1 Chapter 5 Consequences of the First Law. 2 There’s this unhinged physicist who is working on a device that can be used to blow up the world. In this.

1

Chapter 5Consequences of the First Law

2

There’s this unhinged physicist who is working on a device that can be used to blow up the world. In this ground-breaking work he needs to know for a substance. He can find no

tabulations for this quantity. Must he perform an experiment todetermine this quantity? Not necessarily. He could attempt to writethis partial in terms of other quantities that are tabulated.

Such a situation arises very often in thermodynamics andso there is much manipulation of partial derivatives. We beginthis chapter with some useful formulae for partial derivatives.

This then adds to what we have already learned about partials inChapter 2.

In general results for ordinary derivatives cannot be taken over directly when dealing with partial derivatives!

Tv

u

3

Mathematical Relationships:

Suppose a relationship exists among x,y and z. We can then consider z=z(x,y) and write:

dyy

zdx

x

zdz

xy

Write NdyMdxdz

Then yx

x

N

y

M

is the condition that dz is an exact differential. (See Chapter 2)

Suppose f=f(x,y,z) and a relationship exists among x, y and z. The function f may be regarded as a function of any two of x, y or z.Similarly, for example, x may be regarded as a function of f and y or z. If one considers x=x(f,y)

4

dyy

xdf

f

xdx

fy

Using such expressions one can show

1

fffx

z

z

y

y

x

and also

zfyzf

y

y

x

f

x

f

x

(Note!)

5

There are many relationships between thermodynamic quantities which can be obtained by combining the definitions, laws and rules for partial derivatives. In this course we will concentrate on P, V, T systems, but there are similar relationships for other systems. In such systems there are certain quantities which are measured directly and often tabulated:

)(,,,,,, VP ccTVP

Other quantities can often be expressed in terms of the above quantities. A quantity expressed in terms of measurable quantities is often said to be expressed in standard form. Some relationships will be derived in this course and others are given as assignments.

6

The Gay-Lussac-Joule ExperimentWe discussed free expansion earlier and now we consider it

in more detail. Consider u=u(T,v). An attempt was made to determine how the internal energy changes with a change in volume if T is fixed.

VV T

uc

uV

TTuV

Tuv v

Tc

v

u1

u

v

v

Tc1

u

v

v

T

T

u

We introduce the Joule coefficientuv

T

V

T

cv

u

How can we keep u fixed?

This can be accomplished by performing a free expansion under adiabatic conditions.

( a known quantity)Vc

du=đq-đw

So how the internal energy changes with volume if the T is fixed depends on the Joule coefficient. Let us consider this coefficient.

We have

7

adiabatic walls

diaphragm gas vacuum

00 Tv 01 vv

When the diaphragm is punctured the gas undergoes free expansion into the vacuum

vdvdv

Tvd

v

Tdu

u

TdT)v,u(T

uuv

1

0

1

0

1

0

v

v

01

v

v

T

T

vdTTvddT Joule found 01 TT

Much later, more refined experiments showed that with

01 TT

3

1

u m

kmolK001.0

v

T

As mentioned earlier, we define an ideal gas to be one for which0 u=u(T) only and so

=0

so 0cv

uV

T

8

Let us consider an ideal gas. Earlier we showed that vdP

v

udT

T

uq

Tv

vdPv

u

T

1dT

T

u

T

1

T

q

Tv

The reason for dividing by T is that the LHS becomes an exact differential, as will be discussed in detail in later chapters. Employing the relationship for an exact differential:

vTTv

Pv

u

T

1

TT

u

T

1

v

vvTT2

Tv T

P

T

1

v

u

TT

1P

v

u

T

1

T

u

vT

1

Interchanging the order of the partials in term on LHS results in cancellation with a term on the RHS

đ

đ

9

PT

PT

v

u

T

PP

v

u

T

1

vTvT

Since we are considering an ideal gas:

0v

uP

v

RT

TT

v

u

TvT

Hence, for an ideal gas u=u(T) only.

0P

u0

P

v0

P

v

v

u

P

u

TTTTT

(Not zero!)

10

The Joule-Thomson ExperimentThis is a different approach from the one just discussed in

that the internal energy does not stay constant. In this experiment gas is forced through a porous plug and is called a throttling process.

iii TvP fff TvP piston

porous plug adiabatic walls

In this experiment work is obviously done. The pressures are kept constant.

iiff

v

0

f

0

v

i vPvPvdPvdPwf

i

wduq

iiifffiiffif vPuvPuor)vPvP()uu(0

From the definition of enthalpy if hh

đ đ

11

Hence, in a throttling process, enthalpy is conserved.In an actual experiment, there are no pistons and there is a

continuous flow of gas. A pump is used to maintain the pressure difference between the two sides of the porous plug.

12

Throttling Process (Joule-Thomson or Joule-Kelvin expansion) (This process is widely used in refrigerators.)

The pump maintains the pressures Pi and Pf with Pf smaller than Pi. In the experiment Pi, Ti and Pf are set and Tf is measured.

pump

porous plug

Pi Ti Pf Tf

13

As just discussed the enthalpy is the same on the two sides of the porous plug i.e., hf = hi. Consider a series of experiments in which Pi and Ti are constant (hi constant) and the pumping speed is changed to change Pf and hence Tf. Since the final enthalpy does not change, we get points of constant enthalpy. We plot Tf as a function of Pf.

14

A smooth curve is placed through the points yielding an isenthalpic curve. {Note that this is not a graph of the throttling process as it passes through irreversible states.}

•• • • •

••

•

Pf , Tf

isenthalpic curve

Pi, , Ti

Pf

Tf

15

We now change Pi and Ti and obtain another isenthalpic curve.

maximum inversion T

cooling

•d

•c•b

•a

heating

Inversion curve

T

P

inversion curve

ideal gas

16

It is useful to define the Joule-Thomson coefficient since we are interested in the temperature change due to the pressure change.

This is the slope of an isenthalpic curve and hence varies from point to point on the graph. A point at which = 0 is called an inversion point. Connecting all of these points produces the inversion curve.

hP

T

17

If point a on the diagram ( < 0) is a starting point and point b is the final point, then the T of the gas will rise, i.e. we have heating.

If, on the other hand, we start at point c ( > 0) and go to point d, then the T of the gas will drop, i.e. we have cooling.

As higher initial starting temperatures are used, the isenthalpic curves become flatter and more closely horizontal. These curves are horizontal lines for an ideal gas. As noted on the plot there will be a maximum inversion T, the value of which depends on the gas. For cooling to occur, the initial T must be less than the maximum inversion T. For such a T the optimum initial P is on the inversion curve.

18

To make the discussion clear, we have exaggerated the slopes in the above TP diagram. In fact, for most gases at reasonable T’s and P’s the isenthalpic curves are approximately flat and so 0

It can be shown (a problem) that P

h

P

T

cP

Tc

P

h

If )(00 ThhsoandP

hthen

T

We now have )1.5oblem(Pr0P

h

v

u

TT

for an ideal gas.

19

If a throttling process is used to liquefy a gas, the cooled gas is recycled through a heat exchanger to precool the gas moving towards the throttle. The gas continues to cool and when a steady state is reached a certain fraction, y, is liquefied and a fraction (1-y) is returned by the pump. Using the notation:

hi = molar enthalpy of entering gas hf = molar enthalpy of emerging gas hL = molar enthalpy of emerging liquid

Since the enthalpy is constant we havehi = y h L + (1-y)hf

Of course, as some of the gas liquefies, additional gas must be added to the system.

20

Liquifaction of Gases:Some gases can be liquified in a simple process. For

example, carbon dioxide can be liquified at room temperature by a simple isothermal compression to about 60 atm.

To liquify nitrogen or air is not so simple. At room temperature, regardless of any increase in pressure, these gases will not undergo a phase transformation to the liquid state. A method for these gases, using the throttling process, was invented in 1895 and is called the Hampson-Linde Process. The basis idea is to use the gas cooled in the throttling process to precool the gas going towards the throttle until the T is below the maximum inversion T. Starting from room temperature, this cycle can be used to liquify all gases except hydrogen and helium. To liquefy H by this process, it must first be cooled below 200K and to accomplish this liquid N at 77K is used. To liquefy He by this process, it must first be cooled below 43K and to accomplish this liquid H can be used. (A device called the Collins helium liquifier is used to liquify He.)

21

It should be mentioned that Joule-Thomson liquefaction of gases has these advantages:

• No moving parts that would be difficult to lubricate at low T.

• The lower the T , the greater the T drop for a given pressure drop.

22

Heat engines and the Carnot Cycle.

A reservoir is anything large enough so that its temperature does not change as heat enters or leaves in a particular process.

One can easily devise a system in which work is converted to energy into a reservoir with 100% efficiency. Can one reverse this process, that is, use energy from some reservoir to perform work with 100% efficiency? No!! We will study a cyclic operation, which can continue indefinitely, with the intermediary returning to its initial state at the end of each cycle. This intermediary is called an engine.

The next slide gives an example of an engine, the steam engine, which led to the industrial revolution and is still used today.

An idealization of an engine is then given on the following slide.

23

Boiler

Condenser

Tur

bine

Pum

p

Hot Reservoir

Cold Reservoir

High P

Low P

STEAM ENGINE

Water

Steam

Coal, Oil, Nuclear

Air, Body of water

Working Substance Work

24

A heat engine is any device that absorbs “heat” and converts part of that energy to work

2Q

1Q

engine

hot reservoir

cold reservoir

work

Heat into a system is considered positive, while heat leaving a system is negative.

25

The notation that we will use is as follows:

2Q

W

1Q

heat exchange between high T reservoir and system

heat exchange between low T reservoir and system

work exchanged between system and surroundings

For a heat engine, and is the work done by the system 12 QQ W

The thermal efficiency of a heat engine is defined by 2Q

W

Since there is no change in the internal energy of the system,

2

112 1

Q

QsoandWQQ

An efficiency of unity occurs if but this is not possible. Later we will determine the maximum possible efficiency.

01 Q

26

We will be considering engines. We will idealize the engines and the idealization will be severe. Nevertheless we shall be able to answer to the following question:

What basically limits the efficiency of the engine?

27

There are two types of practical engines:internal combustion engine: gasoline, dieselexternal combustion engine: steam, Sterling

We will consider the internal combustion engine.The engine gases are contained in a cylinder with a moveable piston.In part of the cycle the gases are raised to a high T and P.The gasoline engine: In this engine there are six processes, four of which are strokes of the piston. An analysis of a real engine is complicated by, among other things, the presence of friction. We obtain an upper limit to what is possible (the efficiency) by idealizing the cycle. This idealization is called the Otto Cycle and makes the following approximations:

(a) the working substance (air and gasoline in a real engine) is an ideal gas with constant heat capacities(b) no friction or turbulence(c) no heat loss by conduction(d) processes are reversible

28

It should be remarked that, in a gasoline engine, there is no external high temperature reservoir. The thermal energy is produced internally by burning fuel.

adiabats

51

2

3

4

P

0P

V

2V 1V

intake exhaust

compression stroke

power stroke

ignition

Exhaust valve open

In the exhaust stage what actually happens is that the piston pushes the old mixture out through one valve and pulls in a new mixture through another valve.

29

5 -> 1 isobaric intake (at atmospheric pressure)1 -> 2 adiabatic compression (compression stroke)2 -> 3 isochoric increase in T during ignition. Gas combustion is

an irreversible process, here replaced by a reversibleisochoric process in which heat is assumed to flow in

from a reservoir.3 -> 4 adiabatic expansion (power stroke) 4 -> 1 isochoric decrease in T (exhaust valve opened)1 -> 5 isobaric exhaust (at atmospheric pressure)

What is the efficiency? What does it depend upon?

The diagram is not to scale. In a real engine, might be about 10. 21 /VV

123

114

122

111 4321 VTVTVTVT

Subtracting and dividing gives

1

1

2

23

14

V

V

TT

TT

We define the compression ratio which gives21 /VVrc

30

)1(11

23

14

crTT

TT

Dividing the two equations for the adiabats gives so 3

2

4

1

T

T

T

T

2

1

142

141

21

24

14

23

14

)()(

T

T

TTT

TTT

TTTT

TT

TT

TT

From the definition of specific heat at constant volume, and assuming that the specific heat is constant:

giving)TT(CQ)TT(CQ 14V123V2

2

1

23

14 11T

Tor

TT

TT

Using equation (1) gives

1

11

crA higher compression ratio increases the efficiency! For 4.1if56.0,8rc

For a real gasoline engine 3.02.0 to

31

Hence, for a gasoline engine, the efficiency can be increased by reducing friction and making other improvements. However the maximum attainable efficiency is limited by the compression ratio.

32

The Diesel Engine:To increase the efficiency of gasoline engines one could try to increase the compression ratio. Unfortunately there are severe preignition problems associated with high compression ratios. The diesel engine avoids preignition problems by compressing only air, then spraying fuel into the cylinder after the air is hot enough to ignite the fuel. The spraying-ignition is done as the piston begins to move outward, at a rate adjusted to maintain approximately constant pressure. Diesels typically have compression ratios of about 20.

Again we make an idealization, neglecting such problems as friction, all of which will reduce the efficiency.

The idealized diesel cycle is shown on the next slide.

33

P

0P 5 1

2 3

4

intake exhaust

1V2V 3V

1Q

injection-ignition

adiabats

power-stroke: T drops

aircompressed

V

4 -> 1 constant volume so )( 141 TTCQ V

2 -> 3 constant pressure so )( 232 TTCQ P

Using the definition of gamma )()(

123

14

TT

TT

2Q

34

Introduce the expansion ratio:3

1

V

VrE

)1(21 112

122

111

crTTsoVTVT

2 -> 3 ideal gas and P=constant, so3

3

2

2

T

V

T

V

E

c

r

rTT

V

V

V

VT

V

VTT 23

2

1

1

32

2

323

Using (1) )2(13 E

c

r

rTT

3 -> 4 13

41

141

33

Er

TTsoVTVT

and using (2) )3(14

E

c

r

rTT

Using (1), (2), and (3) in the expression for the efficiency

35

1

11

1

cE

c

E

c

rrr

rr

Divide all terms by

cr

11

11

cE

cE

rr

rr

EXAMPLE: If then

4.120r0.10r cE 65.0

Real diesel engines have efficiencies of about 0.4

Both the gasoline and diesel engines take in the working substance at the beginning of the cycle and eject it at the end of the cycle. These are not closed systems and hence not appropriate systems for the thermodynamic principles that we have used. However they can, in fact, be replaced by the closed systems that we have been discussing.

)()(

123

14

TT

TT

36

Since the efficiency of a heat engine cannot be 1.0, the question arises as to the upper limit for the efficiency and what factors influence the efficiency. These questions were investigated by a brilliant French engineer, Sadi Carnot. He considered an engine operating cyclically and reversibly between two heat reservoirs.

A Carnot cycle is a reversible series of adiabatic, isothermal, adiabatic, isothermal processes which returns the thermodynamic system to its original state. The “working substance” can be any thermodynamic material, but here we will consider an ideal gas.The cycle is shown on the next slide.

NOTE: You may wonder what the relevance of this cycle is to the cycles that we have discussed. The answer will be discussed in the next chapter.

37

adiabat

adiabat

1

2

3

4

2T

1T

isotherm

isotherm

P

V

Carnot Cycle

1 -> 2 No heat enters or leaves. As the compression slowly continues work is done on the system and the internal energy increases. Thus the temperature also increases.2 -> 3 The T is constant and hence the internal energy does not change. As the volume slowly increases, the system does work on its surroundings and so heat must enter from the reservoir to keep the T constant.

2Q

1Q

{Not to scale!}

38

3 -> 4 The system is thermally isolated so that no heat enters or leaves. As the slow expansion continues the system does work on its surroundings. The internal energy decreases and so T decreases.

4 -> 1 The T is constant and therefore the internal energy does not change. As the volume is slowly decreased, the surroundings do work on the system, and so heat must go from the system to the low-temperature reservoir to maintain constant internal energy.

A Carnot refrigerator is a Carnot cycle operating in the opposite direction. The magnitudes of the work and heat transfers remain the same.

39

In the Carnot cycle heat transfer occurs during the isothermal processes. For these portions of the cycle there is no change in the internal energy. PdVWandWQ

Since we are considering an ideal gas we obtain:

1

411

2

322 lnln

V

VnRTQ

V

VnRTQ

For the adiabatic processes: 141

132

111

122

VTVTVTVT

Dividing these last two equations givesHence

)/()/( 4132 VVVV )/ln()/ln( 1423 VVVV

Therefore 2

1

2

1

2

1 1T

Tsoand

T

T

Q

Q

We will make use of this equation in the next chapter.

(What a simple and beautiful result!)

40

Real Refrigerator (or air conditioner)

We are concentrating on the Carnot cycle, but it should be noted that, while we talk of a Carnot Refrigerator, the refrigerator in our kitchen does not use the Carnot cycle, but uses a modification of what is called the Rankine Cycle. The COP is more difficult to work out and we will not do so.A schematic diagram of an actual refrigerator is shown on slide 42. The working substance changes back and forth from a gas to a liquid. The most common substance was Freon-12(CCl2F2). However escaping freon destroys the Earth’s ozone layer and so it has been replaced by other substances such asC2H2F4.

Notice that the throttling process is used. On the diagram it is represented by a dashed line (green) because it is an irreversible process. Some liquid is vaporized and the liquid-vapor system cools. (T4<T3)

41

The evaporator consists of a series of pipes connected to the interior of the refrigerator. After the throttling process the working substance is cooler than the interior of the refrigerator, so it absorbs heat from the interior, cooling the interior. (T1=T4). A change of phase takes place.The gas is then compressed so that its pressure and temperature increase and then reaches the condenser. (T2>T1)The condenser consists of a series of pipes connected to the room. (You can see these underneath the refrigerator and it might be necessary to clean this system periodically.) The gas is now at a higher temperature than the outside. In the condenser the gas condenses, giving up heat to the outside. A change of phase again takes place.

42

condenser

evaporator

com

pres

sor

Throttle

room

refrigerator

2Q

1Q

W

liquid gas

liquid + gas

1

23

4

P

V

throttle

Real Refrigerator 21 Gas compressed: P,T raised32 Liquefies, giving up heat to

room via network of pipes43 Through throttle, reducing

P and T14 Absorbs heat and becomes

a gas via a network of pipes

43

EXAMPLE: A Carnot cycle operates between two heat reservoirs at temperatures of 400K and 300K.(a) If the engine receives 1200J from the high-T reservoir in each cyclehow much energy does it reject to the low temperature reservoir?(b) If the engine is operated as a refrigerator and receives 1200J from the low-T reservoir, how much energy does it deliver to the high-T reservoir?(c) How much work is done by the engine in each case?

1200J

400K

300K

1200J

300J

-900J

-1600J

-400J

(a) 250.0400300

112

1 K

K

T

T

212

1

2

1 75.075.01 QQQ

Q

Q

Q

JQ 9001

JWJQQW 30030012

44

(b) Carnot refrigerator

Again JQQQQ

Q1600

34

75.0 2122

1

JWJQQW 400)12001600(12

45

EXAMPLE: (a) Show that for Carnot engines operating between the same high-T reservoirs but different low-T reservoirs, the engine operating over the largest T difference has the greatest efficiency. (b) Is the more effective way to increase the efficiency of a Carnot engine is to increase the temperature of the high-T reservoir keeping the low-T reservoir fixed, or vice-versa?(a) TfixedT

T

T

T

TT

T

T

2

22

12

2

11

(b) 222

1

222

122

1

2

111

11TTT

T

TTT

T

T

T

TTT

1222212121

11

TTTTTTTTTT

Hence decreasing the T of the low-T reservoir causes the largest increase in efficiency.

46

EXAMPLE: The “sadly Carnot” cycle.

adiabat

1

2

T,U max

Q reverses

Ideal gas

P

V

This is called the “sadly Carnot” cycle because, with a cursory analysis, the efficiency appears to be unity. We are used to cycles involving isobaric, isothermal, isochoric or adiabatic processes. We calculate the net work done and use the end points to calculate the change in internal energy. Such a calculation will lead to an efficiency of unity.

The error comes about by assuming that the heat transfer is unidirectional along the linear portion of the cycle. For the linear portion P=aV+b. Along the linear part 1 -> 2, heat both leaves and enters the system. There is a transition point M at which the direction of heat transfer changes. At this point

H

M

0)(

0)(

2

2

dV

VQdand

dV

VdQ

47

There is also a point H at which the temperature, and hence U, is a maximum. At this point 0

)(

dV

VdU

One can solve for the points H and M in a straightforward manner.

48

EXAMPLE: One kilomole of an ideal gas is the working substance of a Carnot engine. During the isothermal expansion the volume doubles. The ratio of the final volume to the initial volume in the adiabatic expansion is 5.7. The work output of the engine in each cycle is Calculate the temperatures of the reservoirs between which the engine operates. Take

J5109RCV )2/3(

P

V

A

B

C

isotherm

adiabat

A -> B isothermal expansion

ABAB WQWQU 2200

)2ln(ln AAA

B

V

V

AAAB VPV

VVPPdVW

B

A

2T

)2ln()2ln( 222 RTQRTWAB

But soT

TT

T

T

Q

W

2

12

2

1

2

1

1T

49

)1(15610314.8)2ln(

109)2ln(

10912

3

5

122

12

2

5

KTT

KJ

JTT

T

TT

RT

J

B -> C adiabatic expansion. From

constantTV sonRTPVandconstantPV 1

)2(19.3)7.5(1

23/2

1

1

211

12

T

T

V

V

T

TVTVT

B

CCB

(2) into (1) KTKTT 2.7115619.3 111

KTKKT 2271562.71 22

50

In this chapter we examined the throttling process, which is isenthalpic. This process is used in refrigerators and to liquify real gases.

The concept of a heat engine was introduced and the efficiency of the gasoline and diesel engines studied.

The most important concept introduced, for this course, was that of a Carnot Cycle.