system, thus causing changes to another property, the volume. The combination of these two changes

results in work being done.

Also observe that the relationship involves an integral. Why? Because pressure and volume describe the state

of a system at a specificinstantin time, whereas work is done over thedurationof a process. It follows that

we must sweep over (i.e. integrate over) the full range of pressures and volumes that occur during the process

in order to find the work done.

!nalo#% $etween Heat Transfer and Work

The question arises: does a similar relationship link the heat transfer, q, to the fluid properties that are involvedin the thermal movement of energy? The answer is, of course, yes. Drawing an analogy with the work

equation, the relationship will involve an integral and q will appear on the left hand side, in place of w. The

analogy also suggests that the temperature, T, should appear in the same place as the pressure, p, because

temperature is the driving force that causes heat to flow.

What appears in place of the specific volume, v? That answer is: a new property, thespecific entropy, s. The

integral for the heat transfer during a reversible process is thus:

&ntrop% and its 'eanin#

For the present, treat the above integral as the definitionof the specific entropy, s. The analogy argument is

very informal, so you should be aware that the q-T-s integral and the existence of entropy may be proved

rigorously using the Second Law of Thermodynamics.

Because it cannot be perceived by the senses or be directly measured by experiment, entropy is a notoriously

difficult property for neophytes understand. We now make some comments that suggest its role in

thermodynamics.

Entropy as a measure of irreversibility. As stated above, the q-T-s integral is valid only for reversible

processes. Suppose a reversible process takes place adiabatically(i.e. with zero heat transfer, q = 0), then theintegral shows that the entropy must remain constant, s 1= s2. Such a process is said to be isentropic. Thus,

isentropic reversible + adiabatic. (Note: The other option, T = 0, is not physically possible.)

It can be shown from the Second Law that if an adiabatic process occurs irreversibly, then the entropy always

increases, so that s2>s1. In fact, the greater the degree of irreversibility (e.g. the greater the friction duringthe process), the greater the inequality. In this sense, entropy acts as a measure of irreversibility. The entropy

at the end of an irreversible process is always greater than that at the end of its reversible counterpart.

Entropy as a measure of disorder. These notes approach thermodynamics from an engineering standpoint. A

number of fundamental results were introduced by reference to engines and the processes that occur within

them. However, it is also possible to take a statistical approach to thermodynamics whereby events at the

molecular level are analysed to discover the macroscopic behaviour. The kinetic theory of gases, which youmay have encountered in your earlier studies, is a well-known example. When interpreted in this way, entropy

is revealed as a measure of the disorder within a system.

=2

1

v

v

dvp-w

dsTq

s

s

=2

1


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Consider a block of ice placed in a hot environment. Initially, the ice is in a very orderly state where the inter-

molecular forces hold the water molecules in fixed positions with respect to each other. As the ice melts, the

degree of disorder increases: the water molecules still interact with each other (this is why we sense water as a

dense substance) but they are not held in fixed positions (this is why water flows). If further heating is

applied, the water evaporates to form steam and the degree of disorder increases yet again. The molecules now

become very diffuse and move almost independently of each other.

Entropy provides an objective measure of this disorder: the entropy of ice is numerically smaller than that of

water, which is much smaller than that of steam. This agrees with the q-T-s integral; a positive heat transfer

(i.e. heat addition) always increases the entropy, s.

Entropy: A measure of the amount of energy in a physical system not available to do work. As a physical

system becomes more disordered, and its energy becomes more evenly distributed, that energy becomes less

able to do work. For example, a car rolling along a road has kinetic energy that could do work (by carrying or

colliding with something, for example); as friction slows it down and its energy is distributed to its

surroundings as heat, it loses this ability. The amount of entropy is often thought of as the amount of disorder

in a system.


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ppendi' !ol&ing thermodynamics problems 45ases6

$ack#round

When you are new to a subject, solving even the simplest problem can seem difficult. The key to making

progress is to acquire a general framework of skills and knowledge that you can use when tackling problems.The purpose of this appendix is to present such information.

All thermodynamics problems share some common characteristics. These are:

Problems involve a system(usually a mass of gas or sometimes a liquid) that undergoes a process.

This means that twosets of equations are used during calculations: one set to describe the gas or

liquid and the other to describe the process

The gas or liquidthat comprises the system may be described by either equations or tables of

properties. In this module, we primarily restrict our attention to perfect gases (usually air) for

which the appropriate equations are:

pv = RT u = CvT h = CpT

Theprocessundergone by the system involves three elements: i) a start state, ii) an interaction in

which work and/or heat transfer change the systems energy, iii) a resulting end state. Problems

generally require you to find one of the three elements given data about the other two. Each

process has three equations that describe the nature of the process, the work done, w, and the heat

transfer, q, during the process.

The word stateis used to refer to the condition of the system at a specified instant of time. The

state is described by means of properties such as pressure, volume, temperature, internal energy,

enthalpy and entropy. Experience shows that a knowledge of two independent properties is

enough to determine the state, and hence all the other properties. For a perfect gas u, h and T are

notindependent properties, as shown by the previous equations.

(irst Steps

Once you have understood the four points listed in the previous section, you will find that solving thermo

problems is largely a question of organising information. None of the equations developed in this module are

complicated (they usually involve just three terms, two of which you are given or can find quite easily) but

they often involve a fairly large volume of data that needs to be applied at different stages of the problem.

Thus, managing and applying data is a key skill that you should aim to develop. The following steps are

recommended:

Draw a graph showing the process or cycle of processes. A p-v chart is a good choice because the

area under the curve is equal to the work done during the process - With a little practice, you willfind the graph to be an invaluable aid for organising your thinking, especially if the problem

involves a sequence of processes. See the following example.

Write down the equations describing the substances properties. In this module, the substance will

nearly always be a perfect gas, which is described by the three equations given above. Remember

thatthese equations always apply,irrespective of whatever the process might be.

Write down the properties describing the start state of the process. Remember, if you know two

independent properties then you can alwayscalculate the others from the property equations.

Write down the properties describing the end state. Remember, if you know two independent

properties then you can alwayscalculate the others from the property equations.

Finally, write down the equations describing the process undergone by the system. Remember, in

general there will be three equations: i) an equation characterising the process - e.g. constant

pressure implies p1= p = p2. Although this is a simple equation, it provides key information and


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is often overlooked by novices; ii) an equation relating the work done, w, to properties at the end

states; iii) an equation relating the heat transfer, q, to properties at the end states.

Remember that every thermodynamic process involves just three things (a start state, a process and an end

state) and that you generally know two and wish to find the third. Having organised the data, check which two

pieces of information you know and calculate the third. Working in this way will help you to organise your

approach to problem solving and reinforce your understanding of thermodynamic theory.

&)aple

A closed system undergoes a sequence of three reversible processes. Starting from an initial pressure of 1 bar

and a specific volume of 1 m3/kg, a mass of air undergoes an isentropic compression until its volume is

reduced to one-fifth the original value. Find the final pressure and temperature and also calculate the specific

work done and the heat transfer during the process.

The air then undergoes an expansion at constant pressure until the volume returns to its original value. Find

the final temperature and calculate the specific work done and the heat transfer during this second process.

Finally, the system returns to its original state by means of a constant volume process. Find the specific work

done and the heat transfer. Also calculate the total work done and the total heat transfer over all three

processes, compare the two values and explain what you find.

Solution

1) Draw a diagram of the processes. A p-v chart is a good choice because the area beneath a curve is equal

to the work done during the process.

2) Write down the equations describing the fluids properties. Fluid is air, so treat as a perfect gas

pv = RT u = CvT h = CpT

For air:

R = 0.287 kJ/kg K Cv= 0.718 kJ/kg K Cp= 1.005 kJ/kg K

= 1.4

1

23

v

p

v2= v

1/ 5 v

3= v

1

p1= 1 bar = 100 kN/m2

v1= 1 m3/kg

p3= p2 = ?v

3= v

1= 1 m3/kg

p2= ?

v2= v

1/5 = 0.2 m3/kg

isentropic

const pressure

const volume

p2

= p3


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3) Write down what you know about each state. Remember, if you know two independent properties, you

may apply the equations given in 2) above to calculate any third property. Add this summary to the p-v chart.

State 1 p1 = 1 bar = 100 kN/m2 v1 = 1 m3/kg

State 2 v2 = v1/ 5 = 0.2 m3/kg

State 3 v3 = v1 = 1 m3/kg

4) Write down and then use the equations describing each process . Remember, these equations may be used

in additionto those that describe the gas: one set of equations describe the gas whilst the other set describe

what happens to the gas.

Because the problem involves three steps, we will use a subscript notation to identify the work and heat

transfer that occurs in each process. Thus, w1,2 is the work done in process that links states 1 to 2, whilst q 3,1is

the heat transfer in the final process that takes the system from state 3 back to state 1.

a)Isentropic process, 1 -> 2 For an isentropic process:

p1v1 = pv = p2v2

The appropriate equations for w and q are:

w1,2 = u2- u1 = Cv(T2 - T1)

q1,2 = 0

Note: By definition, an isentropic process is both reversible and adiabatic, so q = 0, always.

Use the first equation to calculate the pressure at the end of the process:

p2 = p1(v1/ v2) = 1 x 5 1.4 = 9.518 bar = 951.8 kN/m2

We have two end state properties (p2and v2) so we can now find the temperature from perfect gas theory:

T2 = (p2v2)/R = 951.8 x 0.2 / 0.287 = 663.3 K

To calculate the work done, we also require initial temperature:

T1 = (p1v1)/R = 100 x 1 / 0.287 = 348.4 K

Then:

w1,2 = Cv(T2 - T1) = 0.718 x (663.3 - 348.4) = 226.1 kJ/kg

Also:

q1,2 = 0

b) Constant pressure process, 2 -> 3 for which p2 = p = p3 = 951.8 kN/m2. We also told that v3 = v1= 1 m3/kg. The equations for w and q for a constant pressure process are:

w2,3 = -p (v3- v2) q2,3 = h3 h2 = Cp(T3 - T2)


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Since we have two properties at the end state (p3and v3) we can calculate any third in this case T3:

T3 = (p3v3)/R = 951.8 x 1 / 0.287 = 3316.3 K

We can now find the work done and heat transfer during process 2 -> 3:

w2,3 = -951.8 x (1 - 0.2) = -761.4 kJ/kgq2,3 = 1.005 x (3316.3 - 348.4) = 2666.3 kJ/kg

c) Constant volume process, 3 -> 1 for which v3 = v = v1 = 1 m3/kg. The equations for w and q for a

constant volume process are:

w3,1 = 0

q3,1 = u1- u3 = Cv(T1 - T3) = 0.718 * (348.4 - 3316.3) = -2130.9 kJ/kg

d) Total heat transfer and total work done over all processes.

qnet = q1,2 + q2,3 + q3,1 = 0 + 2666.3 - 2130.9 = 535.3 kJ/kg

wnet = w1,2 + w2,3 + w3,1 = 226.1 - 761.4 + 0 = - 535.3 kJ/kg

So qnet + wnet = 0 as expected for a cycle of processes. The correctness of this overall result encourages us

to believe that the detailed calculations are also correct.

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Thermo Handouts 20110