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Physical Chemistry

CHAPTER 1
INTRODUCTION 1.1 What is thermodynamics?
Thermodynamics is the science which has evolved from the original investigations in the 19th century into the nature of “heat.” At the time, the leading theory of heat was that it was a type of ﬂuid, which could ﬂow from a hot body to a colder one when they were brought into contact. We now know that what was then called “heat” is not a ﬂuid, but is actually a form of energy – it is the energy associated with the continual, random motion of the atoms which compose macroscopic matter, which we can’t see directly. This type of energy, which we will call thermal energy, can be converted (at least in part) to other forms which we can perceive directly (for example, kinetic, gravitational, or electrical energy), and which can be used to do useful things such as propel an automobile or a 747. The principles of thermodynamics govern the conversion of thermal energy to other, more useful forms. For example, an automobile engine can be though of as a device which ﬁrst converts chemical energy stored in fuel and oxygen molecules into thermal energy by combustion, and then extracts part of that thermal energy to perform the work necessary to propel the car forward, overcoming friction. Thermodynamics is critical to all steps in this process (including determining the level of pollutants emitted), and a careful thermodynamic analysis is required for the design of fuel-eﬃcient, low-polluting automobile engines. In general, thermodynamics plays a vital role in the design of any engine or power-generating plant, and therefore a good grounding in thermodynamics is required for much work in engineering. If thermodynamics only governed the behavior of engines, it would probably be the most economically important of all sciences, but it is much more than that. Since the chemical and physical state of matter depends strongly on how much thermal energy it contains, thermodynamic principles play a central role in any description of the properties of matter. For example, thermodynamics allows us to understand why matter appears in diﬀerent phases (solid, liquid, or gaseous), and under what conditions one phase will transform to another. 1

CHAPTER 1. INTRODUCTION

2

The composition of a chemically-reacting mixture which is given enough time to come to “equilibrium” is also fully determined by thermodynamic principles (even though thermodynamics alone can’t tell us how fast it will get there). For these reasons, thermodynamics lies at the heart of materials science, chemistry, and biology. Thermodynamics in its original form (now known as classical thermodynamics) is a theory which is based on a set of postulates about how macroscopic matter behaves. This theory was developed in the 19th century, before the atomic nature of matter was accepted, and it makes no reference to atoms. The postulates (the most important of which are energy conservation and the impossibility of complete conversion of heat to useful work) can’t be derived within the context of classical, macroscopic physics, but if one accepts them, a very powerful theory results, with predictions fully in agreement with experiment. When at the end of the 19th century it ﬁnally became clear that matter was composed of atoms, the physicist Ludwig Boltzmann showed that the postulates of classical thermodynamics emerged naturally from consideration of the microscopic atomic motion. The key was to give up trying to track the atoms individually and instead take a statistical, probabilistic approach, averaging over the behavior of a large number of atoms. Thus, the very successful postulates of classical thermodynamics were given a ﬁrm physical foundation. The science of statistical mechanics begun by Boltzmann encompasses everything in classical thermodynamics, but can do more also. When combined with quantum mechanics in the 20th century, it became possible to explain essentially all observed properties of macroscopic matter in terms of atomic-level physics, including esoteric states of matter found in neutron stars, superﬂuids, superconductors, etc. Statistical physics is also currently making important contributions in biology, for example helping to unravel some of the complexities of how proteins fold. Even though statistical mechanics (or statistical thermodynamics) is in a sense “more fundamental” than classical thermodynamics, to analyze practical problems we usually take the macroscopic approach. For example, to carry out a thermodynamic analysis of an aircraft engine, its more convenient to think of the gas passing through the engine as a continuum ﬂuid with some speciﬁed properties rather than to consider it to be a collection of molecules. But we do use statistical thermodynamics even here to calculate what the appropriate property values (such as the heat capacity) of the gas should be.

CHAPTER 1. INTRODUCTION

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1.2 Energy and Entropy
The two central concepts of thermodynamics are energy and entropy. Most other concepts we use in thermodynamics, for example temperature and pressure, may actually be deﬁned in terms of energy and entropy. Both energy and entropy are properties of physical systems, but they have very diﬀerent characteristics. Energy is conserved: it can neither be produced nor destroyed, although it is possible to change its form or move it around. Entropy has a diﬀerent character: it can’t be destroyed, but it’s easy to produce more entropy (and almost everything that happens actually does). Like energy, entropy too can appear in diﬀerent forms and be moved around. A clear understanding of these two properties and the transformations they undergo in physical processes is the key to mastering thermodynamics and learning to use it conﬁdently to solve practical problems. Much of this book is focused on developing a clear picture of energy and entropy, explaining their origins in the microscopic behavior of matter, and developing eﬀective methods to analyze complicated practical processes1 by carefully tracking what happens to energy and entropy.

1.3 Some Terminology
Most ﬁelds have their own specialized terminology, and thermodynamics is certainly no exception. A few important terms are introduced here, so we can begin using them in the next chapter. 1.3.1 System and Environment In thermodynamics, like in most other areas of physics, we focus attention on only a small part of the world at a time. We call whatever object(s) or region(s) of space we are studying the system. Everything else surrounding the system (in principle including the entire universe) is the environment. The boundary between the system and the environment is, logically, the system boundary. The starting point of any thermodynamic analysis is a careful deﬁnition of the system.
Environment System Boundary System

1 Rocket

motors, chemical plants, heat pumps, power plants, fuel cells, aircraft engines, . . .

CHAPTER 1. INTRODUCTION

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Mass Mass

Control Mass

Control Volume

Figure 1.1: Control masses and control volumes. 1.3.2 Open, closed, and isolated systems Any system can be classiﬁed as one of three types: open, closed, or isolated. They are deﬁned as follows: open system: Both energy and matter can be exchanged with the environment. Example: an open cup of coﬀee. closed system: energy, but not matter, can be exchanged with the environment. Examples: a tightly capped cup of coﬀee. isolated system: Neither energy nor matter can be exchanged with the environment – in fact, no interactions with the environment are possible at all. Example (approximate): coﬀee in a closed, well-insulated thermos bottle. Note that no system can truly be isolated from the environment, since no thermal insulation is perfect and there are always physical phenomena which can’t be perfectly excluded (gravitational ﬁelds, cosmic rays, neutrinos, etc.). But good approximations of isolated systems can be constructed. In any case, isolated systems are a useful conceptual device, since the energy and mass contained inside them stay constant. 1.3.3 Control masses and control volumes Another way to classify systems is as either a control mass or a control volume. This terminology is particularly common in engineering thermodynamics. A control mass is a system which is deﬁned to consist of a speciﬁed piece or pieces of matter. By deﬁnition, no matter can enter or leave a control mass. If the matter of the control mass is moving, then the system boundary moves with it to keep it inside (and matter in the environment outside). A control volume is a system which is deﬁned to be a particular region of space. Matter and energy may freely enter or leave a control volume, and thus it is an open system.

CHAPTER 1. INTRODUCTION

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1.4 A Note on Units
In this book, the SI system of units will be used exclusively. If you grew up anywhere but the United States, you are undoubtedly very familiar with this system. Even if you grew up in the US, you have undoubtedly used the SI system in your courses in physics and chemistry, and probably in many of your courses in engineering. One reason the SI system is convenient is its simplicity. Energy, no matter what its form, is measured in Joules (1 J = 1 kg-m2/s2 ). In some other systems, diﬀerent units are used for thermal and mechanical energy: in the English system a BTU (“British Thermal Unit”) is the unit of thermal energy and a ft-lbf is the unit of mechanical energy. In the cgs system, thermal energy is measured in calories, all other energy in ergs. The reason for this is that these units were chosen before it was understood that thermal energy was like mechanical energy, only on a much smaller scale. 2 Another advantage of SI is that the unit of force is indentical to the unit of (mass x acceleration). This is only an obvious choice if one knows about Newton’s second law, and allows it to be written as F = ma. (1.1)

In the SI system, force is measured in kg-m/s2 , a unit derived from the 3 primary SI quantities for mass, length, and time (kg, m, s), but given the shorthand name of a “Newton.” The name itself reveals the basis for this choice of force units. The units of the English system were ﬁxed long before Newton appeared on the scene (and indeed were the units Newton himself would have used). The unit of force is the “pound force” (lbf), the unit of mass is the “pound mass” (lbm) and of course acceleration is measured in ft/s2 . So Newton’s second law must include a dimensional constant which converts from M a units (lbm ft/s2 ) to force units (lbf). It is usually written F= where gc = 32.1739 ft-lbm/lbf-s2. Of course, in SI gc = 1.
2 Mixed unit systems are sometimes used too. American power plant engineers speak of the “heat rate” of a power plant, which is deﬁned as the thermal energy which must be absorbed from the furnace to produce a unit of electrical energy. The heat rate is usually expressed in BTU/kw-hr.

1 ma, gc

(1.2)

(1.3)

CHAPTER 1. INTRODUCTION

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In practice, the units in the English system are now deﬁned in terms of their SI equivalents (e.g. one foot is deﬁned as a certain fraction of a meter, and one lbf is deﬁned in terms of a Newton.) If given data in Engineering units, it is often easiest to simply convert to SI, solve the problem, and then if necessary convert the answer back at the end. For this reason, we will implicitly assume SI units in this book, and will not include the gc factor in Newton’s 2nd law.

CHAPTER 2
ENERGY, WORK, AND HEAT 2.1 Introduction
Energy is a familiar concept, but most people would have a hard time deﬁning just what it is. You may hear people talk about “an energy-burning workout,” “an energetic personality,” or “renewable energy sources.” A few years ago people were very concerned about an “energy crisis.” None of these uses of the word “energy” corresponds to its scientiﬁc deﬁnition, which is the subject of this chapter. The most important characteristic of energy is that it is conserved: you can move it around or change its form, but you can’t destroy it, and you can’t make more of it.1 Surprisingly, the principle of conservation of energy was not fully formulated until the middle of the 19th century. This idea certainly does seem nonsensical to anyone who has seen a ball roll across a table and stop, since the kinetic energy of the ball seems to disappear. The concept only makes sense if you know that the ball is made of atoms, and that the macroscopic kinetic energy of motion is simply converted to microscopic kinetic energy of the random atomic motion.

2.2 Work and Kinetic Energy
Historically, the concept of energy was ﬁrst introduced in mechanics, and therefore this is an appropriate starting point for our discussion. The basic equation of motion of classical mechanics is due to Newton, and is known as Newton’s second law.2 Newton’s second law states that if a net force F is applied to a body, its center-of-mass will experience an acceleration a proportional to F: F = ma. The proportionality constant m is the inertial mass of the body.
1 Thus, energy can’t be burned (fuel is burned), it is a property matter has (not personalities), there are no sources of it, whether renewable or not, and there is no energy crisis (but there may be a usable energy, or availability, crisis). 2 For now we consider only classical, nonrelativistic mechanics.

(2.1)

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CHAPTER 2. ENERGY, WORK, AND HEAT

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Suppose a single external force F is applied to point particle moving with velocity v. The force is applied for an inﬁnitesimal time dt, during which the velocity changes by dv = a dt, and the position changes by dx = v dt. m F v Taking the scalar product3 (or dot product) of Eq. (2.1) with dx gives F · dx = ma · dx dv = m dt = mv · dv = d(mv2 /2).

· [vdt]

(2.2)

Here v = |v| is the particle speed. Note that only the component of F along the direction the particle moves is needed to determine whether v increases or decreases. If this component is parallel to dx, the speed increases; if it is antiparallel to dx the speed decreases. If F is perpendicular to dx, then the speed doesn’t change, although the direction of v may. Since we’ll have many uses for F · dx and mv2 /2, we give them symbols and names. We call F · dx the inﬁnitesimal work done by force F, and give it the symbol dW : ¯ dW = F · dx ¯ (We’ll see below why we put a bar through the d in dW .) ¯ The quantity mv2 /2 is the kinetic energy Ek of the particle: mv2 2 (2.3)

Ek =

(2.4)

With these symbols, Eq. (2.2) becomes dW = d(Ek ). ¯ (2.5)

Equation (2.5) may be interpreted in thermodynamic language as shown in Fig. 2.1. A system is deﬁned which consists only of the particle; the energy
3 Recall that the scalar product of two vectors A = i A + j A + k A and B = i B + j B + i j i j k k Bk is deﬁned as A · B = Ai Bi + Aj Bj + Ak Ck . Here i, j, and k are unit vectors in the x, y, and z directions, respectively.

CHAPTER 2. ENERGY, WORK, AND HEAT

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System dW d(Ek) Environment
Figure 2.1: Energy accounting for a system consisting of a single point particle acted on by a single force for time dt. “stored” within the system (here just the particle kinetic energy) increases by d(Ek ) due to the work dW done by external force F. Since force F is produced ¯ by something outside the system (in the environment), we may regard dW as ¯ an energy transfer from the environment to the system. Thus, work is a type of energy transfer. Of course, dW might be negative, in which case d(Ek ) < 0. ¯ In this case, the direction of energy transfer is actually from the system to the environment. The process of equating energy transfers to or from a system to the change in energy stored in a system we will call energy accounting. The equations which result from energy accounting we call energy balances. Equation (2.5) is the ﬁrst and simplest example of an energy balance – we will encounter many more. If the force F is applied for a ﬁnite time t, the particle will move along some trajectory x(t).

A

F(x,t) m v

B

The change in the particle kinetic energy ∆Ek = Ek (B) − Ek (A) can be determined by dividing the path into many very small segments, and summing Eq. (2.2) for each segment.
∆x i Fi

In the limit where each segment is described by an inﬁnitesimal vector dx,

CHAPTER 2. ENERGY, WORK, AND HEAT

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W2

W

(Ek )

W1

(Ek )

Figure 2.2: Energy accounting for a single particle acted on by (a) a single force (b) multiple forces for ﬁnite time. the sum becomes an integral: dW = ¯ path path

d(Ek )

(2.6)

The right-hand side of this can be integrated immediately: d(Ek ) = ∆Ek . path (2.7)

The integral on the left-hand side deﬁnes the total work done by F:

W = path dW = ¯ path F · dx.

(2.8)

Note that the integral is along the particular path taken. Eq. (2.6) becomes W = ∆Ek . (2.9)

The thermodynamic interpretation of this equation is shown in Fig. 2.2 and is similar to that of Eq. (2.5): work is regarded as a transfer of energy to the system (the particle), and the energy stored in the system increases by the amount transferred in. (Again, if W < 0, then the direction of energy transfer is really from the system to the environment, and in this case ∆Ek < 0.) If two forces act simultaneously on the particle, then the total applied force is the vector sum: F = F1 + F2 . In this case, Eq. (2.9) becomes W1 + W2 = ∆Ek , (2.10)

where W1 = path F1 · dx and W2 = path F2 · dx.4 The generalization to N forces is obvious: the work done by all N forces must be considered to compute ∆Ek .
4 For now we’re considering a point particle, so the path followed is the same for both forces; this won’t be true for extended objects, which will be considered in section 2.4.

CHAPTER 2. ENERGY, WORK, AND HEAT

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2.3 Evaluation of Work
Since in general a force may depend on factors such as the instantaneous particle position x, the instantaneous velocity v, or may depend explicitly on time, the work done by the force will clearly depend on the path the particle takes from A to B, how fast it travels, and the particular time it passes each point. Since there are inﬁnitely many possible trajectories x(t) which start at point A at some time and pass through point B at some later time, there are inﬁnitely many possible values for W = path dW ; we need additional information [i.e., ¯ x(t)] to evaluate W . This is the reason we put the bar through dW but not through d(Ek ). It’s ¯ always true that path d(Q) may be formally evaluated to yield QB − QA , where Q is some function of the state (position, velocity, etc.) of the particle and of time, and QA and QB denote the values of Q when the particle is at endpoints of the path. But dW is not like this: it’s only the symbol we use to denote “a little ¯ bit of work.” It really equals F · dx, which is not of the form d(Q), so can’t be integrated without more information. Quantities like dW are known as ¯ “inexact diﬀerentials.” We put the bar in dW just to remind ourselves that ¯ it is an inexact diﬀerential, and so its integral depends on the particular path taken, not only on the state of the particle at the beginning and end of the path. Example 2.1 The position-dependent force F(x, y, z) = +iC −i2C if y > 0 if y ≤ 0

is applied to a bead on a frictionless wire. The bead sits initially at the origin, and the wire connects the origin with (L, 0, 0). How much work does F do to move the bead along wire A? How much along wire B? Does the contact force of the bead against the wire do any work?

y

A L B x

Solution: W = path F(x, y, z) · dx.

CHAPTER 2. ENERGY, WORK, AND HEAT Since F always points in the x direction, F(x, y, z) · dx = Fx (x, y, z)dx

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Therefore, along path A, W = CL, and along path B, W = −2CL. Along path A, the force does work on the particle, while along path B the particle does work on whatever is producing the force. Of course, for motion along path B to be possible at all, the particle would have to have an initial kinetic energy greater than 2CL. The contact force does no work, since it is always perpendicular to the wire (and therefore to dx), so Fcontact · dx = 0. If we do know x(t), we can convert the path integral deﬁnition of work [Eq. (2.8)] into a time integral, using dx = v(t)dt:

tB

W = tA F(x(t), v(t), t) · v(t) dt

(2.11)

This is often the easiest way to evaluate work. Note that the integrand is F · v. Therefore, F · v is the rate at which force F does work, or in other words the ˙ instantanteous power being delivered by F. We denote the power by W :

˙ W = F·v

(2.12)

L F d Fa

M

x(t)

t Example 2.2 A ball initially at rest at x = 0 in a viscous ﬂuid is pulled in a straight line by a string. A time-dependent force Fa (t) is applied to the string, which causes the ball to move according to

T

x(t) =

L 1 − cos 2

πt T

.

At time t = T , the ball comes to rest at x = L and the force is removed. As the ball moves through the ﬂuid, it experiences a drag force proportional to its

CHAPTER 2. ENERGY, WORK, AND HEAT

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speed: Fd = −C x(t). How much work is done by the applied force to move the ˙ ball from x = 0 to x = L? Solution: Newton’s second law requires Fa + Fd = m¨(t), x so Fa (t) = m¨(t) + C x(t). x ˙ Since we know x(t), we can diﬀerentiate to ﬁnd x(t) = ˙ and L π sin τ 2 T (2.15) (2.14) (2.13)

L π 2 cos τ (2.16) 2 T where τ = πt/T . Substituting these expressions into Eq. (2.14) results in x(t) = ¨ Fa (t) = CL π mL sin τ + 2 T 2 π T
2

cos τ.

To calculate the work done by Fa (t), we need to evaluate
L

Wa = path Fa · dx =
0

Fa dx.

Since we know both Fa (t) and x(t), it is easiest to convert this path integral to a time integral using dx = x(t)dt: ˙
T

Wa =
0

Fa (t)x(t) dt. ˙

Changing the integration variable to τ (dτ = (π/T )dt), Wa = Since π 0

L 2

2

π T

π

C sin2 τ +
0 π 0

π sin τ cos τ dτ. T

sin2 τ dτ = π/2 and

sin τ cos τ dτ = 0, Wa = π 2 CL2 . 8T

If there were no drag (C = 0), then the work would be zero, since the work done to accelerate the ball for t < T /2 would be fully recovered in decelerating the ball for t > T /2. But in the presence of a drag force, a ﬁnite amount of work must be done to overcome drag, even though the ball ends as it began with no kinetic energy.

CHAPTER 2. ENERGY, WORK, AND HEAT
Fext,i i F ji j Fext,j Fij System Boundary

14

Figure 2.3: External and internal forces acting on two masses of a rigid body. Note that the work is inversely proportional to the total time T . It takes more work to push the ball rapidly through the ﬂuid (short T ) than slowly. By carrying out the process very slowly, it is possible to make Wa as small as desired, and in the limit of T → ∞ the process requires no work. This behavior is characteristic of systems which exhibit viscous drag.

2.4 Energy Accounting for Rigid Bodies
Up until now we have only considered how to do energy accounting for point masses. To develop energy accounting methods for macroscopic matter, we can use the fact that macroscopic objects are composed of a very large number of what we may regard as point masses (atomic nuclei), connected by chemical bonds. In this section, we consider how to do energy accounting on a macroscopic object if we make the simplifying assumption that the bonds are completely rigid. We’ll relax this assumption and complete the development of energy accounting for macroscopic matter in section 2.8. Consider a body consisting of N point masses connected by rigid, massless rods, and deﬁne the system to consist of the body (Fig. 2.3). The rods will transmit forces between the masses. We will call these forces internal forces, since they act between members of the system. We will assume the internal forces are directed along the rods. The force exerted on (say) mass j by mass i will be exactly equal in magnitude and opposite in direction to that exerted on mass i by mass j (Fij = −Fji ), since otherwise there would be a force imbalance on the rod connecting i and j. No force imbalance can occur, since the rod is massless and therefore would experience inﬁnite acceleration if the forces were unbalanced. (Note this is Newton’s third law.) Let the masses composing the body also be acted on by arbitrary external forces from the environment. The external force on mass i will be denoted

CHAPTER 2. ENERGY, WORK, AND HEAT Fext,i. The energy balance in diﬀerential form for one mass, say mass i, is   dWext,i +  ¯ j 15

Fji  · dxi = d(Ek,i),

(2.17)

where dWext,i = Fext,i · dxi and of course Fii = 0. Summing the energy ¯ balances for all masses results in an energy balance for the entire system: dWext,i + ¯ i i j

Fji · dxi = d(Ek ),

(2.18)

where d(Ek ) = i d(Ek,i) = i 2 d(mi vi /2)

(2.19)

is the change in the total kinetic energy of the body. Equation (2.18) can be simpliﬁed considerably, since the second term on the left is exactly zero. To see this, recall that the rods are rigid, so d(|xi − xj |) = 0 for all i and j. Equation (2.20) can be written as (xi − xj ) · d(xi − xj ) = 0. (2.21) (2.20)

Now Fij is parallel to (xi − xj ), so multiplying Eq. (2.21) by |Fij |/|xi − xj | results in Fij · d(xi − xj ) = 0. (2.22) Since Fji = −Fij , we can re-write this as Fji · dxi = −Fij · dxj . (2.23)

Therefore, because the body is rigid, the work done by Fji on mass i is precisely equal to the negative of the work done by Fij on mass j. Thus, the internal forces Fij cause a transfer of kinetic energy from one mass within the body to another, but considering the body as a whole, do no net work on the body. Mathematically, the second term on the left of Eq. (2.18) is a sum over all pairs of mass indices (i, j). Because of Eq. (2.23), for every i and j, the (i, j) term in this sum will exactly cancel the (j, i) term, with the result that the double sum is zero. With this simpliﬁcation (for rigid bodies), Eq. (2.18) reduces to dWext,i = d(Ek ). ¯ i (2.24)

CHAPTER 2. ENERGY, WORK, AND HEAT
(a) (b) v Mg F t θ F = Mg sin(θ) t F = M v2 / R R F B q F=qvxB (c) v

16

Figure 2.4: Some forces which do no work: (a) traction force on a rolling wheel; (b) centrifugal force; (c) Lorentz force on a charged particle in a magnetic ﬁeld We see that to carry out an energy balance on a rigid body, we only need consider work done by external forces, not by internal ones. We can always tell which forces are external ones – they are the ones which cross the system boundary on a sketch. A macroscopic solid object is composed of a huge number of essentially point masses (the atomic nuclei) connected by chemical bonds (actually rapidly moving, quantum-mechanically smeared out electrons). If we ignore for the moment the fact that bonds are not really rigid, a solid object can be approximated as a rigid body. If this approximation holds, then the appropriate energy balance equation will be Eq. (2.24). For simplicity, assume that the external forces act only at L discrete locations on the surface of the object, where it contacts the environment. 5 In this case, L the external work term in Eq. (2.24) becomes =1 F · dx , where dx is the displacement of the surface of the object at the point where the force F is applied. The energy balance Eq. (2.24) becomes
L

F · dx = d(Ek ).
=1

(2.25)

It is very important to remember that the displacements to use in this equation are those where the forces are applied, and may diﬀer for each force. Do not make the mistake of using the displacement of some other point (e.g. the center of mass). If a force is applied to a macroscopic object at a point where it is stationary, the force does no work no matter how large the force is. (If you push against a stationary wall, you may exert yourself, but you do no work on it.) Also, a force
5 If the macroscopic force is exerted over some small but ﬁnite contact area, the macroscopic force F in Eq. (2.25) is simply the sum over the atomic-level forces Fext,i in Eq. (2.24) for all atoms i in the contact area.

CHAPTER 2. ENERGY, WORK, AND HEAT

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applied perpendicular to the instantaneous direction of motion of the contact area can do no work. Some common forces which do no work are shown in Fig. 2.4. A traction force |Ft | = mg sin θ in the plane of the surface keeps a rolling wheel from sliding down a hill; but since the wheel is instantaneously stationary where it contacts the ground, Ft · dx = 0 and therefore the traction force does no work. A centrifugal force and the Lorentz force a charged particle experiences in a magnetic ﬁeld are both perpendicular to the direction of motion, and thus can do no work. Example 2.3 A downward force F1 is applied to a rigid, horizontal lever a distance L1 to the right of the pivot point. A spring connects the lever to the ground at a distance L2 to the left of the pivot, and exerts a downward force F2 . An upward force Fp is exerted on the lever at the pivot. Evaluate the work done by each force if end 2 is raised by dy2 , and determine the value of F1 which achieves this motion without changing the kinetic energy of the lever.
System Boundary F 1

F

2

Spring

F

p

L2

L1

Solution: Deﬁne the system to consist of the lever only (a rigid body). The body is acted on by three external forces, and so we must evaluate the work input to the system from each force. Since the lever is rigid, if the height of end 2 changes by dy2 while the height at the pivot point is unchanged, then the height of end 1 must change by dy1 = −(L1 /L2 )dy2 . So the three work inputs are: dW1 ¯ dW2 ¯ dWp ¯ = (−jF1 ) · (−jL1 dy2 /L2 ) = (F1 L1 /L2 )dy2 > 0 = (−jF2 ) · (+jdy2 ) = −F2 dy2 < 0 = (+jFp ) · (0) = 0. (2.26) (2.27) (2.28)

Note that the work due to the pivot force is zero, since the lever does not move at the pivot. Force F1 does positive work on the lever, since the force and displacement are in the same direction. The spring which produces force F2 does negative work on the lever, since the force and displacement are in opposite directions. In this case, we say that the lever does positive work on the spring, since the force exerted by the lever on the spring is oppositely directed

CHAPTER 2. ENERGY, WORK, AND HEAT to F2 (Newton’s third law). The energy balance on the lever is then dW1 + dW2 + dWp ¯ ¯ ¯ (F1 L1 /L2 − F2 )dy2 = d(Ek ) = d(Ek ).

18

(2.29)

If we wish to move the lever without increasing its kinetic energy, then we must choose F1 L1 = F2 L2 . (2.30) This is the familiar law of the lever, but note that we obtained it from an energy balance, not by balancing torques as would be done in mechanics.

2.5 Conservative Forces and Potential Energy
2.5.1 A Uniform Gravitational Field Suppose a point particle near the surface of the earth is acted on by gravity, which exerts a constant downward force Fg = −jmg. It is also acted on by an arbitrary external applied force Fa (x, t).

F (x,t) a Fg

In this case, Eq. (2.10) becomes Wa + Wg = ∆Ek where Wa = path (2.31)

Fa · dx

(2.32)

is the work done by the applied force, and Wg = path Fg · dx

(2.33)

is the work done by the gravitational force. Due to the special character of Fg (a constant force), Wg can be evaluated for an arbitrary path from A to B: Wg = − path jmg · dx

CHAPTER 2. ENERGY, WORK, AND HEAT = − yB 19

mgdy yA = −mg(yB − yA ) = −mg∆y.

(2.34) (2.35)

If ∆y < 0, gravity does work on the particle, and its kinetic energy increases. If ∆y > 0, Wg < 0, which means that the particle must do work against gravity. In this case the kinetic energy decreases. Note that Wg can be expressed solely in terms of the diﬀerence in a property (the height) of the particle at the beginning and end of its trajectory: any path connecting A and B would result in the same value for Wg . This is due to the special nature of the force Fg , which is just a constant. Of course, for an arbitrary force such as Fa (x, t), this would not be possible. The force Fg is the ﬁrst example of a conservative force. Since Wg is independent of the particular path taken, we can bring it to the other side of Eq. (2.31): Wa = (−Wg ) + ∆Ek = mg∆y + ∆Ek = ∆(Ek + mgy) (2.36)

We deﬁne mgy to be the gravitational potential energy Eg of the particle in this uniform gravitational ﬁeld: Eg = mgy. With this deﬁnition, Eq. (2.31) becomes Wa = ∆(Ek + Eg ). (2.38) (2.37)

Equations (2.31) and (2.38) are mathematically equivalent, but have diﬀerent interpretations, as shown in Fig. 2.5. In Eq. (2.31), the gravitational force is considered to be an external force acting on the system; the work Wg it does on the system is included in the energy balance but not any potential energy associated with it. In (b), the source of the gravitational force (the gravitational ﬁeld) is in eﬀect considered to be part of the system. Since it is now internal to the system, we don’t include a work term for it, but do include the gravitational potential energy (which we may imagine to be stored in the ﬁeld) in the system energy. It doesn’t matter which point of view we take – the resulting energy balance is the same because ∆Eg is deﬁned to be identical to

CHAPTER 2. ENERGY, WORK, AND HEAT

20

(a)

Wg (Ek )

(b)

Wa

Wa

(E k + E g )

Figure 2.5: Two energy accounting schemes to handle the eﬀects of a constant gravitational force. In (a), the gravitational ﬁeld is considered to be external to the system, while in (b) the ﬁeld is part of the system. −Wg . But remember not to mix these points of view: don’t include both Wg and ∆Eg in an energy balance! We may generalize this analysis to a macroscopic body. In this case, the gravitational potential energy becomes Eg = body ρ(x)gy dV,

(2.39)

where ρ(x) is the local mass density (kg/m3 ) at point x within the body. This can be re-written as (2.40) Eg = M gycm , where M= body ρ(x) dV

(2.41)

is the total mass of the body and ycm is the y-component of the center of mass, deﬁned by 1 xcm = ρ(x) x dV. (2.42) M 2.5.2 General Conservative Forces A constant force, such as discussed above, is the simplest example of a conservative force. The general deﬁnition is as follows: a force is conservative if and only if the work done by it in going from an initial position xA to a ﬁnal position xB depends only on the initial and ﬁnal positions, and is independent of the path taken. Mathematically, this deﬁnition may be stated as follows: Wc = path Fc · dx = f(xB ) − f(xA ),

(2.43)

CHAPTER 2. ENERGY, WORK, AND HEAT where f is some single-valued scalar function of position in space. For the special case of a closed path (xB = xA ), Eq. (2.43) reduces to Fc · dx = 0,

21

(2.44)

where denotes integrating all the way around the path. Therefore, the work done by a conservative force on a particle traversing any arbitrary closed loop is exactly zero. Either Eq. (2.43) or Eq. (2.44) may be taken as the deﬁnition of a conservative force. Only very special functions F(x, v, t) can satisfy the conditions for a conservative force. First of all, consider the dependence on velocity. The only way Eq. (2.44) can be satisﬁed by a velocity-dependent force for all possible loops, traversing the loop in either direction at arbitrary speed, is if the velocitydependent force does no work. This is possible if F(x, v, t) is always perpendicular to v. Thus, any conservative force can have an arbitrary velocity-dependent force Fv added to it and still be conservative as long as Fv · v = 0 at all times. It seems that in nature there is only one velocity-dependent conservative force, which is the Lorentz force felt by a charged particle moving through a magnetic ﬁeld B. This Lorentz force is given by FL = qv × B, (2.45)

which is always perpendicular to both v and B. Unless stated otherwise, we will assume from here on that conservative forces do not have a velocity-dependent part, keeping in mind that the Lorentz force is the one exception. Having dealt with the allowed type of velocity dependence, consider now the time dependence. It is clear that Fc can have no explicit time dependence (i.e., F(x(t)) is OK but F(x(t), t) is not). If Fc depended explicitly on time, then the result for Wc would too, rather than on just the endpoint positions in space. So we conclude that a conservative force (or at least the part which can do work) can depend explicitly only on position: Fc (x). 2.5.3 How to Tell if a Force is Conservative If we are given a force function F(x), how can we tell if it is conservative? First consider the inverse problem: If we know the function f(x), can we derive what Fc must be? Consider a straight-line path which has inﬁnitesimal length: xB = xA + dx. Then equation 2.43 reduces to Fc (xA ) · dx = f(xA + dx) − f(xA ). (2.46)

CHAPTER 2. ENERGY, WORK, AND HEAT Since dx is inﬁnitesimal, we may expand f(xA + dx) in a Taylor series:6 f(xA + dx) = f(xA ) + where the gradient of f is deﬁned by f =i ∂f ∂f ∂f +j +k . ∂x ∂y ∂z f(xA ) · dx + O(|dx|2),

22

(2.47)

(2.48)

As we let |dx| go to zero, the higher-order terms go to zero rapidly, so Eq. (2.46) becomes Fc (x) · dx = f(xA ) · dx (2.49) The only way this equation can hold for arbitrary xA and dx is if (2.50)

Fc (x) =

f(x).

Therefore, a conservative force which depends only on position must be the gradient of some scalar function of position in space f(x). How can we tell if a given vector function F(x) is the gradient of some unknown scalar function f(x)? The easiest way is to write them both out explicitly: F(x, y, z) = iFi (x, y, z) + jFj (x, y, z) + kFk (x, y, z) ∂f ∂f ∂f f(x, y, z) = i +j +k . ∂x ∂y ∂z If these are equal, then each component must be equal, so Fi(x, y, z) Fj (x, y, z) Fk (x, y, z) = ∂f(x, y, z)/dx = ∂f(x, y, z)/dy = ∂f(x, y, z)/dz. (2.53) (2.54) (2.55) (2.51) (2.52)

Consider now the mixed second derivatives of f(x, y, z). It doesn’t matter which order we do the diﬀerentiation: ∂ ∂x ∂f ∂y = ∂ ∂y ∂f ∂x = ∂2f , ∂x∂y (2.56) f,

with similar results for the partial derivatives involving z. Therefore, if F = we may substitute eqs. (2.53) and (2.54) into Eq. (2.56) and obtain ∂Fj ∂Fi = . ∂x ∂y
6 If

(2.57)

this is not clear to you in vector form, write it out component by component.

CHAPTER 2. ENERGY, WORK, AND HEAT Similarly, ∂Fi ∂Fk = , ∂z ∂x and ∂Fj ∂Fk = , ∂z ∂y

23

(2.58) (2.59)

Equations (2.57)–(2.59) provide a simple test to determine if F(x) is conservative. If F passes this test, it should be possible to integrate equations (2.53)– (2.55) and ﬁnd a function f(x) such that F = f. If F fails the test, then no such f(x) exists. 2.5.4 Energy Accounting with Conservative Forces We can easily generalize the analysis of the mass in a constant gravitational ﬁeld to handle an arbitrary conservative force acting on a particle. The energy balance is Wa + Wc = ∆Ek . (2.60) Since the force is conservative, Wc = f(xB ) − f(xA ) = ∆f. Therefore, we may write the energy balance as Wa = ∆Ek − ∆f = ∆(Ek − f). (2.61)

Now deﬁne the potential energy associated with this conservative force as follows: Ep(x) = −f(x) + C. (2.62) Since only diﬀerences in potential energy have any physical signiﬁcance, we can set the additive constant C to any convenient value. The energy balance now becomes Wa = ∆(Ek + Ep ). (2.63) As with the gravitation example, the energy balances (2.60) and (2.63) are completely equivalent mathematically, and we can use whichever one we prefer. They diﬀer only in interpretation. Using Eq. (2.60), we regard whatever produces the conservative force (e.g. a gravitational, electric, or magnetic ﬁeld, a frictionless spring, etc.) as part of the environment – external to the system. Therefore, we include the work Wc done by this force on our system when we do energy accounting. If we write the energy balance as in Eq. (2.63), we are regarding the source of the conservative force as part of the system. Since in this case the force becomes an internal one, we don’t include the work Wc in the energy balance, but we must account for the potential energy stored in the ﬁeld or spring as part of the system energy.

CHAPTER 2. ENERGY, WORK, AND HEAT

24

2.6 Elementary Forces and Conservation of Energy
Elementary forces are those forces which are part of the basic structure of physics, such as the gravitational force, electromagnetic forces, nuclear forces, etc. These forces are responsible for all atomic-level or subatomic behavior, including chemical and nuclear bonding and the forces atoms feel when they collide with one another. (But quantum mechanics, rather than classical mechanics, must be used to correctly predict these features). As far as we know now, every elementary force of nature is conservative that is, it may be derived from some potential energy function. Considering how special conservative forces are (there are inﬁnitely more functions F(x) which are not the gradient of some f(x) than there are functions which are), this can be no accident – it must be a deep principle of physics. The universe can be thought of as a very large number of elementary particles interacting through conservative, elementary forces. If we do an energy accounting for the entire universe, treating the conservative interactions between particles by adding appropriate potential energy terms to the system energy as discussed in section 2.5.4, we ﬁnd7 ∆(Ek + Ep ) = 0, (2.64)

where Ek and Ep represent the kinetic and potential energies, respectively, of the entire universe. Of course there can be no external work term, since the entire universe is inside our system! Therefore, the total energy of the universe (kinetic + all forms of potential) is constant. Everything that has happened since the birth of the universe — its expansion, the condensation of protons and electrons to form hydrogen gas, the formation of stars and heavy nuclei within them, the formation of planets, the evolution of life on earth, you reading this book — all of these processes simply shift some energy from one type to another, never changing the total. The constancy of the energy of the universe is the principle of conservation of energy. Of course, any small part of the universe which is isolated from the rest in the sense that no energy enters or leaves it will also have constant total energy. Another way of stating the principle of conservation of energy is that there are no sinks or sources for energy — you can move it around or change its form, but you can’t create it, and you can’t destroy it.
7 Of course, to calculate E and E correctly we would have to consider not only quantum p k mechanics but general relativity. These change the details in important ways, but not the basic result that the energy of the universe is constant.

CHAPTER 2. ENERGY, WORK, AND HEAT

25

Why is the energy of the universe constant? This is equivalent to asking why all elementary forces are conservative. Quantum mechanics provides some insight into this question. In quantum mechanics, a system has a well-deﬁned constant total energy if two conditions are met: a) there are no interactions with external forces, and b) the laws governing the elementary forces are constant in time. If this is applied to the whole universe condition a) is automatically satisﬁed, and b) says simply that the basic laws of physics have always been the same as they are now. As far as we know, this is true – the laws of physics don’t depend on time.

2.7 Non-Conservative Forces
Since all elementary forces are conservative, it might be thought that any macroscopic forces between macroscopic objects (which, after all, are composed of elementary particles interacting through elementary forces) should be conservative. This is actually not true, as a simple thought experiment demonstrates. Imagine sliding an object around in a circle on a table, returning to the starting point. If the table were perfectly frictionless, it would take no net work to do this, since any work you do to accelerate the object would be recovered when you decelerate it. But in reality, you have to apply a force just to overcome friction, and you have to do net work to slide the object in a circle back to its original position. Clearly, friction is not a conservative force. If we were to look on an atomic scale at the interface between the object and the table as it slides, we don’t see a “friction force” acting at all. Instead, we would notice the roughness of both the table and the object – sometimes an atomic-scale bump sticking out of the object would get caught behind an atomic-scale ridge on the table. As the object continued to move, the bonds to the hung-up atoms stretch or bend, increasing their potential energy (like springs or rubber bands); ﬁnally, the stuck atoms break free and vibrate violently, as the energy due to bond stretching is released. The increased vibrational kinetic energy of these few atoms is rapidly transferred through the bonds to all of the other atoms in the object, resulting in a small increase in the random, thermal energy of the object.8 If we reverse the direction we slide the object, the apparent friction force
8 Essentially the same process happens in earthquakes as one plate of the earth’s crust attempts to slide past another one along faults (such as the San Andreas fault or the many other faults below the LA basin). The sliding slabs of rock get hung up, and as the plates keep moving, huge strain energy is built up. Eventually, the plates break free, converting the pent-up strain energy (potential energy) into the kinetic energy of ground motion, which we experience as an earthquake. Sliding friction is a microscopic version of an earthquake!

CHAPTER 2. ENERGY, WORK, AND HEAT

26

reverses direction too, always opposing the direction of motion. This means that the friction force depends on the velocity of the object. For sliding friction, the dependence is usually only on the direction of the velocity vector (not its magnitude). But viscous drag in a ﬂuid (also a type of friction) depends on the magnitude also, increasing with speed. This behavior is in sharp contrast to conservative forces, which only depend on position. For example, the gravitational force on an object of mass m is always mg directed in the same direction (toward the center of the earth) no matter what the velocity of the object is. We see then that macroscopic forces which are non-conservative (friction) are actually “eﬀective” forces which result from very complex atomic-level motion. Frictional forces always result in an irreversible conversion of macroscopic kinetic energy (the motion of the object) to disorganized, random thermal energy, and always oppose the direction of motion, so Fnc · dx is always negative.

2.8 The First Law of Thermodynamics
We now wish to do energy accounting for arbitrary macroscopic material systems. We’re already part way there – in Section 2.4 we developed an energy balance equation for macroscopic matter valid if the bonds between atoms were rigid. Unfortunately, this is not really the case. Bonds in solids can stretch and bend like springs, so the atoms are continually vibrating. This means that a solid will have kinetic energy associated with this motion, and potential energy due to stretching bonds. In liquids and gases, molecules can move and rotate, as well as vibrate. In this section, we extend our previous analysis to account for these eﬀects, and develop a purely macroscopic statement of energy accounting, which is the celebrated First Law of Thermodynamics. 2.8.1 The Internal Energy Consider a macroscopic sample of matter (solid, liquid, or gaseous) at rest. Although no motion is apparent, on a microscopic level the atoms composing the sample are in continual, random motion. The reason we don’t perceive this motion, of course, is that all macroscopic measurements we can do average over a huge number of atoms. Since the atomic motion is essentially random, there are just as many atoms travelling to the right with a given speed as to the left. Even though individual atomic speeds may be hundreds of meters per second, the atomic velocities tend to cancel one another when we sum over a large number of atoms. But the kinetic energies due to the atomic motion don’t cancel, since the

CHAPTER 2. ENERGY, WORK, AND HEAT

27

E p (r)

r0

r

Figure 2.6: Potential energy of a chemical bond as a function of bond length r. The unstretched length is r0 . kinetic energies are all positive, scalar numbers. Even a sample of matter at rest (no center-of-mass motion) has microscopic kinetic energy, which we will call the internal kinetic energy: Ek,int = j 2 mj vj /2,

(2.65)

where the sum is over all atoms in the sample. The sample has microscopic potential energy too. As the atoms move, they stretch or compress the bonds holding them together. The bonds may be modeled as springs, although ones with a spring constant which depends on bond length. The potential energy of these “springs” as a function of length typically looks something like the curve in Fig. 2.6. If the bond is compressed so that it is shorter than r0 , the potential energy rises rapidly. If it is stretched, the potential energy rises, too. The bond can be broken (r → ∞) if work ∆ is done to pull the atoms apart. Other types of interactions between atoms can be modeled in a similar way. Molecules in a gas hardly feel any force from other molecules far away, but when two molecules approach closely (collide) the potential energy rises rapidly, causing them to repel one another and move apart again. Similarly, the interaction of two atoms which are charged may be described by a repulsive or attractive electrostatic potential energy which depends on their separation. If the atoms or molecules have a magnetic moment (e.g. iron or nickel atoms, oxygen molecules), then their interaction is described by a potential energy function which depends on their separation and the relative alignment of their magnetic moment vectors. In fact, every atomic-level interaction between atoms can be described in terms of some potential energy function. We know this is possible, since we know atomic-level forces are conservative. At any instant in time, the sample has a microscopic, internal potential energy, which is the sum of all of the potential energy contributions describing

CHAPTER 2. ENERGY, WORK, AND HEAT the interactions between the atoms or molecules: Ep,int = k 28

Ep,k .

(2.66)

The index k runs over all pairs 9 of atoms which exert forces on one another in any way. Of course, since the particles are constantly moving and interacting, both the atomic positions and atomic velocities keep changing. Hence, both Ek,int and Ep,int are constantly changing, as energy shifts from kinetic to potential and back again (for example, as two atoms in a gas collide and rebound). In doing energy accounting for this sample, we will ﬁrst of all choose the system to consist of the sample itself (a closed system). We will furthermore choose to treat the conservative interactions between atoms within it by including Ep,int in the system energy, rather than accounting explicitly for the work done by these forces. Therefore, the only work terms which will appear are those due to external forces. If no external forces act on the atoms of the sample (if it is completely isolated from the environment), then energy accounting leads to the conclusion that the sum of Ek,int and Ep,int must be constant: ∆(Ek,int + Ep,int) = 0. We deﬁne the internal energy U by U = Ek,int + Ep,int, (2.68) (2.67)

For a stationary sample which is isolated from the environment ∆U = 0. The internal energy includes all of the kinetic energy associated with the atomic-level, random motion of the atoms of the system, and all of the potential energy associated with all possible interactions between the atoms. Since the potential energy associated with chemical bonds is included in Ep,int , chemical energy is part of the internal energy. Chemical energy is essentially the energy required to break chemical bonds (∆ in Fig. 2.6). Since ∆ diﬀers for every diﬀerent type of bond, if a chemical reaction occurs which breaks bonds of one type and forms bonds of another type, Ep,int may go up or down. If the system is isolated, U must be constant, and therefore Ek,int must change oppositely to
9 The potential energy of some interactions – for example, bending of chemical bonds – may depend on the positions of three or more atoms. This doesn’t change anything – we simply add these terms too to Ep,int .

CHAPTER 2. ENERGY, WORK, AND HEAT

29

the change in Ep,int . The change in Ek,int would be experienced as a change in temperature. 10 Example 2.4 At suﬃciently low density and a temperature of 300 K, the internal energy of gaseous H2 is -2441 kJ/kmol11 and the internal energy of gaseous I2 is 59,993 kJ/kmol. (We will show later that in the limit of low density the internal energy per mole of a gas is a function only of temperature – assume this limit applies here.) The internal energy of gaseous hydrogen iodide HI is given by the formula UHI = 17, 655 + 21.22T kJ/kmol (2.69)

which is valid for 300 < T < 600 K. If one kmol of H2 is reacted with one kmol of I2 to form two kmol of HI in a closed, constant-volume container with no energy transfer to the environment, what is the ﬁnal temperature if the initial temperature is 300 K? Solution: The internal energy of the initial mixture of H2 and I2 at 300 K is U = (1 kmol)(-2441 kJ/kmol) + (1 kmol)(59,993 kJ/kmol) = 57,552 kJ. (2.70) Since the system is isolated (no energy transfer to the environment), U does not change during the reaction. The ﬁnal state consists of 2 kmol of HI, so the ﬁnal internal energy per kmol of HI is 28,776 kJ/kmol. From Eq. (2.69), the ﬁnal temperature is 524 K. Note that the internal energy of H2 is negative at 300 K. This is not a problem, since only diﬀerences in internal energy matter. It simply reﬂects a particular choice for the arbitrary constant C in the internal potential energy for H2 . Nuclear or even relativistic mass energy (E = mc2 ) could be included in U if we like. A nuclear physicist would certainly want to do this. But since only changes in energy have physical signiﬁcance, we can disregard these forms of energy if we don’t plan to consider processes in which they change. The internal energy is deﬁned in a reference frame in which the sample is at rest. If this frame is moving and/or rotating with respect to the lab frame, then the macroscopic kinetic energy associated with these motions must be added to
10 Temperature will be formally introduced in the next chapter. For now, think of it as a measure of the internal kinetic energy per atom. This would be exactly true if atomic motions were really described by classical mechanics, but when quantum eﬀects are important (and the usually are) it is only approximately true. 11 One kmol is 6.023 × 1026 molecules. The mass of 1 kmol in kg equals the molecular weight of the molecule.

CHAPTER 2. ENERGY, WORK, AND HEAT U to determine the total sample energy E in the lab frame: 1 1 2 E = U + M vcm + Iω2 , 2 2

30

(2.71)

where vcm is the center-of-mass speed and ω is the rotation rate (assumed to be about a principal axis which has moment of inertia I).12 It is important to note that Ep,int does not include any potential energy arising from interactions of atoms in the sample with gravitational, electric, or magnetic ﬁelds produced by external sources. If we choose to include these macroscopic potential energy terms in the sample energy, we have to add them explicitly. If the sample is near the surface of the earth and has charge q, the total energy including potential energy terms would be 1 1 2 E = U + M vcm + Iω2 + M gy + qE, 2 2 (2.72)

where E is the value of the electrostatic potential (volts). With macroscopic kinetic and potential energy modes, the energy balance for an isolated sample is ∆E = 0, (2.73) not ∆U = 0. For example, a rubber ball dropped onto a rigid table will eventually come to rest, even if there is no energy loss to the environment. The gravitational potential energy M gy is converted into an increase in U , which would be experienced as an increase in temperature. E, however, remains constant. 2.8.2 Atomic Level Energy Transfer: Microscopic Work No sample of matter can really be completely isolated from the environment. Usually, it is in contact with other matter (e.g., a container for a gas or liquid; a table a solid rests on). Even if it were ﬂoating in interstellar space it would still exchange energy with the environment through radiation. We now need to consider how to do energy accounting for a macroscopic sample allowing for external work done by forces from the environment. Consider a sample of gas in a container such as shown in Fig. 2.7 which has one movable wall (a piston). We will take the system to be the gas, and container to be part of the environment. On an atomic level, both the gas and container consist of atoms which are in constant motion. The atoms of the gas are moving randomly in all directions,
12 We assume here that all parts of the sample are moving or rotating macroscopically together. If not, then the macroscopic kinetic energy must be determined as (1/2) ρ(x)v(x)2 dV , where the integration is over the sample and ρ is the mass density.

CHAPTER 2. ENERGY, WORK, AND HEAT
System Boundary

31

Gas F

Figure 2.7: A gas in a container, as seen from the macroscopic and microscopic points of view. colliding with one another and occasionally with the container walls. The atoms in the container are vibrating chaotically about their equilibrium positions as they are buﬀeted by the neighboring atoms they are bonded to, or (at the surface) by gas atoms. When a gas atom collides with a wall atom, the gas atom may rebound with either more or less kinetic energy than it had before the collision. If the wall atom happens to be moving rapidly toward it (due to vibration) when they hit, the gas atom may receive a large impulse and rebound with more kinetic energy. In this case, the wall atom does microscopic work on the gas atom: positive microscopic work is done by the environment on the system. On the other hand, the wall atom may happen to be moving away when the gas atom hits it, or it may rebound signiﬁcantly due to the impact. In this case, the gas atom will rebound with less kinetic energy than it had before — therefore, the gas atom does microscopic work on the wall atom: negative microscopic work is done by the environment on the system. We see that collisions between the gas atoms and the walls can do microscopic work even if macroscopically the walls appear stationary. If we let time dt elapse,

CHAPTER 2. ENERGY, WORK, AND HEAT then the energy balance on the gas is dWmicro = dU, ¯

32

(2.74)

where dWmicro is the total work done on the gas by wall collisions during time ¯ dt. 2.8.3 Energy Transfer as Heat Suppose the piston is held ﬁxed, but the container starts out “hotter” than the gas, meaning that the container atoms have more kinetic energy per atom than do the gas atoms.13 Then over time the gas atoms on average will pick up kinetic energy from collisions with wall, and wall atoms will lose kinetic energy: dWmicro will be positive, Ugas will tend to go up, and Ucontainer will tend to go ¯ down. Of course, if the gas started out hotter, then dWmicro would be negative, ¯ and the changes in internal energy would be reversed. Eventually, when their kinetic energies per atom are comparable,14 the number of collisions per unit time which impart extra energy to the gas atoms will just balance the number per unit time which remove energy from the gas atoms, and Ugas and Uwall will stop changing on average. There would still be very rapid statistical ﬂuctuations about these average values, but for a reasonable sized sample these ﬂuctuations are not observable, since it can be shown from statistics that random ﬂuctuations like this have a relative magnitude propor√ tional to 1/ N . For example, if N = 1020, then δU/U ∼ 10−10: the internal energy is constant to one part in 1010 in this case. The process we have just described is energy transfer between the wall (part of the environment) and the gas (the system) due to microscopic work. However, macroscopically it doesn’t appear that any work is being done, since the piston isn’t moving, and we can’t see the microscopic deﬂections due to atomic motion. Therefore, there is no observable, macroscopic F· dx, and no macroscopic work. We call this process of energy transfer by microscopic work without observable macroscopic work energy transfer as heat, or heat transfer for short. The amount of energy transferred in this way is denoted by the symbol Q. For an inﬁnitesimal amount, we use the symbol dQ. As for work, the bar in dQ re¯ ¯ minds us that it is not the diﬀerential of any function, it only means “a little bit of heat.” (Or the other way to say it is that dQ, like dW , is an inexact ¯ ¯ diﬀerential.)
13 Of course, “hotness” is really related to temperature, which we’ll introduce in the next chapter. 14 More precisely, when their temperatures are equal.

CHAPTER 2. ENERGY, WORK, AND HEAT The energy balance for this process is then dQ = dU. ¯ 2.8.4 Energy Transfer as Macroscopic Work

33

(2.75)

Each collision of a gas atom with a wall delivers an impulse to the wall. At typical gas densities, the number of collisions per unit area of wall per unit time is very large. For example, objects sitting in room temperature ambient air experience roughly 1024 collisions per cm2 per second. Macroscopically, it is not possible to detect the individual impulses from so many frequent collisions. Instead, a macroscopic force on the wall is felt, which is proportional to wall area: Fwall = P A. (2.76) The propotionality constant P is the gas pressure. Suppose the piston is now moved slowly toward the gas a distance dx. The macroscopic work required to do this is ¯ dWmacro = F · dx = (P A)dx. (2.77)

The gas atoms which collide with the moving piston have their kinetic energy increased on average slightly more than if the pison had been stationary; therefore, Ugas increases. If dQ = 0, then the energy balance is ¯ dU = dWmacro = P Adx. ¯ (2.78)

Of course, there may also be microscopic work occurring which is not visible macroscopically (heat transfer). To account for this, we must write the energy balance as dU = dQ + dWmacro . ¯ ¯ (2.79) For a more general system, macroscopic kinetic energy and potential energy may also be part of the system energy. If energy is transferred to such a system by macroscopic work and by heat transfer, the most general energy balance for a closed system is

dE = dQ + dW. ¯ ¯

(2.80)

We have to stipulate that the system is closed, since if matter were to enter or leave the system, it would carry energy with it which is not accounted for

CHAPTER 2. ENERGY, WORK, AND HEAT

34

W

E

Q

Figure 2.8: The First Law for a Closed System. in Eq. (2.80). Note that we have removed the subscript “macro” on the work term. In thermodynamics generally, and from here on in this book, the term work means macroscopic work, unless otherwise stated. Equation (2.80) is known as the First Law of Thermodynamics. The First Law simply states that the change in the total energy of a system equals the energy transfer to it as heat, plus the energy transfer to it as work. It is simply a statement of conservation of energy for a macroscopic system. Note that there is no formula for dQ like dW = F · dx. In practice, dQ is ¯ ¯ ¯ determined from equation 2.80 once dW and dE have been evaluated. ¯ We can integrate Eq. (2.80) for some ﬁnite change from an initial state to a ﬁnal one, yielding

∆E = Q + W where W = i (2.81)

f

f

dW = ¯ i f

Fmacro · dxmacro

(2.82)

and Q= i dQ. ¯

(2.83)

The interpretation of Eq. (2.81) is as shown in Fig. 2.8. Both work and heat represent energy transfers across the system boundary; the energy E stored within the system (in whatever form) changes by the amount of energy transferred in. Alternatively, we may divide Eq. (2.80) by the elapsed time dt to obtain

dE ˙ ˙ =Q+W dt

(2.84)

CHAPTER 2. ENERGY, WORK, AND HEAT

35

˙ where the ratio dQ/dt is the heat transfer rate Q and the ratio dW/dt is the ¯ ¯ power input or work rate we’ve deﬁned previously. All three equations (2.80), (2.81), and (2.84) are diﬀerent forms of First Law of Thermodynamics for a closed system. In solving problems involving the First Law, you should carefully consider which form is most appropriate to use. If the process occurs during an inﬁnitesimal time dt, use Eq. (2.80). If you are given initial and ﬁnal states of the system, often Eq. (2.81) is the best choice. If you are given a heat transfer or work rate, then probably Eq. (2.84) would be easiest to use. For many processes, both Q and W will be signiﬁcant, and must be included to correctly calculate the change in the system energy E from the First Law. But in some cases, either Q or W may be very much smaller than the other. In analyzing such processes, it is often acceptable to only include the dominant energy transfer mechanism, although this all depends on how accurate an answer is required for ∆E. For example, if a solid is heated, it usually expands a little bit. But in many cases the work done in the expansion against atmospheric pressure is so small that W Q. In this case, it might be OK to neglect W in calculating ∆E due to heating. The opposite case would occur, for example, if a rubber band were rapidly stretched. Since heat transfer takes some time to occur, if the stretching is rapid enough it might be OK to neglect Q in calculating the increase in internal energy of the rubber band due to the work done to stretch it.15 Processes for which Q = 0 are called adiabatic.

2.9 Reversible and Irreversible Work
An important concept in thermodynamics is the idea of reversible work. Work dW = F(x, v) · dx is reversible if and only if the work done in moving dx is ¯ exactly recovered if the motion is reversed. That is, Forward: Reverse: dWforward ¯ = F(x, v) · dx = F(x, −v) · (−dx). (2.85) (2.86)

dWreverse = − dWforward ¯ ¯

Note that the velocity changes sign when the direction of motion is reversed. This condition is satisﬁed if F(x, v) · v = F(x, −v) · v.
15 You

(2.87)

can verify for yourself that rubber bands heat up when stretched by rapidly stretching one and holding it to your lip.

CHAPTER 2. ENERGY, WORK, AND HEAT

36

(a) Quasi-Static x F F x

(b) Rapid

F

F

F

F

A

x

B

A

x

B

Figure 2.9: (a) Quasi-static and (b) rapid compression and expansion of a gas. Therefore, the condition is that the force component in the direction of v be the same for forward and reverse motion. A force which depends only on position x will satisfy this, and therefore work done by any F(x) is reversible. Friction and drag forces always depend on velocity, and act opposite to the direction of motion, changing sign when the direction of motion is reversed. Therefore, work done on a system by a friction or drag force is always negative (i.e., the system must do work against the friction force). Work done by or against such forces is never reversible – work must always be done to overcome friction, and you can never recover it. Consider compressing a gas in a cylinder by pushing in a piston, as shown in Fig. 2.9. As discussed above, the gas exerts a force on the piston due to collisions of gas atoms with the piston surface. To hold the piston stationary, a force F = P A must be applied. We will assume the piston is lubricated, and is well-insulated so the compression process is adiabatic.

CHAPTER 2. ENERGY, WORK, AND HEAT

37

The piston can be moved very slowly by applying a force just slightly greater than P A. If the piston velocity is suﬃciently small, then the work required to overcome viscous drag in the lubricant will be negligible (example 2.2). Also, if the piston speed is slow enough, the gas molecules which collide with the piston have plenty of time to move away from it and distribute their excess energy with other molecules through collisions before the piston has moved any signiﬁcant distance. In this case, the state of the gas is the same as would occur if the piston were stationary at the instantaneous value of x: The gas molecules are uniformly distributed in the cylinder, and the force on the piston is the same as if the piston were not moving – it is P A. In this limit of zero piston speed, the force on the piston approaches P A, no matter whether the piston is moving in or out. In this limit, the compression or expansion process is called quasi-static, since the force on the piston is the same as if the piston were static. Therefore, the work done during quasi-static compression of a gas is reversible work. If the piston velocity is high, two things happen which make the process irreversible. First, the work to overcome viscous drag in the lubricant may no longer be negligible. Second, if the piston velocity is comparable to the average molecular speed in the gas, then the piston will tend to sweep up molecules near it, forming a high-density region just in front of it (similar to a snowplow). Since the rate at which molecules collide with the piston is proportional to their number per unit volume, the piston will experience more collisions at a given x location than if it were moving slowly. Therefore, the applied force F to move the piston must be greater than the quasi-static force, and thus the work to compress the gas is greater than in the quasi-static limit. A typical plot of F (x) for rapid compression is shown in Fig. 2.9(b). If this process is now reversed and the gas is rapidly expanded, we still have to do work to overcome viscous drag in the lubricant (not only do we not get back the work done to overcome drag during compression, we have to do still more work to overcome it during expansion). Also, there is now a low density gas region near the piston, since the piston is moving away so fast the molecules lag behind. So the gas pushes on the piston with less force than if the expansion were done very slowly. Therefore, the work we get back in the expansion is less than we put in during compression. Since W = F · dx, the work input WAB to move the piston from A to B B is A Fcomp (x)fx, where Fcomp (x) is the force applied along the compression part of the curve. This of course is simply the area under the Fcomp (x) curve. A The work input to expand the gas from B to A is WBA = B Fexp(x)dx =

CHAPTER 2. ENERGY, WORK, AND HEAT −
B A

38

Fexp(x)dx. If we consider the entire process A → B → A, then the total work input W = WAB + WBA . In the quasi-static case Fcomp (x) = Fexp(x) = P A, so W = WAB + (−WAB ) = 0. No net work is required to return the piston to its starting point. From the ﬁrst law for this process (remember Q = 0) W = ∆U = 0. (2.88)

Therefore, the gas internal energy returns to its starting value after quasi-static compression followed by quasi-static expansion. In the non-quasi-static case, Fcomp (x) > Fexp (x). Therefore, W > 0: net work input must be done if the piston is rapidly moved from A to B and then back to A. From the First Law then, ∆U > 0. The gas ends up with more internal energy (hotter) at the end of the process than at the beginning.

2.10 Some Reversible Work Modes
There are several diﬀerent ways of doing reversible (quasi-static) work on matter. A few of these are described here. 2.10.1 Compression We saw in the last section that if a gas is slowly compressed, the work required to move the piston dx is dW = (P A)dx. The same analysis would apply if the ¯ gas in the cylinder were replaced by any compressible substance, so this result is quite general. The volume change of the substance is dV = −Adx, so we may write this as dWqs = −P dV. ¯ (2.89)

We add the subscript “qs” since this only applies if the compression is done quasi-statically. Note this expression is for the work done on the substance (input to the system); For compression, dV < 0 and dWqs > 0, for expansion ¯ dV > 0 and dWqs < 0. ¯ 2.10.2 Stretching a Liquid Surface If a liquid ﬁlm is suspended on a wire frame, as shown in Fig. 2.10, a force is exerted on the wire that is proportional to its wetted length L that results from a tensile force16 per unit length in the surface of the liquid. This is known as the
16 A

tensile force is the opposite of a compression force – it pulls, rather than pushes.

CHAPTER 2. ENERGY, WORK, AND HEAT
Movable Bar

39

L

Liquid Film

F

Side View

Figure 2.10: Surface tension in a liquid. surface tension σ, and has units of N/m. For example, for a water/air interface at 25 ◦ C, σ = 0.072 N/m. The physical origin of the surface tension is that molecules in a liquid exert attractive forces on one another, which hold the liquid together. These forces are much weaker than covalent chemical bonds, but nevertheless have a dependence on distance similar to that shown in Fig. 2.6. A molecule will have lower potential energy in the bulk, where it is surrounded by molecules on all sides, than at the surface, where it feels the attractive force only on one side. Therefore, surface molecules will try to move into the bulk, until as many have crowded into the bulk as possible and there is a shortage of surface molecules left to cover the area. The remaining surface molecules will be spaced slightly further apart than ideal (r > r0 ), and therefore they will pull on their neighboring surface molecules, resulting in the surface tension. Since the ﬁlm has two surfaces, the force required to hold the movable wire stationary is F = 2σL. (2.90) If the wire is now quasi-statically pulled, so that F is only inﬁnitesimally greater than 2σL, the work done to move dx is dWqs = 2σLdx. ¯ (2.91)

During this process the total surface area of the ﬁlm has increased by 2Ldx. Therefore, we may write

dWqs = σdA. ¯

(2.92)

This expression for the work required to quasi-statically increase the surface area of a liquid is valid for arbitrary geometry.

CHAPTER 2. ENERGY, WORK, AND HEAT

40

qE

E p

+ +q

-

-q

qE

Figure 2.11: Forces exerted by an electric ﬁeld on a polar diatomic molecule. 2.10.3 Electric Polarization Many materials are polar, which means that although they are electrically neutral, they are composed of positively and negatively charged atoms. Any ionic crystal (NaCl) is polar, as is water (the hydrogen atoms have positive charge, and the oxygen atom has negative charge). If an electric ﬁeld is applied to a polar material, it is possible to do work on it. Consider the situation shown in Fig. 2.11. A polar diatomic gas molecule is oriented at a particular instant in time at an angle θ with respect to an applied electric ﬁeld. (Due to collisions between the gas molecules, at any instant in time there is a distribution of orientations – they are not all lined up with the ﬁeld, except at absolute zero.) The force on a charge q in an electric ﬁeld E is given by F = qE. (2.93)

Therefore, the positive end of the molecule at position x+ feels a force qE, and the negative end at x− feels a force −qE. The molecule will turn and may be stretched by the forces due to the electric ﬁeld acting on each end. (The center of mass motion is unaﬀected, since there is no net force.) If, due to the ﬁeld, the atoms move by dx+ and dx− , respectively, then the work done on this one molecule is dW1 ¯ = (qE · dx+ ) + (−qE · dx−) = qE · d(x+ − x− ). The electric dipole moment p of the molecule is deﬁned by p = q(x+ − x− ). (2.96) (2.94) (2.95)

CHAPTER 2. ENERGY, WORK, AND HEAT

41

The dipole moment is a vector which points along the direction from the negative charge to the positive charge. In terms of p, the work done is dW1 = E · dp. ¯ (2.97)

This is the work done on one polar molecule; to determine the work done on the entire polar gas, we must sum over all N molecules in the gas:
N N

dW = ¯ n=1 E · dpn = E · d n=1 pn .

(2.98)

The quantity in parenthesis is the total dipole moment of the gas. This may be rewritten in terms of purely macroscopic quantities by deﬁning the polarization P to be the dipole moment per unit volume: P= Then Eq. (2.98) becomes 1 V
N

pn . n=1 (2.99)

dW = E · d(V P). ¯

(2.100)

Although we derived this equation for a polar gas, it is in fact completely general and applies to polar solids and liquids also. For static polarization, P is some function of E and temperature T . Except in extremely strong ﬁelds, the dependence on E is linear, so that P may be expressed as P = 0 χe (T )E. In Rationalized MKSA units (the most common electromagnetic unit system for engineering work), 0 = 8.90 × 10−12 C2 · s2 /kg· m3 . The dimensionless material property χe (T ) is the electric susceptibility. The susceptibility for many materials may be expressed as χe (T ) = A + B/T, (2.101)

where A and B are constants. The A term describes polarization due to stretching of polar chemical bonds, a process which is not strongly temperature dependent. The B/T term describes orientation of polar molecules in a liquid or a gas, as we considered above. It is not hard to see why this orientation process should be temperaturedependent. If E = 0, then the dipoles are oriented randomly, so just as many point up as down. In this case, P = 0. For non-zero E, the ﬁeld will try to align

CHAPTER 2. ENERGY, WORK, AND HEAT

42

p for each molecule with E, but collisions between molecules will upset this alignment, tending to randomize p. The net eﬀect of the competition between alignment by E and randomization by collisions is that molecules point in all directions, but p · E is somewhat more likely to be positive than negative. This means that there is non-zero polarization, with P directed along E. Since higher E increases the tendency to align, and higher T increases the tendency to randomize direction, P typically increases with increasing E, and decreases with increasing T . Example 2.5 A dielectric material is one which may be polarized, but has no mobile free charges, so no electrical currents can ﬂow through it. Dielectrics are often used to ﬁll the space between the plates in capacitors. A particular dielectric liquid, which obeys Eq. (2.101) with A = 0, is quasi-statically polarized at constant temperature starting at E = 0 and ending at E = E1 . For this material, the internal energy depends only on T . Determine the work and heat transfer during this process. Solution: W = dW = E · d(V P). ¯ (2.102) For quasi-static polarization, the static relationship between P, E, and T holds, so E1 2 0 V BE1 Wqs = Ed(V 0 BE/T ) = . (2.103) 2T 0 Since this process is carried out isothermally, and for this particular material U = U (T ), ∆U = 0 for this process. The ﬁrst law applied to this system is ∆U = 0 = Q + Wqs , (2.104)

2 from which we conclude that Q = − 0 V BE1 /2T . Therefore, heat must be removed (Q < 0) to polarize this material at constant temperature; if no heat were removed (adiabatic polarization), U would increase by Wqs , and the temperature would increase.

2.10.4 Magnetization Some materials are magnetic – that is, they contain atoms which have magnetic dipole moments and behave just like atomic-scale magnets. Magnetic atoms are usually ones with unpaired electrons, such as iron, nickel, or rare-earth elements. Some molecules can have unpaired electrons also, for example O2 . An applied magnetic ﬁeld can do work on magnetic materials. The analysis is very similar to that for electric polarization. If a single magnetic dipole with

CHAPTER 2. ENERGY, WORK, AND HEAT

43

dipole moment m is placed in a uniform magnetic ﬁeld H, the ﬁeld exerts a torque on the dipole given by τ = µ0 m × H. (2.105)

where µ0 is a constant which depends on the unit system we use to measure m and H known as the permeability of free space. In Rationalized MKSA units µ0 = 4π × 10−7 kg· m/C2 . The work done by this torque is dW1 = µ0 H · dm. ¯ (2.106)

Summing over all magnetic dipoles in the material and deﬁning the magnetization M by N 1 M= mn , (2.107) V n=1 we obtain an expression for the work required to change the magnetization of a magnetic material: dW = µ0 H · d(V M) ¯ (2.108)

Analogous to the discussion above for dielectrics, if a magnetic material is placed in a static magnetic ﬁeld, it will develop some static magnetization M(H, T ), which results from the balance between the ﬁeld trying to align the dipoles and random thermal motion (collisions, vibrations) upsetting perfect alignment. If the ﬁeld is increased very slowly (quasi-statically), then this relation between M, H, and T will still hold, and we may write dWqs = µ0 H · d [V M(H, T )] ¯ (2.109)

Example 2.6 A Curie substance is one for which the static magnetization is M=C H , T (2.110)

where C is a material-speciﬁc constant. Most magnetic materials behave as Curie substances in the limit of high temperature and low ﬁeld strength. A Curie substance in a uniform magnetic ﬁeld H0 is quasi-statically, isothermally magnetized by slowly increasing the ﬁeld to H1 > H0 . Calculate the work done on the substance, and the change in its internal energy. It may be shown

CHAPTER 2. ENERGY, WORK, AND HEAT

44

that for a Curie substance U = U (T ). Calculate the heat transfer which must occur in this process. Solution: Since the process is quasi-static, the static relationship M(H, T ) holds at every step in the process. Therefore,
H1

W

=
H0

µ0 C H · dH T (2.111) (2.112)

=

µ0 C 2 2 (H1 − H0 ). 2T

The First Law for this process is ∆U = Q + W. (2.113)

Since the process is isothermal and we are given U = U (T ), ∆U = 0. Therefore,
2 2 Q = −W = −(µ0 C/2T )(H1 − H0 ).

(2.114)

Example 2.6 shows that heat must be given oﬀ to the environment in order to quasi-statically (reversibly) magnetize a Curie substance at constant temperature. If the process had been done adiabatically instead (Q = 0), the internal energy and temperature of the substance would have increased. The reason for this is that the microscopic torque exerted on the individual dipoles by the ﬁeld imparts to them some rotational kinetic energy, which is then transferred to the rest of the substance by collisions. The reverse process of quasi-static isothermal demagnetization require heat input to maintain the sample temperature; if no heat is supplied (adiabatic demagnetization), the sample temperature drops. These processes may be combined to produce useful devices, such as magnetic engines or magnetic refrigerators. Magnetic refrigerators are used in practice to achieve very low temperatures (T < 1 K), where conventional refrigerators cannot function. 2.10.5 Generalized Forces and Displacements The expressions for quasi-static work are always of the general form FdX or F · dX: −P dV , σdA, E·dP, µ0 H·dM, etc. We can think of these as generalizations of F· dx. We call the F terms (−P , σ, E, H) generalized forces and the X terms (V , A, P, M) generalized displacements. The work done by any of these for a ﬁnite change is W = path F · dX.

(2.115)

CHAPTER 2. ENERGY, WORK, AND HEAT

45

This is still a path integral, but note it is not an integral in physical coordinate space, but in the space deﬁned by X. For example, polarization work would involve integrating E · dP along some P(t) trajectory.

2.11 Some Irreversible Work Modes
If any of the processes discussed in the last section are done too rapidly, the work done will not be reversible. For example, if the magnetic ﬁeld is increased too rapidly, the induced magnetization will lag behind the static M(H, T ). This will result in µ0 H · dM being greater than the quasi-static value for a given dM. Therefore, more magnetization work must be done to eﬀect a given change in magnetization; less work is recovered during demagnetization. Some other ways of doing work are inherently irreversible – if the direction of the motion is reversed, the force changes sign, so you can’t recover any of the work put in. As we’ve already discussed, work done to overcome friction or viscosity is like this. 2.11.1 Stirring a Viscous Fluid An example of purely irreversible work is stirring a viscous ﬂuid.17 Work must be done to turn the stirrer, no matter which direction it is turned. The ﬂuid will have some macroscopic kinetic energy for a while due to stirring, but eventually it will come to rest, with the energy transfer as work to the system due to stirring appearing ﬁnally as an increase in internal energy U . In fact, the state of the ﬂuid after it is stirred and has come to rest again is no diﬀerent than if the same amount of energy had been added to it as heat. Fully irreversible work is equivalent to heat addition. 2.11.2 Electrical Current Flow Through A Resistor Another common type of fully irreversible work is electrical current ﬂow through a resistor. As electrons move through a resistor with an electrical potential ∆E > across it, they lose electrostatic potential energy in the amount |e∆E| by doing this amount of irreversible work. The work is done by colliding with atomic scattering centers within the resistor, which transfers energy to them, increasing the internal energy of the resistor. ˙ If the number of electrons ﬂowing through the resistor per second is Ne , then the rate at which irreversible work is done is ˙ ˙ We = Ne |e∆E| = I·E.
17 All

(2.116)

ﬂuids, even water and gases, have some viscosity.

CHAPTER 2. ENERGY, WORK, AND HEAT

46

˙ where I = eNe is the electrical current. The current through the resistor is proportional to the voltage across it: I = ∆E/R, where R is the resistance. ˙ ˙ Therefore, Eq. (2.116) can be also written We = I 2 R, or We = (∆E)2 /R. As was true for stirring a ﬂuid, passing a current through a resistor is fully irreversible, since changing the sign of the voltage also reverses the direction of current ﬂow. The irreversible electrical work done on the resistor is equivalent thermodynamically to heat addition to the resistor.

Problems
2.1 A cart is sitting near the edge of a ﬂat, horizontal table. The wheels of the cart have radius r, and a lever of length R > r is attached to one wheel. When the lever is straight down, it is pushed by a force F to the left, causing the wheels to turn a small amount dθ and the cart to move to the right. A traction force is also present on the wheels, which keeps the cart from sliding when the force F is applied. Determine the work done by force F and by the traction force. (Hint: calculate displacements with respect to the table carefully, remembering the cart moves. Make sure your answer is sensible when R = r.)

r R F
2.2 Which of the following forces are conservative? The forces are all twodimensional – the component in the z direction is zero. Here ˆ is a unit ı vector in the x direction, and  is a unit vector in the y direction. ˆ F1 F2 F3 F4 = iy − jx ix + jy = x2 + y 2 = i2xy + jx2 = i2x + jx
2

Ft

(2.117) (2.118) (2.119) (2.120)

CHAPTER 2. ENERGY, WORK, AND HEAT F5 F6 = ia sin(ax) sin(by) − jb cos(ax) cos(by) = i sin(ax) cos(by) + j cos(ax) sin(by)

47 (2.121) (2.122)

2.3 Show that any force with the properties listed below which acts between two particles must be conservative. 1. The force is directed along the line connecting the particles. 2. The force depends only on the distance r between the particles. 2.4 A ball of mass m is dropped at height y = H onto a plate which at the moment the ball hits is at y = 0 and is travelling upward with constant speed vp . Assuming the ball rebounds elastically and assuming air resistance is negligible, how high does the ball bounce? How much work does the plate do on the ball? 2.5 You are a scientist on your way to the new research station on Mars in the year 2020. Your spacecraft has just undocked from the space station in earth orbit, and is beginning to accelerate to cruising speed for the trip to Mars. The ion engines are turned on at t = 0, and the engine thrust begins to increase, producing for a while a spacecraft acceleration linear in time: a = Ct. Within the spacecraft, the eﬀects of the acceleration are identical to those of a linearly-increasing gravitational ﬁeld. Free-ﬂoating objects at t = 0 begin “falling” toward the rear wall. You notice a ball which at t = 0 is a distance H from the rear wall. In terms of the mass of the ball, C, and H, what is the kinetic energy (in the local reference frame within the spacecraft) of the ball when it strikes the wall? Is the concept of potential energy useful to solve this problem? Why or why not? ˆ 2.6 An ideal gas is deﬁned to be one which satisﬁes P V = N RT , where N is the ˆ = 8.3143 kJ/kmol/K. Consider number of kg-moles (kmol) of gas, and R an ideal gas contained in a vertical cylinder, with a piston of mass M at the top. 1. Initially, no external force is applied to the piston, and it comes to some equilibrium height. If the atmospheric pressure is P0 and the cylinder area is A, what is the gas pressure in the cylinder? 2. The gas is now heated quasi-statically at constant pressure until its volume is tripled. How much work does the gas do against the environment (which includes the massive piston) during this process?

CHAPTER 2. ENERGY, WORK, AND HEAT

48

3. Finally, a force is applied to the top of the piston, and the gas is quasi-statically, isothermally compressed back to its initial volume. How much work must be done? 2.7 A Curie substance undergoes a three-step quasi-static process: 1. The substance is isothermally magnetized from (H0 , M0 ) to (H1 , M1 ), where H1 = 2H0 2. The ﬁeld is reduced back to H0 , holding M ﬁxed 3. The magnetization is reduced to M0 , holding H ﬁxed At the end of this process, the substance is back in its initial state. Sketch this process on a plot of H vs. M . For each step, determine 1. Whether work is done on or by the substance, and how much 2. Whether heat is transferred to or from the substance, and how much 3. The temperature at the end of the step. Explain how this process might be used as the basis for a magnetic refrigerator. 2.8 For a particular dielectric material, the susceptibility is given by χe = A + B/T . With E held constant, the material is cooled to half its initial temperature, causing the polarization to increase. How much work is done on the material by the electric ﬁeld in this process? 2.9 Due to surface tension, the pressure inside a small water droplet will be greater than the air pressure outside. Show that ∆P = Pi − Po = 2σ , r

where σ is the surface tension and r is the droplet radius. Hint: consider a force balance in the x direction on the system deﬁned by the dotted line, which cuts the droplet in half.
Po Pi
System x

CHAPTER 2. ENERGY, WORK, AND HEAT

49

Small water droplets in a fog may be only a few microns in size. Evaluate numerically the pressure inside a water droplet for diameter D = 1 µm and for D = 10 µm if the air pressure is 1 atm. (1 atm = 1.01325 × 105 N/m2 ). Express your answer in atmospheres. 2.10 An ideal gas has the properties described in problem 2.6. In addition, for monatomic ideal gases (He, Ar, etc.) the internal energy is related to N and T by 3 ˆ U = N RT. 2 Consider a process in which a monatomic ideal gas is quasi-statically adiabatically compressed from a state with V0 , P0 , T0 to a ﬁnal state with properties V, P, T . 1. Derive an expression for how the temperature varies with volume during this process – that is, ﬁnd the function T (V ). Do your analysis systematically: draw a sketch showing the system and any energy transfers from the environment, state any assumptions, invoke the ﬁrst law, use necessary property data, etc. 2. Using this formula, evaluate the ﬁnal temperature (K), ﬁnal pressure (MPa), and work input (J) if argon at 300 K and 0.1 MPa is quasistatically, adiabatically compressed to one-fourth of its initial volume. (Recall 1 Pa = 1 N/m2.)

CHAPTER 3
EQUILIBRIUM 3.1 Introduction
In this chapter, we introduce the important idea of equilibrium, and some new properties which tell us about equilibrium. We discuss how to measure two of these properties (pressure and temperature), and conclude with some observations about thermodynamic equilibrium states of matter.

3.2 Equilibrium
It’s a universal observation that any real system, if left alone for enough time, reaches a state where the macroscopic properties stop changing. If a ball is dropped onto the ﬂoor, it will bounce for a while but eventually it will sit motionless. If a drop of dye is placed in a glass of water, the dye will slowly diﬀuses away until the dye concentration is the same everywhere in the glass, after which no further change is observed. If a ﬂuid is stirred and then left alone, the motion will eventually cease, due to viscosity. We call the motionless, unchanging state a physical system approaches at long times the equilibrium state. The word equilibrium is deﬁned as “a state of balance between opposing forces or actions.”1 When a physical system is in an equilibrium state, every part of it is in balance (equilibrium) with every other part. There is no net transfer of matter or energy between any two parts of the system. The balance is dynamic, since molecules may leave one region and enter another (in a liquid or gas), but the same number must be going the opposite direction, since otherwise there would be a net transfer of molecules. Some idealized systems studied in introductory physics courses do not approach equilibrium states. Perfectly elastic balls keep bouncing forever, objects sliding across frictionless tables don’t slow down, and a mass hanging from a frictionless spring will oscillate forever. But the motion of every real object is subject to non-zero friction or drag forces. No matter how small these forces are, they will eventually cause the object to stop moving and come into equilibrium.
1 Webster’s

New Collegiate Dictionary

50

CHAPTER 3. EQUILIBRIUM

51

Initial State

Intermediate

Equilibrium

Figure 3.1: Diﬀusion of a drop of dye in water. A general characteristic of equilibrium states is that they are much simpler to describe than non-equilibrium states. For example, to describe the state shortly after a drop of dye has been introduced into a glass of water requires specifying the concentration C(x) throughout the glass. Once equilibrium has been achieved, it is only necessary to know how much dye was added to fully specify the state. The dye/water example illustrates another characteristic of equilibrium states: they depend only on “intrinsic” factors (the amount of dye in this case), not one the time history of the non-equilibrium states which preceded it. In general, the characteristics of systems in equilibrium states depend on intrinsic factors such as the volume available to the system, the energy and mass it contains, and its chemical composition. They do not depend on the time-history of how the state was prepared. Any substance which has properties which depend on past processing history is not in an equilibrium state. For example, the properties of glass depend on the rate it is cooled from the molten state. Therefore, glass is not an equilibrium state. Thermodynamics deals primarily with the properties of matter in the simpleto-describe equilibrium states. To describe matter in non-equilibrium states, we would have to supplement thermodynamics with results from the theories of ﬂuid mechanics, diﬀusion, heat conduction, electromagnetics, or other areas of physics beyond the scope of thermodynamics. The nature of these states and how they evolve with time is the subject of courses in these other ﬁelds. Although the restriction to equilibrium states seems drastic, it’s really not so bad. In many processes, a system begins and ends in an equilibrium state, even if it passes through non-equilibrium states during the process (for example, rapid exansion of a gas). Since the First Law only requires evaluation of ∆U = Uf −Ui , in many cases we only need to evaluate properties (e.g. U ) in equilibrium states. We can deﬁne several diﬀerent types of equilibrium, depending on what type

CHAPTER 3. EQUILIBRIUM
Piston

52

Initial P >P A B

A

B

Equilbrium P =P A B

A

B

Figure 3.2: Mechanical equilibrium: the piston moves until PA = PB . of “force or action” is in balance within a system. For every type of equilibrium, we would like to deﬁne some property which can be used to determine if two parts of a system are in equilibrium. In the following sections, various types of equilibrium, and the properties associated with each type, will be discussed. 3.2.1 Mechanical Equilibrium Suppose a cylinder is divided in two parts (A and B) by a piston, which can move back and forth (Fig. 3.2). The piston is initially locked rigidly in place and some gas is added to each side of the cylinder. The piston is now unlocked, and oscillates back and forth a while. Eventually, friction or viscous forces damp out the piston motion, and it will stop, though perhaps not at its initial position. Once it has stopped, we say A and B are in mechanical equilibrium with each other. If the piston is stationary, the force must be the same on both sides. Therefore, PA = PB if A and B are in mechanical equilibrium. Pressure is seen to be the property which tells us if two adjacent parts of a system are in mechanical equilibrium with one another. 3.2.2 Thermal Equilibrium Consider now two systems connected by a rigid wall which can conduct heat (Fig. 3.3). (We say they are in “thermal contact.”) When they are placed in thermal contact, energy may transfer through the wall as heat, but eventually no net energy transfer will occur. When this condition is reached, the two systems are in thermal equilibrium. What property tells us if two systems are in thermal equilibrium? It’s clearly

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53

Heat conducting wall

Initial T >T A B

A Q

B

Equilbrium T =T A B

A

B

Figure 3.3: Thermal equilibrium: heat ﬂows until TA = TB . not the energy, since the two systems may have very diﬀerent compositions and sizes, and so will have diﬀerent energies even at thermal equilibrium. We introduce a new property, speciﬁcally for the purpose of determining if systems are in thermal equilibrium. We make the following postulate. Postulate: There exists a scalar property called temperature with the following characteristic: two systems placed in thermal contact will be in thermal equilibrium if and only if they have the same value of temperature. The so-called zeroth law of thermodynamics is the postulate that if A is in thermal equilibrium with B, and B is with C, then A is with C. The zeroth law is equivalent to the postulate that the temperature property exists. 3.2.3 Diﬀusive Equilibrium If a dye drop is added to a glass of water, we know that at equilibrium the dye will diﬀuse throughout the glass. This type of equilibrium is known as diﬀusive equilibrium. We would like to deﬁne some property of the water/dye system we could calculate for various regions in the water, to test if they are in diﬀusive equilibrium. We might consider using the dye concentration as this property. But this is not a good choice in general, since there are some situations where the concentration is not equal everywhere, even though the system is in diﬀusive equilibrium. These situations arise if more than one phase is present or if external conservative forces act on the system. For example, in a gasous mixture of heavy

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and light species in a very tall column, the heavy ones will be preferentially concentrated near the bottom of the column, and the light ones near the top. We would like a property that can handle even these cases. Not knowing what else to do, let us postulate that such a property exists: Postulate: There exists a scalar property called chemical potential with the following characteristic: two parts of a system will be in diﬀusive equilibrium if and only if they have the same value of chemical potential. The chemical potential is given the symbol µ. In situations where multiple chemical species are present, we need to test for diﬀusive equilibrium for each one, so each will require its own chemical potential. We will show later that µ for a particular species depends on its concentration2 and on temperature and pressure. If the system has macroscopic potential energy (e.g. due to gravity), the chemical potential will also depend on the local value of the potential energy. We’ll see in Chapter 6 how to calculate the chemical potential. 3.2.4 Phase Equilibrium If a substance is placed in a sealed container, under some conditions it segregates into separate regions with distinct properties (for example, solid and liquid, or liquid and vapor). We call each region with homogeneous properties a phase. When the amount of each phase is constant, the substance is in phase equilibrium. What property tells us when systems are in phase equilibrium? Again it’s the chemical potential. Phase equilibrium is similar to diﬀusive equilibrium, since molecules are moving between the phases, but at equilibrium the rate they move is the same for both directions. For example, in a mixture of ice and liquid water, the rate at which H2 O molecules from the liquid stick onto the solid ice equals the rate molecules detach from the ice and enter the liquid. The chemical potential of any given phase of a pure substance is some function of T and P . For example, for water there exist two functions µice (T, P ) and µliquid (T, P ). The condition µice(Tm , P ) = µliquid (Tm , P ) determines the temperature where the two phases can co-exist (the melting temperature Tm ) for a given P .
2 In

some cases, it depends on the concentration of other species too.

CHAPTER 3. EQUILIBRIUM 3.2.5 Chemical Equilibrium

55

If multiple chemical species are mixed together, they might chemically react. If they do, the concentration of each reactant will decrease, and the concentrations of the products of the reaction will build up. The products may react with one another, too, “reversing” the reaction and re-forming the reactants. Eventually, the concentrations of all species will settle at values such that the forward reaction rate just balances the reverse one. When this happens, the concentrations will no longer change, and we say the system is in chemical equilibrium. As we’ll see later, the chemical potential is again the relevant property to determine when a system is in chemical equilibrium, although the analysis is more complicated than for diﬀusive or phase equilibrium. 3.2.6 Thermodynamic Equilibrium When all parts of a system are equilibrated in all ways (mechanical, thermal, diﬀusive, phase, chemical), we say the system is in thermodynamic equilibrium. The temperature, pressure, and chemical potential will be the same in all parts of a system in thermodynamic equilibrium, and the concentrations of every chemical species will be constant in time. 3.2.7 Restricted Equilibrium Sometimes a system is kept from attaining one type of equilibrium, although it is equilibrated in other ways. For example, a rigid, heat-conducting partition dividing a system in two parts will allow thermal, but not mechanical, equilibrium to be attained. In other cases, there is no partition but one type of equilibrium is attained very slowly compared to other types. When a system is equilibrated in some ways but not others, we say it is in restricted equilibrium. For example, a mixture of methane and air at room temperature is not in chemical equilibrium, although it may be in mechanical, thermal, and diﬀusive equilibrium. The reaction CH4 + 2O2 → CO2 + 2H2 O (3.1)

is running very slowly in the forward direction, while the reverse reaction is hardly occurring due to the very low CO2 and H2 O concentrations. Therefore, some of the methane is continually being oxidized, but the rate is so small at room temperature to be virtually impossible to detect. If the temperature is increased by adiabatically compressing the mixture or a spark is provided, the mixture may rapidly approach chemical equilibrium (it may burn or explode). Almost always the slow component is some type of chemical, phase, or nuclear

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equilibrium. Examples of substances approaching true equilibrium very slowly include radioactive elements with very long half-lives, any type of glass, or coldworked steel. In many cases, the slow process is so slow that it may be neglected entirely – the properties of cold-worked steel do not change measurably over any practical timescale.

3.3 Pressure and Its Measurement
3.3.1 Hydrostatics The pressure at any point in a ﬂuid at rest is the same in all directions. If a ﬂuid is contained in a cylinder and you push on a piston to apply a force in one direction, the ﬂuid transmits the force in all directions, so the pressure at the piston equals the pressure on the cylinder walls. If no external body forces act on a ﬂuid at rest, then the pressure is the same at every point. But in a ﬂuid sitting in a gravitational ﬁeld, the pressure will depend on height y, due to the weight of the ﬂuid above y.
PA 2

y2

Mg

y1 PA 1

A force balance on a vertical column of ﬂuid yields P1 A = P2 A + M g. Since M = ρA(y2 − y1 ) (assuming constant density), we ﬁnd P1 − P2 = ρg(y2 − y1 ). Diﬀerentiating with respect to y1 yields dP = −ρg dy (3.3) (3.2)

(3.4)

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57

which is the basic equation of hydrostatics, and describes how pressure varies with height in a static ﬂuid. Since Eq. (3.4) is a diﬀerential equation, it applies even if the density ρ depends on height y, as would be the case if the ﬂuid is a gas. In this case, Eq. (3.3) would be replaced by P2 − P1 = − ρ(y)g dy. (3.5)

Example 3.1 A dam of length L holds water to depth H. What is the net horizontal force on the dam if L = 200 m and H = 30 m? Solution: Taking the density of water to be constant at ρ = 103 kg/m3 , the pressure distribution on the dam is P (y) = P0 + ρgy, (3.6)

where P0 is atmospheric pressure and y is the distance below the surface. The force due to the water is
H

Fw = L
0

P (y)dy = L(P0 H + ρgH 2 /2).

(3.7)

Since the force due to atmospheric pressure on the other side is P0 LH, the net force is Fnet = ρgLH 2 /2 = 8.8 × 108 N. (3.8) We have neglected the variation in air pressure with height. Since the density of air is about 1.2 kg/m3 , for H = 30 m, ∆Pair = 345 N/m2 , which is much less that P0 , which is about 105 N/m2 . If H were several km, then ∆Pair would have to be considered.

3.3.2 Manometers A simple arrangement to measure the pressure of a gas is shown in Figure 3.4. Mercury ﬁlls a U-shaped tube of constant cross-sectional area, which is closed on the left side and can be connected to a system of unknown pressure on the right. The region on the left above the mercury column is evacuated.3 A device like this is known as a manometer. At the height of the gas/mercury interface in the right-hand column, the pressure is the gas pressure Pgas . The pressure in the left-hand column at the same height must therefore also equal Pgas . From Eq. (3.3), the pressure at
3 Actually it is ﬁlled with mercury vapor. But mercury has a very low vapor pressure, and exerts negligible force on the liquid.

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58

Vacuum

Gas

h

P

Figure 3.4: A manometer. this location is also ρgh. Therefore, the unknown gas pressure is related to the measurable height h by Pgas = ρgh. (3.9) A diﬀerent type of manometer can be used to measure the pressure diﬀerence between two systems. If instead of evacuating the region above the column on the left we attach it to a system with some pressure PA , and attach the other side to a system with PB , then PB − PA = ρgh. (3.10)

The SI unit for pressure is the Pascal, which is simply the name given to 1 N/m2 . A standard atmosphere (1 atm) is deﬁned to be 1.01325 × 105 Pa, or 0.101325 MPa. Other common units for pressure are the Torr (1/760 atm or 133.32 Pa), the bar (105 Pa), and lbf/in2 or psi (1 atm = 14.7 psi). Usually, pressure is measured with other, more convenient instruments which may be calibrated against mercury manometers. In some pressure gauges, gas at the unknown pressure enters a curved, ﬂexible tube. The tube tends to straighten as the gas pressure increases, which causes a needle on a dial to move. Another type of gauge, known as a capacitance manometer, consists of an evacuated disk-shaped cavity, one side of which consists of a thin, ﬂexible membrane. As the membrane deﬂects due to external pressure, the capacitance of the disk changes. Since capacitance can be determined very accurately by electrical measurements, capacitance manometers are designed to produce an electrical signal (usually a voltage) proportional to pressure. Sometimes you will hear a pressure described as a “gauge pressure.” This is deﬁned as the pressure diﬀerence between the actual pressure and the local

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59

atmospheric pressure, and is what is directly measured by some types of pressure gauges.

3.4 Temperature and Its Measurement
3.4.1 Thermometers To measure temperature, we need some system with an easily-measurable property which changes as its temperature changes (a “thermometer”). A simple thermometer can be constructed by ﬁlling a glass tube with some ﬂuid which expands when heated. Mercury is again a convenient choice. The tube is marked at the mercury level corresponding to some reference temperature (maybe the freezing point of water), and marked again at another reference temperature (maybe the boiling point). The interval between the two marks is then divided into an arbitrary number of equal segments, and the mark the mercury rises is used as the measure of temperature. The marks could be assigned numbers, but letters, names, or any other symbols would work too. The Fahrenheit temperature scale assigns the values 32 and 212 to the freezing and boiling points, respectively, of water at a pressure of 1 atm. The Celsius scale (formerly called the centigrade scale) assigns the values 0 and 100. This procedure deﬁnes a practically-useful temperature scale, but has a few shortcomings for accurate work. For example, the freezing and boiling points of water depend on pressure; if you wanted to check the calibration of the thermometer you’d have to know and reproduce the atmospheric pressure when the marks were made. The thermal expansion properties of mercury also depend slightly on pressure. For a temperature reading to be meaningful, you would need to report the pressure at which the measurement was made. If a thermometer were contructed using a diﬀerent ﬂuid (an alcohol, for example), the freezing and boiling points measured by the two thermometers would agree, but other temperatures would not, since the second ﬂuid expands with temperature in a somewhat diﬀerent way than mercury does. So when the mercury thermometer reads, say, 50 marks above the freezing-point mark, an alcohol thermometer might read 52 marks. To record a meaningful temperature, you’d have to always record what sort of thermometer was used. 3.4.2 Ideal Gas Temperature Scales It would be preferable to measure temperature in a way which is completely independent of the particular thermometer used. One interesting possibility is to use the fact that all gases obey the same equation – the ideal gas law – in

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the limit of zero density. (No real gas exactly satisﬁes the ideal gas law at any ﬁnite density – see Chapter 4). The procedure to construct a substance-independent ideal-gas thermometer is as follows. A low density gas is put in a small, ﬁxed-volume closed container. This serves as the thermometer. A pressure gauge of some sort (maybe a capacitance manometer) is connected to the container to measure its pressure. Now a single reproducible reference state is chosen, and the ideal-gas thermometer is put in thermal contact with it and allowed to come to thermal equilibrium.4 The pressure reading Pref is recorded. To measure an unknown temperature, the ideal-gas thermometer is placed in thermal contact with the system of unknown temperature and allowed to come to thermal equilibrium. The pressure in the gas in the container will change to a new value P (remember the thermometer volume and amount of gas inside are ﬁxed). We will deﬁne the temperature on the ideal-gas scale by T = Tref P Pref . (3.11)

Here Tref is an arbitrary constant, which sets the size of a degree. An interesting feature about this temperature scale is that only one reference state is needed (not two, as with the Celsius or Fahrenheit scales). It is best to pick a reference state which can be precisely reproduced. It will be shown in Chapter 4 that for any pure substance there is a single temperature (and pressure) at which solid, liquid, and vapor all co-exist. This point is called the triple point. The most common procedure is to take the triple point of water as the reference point. Then to specify a particular ideal-gas scale, all that is left is is to decide what number we should assign to Tref . It might seem logical to pick a round number (say, 1, 100, 1000, etc.). On the other hand, it might be convenient if the size of a degree was the same as we are accustomed to on the Celsius or Fahrenheit scale. To ﬁnd the necessary Tref value to reproduce the degree size of the Celsius scale, the following experimental procedure can be used. Go to the reference temperatures used to establish the Celsius scale (the freezing and boiling points of water at 1 atm), and measure the pressure ratios (Pfp /Pref ) and (Pbp/Pref ) with the ideal-gas thermometer. If we want Tbp − Tfp to be 100 on the ideal-gas
4 We assume that the ideal gas only exchanges a negligibly small amount of energy with the system in coming to thermal equilibrium, so that it doesn’t aﬀect the temperature being measured. We can make this assumption as good as we like by simply decreasing the total amount of gas in the thermometer.

CHAPTER 3. EQUILIBRIUM scale, then Tref must satisfy Tref Pbp Pref − Pfp Pref = 100.

61

(3.12)

With the measured pressure ratios, the required value for Tref is found to be Tref = 273.16. (3.13)

The temperature scale deﬁned in this way is known as the Kelvin scale. Since the triple point of water occurs at 0.01 ◦ C, T (K) = T (◦ C) + 273.15. (3.14)

Note that the Kelvin temperature scale has an absolute zero, where the pressure in the gas in principle becomes zero. The only freedom we have is in setting the size of a degree, not where the zero point is. To construct an ideal-gas scale with the degree size of the Fahrenheit scale, choose Tref = 1.8 × 273.16 = 491.69. This scale is known as the Rankine temperature scale, and of course has its zero at the same temperature as the Kelvin scale. Since on the Fahrenheit scale the triple point of water is at 32.02 ◦ F, T (R) = T (◦ F) + 459.67. (3.15) In practice, ideal gas temperature scales have some limitations. First of all, the pressure readings not entirely independent of the gas chosen, since P may be small but it is not zero. To remove this dependence, multiple thermometers would have to be used, each with a diﬀerent amount of gas inside, and their readings extrapolated down to zero pressure. Also, gases condense at low temperature. Even using helium it is not possible to measure temperatures below 1 K with an ideal gas thermometer. At the other extreme, at temperatures above a few thousand Kelvin, gases dissociate to form plasmas. This puts an upper limit on the useful temperature range. Outside this range, temperature must be deﬁned and measured some other way. 3.4.3 Thermodynamic Temperature We’ll show in Chapter 6 that there is a more fundamental thermodynamic way to deﬁne temperature, which is not dependent on the behavior of any particular type of thermometer, not even an ideal gas one. This thermodynamic temperature may be deﬁned over the range zero to inﬁnity, and can be measured with good accuracy in various temperature ranges using diﬀerent physical phenomena. Temperatures as low as 10−8 K and higher than 106 K have been

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62

measured using techniques which rely on the deﬁnition of the thermodynamic temperature. In the temperature range where ideal-gas thermometers can be used, the thermodynamic temperature is identical to the temperature deﬁned by the ideal gas scale. Both the Kelvin and Rankine scales may be considered to be thermodynamic temperature scales. The Celsius and Fahrenheit scales are not, however, since they have their zero points in the wrong place. They are acceptable to use for temperature diﬀerences, but not for absolute (i.e. thermodynamic) temperatures.

3.5 Intensive and Extensive Properties
In Chapter 2, we introduced several thermodynamic properties of macroscopic matter, such as the volume V , the mass M , the total energy E, the internal energy U , the surface area A, the total electric dipole moment V P, and the total magnetic moment V M. These properties are deﬁned both for equilibrium and non-equilibrium states of matter. They are also additive: if we have a system composed of two parts A and B, MA+B = MA + MB , EA+B = EA + EB , etc. Thermodynamic properties with these characteristics are called extensive. In this chapter, we have introduced three new properties (T , P , µ) which serve to determine whether or not two systems or parts of a single system are in equilibrium with one another. These properties have a very diﬀerent character. For a system composed of parts A and B, TA+B is meaningful only if TA = TB (thermal equilibrium). Otherwise, there is no single temperature of the whole system. If A and B are in thermal equilibrium, then TA+B = TA (not 2TA ). Pressure and chemical potential have similar characteristics: they are only deﬁned for systems in equilibrium, and are not additive for systems composed of multiple parts. Properties like this are called intensive. All properties in thermodynamics are either extensive or intensive. The extensive properties depend on how much matter the system contains. If a substance in an equilibrium state with mass M , volume V and internal energy U is divided into n equal parts, each part will have mass M/n, volume V /n, and internal energy U/n. Because the extensive properties are proportional to how much matter is in the system, it is usually more convenient to work with quantities normalized to a unit amount. One choice is to normalize to a unit mass. The speciﬁc internal energy is deﬁned to be the internal energy per unit mass: u= U . M (3.16)

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63

We always use the term “speciﬁc” to mean “per unit mass,” and use a lower-case letter to denote a speciﬁc property. We can also deﬁne the speciﬁc volume v= V , M (3.17)

which is simply the reciprocal of the mass density ρ. Alternatively, we could choose to normalize to one mole. The molar internal energy is deﬁned to be U u= , ˆ (3.18) N where N is the number of moles, and the molar volume is v= ˆ V . N (3.19)

We use the hat notation to denote molar properties. Note that both speciﬁc and molar properties are intensive (independent of the size of the system), since they are the ratio of 2 extensive properties. In engineering thermodynamics, it is most common to work with speciﬁc properties (per unit mass), since mass is conserved in any process but moles may not be if chemical reactions occur. Also, important engineering quantities like the momentum of a ﬂowing ﬂuid depend on the rate of mass ﬂow, not the molar ﬂow rate.

3.6 The Thermodynamic State
We’ve now introduced a rather long list of material properties (U , V , M , P , T , σ, A, E, P, H, M) and we will soon introduce more properties. Fortunately, we usually only need to work with a subset of these which relate to the ways energy can be transferred to or from the substance as work or heat.5 For equilibrium states, even the properties in this subset are not all independent. Specifying a small number of them is suﬃcient to determine all the rest. For example, if the mass, volume, and temperature of an equilibrium sample of water are given, then its pressure, internal energy, and phase are fully determined. When enough property values for an equilibrium state are speciﬁed so that all the other relevant properties of that state can be determined, we say that the thermodynamic state is speciﬁed. The number of properties which must be given to specify the thermodynamic state is called the number of degrees of freedom f of the substance.
5 For example, we don’t usually care about the magnetization of water, since for achievable magnetic ﬁelds it is very, very, small, and the possible magnetic work is therefore negligible.

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64

The number of degrees of freedom is also equal to the number of independent ways the equilibrium state can be altered. If there are two independent ways to change the state, two independent properties will be needed to specify the state, etc. Consider a ﬁxed mass M of some substance for which the only reversible work mode is compression, and which is in an equilibrium state with no macroscopic kinetic or potential energy. The substance is in a cylinder with a piston which can be used to change its volume. The substance may be solid, liquid, or gaseous, or may consist of an equilibrium mixture of phases. If we add heat dQ and do quasi-static work dWqs = −P dV , the First Law ¯ ¯ for this process is dU = dQ − P dV. ¯ (3.20) From this, we see that the two extensive properties U and V may be varied independently. To change U by dU holding V constant, just add heat in the amount dQ = dU . To change V by dV holding U constant, add heat in the ¯ amount dQ = +P dV while changing the volume. This is true no matter what ¯ the details are of the substance in the container (a single phase, a mixture of phases, etc.) Suppose U and V are held ﬁxed and we try to vary something else (maybe T or P ). Since V is ﬁxed, no compression work can be done. But since U is ﬁxed, no heat can be added either. So there’s really no way to change any other properties (T , P , etc.) without changing U or V or both. We conclude that specifying U and V (for given M ) fully speciﬁes the thermodynamic state of this substance. Now consider a substance with two reversible work modes – maybe in addition to being compressible, it has a surface area, and non-negligible surface tension. Then the First Law would become dU = dQ − P dV + σdA. ¯ (3.21)

Now it is possible to hold U and V ﬁxed and vary another property: all we have to do is change the shape at constant volume, changing the surface area. This does some work, but as before we just remove some heat to get U back to the original value ( dQ = −σdA). So in this case, A can be varied independently of ¯ U and V . But there would be no way to hold U , V , and A ﬁxed for this system and vary something else. We conclude that specifying U , V , and A in this case fully speciﬁes the thermodynamic state of the system. We begin to see a pattern here: The number of degrees of freedom equals the number of reversible work modes, plus one more (to account for heat addition).

CHAPTER 3. EQUILIBRIUM If the number of reversible work modes is r, then we have

65

f = r+1

(3.22)

The independently-variable properties which serve to specify the thermodynamic state are U plus the generalized displacements corresponding to each reversible work mode. It might be objected that in this analysis we have only considered reversible work, and have ignored irreversible work modes. But as we discussed in Chapter 2, the change of state produced by irreversible work can be reproduced by reversible work plus some amount of heat input. For example, electrical work done on a system by current ﬂowing through a resistor is fully equivalent to heat addition, while rapid compression of a gas results in a state which can be reproduced by quasi-static, reversible compression plus heat input. Irreversible work does not represent an independent way to alter the state of the system, and therefore does not aﬀect the calculation of the number of degrees of freedom.

3.7 Equations of State
From Eq. (3.22), we see that a substance with only one reversible work mode has 2 degrees of freedom. Such substances are known as simple substances. If only compression work can be done, then the substance is a simple compressible substance (SCS); if only magnetic work can be done, it is a simple magnetic substance (SMS), and so on. Often, a substance may really have more than one reversible work mode, but only one is important. In this case, the substance is often approximated as a simple one, considering only the dominant work mode. For example, we usually may neglect compression work in comparison to magnetic work when dealing with magnetic solids. For a simple compressible substance, specifying U and V ﬁxes the thermodynamic state for given M . Since U and V are extensive and simply scale with the total mass, the more important parameters are actually u = U/M and v = V /M . Therefore, properties such as temperature and pressure are ﬁxed once u and v are speciﬁed: there must exist some single-valued function T (u, v) and some single-valued function P (u, v). We call functions like this which express relationships among properties of substances in equilibrium states equations of state. The two independent variables in the equation of state don’t have to be (u, v). For example, if heat dQ is added to an SCS holding v constant, u increases ¯

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66

T(u,v)

Invert

u(T,v)

u

T

Figure 3.5: Inverting T (u, v) to form u(T, v) by reﬂecting across the dashed u = T line for speciﬁed v. by dQ (the First Law); it is also always observed that the temperature change ¯ dT is positive.6 That is, (∂T (u, v)/∂u)v > 0, which means that T (u, v) is a monotonically increasing function of u for ﬁxed v. Since monotonic function can be inverted, it is possible to form the function u(T, v) (Fig. 3.5). Therefore, the pair (T, v) are always independently variable. Now substitute u(T, v) into P (u, v) = P (u(T, v), v) resulting in a new function P (T, v). This new function doesn’t involve energy, and is called the mechanical equation of state. It is also relatively easy to measure, which is why most experimental investigations of equations of state of simple compressible substances focus on measuring P (T, v). We will examine the nature of this function for an SCS in the next chapter. Note that P (T, v) is a very diﬀerent function than P (u, v), and the partial derivative (∂P/∂v) depends very much on whether it is u or T we’re holding constant. For this reason, we always use a subscript in thermodynamics to state what variable is being held constant in a partial derivative: (∂P/∂v)T is a partial derivative of P (T, v), while (∂P/∂v)u is a partial derivative of P (u, v).7 We have to be a little careful in choosing which 2 variables to take as the independent ones. Both (u, v) and (T, v) are always independent, but not all pairs of variables are. In particular, P and T are not always independent. If, for example, the substance consists of an equilibrium solid/liquid mixture, then P and T become coupled – T must equal the melting temperature at the pressure P. Equations of state also may be deﬁned for substances other than simple compressible substances. For a simple magnetic substance, an argument analogous observation will be proven to be true in Chapter 6. is no diﬀerent than with any other functions. For example, consider f (x, y) = x2 y. Diﬀerentiating with respect to x, (∂f /∂x)y = 2xy. Now change variables to (x, z), where z = xy. Then f (x, z) = xz and (∂f /∂x)z = z = xy not 2xy.
7 This 6 This

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67

to the one above leads to the conclusion that u and M are always independently variable, and can be used to deﬁne the thermodynamic state. Therefore, there must exist equations of state T (u, M) and H(u, M). Also, T (u, M) can be inverted to form u(T, M), which can then be used to form the equation of state H(T, M). Like P (T, v), this equation of state can be easily measured. For other simple substances, the procedure is analogous to that outlined here.

CHAPTER 4
THE SIMPLE COMPRESSIBLE SUBSTANCE 4.1 Introduction
Since all matter can be compressed if a large enough pressure is applied, a study of the thermodynamic properties of the simple compressible substance is a good starting point for any description of the macroscopic properties of matter in equilibrium. Simple compressible substances are also by far the most important ones for engineering thermodynamics, since most (but not all) power plants and engines employ compression, heating, and expansion of a ﬂuid to produce power. A substance may be approximated as a simple compressible substance if eﬀects due to other reversible work modes are negligible. For example, if the surface-to-volume ratio of a large body of water is small enough, then surface tension will not measurably aﬀect the properties of the water except very near the surface. On the other hand, surface tension will have a dramatic inﬂuence on the properties of a very small water droplet, and will, for example, cause the pressure inside the droplet to be elevated above the value predicted if surface tension were neglected. Clearly, a very small water droplet can’t be treated accurately as a simple compressible substance, while a large body of water is approximated very well in this way. In this chapter, we examine the properties of simple compressible substances. We will restrict attention to pure substances, which contain only one type of molecule. Mixtures will be considered in a later chapter.

4.2 Phases of a Simple Compressible Substance
A simple compressible substance may exist in diﬀerent phases: solid, liquid, or gas. Some substances have multiple solid phases, some even have multiple liquid phases (helium), but all have only one gas phase. An experimental apparatus is shown in Fig. 4.1 which can be used to measure the properties and phases of a simple compressible substance as a function of temperature and pressure. A cylindrical solid sample is placed in a vertical cylinder of the same diameter, which is ﬁtted with a piston. The ambient pressure is P0 , and the piston weight provides a constant downward force F = 68

CHAPTER 4. THE SIMPLE COMPRESSIBLE SUBSTANCE

69

Piston

Substance Q Resistor Insulated Cylinder

Figure 4.1: Constant-pressure heating experiment. Mp g. The pressure in the cylinder is P = F/A = Mp g/A + P0 . (4.1)

The sample height is small enough that the pressure may be taken to be uniform within the sample. The cylinder and piston are well insulated, so there is no heat loss to the environment. A small amount of heat Q is added by brieﬂy passing current through a resistor mounted in the cylinder wall, after which the system is allowed to reestablish equilibrium. Once the system has come back to equilibrium, both the temperature and the volume may have changed. The new temperature is measured with a thermometer, and the new volume by the piston height. If Q is suﬃciently small, the expansion will occur slowly enough that friction between the piston and cylinder is negligible. In this case, even if the piston oscillates for a while due to the perturbation, once the oscillations have died out and the piston has settled down at a new height, the work done by the substance on the piston will be equal to the work done against atmospheric pressure, plus the change in the gravitational potential energy of the piston:1 W = (Mp g+P0 A)∆y = (Mp g+P0 )(∆V /A) = (Mp g/A+P0 )∆V = P ∆V. (4.2) An energy balance on the substance yields the change in its internal energy: ∆U = Q − P ∆V. (4.3)

1 If friction were not negligible, some kinetic energy of the piston would be converted to internal energy in the piston or cylinder due to friction, and therefore the work would be > P ∆V .

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70

The quantities Q, P , and ∆V are all measured, so we can calculate ∆U . Since P is a constant in this experiment, this equation may be rearranged in the form ∆(U + P V ) = Q. (4.4)

The combination U + P V occurs often in analysis of problems at constant pressure. Since U , P , and V are material properties, so is U + P V . Rather than always write U + P V , we give this property its own name and symbol: the enthalpy!deﬁnition H is deﬁned by H = U + P V. Like U and V , H is an extensive property. In terms of the enthalpy, Eq. (4.4) becomes ∆H = Q. (4.6) (4.5)

For heat addition at constant pressure, the heat added equals the change in enthalpy of the substance. In contrast, recall from the First Law that if heat is added at constant volume (W = 0), then ∆U = Q. Returning to our experiment, the process in now repeated many times, and the resulting property values are recorded at every step: heat Q is added, time is allowed to elapse to re-establish equilibium, the new T and V are measured, H is incremented by Q. After n heat addition steps, the volume and temperature have values Vn and Tn , and enthalpy Hn of the substance relative to its starting value H0 is Hn − H0 = nQ. (4.7)

Since the extensive properties (V and H) depend on how much of the substance was placed in the container, it is preferable to convert them to speciﬁc quantities (v = V /M , h = H/M ). In Fig. 4.2, the measured temperature and change in speciﬁc enthalpy are shown plotted vs. the measured speciﬁc volume (connecting the individual measurements with solid lines). When heat is ﬁrst added to the solid, its temperature increases and it expands slightly (region a-b in Fig. 4.2). At point b, the temperature stops increasing, although the volume still increases. A look inside the cylinder reveals the presence of some liquid – the solid is melting. At point c, all of the solid has melted, and precisely at this point the temperature begins to rise again. But when point d is reached, it stops again and the volume begins to increase signiﬁcantly. Bubbles are observed to begin forming

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71

T d b c e

h-h

0 d b a c v e

a V

Figure 4.2: Measured temperature (a) and speciﬁc enthalpy change (b) vs. measured speciﬁc volume for constant-pressure heating. in the liquid at point d – the liquid is boiling. Moving across from point d to e, the amount of vapor in the cylinder increases, and the amount of liquid decreases. At point e, no liquid remains, and both temperature and volume increase upon further heat addition. Figure 4.2 shows that when two phases are present (solid/liquid or liquid/vapor), h − h0 continues to increase with v even though T is constant, since energy input is required to convert solid to liquid, or liquid to vapor. Note that the only way to measure h is by means of Eq. (4.7), which actually only allows the change in h from the initial state to be determined. There is no experiment we could do to measure the value in the initial state h0 .2 Since h = u + P v, if h0 can’t be determined, then u0 can’t be either. This isn’t a problem, however, since only diﬀerences of energy (or enthalpy) have any physical signiﬁcance. We can start the experiment in some convenient, reproducible state, and simply assign any value we like to h0 (for example, h0 = 0). We call the initial state with arbitrarily-chosen h0 the reference state or datum state. The results shown in Fig. 4.2 are for a single pressure, P = Mp g/A + P0 . By changing the mass of the piston, we can repeat the experiment for diﬀerent pressures, determining T (v) and h(v) − h0 curves for a range of pressures. Typical T (v) curves are shown in Fig. 4.3(a) with the points on diﬀerent curves where the slope is discontinuous connected by dotted lines. The curve corresponding to Fig. 4.2(a) is labeled P1 on this plot. In Fig. 4.3(b), the dashes
2 We could measure it by starting in some other state, but it wouldn’t be the initial state; the measured value would be relative to the enthalpy in this new initial state, which would still be undetermined.

CHAPTER 4. THE SIMPLE COMPRESSIBLE SUBSTANCE
P 4 P 3 P 2

72

P 1

liquid + solid liquid
T P 0

T liquid + gas solid solid + gas
V

gas

V

Figure 4.3: (a) T − v curves for constant-pressure heating; (b) phase diagram. lines are drawn solid, and each region is labeled by which phase or phases are present in the cylinder. Diagrams like this are called phase diagrams. The experimental observations are as follows. As the pressure is increased, the temperature at which liquid appears (the melting point) and the temperature at which vapor appears (the boiling point) both increase. Also, the speciﬁc volume of the liquid increases to a larger value before boiling begins, and the speciﬁc volume of the vapor once the last liquid has evaporated decreases. Therefore, the change in speciﬁc volume upon boiling decreases as the pressure increases. Beyond a particular pressure P3 , the T (v) curve changes character. As P3 is approached, the change in speciﬁc volume upon boiling goes to zero – the liquid and vapor approach the same density, and in fact become identical in all respects at P3 . For P > P3 , there is no longer a meaningful distinction between liquid and vapor, and there is no longer any conventional boiling behavior observed. In this pressure regime, as heat is added the high-density ﬂuid simply expands continuously and homogeneously to a low-density ﬂuid, without ever breaking up into separate liquid and vapor regions within the cylinder. Pressure P3 is known as the critical pressure Pc . Below Pc , the transformation from liquid to vapor upon heating occurs by means of the ﬂuid in the cylinder splitting into two separate regions (high-density liquid and low-density vapor); as more heat is added, the liquid portion shrinks, and the vapor portion grows. Above Pc , the transformation from liquid to vapor occurs continuously, with the ﬂuid remaining uniform throughout the cylinder at all times. As P approaches Pc from below, the limiting value approached by the boiling

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73

Table 4.1: Critical temperature, pressure, and density for a few substances. Tc (K) 5.20 32.94 126.2 154.6 190.6 304.2 647.3 Pc (MPa) 0.2275 1.28 3.4 5.04 4.60 7.38 22.1 ρc (kg/m3) 69.64 31.36 314.03 436.15 160.43 464.00 317.0

Helium-4 Hydrogen Nitrogen Oxygen Methane Carbon Dioxide Water

temperature is known as the critical temperature Tc . The T (v) curve for P = Pc has an inﬂection point at T = Tc : ∂T ∂v =0
P

and

∂2T ∂v2

= 0 at T = Tc , P = Pc.
P

(4.8)

The speciﬁc volume of the ﬂuid at the point where P = Pc and T = Tc is known as the critical speciﬁc volume vc ; the reciprocal of vc is the critical density ρc . The quantities Tc , Pc, and vc (or ρc ) deﬁne the critical point. The critical point quantities for a few substances are listed in Table 4.2. If the pressure is now lowered below P1 , another change in the character of the T (v) curve is observed. On the P0 curve, there is only one segment where T is constant, not two. On the ﬂat segment, a solid/vapor mixture is found in the cylinder, rather than a solid/liquid or liquid/vapor mixture. Evidently, at suﬃciently low pressure no liquid phase forms – instead, the solid transforms directly to vapor. This process is known as sublimation.

4.3 P − v − T Surfaces
Since any two independent properties serve to deﬁne the thermodynamic state for a simple compressible substance, we may regard any other property as a function of these two. For example, at every (T, v) point in Fig. 4.3 we know the pressure, so we can construct P (T, v). The function P (T, v) deﬁnes a surface over the T − v plane. A typical P − v − T surface is shown in Fig. 4.4. Every equilibrium state of the substance corresponds to some point on the P −v −T surface. In the pure solid region and in the pure liquid region below Tc , the slope of the surface is very steep, since compressing a solid or liquid even a little requires a huge increase in pressure. In the gas or vapor region, the surface is gently sloped, since gases are easily

CHAPTER 4. THE SIMPLE COMPRESSIBLE SUBSTANCE

74

Figure 4.4: A P − v − T surface for a substance which expands upon melting. compressed. Of course, above the critical point the liquid and gas regions merge smoothly. In Fig. 4.4 the term “gas” is used above the critical point, and “vapor” below it. This convention dates back to the early 19th century, when it was thought that “gases” like oxygen were diﬀerent than “vapors” like steam. Vapors could be condensed to liquid, but gases (it was believed) could not be. When it was demonstrated in 1877 that oxygen and nitrogen could be liquiﬁed at suﬃciently low temperature, it became clear that there was no fundamental distinction between gases and vapors; the only diﬀerence is that substances such as oxygen have critical temperatures well below room temperature, while the critical temperature of water is above room temperature. Thus, liquid water is commonplace, but liquid oxygen, nitrogen, hydrogen, or helium are not. However, processes to produce these as liquids are now straightforward, and liquid oxygen, nitrogen, and helium have very signiﬁcant technological applications.3 The slope of the P − v − T surface is discontinuous on the boundary between the single-phase and the two-phase regions. Note that since T remains constant
3 For example, liquid hydrogen and oxygen as used as propellants in the Space Shuttle Main Engine; liquid nitrogen is widely used to cool electronic equipment and photodetectors; liquid helium is used to cool large superconducting magnets to temperatures a few degrees above absolute zero.

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75

Table 4.2: Triple-point temperatures and pressures. Tt (K) 2.18 13.8 63.15 54.34 90.68 216.54 273.16 Pt (kPa) 5.1 7.0 12.5 0.14 11.7 517.3 0.61

Helium-4 Hydrogen Nitrogen Oxygen Methane Carbon Dioxide Water

in the two-phase regions during constant-pressure heating, any horizontal slice through the surface in a two-phase region must produce a line perpendicular to the T axis; in other words, the slope of the surface in the T direction is zero in the two-phase regions. The solid-vapor two-phase region intersects the solid-liquid and liquid-vapor regions in a single line parallel to the volume axis. This line is known as the triple line, since along this line all three phases may coexist in equilibrium. The pressure and temperature are the same everywhere along the triple line. Therefore, for a given substance, there is only one pressure Pt and one temperature Tt at which solid, liquid, and vapor may coexist in equilibrium. The combination (Tt , Pt ) is known as the triple point. Of course it is only a point in the P − T plane; when the volume axis is considered, it is a line. This is in contrast to the critical point, which is really a point in (P, v, T ) space. The triple points for several substances are listed in Table 4.3. Note that of the ones listed, only carbon dioxide has Pt > 1 atm (1 atm = 101.325 kPa). Therefore, at 1 atm pressure, solid carbon dioxide (“dry ice”) sublimates, while solid water melts. Th P − v − T surface shown in Fig. 4.4 is appropriate for a substance which expands upon melting, which we assumed implicitly in our discussion above. However, a few substances – including water – contract when they melt. For these substances, the P − v − T surface looks like that shown in Fig. 4.5. If a substance has multiple solid phases, the P − v − T surface can become very complex. A portion of the actual surface for water is shown in Fig. 4.6, showing the diﬀerent phases of ice. Two-dimensional phase diagrams may be obtained by projecting the P −v−T surface onto the P − T , T − v, or P − v planes. We have already looked at the T − v projection in constructing the P − v − T surface [Fig. 4.3(b)]. The P − T

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76

Figure 4.5: A P − v − T surface for a substance which contracts upon melting.

Figure 4.6: A portion of the P − v − T surface for water.

CHAPTER 4. THE SIMPLE COMPRESSIBLE SUBSTANCE
(a)
c c

77
(b)

P solid t

liquid

P solid liquid

vapor

t

vapor

T

T

Figure 4.7: P −T phase diagrams for (a) a substance which expands on melting; (b) a substance which contracts on melting. projection is shown in Fig. 4.7 for a substance which expands on melting (a) and for one which contracts on melting (b). The critical point (c) and the triple point (t) are both shown. The lines separating the single-phase regions on a P − T plot are known as coexistence lines, since on these lines two phases may coexist. The liquid-vapor coexistence line terminates at the critical point. In contrast, the solid-liquid coexistence line never terminates, no matter how high the pressure. This is because solid and liquid are fundamentally diﬀerent – in a solid, atoms are arranged in a highly regular, periodic way, while in a liquid they are arranged randomly. There is no way for these states with very diﬀerent symmetry to transform into one another continuously, and so it is not possible for a critical point to exist on the solid-liquid coexistence line. Each of the coexistence lines in a P − T phase diagram can be described by some function P (T ), so clearly P and T are not independent when two phases are simultaneously present. On the liquid-vapor and solid-vapor coexistence lines, the term vapor pressure is used to denote P (T ), since this is the pressure of the vapor in equilibrium with the solid or liquid. An equivalent term is saturation pressure Psat (T ). If pressure is speciﬁed, the saturation temperature Tsat (P ) is deﬁned to be the temperature on the coexistence curve where the pressure is P . The saturation temperature is just another name for the boiling temperature. For example, for water, Tsat (1 atm) is 373.15 K (100 ◦ C), and Psat (373.15 K) = 1 atm.

4.4 Determining Properties in the Mixed-Phase Regions
Under conditions where two phases coexist in equilibrium, some care must be taken to correctly determine the properties and amount of each phase from a

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78

P T

v

f

v

v

g

Figure 4.8: In the liquid/vapor two-phase region, the liquid has speciﬁc volume vf and the vapor has speciﬁc volume vg . phase diagram. The two phases have very diﬀerent properties. For example, the speciﬁc volume of a liquid is much less than that of a gas. How can we determine both values from a phase diagram? The key is to realize that in a two-phase region, the properties of each phase present are those at the “edges” of the region. For example, consider boiling a liquid at constant pressure. Just before the temperature where gas bubbles ﬁrst appear, the cylinder is still ﬁlled with liquid. Call the speciﬁc volume at this point vf .4 As more heat is added at constant pressure, the only thing that happens is that some liquid becomes vapor – the properties (per unit mass) of the remaining liquid don’t change. Although there is less liquid, the liquid remaining still has speciﬁc volume vf . What is the speciﬁc volume of the vapor which has been created? It too is constant during the constant-pressure boiling process, and thus must equal the value of v obtained once the cylinder contains only vapor. Call this value vg . Suppose now that the system is somewhere in the two-phase region on the isobar5 labeled P in Fig. 4.8 and the measured total volume V results in a value for v = V /M as shown in this ﬁgure. This v does not actually correspond to the speciﬁc volume of either the liquid or the vapor in the container. These are vf and vg , respectively. Instead, v is an average of vf and vg , weighted by the mass of each in the container. We use the term “saturated” to denote the states on either side of the vapor dome (Fig. 4.9). Thus, “saturated liquid” has speciﬁc volume vf , and “saturated it is not entirely logical, it is conventional to use the subscript “f” to denote properties of the liquid and “g” to denote properties of the vapor in an equilibrium liquid/vapor mixture. 5 An isobar is a line of constant pressure. The P in a circle simply labels the pressure of this isobar.
4 Although

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79

P T subcooled liquid superheated vapor saturated liquid saturated vapor

v
Figure 4.9: Deﬁnition of saturated, superheated, and subcooled states. vapor” has speciﬁc volume vg . When liquid and vapor are present together in equilibrium, the liquid is always saturated liquid, and the vapor is saturated vapor. Vapor at a temperature above Tsat (P ) is called superheated vapor, and liquid at a temperature below Tsat (P ) is called subcooled liquid. Subcooled liquid is also called compressed liquid, since it is at a higher pressure than Psat (T ). Let the mass of the vapor in the container be M x, where 0 ≤ x ≤ 1. Then the mass of the liquid must be M (1 − x), since the total mass is M . The total volume is then V = M (1 − x)vf + M xvg , (4.9) or V = (1 − x)vf + xvg . M If we know v, we can then solve for x: v= x= v − vf . vg − vf (4.10)

(4.11)

The vapor mass fraction x is an intensive thermodynamic property of a liquid/vapor mixture. The common name in engineering thermodynamics for x is the quality. This name was given to x by engineers developing steam engines and power plants: the presence of liquid droplets in the steam damages engine parts such as turbine blades, hence from the engineer’s perspective higher “quality” steam had less liquid content. Of course, for other applications the relative merits of liquid and vapor might be reversed. In this book, we will usually refrain from making a value judgement about liquid vs. vapor, and simply call x the vapor mass fraction.

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80

Equation (4.11) is known as the lever rule. It may be interpreted in terms Fig. 4.8 as follows. To determine the mass fraction of a phase in a saturated vapor/liquid mixture, locate the system point in the two-phase region on a T −v or P − v plot. Now take the length of the horizontal line segment from v to the saturation line corresponding to the other phase, and divide this length by the total width of the 2-phase region (vg − vf ). Note that this rule works for any 2-phase region, for example for liquid/solid mixtures. Example 4.1 A bottle contains 10 kg of carbon dioxide at 260 K. If the volume of the bottle is 100 liters, does the bottle contain liquid, solid, gas, or a mixture? How much of each? What is the pressure? Solution: Since 260 K is greater than the triple-point temperature for CO2 , no solid will be present. Calculate v = V /M = (100 liters)/(10 kg): v = 0.01 m3 /kg. From a phase diagram for CO2 at 260 K, we ﬁnd that vf = 0.001001 m3 /kg, and vg = 0.01552 m3 /kg. Since v is between these two values, the bottle contains a mixture of liquid and gaseous CO2 . The vapor mass fraction is 0.01 − 0.001001 x= = 0.62. 0.01552 − 0.001001 Since it is in the mixed phase region, the pressure is the saturation pressure at 260 K, which is 2.421 MPa.

4.5 Software for Property Evaluation
To solve problems involving real substances (such as the last example), some source of property data is required. Traditionally in thermodynamics courses, properties were looked up in tables, or estimated from detailed phase diagrams. A typical diagram for oxygen is shown in Fig. 4.10. Here the pressure is plotted against the enthalpy, and many curves representing particular values of other properties are shown. Since P and h are two valid, independent properties, the thermodynamic state is represented by a point on this plot. The point may be ﬁxed by interpolation if any two properties are known for which curves are plotted on the chart. Once the state point is found, any other properties can be read oﬀ (with some care and practice). Most thermodynamics textbooks now come with at least some rudimentary software to evaluate properties. In many cases, the programs are simply electronic versions of tables, which print to the screen the property values. In other cases, more elaborate software is provided which allows you to do a complete thermodynamic analysis. But even these packages are specialized applications, which you use for thermodynamics but nothing else.

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81

Figure 4.10: A pressure-enthalpy plot for oxygen.

CHAPTER 4. THE SIMPLE COMPRESSIBLE SUBSTANCE

82

A new software package is provided as a supplement to this book which implements thermodynamic property functions in Microsoft Excel (a spreadsheet program). The thermodynamic property functions are provided by an “add-in” module called TPX (“Thermodynamic Properties for EXcel”). Details of how to load it into Excel and use it are given in Appendix A. Calculating properties is easy with TPX. For example, if you want to know the speciﬁc enthalpy of oxygen at 1 MPa and 500 K, you simply type into a cell: =h("o2","PT",1,500) The parameters are: the substance name (case is unimportant); a string stating the properties which will be used to ﬁx the state (here P and T ); the value of the ﬁrst parameter (P ); the value of the second parameter (T ). The value of the function returned in the cell is 655.83 kJ/kg. You can select any system of units you like; all inputs and outputs will then be in those units. (This example assumes the user selected units of Kelvin for temperature, MPa for pressure, kJ for energy, and kg for mass.) The real power of using a spreadsheet becomes apparent in more complex analyses. For example, the temperature and pressure may not be speciﬁed inputs, but are themselves the result of calculations in other cells. In this case, simply replace the numerical value in the function parameters by the appropriate cell address (e.g. B4). The same functions implemented by TPX are also available in a WWW property calculator. The calculator is convenient for simple calculations, if you have access to the Web but not to Excel. Example 4.2 One kg of water is placed in a closed container and heated at constant volume. The initial temperature is 300 K. If the desired ﬁnal state is the critical point, determine the necessary container volume, initial pressure, initial vapor mass fraction, and energy transfer as heat. Solution: Two properties are needed to specify the initial state. The temperature is given, so one more is required. Since the process occurs at constant volume, the initial volume V1 must equal the ﬁnal volume V2 . The ﬁnal state is the critical point, so V2 = (1 kg) × vc . The critical speciﬁc volume of water may be calculated using TPX: vc = vcrit("h2o") = 0.003155 m3 /kg. Therefore, state 1 is ﬁxed by T1 = 300 K and v1 = vc . Using TPX, any other

CHAPTER 4. THE SIMPLE COMPRESSIBLE SUBSTANCE desired property for state 1 may be computed: P1 = 3528.2 Pa X1 = 5.49 × 10
−5

83

P("H2O","TV",300,vcrit("H2O")) x("H2O","TV",300,vcrit("H2O")) u("H2O","TV",300,vcrit("H2O")) h("H2O","TV",300,vcrit("H2O"))

u1 = 112.727 kJ/kg h1 = 112.738 kJ/kg

Note that the initial state is a mixed liquid/vapor state, and so P1 = Psat (300 K). The vapor fraction is very small, since vc is only slightly greater than the speciﬁc volume of the saturated liquid. The required heat transfer Q is determined from the First Law: ∆U = Q + W. Since the volume is constant, W = 0, so Q = ∆U = M (u2 − u1 ). Using TPX, u2 = u("h2o","tv",tcrit("h2o"),vcrit("h2o")) = 2029.6 kJ/kg. so Q = (1 kg)(2029.6 kJ/kg - 112.7 kJ/kg) = 1916.9 kJ.

4.6 More Properties: Partial Derivatives of Equations of State
Consider an equation of state like P (T, v). Clearly, a partial derivative of this function, for example (∂P/∂T )v , is some new function of (T, v) and may rightly be regarded as a thermodynamic property of the system. Some useful derivative properties are deﬁned here. 4.6.1 Thermal Expansion Coeﬃcient Most substances expand when heated. The property which tells us how much a substance expands when heated at constant pressure is the thermal expansion coeﬃcient β, deﬁned by

β=

1 v

∂v ∂T

(4.12)
P

To calculate β, the equation of state v(T, P ) would be diﬀerentiated with respect to T .

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84

Note the 1/v in the deﬁnition – β is deﬁned as the fractional change in volume per degree of temperature increase. Also, note that this deﬁnition is only really meaningful if a single phase is present, since otherwise T can’t be increased holding P constant. Example 4.3 What is the thermal expansion coeﬃcient of liquid water at 300 K and 1 atm? Solution: β may be calculated approximately using TPX, evaluating the partial derivative by a ﬁnite-diﬀerence approximation. Taking a small increment of, say, 0.1 K, use TPX (with units set to K and atm) to evaluate (∂v/∂T )P : ∂v ∂T ≈ (v("h2o","tp",300.1,1) - v("h2o","tp",300,1)/0.1)
P

= 2.75 × 10−7 m3 /kg-K. (4.13) Since v = v("h2o","tp",300,1) = 0.001003378 m3 /kg, β ≈ 2.74 × 10−4 K−1 .

4.6.2 Isothermal Compressibility All matter decreases slightly in volume if the pressure is increased at constant temperature. The property how the volume varies with pressure at constant temperature is the isothermal compressibility κ, deﬁned by

κ=−

1 v

∂v ∂P

(4.14)
T

As with β, if the equation of state v(T, P ) is known, it can be diﬀerentiated to ﬁnd κ(T, P ). And as with the thermal expansion coeﬃcient, κ is only meaningful if the substance is in a single phase. 4.6.3 Speciﬁc Heats Suppose a unit mass of a substance absorbs an amount of heat dQ; how much ¯ does the temperature increase? It depends in part on how the heating is done. From the First Law, du = dQ + dW = dQ − P dv. ¯ ¯ ¯ (4.15)

CHAPTER 4. THE SIMPLE COMPRESSIBLE SUBSTANCE Constant Volume If the heating is done at constant volume, the work term is zero. Therefore du = dQ. ¯

85

(4.16)

Diﬀerentiating the equation of state u(T, v) at constant v, we may also write for du ∂u(T, v) du = dT. (4.17) ∂T v Equating these two expressions for du produces the desired relationship between dQ and dT : ¯ ∂u dQ = ¯ dT. (4.18) ∂T v The speciﬁc heat at constant volume cv is deﬁned by ∂u ∂T

cv =

(4.19) v Then du = cv dT for constant-volume heating. The SI units of cv are J/kg/K. Constant Pressure At constant pressure, we have to be careful to remember to account for the work done against the environment as the sample expands. Rearranging Eq. (4.15), we ﬁnd dQ = du + P dv. But since P is constant in the process, du + P dv = ¯ d(u + P v). We recognize u + P v to be the speciﬁc enthalpy h. Therefore, for constant pressure heating, dQ = dh. We can also write ¯ dh = ∂h(T, P ) ∂T dT,
P

(4.20)

(4.21)

since P is constant. Therefore, for constant pressure heating, dQ and dT are ¯ related by ∂h dQ = ¯ dT. (4.22) ∂T P The speciﬁc heat at constant pressure cp is deﬁned by ∂h ∂T

cp =

(4.23)
P

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86

The units are the same as those for cv . Another common term for cv and cp is heat capacity. Heat capacities may ˆ also be deﬁned per mole of substance (ˆv and cp ) or for a total amount of c substance of mass M (Cv and Cp ). Finally, it is worth noting that the names “heat capacity” and “speciﬁc heat” both derive from the old idea that a body can store “heat.” The names shouldn’t be taken too literally – cv and cp are simply deﬁned in terms of the derivatives given above.

4.7 Model Equations of State
A software package like TPX evaluates properties using some mathematical functions which have been ﬁt to a large number of experimental measurements. The measured data might be P (T, v), cp (T, v), or more indirect quantities like the speed of sound. The functions are empirical, and contain many adjustable constants which may be set to ﬁt the measurements as accurately as possible. For example, the functions used by TPX to represent the properties of water contain more than 60 constants, with values chosen to provide the best ﬁt to measurements. 6 Sometimes it is desirable to work with much simpler approximate equations of state. The price for simplicity is lack of accuracy, but in some cases we are interested in examining qualitative behavior rather than calculating precise numbers. Here we examine some common approximate or model equations of state, starting with the simplest and working up to more complex but more accurate ones. 4.7.1 The Ideal Gas The quantity P v/T (where v is the molar volume) is found to approach the ˆ ˆ same value for all ﬂuids in limit of low density when the molecules are far from one another (Fig. 4.11). The limiting value is found to be
P →0

lim

Pv ˆ ˆ = R = 8.314 kJ/mol-K T

(4.24)

ˆ The constant R is known as the universal gas constant. If the pressure is low enough that P v/T has reached this limiting value, then the equation of state is ˆ approximated by ˆ P v = RT. ˆ (4.25)
6 J.

H. Keenan et al., Steam Tables, John Wiley and Sons, New York, 1969.

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87

14 Pv T
3

H 2, 500 K CH , 300 K
4

12 10

H2O, 800 K

(10 kJ / 8 kg-mol K) 6 4 1 10 P (bar)
Figure 4.11: Demonstration that P v/T approaches the same value for all ﬂuids ˆ as P → 0. This equation of state is known as the ideal gas law, and represents the lowdensity limit of the equilibrium equation of state of all real simple compressible substances. An ideal gas (or perfect gas) is deﬁned to be any gas which obeys Eq. (4.25). Since v = V /N , the ideal gas law can be written in terms of the total volume ˆ V and total number of moles N : ˆ P V = N RT. (4.26)

102

103

ˆ The value of R depends on how N is expressed. If N is expressed as a number of ˆ gram-moles (mol), then R = 8.3143 kJ/mol-K; if N is expressed in kg-moles (1 ˆ kmol = 1000 mol), then R = 8314.3 kJ/kmol-K. Alternatively, we could dispense with using mole numbers, and express N as the actual number of molecules. In ˆ this case, R needs to be converted from mol or kmol units to molecule units (1 mol = 6.023 × 1023 molecules; 1 kmol = 6.023 × 1026 molecules). In this case, ˆ R = (8.3143 kJ/mol-K)(1000 J/kJ)(1 mol/6.023 × 1023) = 1.38 × 10−26 J/K. (4.27) ˆ is usually called Boltzmann’s constant kB : In these units R kB = 1.38 × 10−26 J/K. (4.28)

CHAPTER 4. THE SIMPLE COMPRESSIBLE SUBSTANCE In this case, the ideal gas law becomes P V = N kB T.

88

(4.29)

ˆ It’s important to remember that kB and R are really the same thing – just diﬀerent units. We’ll see Boltzmann’s constant again in Chapter 6 when we discuss entropy. We can also divide Eq. (4.26) by the total mass to obtain Pv = ˆ NR M T = ˆ R ˆ M = RT (4.30)

ˆ ˆ ˆ ˆ where M = M/N is the molecular weight, and R = R/M. Unlike R, R is diﬀerent for every gas. For helium, R = (8314.3 J/kmol-k)(1 kmol/4.0026 kg) = 2077.2 J/kg-k, while for water R = (8314.3 J/kmol-K)(1 kmol/18.016 kg) = 461.5 J/kg-K. It is often more convenient to work per unit mass rather than per mole, and so the form of the ideal gas law we will most often use is Eq. (4.30), P v = RT . It is important to bear in mind that the ideal gas law is not rigorously true, but becomes a good approximation at “suﬃciently low” density. Some students reﬂexively invoke P v = RT if a problem involves a substance they normally think of as a gas (e.g. helium, oxygen), forgetting that these substances not only can behave as non-ideal gases, but also can be liquid or solid – it all depends on the conditions. It is simple to check whether the ideal gas law is a suitable approximation under speciﬁed conditions. Simply compute the compressibility factor Z, deﬁned by Pv . (4.31) Z= RT If the ideal gas law is valid, then Z should equal 1. The deviation from Z = 1 is a measure of the error made in assuming ideal-gas behavior under the speciﬁed conditions. Example 4.4 How much error is made in using the ideal gas law to calculate the speciﬁc volume of hydrogen at 20 MPa and 90 K? Solution: For hydrogen, R = 8314.3/2.016 = 4124 J/kg-K. Using TPX, v = v("h2","tp",90,20) = 0.023 m3 /kg under these conditions, so Z = (2 × 107 )(0.023)/4124 × 90) = 1.24. Therefore, the actual volume is 24% larger than would be predicted by the ideal gas law.

CHAPTER 4. THE SIMPLE COMPRESSIBLE SUBSTANCE Internal Energy, Enthalpy, and Speciﬁc Heats

89

The speciﬁc internal energy of any simple compressible substance may be expressed as a function of T and v: u(T, v). Alternatively, we could take the independent variables to be (T, ρ) instead of (T, v), in which case we would write the speciﬁc internal energy as u(T, ρ). Since every simple compressible substance approaches ideal gas behavior as ρ → 0, the speciﬁc internal energy in the ideal-gas limit is u(T, ρ = 0), which is a function of temperature alone: u(T, 0) = u0 (T ). We shall use a superscript “0” to denote properties in the ideal-gas, ρ → 0 limit. The speciﬁc enthalpy of an ideal gas is h0 = u0 (T ) + P v = u0 (T ) + RT. (4.32)

Therefore, the speciﬁc enthalpy of an ideal gas is also a function only of temperature (independent of pressure). From the deﬁnitions of cv and cp , in the ideal gas limit c0 (T ) = v and c0 (T ) = p Since h0 (T ) = u0 (T ) + RT , du0 (T ) , dT dh0 (T ) . dT (4.33)

(4.34)

c0 (T ) = c0 (T ) + R. p v

(4.35)

ˆ The analogous equation on a molar basis would be c0 (T ) = c0 (T ) + R. Of ˆp ˆv course, Eq. (4.35) holds only in the ideal gas limit, where P v = RT applies. The function c0 (T )/R is shown in Fig. 4.12 for several gases. Since c0 and p p 0 cv have the same units as R, the ratio c0 /R(= c0 /R) is dimensionless. By ˆp ˆ p ˆ multiplying the non-dimensional c0 /R by the appropriate appropriate R (or R) p 0 0 value, cp or cp may be determined in any desired unit system. ˆ 0 If cp (T ) is known, Eq. (4.34) may be integrated between any two temperatures to ﬁnd the change in enthalpy:
T1

h0 (T1 ) = h0 (T0 ) +
T0

c0 (T )dT . p

(4.36)

CHAPTER 4. THE SIMPLE COMPRESSIBLE SUBSTANCE
15.0

90

12.5 CH4 10.0

cp/R

7.5

CO2 H2O

Figure 4.12: The ideal-gas speciﬁc heat c0 (T ) for several gases. p If u0 (T1 ) is desired, it could be computed by integrating Eq. (4.33), or from u0 (T1 ) = h0 (T1 ) − RT . For the special case of c0 independent of T , Eq. (4.36) becomes p h0 (T1 ) = h0 (T0 ) + c0 (T1 − T0 ), p and the analogous expression for u0 is u0 (T1 ) = u0 (T0 ) + c0 (T1 − T0 ). v (4.38) (4.37)

For some gases, c0 and c0 are truly independent of temperature. This is the p v case for any monatomic gas, such as argon, helium, neon, etc. In other cases, it may be approximately true for a limited temperature range (e.g., for N2 from 300-500 K or from 2000-3000 K). Physics of the Temperature Dependence of cp (T ) From Figure 4.12, it is apparent that the behavior of the function c0 (T )/R p depends on the molecular structure of the gas. Using the methods of statistical physics, it is possible to calculate this dependence exactly. We will do this later, but for now we will discuss qualitatively the physics governing the idealgas speciﬁc heat, and learn how to compute at least the high-temperature limit of c0 (T ) from molecular structure. p

0

5.0 N2 2.5 Ar

0 500 1000 1500 2000 2500 3000

Temperature [K]

CHAPTER 4. THE SIMPLE COMPRESSIBLE SUBSTANCE

91

A principle of classical statistical physics known as the principle of equipartition of energy says that every “square term” in the classical expression for the energy of a molecule will contribute a factor of (1/2)R to c0 . For example, the v kinetic energy of a single atom contains three square terms: 1 2 2 2 = m vx + vy + vz . 2 (4.39)

So c0 = (3/2)R and c0 = c0 + R = (5/2)R. v p v For a diatomic molecule (assuming it can be approximated as two point masses joined by a linear spring), the energy is 1 1 1 1 2 2 2 2 2 2 = (m1 + m2 ) vx + vy + vz + m12 r2 (ωy + ωz ) + m12 vr + k(r − r0 )2 2 2 2 2 translation rotation vibration (4.40) where m12 = m1 m2 /(m1 + m2 ) is the reduced mass and vr = dr/dt.

Translation

Rotation

Vibration

Center-of-mass motion (translation) contributes 3 square terms, rotation about 2 mutually orthogonal axes contributes 2 square terms, and vibration contributes 2 square terms (one for kinetic energy of vibration, and one for potential energy due to stretching the bond). The principle of equipartition of energy would predict c0 /R = 1/2(3 + 2 + 2) = 7/2, and thus c0 /R = 9/2. v p This is in fact observed at high temperatures for diatomic gases (see the N2 curve in Fig. 4.12). But at lower temperatures, c0 is found to be less than p the value predicted by the equipartition principle. The reason for this is that equipartition of energy is a classical principle, which is valid only if classical physics provides an acceptable description of the molecular motion. If we had calculated the energy of the molecule using quantum mechanics, we would have found that the possible energies for translation, rotation, and vibration are quantized. According to the correspondence principle, whenever the spacing between quantum levels is very small compared to the average energy a molecule possesses, the predictions of quantum mechanics approach those of classical mechanics. An appropriate comparison would be the average translational energy per molecule, which is just (3/2)kB T . Actually, we’re interested only in orders of magnitude so we can drop the 3/2 and just compare to kB T .

CHAPTER 4. THE SIMPLE COMPRESSIBLE SUBSTANCE Mode Translation Rotation Vibration /kB (K) 10−15 2.9 3390

92

Table 4.3: Characteristic quantum level spacings divided by kB , in Kelvin. The value for translation assumes a 1 cm3 box.

Typical numbers for the energy level spacings of N2 are shown in Table 4.3. These energies are shown divided by kB , giving them the units of Kelvin and making the comparison to kB T easy. We see that a classical description of translation is essentially always valid, and for rotation it is valid except at very low temperatures, when in any case N2 wouldn’t be a gas. But vibration is another matter entirely. The spacing between vibrational levels for N2 predicted by quantum mechanics is large compared to the average energy per molecule at room temperature. For this reason, collisions of a room-temperature N2 molecule with others do not have enough energy to “excite” vibration, and so the molecule can’t acquire vibrational energy. Higher temperatures are needed for vibration to become “fully excited.” This explains the c0 (T ) behavior for N2 . At room temperature, c0 ≈ 7/2. p p This suggests that the contribution from vibration is missing. The molecule is translating and rotating, but essentially not vibrating at all, since the lowest vibrational level is too high to be reached by collisions with other roomtemperature molecules. As the temperature increases, c0 approaches 9/2 as p expected, as vibration becomes excited. High Temperature Limit for Polyatomic Molecules Molecules containing more than 2 atoms have several diﬀerent vibration modes, with diﬀerent frequencies and therefore diﬀerent spacings between vibrational levels. The vibrational modes become excited at diﬀerent temperatures in general. As each mode becomes active, the value of c0 increases by R. p The vibrational modes of CO2 are shown below, along with the vibrational frequencies of each. (The units used are the standard spectroscopic units of wavenumbers. To convert to Hz, multiply by the speed of light.) There are actually two bending modes with the same frequency, since the molecule may bend in the plane of the paper (as shown) or out of the plane, so CO2 has a total of four vibrational modes.

CHAPTER 4. THE SIMPLE COMPRESSIBLE SUBSTANCE

93

Symmetric Stretch 1,388 cm -1

Asymmetric Stretch 2,349 cm
-1

Bending (2 modes) 667 cm
-1

We can determine the high-temperature limit for c0 for any polyatomic p molecule using the equipartition principle. Suppose the molecule contains N atoms, has NR rotational modes, and NV vibrational modes. If the molecule is linear (such as CO2 ), it has only 2 possible rotations, just like a diatomic, so NR = 2. (It is impossible to rotate about the molecular axis, since the nuclei are eﬀectively point masses.) But a nonlinear, bent molecule (such as H2 O) has 3 possible independent rotations (NR = 3). The number of vibrational modes is given by 7 NV = 3N − 3 − NR . (4.41)

From the discussion of the equipartition principle above, we see that each rotational mode contributes (1/2)R to c0 , while each vibrational mode conp tributes a full R. (Vibration has 2 square terms – one for kinetic energy, one for potential energy.) So the general formula for a polyatomic molecule is cp (T → ∞) 5 1 = + NR + NV . R 2 2 Therefore, cp (T → ∞) = R 3N − 3/2 linear 3N − 2 nonlinear (4.42)

(4.43)

4.7.2 The van der Waals Equation of State In 1873, The Dutch physicist van der Waals proposed (as part of his doctoral thesis) two simple, empirical modiﬁcations to the ideal gas law, in an attempt to ﬁnd a gas law valid over a wider range of conditions. The ﬁrst modiﬁcation was to replace v in the ideal gas law by (v − b), where b is a small positive constant. This accounts for the fact that real gases are not inﬁnitely compressible, since the
7 It takes 3N numbers to specify the instantaneous conﬁguration of the molecule (x, y, and z for each atom). Therefore, the molecule has 3N “degrees of freedom.” We can alternatively describe the molecular conﬁguration by specifying its center-of-mass position (3 numbers), angular orientation (1 angle for each rotational mode), and amplitude of each vibrational mode (1 number for each mode). The sum of these must equal 3N .

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94

molecules have ﬁnite volume (v does not approach zero as P increases holding T constant). The constant b corresponds roughly to the speciﬁc volume of the ﬂuid when the molecules are packed together — essentially the liquid or solid speciﬁc volume. The second modiﬁcation accounts approximately for the attractive forces between molecules. Molecules about to strike the wall feel a net restraining force due to their attraction to the other molecules in the gas behind them; this lowers the speed with which they strike the wall and the impulse delivered to the wall in the collision. Since any pressure measurement can be thought of as measuring the force on a wall due to the impulse of many collisions per second, the measured pressure P will be somewhat lower than the value which would be measured if the attractive forces were absent. The magnitude of the lowering, van der Waals argued, should be proportional to 1/v2 , since the wall collision frequency per unit area is proportional to 1/v, and to ﬁrst approximation the net restraining force should be proportional to the number of “nearby” molecules, which also scales with 1/v. Based on these considerations, van der Waals proposed replacing P in the ideal gas law by P + a/v2 , since the measured P is lower than it “should be” in the ideal gas law by a factor a/v2 , where a is a positive constant. With these modiﬁcations to the ideal gas law, the van der Waals equation (P + a/v2 )(v − b) = RT (4.44)

is obtained. For large v (low density), this equation reduces to the ideal gas law, as it should. Unlike the ideal gas equation of state, the van der Waals equation has a critical point, where (∂T /∂v)P = 0 and (∂ 2 T /∂v2 )P = 0. It is left as an excercise to show that setting these two partial derivatives to zero results in the solution for Tc and Pc RTc Pc = = 8a 27b a . 27b2 (4.45) (4.46)

Solving for a and b in terms of Tc and Pc, a b = = 27 (RTc )2 64 Pc RTc . 8Pc (4.47) (4.48)

CHAPTER 4. THE SIMPLE COMPRESSIBLE SUBSTANCE

95

If we pick a and b using these expressions, the van der Waals equation is guaranteed to reproduce the correct critial temperature and pressure, although this certainly does not mean it must produce any other accurate property values. In Fig. 4.13, the predictions of the van der Waals equation, the ideal gas equation, and TPX are compared near the critical point of methane. Three isotherms at 0.94Tc, Tc , and 1.06Tc are shown, along with the actual vapor dome of methane. All three agree fairly well at large v, where the ideal gas law holds approximately. For smaller v, the ideal gas law does not adequately approximate the actual P (v, T ) at all. The van der Waals equation does much better. For T > Tc , the shape of the isotherm is qualitatively right, although quantitatively it overpredicts P (v, T ) at small v. It also overpredicts the critical volume vc slightly. An interesting thing happens below Tc : the van der Waals P (v, T ) is no longer monotonic in v – instead, the isotherm exhibits a local maximum and a local minimum. Stability and the Liquid-Vapor Phase Transition Suppose we prepare a van der Waals gas in a cylinder at T0 < Tc , with v0 chosen such that the state lies on the portion of the T0 isotherm where (∂P/∂v)T > 0 (Fig. 4.14). The piston weight is chosen to balance P0 = P (v0 , T0 ) exerted by the gas. But due to very small random ﬂuctuations in the number of molecules striking the piston from both sides, the piston position (and therefore v) will ﬂuctuate very slightly. Suppose due to a small ﬂuctuation v increases very, very slightly. Then since (∂P/∂v)T > 0 on this portion of the isotherm, P will increase. This will create a net upward force on the piston, so v will increase even more. The process will stop only when point b is reached, where the pressure again balances the piston weight. Now consider the other case: suppose v decreases very slightly. Now the pressure in the gas drops, so there is a net downward force on the piston, and it falls until point a is reached. We conclude from this thought experiment that the entire portion of the isotherm with (∂P/∂v)T > 0 is unstable. A small ﬂuctuation in v, no matter how small, results in the system changing state: it will either go to the lowvolume state a, or the high-volume state b (both of which are stable to small ﬂuctuations, since (∂P/∂v)T < 0 for these states). Thus, the van der Waals equation predicts that for a given T < Tc , there is a range of pressures for which there are two possible states which are stable to small ﬂuctuations. One state has v < vc , and may be regarded as the liquid

CHAPTER 4. THE SIMPLE COMPRESSIBLE SUBSTANCE

96

14 12 10 8 6 4 2 0 14 12 10 8 6 4 2 0 14 12 10 8 6 4 2 0 0.001

Actual CH4

P (MPa)

van der Waals

P (MPa)

ideal gas law

P (MPa)

0.01 v (m /kg)
3

0.1

Figure 4.13: Isotherms for methane predicted by TPX, the van der Waals equation, and the ideal gas equation. The temperatures are 179.1, 190.5, and 202.0 K.

CHAPTER 4. THE SIMPLE COMPRESSIBLE SUBSTANCE
T0

97

Unstable

P

0

v a

v 0

v b

Figure 4.14: An isotherm of the van der Waals equation for T < Tc .
T0

actual equilibrium behavior vdw isotherm

P sat v f v g

Figure 4.15: Comparison of actual behavior during isothermal compression to prediction of van der Waals equation. state, and the other has v > vc , and may be regarded as the vapor state. In reality, of course, we don’t observe isotherms which have the shape in Fig. 4.14 for T < Tc . Instead, at some pressure Psat (T ) the ﬂuid abruptly switches from the low-density (vapor) state to the high-density (liquid) state or visa versa, as shown in Fig. 4.15. The reason this happens is that although both the liquid and vapor states for pressures within the shaded region are stable to small perturbations, only one is stable to “big” perturbations. For P > Psat (T ), it is the liquid state which is most stable, and for P < Psat(T ) it is the vapor state. We’ll see how to predict Psat (T ) soon. This qualitative behavior applies to real ﬂuids, not only to idealized van der Waals ﬂuids: the liquid-vapor phase transition results from an instabil-

CHAPTER 4. THE SIMPLE COMPRESSIBLE SUBSTANCE

98

ity of an “underlying” smooth P (v, T ) function. By carefully avoiding “big” perturbations (like dirt particles or scratches on the container wall where vapor bubbles can get a start) it is actually possible experimentally to prepare liquid at P < Psat (T ). Similarly, by carefully compressing a vapor in very clean conditions pressures greater than Psat (T ) can be reached without liquid droplets forming. Such states are called metastable states. Metastable states are not thermodynamic equilibrium states, since if the system is given a big enough perturbation (a dust particle, a scratch, gently shaking the container), it switches to the more stable state. To experimentally prepare metastable states, it is actually more common to vary temperature at a given P . In this case, liquid is stable below Tsat (P ) (the boiling point), and metastable above it. Liquid heated above the boiling point without boiling is called superheated liquid and vapor cooled below the boiling point without condensing is called supercooled vapor. A Generalized Equation of State The van der Waals equation of state may be put in non-dimensional form by deﬁning the nondimensional reduced pressure P ∗ = P/Pc and reduced temperature T ∗ = T /Tc . If we write Eq. (4.44) in terms of P ∗ and T ∗ , and substitute for a and b from Eq. (4.47) and Eq. (4.48), it reduces to Z3 − P∗ + 1 Z2 + 8T ∗ 27P ∗ 64T ∗2 Z− 27P ∗2 = 0, 512T ∗3 (4.49)

where Z is the compressibility factor P v/RT . The roots of this cubic equation determine Z(T ∗ , P ∗). Depending on T ∗ and P ∗, this equation will have either one root, or three. (If three, then as discussed above the middle one is unstable.) Note that a and b do not appear in Eq. (4.49). Thus, the van der Waals equation of state reduces to a single equation for all ﬂuids, independent of a and b, as long as T and P are expressed in reduced form. Equations of state which depend only on T ∗ and P ∗ are called generalized equations of state. The van der Waals equation is the simplest example of a generalized equation of state. At the critical point (P ∗ = 1, T ∗ = 1) the solution to Eq. (4.49) is Z = 3/8. Thus, the van der Waals equation predicts that all ﬂuids should have Zc = Pc vc /RTc = 0.375. Unfortunately, Zc for real ﬂuids is not 0.375, and diﬀers from one ﬂuid to the next. Most ﬂuids have Zc in the range from about 0.23 to about 0.33 – less than the van der Waals equation of state would predict. So clearly the van der Waals equation of state is not very accurate for real ﬂuids. This is also clear from Fig. 4.13: the critical speciﬁc volume is overpredicted, as is the liquid speciﬁc

CHAPTER 4. THE SIMPLE COMPRESSIBLE SUBSTANCE

99

Figure 4.16: Compressibility factor Z vs. P ∗ (Pr ) for several values of T ∗ (Tr ). volume at high pressure. 4.7.3 The Principle of Corresponding States Even though the van der Waals equation is not particularly accurate, the ability to write it as a universal function Z(T ∗ , P ∗) is intriguing — maybe Z really can be expressed as a function of only T ∗ and P ∗, but the function resulting from the van der Waals equation simply isn’t the right one. To test this hypothesis, we can take measured P − v − T data for diﬀerent ﬂuids, calculate Z = P v/RT , and the plot Z as a function of P ∗ = P/Pc and T ∗ = T /Tc , where of course Tc and Pc diﬀer for each ﬂuid. If Z = Z(T ∗ , P ∗), plotting Z in this way should collapse the experimental P − v − T data onto the same set of curves for all ﬂuids. A plot of this type is shown in Fig. 4.16, in which Z is plotted vs. P ∗ for T ∗ values from 1 to 2. It is observed that the experimental values for diﬀerent ﬂuids do tend to fall onto the same curves for a given T ∗ , independent of the particular ﬂuid. However, it is also clear that this result is only approximate. For example, on the T ∗ = 1.5 curve in Fig. 4.16, the propane data points are systematically above the mean, and the methane points systematically below. Also, we have already stated that Zc varies slighly from one ﬂuid to another, which could not be true if Z = Z(T ∗ , P ∗) were exactly true, since Zc = Z(1, 1).

CHAPTER 4. THE SIMPLE COMPRESSIBLE SUBSTANCE

100

The statement that Z = Z(T ∗ , P ∗) is known as the Principle of Corresponding States. As we have already seen, it an approximate principle, not an exact one. Several empirical ﬁts for Z(T ∗ , P ∗) have been proposed which seek to reproduce the “average” behavior shown in Fig. 4.16. The earliest was a set of charts due to Nelson and Obert in 1954. A more recent, popular generalized equation of state is one proposed by Lee and Kesler in 1975. The function they propose for Z(T ∗ , P ∗) for “simple” ﬂuids is shown graphically in Fig. 4.17. It can be seen that this function corresponds fairly closely to the experimental data shown in Fig. 4.16. Simple ﬂuids are basically ones composed of small, simple molecules, but the precise deﬁnition of a simple ﬂuid is circular: it is one for which Z(T ∗ , P ∗) is as shown in Fig. 4.17. Lee and Kesler (following earlier work by Pitzer) also introduced a third parameter, the so-called acentric factor ω, to allow better correlation of experimental data than can be achieved using only the parameters T ∗ and P ∗ . They observe that simple ﬂuids have Psat (T )/Pc = 0.1 when T /Tc = 0.7. To parameterize deviations from the behavior of Fig. 4.17, they deﬁne ω as ω = − log10 Psat (0.7Tc) Pc − 1.0. (4.50)

The acentric parameter is zero for a simple ﬂuid, and usually positive for other ﬂuids. Lee and Kesler assume that the eﬀects of non-zero ω can be accounted for by adding a correction term linear in ω: Z = Z (0) (T ∗ , P ∗) + ωZ (1) (T ∗ , P ∗). (4.51)

Here Z (0) is the function shown in Fig. 4.17, and Z (1) is a correction factor to account for non-simple-ﬂuid eﬀects. Speciﬁcally, Z (1) is computed so that an accurate Z(T ∗ , P ∗) function is obtained for octane (C8 H18 ), which is not a “simple” ﬂuid and has ω = 0.3978. Of course, Eq. (4.51) is still approximate, but it is found to be accurate to within 2% or 3% for most non-polar or slightly polar ﬂuids. For highly-polar ﬂuids (e.g. water) or very light ones (e.g. hydrogen, helium, or neon) for which quantum eﬀects are important, it is less accurate. Equation 4.51 is implemented in TPX as the function ZLK(T*, P*, Omega). The Omega parameter is optional – if omitted, the result will be calculated for ω = 0 (a simple ﬂuid). 4.7.4 The Incompressible Substance Finally, now that we have considered some rather complex model equations of state, we turn to the simplest possible equation of state. The compressibility

CHAPTER 4. THE SIMPLE COMPRESSIBLE SUBSTANCE

101

Figure 4.17: The Lee-Kesler generalized compressibility function Z (0) for a simple ﬂuid.

CHAPTER 4. THE SIMPLE COMPRESSIBLE SUBSTANCE

102

of most liquids and solids is quite low. Unless very large pressures are applied, the speciﬁc volume hardly changes at all. If we wish to ignore entirely the compressibility of a liquid or a solid, we can replace the real P (v, T ) behavior by a simple model: v = v0 , (4.52)

where v0 is a constant. A idealized substance which obeys Eq. (4.52) is called an incompressible substance. It is not possible to do compression work on an incompressible substance, since by deﬁnition dv = 0. Therefore, the First Law for an incompressible substance is du = dQ. ¯ (4.53)

Therefore, the only way to change u is by heat addition. This implies that an incompressible substance has only one degree of freedom, so u = u(T ). If we choose to approximate a real liquid as incompressible in solving a particular problem, we simply neglect the small pressure dependence of v and u. A common choice is to evaluate v and u at the saturation pressure at the local temperature, and use these values no matter what the real liquid pressure is. Note that by deﬁnition h = u + P v, so h will still depend on P for an incompressible substance, even though u does not. Therefore, for a process in which T and P change, ∆h = ∆u + v∆P . For most liquids and solids, v is small enough that the v∆P term is small compared to ∆u. Note also that (∂h/∂T )P = du/dT in this case, so cp (T ) = cv (T ) (4.54)

for an incompressible substance. This relationship is very diﬀerent than the one for ideal gases [Eq. (4.35)].

Problems
4.1 Some properties of ice, liquid water, and water vapor at the triple point T = 273.16 K are given below. Phase Ice Liquid Vapor Density (kg/m3) 917.0 999.8 4.84 × 10−3 Speciﬁc enthalpy (kJ/kg) -333.5 0.0 2501.4

CHAPTER 4. THE SIMPLE COMPRESSIBLE SUBSTANCE The enthalpy values are relative to the liquid enthalpy.

103

A closed, constant volume 1 liter container initially is at a temperature inﬁnitesimally below 273.16 K. It holds solid and vapor in equilibrium, with 80% solid by mass. Heat is now added until equal masses of liquid and solid are present. Write down equations expressing energy, mass, and volume balance. Determine numerically how much heat must be added. 4.2 The air pressure in aircraft cabins is kept lower than sea level atmospheric pressure, since otherwise the pressure force on the airframe would be too great at high altitude. A typical value for a transatlantic ﬂight would be a cabin pressure equivalent to atmospheric pressure at an elevation of 8,000 ft. On such a ﬂight, to what temperature should the ﬂight attendants heat water for it to boil? How much heat must be added to boil a unit mass of water, and how does it compare to the heat needed at sea level? Make any reasonable assumptions you need for the analysis, but state what you are assuming. 4.3 Five kg of methane are contained in a closed, 150 liter container. If the pressure is 1 MPa, determine: 1. Whether any liquid is present, and if so how many kg 2. The temperature 3. The internal energy U of the methane (J) 4.4 A particular substance has an isothermal compressibility κ = aT 4 /P 3 and a thermal expansion coeﬃcient β = bT 3 /P 2 . Determine the equation of state v(T, P ) to within an additive constant and the ratio a/b. 4.5 Ten kg of saturated nitrogen vapor at 90 K is heated at constant pressure until its volume is 3 m3 . Determine 1. The pressure 2. The ﬁnal temperature 3. The heat added 4. The work done by the nitrogen on the environment. 4.6 Using TPX, plot the compressibility factor Z for hydrogen and for oxygen at 300 K vs. pressure. For hydrogen, plot Z over the pressure range of 1 atm to 100 MPa, and for oxygen over the pressure range of 0.02 MPa to 20 MPa. Use a logarithmic scale for pressure.

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104

For a particular gas storage tank design, it is necessary to estimate to within 1% accuracy the tank volume required for a given mass of gas and pressure at T = 300 K. For both hydrogen and oxygen, determine the pressure below which use of the ideal gas equation provides acceptable accuracy. 4.7 Estimate the temperature water at the bottom of a 500 ft deep lake would have to be heated before it begins to boil. 4.8 The speciﬁc heat at constant volume cv for insulating solids at low temperature is given by the equation cv = R 12π 4 5 T θ
3

,

which is known as the Debye T 3 law. The constant θ is known as the Debye temperature, and is a characteristic of the material in question. For diamond, θ = 2200 K. Treating diamond as incompressible, how much heat in Joules must be added to raise the temperature of 1 kg of diamond from 1 K to 50 K? 4.9 Show that for any simple compressible substance ∂P ∂T and ∂β ∂P = v β κ ∂κ ∂T

=−
T

P

4.10 Derive the results for a van der Waals gas RTc = and Pc = 8a 27b

a . 27b2

4.11 A particular substance is found to obey the Dieterici equation of state P (v − b) exp(a/vRT ) = RT. 1. Derive expressions for the properties at the critical point Pc, vc , and Tc in terms of the constants a, b, and R.

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105

2. Compare the values for Zc = Pc vc /RTc predicted by this equation with the values from Table 4.2 or calculated using TPX for hydrogen, carbon dioxide, and water. 4.12 A particular ﬂuid has critical state parameters Tc = 300 K and Pc = 6 MPa, and a molecular weight of 30. Estimate its speciﬁc volume at T = 330 K and P = 12 MPa using 1. the van der Waals equation of state 2. the Lee-Kesler generalized compressibility function, assuming it is a simple ﬂuid. 4.13 Determine the high-temperature limit for cp for Ar, N2 , H2 O, CO2 , and CH4 , and compare to the results in Figure 4.12. What is the value of cp (kJ/kg/K) in the high-temperature limit for C60 ? 4.14 Calculate the acentric parameter ω for N2 , CH4 , and CF4 H2 (“HFC134a”) using TPX. Compare the predictions for Z at T ∗ = 1.2, P ∗ = 1.5 using 1. TPX 2. Lee-Kesler with ω = 0 3. Lee-Kesler with the actual ω.

Index adiabatic, 35 Boltzmann’s constant, 87 center of mass, 20 chemical potential, 54 coexistence lines, 77 compressibility factor, 88 conservative force, 19 deﬁnition, 20 control mass, 4 control volume, 4 Corresponding States Principle of, 100 critical point, 73 Curie substance, 43 datum state, 71 degrees of freedom, 63 dielectric, 42 electric dipole moment, 40 electric susceptibility, 41 energy, 3 accounting, 9 balance, 9 conservation of, 24 enthalpy deﬁnition, 70 entropy, 3 environment, 3 equations of state, 65 equilibrium, 50 chemical, 55 diﬀusive, 53 106 mechanical, 52 phase, 54 restricted, 55 thermal, 52 equipartition of energy, 91 extensive, 62 First Law of Thermodynamics, 34 force centrifugal, 17 Lorentz, 17, 21 traction, 17 gas constant, 86 generalized displacements, 44 generalized equations of state, 98 generalized forces, 44 heat capacity, 86 heat transfer, 32 rate, 35 hydrostatics, 57 ideal gas, 87 incompressible substance, 102 inertial mass, 7 intensive, 62 internal energy, 28 isobar, 78 isothermal compressibility, 84 kinetic energy, 8 Lee-Kesler generalized compressibility, 100 lever rule, 80

INDEX liquid compressed, 79 subcooled, 79 magnetization, 43 manometer, 57 mass density, 20 phase, 54 phase diagrams, 72 polar, 40 polarization, 41 potential energy deﬁnition, 23 gravitational, 19 power, 12 pressure, 33 critical, 72 reduced, 98 quality, 79 quasi-static, 37 reference state, 71 saturation, 78 saturation pressure, 77 saturation temperature, 77 speciﬁc, 62 speciﬁc heat cp , 85 cv , 85 sublimation, 73 supercooled vapor, 98 superheated liquid, 98 surface tension, 39 system, 3 classiﬁcation, 4 temperature, 53

107 critical, 73 reduced, 98 thermodynamic, 61 thermal energy, 1 thermal expansion coeﬃcient, 83 thermodynamic state, 63 triple line, 75 triple point, 75 van der Waals equation of state, 94 vapor pressure, 77 work, 8, 10 deﬁnition of, 10

CHAPTER 5
THE FIRST LAW FOR OPEN SYSTEMS 5.1 Introduction
An open system is one which matter may enter or leave. An uncovered cup of coﬀee is an open system, since water evaporates from it. Open systems are very common in engineering thermodynamics. A simple example is a jet engine, which takes in air and fuel, mixes and burns them, and then exhaust the hot combustion products at high speed to produce thrust. Design of an eﬃcient jet engine requires a careful thermodynamic analysis, which we will discuss in more detail soon. In this chapter, we develop one of the major tools needed to carry out such an analysis — an expression of the First Law of Thermodynamics suitable for open systems. Following this development, we will apply it to examine the characteristics of some very useful devices for changing the state of a ﬂowing ﬂuid, and take a ﬁrst look at how such simple devices may be combined to do useful tasks, such as generate electrical power.

5.2 Some Approximations Involving Flowing Fluids
In most cases of interest to us, the matter which enters or leaves an open system is a ﬂuid. Problems involving ﬂowing ﬂuids are more complicated than the problems we have considered up to now, since strictly speaking a ﬂowing ﬂuid is not in equilibrium — unbalanced forces are causing it to move. Depending on the particular features of the ﬂow, it may be turbulent, and properties such as temperature, pressure, velocity, and density may vary from point to point within the ﬂuid. To make the analysis of problems with ﬂowing ﬂuids tractable, we need to make some approximations. 5.2.1 Equilibrium Properties What property relationships should we use to determine the properties of a ﬂowing ﬂuid? We now know a great deal about the properties of ﬂuids in equilibrium (at rest). But if the velocity varies from point-to-point in a ﬂuid, as it does in all real viscous ﬂuids, then there is no reference frame in which it 106

CHAPTER 5. THE FIRST LAW FOR OPEN SYSTEMS

107

appears stationary. Fortunately, for most situations of engineering interest the relationships among properties in a ﬂowing ﬂuid are indistinguishable from the equilibrium equations of state we discussed in the last chapter, as long as the thermodynamic properties are interpreted as “local” values at a point in a ﬂuid. When this is true, we say the ﬂow is in local thermodynamic equilibrium, sometimes abbreviated LTE. For example, a very low density ﬂowing gas will usually still obey P v = RT to a very good approximation at every point in the gas, even if P , v, and T diﬀer from point to point, as long as the P , v, and T values are all measured at the same point. Similarly, when LTE holds, the local speciﬁc internal energy u will have the same dependence on the local temperature as if the gas were truly in equilibrium. There can be some extreme situations where LTE does not hold and equilibrium equations of state break down entirely. For example, in a shock wave the properties of a gas change dramatically over a distance comparable to a molecular mean free path, which in air at 1 atm is about 0.15 µm. When properties change signiﬁcantly over a distance of a mean free path, then the equilibrium property relationships no longer hold, and properties such as pressure and temperature may not even be meaningful. However, the properties of the gas just a few mean free paths on either side of the shock wave would be found to be well-described by equilibrium equations of state. Fortunately, we will not need to evaluate any gas properties right at a shockwave. Another type of system where non-LTE behavior is often found is a plasma, which is simply a gaseous mixture of ions and electrons (and often neutral atoms or molecules too). Plasmas occur naturally in the upper atmosphere and in outer space, and also can be created in the laboratory. Plasmas have been investigated actively since the 1950’s as a means to achieve the conditions necessary for nuclear fusion. Smaller scale plasmas are also widely used in industry, for example to etch patterns in semiconductors, to clean surfaces, and to deposit thin coatings. Plasmas can have strange behavior – often the electrons have much more kinetic energy on average than do the heavy ions or atoms. In plasmas used in the semiconductor industry, the heavy particles are at room temperature, while the electrons behave as if they had a much higher temperature, often greater than 10,000 K. Clearly a gas which behaves as if diﬀerent species have diﬀerent temperatures is not close to equilibrium, and equilibrium gas properties could not predict the correct properties of such a plasma.

CHAPTER 5. THE FIRST LAW FOR OPEN SYSTEMS

108

1

2 . m = VA) = 1 A V 2 1 VA) 2

x

Figure 5.1: Approximate one-dimensional ﬂow of a river. Another situation where deviations from equilibrium equations of state can be important is if the ﬂuid properties are changing too rapidly for the ﬂuid to adjust. For example, if the pressure in a liquid is rapidly reduced, it may drop below Psat (T ) without vapor forming, since formation of vapor bubbles is not instantaneous. When Psat − P reaches some critical value, vapor bubbles will form explosively, which is known as cavitation. Some “ﬂuids” like glasses and polymer melts can take a very long time to re-establish equilibrium when perturbed, and thus often exhibit non-equilibrium properties. Situations like these in which the ﬂuid properties diﬀer signiﬁcantly from the equilibrium properties are relatively rare. In this book, we will usually assume that equilibrium equations of state may also be used to compute the properties of ﬂuids slightly out of equilibrium (due, for example, to ﬂow). We will call this the assumption of equilibrium properties. 5.2.2 One-Dimensional Flow Another often-useful assumption is the assumption of one-dimensional ﬂow. Contrary to what its name seems to imply, it does not mean the molecules follow one after another along a mathematical line. Instead, it only means that the ﬂow velocity is assumed to depend on only a single co-ordinate (not two or three, which would deﬁne two-dimensional or three-dimensional ﬂow, respectively). The one-dimensional ﬂow assumption is easily understood by considering the ﬂow of a river (Fig. 5.1). A detailed description of a river ﬂow would be quite complicated, since it would have to describe in detail the turbulent ﬂow around over and around rocks, etc. But on a more global scale, a river

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109

ﬂow is simple: the water ﬂows basically downstream. When we make the onedimensional ﬂow assumption, we ignore the details of the actual ﬂuid ﬂow and focus on the principal direction the ﬂuid is moving. We approximate the ﬂow velocity as being the same everywhere on a cross-section normal to the local downstream direction. Essentially, we are assuming that all ﬂuid “packets” at a given downstream location are moving with the average velocity of the real three-dimensional ﬂow at that downstream location. One-dimensional ﬂow does not necessarily mean the velocity is constant. In a river, the ﬂow is rapid where the river is narrow, and slow where the river is wide. What is constant is the water mass ﬂow rate m (kg/s). The mass ﬂow ˙ rate of water is mw = ρw V (x)A(x), ˙ (5.1) where x is the distance downstream, ρw is the water density (constant), V (x) is the one-dimensional ﬂow speed at x, and A(x) is the cross sectional area of the river at x. Therefore, for a river with constant m, the product V (x)A(x) is constant. In ˙ practice, Eq. (5.1) would be the equation used to calculate V (x), and so serves to deﬁne the one-dimensional ﬂow speed. We will be interested in many situations in which a gas or vapor ﬂows through a tube of variable cross-sectional area A(x). As long as the changes in area are gradual, the ﬂow down the tube is approximately one-dimensional. The gas mass ﬂow rate through the tube is given by the same equation as for the river ﬂow, except that we should allow for the possibility that the gas density may also depend on x: m = ρ(x)V (x)A(x). ˙ (5.2) The density would depend on x if, for example, the gas heats up or cools down as it ﬂows down the tube (perhaps the tube walls are heated or cooled). As we’ll see later, the density can also vary greatly if the ﬂow speed is comparable to or greater than the speed of sound and the tube area changes. Of course, Eq. (5.2) also applies to liquids, for which ρ(x) ≈ constant. 5.2.3 Steady Flow Equation 5.2 always holds in one-dimensional ﬂow, even if m is time-dependent. ˙ In this case, the ﬂow is unsteady. If all quantities appearing in Eq. (5.2) do not depend on time, then the ﬂow in the tube is by deﬁnition steady.

CHAPTER 5. THE FIRST LAW FOR OPEN SYSTEMS
Vin dt dWext

110 dWext CV
Vin dWflow

CV CM

Vin dWflow

CM

dQ

dQ

Figure 5.2: Control mass and control volumes for analysis of a system with ﬂuid entering through an inlet pipe. (a) time t; (b) time t + dt.

5.3 Energy Accounting for an Open System
In order to develop the First Law for an open system, consider the simple situation shown in Fig. 5.2. A container is connected to an inlet tube, and ﬂuid (either liquid or gas, it doesn’t matter) is ﬂowing in. There is no outlet, so ﬂuid builds up in the container. The ﬂow into the container is assumed to be one-dimensional, with speed Vin . Since we know how to write the First Law for closed systems, let’s begin by analyzing the control mass system labeled CM. Suppose we start the analysis at time t and allow an inﬁnitesimal time dt to elapse, during which energy ﬂows from the environment to the system in the form of heat dQ and work dW . ¯ ¯ The control mass boundary at time t in the inlet pipe is chosen to include the ﬂuid which ﬂows into the container in time dt. At time t + dt, this ﬂuid has moved into the container, so the control mass boundary has moved ﬂush with the end of the inlet pipe. The work done on the system in time dt contains two contributions, which ¯ we will call dWext and dWflow . The ﬁrst is the work done by external forces ¯ (such as a stirrer, or electrical work from a battery, or any other type of work). The second contribution is the force on the moving system boundary in the inlet pipe. Remember any time a system boundary moves by dx under action of an external force F, work in the amount F · dx is done by the external force on the system. In the present case, the external force is provided by the ﬂuid in the inlet

CHAPTER 5. THE FIRST LAW FOR OPEN SYSTEMS

111

pipe just outside the control mass, pushing on the ﬂuid just inside the control mass. The ﬂow work done in time dt is dWflow = F dx = (Pin Ain )dx = Pin Ain Vin dt. ¯ (5.3)

This expression may be re-written in terms of the mass ﬂow rate min in the ˙ inlet pipe. The mass ﬂow rate (kg/s) is min = ρin Vin Ain . ˙ Therefore, in terms of min , ˙ ˙ ˙ dWflow = min (P/ρ)in dt = (mP v)in dt. ¯ The ﬁrst law for system CM is: dECM = = dQ + dW ¯ ¯ dQ + dWext + (mP v)in dt. ¯ ¯ ˙ (5.6) (5.5) (5.4)

The term ECM includes all energy in the control mass, whether internal, kinetic or potential: ECM = (U + Ek + Ep )CM . (5.7) Now consider the control volume CV, which consists only of the container. At time t, CV does not contain the ﬂuid in CM which is still in the inlet tube. Therefore, ˙ (5.8) ECM (t) = ECV (t) + (min dt)(ein ) where ein is the total speciﬁc energy of the ﬂuid in the inlet: ein = uin + ek,in + ep,in . (5.9)

If the inlet ﬂow is one-dimensional and if the potential energy is due to gravitation then V2 ein = uin + in + gyin , (5.10) 2 where yin is the elevation of the inlet tube. At time t + dt, CM and CV coincide, so ECM (t + dt) = ECV (t + dt). Subtracting Eq. (5.8) from Eq. (5.11), we ﬁnd dECM = dECV − (min dt)(ein ). ˙ (5.12) (5.11)

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112

Flow Work

Convected Energy

. (m Pv)in . Wext dE CV dt
External Work

. (m e)in . Q
Heat

Figure 5.3: Energy accounting for a control volume with ﬂuid entering through one inlet. Putting this into Eq. (5.6) yields ˙ ¯ ¯ ˙ dECV − (min dt)(ein )E = dQ + dWext + (mP v)in dt. Rearranging, we have dECV = dQ + dWext + [m(e + P v)]in dt. ¯ ¯ ˙ (5.14) (5.13)

We may divide this equation by the elapsed time dt to express this equation on a rate basis (per unit time): d ˙ ˙ ˙ ECV = Q + Wext + [m(e + P v)]in , dt where dQ ¯ ˙ Q= dt is the rate at which heat is added to CV, and dWext ¯ ˙ Wext = dt (5.15)

(5.16)

(5.17)

is the rate at which external work is done on CV. Note that dQ/dt should be ¯ interpreted as a fraction ( dQ divided by dt), and not as a derivative; the same ¯ holds for dW/dt. ¯ Both Eq. (5.14) and Eq. (5.15) are statements of the First Law for an open system with one ﬂuid inlet. They may be interpreted as shown in Fig. 5.3: the rate of increase of energy stored in the control volume (dECV /dt) is equal to the sum of the rates of energy transfer into the system from the environment. The energy transfer terms are due to

CHAPTER 5. THE FIRST LAW FOR OPEN SYSTEMS ˙ 1. external work: Wext ˙ 2. heat transfer: Q

113

3. ﬂow work required to push the ﬂuid into the control volume: (mP v)in ˙ 4. energy which is carried in (or convected in) by the entering ﬂuid: (me)in . ˙ Equation (5.15) may be easily generalized to handle an arbitrary control volume with multiple ﬂuid inlets and multiple ﬂuid outlets. For the ith inlet, there is a term [m(e + P v)]i in the energy balance to account for the energy ˙ convected in with the ﬂuid, and the ﬂow work done by the environment to push it in. For outlets, the analysis is the same as for inlets except V < 0. This ˙ simply changes the sign: for the oth outlet, a term −[m(e + P v)]o must be added. Therefore, the general statement of the First Law for an open system is

d ˙ ˙ ECV = Q + Wext + dt

[m(e + P v)]i − ˙ inlets outlets

[m(e + P v)]o ˙

(5.18)

Note that m, e, P , and v may diﬀer for each inlet and outlet, and may diﬀer ˙ from the state of the ﬂuid inside the control volume. Also, all quantities in this equation may be time-dependent. We have not said anything about the state of the matter inside the control volume. It might be near equilibrium, or might be very far from it, with shock waves, plasmas, chemical reactions, or other complex phenomena occurring. Of course, if we wish to relate ECV to other properties (e.g. pressure or temperature), we would need to know more about just what is happening inside CV. We do need to evaluate (e+P v) for the ﬂuid in each inlet and in each outlet. Typically, we will make the equilibrium properties assumption for the inlet and outlet streams to allow evaluating e + P v. Since e = u + ek + ep , e + Pv = (u + P v) + ek + ep = h + ek + ep . (5.19)

Therefore, e + P v is simply the speciﬁc enthalpy h plus any macroscopic kinetic and potential energy per unit mass. For the common case of a one-dimensional

CHAPTER 5. THE FIRST LAW FOR OPEN SYSTEMS ﬂow where the only potential energy is gravitational, |V|2 + gy 2

114

e + Pv = h +

(5.20)

Another way to write Eq. (5.18) for this case is d ˙ ˙ ECV = Q + Wext dt + inlets m h+ ˙

|V2 | + gy 2

i

− outlets |V2 | m h+ ˙ + gy 2

(5.21) o Example 5.1 An insulated tank is to be pressurized with air. A high-pressure air line is connected to the tank inlet through a ﬂow-regulating valve which produces a constant mass ﬂow of air m. The temperature of the air in the high-pressure ˙ line is the ambient temperature T0 . Assuming that air may be idealized as an ideal gas with constant speciﬁc heats, determine the temperature in the tank as a function of time. CV

. m T0
Solution: Since mass is entering the tank, this problem is most easily solved by deﬁning a control volume and using the First Law in the form of Eq. (5.18). The surface of the control volume should always be placed where we know some information. Since we don’t know the state of the air at the outlet of the valve, but we do know the state at the valve inlet, let the control surface cut through the inlet before the valve. We will assume the following: 1. One-dimensional, steady ﬂow in air line 2. Air inside tank is uniform, with negligble kinetic energy 3. Potential energy may be neglected 4. Kinetic energy in air line may be neglected

CHAPTER 5. THE FIRST LAW FOR OPEN SYSTEMS 5. Equilibrium properties in air line and in tank ˙ 6. Adiabatic process (Q = 0) 7. Ideal gas with constant cv For this problem, Eq. (5.18) becomes d ECV = [m(e + P v)]in . ˙ dt

115

(5.22)

˙ Note that Wext = 0, since there is no external work in this problem. Since we are assuming kinetic and potential energies are negligible, ECV = UCV and (e + P v)in = hin . Therefore, d UCV = [mh]in . ˙ (5.23) dt Since the ﬂow is steady, the right-hand side of this equation is a constant, and therefore it is easily integrated: ˙ UCV (t) = UCV (0) + mhin t. We can also write a mass balance on the control volume: MCV (t) = MCV (0) + mt ˙ (5.25) (5.24)

The rate of these two equations gives the speciﬁc internal energy of the air in the tank as a function of time: UCV (t) ˙ UCV (0) + mhin t uCV (t) = = . (5.26) MCV (t) MCV (0) + mt ˙ At long times, the terms linear in t will dominate over the constant terms, so t→∞ lim uCV (t) = hin .

(5.27)

For an ideal gas with constant cv , we may write u = cv T and h = cp T , taking the reference state at T = 0. Then lim TCV (t) = cp Tin cv (5.28)

t→∞

Thus, the air in the tank ends up hotter than the air in the high-pressure line. The physical reason for this is that to put more air in the tank, the air already there has to be compressed to make room. We’ve seen before that adiabatic compression of a gas causes it to heat up, due to the compression work done on it. This is one more example of this eﬀect. An interesting feature of this problem is that the temperature in the tank depends only on Tin and cp /cv , and not on the ﬁnal pressure. For air, cp /cv ≈ 1.4, so if Tin = 300 K, the ﬁnal temperature in the tank is 420 K.

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116

5.4 Steady State
If the energy and mass contained within the control volume are not changing with time, we say the control volume is in steady state. In this case, dECV /dt = 0, so Eq. (5.18) becomes ˙ ˙ Q + Wext + inlets [m(e + P v)]i = ˙ outlets [m(e + P v)]o . ˙

(5.29)

This equation simply states that at steady state the energy inﬂow rate to the control volume (the left-hand side) must equal the energy outﬂow rate (the right-hand side). Since no mass can be accumulating inside the control volume at steady state, we also require mi = ˙ mo . ˙ (5.30) inlets outlets

It is important to remember the diﬀerence between the assumption of steady ﬂow and the assumption of steady state. Steady ﬂow applies to an individual inlet or outlet stream, and means that m and the ﬂuid properties of the stream ˙ are constant. Steady state applies to the entire control volume, and means that energy and mass inﬂows and outﬂows balance, so that there is no net change in the energy or mass contained in the control volume. In the last example, the inlet ﬂow was a steady ﬂow, but the system was not in steady state, since there was no outlet and both mass and energy continued to accumulate in the tank. The steady state assumption is very useful to analyze many real engineering systems, including most power plants, chemical plants, jet engines, rocket motors, refrigeration systems, etc. Although they are not always truly in steadystate (for example, the thrust of an aircraft engine changes from take-oﬀ to cruise), the transient terms in the energy and mass balances (dECV /dt and dMCV /dt, respectively) are usually small enough compared to the other terms that they may be neglected, and the problem treated as a steady state one. Most of these complex engineering systems are constructed from a set of simple steady-ﬂow, steady-state devices which alter the state of a ﬂowing ﬂuid in some way, and may exchange energy as heat or work with the environment. We will introduce some of the major ones in the next section. Analyzing these simple systems shows how the First Law for open systems is used, and it allows us to begin assembling a “toolkit” of devices which we will use to build more complex systems in later chapters.

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117

5.5 Some Steady-Flow, Steady-State Devices
5.5.1 Nozzles A nozzle is a device which is accelerates a ﬂuid ﬂowing steadily through it. For liquids or subsonic gases, a nozzle simply consists of a converging tube (Fig. 5.4.1 To ﬁrst approximation, nozzles are often idealized as adiabatic. In a real nozzle, there may be some heat transfer to or from the environment if the ﬂuid is hotter or colder than the surroundings. But often the ﬂuid ﬂows through the nozzle so rapidly that there is little time for a signiﬁcant amount of heat transfer to occur. We will make the following assumptions to analyze ﬂow through an adiabatic nozzle: 1. One-dimensional, steady ﬂow at inlet and outlet (states 1 and 2) 2. Equilibrium properties at states 1 and 2 ˙ 3. Q = 0 4. Negligible change in potential energy 5. Steady state With these assumptions, the energy balance is m h+ ˙ V2 2 = m h+ ˙
1

V2 2

.
2

(5.31)

Because of the steady-state assumption, m1 = m2 , so ˙ ˙ h1 + V12 V2 = h2 + 2 . 2 2 (5.32)

Since the ﬂuid accelerates through a nozzle, the pressure must decrease in the direction of ﬂow. Thus, P2 < P1 . The action of an adiabatic, steady nozzle on the thermodynamic state of the ﬂuid is as shown in the P −h plot in Fig. 5.4: both the pressure and the enthalpy are lower at the outlet than at the inlet. The horizontal distance h1 − h2 between the inlet and outlet state points equals the increase in kinetic energy. The line connecting states 1 and 2 is shown as dashed in Fig. 5.4. A dashed, straight line is used simply to indicate that state 1 is transformed to state 2 by
1 A nozzle for a supersonic gas consists of a diverging tube, as we’ll discuss in more detail later. To accelerate a gas from subsonic to supersonic speed, a converging-diverging tube is required (an hourglass shape).

CHAPTER 5. THE FIRST LAW FOR OPEN SYSTEMS
1
V 1 V2

118

log P

2

h

(V / 2)

2

Figure 5.4: An adiabatic nozzle. the nozzle. It does not imply that the ﬂuid state really followed a straight-line trajectory in (P, h) in going from state 1 to state 2. In fact, we have only made the equilibrium properties assumption for states 1 and 2. Inside the nozzle, the ﬂuid might not be describable by any equilibrium state – we don’t know. Of course, the positions of state 1 and state 2 do not need to be as shown in this diagram. States 1 and 2 could be located anywhere, including in the liquid or two-phase regions. The only requirement is that P2 < P1 and h2 < h1 . Example 5.2 Nitrogen at 300 K and 2 atm ﬂows at 1 m/s into an adiabatic nozzle which exhausts into the ambient air at 1 atm. If the exit speed is 300 m/s, what is the temperature at the exit? Solution: Using TPX, h1 = h("n2","tp",300,2) = 4.613 × 105 J/kg. The initial speciﬁc kinetic energy V12 /2 = 0.5 J/kg, and the ﬁnal speciﬁc kinetic energy V22 /2 = 4.5 × 104 J/kg. Therefore, h2 = 4.613 × 105 + (0.5 − 4.5 × 104 ) = 4.163 × 105 J/kg. Since P2 is given, P2 and h2 ﬁx the exit state. From TPX, T2 = Temp("N2","PH",1,4.163E5) = 256.57 K.

Note that in the last example the initial kinetic energy of the gas is very small compared to the ﬁnal kinetic energy, and thus ∆(V 2 /2) ≈ V22 /2. When this is the case, Eq. (5.32) reduces to V2 = 2(h1 − h2 ). We will often assume that the initial kinetic energy is negligible when analyzing ﬂow through nozzles.

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V 1

V2

log P 1 h

2

(V / 2)|

2

Figure 5.5: An adiabatic diﬀuser. 5.5.2 Diﬀusers A diﬀuser is in a sense the opposite of a nozzle. In a diﬀuser, a ﬂow decelerates, with a corresponding pressure rise. (The pressure must rise, since ﬂuid elements will only decelerate if they experience a net force directed opposite to the ﬂow direction.) For liquids and subsonic gases, a diﬀuser simply consists of a diverging tube (Fig. 5.5. The energy balance for a steady-state adiabatic diﬀuser is the same as that for an adiabatic nozzle (with the same assumptions): h1 + V2 2 = h2 +
1

V2 2

.
2

(5.33)

Now, however, V2 < V1 , so h2 > h1 . The state of the ﬂuid is transformed by the diﬀuser as shown in Fig. 5.5. If the exit area of the diﬀuser is much larger than the inlet area and the ﬂow is one-dimensional, then V2 V1 and ∆(V 2 /2) ≈ −V12 /2. In this limit, h2 = h1 + V12 /2. The quantity h∗ = h + V2 2 (5.34)

is known as the stagnation enthalpy of a ﬂuid with enthalpy h and speed V , since it is the enthalpy the ﬂuid would have if it were brought to rest (stagnation) in an adiabatic diﬀuser. In terms of the stagnation enthalpy, the energy balance for an adiabatic nozzle or diﬀuser may be written h∗ = h∗ . 1 2 (5.35)

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1 1 2 log P 2

h
Figure 5.6: An adiabatic valve. 5.5.3 Valves A valve is a device which drops the pressure in a ﬂowing ﬂuid without significantly increasing its kinetic energy. The interior of a valve looks something like a nozzle: there is a constriction in the ﬂow passage, requiring the ﬂuid to accelerate to pass through. But in a valve, the ﬂow passage opens up abruptly after the constriction, causing one-dimensional ﬂow to break down. A highly three-dimensional turbulent ﬂow develops, in which most of the kinetic energy of the ﬂow is converted into internal energy through viscous (friction) forces. Valves are highly non-equilibrium, irreversible devices. But all we require to carry out a steady-state analysis of a valve is the ability to evaluate properties just upstream and just downstream of the valve. As long as the ﬂow is reasonably one-dimensional and describable with equilibrium properties at these points, what is going on within the valve is irrelevant for the analysis. Like nozzles and diﬀusers, valves are often approximated as adiabatic, since usually the ﬂuid does not spend enough time in the valve for appreciable heat transfer to occur. Therefore we will make the following assumptions for the analysis of a valve: steady, one-dimensional ﬂow at the inlet and outlet (states 1 and 2); equilibrium properties at 1 and 2; adiabatic; steady state; negligible change in kinetic and potential energy. With these assumptions, the energy balance is simply h1 = h2 . (5.36)

Since P2 < P1 , the action of a valve on the state of the ﬂuid is as shown in Fig. 5.6. Of course, state 1 can be anywhere in the (P, h) plane, and state 2

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can be any point directly below state 1. Depending on conditions, the ﬂuid emerging from the valve could be liquid, vapor, or mixed. Example 5.3 An adiabatic valve in a liquid oxygen line causes a pressure drop of 0.5 MPa. If saturated liquid at 1 MPa enters the valve, what is the state of the oxygen emerging from the valve? Solution: Assume: 1. steady, 1D ﬂow, equilibrium properties at inlet and outlet 2. steady state 3. adiabatic 4. negligible change in P.E. or K.E. Then the energy balance is h1 = h2 . From TPX, h1 = hf (P1 ) = h("o2","px",1,0) = 114.16 kJ/kg. The inlet temperature T1 = Tsat (P1 ) = temp("o2","px",1,0) = 119.68 K. State 2 is ﬁxed by P2 = 0.5 MPa and h2 = h1 . Therefore, T2 = temp("o2","ph",0.5,114.16) = 108.86 K X2 = x("o2","ph",0.5,114.16) = 0.106.

log P
1 MPa 0.5 MPa

1 2 h

5.5.4 Compressors and Pumps Compressors and pumps have the same function – they increase the pressure of the ﬂuid ﬂowing through them. We usually use the term pump if the ﬂuid is a liquid, and compressor if the ﬂuid is a gas. Work input is required to push the ﬂuid from the low-pressure inlet to the high-pressure outlet. (Unlike a diﬀuser, the ﬂow entering a compressor or pump does not have signiﬁcant kinetic energy at the inlet.) Internally, a compressor or pump has a set of blades or vanes, which are mounted on a central rotating shaft. The blades are designed so that

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2 W c log P 1 1 h
Figure 5.7: An adiabatic compressor. as the shaft rotates the blades impart momentum to the ﬂuid, pushing it toward the outlet. Again, compressors and pumps are usually idealized as adiabatic devices. Making appropriate assumptions about the system, the energy balance for a compressor is ˙ mh1 + Wc = mh2 . ˙ ˙ (5.37) See if you can list the assumptions implied by this equation.2 Since m is a constant, it is convenient to divide through by it to obtain ˙ h1 + Wc = h2 . (5.38)

2

W c

˙ ˙ Here Wc = Wc /m is the input compressor work per unit mass of ﬂuid ﬂowing through the compressor (J/kg). Of course, for a pump the analogous energy balance is h1 + Wp = h2 , ˙ where Wp = Wp /m. ˙ 5.5.5 Turbines A turbine is a device which produces continuous power by expanding a ﬂuid ﬂowing through it. The most familiar example is a windmill, or wind turbine. A wind turbine has blades mounted on a rotating shaft. As air ﬂows over the blades, the shaft turns. If the shaft is connected to a load, such as an electrical generator, power is delivered to the load.
2 At inlet and outlet: 1D, steady ﬂow, equilibrium properties, negligible K.E. and P.E.; steady state; adiabatic.

(5.39)

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2 W p 1 log P 2

1 h

W p

Figure 5.8: An adiabatic pump. Although a wind turbine extracts the power from the kinetic energy of the air stream, an industrial turbine incorporates internal nozzles to accelerate a highpressure input stream to high velocity before it impacts the turbine blades. Many turbines incorporate several rows of blades, one behind the other, to convert as much of the kinetic energy at the outlet of the nozzle to useful shaft power as possible. The ﬂow emerges from the turbine at lower pressure than it went in, with negligible kinetic energy. Turbines too are usually idealized as adiabatic. Therefore, to analyze a turbine we will assume: steady, 1D ﬂow at inlet and outlet; negligible K.E. and P.E. and inlet and outlet; adiabatic; steady state. Up until now, when writing down energy balance equations, we have always regarded heat and work as positive if the energy transfer is from the environment to the system. Energy transfer in the other direction can be treated as negative energy transfer to the system. Since a turbine delivers power to an external load, if we do the energy balance this way we would have to regard the turbine power as negative. However, it is usually more convenient to regard heat and work transfers as positive in doing an analysis. It is simple to modify our procedure to do this. ˙ ˙ We only need to remember whether we are deﬁning W and Q as an energy inﬂow or outﬂow, and put it in the appropriate place in the energy balance equation (i.e., on the same side of the equation as the other energy inﬂows or outﬂows). To avoid confusion, it is best to always draw a sketch of the process, with the direction of energy ﬂows clearly labeled. With the assumptions listed above, and the turbine power regarded as an

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2 W t log P 2 h
Figure 5.9: An adiabatic turbine. energy outﬂow (as shown in the sketch in Fig. 5.9), the energy balance is ˙ mh1 = mh2 + Wt . ˙ ˙ Dividing by m, ˙ h1 = h2 + Wt , (5.41) ˙ ˙ where Wt = Wt /m. An adiabatic turbine changes the state of the ﬂuid as shown in Fig. 5.9. 5.5.6 Heat Exchangers The last devices we will introduce are heat exchangers. As the name implies, these are devices which transfer heat from one ﬂuid stream to another. An example of a heat exchanger is a car radiator, which transfers heat from the hot engine coolant to air which is forced to ﬂow over the radiator coils by the fan. The simplest heat exchanger is a tube which one ﬂuid (“A”) ﬂows through, with the other one (“B”) ﬂowing over the tube. If A and B are at diﬀerent temperatures and the tube wall is a good conductor of heat (usually a metal) then heat will ﬂow between the two ﬂuids. If ﬂuid B is the atmosphere (as in a car radiator) or a body of water such as a river, this open design is ﬁne. Otherwise, ﬂuid B should be enclosed in a tube also. One common design is to use concentric tubes, with A ﬂowing one way through the center tube and B ﬂowing the other way through the annulus surrounding the center tube. The change of state for each ﬂuid can be analyzed separately by deﬁning a control volume which encompasses only that ﬂuid. For example, the control volume shown in Fig. 5.10 is appropriate to determine the change in state of ﬂuid A. (5.40)

1

1

W

t

CHAPTER 5. THE FIRST LAW FOR OPEN SYSTEMS
Fluid A

125

B2

log P A1 . Q A2

A1

A2 Fluid A

. m

A
Fluid B B1

. . Q/ m

A

h
Fluid B

. mB log P

B2

B1

. . Q/ m

h
B

Figure 5.10: A simple constant-pressure heat exchanger. Note that A and B may be diﬀerent ﬂuids, with diﬀerent mass ﬂow rates. Heat exchangers are usually constructed of such that the pressure drop due from the inlet to the outlet for either ﬂuid due to viscous drag is small. For this reason, the ﬂow through a heat exchanger is usually modeled (at least to ﬁrst approximation) as occurring at constant pressure. An appropriate set of assumptions to analyze ﬂow through a heat exhanger is: 1. Steady, 1D ﬂow at inlet and outlet; equilibrium properties at these points 2. Steady state 3. Constant pressure For stream A in Fig. 5.10, with the direction of heat transfer as shown, the energy balance is ˙ mA hA1 + Q = mA hA2 . ˙ ˙ (5.42) Dividing by mA , ˙ hA1 + QA = hA2 , ˙ ˙ where QA = Q/mA . (5.43)

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Of course, the heat added to stream A came from stream B, so the energy balance on B is ˙ mB hB1 = Q + mB hB2 , ˙ ˙ (5.44) or hB1 = QB + hB2 , (5.45) ˙ ˙ where QB = Q/mB . Note that in general mA = mB , so QA = QB , even though ˙ ˙ ˙ Q is the same for both. The inlet state for either ﬂuid may be anywhere on the P − h plot, and the outlet state may be anywhere at the same pressure. Some particular types of heat exchangers are given names, depending on their primary function. An evaporator or boiler is as heat exchanger which takes in liquid and produces vapor (as shown for stream A in Fig. 5.10). The minimum heat input required to do this is for the case where the input is saturated liquid, and the output saturated vapor. In this case, Qin = hg − hf . (5.46)

The symbol hfg is used to denote hg − hf , and is called the enthalpy of vaporization or heat of vaporization. If the output of the evaporator is superheated vapor, the last segment of the heat exchanger where liquid is no longer present is often called the superheater. A condenser is just the opposite: it takes in vapor, removes heat causing the vapor to condense, and outputs liquid. A condenser must transfer at least hfg of heat to the other ﬂuid stream in order to condense the vapor.

CHAPTER 6
ENTROPY AND THE SECOND LAW 6.1 Introduction
We have now developed the First Law of Thermodynamics for both closed and open systems, and shown how it may be used to solve practical problems. But the First Law is not the end of the story. There are many imaginable processes which satisfy the First Law, but are nevertheless don’t seem to happen. We need another principle, or “law,” to explain why not. In this chapter, we introduce this principle – the second law of thermodynamics and the new property associated with it, the entropy.

6.2 Possible and Impossible Processes
Of all processes which satisfy the First Law, some actually happen and some never do. If a few ice cubes are added to a thermos bottle containing boiling water, the ice spontaneously melts and the initially-hot water cools. When equilibrium is reached, only warm water is left. The reverse, however, never happens: warm water has never been observed to spontaneously transform into hot water + ice cubes, even though it is possible to do so in a way which conserves energy. If a rubber ball is held above a table and then dropped, it bounces a few times and comes to rest. The initial gravitational potential energy of the ball is converted to internal energy of the ball (and possibly of the table): the ball ends up slightly warmer than it began. The reverse process — a ball cooling oﬀ slightly and jumping oﬀ a table – has never been observed, although it could conserve energy. In fact, for every process that really happens, the time-reversed version — in which the initial and ﬁnal states of the system + environment are switched — never seems to. Eggs break when dropped on the ﬂoor; broken eggs have never been seen to “unbreak” and rise oﬀ the ﬂoor. Heat ﬂows spontaneously from a high temperature body to a low temperature one it contacts; it has never been observed to ﬂow spontaneously from low temperature to high. A helium balloon slowly deﬂates due to diﬀusion of helium atoms through the balloon 127

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128

skin; a deﬂated balloon has never been observed to spontaneously inﬂate due to helium atoms diﬀusing from the surroundings to the interior. Despite the observed one-way nature of real processes, the First Law makes no distinction between possible process and their impossible time-reversed versions. Consider, for example, a closed system which undergoes a process during which heat Q and work W are transferred to it from the environment. The First Law requires Efinal − Einitial = Q + W. (6.1) For the time-reversed version, the energy transfers occur in the other direction, so (rev) (rev) Efinal − Einitial = (−Q) + (−W ). (6.2) But since Einitial = Efinal and Efinal = Einitial, this is equivalent to −(Efinal − Einitial) = (−Q) + (−W ), (6.3)
(rev) (rev)

which is of course equivalent to Eq. (6.1). We see that the First Law is satisﬁed equally by the forward and reverse processes. This must be true, since the First Law takes the form of an equality: if A = B, then −A = −B. What we would need to distinguish forward from backward would be an inequality: if A and B satisfy A > B, then (−A) and (−B) do not. Evidently, the reason some energy-conserving processes occur spontaneously and others don’t has nothing to do with the First Law.

6.3 The Microscopic View of Spontaneous Processes
When we consider what happens on an atomic level, it is clear why some processes happen spontaneously and others don’t. We’ll consider two speciﬁc examples — dropping a ball, and letting a gas expand into vacuum. The general principles apply to any spontaneous process. 6.3.1 Dropping A Ball Consider what happens on an atomic level when a rubber ball is dropped on a table (Fig. 6.1). To simplify matters, let’s neglect the thermal energy the ball has before being dropped, ignore any energy transfer to the table, and assume the ball is dropped in vacuum, so there is no air resistance. Let’s also imagine we have some way of taking an atomic-level “snapshot” of the positions and velocities of all atoms in the ball at any desired time – we can’t do this in a real experiment, but it’s quite easy in a thought experiment.

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129

A

B

E = Mgh h

2 E = (1/2)MV

Gravitational Potential Energy (Organized)

Directed Kinetic Energy (Organized)

C
Thermal Energy (Disorganized)
Figure 6.1: A rubber ball dropped on a table.

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130

The ball has mass M and is dropped from a distance h above the table. Three diﬀerent macroscopic states of the ball are shown in Fig. 6.1. In state A, the ball hasn’t yet been dropped, so is stationary at a height h above the table. In state B, the ball is just about to strike the table the ﬁrst time. In state C, the ball has stopped bouncing and reached the equilibrium state, resting on the table. A set of atomic-level snapshots are taken in each state A, B, and C. In state A, all atoms are initially stationary (no thermal energy), so repeated snapshots always ﬁnd the same atomic-level conﬁguration. In state B, all atoms moving √ downward in unison at speed V = 2gh. Still, every snapshot is the same. But in state C, every snapshot is diﬀerent. The atoms are vibrating chaotically, and the velocity of each atom changes in an apparently random way from one snapshot to the next as the atom is pulled or pushed by the stretched or compressed bonds to neighboring atoms. Of course, energy is still conserved. If we computed for any snapshot the sum of the kinetic energies of all atoms plus the potential energies of all stretched or compressed bonds, the result would always be M gh. Each snapshot fully determines the instantaneous microscopic state of the ball, which we will call the microstate. In contrast, the macroscopic state of the ball is speciﬁed by only a few numbers, such as the height, center-of-mass velocity, temperature, etc. A macroscopic description of a ball (or any other system) is certainly simpler than a detailed, atomic-level speciﬁcation of the microstate. But it is also much less precise, since a system in a given macroscopic state (sometimes called a macrostate) might be found in any one of a very large number of microstates. If we tried to compile a list of all observed microstates for the ball in macrostate C, it would be a very long list. In fact, if position and velocity are measured with arbitrary precision (and therefore speciﬁed by real numbers), we would never ﬁnd exactly the same value for any atomic position or velocity; there would be an uncountably inﬁnite number of microstates. But usually there is some uncertainty in measurements. For example, if the measurements are acquired by a computer or stored in one, they need to be represented by a ﬁnite number of bits, leading to round-oﬀ error. In this case, positions which diﬀer by less than the round-oﬀ error ∆x would be recorded as the same position, and velocities which diﬀer by less than the velocity round-oﬀ error ∆v would be recorded as the same velocity. In this case, the number of microstates observed in state C would be ﬁnite, but extremely large. If we were extraordinarily patient, we could take enough snapshots to build

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up statistics on the number of times each microstate is found. If after taking Ns snapshots, the ball was found in the j th microstate in Nj,s snapshots, then as Ns → ∞ the probability pj of ﬁnding the ball in microstate j is pj = lim Nj,s . Ns →∞ Ns (6.4)

There is no particular reason to think that any one microstate of the ball in macrostate C should be found more often than any other one. Remember, we said the ball is in equilibrium in state C, which means it has been sitting on the table a very long time (at least compared to atomic vibration time scales). After enough time has elapsed, the atoms of the ball have “forgotten” about any coordinated motion (bouncing) they were executing previously, and are all moving more or less randomly. Of course, we’re only considering microstates which are “accessible,” given the initial conditions and the conserved quantities (total energy, number of atoms, etc.). There are microstates of the ball which have total energy diﬀerent than M gh, or with a few atoms missing or bonded diﬀerently, but with the given initial conditions the ball will never be found in one of these inacessible states. With the assumptions we’ve made, the ball does not interact in any way with the environment, which is why its total energy must be constant. This means we are treating the ball as an isolated system. The assumption that all microstates of the ball are equally probable at equilibrium can be generalized to any isolated system. In fact, this is the basic postulate of statistical mechanics: Postulate: At equilibrium, all accessible microstates of an isolated system are equally probable. We can’t prove this postulate, but we can ask what consequences follow from it, and whether they agree with experiment. Let’s now return to the question of why some processes occur, and some don’t. On a microscopic level, all accessible microstates are equally likely at equilibrium. A particular microstate of the ball with atoms moving in various directions such that there is no center of mass velocity is just as likely as another one which has all atoms moving upward with the same speed, or one with the ball hovering completely stationary a distance h above the table. At ﬁrst glance, this might lead you to conclude that a ball should be just as likely to jump oﬀ a table as to remain sitting there.

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But of course, once the atoms are moving randomly in state C, it is highly unlikely that they would just happen to all move in such a way to push oﬀ from the table and rise up in unison, although there is no physical law preventing it. Another way of saying this is that the number of microstates in which the ball is sitting on the table with all atoms moving more or less randomly (and thus with negligible net center-of-mass velocity) is huge compared to those few special microstates with the atoms all moving in one direction, or all displaced above the table. All accessible microstates — even ones corresponding to bizzare macroscopic behavior — are equally probable, but there are far, far more microstates which correspond to the ball simply sitting on the table. Therefore, we would expect to hardly ever observe one of the microstates corresponding to the ball doing anything but sitting on the table. If we worked out the numbers, we would ﬁnd that we would have to wait much longer than the age of the universe to see the ball spontaneously move – during which time, of course, other processes would occur which would make the whole experiment moot (disintegration of the ball and table, the end of life on earth, etc.) So the basic idea is as follows. Denote by Ω the number of microstates which a system might be in when it is in some speciﬁed macroscopic state. Suppose an isolated system starts out in some equilibrium macrostate 1 which has a certain number of microstates Ω1 . Now some process occurs within the system (e.g. the ball is dropped) which changes the number of available microstates to Ω2 Ω1 . Once enough time has elapsed, the system is again in equilibrium, so it could be in any one of its Ω2 microstates with equal probability. It is still possible that it could be found in one of the original microstates, which would mean that macroscopically it would appear to be in macrostate 1. But the probability of this happening is Ω1 /Ω2 , which is very small if Ω2 Ω1 . 6.3.2 Irreversible Expansion of an Ideal Gas To see how the numbers work out, we need a system so simple that we can actually compute the number of microstates. (This would be hard to do in practice for a rubber ball.) Consider a gas of N identical point atoms in a container of volume V . For the purposes of this discussion, we’ll only specify the microstate by the spatial location of the atoms, and not consider how many ways there are to assign velocities to the atoms consistent with a speciﬁed energy. We’ll come back to the question of how to partition the energy in Section 6.7. If we specify the microstate by the exact position of each atom ({x1 , . . . , xN }), then there are an uncountably inﬁnite number of microstates, since each coordi-

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nate position is speciﬁed by a real number. To make the number of microstates countable, let’s assume each position measurement has some ﬁnite round-oﬀ error ∆x. This is equivalent to dividing the volume into small cubes of volume (∆x)3 , and only recording which small cube an atom is in, rather than the exact co-ordinates within the cube. This makes the number of microstates ﬁnite and countable, although the answer will depend on ∆x. The number of small cubes nc is nc = V . (∆x)3 (6.5)

We will assume that nc N , so that if we place the N atoms in cubes randomly, the odds of ﬁnding two or more in the same cube are very low. The vast majority of states do not have any cubes multiply-occupied. We’ll allow multiple-occupancy in counting states, but it won’t aﬀect the result much one way or the other as long as nc N. The number of spatial arrangements Ωx is the number of ways of placing N identical atoms in nc cubes. Since the atoms are identical,1 we won’t count arrangements which diﬀer only by a permutation of atom labels as being diﬀerent. That is, for N = 2, the arrangement with atom 1 in cube 47 and atom 2 in cube 129 is no diﬀerent than arrangement with atom 2 in cube 47 and atom 1 in cube 129. We should count this arrangement once, but not twice. Therefore, for N = 2, Ωx (N = 2) = n2 /2. For N = 3, we have to divide the number of c arrangements of 3 labeled atoms (n3 ) by the number of ways to permute labels c among 3 atoms (3! = 6). So Ωx (N = 3) = n3 /6. The general result is c 1 N 1 V n = N! c N ! (∆x)3
N

Ωx(V, N ) =

.

(6.6)

Let’s use this result to analyze the process shown in Fig. 6.2. A container is divided into two compartments, of volume VA and VB , respectively. Initially, compartment A contains an equilibrium gas of N identical point atoms, and compartment B is empty. A shutter is now opened in the partition, allowing atoms to move freely between A and B. Enough time is allowed to elapse so that the gas comes to equilibrium in the new volume VA + VB . Now a sensitive detector is turned on, which can detect the presence of even one atom in compartment B. The detector looks for atoms in B every ∆t
1 Quantum mechanics requires that identical atoms (the same isotope of the same element) are indistinguishable even in principle. So just exchanging two identical atoms, keeping everything else the same, can’t yield a new microstate.

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134

A

B

A

B

initial state

final state

Figure 6.2: Irreversible expansion of a gas. seconds. If the detector ever detects that B is empty, it sends a signal which closes the shutter, which returns the gas to its original state of being conﬁned in compartment A. How much time elapses on average before B is found to be empty and the shutter closes? Once the shutter has opened and the atoms have had ample time to move back and forth between A and B, the number of available microstates has increased by the factor N VA+B Ωfinal = . (6.7) Ωinitial VA Since all microstates are equally probable at equilibrium, the probability pempty that the detector will ﬁnd none of the N atoms in B in any one measurement is simply the ratio of the number of microstates which have all of the atoms in A to the total number of microstates: pempty = Ωinitial = Ωfinal VA VA+B
N

.

(6.8)

If N is small, then pempty is fairly large, and only a few measurements would be needed, on average, before the detector ﬁnds B empty. For example, if VA = VB and N = 2, then pempty = 1/4, so the detector would be expected to ﬁnd B empty after only about 4 measurements. If N = 20, pempty = 9.5 × 10−7. Now it is unlikely to ﬁnd B empty on any given measurement, but statistically once every million or so times it should happen, and the shutter would then close, trapping the gas in A. If the detector took a measurement, say, every second, then it would take about 12 days (a million seconds) before we would expect the shutter to close and eﬀectively “reverse” the eﬀect of opening the shutter. We conclude that if N is small (say, 20 or so) then the process of opening the shutter can be reversed, if we are willing to wait a little while.

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135

Suppose, however, that we have a macroscopic gas, with N ∼ 1023 . Then pempty = If VA = VB , p1023 = 2−10
23

VA VA+B

1023

.

(6.9)

= 10−10

23

log10 2
22

≈ 10−3×10 .

(6.10)

Now that is a small probability — its reciprocal would be written as a “1” followed by about 3 × 1022 zeros. The universe is of order 1010 years old, which is 3 × 1017 seconds. Therefore, the chance of detecting no atoms in B during the age of the universe, checking every second, is (3 × 1017)(10−3×10 ) = 3 × 10−(3×10
22 22

−17)

.

Thus, it is overwhelmingly unlikely to occur in the age of the universe. Even if the detector looked every microsecond, this would only change the probability 22 to 3 × 10−(3×10 −23) — still far too small. Even if we could wait 1010 universe lifetimes it would still be overwhelmingly unlikely. We conclude therefore that if N ∼ 1023 , the process of opening the shutter is truly irreversible: it won’t spontaneously reverse, even over times much longer than the age of the universe. For this reason, the process of a macroscopic gas expanding into vacuum is called irreversible expansion. It is also called unrestrained expansion, since there is no piston restraining the gas expansion that the gas must do work against. This example illustrates a common feature of macroscopic systems. Since the number of atoms is very large (say, > 1020 ), and the formula for Ω typically has N in an exponent, making a minor change like opening a shutter to increase the volume expands the number of available microstates by an unimaginably huge degree. Once the system can be found in any of these with equal probability, the probability of ﬁnding it in one of its original allowed microstates is eﬀectively zero, even if we watch it for the age of the universe. When this is the case, we say that the “minor change” which was made is irreversible. Of course, it is possible to restore a system to its initial state, but only at the cost of altering the environment irreversibly. In the previous example, a piston could be used to compress the gas back into compartment A. However, this would require work input, which would increase the energy of the gas.

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This extra energy would have to be removed to the environment as heat, which would irreversibly alter the microscopic state of the environment. In particular, the energy added to the environment would increase its number of possible microstates by a huge factor.

6.4 The Second Law of Thermodynamics
The Second Law of Thermodynamics simply expresses as a general “law” the characteristics of spontaneous, irreversible processes discussed in the last section. Consider ﬁrst an isolated system (constant energy), since in this case all accessible microstates have the same energy, and we don’t have to be concerned about what is happening in the environment. The Second Law states that No process will occur within an isolated, macroscopic system that decreases the number of accessible microstates. All this says is that isolated, macroscopic systems will proceed in the direction of increasing Ω if allowed to (say, by letting go of a ball, or opening an internal shutter in a gas container, etc.), but will never proceed in the direction of decreasing Ω. Of course, we mean “never” in the sense of the irreversible expansion example of the last section. The second law says only that the number of possible microstates of an isolated system cannot decrease; it does not require the number to increase by any speciﬁc amount. In particular, processes which do not change the number of microstates are allowed. Such processes are called reversible, since they are the only ones which may be reversed without violating the Second Law. For non-isolated systems which interact with the environment, the system + environment together constitute an isolated system, so the Second Law may be state more generally: No process that decreases the number of accessible microstates of the system + environment will occur.

6.5 The Entropy
The number of microstates Ω associated with some macroscopic, equilibrium state of matter is a useful quantity to know. As we saw in the irreversible expansion example, knowing Ω and its dependence on macroscopic parameters such as the volume allows us to determine what sort of processes can occur (expansion of gas through an opening in a partition), and what sort can’t (spontaneous collection of gas atoms in one part of a container).

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137

However, there are two reasons why working with Ω directly is inconvenient. 23 First, it is diﬃcult to work with numbers as large as, say, 1010 – we don’t have much intuitive feel for them. The second problem becomes clear if we consider a system which is composed of two non-interacting parts, A and B. If A and B are independent, and A has ΩA microstates, and B has ΩB microstates, then ΩA+B = ΩA ΩB . (6.11)

Therefore, the property Ω is multiplicative for systems composed of separate parts. Since other properties we’ve worked with like mass, volume, and energy are additive (extensive), it is inconvenient to have to work with a multiplicative property. Both problems can be solved by working with ln Ω, rather than Ω. The logarithm of Ω typically is proportional to N , which is still a big number but is quite manageable compared to 10N . Also, ln Ω is additive: ln ΩA+B = ln ΩA + ln ΩB . We will deﬁne the entropy S as S = kB ln Ω. (6.13) (6.12)

The constant kB is “Boltzmann’s constant,” named after Ludwig Boltzmann, who ﬁrst proposed this equation. It is introduced only for convenience, and can be set to any desired value. Given S, it is always possible to recover Ω = exp(S/kB ). Sometimes it is most convenient to use kB = 1, in which case the entropy is dimensionless. But we’ll soon see that the choice of kB aﬀects the units of other properties, in particular temperature. If we have already decided what units we want to use for temperature (e.g. Kelvin), then this will determine the value for kB . We’ll discuss this point more in Section 6.9. It is important to remember that we have been considering equilibrium states of isolated systems (constant energy U , volume V , number of atoms N ). For isolated systems, the only accessible microstates are the ones with the right values of U and N , and for which all atoms are contained within V . The number of these will depend on U , V , and N : Ω(U, V, N ). Therefore, the entropy has the same functional dependence: S(U, V, N ). Since the system is in equilibrium, all accessible microstates are assumed to be equally likely, and therefore all that really matters is how many of them there are. If it were not in equilibrium, then some microstates would be more

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138

probable than others. For example, in the irreversible expansion example, just after the shutter is opened microstates with most atoms in A are more probable than ones with most atoms in B. It can be shown that in this case the entropy should be deﬁned as S = kB pj ln pj , (6.14) j where the sum is over all accessible microstates. If all microstates are equally probable, (pj = 1/Ω) Eq. (6.14) reduces to Eq. (6.13). For our purposes, we won’t need to evaluate S for non-equilbrium states, so we won’t need Eq. (6.14). The only sort of non-equilibrium states we’ll need to consider explicitly are ones composed of separate parts, each of which is in local thermodynamic equilbrium. Since S is extensive, in this case S can be written as
K

S= k=1 Sk (Uk , Vk , Nk ).

(6.15)

Here the system is divided into K parts, each of which is internally in equilibrium (all of its microstates are equally probable), but which may not be in equilibrium with one another. Each term Sk is evaluated as the entropy of the equilibrium state with energy Uk , volume Vk , and Nk atoms. Once we have S(U, V, N ), we could change variables from number of atoms to the total mass, since M = mN , where m is the mass of one atom. This would produce the function S(U, V, M ), which depends only on macroscopicallymeasurable quantities. Now we also know that S is extensive, like U , V , N , and M . The extensive properties have the characteristic that they all scale with the amount of the substance present. Therefore, if we scale up the size of the system by some factor λ, all extensive variables must be scaled by λ. The entropy function evaluated for the scaled-up system S(λU, λV, λM ) must be λ times larger than the entropy function evaluated for the original system S(U, V, M ): λS(U, V, M ) = S(λU, λV, λM ). Taking λ = 1/M , S =S M U V , ,1 . M M (6.17) (6.16)

Therefore, as for the other extensive properties, we deﬁne the speciﬁc entropy s by S s= = s(u, v), (6.18) M

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139

which from Eq. (6.17) depends only on (u, v). Therefore, for a simple compressible substance in equilibrium, there exists some entropy equation of state s(u, v). Of course, as we discussed in Chapter 3, any two independent properties can be used to specify the thermodynamic state. To construct, for example, s(h, P ) we only need to know the equations of state u(h, P ) and v(h, P ): s(h, P ) = s(u(h, P ), v(h, P )). The speciﬁc entropy can also be used as one of the properties to ﬁx the state. For example, s(h, P ) could be inverted to form h(s, P ). For most substances, ﬁguring out all of the microstates consistent with speciﬁed (U, V, N ) is too diﬃcult to do in practice, although in principle it can always be done. But the function s(u, v) still exists, even if we have diﬃculty calculating it directly from S = kB ln Ω. Fortunately, as we’ll discuss below, there are ways to determine s(u, v) purely from macroscopic measurements except for an arbitrary constant of integration.

6.6 Entropy Production
If some process occurs within an isolated system, the Second Law requires Ωfinal ≥ Ωinitial. Therefore, ∆S = Sfinal − Sinitial = kB ln Ωfinal Ωinitial ≥ 0. (6.19)

Since this extra entropy wasn’t transferred into the system from the environment (the system is isolated), we must regard the extra entropy to have been produced inside the system during this process. Thus, irreversible processes produce entropy – unlike energy and mass, entropy is not conserved. We’ll call the amount of entropy produced by a process the entropy production Ps . Every irreversible process produces entropy (Ps > 0). Besides unrestrained expansion, some other processes which produce entropy include motion against frictional forces, combustion, electrical current ﬂow through a resistor, heat transfer through a ﬁnite temperature diﬀerence, mixing of two diﬀerent ﬂuids, and even clearing data in the memory of a digital computer. For a process occurring in a non-isolated system, the entropy produced by the process equals the increase in the entropy of the system + environment. A very compact, general way of writing the Second Law is

Ps ≥ 0

(6.20)

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140

That is, entropy can be produced, but it can never be destroyed. Since this form of the Second Law applies to all types of systems, it is the form we will use most often. The special class of processes for which Ps = 0 are reversible, since reversing them would not result in destruction of entropy, as it would for a process with Ps > 0. Reversible processes can’t be actually achieved in practice, since friction or other irreversibilities are always present, even if only in very small amounts. However, real processes can approach reversible ones as a limiting case, as we make irreversibilities (electrical resistance, friction, etc.) smaller and smaller.

6.7 The Entropy of a Monatomic Ideal Gas
One of the few systems which is simple enough to allow computing s(u, v) from ﬁrst principles is the monatomic ideal gas. We’ve introduced the ideal gas previously as the low-density limit of a simple compressible substance, and stated that it satisﬁes P v = RT . From a microscopic viewpoint, an ideal gas is a gas of particles of negligible size, which have kinetic energy but have negligible interaction potential energy. A real gas approximates this when the container volume per atom is much larger than the atomic size, and the average distance between atoms is much greater than the distance over which they exert appreciable attractive or repulsive forces on one another. A monatomic ideal gas (e.g. He, Ar, Ne) is particularly simple, since in this case the energy of each particle is simply (1/2)m|v|2 – there is no rotational or vibrational energy, as there would be in a molecular ideal gas. Consider an isolated system consisting of a monatomic ideal gas of N atoms contained in volume V . For simplicity, assume the volume is a cube with side length L. Since it is isolated, the total energy is ﬁxed at some value; call it U . Then every possible microstate {x1 , . . . , xN ; v1 , . . . , vN } of the system must satisfy N 2 mvn = U. (6.21) 2 n=1 It will turn out to be more convenient to work with the momentum of an atom pn = mvn , rather than velocity. In terms of momentum, this equation becomes p2 n = U. 2m n=1
N

(6.22)

Also, all N atoms must be in the container, so the microstate must satisfy 0 ≤ xn ≤ L, n = 1, . . . , N

CHAPTER 6. ENTROPY AND THE SECOND LAW 0 ≤ yn ≤ L, 0 ≤ zn ≤ L, n = 1, . . . , N n = 1, . . . , N.

141

(6.23)

How many microstates are there which satisfy these constraints? Clearly, if we can really measure position and momentum with arbitrary accuracy, there is an uncountably-inﬁnite number. So let’s introduce some small round-oﬀ error to make the states countable. Say the position round-oﬀ error is ∆x, and the momentum round-oﬀ error is ∆p. We already solved part of this problem when we calculated how many ways there are to arrange N atoms in volume V [Eq. (6.6)]. Since for an ideal gas the energy depends only on the atomic momenta, not on their positions, the calculation of the number of ways to arrange the atoms in space (Ωx ) and the number of ways to distribute the energy as kinetic energy among the atoms (Ωp ) can be done independently. The total number of microstates is simply the product: Ω = Ωx Ωp . (6.24) The details of the calculation of Ωp are relegated to Appendix A. The result is Ωp = (2πmU )3N/2 . (∆p)3N 3N ! 2 (6.25)

Therefore, using Eq. (6.6), the total number of microstates is Ω = Ωx Ωp = 1 V N (2πmU )3N/2 . N ! (∆x∆p)3N 3N ! 2 (6.26)

Classically, the choice of ∆x∆p is arbitrary. Nevertheless, when S = kB ln Ω is evaluated, the term involving ∆x∆p just becomes an additive constant equal to −3N kB ln(∆x∆p), so this classical treatment determines the entropy of an ideal gas to within an arbitrary additive constant. Since we only need to know diﬀerences in entropy to determine if a process satisﬁes the Second Law, any additive constant will cancel when the diﬀerence is taken. We can do a little better if we supplement this classical analysis with the uncertainty principle of quantum mechanics. The uncertainty principle states that a particle (which in quantum mechanics is a wave packet) cannot simultaneously have precise values of position and momentum — x or p or both must have some “fuzziness” or uncertainty. The momentum uncertainty in any direction δp and the position uncertainty in that direction δx are related by (δx)(δp) ≥ h, (6.27)

CHAPTER 6. ENTROPY AND THE SECOND LAW where h = 6.626 × 10−34 J-s

142

(6.28)

is Planck’s constant. It seems natural to choose ∆x∆p = h in Eq. (6.26). In fact, it can be shown that this choice reduces Eq. (6.26) to what would result if we had done the analysis quantum-mechanically from the start. Making this substitution, Ω= 1 V N (2πmU )3N/2 . N! h3N 3N ! 2 (6.29)

The entropy is kB ln Ω. If N is larger than about 100, ln N ! is well-approximated by ln N ! ≈ N ln N − N , which is known as Stirling’s approximation. It is left as an exercise to show that in this case the entropy is S = N kB 3 ln 2 U N + ln V N + 3 ln 2 4πm 3h2 + 5 . 2 (6.30)

This expression for the absolute entropy of a monatomic ideal gas.

6.8 The Second Law and Equilibrium
The Second Law has signiﬁcant implications regarding the nature of equilibrium. In this section, we’ll take a look at equilibrium from the point of view of the Second Law. This will extend our analysis of Chapter 3, and result in relationships between the properties deﬁned there (temperature, pressure, and chemical potential) and the entropy. 6.8.1 The Entropy Maximum Principle Consider the process shown below. Two solid blocks, A and B, are brought into contact brieﬂy and then separated. Heat may ﬂow from one to the other during the contact time, but there is no energy or mass transfer to the environment. Therefore, the two blocks together constitute an isolated system.

isolated system A B

The Second Law (Ps ≥ 0) prohibits any process occurring within an isolated system which would decrease its entropy. We will also exclude the possibility of

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143

a reversible process occurring (Ps = 0), since every real process which occurs in ﬁnite time produces some entropy, although Ps can be made arbitrarily small. Therefore, if energy transfers as heat between the blocks, it must do so in a way which increases the entropy S = SA (UA ) + SB (UB ) of the whole system, measured after the blocks have been separated and each has again come into internal equilibrium (uniform temperature). If this process is repeated many times, the entropy of the whole system must increase each time. But the entropy can’t go on increasing without bound, since there is a ﬁnite number of ways to distribute the total energy UA + UB among all the atoms in block A and block B. Since Ω is bounded, so is S. Therefore, the entropy must approach some limiting value as the number of contact periods goes to inﬁnity.
S* S + SB A

0

1

2

3

4 n

5

6

7

In the limit of inﬁnite contact periods (inﬁnite contact time) the state is no longer changing, and by deﬁnition this is the condition of thermal equilibrium between blocks A and B. The same analysis would apply to any other process occurring in an isolated system, not only heat ﬂow (chemical reaction, mixing of two diﬀerent species, ﬂuid motion, etc.). The process would occur for a while, continually increasing the entropy of the system, until it has maximized the entropy as much as it can, then no further change will be observed. The system will be in equilibrium. So we have a general criterion for the attainment of equilibrium in an isolated system: The equilibrium state of an isolated macroscopic system has the maximum entropy, compared to all other possible states of the system. This is known as the Entropy Maximum Principle. By “state” we mean of course the macroscopic state, and by “possible” states, we mean ones with the same total energy, volume, mass, and elemental composition as the system began with, since these quantities are ﬁxed for an isolated system. A few examples will make clear how the Entropy Maximum Principle is used.

CHAPTER 6. ENTROPY AND THE SECOND LAW
UB U tot * UB S + S A B 0

144

S

SB

S A

0

* UA UA

U tot

Figure 6.3: The equilibrium state of an isolated system maximizes the total entropy SA + SB with respect to the quantity being varied (here UA ). 6.8.2 Thermal Equilibrium Consider again the example of the last section. Suppose block A initially has internal energy UA,0 , and block B has UB,0 . After they have come to thermal equilibrium, energy may have transferred as heat from one to the other, changing ∗ ∗ UA and UB . What are the equilibrium values UA and UB ? To answer this question, we can use the Entropy Maximum Priciple. First note that the total energy Utot = UA + UB is ﬁxed, since the system is isolated. So at any time, U2 = Utot − UA . The entropy of each block depends on its energy, volume, and number of atoms. So the total system entropy is S = SA (UA , VA , NA) + SB (UB , VA , NB ), where UB = Utot − UA . (6.32) (6.31)

Note that in this problem the only independent variable we can change is UA ; everything else is ﬁxed. The Entropy Maximum Principle says that at equilibrium, UA will have the value which maximizes S. Therefore, by plotting S vs. UA , it is possible to ∗ determine the equilibrium value UA (Fig. 6.3). There is another way to ﬁnd the maximum of S. Instead of plotting it, we

CHAPTER 6. ENTROPY AND THE SECOND LAW
∗ could calculate dS/dUA , and set this derivative to zero to ﬁnd UA :

145

dS = dUA

∂SA ∂UA

+
VA ,NA

dUB dUA

∂SB ∂UB

=0
VB ,NN

(6.33)

at equilibrium. From Eq. (6.32), dUB = −1. dUA Putting this into Eq. (6.33), we ﬁnd dS = dUA ∂SA ∂UA −
VA ,NA

(6.34)

∂SB ∂UB

.
VB ,NB

(6.35)

At the maximum (thermal equilibrium), dS/dUA = 0, so we conclude that at thermal equilibrium ∂SA ∂UA =
VA ,NA

∂SB ∂UB

(6.36)
VB ,NB

or changing the notation slightly ∂S ∂U =
V,N A

∂S ∂U

.
V,N B

(6.37)

This equation states that at thermal equilibrium, A and B will have the same value of the property (∂S/∂U )V,N . The functions SA and SB may diﬀer, since A and B may be diﬀerent substances. But when equilibrium is attained, the partial derivatives of these two diﬀerent functions with respect to internal energy will be equal. Clearly, this is an important and useful partial derivative property. Note that (∂S/∂U )V,N = (∂s/∂u)v , so we can state the thermal equilibrium condition more simply as ∂s ∂u = v A

∂s ∂u

. v B

(6.38)

In chapter 3, we already introduced the property temperature to tell us if two systems are in thermal equilibrium (they are if TA = TB ; otherwise they are not). Evidently, there must be a close relationship between temperature and (∂s/∂u)v . We could in fact use any monotonic function of this derivative f((∂s/∂u)v ) to deﬁne a thermodynamic temperature scale. What function should we pick? Let’s ﬁrst look at how heat ﬂows, from (0) ∗ Fig. 6.3. Suppose we start with UA < UA (to the left of the maximum in

CHAPTER 6. ENTROPY AND THE SECOND LAW

146

Fig. 6.3. Then dS/UA > 0, which from Eq. (6.35) means that [(∂s/∂u)v ]A > ∗ [(∂s/∂u)v ]B . As the system approaches equilibrium, UA increases to UA . Thus, heat must ﬂow into A from B. (0) ∗ Now consider the other possibility: suppose we start with UA > UA (to the right of the maximum in Fig. 6.3). Then dS/UA < 0, which from Eq. (6.35) means that [(∂s/∂u)v ]A < [(∂s/∂u)v ]B . As the system approaches equilibrium, ∗ UA decreases to UA . Thus, heat must ﬂow from A into B. So it seems that heat ﬂows from the system with lower (∂s/∂u)v to the system with higher (∂s/∂u)v . By convention, we choose to deﬁne temperature so that heat ﬂows from high to low temperature. So (∂s/∂u)v is not a suitable deﬁnition of thermodynamic temperature, if we want heat to ﬂow from high to low temperature. But we could ﬁx this up by taking the reciprocal: if [(∂s/∂u)v ]A > [(∂s/∂u)v ]B , then [(∂s/∂u)v ]−1 < [(∂s/∂u)v ]−1 . Therefore, we A B deﬁne the thermodynamic temperature as

1 = T

∂s ∂u

(6.39) v Of course, Eq. (6.39) is not the only possibility. We could have put 1/T 2 on the left instead of 1/T , for example. What we need to show is that Eq. (6.39) is equivalent to temperature as measured on the ideal-gas temperature scale. It is, but we’ll save the proof for the exercises. 6.8.3 Mechanical Equilibrium Consider again the problem of mechanical equilibrium, ﬁrst discussed in Chapter 3. Suppose a cylinder is divided into two parts A and B by a piston, which may freely move back and forth (Fig. 3.2). The cylinder is isolated from the environment, and has ﬁxed total volume Vtot = VA + VB and total energy Utot = UA + UB . (6.41) (6.40)

The piston will be assumed to not only move, but also conduct heat. Therefore, at equilibrium, A and B will be in both thermal and mechanical equilibrium. As before, according to the Entropy Maximum Principle, the equilibrium state will maximize S. But now we have two independent parameters to vary:

CHAPTER 6. ENTROPY AND THE SECOND LAW UA and VA . We seek to maximize the function S = SA (UA , VA, NA ) + SB (Utot − UA , Vtot − VA , NB ), 0 ≤ UA ≤ Utot ; 0 ≤ VA ≤ Vtot . At the maximum (the equilibrium state), ∂S ∂UA = 0 and ∂S ∂VA = 0.

147

(6.42) (6.43)

(6.44)

The derivative with respect to UA is the same condition as we discussed above; it leads to the conclusion that TA = TB at equilibrium. But the derivative with respect to VA gives us some new information: ∂S ∂VA = = ∂SA ∂VA ∂SA ∂VA +
UA ,NA

dVB dVA

∂SB ∂VB ∂SB ∂VB

(6.45)
UB ,NB

+ (−1)
UA ,NA

.
UB ,NB

(6.46)

Here we used Eq. (6.40) to set dVB /dVA = −1. Setting (∂S/∂VA ) = 0 at equilibrium, we ﬁnd ∂S ∂V or =
U,N A

∂S ∂V ∂s ∂v

.
U,N B

(6.47)

∂s ∂v

= u A

. u B

(6.48)

Thus, when the piston stops moving and A and B are in both thermal and mechanical equilibrium, A and B will have the same value for (∂s/∂v)u . We deﬁne the thermodynamic pressure P as follows:

P = T

∂s ∂v

(6.49) u Then at equilibrium, P T =
A

P T

.
B

(6.50)

Since the wall is heat-conducting, TA = TB . Therefore, if the piston can move, then we must have PA = PB at equilibrium. The proof that the thermodynamic pressure equals the mechanical pressure is left as an exercise.

CHAPTER 6. ENTROPY AND THE SECOND LAW 6.8.4 Diﬀusive Equilibrium

148

Finally, let’s consider diﬀusive equilibrium. Imagine an isolated system divided into two parts A and B by a partition which is rigid but allows molecules to pass through it. It also conducts heat. For now, we will say that substances A and B are the same chemically, but may be in diﬀerent states (e.g. diﬀerent pressures, temperatures, and/or phases). For example, we might place in A a cold, low-pressure mixture of liquid water and ice, and in B hot, high pressure water vapor. The analysis to ﬁnd the equilibrium state is very similar to what we did in the previous section, except that the two variable quantities are UA and NA . The total energy and number of molecules are ﬁxed, so UB = Utot − UA and NB = Ntot − NA . (6.52) Following the same procedures as above, we ﬁnd that in the equilibrium state, ∂S ∂N =
U,V A

(6.51)

∂S ∂N

(6.53)
U,V B

We deﬁne the chemical potential µ by

−

µ = T

∂S ∂N

(6.54)
U,V

and therefore at equilibrium µ T =
A

µ T

.
B

(6.55)

Since the partition is heat-conducting, TA = TB , so the condition for equilibrium with respect to mass transfer is µA = µB . The negative sign is included in the deﬁnition so that mass will ﬂow from the system with higher µ to the one with lower µ. The proof is left as an exercise. Note that in this deﬁnition, the total values of U and V are being held constant (not u and v, or u and v). So as more matter is added (N increases) ˆ ˆ at constant U and V , u = U/N and v = V /N must decrease. ˆ ˆ It is not necessarily the case that PA = PB for diﬀusive equilibrium, even though TA = TB and µA = µB . For example, if region A is a very small water

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149

droplet and region B is the space around it containing water vapor, the liquid and vapor may be in equilibrium even though there is a pressure diﬀerence due to surface tension.

Virtual Partition

Liquid Water

Solid Ice
But in most cases other than small droplets or bubbles, the pressure is the same in all phases present in equilibrium. In this case, the phase-equilibrium conditions between, say, co-existing liquid and solid phases would be T P µ = Ts = Ps = µs . (6.56) (6.57) (6.58)

Clearly, these conditions would hold for any other type of phase equilibrium also (liquid/vapor, solid/vapor).

6.9 The Value of Boltzmann’s Constant
We have not yet said what value should be assigned to kB in S = kB ln Ω. We have seen now that 1/T = (∂S/∂U )V,N , so the units of entropy must be [energy]/[temperature]. If we wish to use SI units, then, entropy must have units of J/K, and speciﬁc entropy J/kg-K. Since ln Ω is dimensionless, we must then give kB the units of J/K. Using the expression we have derived for the entropy of a monatomic, ideal gas along with the deﬁnitions of thermodynamic temperature and pressure, it is possible to prove that for an ideal gas P V = N kB T. Converting from units of molecules to units of moles, P V = (N/NA )(NA kB )T. (6.60) (6.59)

where NA = 6.023 × 1026 kmol−1 is Avagadro’s Number. This may be written in terms of molar volume as Pv ˆ = NA kB (6.61) T Experimentally, we know that this quantity equals 8314.4 J/kmol-K in the lowdensity, ideal gas limit (Chapter 4).

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150

The experimental result uses temperature measured on the Kelvin scale (an ideal gas scale). If we wish the thermodynamic temperature to coincide with the Kelvin scale, then we must choose kB = (8314.3 J/kmol-K)(1 kmol/6.023 × 1026 ). Therefore, kB = 1.38 × 10−23 J/K (6.63) (6.62)

6.10 The Gibbs Equation
Often, we want to know how the entropy of a substance in equilibrium changes if its internal energy, volume, and number of molecules change by a small amount. Diﬀerentiating the function S(U, V, N ) yields dS = ∂S ∂U dU +
V,N

∂S ∂V

dV +
U,N

∂S ∂N

dN.
U,V

(6.64)

Using the deﬁnitions for thermodynamic temperature, pressure, and chemical potential, this equation becomes

dS =

1 T

dU +

P T

dV −

µ dN. T

(6.65)

This equation is known as the Gibbs equation. It expresses a relationship among equilibrium properties of a substance, and doesn’t depend on any particular type of process for its validity (e.g. reversible, etc.). We can write the Gibbs equation on a per-mole or per-kg basis also. In this case, dN = 0, so dˆ = s 1 P dˆ + dˆ, u v T T (6.66)

and ds = 1 P du + dv. T T (6.67)

CHAPTER 6. ENTROPY AND THE SECOND LAW We can also solve Eq. (6.65) for dU :

151

dU = T dS − P dV + µdN.

(6.68)

This equation is completely equivalent to Eq. (6.65). Now the property U exists, so it must be that Eq. (6.68) is what we would get if we diﬀerentiated the function U (S, V, N ). Therefore, we can identify T = ∂U ∂S ;
V,N

P =−

∂U ∂V

;
S,N

µ=

∂U ∂N

.
S,V

(6.69)

These equations are alternative ways to deﬁne the thermodynamic temperature, pressure, and chemical potential. Applying Eq. (6.68) to a unit mass du = T ds − P dv, and so T = ∂u ∂s P =− ∂u ∂v (6.70)

; v . s (6.71)

Adding a “hat” would give the corresponding equations on a molar basis.

6.11 Macroscopic Determination of the Entropy
For most substances, it is virtually impossible to ﬁgure out Ω(U, V, N ). Fortunately, the Gibbs equation gives us another way to determine S. If the equations of state T (u, v) and P (u, v) are known, then these may be inserted into Gibbs equation. For a unit mass, ds = 1 P (u, v) du + dv. T (u, v) T (u, v) (6.72)

This can be integrated along some path in the (u, v) plane, starting from some point (u0 , v0 ): s(u, v) − s(u0 , v0 ) = path 1 P (u, v) du + dv . T (u, v) T (u, v)

(6.73)

Since we know the function s(u, v) exists, the value of the integral won’t depend on the particular path chosen. We can choose any convenient path. This procedure, in slightly modiﬁed form, is how the entropy is usually determined for real substances. We will discuss this procedure more in Chapter 8.

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We cannot determine the entropy of the initial state s0 = s(u0 , v0 ). We have two choices. We can simply choose a convenient “reference” state (u0 , v0 ) and assign the entropy in this state an arbitrary value (for example, zero). Then all entropies s(u, v) are really relative to s0 , the entropy in the reference state. Or we can choose the reference state so that the ideal gas approximation holds for the reference state. In this case, we can use expressions from statistical mechanics for the absolute entropy of an ideal gas to get s0 . If the substance is monatomic, then Eq. (6.30) would be the right expression to evaluate s0 . If it is a molecular gas, the expression is slightly more complicated, but s0 still can be computed. The choice of how to get s0 diﬀers from one source of thermodynamic data to another. Care must be taken when comparing s (or h or u) from diﬀerent sources. In TPX, for most substances the reference state is taken to be saturated liquid at the lowest temperature for which the data are valid. Both the entropy and the enthalpy are assigned the value zero in the reference state. Of course, the value of s0 is irrelevant in computing entropy diﬀerences, as long as both entropy values are taken from the same source.

6.12 Entropy Flow with Heat
Consider two systems, each composed of some simple compressible substance which is internally in equilibrium. System 1 has temperature T1 , and system 2 has temperature T2 = T1 + ∆T . We bring the into contact, and heat Q ﬂows from 2 to 1. Let’s take the limit ∆T → 0. In this limit, the temperature diﬀerence becomes inﬁnitesimal (dT ) and the process approaches heat transfer at constant T . Of course, as ∆T → 0, the time t required to transfer a ﬁnite amount of heat Q increases proportional to Q/∆T . We’ll transfer only an inﬁnitesimal amount of heat dQ (which only takes ﬁnite time). ¯ The two systems start out internally in equilibrium. Since the heat transferred is inﬁnitesimal, we can assume that the systems stay in internal equilibrium during this process, with the result that their intensive properties (T and P ) don’t change, and their extensive properties (U , V , and S) change only inﬁnitesimally. This also means that there are no internal irreversibilities (viscous forces, electrical power dissipation, etc.) since these only exist when a system is not in internal equilibrium. Therefore, no entropy is being produced in either system, and this process is reversible.

CHAPTER 6. ENTROPY AND THE SECOND LAW dW1 = -P1 dV1
T, P 1 dU1 dS 1

153

dW2 = -P2 dV2 dQ
T, P 2 dU2 dS 2

Writing the ﬁrst law for the process, ¯ dU1 = dQ − P1 dV1 , and dU2 = − dQ − P2 dV2 . ¯ (6.75) (6.74)

Since 1 and 2 are not necessarily in mechanical equilibrium with one another, we allow P1 and P2 to diﬀer. (For example, they may be sitting in environments at diﬀerent pressures, and are connected by a copper wire allowing heat to ﬂow.) Note that we have used the assumption of reversibility in writing the work terms as −P dV . We can also write the Gibbs equation for each substance: dS1 = and 1 P1 dU1 + dV1 T T (6.76)

1 P2 dU2 + dV2 . (6.77) T T Since dN1 = dN2 = 0, we have left oﬀ the −(µ/T )dN term. Substituting for dU from the ﬁrst-law expressions, we ﬁnd dS2 = dS1 = and dQ ¯ T (6.78)

dQ ¯ . (6.79) T Although no entropy was produced by this process of heat transfer at constant T , the entropy of system 1 increased, and that of system 2 decreased by exactly the same amount. It must be, then, that entropy was transferred from 2 to 1 along with the energy transfer dQ. Therefore, ¯ dS2 = − When heat dQ enters or leaves a system at temperature T , it carries ¯ with it entropy in the amount dQ/T . ¯

CHAPTER 6. ENTROPY AND THE SECOND LAW
Energy Transfer dQ T, P 2 dU2 dS 1 Entropy Transfer dQ/T 2

154

T, P 1 dU1 dS

Note that the work terms canceled out of the above equations. Evidently, entropy transfer is associated only with energy transfer as heat – when energy is transferred as work, it does not carry any entropy with it.

dW

dQ

dQ T

Energy transfer

Entropy transfer

Problems
6.1 A rectangular box contains N molecules. Suppose that each molecule can be on either side of the box with equal probability. 1. What is the probability that all N molecules are on one side? Compute this number for N = 1, 2, 20, and 1020. 2. Astronomers believe the universe is between 8 and 13 billion years old. Suppose you examine this system once every second, and the molecules are distributed completely randomly each time. Estimate the largest value N can have so that on average you will ﬁnd all molecules on one side at least once in the age of the universe. 6.2 Using Eq. (6.30) and the deﬁnitions of thermodynamic temperature and pressure, show that for a monatomic gas of N atoms U= and P V = N kB T. 6.3 Prove that the thermodynamic pressure P equals the mechanical pressure Pmech = F/A. To do so, consider a quasi-static, reversible, adiabatic compression in which the volume changes by dV . The work done on the system is dWqs = −Pmech dV . Use the ﬁrst law, and the fact that dS = 0 ¯ for this process. (Hint: write ds(u, v) in terms of du and dv.) 3 N kB T 2

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155

6.4 Show that if µA = µB , mass ﬂows from the one with higher µ to the one with lower µ. 6.5 Using ˆu ˆ ˆ S(U, V, N ) = N s(ˆ, v) = N s(U/N, V /N ) show that µ = u + P v − T s. ˆ ˆ ˆ The right-hand side of this equation is the deﬁnition of the molar Gibbs free energy, and thus µ = g. ˆ (N may be interpreted as either the number of molecules or the number of moles, and all “molar” quanties as per molecule or per mole, whichever is more convenient.) 6.6 Consider a crystal composed of N atoms, each of which can occupy one of two quantum states. The ﬁrst state (the ground state) has energy zero, and the other one (the upper state) has energy > 0. 1. Suppose the total internal energy of the crystal is U , where U is an integral multiple of . Show that the number of diﬀerent microscopic quantum states of the crystal which have this energy U is given by Ω=
U

N! ! N−

U

!

(6.80)

Neglect any other forms of energy, such as vibration. 2. In the limit N → ∞, determine the entropy of the system at equilibrium. (Note that for large n, ln(n!) ≈ n ln n − n.) 3. From the entropy and the deﬁnition of thermodynamic temperature, determine the thermal equation of state U (T, N ). 6.7 We showed that when two phases 1 and 2 of a substance are in equilibrium with one another, P1 = P2 , T1 = T2 , and µ1 = µ2 . Using TPX, verify that µ1 = µ2 for saturated liquid water and water vapor at 300 K. 6.8 A mixture of saturated liquid nitrogen and saturated vapor at 1 MPa is contained on one side of an insulated container which is divided in half by a partition. The initial vapor mass fraction is 0.7. The other side is evacuated. Now the partition is broken, and the nitrogen expands unrestrained to ﬁll the container, doubling its volume. Using TPX:

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156

1. Find the new pressure and temperature of the system once equilibrium has been re-established. 2. Determine how much entropy was produced by this process, per unit mass of nitrogen.

CHAPTER 7
ENTROPY ACCOUNTING AND APPLICATIONS 7.1 Introduction
In the last chapter, we found two important properties of the entropy: 1. Entropy can be produced by irreversible processes, but can never be destroyed: Ps ≥ 0 for every possible process. 2. When heat Q enters or leaves a system at temperature T , it carries entropy in the amount Q/T with it. With these two results, it is possible to write down entropy accounting expressions for closed or open systems. In this chapter, we will develop these expressions, and look at several types of problems which can be solved by combining energy and entropy accounting.

7.2 Some General Accounting Principles
We must take some care in developing entropy accounting expressions, since, unlike energy, entropy is not conserved. First let’s consider in general how to do accounting for things which can be produced. Consider the monthly statement you might receive from a bank if you have an interest-bearing bank account. The account balance at the end of a month is related to the previous month’s balance by (new balance) = (old balance) + (deposits) − (withdrawals) + (interest). (7.1) The account balance is the amount of money “stored” in the account; the deposits represent an “inﬂow” of money, the withdrawals an “outﬂow” of money, and we can think of the interest as money “produced” within the account during the month, since it appears in the account even though you didn’t deposit it. Therefore, the accounting principle for a bank account is ∆(stored money) = (money inﬂow) − (money outﬂow) + (money production). (7.2) 157

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158

Of course, this basic accounting principle applies to anything which can be stored, transported, and produced. Suppose a factory manufactures a product. The change in the inventory stored at the factory over some period of time equals the number of products produced in that period, plus the number returned from stores (inﬂow to the factory), minus the number shipped to stores (outﬂow from the factory). So Eq. (7.2) can be written more generally:

∆(storage) = (inﬂow) − (outﬂow) + (production).

(7.3)

This general accounting principle can be applied to the storage, transportation, and production of physical quantities like mass, momentum, energy, and entropy. For mass and energy, the production term is always zero, since these are conserved. Therefore, the energy balances we developed in Chapters 2 and 4 were always of the form ∆(energy stored in system) = (energy inﬂow) − (energy outﬂow). For entropy, the production term is non-zero, so we have ∆(entropy stored in system) = (entropy inﬂow) − (entropy outﬂow) + (entropy production). (7.5) (7.4)

7.3 Entropy Accounting for a Closed System
For a closed system, the only way entropy can enter or leave the system is with heat which enters or leaves. The simplest case occurs when heat enters at a single temperature, as shown below.

dS dP s

dQ T

During elapsed time dt, heat dQ crosses the system boundary where the ¯ temperature is T . We saw in the last chapter that when heat dQ enters a system ¯ at temperature T , it brings with it entropy in the amount dQ/T . Therefore, ¯ the entropy inﬂow is dQ/T , and there is no entropy outﬂow. ¯

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During the elapsed time dt, entropy in the amount dPs may have been also ¯ produced within the system due to internal irreversible processes (friction, viscous ﬂuid ﬂow, electrical current ﬂow, unrestrained expansion, etc.). Therefore, according to Eq. (7.5), the increase in the amount of entropy stored within the system is dQ ¯ + d Ps . ¯ dS = (7.6) T Note that work dW might have also been done on or by the system during dt, ¯ but since energy transfer as work does not carry entropy, it does not enter into the entropy accounting expression. Since the second law requires dPs ≥ 0, ¯ dS ≥ dQ ¯ . T (7.7)

Thus, the system entropy increases by more than the amount which enters with heat; the diﬀerence is what is produced inside. Equation (7.7) also holds if dQ < 0, in which case it says the the entropy decreases by less than the amount ¯ of entropy which leaves with heat. For a reversible process, Eq. (7.7) reduces to dQrev = T dSrev . ¯ (7.8)

From Eq. (7.6), a process which is both adiabatic ( dQ = 0) and reversible ¯ ( dPs = 0) would not change the entropy (dS = 0). Therefore, ¯ the entropy is constant during any reversible, adiabatic process in a closed system. We call processes which occur at constant entropy isentropic. Equation (7.6) can be easily generalized to the case where heat ﬂows across the boundary at various spots, each of which may have a diﬀerent temperature (which might even be time-dependent).

dQ i Ti dQ k Tk Tk dS dP s Tj dQ j Tj Ti

In time dt, the total entropy S of the system changes due to the entropy entering or leaving with heat, and the entropy produced by irreversible processes

CHAPTER 7. ENTROPY ACCOUNTING AND APPLICATIONS inside. If heat enters through I places, and leaves through O places, then

160

dS = in

dQin ¯ Tin

− out

dQout ¯ Tout

+ dPs . ¯

(7.9)

Note that the dQ/T terms are evaluated at the temperature on the system ¯ boundary where the heat ﬂow is occurring. Equation (7.9) can also be written on a per-unit-time basis (a rate basis). Dividing by dt,

dS = dt

in

˙ Qin Tin

− out

˙ Qout Tout

˙ + Ps .

(7.10)

If the temperatures on the boundaries where heat is entering or leaving are constant in time, this equation can be integrated for a ﬁnite time interval (0, t): ∆S = S(t) − S(0) = in Qin Tin − out Qout Tout t + Ps .

(7.11)

Here each Q is just the total heat which enters: Q = 0 dQ. ¯ Equations (7.9) – (7.10) all are simply statements of entropy accounting. They equate the net change in the total entropy of a system to the net entropy inﬂow (inﬂow - outﬂow) plus the amount of entropy produced inside the system. They apply to any closed (control mass) system. In many (but not all) processes we will consider, the system is in steady state. In this case, the entropy contained within the system is not changing, so on a rate basis dS/dt = 0, or on a total time basis ∆S = 0. For a steady-state problem, Eq. (7.10) reduces to ˙ Qin Tin − out ˙ Qout Tout ˙ + Ps = 0. (7.12)

in

A great deal can be learned about real processes by carrying out energy and entropy accounting. Let’s consider a few examples of processes occurring in closed systems.

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161

7.4 Heat Transfer Through Finite ∆T
Consider the situation shown below. Heat is ﬂowing from a high-temperature region at T1 to low-temperature region at T2 through a thin slab of insulation material, which has thickness L and area A. The temperature is everywhere constant in time, and the insulation is in steady state.

T1

T(x)

. Q

1

. Q2

T2

From the theory of heat conduction in solids, the heat ﬂow through the insulation is related to the temperature diﬀerence across it by κA ˙ Q= (T1 − T2 ) , L (7.13)

where κ is a material property called the thermal conductivity (SI units: W/mK). This equation is analogous to I = V /R for current ﬂow through a resistor, ˙ with Q playing the role of current, ∆T the potential diﬀerence, and L/κA the resistance. For this reason, L/κA is called the thermal resistance Rth of the insulation. We will deﬁne a system to consist of the insulation, with the left system boundary at T1 and the right one at T2 . We will assume the heat ﬂow is one dimensional, so heat only crosses the system boundaries perpendicular to x. Since the insulation is in steady state, no energy or entropy are building up in the system: dU/dt = 0 and dS/dt = 0. Therefore, the energy balance reduces to ˙ ˙ Qh = Qc, (7.14) ˙ and so we may drop the subscripts and call the heat transfer rate Q. Applying Eq. (7.10) with dS/dt = 0 and one heat inﬂow and one heat outﬂow, the entropy accounting expression becomes 0= ˙ Solving for Ps , ˙ ˙ Ps = Q 1 1 − T2 T1 . (7.16) ˙ ˙ Q Q ˙ − + Ps . T1 T2 (7.15)

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162

˙ Since T1 > T2 , Ps > 0, in accord with the Second Law. Note that as ˙ ˙ T2 → T1 , the ratio Ps /Q → 0. Therefore, the amount of entropy produced per unit of heat transferred can be made arbitrarily small by making ∆T = T1 − T2 suﬃciently small. We conclude that heat transfer through a ﬁnite ∆T is irreversible (produces entropy), but in the limit as T2 approaches T1 , the process approaches reversibility. Of course, from Eq. (7.13), decreasing ∆T decreases the heat transfer rate, ˙ and as T2 → T1 Q → 0. Therefore, in the reversible limit transferring a ﬁnite amount of heat would take inﬁnite time.

7.5 Heat Engines
The original engineering problem which led to the development of thermodynamics was how to eﬃciently use the thermal energy of a ﬁre (produced by burning wood or coal) to do useful work, for example to turn the wheels of a locomotive. It’s easy to design a process which can take in heat from a high-temperature source and produce work for a little while, but much harder to design one which can go on doing it indeﬁnitely, which is what is really desired. For example, gas in a cylinder ﬁtted with a piston will expand when heated, pushing the piston, doing work against an external load. This is ﬁne until the piston has reached the top of the cylinder, at which time the process must stop. Figuring out how to get the piston back to its initial position to start the “cycle” again without having to put more work in than you get out is the challenge. A heat engine is a device which operates either steadily or in a cycle, which takes in energy as heat and converts at least part of it into useful work. Heat engines are a vital part of modern technology. Most current and proposed electric power plants are heat engines, which take in heat at high temperature given oﬀ by a chemical or nuclear reaction, and convert part of this heat into electrical work delivered to the grid. Even very sophisticated concepts such as nuclear fusion are simply diﬀerent sources of heat — a heat engine of some sort is still needed to convert this to useful work.1 Automobile, aircraft, and rocket engines are other examples of practical heat engines. Since a heat engine operates either continuously or in a cycle, in which it returns to its initial state after some time period T , we can use a steady1 One promising power plant concept which is not a heat engine is a fuel cell, which converts the chemical energy in fuel directly into electrical energy, without ﬁrst converting it into thermal energy in a ﬂame. Thermodynamics still governs fuel cell performance, however, as we’ll discuss later.

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163

Th Qh W
Heat Engine

Figure 7.1: A hypothetical one-temperature heat engine. state analysis to study heat engines. For continuous operation, dE/dt = 0 and dS/dt = 0, while for cyclic operation we will choose the time period T for the analysis, in which case ∆E = 0 and ∆S = 0. In the following, we will assume continuous operation, but the results apply equally for cyclic operation. To analyze heat engines, it is useful to introduce the conceptual device of a thermal reservoir. A thermal reservoir is deﬁned as a substance with a very large heat capacity, so that heat can be taken from it or added to it without signiﬁcantly perturbing its temperature. Furthermore, it has a very large thermal conductivity, so that transferring heat to or from it causes only negligible temperature diﬀerences within it [Eq. (7.13)]. A large block of copper or a large lake would approximate a thermal reservoir. In the limit where the heat capacity and thermal conductivity become inﬁnite, the temperature of a thermal reservoir remains perfectly uniform, and constant in time, no matter how much heat is added or removed. In this limit, heat ﬂow within the reservoir is perfectly reversible. (The electrical analog of a thermal reservoir would be an inﬁnitely-large perfect conductor, which would have constant electrical potential even if current were being drawn from it.) 7.5.1 A (Hypothetical) One-Temperature Heat Engine ˙ Consider the heat engine shown in Fig. 7.1. The engine takes in heat Qh from ˙ . Since the engine a thermal reservoir at temperature Th , and outputs power W is in steady state, energy accounting yields ˙ ˙ Qh = W (7.17)

CHAPTER 7. ENTROPY ACCOUNTING AND APPLICATIONS and entropy accounting [Eq. (7.12)] yields

164

˙ Qh ˙ + Ps = 0. (7.18) Th ˙ Since the Second Law requires Ps ≥ 0 and also Th > 0, Eq. (7.18) requires ˙ ˙ Qh < 0, and therefore W < 0. Thus, it is impossible to produce positive power by any process which only takes in heat at a single temperature Th . It is not hard to see why this doesn’t work. Entropy is entering the engine ˙ at a rate Qh /Th , but no entropy is leaving. (Remember work does not carry entropy with it.) Thus, operation of this engine would require entropy to be destroyed inside the engine, but that violates the Second Law. Therefore, we conclude a one-temperature heat engine cannot be built. This statement is sometimes called the Kelvin-Planck statement of the Second Law. In some classical treatments of thermodynamics which make no reference to the existence of atoms or microstates, this is introduced as a postulate. With this postulate, it is possible to work “backwards” (from our point of view) using some very clever thought experiments to prove that there must exist a property, the entropy, which has all of the macroscopic characteristics we’ve discussed already. 7.5.2 A Two-Temperature Heat Engine To allow production of positive power, some means of continually removing entropy from the engine must be provided. The only way to do this in a closed system is to remove some heat from the engine, which will carry entropy with it. ˙ ˙ Since we want to convert at least part of Qh to W , we can only aﬀord to remove ˙ h . But we need to remove at least as much entropy as entered a portion of Q (since some might have also been produced inside), so we need to remove heat in such a way that the ratio (entropy ﬂow)/(heat ﬂow) is larger going out than coming in. Since the amount of entropy carried with a unit amount of heat ﬂow is 1/T , this suggests we should try rejecting some heat to a reservoir at a cold temperature Tc (large 1/Tc). Doing energy and entropy accounting for the two-temperature heat engine in Fig. 7.2, we have ˙ ˙ ˙ Qh = W + Qc , (7.19) and ˙ ˙ Qc Qh ˙ + Ps = . Th Tc (7.20)

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165

Th Qh W
Heat Engine

Qc Tc
Figure 7.2: A two-temperature heat engine. ˙ Eliminating Qc between these equations, Tc ˙ ˙ W = Qh 1 − Th ˙ − Tc Ps . (7.21)

We see that positive power can be produced by a two-temperature heat engine. ˙ ˙ Since the second law requires Ps ≥ 0, the maximum power for a given Qh is ˙ produced for Ps = 0. Therefore a reversible heat engine produces the most ˙ power possible for a given heat transfer rate Qh , and given Th and Tc . For a ˙ real engine, the term −Tc Ps represents power “lost” due to irreversibilities. The thermal eﬃciency η of a heat engine is deﬁned as η= ˙ W . ˙h Q (7.22)

˙ This deﬁnition makes sense, since the desired output is W , and typically it is ˙ Qh which you must “pay for” in the cost of fuel burned to produce it. A reversible heat engine has the highest thermal eﬃciency of any heat engine operating between Th and Tc .From Eq. (7.21), this limiting value of the eﬃciency is Tc η max = 1 − , (7.23) Th This upper-limit eﬃciency is often called the Carnot eﬃciency, after Sadi Carnot, a French engineer who was the ﬁrst person to systematically analyze heat engines in the 19th century. Note that it depends only on Th and Tc , and is the same for all possible heat engines operating between these temperatures. Most practical engines and generators, including automotive engines, aircraft engines, and electric power generation facilities, can be approximated as

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166

Th Qh W
Heat Pump

Qc Tc
Figure 7.3: A heat pump or refrigerator “pumps” heat from low temperature to high temperature. two-temperature heat engines. In order to produce a fuel-eﬃcient engine, η should be as large as practically possible. Of course, the ﬁrst thing to do is to try to eliminate as much unnecessary entropy production within the engine as ˙ possible, to minimize the term −Tc Ps . But once this has been done to the extent practical, further increases in η require either increasing Th , or decreasing Tc . For most engines, Tc is the temperature of the ambient surroundings and can’t be easily varied. This leaves increasing Th . A ﬂame can easily be used to produce a temperature of, say, 2000 K. If we use Th = 2000 K and Tc = 300 K, we ﬁnd η max = 1 − 300/2000 = 0.85. This is a very high eﬃciency, but unfortunately not many materials can withstand this temperature for long. Metals (which turbines are usually made of) fail at much lower temperatures, so the peak temperature in a practical engine is usually limited by materials constraints. Some of the most signiﬁcant developments in recent years have been ceramic components in engines which can take much higher temperatures and allow designing engines with higher Th .

7.6 Heat Pumps and Refrigerators
Although heat will not spontaneously ﬂow from low temperature to high, a process can be designed which takes in heat at low temperature and expels heat at high temperature. However, such a process will not run by itself — it requires work input to “pump” the heat from low to high temperature. A heat pump is a device which operates either continuously or in a cycle,

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167

which takes in heat at low temperature and expels it at higher temperature (Fig. 7.3). A refrigerator is a common example of a heat pump. Heat continually leaks into the refrigerated compartment from the surroundings through the insulation, or when the door is opened. This heat must be removed and expelled to the surroundings at higher temperature to keep the refrigerated space cold. We may analyze heat pumps assuming steady-state, as we did to analyze heat engines, since they operate continuously (dS/dt = 0) or in a cycle (S(T )−S(0) = 0). For the heat pump shown in Fig. 7.3, energy accounting yields ˙ ˙ ˙ W + Qc = Qh , and entropy accounting yields ˙ ˙ Qc Qh − + Ps = 0. Tc Th (7.25) (7.24)

If the heat pump is used as a refrigerator, what we really care about is how ˙ much power W we have to supply to run the heat pump for a given rate of heat ˙ c from the cold space. In this case, eliminating Qh between equations ˙ removal Q (7.24) and (7.25), Th ˙ ˙ ˙ W = Qc − 1 + Th Ps . (7.26) Tc ˙ ˙ Since the Second Law requires Ps ≥ 0, clearly a reversible heat pump (Ps = ˙ ˙ 0) requires the least power input for given Qc , Tc , and Th . Note that if W were ˙ set to zero in Eq. (7.26), there would be no solution with Qc > 0, even if the heat pump were reversible. Therefore, a zero-work heat pump is impossible to build. This statement is sometimes called the Clausius statement of the Second Law. Like the Kelvin-Planck statement, it is sometimes introduced as a postulate in purely macroscopic treatments of thermodynamics, from which the existence of entropy is inferred. ˙ ˙ The second law requires Ps ≥ 0, so for a given Qc , the minimum power ˙ required is obtained when Ps = 0. Therefore, a hypothetical heat pump which produces no entropy (a reversible heat pump) would require the least work: ˙ ˙ Wmin = Qc Th −1 . Tc (7.27)

Since no real refrigerator is truly reversible, the actual power required to run ˙ it will be greater than this value. The term Th Ps in Eq. (7.26) represents the

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168

“extra” work a real refrigerator does, due to irreversibilities such as friction and ﬁnite-temperature-diﬀerence heat transfer. Design of an energy-eﬃcient refrigerator would require a careful thermodynamic analysis to ﬁnd where entropy is being produced, and how entropy production could be minimized. The performance of a refrigerator is characterized by the coeﬃcient of performance, deﬁned as the watts of heat removed from the cold space per watt of ˙ ˙ power input: COPref = Qc /W . From Eq. (7.27), the maximum value is COPref
(max)

=

1 . Th /Tc − 1

(7.28)

Note that this can be greater than one: more energy can be removed as heat than is required as input to run the process. An interesting thing about this equation is that it only depends on Th and Tc , not on the details of how we implement the heat pump – every possible refrigerator design has the same limiting COP as it approaches reversibility, no matter whether the working substance is a ﬂuid, or a magnetic crystal, or anything else. ˙ If a heat pump is used to heat a house by pumping heat Qh into the house at Th from the outside surroundings at Tc , then eqs. (7.24) and (7.25) still apply, ˙ ˙ ˙ but we are most concerned with the ratio Qh /W . If we eliminate Qc between equations (7.24) and (7.25), we ﬁnd Tc ˙ ˙ W = Qh 1 − Th ˙ + Tc Ps . (7.29)

The performance parameter which is used to characterize heat pumps used for heating is also called the coeﬃcient of performance, but it is deﬁned slightly diﬀerently: ˙ Qh (COP)hp = . (7.30) ˙ W For a reversible heat pump the coeﬃcient of performance is maximized: (COP)hp 7.6.1 The Carnot Cycle We’ve shown based on very general considerations that no heat engine can have an eﬃciency greater than 1 − Tc /Th . In deriving this result, we didn’t have to say anything about how the engine operates – it is a general result, which applies to all conceivable heat engines.
(max)

=

1 . 1 − Tc /Th

(7.31)

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169

Th Qh Isothermal Expansion
P Th Tc V

Adiabatic Expansion

P Th Tc V

Tc Qc
P Isothermal Compression V P Adiabatic Compression V Th Tc

Th Tc

Figure 7.4: The Carnot Cycle two-temperature heat engine. Each step is performed quasi-statically, and therefore the process is reversible. By reversing the cycle, it also can function as a heat pump. It’s instructive to consider a simple idealized, reversible cycle which achieves this limiting eﬃciency. While not a very practical design for an engine, this cycle, known as the Carnot cycle, is conceptually the simplest possible reversible two-temperature heat engine. By running it in reverse, it functions as a heat pump. In a Carnot-cycle engine, a ﬂuid (not necessarily an ideal gas) is placed in a piston/cylinder system. The following 4 steps are executed in a reversible manner in sequence: 1. The cylinder is placed in contact with a thermal reservoir at Th , and the ﬂuid is expanded isothermally, drawing in heat Qh . 2. Now the thermal reservoir is removed, and the expansion continues adiabatically. Since no heat is being added in this step, but the ﬂuid is doing

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170

Th Tc

a 4 d

1

b 2

3 s

c

Figure 7.5: The Carnot cycle as represented on a T − S plot. work, its temperature drops. This step continues until the temperature drops to Tc . 3. Now a thermal reservoir at Tc is placed in contact with the cylinder, and the the ﬂuid is isothermally compressed, expelling heat Qc . 4. Finally, the thermal reservoir is removed, and the compression continues adiabatically, ending with the piston reaching its initial position and the ﬂuid temperature back at Th . In Fig. 7.4, this process is represented on a P − V plot, assuming the ﬂuid is a gas. Since the process is carried out quasi-statically, the ﬂuid remains in equilibrium at all times during the cycle. Therefore, its state may be represented at every time by a point in the P − V plane, and the entire cycle is a closed curve in the P − V plane. A simpler representation of this cycle is a temperature-entropy (T − S) plot, as shown in Fig. 7.5. The Carnot cycle traces out a rectangle on a T − S plot, and this is true no matter what the ﬂuid is. Consider ﬁrst steps 2 and 4, which are both reversible and adiabatic. We have shown that reversible, adiabatic processes are isentropic, so the entropy of the ﬂuid does not change during steps 2 and 4. They must appear as vertical lines on a T − S plot. Steps 1 and 3 are isothermal, so they must appear as horizontal lines on a T − S plot. Step 1 draws in heat Qh , which brings in entropy Qh /Th with it. Since the process is reversible, Ps = 0 and therefore ∆S1 = Sb − Sa = Qh . Th (7.32)

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171

For step 3, heat Qc is rejected and the entropy of the system decreases by Qc /Tc : Qc ∆S3 = Sa − Sb = − . (7.33) Tc Since step 3 brings the system entropy back to its initial value (it is unchanged by step 4), we must have Qh Qc = = ∆S1 . (7.34) Tc Th The net heat taken in by one complete cycle is then Qh − Qc = Th ∆S − Tc ∆S = (Th − Tc )∆S, which is simply the area enclosed by the cycle on the T − S plot. An energy balance for one complete cycle is Qh = Wnet + Qc , (7.36) (7.35)

where Wnet is the net work output for the entire cycle. (If we denote the work output done in step n by Wn , then Wnet = W1 + W2 + W3 + W4 , where W1 and W2 are positive, and W3 and W4 are negative.) Therefore, Wnet = Qh − Qc = (Th − Tc )∆S is also given by the area enclosed on a T − S plot. We can also relate Qh , Qc, and W to areas on the T − S plot: (7.37)

Th Q Tc h

Th Tc

Th Tc

Q net =W

Qc s

s

s

Then the eﬃciency η = W/Qh is simply the ratio of the areas of two rectangles, and is easily seen to be (1 − Tc /Th ). Temperature-entropy plots are very useful for analyzing any reversible cycle. For a reversible process, dQ = T dS. Integrating this around a complete cycle, ¯ returning to the initial state, Qnet = dQ = ¯ T dS. (7.38)

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Thus, the net heat taken in during one complete cycle equals the area enclosed by the cycle on a T − S plot, no matter what the shape of the curve is (not only for a rectangle). From the First Law for a complete cycle, dU = 0 = dQ − ¯ dW ¯ (7.39)

where we are taking work to be positive which is done by the system. Therefore, the net work done during one cycle is also given by the enclosed area: Wnet = Qnet = T dS. (7.40)

7.6.2 Maximum Power Output of a Heat Engine Anyone who has compared the performance and fuel eﬃciency of a Geo Metro and a Corvette will not be surprised to learn that there is a trade-oﬀ between the eﬃciency of an engine and its power output. Let’s look at a simple example which shows this clearly. Suppose we have a two-temperature heat engine operating between speciﬁed Th and Tc , and we want to maximize the power output ˙ ˙ W = η Qh . (7.41)

˙ The highest η will be achieved by minimizing Ps , in which case η will approach the Carnot eﬃciency. ˙ To minimize Ps , the engine must be run very slowly — in the limit, quasistatically. This minimizes the work lost to friction, and also insures that heat transfer from the hot reservoir to the engine and from the engine to the cold reservoir is slow enough that large temperature diﬀerences (needed to “drive” ˙ the heat transfer) don’t develop. But if we run the engine that slowly, Qh will be very small! The engine will be very eﬃcient at converting Qh into W , but the rate of ˙ ˙ ˙ heat extraction Qh will be very low, as will W = η Qh , and in the limit of a reversible process (inﬁnitely slow) both will go to zero. So it seems that building a reversible engine is not what we want to do if we want to maximize power ˙ output. The conditions which maximize η don’t maximize the product η Qh . To maximize power instead of eﬃciency, we must allow some irreversibility. Let’s consider a simple case which only has one type of irreversibility – ﬁnite ∆T heat transfer. In reality, this source of irreversibility is inescapable. For example, suppose heat is taken from a reservoir at Th by passing a ﬂuid through tubes immersed in it. In order for heat to ﬂow from Th into the ﬂuid in the tubes,

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173

Th Qh T1 Qh W
Reversible Heat Engine

Rh

Qc T2 Qc Tc
Figure 7.6: An “endoreversible” engine, in which the only irreversibility is due to the thermal resistances for heat transfer.

Rc

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174

the ﬂuid temperature must be less than Th ; call it T1 . The heat transfer rate is generally proportional to the temperature diﬀerence: Th − T1 ˙ Qh = Rh (7.42)

where Rh is the overall thermal resistance, which depends on factors such as the total tube area, the ﬂuid ﬂow rate, and the thermal conductivities of the materials involved. But if these are constant, then Qh ∝ (Th − T1 ). Also, to reject heat to a reservoir at Tc , the ﬂuid (again ﬂowing through tubes) must be hotter than Tc ; let’s say it has temperature T2 > Tc . Then the heat rejection rate will be T2 − Tc ˙ Qc = , (7.43) Rc where Rc is the thermal resistance for heat transfer to the cold reservoir. Assume for simplicity that there are no other irreversibilities – the engine is a reversible one, operating between its own maximum and minimum temperatures T1 and T2 . An hypothetical engine which is internally reversible, but must transfer heat in or out through thermal resistances is called endoreversible, meaning that the interior (endo) is reversible. Since the engine is internally reversible, its eﬃciency is η= ˙ W = ˙ Qh 1− T2 T1 . (7.44)

But since T1 < Th for Qh > 0, and T2 > Tc for Qc > 0, this eﬃciency is less than the value it would have if it were operating reversibly between Th and Tc . Temperatures T1 and T2 are not ﬁxed, but change depending on the heat extraction rate. We can use Eq. (7.42) and Eq. (7.43) to substitute for T1 and T2 in terms of the constant reservoir temperatures Th and Tc . If we also invoke ˙ ˙ ˙ the ﬁrst law (Qh = W + Qc ), we ﬁnd (after a bit of algebra): η =1− Tc . ˙ h (Rh + Rc ) Th − Q (7.45)

As we increase the rate at which we take heat from the hot reservoir, the engine eﬃciency goes down, since T1 decreases and T2 increases. If we keep ˙ increasing Qh , eventually T1 and T2 approach each other, and η → 0. From Eq. (7.45), this occurs when ˙ (max) = Th − Tc . Qh Rh + Rc (7.46)

CHAPTER 7. ENTROPY ACCOUNTING AND APPLICATIONS
1.0 0.8 0.6 0.4 0.2 0 0 0.2 0.4
. .

175

1.0 0.8 η 0.6 η 0.4
. .

Qh

Qh

W
.

0.2 0 0.6 0.8 1.0 0 0.2 0.4
.

W 0.6 0.8 1.0

Qh

Qh

Figure 7.7: Eﬃciency, heat input, and power output for an endoreversible en˙ ˙ ˙ (max) [Eq. (7.46)]. Left: gine. Both Qh and W are shown normalized by Qh Tc /Th = 0.5; right: Tc /Th = 0.2. Equation (7.45) may be written as η =1− (Tc /Th ) . ˙ h /Q(max) )(1 − Tc /Th )) ˙ 1 − (Q h ˙ ˙ W = η Qh . (7.47)

The power output is given by (7.48) ˙ ˙ ˙ The quantities η, Qh , and W shown below in Fig. 7.7 as a function of Qh for two values of Tc /Th (0.2 and 0.5). The maximum eﬃciency is obtained for ˙ ˙ Qh → 0, but the maximum power output occurs at a value of Qh > 0, where the engine eﬃciency is less than its maximum value. ˙ To ﬁnd the maximum power point, we can diﬀerentiate W with respect to ˙ Qh and set the resulting expression to zero. Doing this, the maximum power condition is found to occur for ˙ (maxpower) = Q(max) ˙ Qh h 1 . 1 + Tc /Th (7.49)

Substituting this expresssion into Eq. (7.47), the engine eﬃciency at the maximum power condition is η (maxpower) = 1 − Tc . Th (7.50)

Note that this (unlike the Carnot eﬃciency) is not an upper limit on the eﬃciency of an engine. It’s simply the eﬃciency an internally reversible engine will

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176

have in the presence of thermal resistance for heat transfer, when it is operating at the maximum-power-output condition. The surprising thing about this result is that η (maxpower) depends only on Tc and Th , not on Rc and Rh . This analysis was ﬁrst done by Curzon and Ahlborn at the University of British Columbia in 1975 and for this reason 1 − Tc /Th is sometimes called the Curzon and Ahlborn eﬃciency. Curzon and Ahlborn pointed out that this eﬃciency is in practice quite close to that of many large power plants. This suggests that ﬁnite-∆T heat transfer is one of the largest irreversibilities encountered in real power plants, and that they are in eﬀect designed for maximum power output achievable for given Th , Tc , Rh , and Rc . This in turn suggests that capital cost, not fuel cost, is the dominant factor in determining how many power plants are designed. (If fuel cost were dominant, it would make sense to operate below maximum power, where the eﬃciency is higher. To produce the required power, a larger (and more expensive) power plants could be built.)

7.7 Entropy Accounting for Open Systems
To extend our analysis to open systems, we have to modify the entropy accounting expressions of Section 7.3 to account for entropy carried by matter entering or leaving the system. We’ve already seen that when matter enters a system, it carries with it energy of e units per kg. Recall e is the total speciﬁc energy: e = u + |V |2 /2 + gz. It should come as no surprise that matter also carries entropy into a system when it enters. After all, entropy is an extensive property like energy or mass. The amount brought in is s units of entropy per kg, where s is the speciﬁc entropy (J/K-kg) of the matter entering. So if a ﬂuid is ﬂowing into a control volume at rate m, and at the inlet the ﬂuid has speciﬁc entropy sin , the rate at ˙ which entropy is carried into the system by the ﬂuid will be msin . ˙ We can easily add terms to our entropy accounting equations to account for this convected entropy. For example, Eq. (7.10) can be generalized as: dS = dt ˙ Qin Tin − heat out + ˙ Qout Tout (7.51)

heat in

˙ (ms)in − ˙ (ms)out + Ps . (7.52) ˙ inlets outlets

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177

T

0

Q

.

m

.

state 1

Steady Flow Device

m

.

state 2

W
Figure 7.8: A steady-ﬂow device.

.

7.8 Steady-Flow Devices
In Chapter 5, a set of useful steady-ﬂow devices was introduced: turbines, compressors, pumps, nozzles, diﬀusers, and heat exchangers. These devices change the state of a ﬂuid ﬂowing steadily through them. In Chapter 5, we did an energy accounting on each of them. Let’s now combine this with an entropy accounting. We’ll make the assumptions of steady state (no change in the amount of energy or entropy contained within the device), steady ﬂow (constant m in ˙ and out), and equilibrium properties at the inlet and outlet, where we need to ˙ evaluate them. We will assume that heat transfer occurs at a steady rate Q to ˙ is being produced at a steady rate. With the surroundings at T0 , and power W these assumptions, the system is as shown in Fig. 7.8. With the system as shown in the sketch, the entropy accounting is ˙ msin + Ps = msout + ˙ ˙ ˙ Q T0 (7.53)

Note that we are taking the system boundary just outside the device, where the temperature is T0 . In that way, the irreversibility due to any ﬁnite-∆T heat ˙ transfer occurs within the system, and is contained in the Ps term. If we put the system boundary inside the device, then entropy would be being produced in the environment – we could do the analysis that way, too, but it would be more complicated. Let’s apply this equation together with an energy balance to some speciﬁc devices.

CHAPTER 7. ENTROPY ACCOUNTING AND APPLICATIONS 7.8.1 Adiabatic steady-ﬂow devices

178

In Chapter 5, we introduced ﬁve steady-ﬂow devices which are often modeled as adiabatic. These were (1) compressors, (2) turbines, (3) nozzles, (4) diﬀusers, and (5) valves. For any adiabatic steady-ﬂow device, Eq. (7.53) reduces to ˙ msin + Ps = msout . ˙ ˙ ˙ Since the second law requires Ps ≥ 0, sout ≥ sin : the speciﬁc entropy of the ﬂuid leaving an adiabatic steadyﬂow device is greater than or equal to the speciﬁc entropy of the ﬂuid entering. Equation (7.54), together with the steady-ﬂow energy equation introduced in Chapter 4, is what we need to analyze any of these devices. Recall the forms of the steady-ﬂow energy equation appropriate for adiabatic compressors, turbines, nozzles, and diﬀusers: ˙ ˙ Wc = Wc /m = h2 − h1 ˙ ˙ Wt = Wt /m = h1 − h2 ∆ V /2
2

(7.54)

(compressor) (turbine) (nozzle or diﬀuser)

(7.55) (7.56) (7.57) (7.58)

= h1 − h2

Since h and s appear in the energy and entropy accounting expressions, an h − s plot is a convenient way to represent the change in the state of the ﬂuid as it passes through any of these devices, as shown in Fig. 7.9. In Fig. 7.9, the inlet state 1 and the outlet state 2 are shown connected by a dashed line. This is because the process may be irreversible, and so the ﬂuid inside the device may not be in an equilbrium state, and therefore would not correspond to any point on the plot. Since the second law requires s2 ≥ s1 , state 2 can be anywhere to the right of point 1, but not to the left. In these plots, another state “2s” is also shown. This is the outlet state which would result for an “ideal” reversible adiabatic device operating between P1 and P2 . The ﬂuid coming out of the idealized device ends up in a diﬀerent state than the outlet state of the real, irreversible device. State 2s has the pressure of state 2, but the entropy of state 1. The turbine work output, compressor work input, or nozzle/diﬀuser change in kinetic energy is given by |h2 −h1 |. This is shown on the plot. Since constant-

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179

h
1

P P 1 2
2 2s

h Wc

P 2
2s 2

P 1

Wt

1

s Turbine h
1

s Compressor h (V 2/ 2) P 2
2s 2

P P 1 2
2

P 1

(V / 2)
2s

2

1

s Nozzle Diffuser

s

Figure 7.9: Enthalpy-entropy process representations for some adiabatic steadyﬂow devices. Isobars at P1 and P2 are also shown.

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180

pressure lines have positive slope on an h − s plot [i.e., (∂h/∂s)P > 0],2 we see that the reversible device always has the best performance. Speciﬁcally, 1. A reversible turbine gives the maximum work output Wt 2. A reversible compressor requires the least work input Wc 3. A reversible nozzle gives the maximum change in kinetic energy.

Example 7.1 Steam enters an adiabatic turbine at 900 K and 10 MPa and leaves at 1 MPa. What is the maximum work output per unit mass of steam, and what is the exit temperature under maximum work conditions? Solution: at the inlet, h1 s1 = h("h2o","tp",900,10) = 3691.5 kJ/kg = s("h2o","tp",900,10) = 6.977 kJ/kg-K.

Since the Second Law requires s2 ≥ s1 , s2 ≥ 6.977 kJ/kg/K, and we’re given P2 = 1 MPa. From TPX, the state with s2 = s1 at 1 MPa has temperature 535.87 K and enthalpy 2970.6 kJ/kg. Any state at this pressure with greater entropy has a higher temperature, and a higher enthalpy (try it and see). The work output per kg of steam is Wt = h1 − h2 . Thus, the most work is produced when h2 is as small as possible; the smallest h2 is achieved when s2 = s1 . Therefore, the greatest work output possible is Wmax = 3691.5−2970.6 kJ/kg = 720.9 kJ/kg, and the exit temperature is 535.87 K. The performance of an adiabatic steady-ﬂow device is often characterized by its isentropic eﬃciency ηs . The deﬁnition of ηs is diﬀerent for each type of device, but in each case is a comparison of the actual performance (work input or output, or change in kinetic energy) with the performance of an ideal isentropic device operating between the same pressures. The isentropic eﬃciency is always deﬁned so that it is ≤ 1. For a turbine, Wt h1 − h2 = . (7.59) ηs,t = Wt,s h1 − h2s Here Wt is the actual work output, and Wt,s is the work output of the ideal isentropic turbine. Values of ηs for real turbines depend on size, varying from perhaps 60% for a very small turbine to 95% for a very large, well-designed turbine. A typical value would be around 90%.
2 Since dh = d(u + P v), dh = (T ds − P dv) + vdP + P dv = T ds + vdP . (∂h/∂s)P = T , which is positive.

Therefore,

CHAPTER 7. ENTROPY ACCOUNTING AND APPLICATIONS For a compressor,

181

Wc,s h2s − h1 = . (7.60) Wc h2 − h1 Compressor eﬃciencies are typically lower than turbine eﬃciencies, with a typical value of about 85%. A very large compressor might approach 90% eﬃciency. ηs,c =

7.9 Steady-Flow Availability
A question which is often of interest is how much power can be produced by any conceivable steady-ﬂow process which takes a steadily-ﬂowing ﬂuid from a speciﬁed inlet state to a speciﬁed outlet state. To answer this question, consider again Fig. 7.8. From Eq. (7.53), the entropy accounting is ˙ Q ˙ ms1 + Ps = ms2 + . ˙ ˙ T0 The energy accounting for this steady ﬂow process is ˙ ˙ m(e + P v)1 = W + Q + m(e + P v)2 . ˙ ˙ (7.62) (7.61)

(We are allowing the ﬂuid to have kinetic and/or potential energy in addition to internal energy in states 1 and 2.) ˙ ˙ Eliminating Q between these equations and solving for W , ˙ ˙ ˙ W = m [(e + P v)1 − (e + P v)2 ] − T0 m(s1 − s2 ) + T0 Ps , ˙ or ˙ ˙ W = m(b1 − b2 ) − T0 Ps , ˙ where b = e + P v − T0 s (7.65) is the steady ﬂow availability function. (Note that it is the environment temperature T0 , not the ﬂuid temperature T , which appears in this expression.) From Eq. (7.64), we see that a real, irreversible process always produces less power than a reversible process acting between the same inlet and outlet states, ˙ for the same mass ﬂow rate. The term T0 Ps represents power which is “lost” due to the irreversibilities in the real process. For this reason, it is called the ˙ irreversibility rate I: ˙ I˙ = T0 Ps . (7.66) (7.64) (7.63)

CHAPTER 7. ENTROPY ACCOUNTING AND APPLICATIONS ˙ The maximum power produced occurs for I = 0, and is ˙ W (max) = m(b1 − b2 ). ˙

182

(7.67)

In some cases, the outlet state 2 is not ﬁxed, and therefore we may ask ˙ what state 2 conditions maximize W (max) . Clearly, this will be the state which minimizes b2 . To minimize b2 = u2 + ek,2 + ep,2 + P2 v2 − T0 s2 , the ﬁrst thing to do is pick state 2 to have zero kinetic energy, and the lowest achievable potential energy. For example, if the potential energy is due to gravitation, pick state 2 at the lowest achievable elevation. (Elevations below the surface of the earth result in lower potential energy, but are not ususally achievable.) Once this is done, we are left with the task of minimizing h2 − T0 s2 . It can be shown (using methods to be discussed in the next chapter) that this is minimized for T2 = T0 and P2 as small as possible. The lowest possible P2 is P0 , since eventually the ﬂuid must ﬂow to the environment. Thus, the optimal state 2 is the one which is in thermal and mechanical equilibrium with the environment. Example 7.2 A certain industrial process exhausts 10 kg/s of steam at 2 atm and 400 K to the atmosphere through an exhaust valve. It has been suggested that the valve should be replaced with some process which can generate electrical power. Before deciding whether to pursue this, it is desired to know how much power could be produced in principle. The environment conditions are P0 = 1 atm and T0 = 300 K. Solution: The maximum power output is m(b1 − b2 ). We will take state 2 to ˙ be liquid water at 1 atm and 300 K. Using TPX, the properties are State 1 2 T (K) 400 300 P (atm) 2 1 h (kJ/kg) 2720.4 112.7 s (kJ/kg-K) 7.156 0.393 b (kJ/kg) 573.6 -5.3

˙ Therefore, W (max) =(10 kg/s)(578.9 kJ/kg) = 5.8 MW. Of course, any real process will produce less power, but a well-designed process should be able to produce at least a couple of megawatts. Whether it makes sense to pursue this would depend on how much the process costs to build and run, and on the cost of electricity saved.

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183

Problems
7.1 Gaseous water at 5 MPa in a piston/cylinder system is adiabatically expanded until its temperature is 400 K. If it is desired that there be no liquid droplets in the ﬁnal state, what is the minimum possible initial temperature? 7.2 An insulated container contains saturated liquid nitrogen at 1 atm. You wish to design a process to produce work from this liquid nitrogen. 1. If the surroundings are at 300 K and 1 atm, what is the maximum work which could be produced in principle? 2. Suppose you use a simple process in which you transfer the liquid to a piston-cylinder system, and let it warm up to room temperature at constant pressure. Sketch this process on an appropriate process diagram, and calculate the work produced. 3. If the work produced in (b) is less than the maximum calculated in (a), discuss why. 4. Can you suggest ways to improve this process? 7.3 The Cassini mission to Saturn was recently launched by NASA, and should arrive at Saturn in 2004. Like all NASA deep space missions, on-board electrical power on Cassini is generated by RTGs (Radioisotope Thermoelectric Generators). These consist of pellets of radioactive plutonium fuel which produce heat, bundles of silicon-germanium thermocouples, and radiator ﬁns which radiate waste heat into space. The thermocouples produce electrical power directly when connected between the hightemperature “General Purpose Heat Source” containing the plutonium and the cold radiator ﬁns.

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The heat rejected from the radiator ﬁns follows the Stefan-Boltzmann law:
4 ˙ Qc = AσTc .

Here Tc is the temperature of the radiator ﬁns, A is the total ﬁn area, and is a material property known as the emissivity (0 < < 1). The array of thermocouples constitutes a heat engine operating between the plutonium fuel temperature (Th ) and the radiator temperature Tc . To minimize the weight and spacecraft size, it is desired to keep the radiator ﬁns as small as possible. For a given power generation requirement ˙ W , , and Th , determine 1. the minimum radiator ﬁn area required, if the thermocouples act as “ideal” heat engines. ˙ ˙ 2. the thermodynamic eﬃciency η = W /Qh and ﬁn temperature Tc under the conditions of part (a). Assume that there is no heat transfer limitation on the rate at which heat may be taken from the radioactive source. 7.4 You have 2 identical copper blocks of mass M . One is initially at temperature T1 (0), and the other is initially at temperature T2 (0) < T1 (0). Both may be assumed to be incompressible substances with constant c. You now let them interact through some process which extracts heat from block 1, generates work, and dumps some heat into block 2. Since the blocks have ﬁnite mass, T1 (t) and T2 (t) change with time. Eventually, they come to the same temperature T1 (∞) = T2 (∞) = Tf and no more work can be done. 1. Taking both blocks together as the system and assuming no heat is transferred to or from the environment, write down expressions for ∆U = U (∞) − U (0) and ∆S = S(∞) − S(0). Assume that the total entropy produced during the process is Ps . 2. Combine these equations to obtain expressions for W and Tf . Your expression should contain only M , c, T1 (0), T2 (0), and Ps . 3. Show that the maximum possible work is Wmax = M c and determine Tf in this case. T1 −
2

T2

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185

7.5 Nitrogen is held in a container at its critical pressure and temperature. If the environment is at 300 K and 1 atm, what is the maximum work which could be done per kg of N2 ? 7.6 A concept for a positioning thruster for a very small (1 kg) spacecraft is shown below. micro-spacecraft 340 K liquid reservoir Q capillary tube micro-nozzle

Liquid HFC-134a at T = 250 K and 0.8 MPa ﬂows from a small on-board reservoir into a capillary tube 0.5 mm in diameter and 1 cm long, where it absorbs waste heat given oﬀ by the on-board electronic equipment, causing it to evaporate. It emerges from the tube as vapor at 320 K at a pressure of 0.65 MPa. (Due to wall friction, there is a substantial pressure drop down the tube.) The vapor emerging from the tube enters an adiabatic micro-nozzle and is accelerated to produce thrust. The micro-nozzle has an isentropic eﬃciency of 0.9, and the pressure at its exit is 0.05 MPa. Determine: ˙ 1. The irreversibility rate I in the capillary tube, if the environment surrounding the tube is at 340 K. 2. The velocity and temperature at the exit of the micro-nozzle 3. The mass ﬂow rate in mg/s required to produce a thrust of 1 mN. (Thrust is mass ﬂow rate multiplied by exit velocity.) 7.7 What is the minimum power required by an adiabatic compressor which compresses 10 kg/min of saturated water vapor at 0.2 MPa to 0.4 MPa? 7.8 What is the minimum power required by any process (not necessarily adiabatic) which compresses 10 kg/min of saturated water vapor at 0.2 MPa to 0.4 MPa? The surroundings are at P0 = 1 atm and T0 = 300 K. Compare this process to that in problem 1. Are the ﬁnal states the same? Does this process involve heat transfer? If so, in which direction?

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186

7.9 An industrial process to produce liquid nitrogen starts with gaseous nitrogen at 300 K and 1 atm, which enters at a steady rate of 1 kg/s. The product is saturated liquid nitrogen at 1 atm. 1. If the environment is at 300 K and 1 atm, does this process require work input, or can it produce work? 2. Determine the minimum work input, or maximum work output (depending on your answer to (a)). 3. Determine the heat transfer rate for the conditions of part (b). 7.10 (NOTE: Requires Helium data; can’t be done with current version of TPX.) Two cylinders with pistons each contain 10 kg of helium. The weight of piston 1 keeps P1 constant at 0.05 MPa, and the weight of piston 2 keeps P2 constant at 0.2 MPa. Initially, both cylinders are at a temperature of 5.2 K. They are now thermally connected to a reversible heat pump, which heats cylinder 1 and cools cylinder 2 with no heat transfer to the surroundings. When cylinder 1 reaches 8 K, the process stops. 1. Write equations expressing energy and entropy accounting for this process. 2. Find the ﬁnal state of the helium in cylinder 2. 3. Find the work input to the heat pump.

1 Reversible Heat Pump

2

7.11 A heat-powered portable air compressor consists of three components: (a) an adiabatic compressor; (b) a constant pressure heater (heat supplied from an outside source); and (c) an adiabatic turbine. The compressor and turbine both may be idealized as reversible, and air may be assumed to be an ideal gas with k = cp /cv = 1.4. Ambient air enters the compressor at 0.1 MPa and 300 K, and is compressed to 0.6 MPa. All of the power from the turbine goes to run the

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187

compressor, and the turbine exhaust is the supply of compressed air. If this pressure is required to be 0.2 MPa, what must the temperature be at the exit of the heater? 7.12 For a particular industrial process, a stream of gaseous oxygen at 2 MPa and 1000 K is required. You have available a stream of saturated liquid oxygen at 100 K, and you wish to design a process to continuously produce the desired gaseous oxygen from this liquid oxygen. The surroundings are at 300 K. Determine whether such a process requires work input or can produce work output, and the minimum work input or maximum work output per kg of O2 . 7.13 An adiabatic pump pumps 200 kg/hr of saturated liquid HFC-134a at 320 K to 2.5 MPa. The temperature of the emerging liquid is 321.26 K. Determine: 1. The power required to run the pump 2. The isentropic eﬃciency of the pump 3. The entropy production rate 4. The irreversibility rate if the environment is at 300 K.

CHAPTER 9
POWER GENERATION AND PROPULSION 9.1 Introduction
One of the most signiﬁcant engineering applications of thermodynamics is the design and optimization of energy conversion devices which take in heat and produce useful work (“heat engines”). In Chapter 7, we discussed in very general terms the principles governing the performance of heat engines. In this chapter, we look at some speciﬁc processes which are actually used for power generation and propulsion.

9.2 Cycles
Any thermodynamic process which takes a substance through a sequence of states eventually returning it to the original state is known as a cycle. A familiar example is the Carnot cycle, discussed in Chapter 7. Virtually all heat engines are implemented as cycles. Usually, the substance the cycle operates on is a ﬂuid, although in principle any substance could be used, as long as it has some reversible work mode. In this chapter, we’ll only consider cycles which operate on ﬂuids, since all important engineering cycles are of this type. The ﬂuid used in the cycle is known as the working ﬂuid. During the cycle, the working ﬂuid may or may not change phase. In vapor power cycles, the working ﬂuid changes phase from liquid to vapor and back to liquid over the course of the cycle; in gas power cycles, the working ﬂuid remains gaseous during the entire cycle. As we will see, they each have certain advantages and disadvantages. In many types of cycles, the working ﬂuid circulates continuously in a loop. For example, the major cycles used for electric power generation are of this type. In these cycles, the thermodynamic state of the working ﬂuid at any one location in the loop is constant in time. But if we focus our attention on a single “packet” of ﬂuid (say, a unit mass) moving around the loop, its state changes as it passes through various components such as pumps, turbines, heat exchangers, etc. To analyze such cycles, we follow a ﬂuid packet around the loop, applying 228

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229

the steady-ﬂow equations expressing energy and entropy accounting to each device the ﬂuid packet encounters to determine how its state changes. In another type of cycle, a ﬁxed mass of ﬂuid is made to undergo a sequential set of operations. The Carnot cycle is an idealized example of such a cycle. Practical examples include spark-ignition and Diesel internal combustion engines. To analyze these, we apply the control-mass energy and entropy accounting equations to each step in the cycle. To analyze real cycles, we need to account for the fact that they are irreversible (they produce entropy). As we saw in Chapter 7, some irreversibility (for example due to ﬁnite-∆T heat transfer) is inevitable if we want to produce power, since reversible processes operate inﬁnitely slowly. In addition to ﬁnite∆T heat transfer, some other common irreversibilities are those due to friction, ﬂuid viscosity, shock waves, and combustion. Nevertheless, we can learn a lot about real cycles by ﬁrst considering ones which are fully reversible, and then considering how the results for reversible cycles need to be modiﬁed to account for irreversibilities. We’ll do this in the next two sections, and then apply these ideas to look at the characteristics of several diﬀerent cycles. 9.2.1 Process Representations To carry out thermodynamic analysis of cycles, we need to keep track of how the thermodynamic state of the ﬂuid changes as the cycle proceeds. Since for a ﬂuid in equilibrium only two properties are required to specify its state, a ﬂuid in thermodynamic equilibrium may be represented as a point on a plot of one thermodynamic property vs. another one (e.g. T vs. s, P vs. v, etc.) Any such plot of the working ﬂuid state at various points in the cycle is known as a process representation. If the ﬂuid remains in equilibrium at every point during the cycle, then the ﬂuid state always corresponds to some point on the process representation and as the cycle proceeds, a closed curve is traced out. Since the ﬂuid is always in equilibrium, no internal irreversible processes (which act only when a system is perturbed from equilibrium) can be occurring. Therefore, such a cycle is reversible. We call reversible cycles ideal cycles. Of course, real cycles can approach ideal cycles, but there will always be some entropy production in reality and the ﬂuid will always be at least a little out of equilibrium. To analyze real, irreversible cycles, we usually assume the ﬂuid is approximately in equilibrium at the inlet and outlet of devices like valves, turbines, pumps, compressors, etc., but not necessarily within them. Points in the pro-

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230

cess representation plane corresponding to the inlet and outlet states are simply connected with a dashed line to represent the action of the device on the ﬂuid. 9.2.2 Analysis of Ideal Cycles The T − s process representation is particularly useful to represent ideal cycles. Since for any reversible process dQ = T ds, the area under a curve traced out ¯ by a reversible process which takes the working ﬂuid from some state 1 to some state 2 is the total heat added:
2

Q12 =
1

T ds.
2 T 1 Q net

(9.1)

2 T 1 Q 12 s s

If we consider a complete cycle, then the net heat taken in by the cycle is the area enclosed by the curve in the T − s plane: Qnet = T ds. (9.2)

Of course, on the portion of the cycle where ds < 0, dQ < 0 so heat is actually ¯ being transferred out of the working ﬂuid to the environment. It is convenient to break up the path of a closed cycle into the portion where heat is being added, and the portion where heat is being rejected:

T A Q in B

T A B Q out s s

T A

Q net

B

Then the heat absorbed per unit mass by one complete cycle is the area under the upper portion of the curve (left), and the heat rejected to the environment per unit mass is the area under the lower portion (center). The net heat taken in is the diﬀerence, which is the enclosed area (right). From the ﬁrst law, the net heat taken in must equal the net work output, since for a complete cycle ∆u = 0.

i

s

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231

Therefore, the area enclosed by an ideal cycle in the T − s plane is the net work produced per unit mass of working ﬂuid for one complete cycle. The net work per unit mass is an important quantity, since the total power produced by the cycle is the product of Wnet and the mass ﬂow rate m: ˙ ˙ W = mWnet . ˙ (9.3)

The mass ﬂow rate needed to produce a given desired power scales inversely with Wnet . Since the overall size of a powerplant scales roughly with the working ﬂuid mass ﬂow rate, higher Wnet results in smaller, more-compact (and usually less expensive) power plants. This is especially important in applications such as aircraft propulsion, where the weight of the engine must be minimized. The thermodynamic eﬃciency η of any power cycle is deﬁned as the net work output divided by the heat input: η= Wnet . Qin (9.4)

Thus, the eﬃciency of an ideal cycle is the ratio of the area enclosed on a T-s plot to the total area under the curve. Two important quantities for ideal cycles are the average temperature at which heat is added (T h ), and the average temperature at which heat is rejected (T c). These are deﬁned by Qin Th = , (9.5) ∆Smax and Qout Tc = , (9.6) ∆Smax where ∆Smax = sB − sA is the diﬀerence between the maximum entropy and minimum entropy points in the cycle. Then the rectangles shown below have area Qin and Qout , respectively.

T Th Q in

T

Tc Q out s ∆s max s

∆s max

The thermodynamic eﬃciency may be expressed in terms of T h and T c : η = Wnet Qin − Qout = Qin Qin (9.7)

CHAPTER 9. POWER GENERATION AND PROPULSION Qout Qin T c∆smax = 1− . T h ∆smax = 1− Therefore, η =1− Tc . Th

232 (9.8) (9.9)

(9.10)

This expression holds for any ideal cycle. For the special case when all heat is taken in at a single temperature Th and all heat is rejected at a single temperature Tc (as in the Carnot cycle), T h = Th and T c = Tc and this expression reduces to the Carnot eﬃciency. We may regard the cycle as being equivalent to a Carnot cycle with temperatures T h and T c . Equation (9.10) shows that the thermodynamic eﬃciency of an ideal cycle can be increased either by increasing the average temperature at which heat is added or by lowering the average temperature at which heat is rejected. This insight will prove to be vital for improving the eﬃciency of real cycles, although of course with real cycles we also have to be concerned with minimizing internal irreversibilities. 9.2.3 Component Models To analyze real cycles, we need models for the action of various components the working ﬂuid encounters (pumps, compressors, heaters, turbines, nozzles, etc.) which include the irreversible character of these components. Several such models of common steady-ﬂow devices were discussed in Section 7.8. We will use these in the analysis below, and introduce some new component models also.

9.3 Vapor Power Cycles
The ﬁrst power cycle to be developed was the Rankine cycle, and in modiﬁed form the Rankine cycle is still the most common vapor power cycle. It is in widespread use for electric power generation in large central power plants, and will be for the forseeable future. Whatever the type of fuel burned in the power plant (coal, natural gas, oil, or nuclear fuel), a modiﬁed Rankine cycle using water as the working ﬂuid is invariably used to convert the heat produced into useful power (sometimes in conjunction with other cycles we’ll discuss later). In the sections below, we work through some speciﬁc examples of the Rankine cycle, progressively adding modiﬁcations which improve performance and lead

CHAPTER 9. POWER GENERATION AND PROPULSION
Q b 3 Boiler 2 Pump W p 1 4 W t Turbine

233

Figure 9.1: The Open Rankine Cycle to the types of cycles actually used in modern power plants. 9.3.1 The Open Rankine Cycle The simplest version of the Rankine cycle is the open Rankine cycle, shown in Figure 9.1. In the open Rankine cycle, a ﬂuid (usually water) is pumped to high pressure, and then heat from a ﬂame is added in a boiler to boil the water at constant pressure. Saturated steam is extracted from the boiler, and expanded back to atmospheric pressure, producing work. In Figure 9.1, the expansion is shown occurring in a turbine, which is how modern implementations of the Rankine cycle operate. The open Rankine cycle was originally developed in the 19th century to power steam locomotives and other steam-operated heavy equipment. The development of steam engines made possible the industrial revolution and everything which followed from it. In a steam locomotive, the high pressure steam entered a cylinder and pushed a piston, which propelled the train forward. At the end of the stroke, an exhaust valve opened and on the reverse stroke the low-pressure steam was expelled to the atmosphere. Let’s analyze this cycle using some typical numbers. We’ll take state 1 to be water at 30 ◦ C and 1 atm, P2 = 0.8 MPa, and assume that state 3 is saturated vapor. The pump and turbine are both adiabatic, with isentropic eﬃciencies of 0.6 and 0.8 respectively. To analyze the cycle, we simply work our way around it, putting a control volume around each component and writing down expressions for energy and entropy accounting. The Initial State We start the analysis in state 1, since enough information is given to fully determine the state here. With T = 303.15 K and P = 1 atm, we using TPX

CHAPTER 9. POWER GENERATION AND PROPULSION that v1 = 0.00100 m3 /kg h1 = 125.2 kJ/kg s1 = 0.4348 kJ/kg-K The Pump

234

To analyze the pump, it is simplest to assume that the liquid water is incompressible. In this case, the work of an ideal (isentropic) pump is Wp,s = v1 (P2 − P1 ). Since these quantities are known, we ﬁnd Wp,s = 0.7 kJ/kg. By deﬁnition of the isentropic eﬃciency for a pump, the actual work Wp is Wp = Wp,s = 1.2 kJ/kg. ηs (9.11)

Knowing the pump work, we can calculate the enthalpy of state 2: h2 = h1 + Wp = 126.4 kJ/kg. Knowing h2 and P2 is suﬃcient to ﬁx all other properties in state 2, and so we ﬁnd T2 = 303.26 K s2 = 0.4366 kJ/kg-K. As required by the second law, s2 > s1 . We see that pumping the liquid produces only a very small temperature rise (0.1 K). The Boiler Let us assume the heat addition in the boiler occurs at constant pressure. Then P3 = P2 = 0.8 MPa. We are not given how much heat is added in the boiler, but we know that state 3 is saturated vapor. Calculating the properties for saturated vapor at 0.8 MPa, we ﬁnd T3 = 443.58 K h3 = 2768.7 kJ/kg

CHAPTER 9. POWER GENERATION AND PROPULSION s3 = 6.662 kJ/kg-K.

235

From an energy balance on the boiler, h2 + Qb = h3 , so the heat added in the boiler is Qb = 2642.3 kJ/kg. The Turbine To analyze the turbine, we ﬁrst must ﬁnd the outlet state which would result from an isentropic turbine. Calling this state 4s, we know P4s = 1 atm s4s = s3 = 6.662 kJ/kg-K. This is enough to fully specify state 4s, so we ﬁnd T4s = 373.144 K h4s = 2417.3 kJ/kg s4s = 6.662 kJ/kg-K x4s = 0.885 Note that state 4s lies within the vapor dome, with nearly 12% liquid content by mass. Having found 4s, we now can get state 4 from the given isentropic eﬃciency of the turbine: h3 − h4 . ηs = h3 − h4s Putting in ηs = 0.8 and solving for h4 , h4 = 2487.6 kJ/kg. With h4 and P4 known, we can ﬁnd the remaining properties: s4 = 6.850 kJ/kg-K x4 = 0.917 We see s4 > s3 , consistent with the second law, and the actual turbine outlet state also contains some liquid. The vapor mass fraction x at the turbine outlet is an important parameter, since if it is too low the water droplets which form within the turbine impact on the turbine blades and cause serious erosion. Typically, x of about 0.9 is the

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236

800

0.8 MPa
3000

0.8 MPa 1 atm 3

600 1 atm
4 4s

h [kJ/kg]

T [K]

3 400 2 1 4s 4

2000

1000

200

1,2

0 0 2 4 6 8 10

0 0 2 4 6 8 10

s [kJ/kg-K]

s [kJ/kg-K]

Figure 9.2: An open Rankin cycle (shown as closed by the environment) lowest tolerable value. Thus, the conditions at state 4 are probably acceptable, but just barely. Since the vapor mass fraction is so important, it is often called the steam quality. Process representations for this cycle in the T − s and h − s planes are shown in Figure 9.2. Note that states 1 and 2 are diﬀerent, but overlap on this scale. Also, the cycle is shown in these process representations as being closed by a constant-pressure line connecting state 4 to state 1. This allows us to regard an open Rankine cycle as being closed by the environment, and the step from 4 to 1 is the heat rejection to the environment required to condense the steam and cool the liquid down to T1 . Having now found all of the states of the cycle, we can calculate its performance. The work produced by the turbine per unit mass of steam ﬂowing through it is Wt = h3 − h4 = 281.1 kJ/kg. Some of this work (1.2 kJ/kg) must be used to run the pump, so the net work output is Wnet = Wt − Wp = 279.9 kJ/kg. To produce this work, Qb = 2642.3 kJ/kg of heat had to be added to the steam

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237

in the boiler. Therefore, the eﬃciency of this process at converting heat into work is Wnet η= = 10.6%. Qb We see in this example that the work output from the turbine is much greater than the work required to run the pump. Their ratio is known as the back work ratio BWR: Wp BWR = . (9.12) Wt For this example, BWR = 0.004. This is one of the major advantages of vapor power cycles: the work required to compress the liquid is negligible compared to the work produced by the vapor expanding through the same pressure diﬀerence. The Carnot eﬃciency for an ideal heat engine operating between the same maximum and minimum temperatures encountered in this cycle is ηCarnot = 1 − 303.15 = 31.6% 443.58 (9.13)

Therefore, the cycle in this example has an eﬃciency only 1/3 of the theoretical maximum for these temperatures. One reason the eﬃciency is low is the fact that the turbine and the pump are not ideal. But if we recalculated the performance of an ideal cycle with ηs,p = ηs,t = 1, we would ﬁnd an eﬃciency of 13.2%. So even the ideal cycle has an eﬃciency far below the Carnot eﬃciency. By examining the T − s plot in Fig. 9.2, we can see some of the reasons for the low eﬃciency. The turbine exhaust is steam at 373.15 K. Much of the total heat rejection to the atmosphere occurs as this steam condenses to liquid at this temperature. Therefore, T c is much greater than 30 ◦ C. Thus, this cycle needlessly throws away hot steam, from which more work could still be extracted. Another factor limiting the eﬃciency is the heat addition from state 2 up to the saturation point. This heat addition in the boiler occurs at less than Tsat , so T h < 443.58 K. In the next example, we consider a modiﬁcation to allow rejecting heat at lower temperature. We’ll look at how to raise T h in the following example. 9.3.2 The Closed Rankine Cycle In the last example, state 4 had a temperature of 373.15 K, which is the result of exhausting at 1 atm. But suppose we didn’t simply exhaust to the atmosphere, but instead provided a low-pressure region for the turbine to exhaust into. For P4 < 1 atm, T4 = Tsat (P4 ) will be less than 373.15 K, and we will be able to lower the temperature at which heat is rejected.

CHAPTER 9. POWER GENERATION AND PROPULSION
Q b 3 Boiler 2 Pump W p 1 4 Condenser Q c W t Turbine

238

Figure 9.3: The Closed Rankine Cycle. To do this, we must close the cycle by adding a condenser, in which heat will be removed from the turbine exhaust, leaving saturated liquid water at state 1. This is shown in Fig. 9.3. We can choose the temperature at which we will condense the steam based on the temperature of the reservoir into which we will be dumping the heat. In many cases, the heat will be removed by cooling water supplied from a nearby lake, river, or ocean. If a source of cooling water is available at, say, 20 ◦ C (293 K), then we could choose a condensing temperature of, say, 300 K. The 7 degree temperature diﬀerence should be enough to drive heat transfer in the condenser. The condenser tube area is determined by the total amount of heat which must be rejected, and by the temperature diﬀerence between the steam and the cooling water. There is always a trade-oﬀ between ∆T and total condenser area. For given Qc , smaller ∆T means larger A. The optimum values would be determined in reality by balancing the eﬃciency gain by lowering the condensing temperature with the cost required to build a bigger condenser. A ∆T value of a few degrees is would be typical. Let us specify T4 = 300 K. Then since state 4 will be in the vapor dome, P4 = Psat (300) = 0.003536 MPa. This is about 1/29 atm. To start the cycle running (for example, after shutting down for maintainance), a vacuum pump is needed to pump out the air. Also, since the pressure is less than 1 atm, any small leaks will result in air entering the condenser. If enough air enters, the pressure will begin to rise, causing Tsat to rise, and lowering the cycle eﬃciency. So the vacuum pump must be used during operation also to remove any air entering through leaks. Thus, the system will be somewhat more complex than the open Rankine cycle. But the gain in eﬃciency will more than oﬀset the increased complexity.

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239

800

0.8 MPa
3000

0.8 MPa 0.0035 MPa 3

600 0.0035 MPa

h [kJ/kg]

T [K]

3 400 2 1 200 4s 4

2000

4 4s

1000

1,2

0 0 2 4 6 8 10

0 0 2 4 6 8 10

s [kJ/kg-K]

s [kJ/kg-K]

Figure 9.4: A closed Rankin cycle With T4 set to 300 K, and state 1 set to saturated liquid at 300 K, we can re-do the analysis of Example 1. The numbers are summarized in the table below. (You should verify these.) State 1 2 3 4s 4 P (MPa) 0.003536 0.8 0.8 .003536 .003536 T (K) 300 300.08 443.58 300 300 h (kJ/kg) 111.7 113.6 2768.7 1993.3 2148.4 s (kJ/kg-K) 0.3900 0.3924 6.662 6.662 7.179

These states are shown in T − s and h − s plots in Fig. 9.4. The pump work is now Wp = h2 − h1 = 1.33 kJ/kg. This is slightly greater than in Example 1, which is not surprising since now state 1 has a lower pressure, so P2 − P1 is greater than before. The turbine work output is Wt = h3 − h4 = 620.3 kJ/kg

CHAPTER 9. POWER GENERATION AND PROPULSION
Q b Q Superheater s 3 Turbine Pump W p 1 4 Condenser Q c W t

240

Boiler 2

Figure 9.5: A Rankine Cycle with Superheat. This is a factor of 2.2 greater than the turbine work in Example 1. Lowering the outlet pressure allowed the steam to expand much more, producing more work. This is clear in comparing the h − s plots in Figures 9.2 and 9.4. The heat added in the boiler Qb is unchanged, so the cycle eﬃciency is η = 23.3% This is quite an improvement over the open Rankine cycle eﬃciency. 9.3.3 Addition of Superheat Having added the condenser, we’ve lowered the temperature at which heat is rejected about as much as possible. We now need to consider how to raise T h if we wish to improve the eﬃciency more. The temperature of the steam in the boiler is set by the pressure, since T = Tsat (P ). So the next improvement to make is to increase the boiler pressure. The limits on boiler pressure are determined by how much pressure the material the tubes carrying the high-pressure steam can withstand at high temperature for long durations. Originally, boilers were constructed from cast iron, which limited the safe operating pressures. Now with high-strength steel boilers, much higher pressures can be used. Let’s take a typical value of 10 MPa. If this is all we do, we have a problem. If we expand saturated vapor at 10 MPa to 0.0035 MPa in a turbine, a very large liquid fraction would result. But there is no reason to restrict ourselves to saturated vapor. We can extract saturated steam from the boiler, but add more heat to it before sending it through the turbine. Doing this is known as superheating the steam, and all modern power plants employ superheat. The upper limit on how hot the steam may be heated is determined by the materials constraints of the turbine. Hot steam is corrosive, and if it is too hot

CHAPTER 9. POWER GENERATION AND PROPULSION

241

800

10 MPa

3

3
3000

600 0.0035 MPa

h [kJ/kg]

T [K]

2000

4 4s

400 2 1 200 4s 4

1000

2 1

0 0 2 4 6 8 10

0 0 2 4 6 8 10

s [kJ/kg-K]

s [kJ/kg-K]

Figure 9.6: A Rankin cycle with superheat. when it enters the turbine it will rapidly corrode the turbine blades. Advances in high temperature materials for turbine blades and elaborate cooling methods have resulted in allowable turbine inlet temperatures of more than 800 K. We’ll take this value for this example. Finally, we can work on improving the isentropic eﬃciencies of the pump and turbine, to more closely approach the ideal cycle. Large-scale turbines can now be manufactured with isentropic eﬃciency above 90%; let’s take a value of ηs,t = 0.9. The isentropic eﬃciency of the pump hardly matters, due to the low BWR, but these have gotten better too. We’ll take ηs,p = 0.8. The process representation is shown in Fig. 9.6. The analysis is very similar to what we’ve done in the last two examples, except that state 3 is now ﬁxed by the speciﬁed T3 and P3 . The numbers work out as shown below. State 1 2 3 4s 4 P (MPa) 0.003536 10 10 .003536 .003536 T (K) 300 300.72 800 300 300 h (kJ/kg) 111.7 124.3 3442.0 1999.7 2143.9 s (kJ/kg-K) 0.3900 0.3993 6.683 6.683 7.164 x 0 0 1 0.774 0.833

CHAPTER 9. POWER GENERATION AND PROPULSION
Q b Q Superheater s 3 4 Pump W p 1 Condenser Q c HP Turbine Q r

242

Boiler 2

Reheater 5 LP Turbine W 6

W

t,1

t,2

Figure 9.7: A Rankine cycle with superheat and one stage of reheat. Heat added: Turbine work: Pump work: Eﬃciency: BWR: 3317.7 kJ/kg 1298.1 kJ/kg 12.5 kJ/kg 38.7% 0.0096

These modiﬁcations have produced a marked increase in eﬃciency, and this value is in the ballpark of real steam power plant eﬃciencies. The work output per unit mass of steam has also signiﬁcantly increased. The steam quality leaving the turbine is too low, however, so we’re not done yet. 9.3.4 Addition of Reheat To improve the steam quality at the turbine exit, we need to move to the right on the T − s plot. We can do this and also increase the average temperature of heat addition by employing reheat, as shown in Fig. 9.7. Due to the large pressure range we’re expanding the steam through, the turbine must be constructed of multiple stages or else multiple separate turbines must be used (small ones for high pressure and progressively larger ones as the pressure becomes lower). In the reheat cycle, we expand the steam ﬁrst through a high pressure turbine to some intermediate pressure, extract it and add more heat, and then expand it through a second low-pressure turbine. The eﬀect of this is seen in the process representations (Fig. 9.8). The work output (the sum of both turbines) per unit mass of steam is increased, as is the eﬃciency. For this example, we will take the intermediate reheat pressure to be 1 MPa. Then carrying out an analysis similar to what we’ve done before, we ﬁnd

CHAPTER 9. POWER GENERATION AND PROPULSION

243

800

3

5

3

5

3000 4 4s

600 10 MPa
6 6s

h [kJ/kg]
6s 6

T [K]

4 4s 1 MPa 400 2 1 200 0.0035 MPa

2000

1000

2 1

0 0 2 4 6 8 10

0 0 2 4 6 8 10

s [kJ/kg-K]

s [kJ/kg-K]

Figure 9.8: A Rankin cycle with superheat and one stage of reheat. State 1 2 3 4s 4 5 6s 6 P (MPa) 0.003536 10 10 1 1 1 .003536 .003536 T (K) 300 300.72 800 471.2 497.6 800 300 300 h (kJ/kg) 111.7 124.3 3442.0 2822.7 2884.7 3536.4 2345.4 2464.5 s (kJ/kg-K) 0.3900 0.3993 6.683 6.683 6.811 7.835 7.835 8.232 x 0 0 1 1 1 1 0.916 0.965

Heat added in boiler / superheater: Heat added in reheater: HP Turbine work: LP Turbine work: Pump work: Eﬃciency: BWR: 9.3.5 Regeneration

3317.7 kJ/kg 651.7 kJ/kg 557.3 kJ/kg 1071.9 kJ/kg 12.5 kJ/kg 40.7% 0.0076

One remaining source of ineﬃciency is the heating of the water when it ﬁrst enters the boiler. Since the water is initially cold (300 K), the heat addition required to bring the water to the boiling point constitutes low-temperature

CHAPTER 9. POWER GENERATION AND PROPULSION
Q b Q Superheater s 3 Q r

244

8 Pump 2 W p,2 2 W p,1 1 7 1-f

Boiler

Reheater 5 LP Turbine W t,1 6 1-f

Open Feedwater Heater

HP Turbine

W 4 1-f

t,2

f Pump1 Condenser Q c

Figure 9.9: A Rankine cycle with superheat, one stage of reheat, and one open feedwater heater. heat addition. If we can eliminate some of the low-temperature heat addition, we can raise the average temperature at which heat is added, and thus the eﬃciency of the cycle. A technique known as regeneration uses heat from the hot steam to preheat the water going into the boiler (the “feedwater”). There are several ways to implement regeneration. A particularly simple way (which however only partially eliminates low-temperature external heat addition) is shown in Fig. 9.9. A portion f of the steam leaving the high-pressure turbine is extracted and simply mixed with the cold water before it enters the boiler. The remaining fraction 1 − f goes through the reheater, low-pressure turbine, and condenser as usual. The chamber where the mixing occurs is known as an open feedwater heater. Since the mixing must occur at the pressure at the outlet of the high-pressure turbine, this scheme requires two pumps, as shown in Fig. 9.9. The water which leaves the feedwater heater is pumped up to the boiler pressure in the second pump. Since it is now hotter than it would have been without the feedwater heater, less heat must be added in the boiler to reach the boiling point. Let’s consider the eﬀect of adding an open feedwater heater to the previous example. We will specify that just enough steam is extracted at state 4 to produce saturated liquid at the outlet of the feedwater heater (state 7). Most of the states remain the same as in example 4. The only changes are state 2 (P2 = 1 MPa now, not 10 MPa) and the two new states 7 and 8. State 7 is determined by our speciﬁcation that it is saturated liquid at 1 MPa, and state 8 is determined by analyzing pump 2, in the same way we found state 2 earlier. The properties in each state are listed below, and the T − s plot

CHAPTER 9. POWER GENERATION AND PROPULSION

245

800

3

5

600 10 MPa

T [K]

7,8 1 MPa 400 2

f

4 4s

1-f 1 200 0.0035 MPa 6s 6

0 0 2 4 6 8 10

s [kJ/kg-K]

Figure 9.10: A Rankin cycle with superheat, one stage of reheat, and one open feedwater heater. Note that only a fraction f of the total mass ﬂow rate ﬂows from 4 to 7, and a fraction 1 − f ﬂows in the loop 4-5-6-1. for this cycle is shown in Fig. 9.10. State 1 2 3 4s 4 5 6s 6 7 8 P (MPa) 0.003536 1 10 1 1 1 .003536 .003536 1 10 T (K) 300 300.01 800 471.2 497.6 800 300 300 453.06 454.92 h (kJ/kg) 111.7 113.0 3442.0 2822.7 2884.7 3536.4 2345.4 2464.5 762.5 775.2 s (kJ/kg-K) 0.3900 0.3915 6.683 6.683 6.811 7.835 7.835 8.232 2.138 2.144 x 0 0 1 1 1 1 0.916 0.965 0 0

Knowing all the states, we now need to calculate the extraction fraction f which is required to produce the desired state 7 (saturated liquid at 1 MPa). To ﬁnd the extraction fraction, consider an energy balance on the feedwater heater: m4 h4 + m2 h2 = m7 h7 . ˙ ˙ ˙ (9.14)

CHAPTER 9. POWER GENERATION AND PROPULSION

246

Note that we have to keep track of the mass ﬂow rate in each stream, since they are not equal. Dividing by m7 , ˙ fh4 + (1 − f)h2 = h7 . Solving for f, f= h7 − h2 h4 − h2 (9.15)

(9.16)

which works out to f = 0.234. Now we can calculate the energy transfer for each component. For the boiler, superheater, and pump 2, the full m ﬂows through them, and we calculate ˙ the work per unit mass in the usual way. But note that only (1 − f)m ﬂows ˙ through pump 1, the reheater, and turbine 2. It is most convenient to express all energy transfers as the power divided by the total m (not the actual mass ﬂow ˙ rate through a given component). Therefore, we must multiply the enthalpy diﬀerence by the fraction of the mass ﬂowing through a given component: Qb Wp,2 Wt,1 Qr Wp,1 Wt,2 = h3 − h8 = h8 − h7 = h3 − h4 = (1 − f)(h5 − h4 ) = (1 − f)(h2 − h1 ) = (1 − f)(h5 − h6 ) (9.17) (9.18) (9.19) (9.20) (9.21) (9.22)

Evaluating these expressions, we ﬁnd Heat added in boiler / superheater: Heat added in reheater: Turbine work 1: Turbine work 2: Pump work 1: Pump work 2: Extraction fraction: Eﬃciency: BWR: 2666.8 kJ/kg 499.0 kJ/kg 557.3 kJ/kg 820.7 kJ/kg .96 kJ/kg 12.68 kJ/kg 23.4% 43.1% 0.0099

Using one open feedwater heater is seen to increase the eﬃciency of the cycle from 40.7% to 43.1%. While this may seem not much considering the added complexity, in fact a gain of 2.4 percentage points in eﬃciency is quite signiﬁcant, and well-worth any added costs due to complexity in even a moderately-sized power plant.

CHAPTER 9. POWER GENERATION AND PROPULSION

247

A typical real power plant would use multiple feedwater heaters, with steam extracted from the turbine at various pressures, in order to achieve even more regeneration. For example, the largest unit at the Pasadena power plant makes use of 5 feedwater heaters.

9.4 Gas Power Cycles
A simple gas power cycle is shown below. A steady ﬂow of air is compressed and enters a combustor, where fuel is injected into the high-pressure air stream and the fuel/air mixture burns. The hot products of combustion1 expand through ˙ a turbine, producing power Wt . Some fraction of the turbine power is used ˙ ˙ ˙ to drive the compressor, and the remaining power Wnet = Wt − Wc may be delivered to an external load.
Fuel Combustor Compressor Turbine

Wc Air

.

W net Exhaust

.

Variations on this basic gas turbine cycle are widely used both for power generation and propulsion. Aircraft jet engines use a modiﬁed version of this cycle, which we will discuss in more detail shortly. Gas turbines have become increasingly popular for ground-based power generation also, as the eﬃciency of these cycles has improved. They are particularly attractive as topping cycles in conjunction with vapor power cycles, or when relatively small, modular power generation units are needed. The two most important performance parameters for a gas turbine cycle are the thermal eﬃciency η and the net work developed per unit mass of air ﬂow, ˙ Wnet = Wnet /mair . The eﬃciency determines the fuel economy; a more eﬃcient ˙ gas turbine cycle for electric power generation consumes less fuel per kW-hr of electricity generated. This lowers both the cost of electricity and the pollutant emissions. For an aircraft engine, higher eﬃciency translates into lower cost, longer range, and lower pollution. Wnet is especially important for transportation applications, since the size
1 for lean combustion, a mixture composed primarily of nitrogen, oxygen, water vapor, carbon dioxide, and smaller amounts of other species, including pollutants such as oxides of nitrogen (NOx ).

CHAPTER 9. POWER GENERATION AND PROPULSION

248

and weight of the engine scale approximately with the air ﬂow rate. For a ﬁxed power output requirement, a cycle with higher Wnet will be lighter and smaller. Gas turbine engines are almost always run with signiﬁcantly more air than required to burn all of the fuel injected (so-called “lean combustion”). A major reason lean combustion is used is to limit pollutant emissions. By burning lean, combustion is complete, so there is no unburned fuel vapor in the exhaust, nor is there much CO. Also, lean combustion limits the temperature of the gas leaving the combustor. This is important both to minimize the production of smogproducing oxides of nitrogen (NOx ) and to protect the turbine blades, which cannot survive excessive temperatures. A complete analysis of a gas turbine cycle would require considering the combustion process, including calculating the amount of heat released, and the composition of the combustion products. A design of a real gas turbine cycle would consider this in some detail, with particular focus on pollutant formation, since the emissions from both ground-based gas turbines and aircraft engines are strictly regulated by government agencies. 9.4.1 The air-standard Brayton cycle We can learn a lot about gas turbine cycles by considering a simple model which captures their essential features, without requiring full consideration of the combustion process. Since the gas mixture is mostly air, even downstream of the combustor, as a ﬁrst approximation we may take the gas to be air throughout the cycle. In this approximation, the combustor is modeled as an air heater, with the heat of combustion now an external heat input. Since in this airstandard cycle the ﬂuid is the same at the turbine exhaust as at the compressor inlet, we may close the cycle by adding a cooler, which removes heat from the air to the environment. The cycle which results from these approximations is called the air-standard Brayton cycle, shown in Fig. 9.11. It is also called the simple Brayton cycle, to distinguish it from modiﬁed versions we will discuss below. To analyze the air-standard Brayton cycle, we will assume the compressor and turbine are both adiabatic, with isentropic eﬃciencies ηs,c and ηs,t , respectively. We will neglect any pressure drop in the heater and cooler, so these are modeled as constant-pressure heat addition or removal processes. With these assumptions, the T -s process representation is as shown in Fig. 9.11. Let us put in some “typical” numbers. Suppose the inlet to the compressor is air at 1 atm and 300 K, the pressure ratio P ∗ = P2 /P1 across the compressor is 4, and the temperature at the inlet to the turbine is ﬁxed at 800 K by materials

CHAPTER 9. POWER GENERATION AND PROPULSION
. Q 2 Compressor

249

h 3 Turbine . W net 4 Exhaust

Heater

. Wc 1 Cooler . Q c

P

3 T 4s 4 2s 2 1 s

P

Figure 9.11: The air-standard Brayton cycle and T -s process representation. constraints. The compressor has isentropic eﬃciency ηs,c = 0.7, and the turbine has isentropic eﬃciency ηs,t = 0.8. To determine the performance of this cycle, we ﬁrst must ﬁnd all states, and their properties. The procedure is very much like what we did to analyze vapor power cycles: start with a known state (state 1), and simply work around the loop. Since this procedure is now familiar to you, we will simply summarize the set of relations needed in Table 9.1. The equations in Table 9.1 may be easily implemented in Excel, using TPX to calculate the properties. Since TPX does not calculate properties for air, we will calculate properties for nitrogen, which of course is the major constituent (78%) of air, with most of the remainder being oxygen. For the parameter values speciﬁed above, the states are as follows: state 1 2s 2 3 4s 4 T (K) 300 445.5 507.1 800 548.2 599.6 P (atm) 1 4 4 4 1 1 h (kJ/kg) 461.5 613.0 677.9 996.4 721.6 776.6 s (kJ/kg-K) 4.41 4.41 4.55 5.05 5.05 5.14

CHAPTER 9. POWER GENERATION AND PROPULSION

250

Table 9.1: Relations needed to solve for the states in a simple air-standard Brayton cycle. state T P h s 1 2s 2 3 4s 4 given T (s, P ) T (h, P ) given T (s, P ) T (h, P given P2 given P2 P1 P1 h(T, P ) h(s, P ) h1 + (h2s − h1 )/ηs,c h(T, P ) h(s, P ) h3 − ηs,t(h3 − h4s ) s(T, P ) s1 s(h, P ) s(T, P ) s3 s(h, P )

The turbine work output per kg of air is given by Wt = h3 − h4 = 219.8 kJ/kg. The compressor work input is Wc = h2 − h1 = 216.4 kJ/kg. We see that almost all of the turbine work output must be used simply to run the compressor! Only 3.4 kJ/kg is left over to deliver to an external load. In terms of the back work ratio, BWR = 0.98. This is in sharp contrast to the results we obtained for vapor power cycles, which had BWR < 0.01. This is a major diﬀerence between these two classes of cycles. The heat input for this example is Qh = h3 − h2 = 318.5 kJ/kg, so the overall thermal eﬃciency is η = (Wt − Wc )/Qh = 0.011. Therefore, this cycle only converts 1.1 % of the heat input into useful work — terrible performance, considering that the Carnot eﬃciency evaluated between the minimum and maximum temperatures in this cycle is 1 − 300/800 = 0.625, or 62.5%. Actually, the numbers used in this example are characteristic of the state of the art several decades ago, when gas turbine engines were ﬁrst being developed. Due to intensive research, it is now possible to build compressors with isentropic eﬃciency in the range 0.8–0.9, and turbines with isentropic eﬃciency as high as 0.95. If we re-work the above example with, say, ηs,c = 0.85 and ηs,t = 0.9, we ﬁnd Wnet = 69.1 kJ/kg and η = 0.19 — quite a substantial improvement.

CHAPTER 9. POWER GENERATION AND PROPULSION
Wnet
0.6 125 0.5 100 75 50 25 0 0 5 10 P* 15 20

251

Efficiency

ideal
0.4 0.3 0.2 0.1 0 0

ideal

5

10 P*

15

20

Figure 9.12: Net work per kg of air and eﬃciency vs. P ∗ . T1 = 300 K, P1 = 1 atm, T3 = 800 K. The dashed curves are for an ideal cycle, and the solid curves for ηs,c = 0.85, ηs,t = 0.9. Eﬀect of Pressure Ratio on Performance The eﬀect of pressure ratio P ∗ = P2 /P1 on Wnet and on η for these conditions is shown in Fig. 9.12. For an ideal cycle with ηs,c = ηs,t = 1 (dashed curves), the net work per kg of air is maximized for P ∗ near 6, while the eﬃciency continues to increase with P ∗. Taking the more realistic values ηs,c = 0.85, ηs,t = 0.9, the solid curves are obtained. In this case, the net work is maximized near P ∗ = 4. The eﬃciency reaches a maximum of about 20% near P ∗ = 5, and falls oﬀ at higher P ∗ , unlike the ideal case. A T -s plot for this cycle for P ∗ = 5 is shown in Fig. 9.13. The dependence of Wnet and η on P ∗ shown in Fig. 9.12 for the ideal case can be explained as in Fig. 9.14. For the ideal cycle, the enclosed area is Wnet . Since Tmax = T3 is ﬁxed, the area is maximized for an intermediate pressure ratio. The eﬃciency of an ideal cycle, on the other hand, is given by η = 1 − Tc/Th . This is an increasing function of P ∗ , and approaches the Carnot eﬃciency 1 − T1 /Tmax ∗ as P ∗ approaches Pmax , which is the pressure ratio at which the air emerges from the compressor at Tmax . The Eﬀect of Turbine Inlet Temperature on Performance Another important development in the last couple of decades has been in advanced high-temperature materials for turbine blades, and development of blade cooling techniques. In the most advanced current gas turbines, turbine inlet temperatures up to 1700 K can be tolerated. If we recompute the above cycle performance using, say, T3 = 1400 K, we ﬁnd BWR = 0.41, Wnet = 261.4 kJ/kg,

CHAPTER 9. POWER GENERATION AND PROPULSION

252

800

3
P= 5 at m

600

2s
T [K]
400

2 4s 4
P tm =1a

1
200

0 4.4 4.6 4.8 5.0

s [kJ / kg-K]

Figure 9.13: Process representation for simple Brayton cycle with T1 = 300 K, P1 = 1 atm, P ∗ = 5, T3 = 800 K, ηs,c = 0.85, ηs,t = 0.9.

Tmax

high P*

medium P*

low P*

Wnet

s
Figure 9.14: An intermediate P ∗ maximizes the enclosed area Wnet for ﬁxed Tmax .

CHAPTER 9. POWER GENERATION AND PROPULSION
Wnet
500 400 300 0.3 200 0.2 100 0 0.1 0 0.6

253

Efficiency ideal 0.5 0.4

ideal

0

5

10 P*

15

20

0

5

10 P*

15

20

Figure 9.15: Net work per kg of air and eﬃciency vs. P ∗ . T1 = 300 K, P1 = 1 atm, T3 = 1400 K. The dashed curves are for the ideal cycle, and the solid curves for ηs,c = 0.85, ηs,t = 0.9. and η = 0.25. For this higher value of T3 , the eﬀect of pressure ratio is as shown in Fig. 9.15. The optimal pressure ratio is higher — a value of P ∗ = 10 would be a reasonable choice. This choice would result in an eﬃciency of 34.9%, and Wnet = 315.3 kJ/kg. The T -s plot for this case is shown in Fig. 9.16. Analysis assuming ideal gas with constant cp Since the peak pressure in a gas-turbine cycle is typically not too high (say 10 atm), air-standard cycles are often analyzed by treating air as an ideal gas. For even greater simplicity, the speciﬁc heat is often assumed constant. The ideal gas approximation should be fairly good at these pressures (cf. Fig. 4.11), but the assumption of constant cp is more questionable, due to the large temperature variation around the cycle. Let’s examine how much error is made for the conditions of Fig. 9.15. If we assume an ideal gas with constant cp , then h(T ) − h(T0 ) = cp (T − T0 ) and s(T, P ) − s(T0 , P0 ) = cp ln T T0 P P0 (9.23)

− R ln

.

(9.24)

Therefore, for the isentropic steps such as 1 → 2s, cp ln T2s T1 = R ln P2s P1 . (9.25)

Since R/cp = (cp − cv )/cp = 1 − cv /cp and P2s /P1 = P ∗, this equation may be

CHAPTER 9. POWER GENERATION AND PROPULSION
1500 1250 1000
P= 10 atm

254

3

T [K]

750 2s 500 250 0 1

4 2
P= 1 at m

4s

4.50

4.75

5.00

5.25

5.50

s [kJ / kg-K]

Figure 9.16: Process representation for a simple Brayton cycle with T1 = 300 K, P1 = 1 atm, P ∗ = 10, T3 = 1400 K, ηs,c = 0.85, ηs,t = 0.9. Performance: Wnet = 315.3 kJ/kg, η = 0.349. written T2s = T1 (P ∗ )1−1/k , where the speciﬁc heat ratio k is deﬁned as cp k= . cv Similarly, T4s = T3 (1/P ∗)1−1/k . (9.28) These expressions may be used instead of TPX to solve for the temperatures and enthalpies in Table 9.1. For nitrogen at room temperature, cp = 1.04 kJ/kg, and k = (7/2)/(5/2) = 1.4 (for air, cp = 1.004 kJ/kg, and k is also 1.4). If we re-do the cycle analysis as a function of P ∗ with the ideal gas, constant cp approximation, we come up with the dashed curves in Fig. 9.17, which are compared to the results using TPX (solid curves). The approximation is seen to be not too bad; certainly the trends with P ∗ are well-predicted. The major eﬀect is an underprediction of Wnet by roughly 10%, and a slight over-prediction of η. 9.4.2 The Regenerative Brayton Cycle Although the cycle of Fig. 9.16 is far more eﬃcient than the one we ﬁrst considered (34.9% vs. 1.1%), more improvements are still possible. Note that due to the high turbine inlet temperature of 1400 K, the turbine exhaust (state 4) is still quite hot — 847 K. Since this temperature is higher than the air temperature at the exit of the compressor (state 2; 624 K), it should be possible to (9.26)

(9.27)

CHAPTER 9. POWER GENERATION AND PROPULSION

255

500 400

Wnet

0.5 0.4

Efficiency ideal gas, constant cp TPX

TPX kJ / kg

300 200 100 0

0.3 0.2 0.1 0

ideal gas, constant cp

0

5

10 P*

15

20

0

5

10 P*

15

20

Figure 9.17: Comparison of results calculated assuming ideal gas with constant cp to results calculated using TPX, for cycle conditions of Fig. 9.15.

Compressor . Wc

Regenerator 3 2 . Qr 6 5

. Qh Heater 4 Turbine . W net

1

Cooler . Qc

Figure 9.18: The regenerative air-standard Brayton cycle.

CHAPTER 9. POWER GENERATION AND PROPULSION

256

Table 9.2: Relations needed to solve for the states in a regenerative air-standard Brayton cycle. state T P h s 1 2s 2 3 4 5s 5 6 given T (s, P ) T (h, P ) T (h, P ) given T (s, P ) T (h, P ) T (h, P ) given P2 given given P2 P1 P1 P1 h(T, P ) h(s, P ) h1 + (h2s − h1 )/ηs,c h2 + r (h5

s(T, P ) s1 s(h, P ) s(h, P ) s(T, P ) s4 s(h, P ) s(h, P ) − h2 )

h(T, P ) h(s, P ) h4 − ηs,t(h4 − h5s ) h5 − (h3 − h2 )

use the turbine exhaust to pre-heat the air entering the heater, thus requiring less external heat input Qh to reach 1400 K. This can be done using a heat exchanger, as shown in Fig. 9.18. Since the work output is unchanged by this modiﬁcation, the eﬃciency Wnet /Qh will increase. The heat exchanger, called the regenerator, transfers heat from the hot turbine exhaust to the air emerging from the compressor. The heat transfer in the regenerator per kg of air is Qr = h3 − h2 = h5 − h6 . (9.29) (9.30)

If the regenerator were inﬁnitely long, the air leaving it at state 3 would have the same temperature as the hot air entering at state 5. Thus, the most heat transfer which could occur in the regenerator is given by Qr,max = h5 − h2 . (9.31)

It is impractical to build a regenerator so large that T3 closely approaches T5 — in practice, T3 will always be less than T5 , and therefore Qr < Qr,max . The regenerator eﬀectiveness is deﬁned by r =

Qr Qr,max

=

h3 − h2 . h5 − h2

(9.32)

A typical realistic value for r would be about 0.8. With r speciﬁed, this equation may be used to solve for h3 . The complete set of relations needed to solve the regenerative air-standard Brayton cycle is given in Table 9.2.

CHAPTER 9. POWER GENERATION AND PROPULSION Wnet ideal 257

Efficiency
1.0 0.8 0.6 0.4 0.2 0

500 400 300 200 100 0

ideal

kJ / kg

η

0

5

10 P*

15

20

0

5

10 P*

15

20

Figure 9.19: Net work per kg of air and eﬃciency vs. P ∗ . T1 = 300 K, P1 = 1 atm, T3 = 1400 K, r = 0.8. The dashed curves are for the ideal cycle, and the solid curves for ηs,c = 0.85, ηs,t = 0.9. If we add a regenerator with r = 0.8 to the simple Brayton cycle of Fig. 9.15, and use TPX to calculate the state properties, we ﬁnd the performance characteristics shown in Fig. 9.19. The net work is unchanged from Fig. 9.15, but the eﬃciency is much diﬀerent. For the ideal case ( r = ηs,c = ηs,t = 1), the eﬃciency is maximized at P ∗ = 1 (although of course Wnet is zero there). For the non-ideal case, the eﬃciency is maximized near P ∗ = 5, at a value of 46.9%. This is much better than the simple Brayton cycle, which from Fig. 9.15 has an eﬃciency of 27.5% at P ∗ = 5. Adding a regenerator is seen to produce a big increase in eﬃciency, particularly at lower P ∗ values, where the diﬀerence (T5 − T2 ) is largest. Based upon the performance results shown in Fig. 9.19, the best choice for P ∗ would likely be between 5 and 10, depending on whether maximizing Wnet or η was more important. The T -s plot for P ∗ = 10 is shown in Fig. 9.20. Note that this is the same as Fig. 9.16, except for the location of the state points. 9.4.3 Intercooling and Reheat Two more modiﬁcations are often used. In order to reduce the compressor work requirement, the compression process may be broken up into two or more stages, with heat removal between stages. This process is known as intercooling.

CHAPTER 9. POWER GENERATION AND PROPULSION
1500 1250 1000
4

258

1 P=
3 2

0a

tm

T (K)
750
2s

5 5s

500 250 0 4.50 4.75
1

P=1

atm

6

5.00

5.25

5.50

s (kJ / kg-K)

Figure 9.20: Process representation for a regenerative Brayton cycle with T1 = 300 K, P1 = 1 atm, P ∗ = 10, T3 = 1400 K, ηs,c = 0.85, ηs,t = 0.9, r = 0.8. Performance: Wnet = 315.3 kJ/kg, η = 0.447.

1

2

Intercooler Qi

That this lowers the work requirement is easiest to see by considering an and h-s process representation of an ideal, 2-stage compressor (below). With no heat removal between stages, the total compressor work to compress from P1 to P2 is WA + WB . With intercooling the work is WA + WC . Since constant-pressure lines on an h-s plot diverge (slope = T ), WC < WB and thus the compressor work is lower with intercooling.
WB h WC W A

s

Reheat is the same process as for vapor power cycles — heat is added between turbine stages. Both intercooling and reheat increase Wnet , but may or may not

CHAPTER 9. POWER GENERATION AND PROPULSION

259

Wnet
1.0 0.8 0.6 400 0.4 200 0.2 0 0 5 10 15 20 0 5

Efficiency

600

kJ / kg

0

10

15

20

P*

P*

Figure 9.21: Net work per kg of air and eﬃciency vs. P ∗ for regenerative Brayton cycle with intercooling and reheat. improve the eﬃciency. If they are used together with regeneration, signiﬁcant gains in eﬃciency are possible, since they increase the potential for regeneration. In Fig. 9.21, the performance of a regenerative cycle with one stage of intercooling and one stage of reheat is presented. Most of the parameters are the same as in Fig. 9.19. The intercooler eﬀectiveness is 0.8, and the reheat temperature is 1400 K. The intercooling and reheat pressures are both taken to be (P ∗ )1/2 . (For an ideal cycle and assuming constant cp , it can be shown this is the optimal value.) It can be seen that the performance is quite good. The net work is above 400 kJ/kg, and above P ∗ = 5 the eﬃciency is slightly greater than 50%.

CHAPTER 9. POWER GENERATION AND PROPULSION

260

1500
6

8

1250 1000
5 7 9s 9

7s

T (K) 750
4s 4 2s 3 2 10

500 250 0 4.0

1

4.5

5.0

5.5

6.0

s (kJ/kg-K)

Figure 9.22: Process representation for a regenerative Brayton cycle with intercooling and reheat. T1 = 300 K, P1 = 1 atm, P ∗ = 10, Pi = Pr = 3.16 atm, T3 = 1400 K, ηs,c = 0.85, ηs,t = 0.9, r = i = 0.8.

CHAPTER 10
CHEMICAL REACTIONS AND CHEMICAL EQUILIBRIUM 10.1 Introduction
(NOTE: This chapter is incomplete; consult the hand-written notes I passed out in class.) Chemically-reacting ﬂows are vital to many engineering applications, including almost all aerospace and terrestrial propulsion systems, and all systems which produce power by burning fossil fuels. Solving current technological problems, such as how to reduce undesirable pollutant emission from these devices, requires an understanding of some of the chemistry occurring in these systems.

10.2 The Reaction A + B

AB

Consider a mixture of 2 chemically-reactive elemental species, A and B. Suppose that these can react with one another to form a molecule AB: A + B → AB. Once an AB molecule is formed, it may break up again via AB → A + B. We can write a short-hand notation for these two reactions as follows: A+B AB. (10.3) (10.2) (10.1)

Suppose we introduce NA,0 moles of A and NB,0 moles of B into a container with volume V at time t = 0. We allow the system to evolve with no interaction with the surroundings (isolated). As time proceeds, we ﬁnd that the mole numbers change, due to formation of AB molecules. If we wait long enough, the ∗ ∗ mole numbers come to steady-state values, which we will denote NA , NB , and ∗ NAB . Evidently, once steady state is reached, the rate at which AB is formed by reaction (10.1) just balances the rate at which it is destroyed by reaction (10.2). When this condition is reached, we say the system is in chemical equilibrium. Depending on the nature of the chemical reaction, and also on the temperature 263

CHAPTER 10. CHEMICAL REACTIONS AND CHEMICAL EQUILIBRIUM264 and pressure, chemical equilibrium may be reached in microseconds (explosions) or may take thousands of years. Determining the rate at which the reaction proceeds is the subject of chemical kinetics. However, the ﬁnal state achieved at long times does not depend on kinetics, but only on the thermodynamic properties of the atoms or molecules involved. To determine the condition for chemical equilibrium, we apply the same basic principle we have used to determine other equilibrium conditions: the system will attain the state for which the entropy is maximal, compared to all other possible states with the same total U , V , NA,0 , and NB,0 . In our previous discussions of equilibrium, we have imagined a system divided by a partition, which can pass energy, volume, or molecules. For the present problem, the “constraint” we will relax is somewhat diﬀerent. Let us imagine that we can turn a knob on the system which can (somehow) stop the reaction from occurring, or can run it in the forward direction [reaction (10.1)] or in the reverse direction [reaction (10.2)]. We will turn the knob back and forth, calculating the entropy of the system for each set of conditions, and ﬁnd the state which maximizes S. This then must be the equilibrium state the system will come to when “unconstrained.” Suppose we initially set the knob so that the reaction is stopped, and the mole numbers are NA , NB , and NAB . We now turn the knob to allow the reaction to proceed inﬁnitesimally in the forward direction, and then reset it to stop the reaction at this new condition. We ﬁnd that NAB has increased by an amount dNAB . Since the only way to increase NAB is by consuming an equivalent amount of A and B atoms, we must have that dNA dNB = −dNAB = −dNAB . (10.4) (10.5)

How much has S(U, V, NA , NB , NAB ) changed? Using the chain rule, dS = ∂S ∂NA + dNA +
U,V,NB ,NAB

∂S ∂NB

dNB
U,V,NA ,NAB

∂S ∂NAB

dNAB .
U,V,NA,NB

(10.6)

We know that the chemical potential is deﬁned by ∂S ∂Ni =−
U,V,Nj

µi . T

(10.7)

CHAPTER 10. CHEMICAL REACTIONS AND CHEMICAL EQUILIBRIUM265 Therefore, if we also use equations (10.4) and (10.5) to substitute for dNA and dNB , Eq. (10.6) becomes dS = 1 [µA + µB − µAB ] dNAB . T (10.8)

At the equilibrium state, S will be maximal with respect to NAB , and therefore dS = 0. The chemical equilibrium condition for this reaction is then µAB − µA − µB = 0. (10.9)

This equilibrium condition is very general, and applies whatever type of solution the species A, B, and AB form. For example, it could be a liquid or even a solid solution. By using the form for the chemical potential appropriate for the phase of the mixture, we can develop an explicit condition for equilibrium in terms of experimental parameters, such as the mole numbers, temperature, and pressure. It is worth noting that, although we imagined an isolated system to derive Eq. (10.9), now that we have this equation we no longer need to restrict our attention only to isolated systems. The condition expressed by Eq. (10.9) must be satisﬁed at the equilibrium state, no matter how this state was reached. It could have been reached in an isolated system (constant U and V ) or, for example, by allowing the reaction to proceed at constant T and P , adding or removing heat as necessary to keep T constant, and allowing the volume to change to keep P constant.

10.3 Reactions in Ideal Gases
If the mixture of A, B, and AB is an ideal gas mixture, then we have already derived the form of the equation for the chemical potential. In this case, recall that Pi ˆ ˆ ˆ (10.10) µi = g(T, Pi ) = g0 (T ) − RT ln P0 That is, the chemical potential of species i in the mixture is the molar Gibbs free energy (of the pure species i) evaluated at the partial pressure Pi . Substituting this expression into Eq. (10.9), we have ln or (PA /P0 )(PB /P0 ) g0 B(T ) − gA (T ) − gB (T ) ˆ ˆ0 ˆ0 =− A ˆ (PAB /P0 ) RT
0 PA PB ˆ = P0 e−∆G (T )/RT , PAB

(10.11)

(10.12)

CHAPTER 10. CHEMICAL REACTIONS AND CHEMICAL EQUILIBRIUM266 where ∆G0 (T ) = gAB (T ) − gA (T ) − gB (T ) ˆ0 ˆ0 ˆ0 (10.13) is the standard free energy of reaction for this reaction. We can re-write this equation in terms of mole numbers as
∗ ∗ NA NB = ∗ NAB

P0 V ˆ RT

e−∆G

0

(T )/RT

.

(10.14)

∗ ∗ This equation provides us with one equation for the 3 unknowns NA , NB , and ∗ NAB . The remaining 2 equations come from the atom conservation conditions: ∗ ∗ NA + NAB = NA,0

(10.15)

and
∗ ∗ NB + NAB = NB,0 .

(10.16)

These equations simply state that the number of moles of A atoms (independent of whether they appear as free A atoms or in AB molecules) is the same as initially placed in the container, and similarly for B. These three equations together ∗ ∗ ∗ allow solving uniquely for the values of NA , NB , and NAB at equilibrium.

10.4 A General Reaction
The above results are easily generalized to any chemical reaction. Let’s consider a reaction such as 2H2 + O2 2H2 O. (10.17) We may write this reaction (or any other one) symbolically as νj Aj = 0, j (10.18)

where Aj stands for the chemical symbol of the jth species (e.g., “H2 ”, “O2 ”, or “H2 O”) and νj is the stoichiometric coeﬃcient for species j. The stoichiometric coeﬃcient is deﬁned to be positive for products (H2 O) and negative for reactants (H2 and O2 ). Therefore, for this reaction ν(H2 ) ν(O2 ) ν(H2 O) = −2 = −1 = +2 (10.19) (10.20) (10.21)

To ﬁnd the equilibrium condition for a general reaction such as reaction (10.18), we again maximize the entropy at ﬁxed U and V as we did above. For

CHAPTER 10. CHEMICAL REACTIONS AND CHEMICAL EQUILIBRIUM267 the general case, the stoichiometric relation among the mole numbers as the reaction proceeds forward or backward is: dNj ˜ = dN. νj (10.22)

That is, each mole number changes by an amount equal to νj multiplied by ˜ the change in a species-independent “extent of reaction” variable N . Note that for the simple A + B AB reaction considered above, Eq. (10.22) reduces to equations (10.4) and (10.5). ˜ The condition that S is maximal with respect to N then leads to r µj νj = 0. j=1 (10.23)

This is the general form of the condition for chemical equilibrium for a single reaction. (Note that νj is non-zero only for those species in the mixture which participate in the reaction.) For ideal gases, we can again substitute in the known form of the chemical potential. This leads us to the generalization of Eq. (10.12): r j=1

Pj P0

νj

= e−∆G

0

ˆ (T )/RT

,

(10.24)

where ∆G (T ) =
0

r

gj (T )νj ˆ0 j=1 (10.25)

is again the standard free energy of reaction. Equation (10.24) is known as the law of mass action. Since the right-hand side of Eq. (10.24) is a function only of temperature, it is convenient to give it a symbol. We deﬁne the equilibrium constant Kp (T ) as Kp (T ) = e−∆G
0

ˆ (T )/RT

.

(10.26)

We can write the law of mass action in terms of mole fractions, since Pj = Xj P . Making this substitution into Eq. (10.24), we have r ν Xj j j=1 r =

P0 P

j=1

νj

Kp (T ).

(10.27)

We can also write the law of mass action in terms of mole numbers or concentrations Cj = Nj /V as follows: r ν Cj j j=1 r =

P0 ˆ RT

j=1

νj

Kp (T ),

(10.28)

CHAPTER 10. CHEMICAL REACTIONS AND CHEMICAL EQUILIBRIUM268 and r ν Nj j j=1

r

=

P0 V ˆ RT

j=1

νj

Kp (T ).

(10.29)

All of the forms (10.27), (10.28), and (10.29) of the law of mass action have an additional factor in front of Kp (T ), which arises due to the conversion from pressure to mole fraction, concentration, or mole number units. The exponent this factor is raised to is the net change in mole numbers as the reaction proceeds from pure reactants to pure products. For example, for the reaction A+B AB, r C + D, j=1 νj = −1. On the other hand, for a reaction of the type A + B this sum is zero, and the factor in front of Kp (T ) does not appear. Remember that P0 is the reference pressure at which the thermodynamic properties are computed, not the actual pressure of the problem. It might seem odd that P0 appears in the law of mass action, since the equilibrium composition can’t depend on our choice of the reference pressure. In fact, it does not, since ∆G0(T ) depends on P0 in just such a way that it cancels the P0 ˆ factor multiplying Kp (T ). (Recall that g(T, P ) has an RT ln P term in it.) ˆ If the reference pressure P0 = 1 atm, then P0 300 = 40.62 ˆ T RT moles/m3 . (10.30)

Let us see how this works in practice by looking at a simple example. Example 10.1 Suppose oxygen gas (O2 ) is heated to 2500 K. We wish to know the mole fraction of atomic oxygen O which forms at chemical equilibrium. The reaction forming O is O2 2O. (10.31) Therefore, ∆G0 = 2ˆ(O) − g(O2 ). g ˆ (10.32) We can look up the molar Gibbs functions in a table such as the JANAF tables, or calculate them from polynomial ﬁts (using g = ˆ − T s). Doing this, we ﬁnd ˆ h ˆ that for T = 2500 K, g(O) ˆ g(O2 ) ˆ Therefore, ∆G0 (2500) = 176.41 kJ/mole and Kp (2500) = 2.06 × 10−4 . (10.35) (10.36) = −219.13 kJ/mole = −614.67 kJ/mole. (10.33) (10.34)

CHAPTER 10. CHEMICAL REACTIONS AND CHEMICAL EQUILIBRIUM269 The law of mass action then becomes XO 2 = XO2 P0 P
+1

Kp (T ).

(10.37)

The atom conservation condition, expressed in mole fraction units, is simply XO + XO2 = 1. Therefore, XO 2 = 2.06 × 10−4 1 − XO P0 P . (10.39) (10.38)

We can solve this simple quadratic equation for the atomic oxygen mole fraction if the pressure is speciﬁed. As P increases, the atomic oxygen mole fraction decreases. For example, at P = 1 atm at 2500 K, we have that XO = 0.0142. But at P = 0.01 atm, XO = 0.133. The pressure dependence seen in Example 10.1 occurs because the mole numbers change by +1 in this reaction. A reaction of the type A + B C + D, in which the mole numbers do not change, would have no pressure dependence to the equilibrium mole fractions, since the term (P0 /P ) in the law of mass action would not appear (the exponent would be zero). This is one example of Le Chatlier’s Principle, which states that when a change is imposed on a system (such as increasing the pressure), the system will respond in such a way to try to counteract the change. In this case, as the pressure is increased, some O atoms recombine to O2 in order to reduce the total number of moles, and therefore reduce the system pressure compared to what it would be if the composition remained constant.

CHAPTER 11
ONE-DIMENSIONAL COMPRESSIBLE FLOW 11.1 Introduction
Thermodynamics plays an important role in ﬂuid mechanics, particularly when the ﬂow speed is high (comparable to the speed of sound). High-speed ﬂow in gases is called compressible ﬂow, since it turns out that only in this case does the density vary much from point to point in the ﬂow; if the velocity is much less than the speed of sound in a gas, then the density is nearly constant everywhere. In this chapter, we introduce some of the basic ideas of compressible ﬂow.

11.2 The Momentum Principle
To analyze ﬂuid ﬂow, we need a principle from mechanics to supplement the laws of thermodynamics. This is the momentum principle. When ﬂuid enters or leaves a control volume, it carries momentum with it. Since a packet of ﬂuid of mass δm moving with velocity V has momentum (δm)V , the speciﬁc momentum (momentum per unit mass) of the ﬂuid packet is simply equal to the velocity V . Therefore, if ﬂuid is ﬂowing into a control volume with a mass ﬂow rate of m kg/s and each kg brings momentum V with it, the momentum ˙ inﬂow rate is mV . ˙ The momentum principle is simply Newton’s second law applied to a control volume. It states that the net force on a control volume is equal to the net rate of momentum outﬂow from the control volume, plus the rate of change of momentum stored inside. The momentum stored inside the control volume only

V1 . m F . m V2

Figure 11.1: Flow in a curved tube. 270

CHAPTER 11. ONE-DIMENSIONAL COMPRESSIBLE FLOW

271

changes in unsteady ﬂows, which we will not consider here. Consider the steady ﬂow in a curved tube shown in Fig. 11.1. The ﬂow is assumed to be one-dimensional at points 1 and 2, which means that the velocity is constant over any cross-section and is oriented perpendicular to the crosssection. In this ﬂow, the momentum inﬂow rate is mV1 , and the momentum ˙ outﬂow rate is mV2 . The net momentum outﬂow rate is m(V2 − V1 ). Therefore, ˙ ˙ the momentum principle states that the net force F on the control volume is F = m V2 − V1 . ˙ (11.1)

Note that F is the resultant of all forces applied to the control volume. The momentum principle doesn’t tell us anything about where these forces appear, just their vector sum. In general, the forces on the control volume result both from pressure in the ﬂuid where the control surface intersects the ﬂuid, and from stresses in the tube walls where the control surface intersects the walls.

11.3 The Speed of Sound
Sound waves are small pressure disturbances, which travel through a ﬂuid with a characteristic speed. As long as the wavelength of the sound wave λ is long compared to the mean free path in the ﬂuid,1 the speed of propagation does not depend on the wavelength. (If you listen to a concert from far away the high notes and the low notes both reach you at the same time.) To calculate the speed of sound, consider the situation shown in Fig. 11.2. A piston at the end of a long tube is suddenly displaced a small distance to the left, locally compressing the gas next to it. This launches a wave down the tube, which travels at the speed of sound. If the piston displacement is very sudden, then the density and pressure will change abruptly across the wavefront, from the undisturbed values ρ and P to the perturbed values ρ + ∆ρ and P + ∆P behind the wave. (If the piston were replaced by a loudspeaker, then both the diaphram displacement and the disturbances would instead be periodic.) Since the piston displacement is rapid, there is little time for heat transfer, and consequently the compression process may be taken as adiabatic. Therefore, the temperature rises to T + ∆T behind the wave. Since the gas behind the wave is expanding, it is in motion; let its velocity be ∆V . (Note that ∆V c; the wave propagates into the undisturbed gas, much like a water wave moves without transporting water with it.) We want to determine the speed c the disturbance propagates with. To do
1 The

mean free path in air is about 10−7 m, so this condition is easily satisﬁed in air.

CHAPTER 11. ONE-DIMENSIONAL COMPRESSIBLE FLOW

272

P+ P T+ T + undisturbed gas P, T,

c

Figure 11.2: The piston is rapidly pushed in a small distance, launching a sound wave travelling to the left.

Control Volume undisturbed gas P, T, c c- V P+ P T+ T +

Figure 11.3: The situation as seen by an observer moving with the wave at speed c.

CHAPTER 11. ONE-DIMENSIONAL COMPRESSIBLE FLOW

273

this, it is convenient to switch to a reference frame travelling with the wave (Fig. 11.3). Deﬁning a control volume surrounding the wave, undisturbed ﬂuid enters the control volume from the left at speed c, passes through the wave, and leaves the control volume at speed c − ∆V . If this wave propagates steadily, then there can be no mass buildup within the control volume. The mass ﬂow rate in the left is min = ρcA, ˙ (11.2)

where A is the tube cross-sectional area. The mass ﬂow rate out the right side is mout = (ρ + ∆ρ)(c − ∆V )A. ˙ (11.3) ˙ Equating min and mout yields ˙ c∆ρ − ρ∆V − ∆ρ∆V = 0. (11.4)

The magnitudes of the pressure, density, and velocity disturbances in real sound waves are very small. In fact, the sound speed is formally deﬁned as the speed inﬁnitesimal disturbances propagate with. We will soon let all disturbance quantities become inﬁnitesimal, which means we can ignore the very small ∆ρ∆V term in this equation. With this approximation, we have ∆V = c ∆ρ . ρ (11.5)

Since ∆ρ/ρ 1, this justiﬁes our claim above that ∆V c. Equation (11.5) resulted from doing a mass balance on the control volume; let’s now see what results from applying the momentum principle. The x momentum leaving the control volume on the right is m(c−∆V ); the x momentum ˙ entering on the left is mc. Therefore, the net x momentum leaving the control ˙ volume is −m∆V . ˙ The momentum principle says that this equals the net force in the x direction on the control volume. The net force is due to pressure: Fnet = AP − A(P + ∆P ) = −A∆P. Therefore, the momentum principle requires that A∆P = m∆V. ˙ Using m = ρAc, this becomes ˙ ∆P = ρc∆V. (11.8) (11.7) (11.6)

CHAPTER 11. ONE-DIMENSIONAL COMPRESSIBLE FLOW Substituting for ∆V from Eq. (11.5), we ﬁnd ∆P = c2 ∆ρ.

274

(11.9)

Finally, an energy balance on the control volume (including the kinetic energy in this reference frame) yields h+ (c − ∆V )2 c2 = (h + ∆h) + . 2 2 (11.10)

Since the propagation is assumed to be adiabatic, there is no heat transfer term Q in this equation. To ﬁrst order in ∆V , the energy balance becomes ∆h = c∆V. Using Eq. (11.5), this can be written ∆h = c2 ∆ρ . ρ (11.12) (11.11)

Now let us take the limit as the disturbances become inﬁnitesimal. Equations (11.9) and (11.12) become dP = c2 dρ, dρ dh = c2 . ρ (11.13) (11.14) (11.15) At this point, it is interesting to ask what happens to the entropy of the gas as it passes through the wave. We know dh = T ds + vdP = T ds + dP/ρ, so solving for ds yields ds = 1 T dh − dP ρ . (11.17) (11.16)

Substituting from above for dh and dP , ds = Therefore, the propagation of an inﬁnitesimal sound wave is isentropic. 1 T (c2 dρ) − (c2 dρ/ρ) ρ = 0. (11.18)

CHAPTER 11. ONE-DIMENSIONAL COMPRESSIBLE FLOW

275

(This would not be true for ﬁnite-sized disturbances, which are shock waves.) With this result, we can use Eq. (11.13) to solve for c. Since s is constant, dP/dρ for this process is equivalent to (∂P/∂ρ)s . Therefore,

c=

∂P ∂ρ

(11.19) s Example 11.1 Estimate the speed of sound in liquid water at 400 K and 1 MPa. Solution: using the WWW thermodynamic property calculator or TPX, for these conditions ρ = 937.88 kg/m3 . Now let’s perturb ρ by a small amount, say 1.0 kg/m3 . Holding s constant and setting ρ = 938.88 kg/m3 , we ﬁnd P = 3.2676 MPa. Thus, c≈ 2.2676 × 106 1.0
1/2

= 1506 m/s.

11.3.1 Speed of sound in an ideal gas In an ideal gas with constant cp , P vk is constant for any isentropic process. Therefore, since ρ = 1/v, P = Cρk (11.20) for an isentropic process. Diﬀerentiating this expression, ∂P ∂ρ Using P = ρRT , ∂P ∂ρ = kRT s = kCρk−1 = s kP . ρ

(11.21)

(11.22)

and therefore the speed of sound in an ideal gas is √ kRT .

c=

(11.23)

ˆ ˆ Recall that R is the gas constant per unit mass: R = R/M . Example 11.2 Calculate the speed of sound in air at 300 K.

CHAPTER 11. ONE-DIMENSIONAL COMPRESSIBLE FLOW

276

P 0 P
V2 2

h0 h

s
Figure 11.4: The stagnation state on an h − s plot ˆ Solution: For air, k = 1.4 and M = 28.97. Therefore, c= (1.4)(8314/28.97)(300) = 347 m/s. (11.24)

Note that the speed of sound in air is much less than in water. The Mach Number M is deﬁned to be the ratio of the ﬂuid velocity to the local speed of sound: V M= . (11.25) c If M < 1, the ﬂow is subsonic, and if M > 1 the ﬂow is supersonic.

11.4 Stagnation Properties
When a high-speed ﬂowing gas is decelerated to zero velocity (for example in a diﬀuser, or when the gas must deﬂect around an object inserted into it), its temperature, pressure, and density increase, since as a ﬂuid packet decelerates the kinetic energy in the ﬂow is expended by doing compression work on the ﬂuid packet. The state reached by decelerating a ﬂuid moving at speed V reversibly and adiabatically to zero velocity is called the stagnation state. The stagnation process is shown in Fig. 11.4. The stagnation state is deﬁned by the two equations h0 = h + and s0 = s. (11.27) V2 2 (11.26)

CHAPTER 11. ONE-DIMENSIONAL COMPRESSIBLE FLOW

277

Here the subscript 0 denotes the stagnation state, and h0 is known as the stagnation enthalpy. The pressure attained in the stagnation state is the stagnation pressure P0 = P (h0 , s0 ), and the temperature T0 = T (h0 , s0 ) is the stagnation temperature. Note that given h, s, and V it is always possible to calculate h0 , s0 , P0 , and T0 , whether or not the problem at hand actually involves decelerating the ﬂuid to a stationary state; the stagnation state is the state which would result if we were to do it. We may regard the stagnation properties as properties of the ﬂowing ﬂuid. Example 11.3 Determine the stagnation temperature and pressure for nitrogen at 10 MPa and 300 K, with speed V = 200 m/s. Solution: Using the WWW calculator, we ﬁnd h = 442.14 kJ/kg (4.4214 × 105 J/kg) and s = 2.988 kJ/kg-K. Therefore, h0 = 4.4214 × 105 + (300)2 /2 = 4.8714 × 105 J/kg. Using h0 = 487.14 kJ/kg, and s0 = s = 2.988 kJ/kg-K, the WWW calculator ﬁnds T0 = 343.44 K, and P0 = 15.83 MPa. For the special case of an ideal gas with constant cp , Eq. (11.26) becomes cp T0 = cp T + V2 . 2 (11.28)

We can re-write this equation in terms of k and Mach number, using cp = √ kR/(k − 1), c = kRT , and M = V /c: T0 k −1 2 =1+ M T 2 For any isentropic process in an ideal gas with constant cp , P2 = P1 T2 T1 k/k−1 (11.29)

.

(11.30)

Taking state 1 to be the state with velocity V and state 2 to be the stagnation state, P0 = P T0 T k/k−1 = 1+

k−1 2 M 2

k/k−1

(11.31)

If we consider the reverse situation, in which a stationary gas at T0 and P0 is accelerated in a nozzle to speed V , Eq. (11.28) still applies, and sets an upper limit on the attainable speed, since T cannot be negative.

CHAPTER 11. ONE-DIMENSIONAL COMPRESSIBLE FLOW

278

T 0 , P0
V=0 V, P, T

Setting T = 0, Vmax = This can be rewritten as Vmax = √ where c0 = kRT0 is the sound speed upstream of the nozzle. For k = 1.4, this becomes Vmax = 2.2c0. Therefore, the maximum velocity air can be accelerated to using an adiabatic nozzle is a little more than twice the upstream sound speed. Of course, since T is dropping as the gas accelerates, the local sound speed c is going down, and the local Mach number M = V /c diverges to inﬁnity as V → Vmax . 2 c0 , k−1 (11.33) 2cp T0 . (11.32)

11.5 Isentropic Flow with Area Change
Consider one-dimensional, steady, ﬂow down a tube with an area A(x) which changes with distance. We will assume the irreversible processes of viscous friction and heat conduction are negligible, in which case the ﬂow is isentropic. First of all, conservation of mass requires that ρ(x)V (x)A(x) = constant. Diﬀerentiating this and dividing by ρV A yields dρ dV dA + + = 0. ρ V A Conservation of energy requires that h(x) + Diﬀerentiating yields dh = −V dV. We also know from basic thermodynamic principles that dh = T ds + vdP = T ds + (1/ρ)dP. (11.38) (11.37) V 2 (x) = constant. 2 (11.36) (11.35) (11.34)

CHAPTER 11. ONE-DIMENSIONAL COMPRESSIBLE FLOW Since the ﬂow is isentropic, ds = 0, and therefore dh = Equating Eq. (11.37) and Eq. (11.39), dP = −ρV dV. dP ρ

279

(11.39)

(11.40)

This relationship makes sense qualitatively, since if the pressure decreases down the tube (dP/dx < 0), we expect the ﬂuid to accelerate (dV /dx > 0), and visa versa. Let’s now ﬁnd how the density changes with distance. Since the ﬂow is isentropic, ∂P dρ. (11.41) dP = ∂ρ s But (∂P/∂ρ)s = c2 , so dP = c2 dρ. Putting this into Eq. (11.39), we ﬁnd dρ V dV dV = − 2 = −M . ρ c c (11.43) (11.42)

Therefore, the density decreases if the ﬂow accelerates and visa versa (dρ < 0 ⇐⇒ dV > 0). Note that the relative change in density for a given dV is greater at high Mach number than at low. We may eliminate density from Eq. (11.35) by substituting from Eq. (11.43): − or dA dV =− A V In terms of the Mach number, 1− V c V dV dV dA + + =0 2 c V A
2

(11.44)

.

(11.45)

dA dV =− 1 − M2 A V

(11.46)

Note that we have not made any ideal gas assumption, so this equation applies for one-dimensional, isentropic ﬂow of any ﬂuid.

CHAPTER 11. ONE-DIMENSIONAL COMPRESSIBLE FLOW

280

P0

PE

PB

Figure 11.5: Flow in a converging nozzle. Equation (11.46) tells us how the velocity of the ﬂuid changes as it passes down the tube. An interesting result of this equation is that whether the ﬂuid accelerates (dV > 0) or decelerates (dV < 0) as the tube area changes depends on whether the Mach number is greater than or less than 1.0 (supersonic or subsonic). We may identify 3 cases: Subsonic Flow: M < 1. In this case, a converging tube (dA < 0) causes the gas to accelerate (dV > 0), and a diverging tube causes the gas to decelerate. Therefore, a subsonic nozzle is a converging tube, and a subsonic diﬀuser is a diverging tube. Supersonic Flow: M > 1. Now the sign of the right-hand side of Eq. (11.46) is positive, so a converging tube causes the gas to decelerate, and a diverging tube causes the gas to accelerate. Therefore, a nozzle to accelerate a supersonic ﬂow must be constructed as a diverging tube, and a supersonic diﬀuser must be converging. Sonic Flow: M = 1. If the Mach number equals one, then the right-hand side becomes zero for any dV . Therefore, sonic ﬂow requires dA = 0: it is not possible to have M = 1 in a portion of the tube where the area is changing with x.

11.6 Flow in a Converging Nozzle
Suppose gas is escaping from a gas tank at pressure P0 through a small converging nozzle, as shown in Fig. 11.5. The pressure of the surroundings is PB < P0 (the “back pressure”). The pressure at the exit plane of the nozzle is PE . Since the nozzle is converging, dA/dx < 0 and therefore the accelerating ﬂow in the nozzle must be subsonic – it is impossible for supersonic ﬂow to exist in this nozzle. At the end, the ﬂow area expands abruptly. We will assume

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281

that dA/dx = 0 at some point very close to the nozzle exit, since the lip will inevitably be at least a little rounded. Let us consider the ﬂow in the nozzle as a function of PB for given P0 . If PB = P0 , then the system is in mechanical equilibrium, and there is no ﬂow, so m = 0. For PB slightly less than P0 , gas will ﬂow, and the pressure through the ˙ nozzle will fall continuously from P0 to PE = PB . As PB is lowered further, m ˙ increases, and still PE = PB . But at some PB the ﬂow becomes sonic at the exit; call this value P ∗ . What happens if PB is now reduced below P ∗? The way the ﬂow “communicates” the change in PB to the ﬂow upstream is through acoustic waves, which travel at the sound speed. As long as the ﬂow is everywhere subsonic, it is possible for sound waves to propagate upstream, and the ﬂow everywhere in the nozzle adjusts to the change in PB . But at PB = P ∗, the exit ﬂow is coming out at the local speed of sound. Sound waves propagate upstream with this speed relative to the moving gas. In the lab frame, the sound waves get nowhere – it is analogous to a ﬁsh trying to swim upstream in a river when the river is ﬂowing downstream as fast as the ﬁsh can swim upstream. Therefore, the information that PB is now below P ∗ can’t be conveyed to the ﬂow in the nozzle; it is unaware of the change. Therefore, everything about the ﬂow in the nozzle – the total mass ﬂow rate, P (x), V (x), etc. – is identical to the results for PB = P ∗, even though now PB < P ∗. In particular, PE = P ∗ , not PB . Once the gas emerges from the nozzle, it ﬁnds itself suddenly at higher pressure than the surroundings. It rapidly adjusts to PB through a set of complex, three-dimensional expansion waves. The pressure distribution would look something like that shown in Fig. 11.6.

11.7 Choked Flow
Once the ﬂow at the exit of a converging nozzle becomes sonic, the downstream conditions no longer aﬀect the ﬂow upstream of the shock wave. Therefore, the mass ﬂow rate through the nozzle is unaﬀected by downstream pressure once M = 1 at the nozzle exit. When this occurs, we say the ﬂow is choked. We can calculate the pressure ration P0 /PB above which the ﬂow becomes choked, and the mass ﬂow rate under choked conditions for the case of an ideal gas with constant cp . Let the upstream conditions be T0 and P0 , and denote the conditions where M = 1 at the exit by T ∗ and P ∗. Then from Eqs. 11.29

CHAPTER 11. ONE-DIMENSIONAL COMPRESSIBLE FLOW
Nozzle exit PB = P0 PB < P0 PB = P* P(x) PB < P*

282

P0

x

Figure 11.6: Pressure vs. distance for ﬂow through a converging nozzle. and 11.31,

T0 k−1 =1+ T∗ 2 P0 = P∗ 1+ k−1 2 k/k−1 (11.47)

and

.

(11.48)

For k = 1.4, these expressions become T0 /T ∗ = 1.2 and P0 /P ∗ = 1.89. Thus, a pressure ratio from inlet to outlet of the tube of only 1.9 is suﬃcient to choke the ﬂow for air. To solve for the choked mass ﬂow rate, we may use the result that m is equal ˙ to ρV A at any cross-section of the nozzle. For simplicity let’s use the throat: m = ρ∗ A∗ c. ˙ (11.49)

Here A∗ is the area of the throat, and we have used the fact that the ﬂow is sonic at the throat. Using the expression for c for an ideal gas, we have m = A∗ ˙ = P∗ √ kRT ∗ RT ∗ P ∗ k T0 P0 A∗ √ P0 R T ∗ T0 (11.50) (11.51)

ˆ The terms in brackets can be evaluated for a given k and M (needed to calculate R). For air, this expression reduces to A∗P0 m = 0.0404 √ ˙ T0

(11.52)

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283

T

0 V

P B

P 0

Figure 11.7: A converging-diverging nozzle. in SI units. Therefore, for a given gas, the maximum (choked) mass ﬂow rate through a nozzle depends only on three quantities: the area of the nozzle where the ﬂow is sonic, and the stagnation temperature and pressure upstream. Example 11.4 Air enters a converging nozzle with an entrance diameter of 15 cm and an exit diameter of 2 cm. The entrance conditions are 6 MPa and 300 K. Treating air as an ideal gas with constant cp , determine the maximum mass ﬂow rate possible through the nozzle. Solution: The maximum ﬂow rate is the choked ﬂow rate. From Eq. (11.52), mmax = 0.0404 ˙ (π)(0.01 m)2 (6 × 106 Pa) √ = 4.39 kg/s. 300 K

11.8 Flow in a Converging-Diverging Nozzle
Consider now attaching a diverging section onto the converging nozzle of the previous section, producing the situation shown in Fig. 11.7. As before, upstream of the nozzle the ﬂow velocity is negligible, and the pressure and temperature are the stagnation values P0 and T0 , respectively. Outside the nozzle, the pressure is PB . As PB is lowered below P0 , gas begins to ﬂow through the nozzle. In the converging section, the ﬂow accelerates and the pressure and temperature drop. If the ﬂow is still subsonic when it reaches the minimum-area position in the nozzle (the “throat”), then as it passes into the diverging portion it decelerates, with a corresponding rise in P and T . It emerges with Pe = PB , and Te = T0 − Ve2 /2cp. Note that this behavior is completely consistent with Eq. (11.46). Since M = 1 at the throat where dA/dx = 0, it must be that dV /dx = 0 here, as it is. If PB is reduced further to P ∗ , the ﬂow at the throat will become sonic. From Eq. (11.46), in this case it is no longer necessary for dV /dx = 0 at the

CHAPTER 11. ONE-DIMENSIONAL COMPRESSIBLE FLOW
Nozzle throat M* = 0 M* < 1 M* = 1 -

284

P0

decreasing PB P(x) supersonic flow

shock waves

x

Figure 11.8: Flow in a converging-diverging nozzle. If PB is low enough that the ﬂow at the throat is sonic (M ∗ = 1), then the ﬂow continues to expand supersonically in the diverging portion (red line). In order to match PB at the nozzle exit, at some point the ﬂow suddenly transitions from supersonic to subsonic in a shock wave, with a sudden rise in pressure. After the shock, the diverging nozzle acts as a diﬀuser, and the pressure rises to PB . The location of the shock depends on PB , beginning at the throat when PB is just low enough to produce supersonic ﬂow, and moving out toward the nozzle exit as PB decreases. throat, so the ﬂow can continue to accelerate into the diverging portion, where (now that it is supersonic) it continues to be accelerated. As the supersonic ﬂow accelerates in the diverging portion, the pressure continues to drop. The value of PB has no eﬀect on the ﬂow in this portion of the nozzle, since any information about PB would propagate at the speed of sound, and so would be swept downstream and would not reach the upstream ﬂow. Therefore, if the gas continues to accelerate isentropically to the exit of the nozzle the pressure of the gas at the nozzle exit would not generally match PB . But somehow the pressure must come to PB at or after the nozzle exit. What really happens is that if the pressure in the nozzle drops below PB , then at some point in the diverging portion the gas undergoes a sudden and irreversible transition to subsonic ﬂow. This transition is known as a shock wave, and will be discussed in the next section. Typically a shock wave normal to the ﬂow will stand somewhere in the diverging portion of the nozzle, at a position such that Pe = PB . The pressure vs. distance plot for various PB values now looks something like the plot shown in Fig. 11.8. In the isentropic portion of the ﬂow (before the shock wave), we can use the condition ρV A =constant to solve for the velocity at any position in the nozzle.

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285

5

4

A/A*

3

2

1

0 0 1 2 3

Mach Number

Figure 11.9: Area ratio A/A∗ for k = 1.4. We may write √ which using the ideal gas equation of state and c = kRT becomes2 1 P∗ A = A∗ M P If we now write this as 1 P ∗/P0 A = A∗ M P/P0 and use eqs. (11.29) and (11.29), we ﬁnd T /T0 T ∗ /T0 (11.55) T . T∗ ρV A = ρ∗ cA∗ (11.53)

(11.54)

1 A = A∗ M

2 k+1

1+

k−1 2 M 2

(k+1)/2(k−1)

.

(11.56)

This equation is plotted in Fig. 11.9 for k = 1.4. Note that for a given area ratio, there are two possible Mach numbers — one subsonic, and the other supersonic.

11.9 Normal Shock Waves
Shock waves which stand perpendicular to the ﬂow direction are called normal shock waves, in contrast to oblique shock waves which can occur when two- and
2 You

should verify this.

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286

Shock Wave P+ P T+ T P, T, V1 V2

Figure 11.10: A steady shock wave. Note that ∆P , ∆ρ, and ∆T are not assumed to be inﬁnitesimal. three-dimensional ﬂow is considered. A shock wave is a discontinuity in the pressure, velocity, density, and temperature of the gas. It may seem suprising to you that these properties can be discontinuous, but in fact the equations describing conservation of mass, momentum, and energy, along with the ideal gas law, do admit discontinuous solutions.3 By carrying out balances of mass, momentum, and energy on a shock wave, we may derive relationships between the gas properties on either side of the shock. The approach is very much like that we did to derive the speed of sound, except that now the perturbations are not inﬁnitesimal. Consider a situation in which gas with velocity V1 in the x direction passes through a stationary normal shock and leaves with velocity V2 . Conservation of mass requires ρ1 V1 = ρ2 V2 . (11.57) We have divided through by the tube area, since it is the same on each side of the shock. The momentum principle states (P1 − P2 ) = (m/A)(V2 − V1 ), ˙ which can also be written P1 + ρ1 V12 = P2 + ρ2 V22 , since m/A = ρ1 V1 = ρ2 V2 . The energy balance is ˙ h1 + V12 V2 = h2 + 2 , 2 2 (11.60) (11.59) (11.58)

3 A more detailed treatment considering viscosity and heat conduction would show that a shock wave has a ﬁnite thickness, which is of order the mean free path.

CHAPTER 11. ONE-DIMENSIONAL COMPRESSIBLE FLOW which is equivalent to h0,1 = h0,2 .

287

(11.61)

Therefore, the stagnation enthalpy is constant across a shock. Also, we can write equations of state that have to be satisﬁed in both states. If the ﬂuid may be treated as an ideal gas in both states, then P1 = ρ1 RT1 , P2 = ρ2 RT2 (11.62) (11.63)

If state 1 is fully speciﬁed, the four equations (11.57), (11.59), (11.60), and (11.63) may be used to solve for the four unknowns P2 , T2 , ρ2 , and V2 . For the case of an ideal gas with constant cp , this set of equations may be solved analytically. Equation (11.60) becomes T0,1 = T0,2 , which, using Eq. (11.29), may be written k−1 2 2 M1 k−1 2 M2 2

(11.64)

1+ T2 = T1 1+ √

(11.65)

Using ρ = P/RT , c =

kRT , and M = V /c, Eq. (11.59) reduces to
2 1 + kM1 P2 = 2 P1 1 + kM2

(11.66)

Equation (11.57) becomes P2 T2 M2 = . (11.67) P1 T1 M1 Substituting eqs. (11.65) and (11.66) into Eq. (11.67), we can solve for M2 :
2 k−1 2k 2 k−1 M1 − 1 2 M1 +

2 M2 =

(11.68)

Since there is no heat transfer in the shock, the entropy production rate as the gas ﬂows through the shock is simply ˙ Ps = m(s2 − s1 ) = m(cp ln(T2 /T1 ) − R ln(P2 /P1 )). ˙ ˙ (11.69)

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288

It may be shown that this is positive only if M1 > 1. Therefore, the ﬂow entering the shock must be supersonic. Or, in the frame of reference of the gas upstream of the shock, this is equivalent to saying the shock wave must propagate at a supersonic (not subsonic) speed.

CHAPTER 12
THERMODYNAMICS OF MAGNETIC SYSTEMS 12.1 Introduction
Thermodynamics applies to much more than just simple compressible substances. We’ve dealt brieﬂy with magnetic work. In this chapter, we’ll discuss in more depth the thermodynamics of materials on which magnetic work can be done.

12.2 The Simple Magnetic Substance
Magnetic ﬁelds can interact with some materials, causing work to be done on or by them. Here we consider the simplest case, when magnetic work is the only possible reversible work mode. If a substance contains molecules or ions which have permanent magnetic dipole moments (for example, O2 or rare-earth ions), then an external magnetic ﬁeld can do work on the sample by turning these dipoles to align with the ﬁeld. Such materials are known as paramagnetic. A magnetic ﬁeld can also do work on a substance by inducing electrical currents to ﬂow. These currents may be macroscopic ones, like those induced in the surface layer of a perfect conductor when a magnetic ﬁeld is applied, or atomic-scale ones due to slight changes in the motion of the electrons orbiting about the nuclei due to the magnetic ﬁeld. In either case, materials in which currents are induced when a magnetic ﬁeld is applied are called diamagnetic. The net magnetic dipole moment of a sample per unit volume deﬁnes the magnetization M . Note that M is a vector. For the paramagnetic case, it is the vector sum of all individual magnetic dipole moments. If they point in random directions (as they do with no applied ﬁeld), then their vector sum is zero and M = 0. But if a ﬁeld is applied so they are at least partially aligned, then there is a net magnetization pointing in the direction of the applied ﬁeld. (They are not perfectly aligned since collisions with other molecules or with the lattice in a solid tend to upset the alignment.) For diamagnetic materials, there is also a magnetization, since current loops have a magnetic moment associated with them. It may be shown that the 289

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290

magnetization of a diamagnetic material points opposite to the direction of the magnetic ﬁeld. All materials have some diamagnetic response, but it is usually weak. Except in extremely strong ﬁelds, most diamagnetic materials can be considered nonmagnetic. If the material contains permanent magnetic dipoles, then the paramagnetic response is usually much larger and the diamagnetism can be neglected. Finally, the most familiar form of magnetism is ferromagnetism. Permanent magnets, which by deﬁnition are magnetized even in zero applied ﬁeld, are made of ferromagnetic materials. Only the elements Fe, Co, Ni, and Gd exhibit ferromagnetism, and all permanent magnets contain at least one of these elements. The ferromagnetic elements have permanent magnetic dipoles, like paramagnetic materials. The diﬀerence is that the dipoles exert forces on one another which tend to lock them into alignment even without an external magnetic ﬁeld. If a ferromagnetic material is heated, beyond a certain temperature the forces between dipoles are overcome, and the material becomes paramagnetic. This temperature is known as the Curie Point, and for iron it is 770 ◦ C. In general, a paramagnetic, diamagnetic, or ferromagnetic substance may have other work modes too. For example, oxygen gas is paramagnetic but also is compressible. A liquid oxygen droplet will have a surface tension, so deforming the droplet is another work mode. But in most cases, magnetic materials are solid, and reasonably incompressible. In this case, the only important work mode is magnetic work, and we may then treat the substance as a simple magnetic substance. This is the only case we will consider here. When an external magnetic ﬁeld is applied to a sample, the response of the sample (turning dipoles or induced currents) modiﬁes the local value of the ﬁeld within the sample. We deﬁne H to represent the applied ﬁeld which would exist without the sample, and B to represent the true ﬁeld value in the sample. The quantities B, H, and M are related by 1 B = µ0 (H + M ). (12.1)

The value of the ﬁeld without the sample is really µ0 H, but since µ0 is just a constant (4π × 10−7 tesla-m/amp) we can treat H as representing the applied magnetic ﬁeld.
1 This equation depends on the unit system. The form here is for the Rationalized MKS system, which is the one used most commonly in applied science and engineering. In another unit system widely used in physics (cgs), B and H have the same units and equation 12.1 is replaced by B = H + 4πM

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291

We’ve shown previously that the work required to reversibly magnetize a magnetic substance is dW = µ0 H · d(M V ). ¯ (12.2) The Gibbs equation for a simple magnetic substance may be derived by considering a reversible process in which magnetic work dW is done and heat ¯ dQ = T dS is added. Since dU = dQ + dW , ¯ ¯ ¯ dU = T dS + µ0 H · d(M V ). On a unit mass basis, du = T ds + µ0 H · d(Mv). We can simplify the notation by deﬁning m = µ0 vM . (12.5) (12.4) (12.3)

Also, the magnetization is generally either parallel to H (paramagnetic substances) or anti-parallel (diamagnetic substances). Therefore, H · dm = Hm, where m < 0 in the diamagnetic case. The magnetic Gibbs equation then becomes du = T ds + Hdm. (12.6)

Starting from this equation, we can carry out an analysis very similar to the one in chapter 6 for a simple compressible substance. The only diﬀerence is that we replace −v by m, and P by H. For example, we can deﬁne a magnetic enthalpy, Helmholtz free energy, and Gibbs free energy by h f g = u − Hm = u − Ts = h − T s = u − T s − Hm. (12.7) (12.8) (12.9)

Although we use the same symbols for these as we did for a simple compressible substance, it’s important to keep in mind the diﬀerence in deﬁnition (e.g. h = u − Hm, not h = u + P v) since the reversible work mode is now magnetization, not compression. The diﬀerential equations of state for these properties are: dh = T ds − mdH df dg = −sdT + Hdm = −sdT − mdH (12.10) (12.11) (12.12)

CHAPTER 12. THERMODYNAMICS OF MAGNETIC SYSTEMS

292

Integratings these equations would yield u(s, m), h(s, H), f(T, m), and g(T, H), all of which are fundamental relations. Experimentally, it is usually easiest to work with temperature and magnetic ﬁeld strength as independent variables, so g(T, H) is in many cases the most useful of these fundamental relations. Of course, analogs of the Maxwell relations may be easily written down. The above diﬀerential equations of state are general and apply to any simple magnetic substance. To determine the speciﬁc properties of a particular substance, we need to know the functional form of one of the fundamental relations [say, g(T, H)] or else a set of experimental equations of state like m(T, H) and cH (T, H). Let’s look at a couple of important magnetic substances. 12.2.1 The Curie Substance For paramagnetic materials, the tendency for the dipoles to be aligned with the ﬁeld is balanced by the eﬀects of random collisions (in a gas) or vibrations of the lattice (in a solid), which tend to randomize the direction the dipole points. Since the energy associated with the random, thermal motion scales with temperature, we expect that the magnetization of a paramagnetic material will increase with H at constant T , and decrease with T at constant H. In fact it is found that M = f(H/T ). The function f(H/T ) is linear for small H/T , and saturates at an asymptotic value for large H/T , since in this limit the dipoles are perfectly aligned and therefore M can’t increase further. A Curie substance is deﬁned as any substance which obeys M =C H . T (12.13)

All paramagnetic substances have this behavior for small H/T . In terms of m, or H m=C , (12.14) T where C = µ0 vC. Like an ideal gas, a Curie substance has an internal energy which depends only on temperature. The proof follows exactly the procedure we used to prove u = u(T ) for an ideal gas. First solve Eq. (12.6) for ds: ds = Substituting for H/M , ds = 1 1 mdm. du − T C (12.16) 1 H du − dm. T T (12.15)

CHAPTER 12. THERMODYNAMICS OF MAGNETIC SYSTEMS Equating mixed partial derivatives, ∂ 1 ∂m T =− u 293

1 C

∂m ∂u

(12.17) m But (∂m/∂u)m = 0, so 1/T depends only on u, not on m. Therefore, T = T (u), and inverting this we have u = u(T ). In fact, it can be seen from this analysis that any material which obeys m = f(H/T ) (12.18)

will have an internal energy which is a function of temperature alone. Knowing that u = u(T ), we may now integrate the Gibbs equation and ﬁnd the entropy of a Curie substance. ds = 1 T du dT dT − H dm. T (12.19)

The coeﬃcient multiplying dT is (∂s/∂T )m , which by deﬁnition is cm /T , where cm is the speciﬁc heat at constant magnetization. Therefore, we see cm (T ) = du(T ) dT (12.20)

which is a function only of temperature for a Curie substance. The Gibbs equation then is cm (T ) H ds = dT − dm. (12.21) T T Since m = C H/T , cm (T ) m ds = dT − dm. (12.22) T C Integrating this,

s(T1 , m1 ) − s(T0 , m0 ) =

T1 T0

cm (T ) 1 dT − T 2C

m2 − m2 , 1 0

(12.23)

or in terms of M ,

s(T1 , M1 ) − s(T0 , M0 ) =

T1 T0

cm (T ) µ0 v 2 2 dT − M1 − M0 , T 2C

(12.24)

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294

Note that the entropy decreases if a Curie substance is magnetized isothermally. This makes sense physically: as individual atomic magnetic dipoles become more aligned with the ﬁeld (increasing M ), the sample becomes more ordered. Therefore, we would expect s to decrease, as it does. This also means that heat must be removed from the sample during isothermal magnetization. If a sample is magnetized reversibly at T from zero magnetization to some value M , µ0 vM 2 Q= (12.25) 2C must be rejected to the environment. If instead the sample is magnetized adiabatically, then it will increase in temperature. Similarly, if a Curie substance is demagnetized (M decreased) at constant temperature, heat must be supplied. If demagnetization is carried out adiabatically, the temperature will decrease. Thus, magnetizing a Curie substance is similar to compressing a gas, and demagnetizing a Curie substance is similar to expanding a gas. This analogy suggests it should be possible to construct a heat pump or a heat engine by taking a Curie substance through a sequence of states in the (T, m) plane (or equivalently, the (T, H) plane) returning to the initial state. While no known practical heat engine operates on this cycle, it is in fact used as a heat pump (refrigerator) to reach very low temperatures near absolute zero. Temperatures as low as 10−8 K have been reached using magnetic refrigerators. To reach these very low temperatures, nuclear magnetic moments (rather than atomic ones) are employed.

12.3 Superconductors
Many metals exhibit superconductivity when cooled to very low temperatures. This eﬀect was ﬁrst discovered by Onnes in 1911, shortly after he succeeded in liquifying helium. Until 1986, all known superconductors required cooling below 20 K before they would become superconductors. Beginning in 1986, some ceramic materials which become superconductors at abnormally high temperatures (above 100 K) were discovered. One of the most useful ones is yttrium barium copper oxide: YBa2 Cu3 O7−x . The discovery of high temperature superconductors has enormous practical consequences, since their transition temperatures are above the boiling point of N2 at 1 atm. Therefore, inexpensive liquid nitrogen can be used to keep these materials superconducting, rather than expensive liquid helium which is required for most lower-temperature superconductors. Superconductors exhibit several strange properties. As they are cooled below

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295

the transition temperature Tc , the electrical resistance drops suddenly to exactly zero. If a current is set up in a superconducting ring (for example by cooling it below Tc in a magnetic ﬁeld and then removing the ﬁeld), it persists indeﬁnitely. Measurements have been done which show currents remaining unchanged for months. This can only be possible if the resistance is really zero, not just very small. If a superconductor is brought into a magnetic ﬁeld, currents will ﬂow in the surface layer of the superconductor, which perfectly shield the the magnetic ﬁeld from the interior, so that the magnetic ﬁeld inside is zero. This is the behavior predicted from EM theory for any perfect conductor. In contrast, for a real conductor with ﬁnite resistance, the induced currents would eventually die out and the ﬁeld would penetrate. But superconductors display an additional property which is not simply a consequence of zero resistance. If a superconducting material is cooled from above Tc to below Tc in a magnetic ﬁeld, the magnetic ﬁeld is completely expelled from the interior at Tc . Since the magnetic ﬁeld penetrates above Tc , there would be no reason for currents to start ﬂowing in the surface to shield the interior from the magnetic ﬁeld, if a superconductor were just like a metal with zero resistance. But the currents do start ﬂowing in the surface layer at Tc , and the ﬁeld is expelled. Evidently, there is something intrinsic about the superconducting state which is incompatible with the presence of a magnetic ﬁeld inside. This effect, ﬁrst discovered in 1933 by Meissner and Ochsenfeld, is called the Meissner eﬀect. This is what causes a permanent magnet to levitate above a superconducting dish, and is employed in several schemes for magnetic levitation trains. Thermodynamics can be used to understand some of the properties of superconductors. When a superconductor expels a magnetic ﬁeld it does magnetic work on the external currents producing the ﬁeld, so magnetic work is a relevant work mode for a superconductor. In principle, we could include −P dv work too, but it is usually acceptable to assume the superconductor is incompressible. In this case, a superconductor is a simple magnetic substance. It is found that applying a suﬃciently strong magnetic ﬁeld to a superconductor can destroy the superconductivity, causing it to revert to the “normal” state. We may regard the superconducting state exactly like a thermodynamic phase. The phase diagram for a superconductor looks something like shown below.

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296

H c,0 normal H superconducting

Tc T

What is the magnetic equation of state of a superconductor? Since we know B = 0 in a superconductor, and also B = µ0 (H + M) the magnetic equation of state in the superconducting state is M = −H. (12.27) (12.26)

In contrast, in the normal state, M = 0 since typically metals which become superconducting do not contain paramagnetic ions. Since the Gibbs free energy is particularly useful if T and H are held constant, let’s calculate g for the normal and superconducting phases. We can do this by integrating the diﬀerential equation of state dg = −sdT − mdH, (12.28)

using the appropriate magnetic equation of state for the normal and superconducting phases (m = 0 or m = −µ0 vH). For the normal phase, m = 0, and therefore Hdm = 0 so it is impossible to do magnetic work on the substance when it is in the normal phase. As in the case of an incompressible ﬂuid, this means that the properties of the substance in the normal state are ﬁxed by only one quantity, the temperature. For the free energy we have dgn = −sn (T )dT. (12.29) Therefore, gn (T1 ) − gn (T0 ) = −
T1

sn (T ) dT.
T0

(12.30)

CHAPTER 12. THERMODYNAMICS OF MAGNETIC SYSTEMS For the superconducting phase, dgs = −ss dT − ms dH = −ss dT + (µ0 vH)dH.

297

(12.31)

If we equate the mixed partial derivatives of the coeﬃcients in this equation, − ∂ss ∂H = µ0 v
T

∂H ∂T

= 0,
H

(12.32)

we see that the entropy in the superconducting phase has no dependence on H, and therefore depends only on T . Thus, s is a function just of T for both the superconducting and normal phases. (Note that this was not true for the Curie substance.) With this information, we can proceed to integrate Eq. (12.31) to ﬁnd the Gibbs free energy of a superconductor. gs (T1 , H1) − gs (T0 , H0 ) = −
T1

ss (T ) dT +
T0

µ0 v 2 2 H1 − H0 . 2

(12.33)

We showed in Chapter six that the stable phase of a simple compressible substance at speciﬁed (T, P ) is the one with the lowest Gibbs free energy g = u + P v − T s. The same analysis can be repeated for the magnetic case, simply replacing P with H and v with −m to show that the stable phase of a simple magnetic substance at speciﬁed (T, H) is the one with lowest magnetic Gibbs free energy g = u − Hm − T s. Therefore, in the region of the (H, T ) plane where the superconducting phase is thermodynamically stable, gs < gn ; outside this region gn < gs . On the coexistence line where they can exist in equilibrium with one another gs = gn . Let’s consider a temperature T < Tc , where the superconducting phase is stable for H = 0, and so gs (T, 0) < gn (T ). Consider what happens as the magnetic ﬁeld is increased from zero. The Gibbs free energy of the normal phase (metastable at H = 0) is not aﬀected by H. But the Gibbs free energy of the superconducting phase increases quadratically with H. As shown in Figure 12.1, gs (T, H) becomes equal to gn (T ) at some value of H; this deﬁnes the value of the critical magnetic ﬁeld strength Hc(T ), above which the superconductor reverts to the normal phase: gs (T, Hc ) 2 µ0 vHc gs (T, 0) + 2 = gn (T ) = gn (T ). (12.34) (12.35)

CHAPTER 12. THERMODYNAMICS OF MAGNETIC SYSTEMS

298

g n (T)

g s (T)

H

H c (T)

Figure 12.1: Gibbs free energy for the superconducting and normal phases. Therefore,
2 Hc (T ) =

2 (gn (T ) − gs (T, 0)) . µ0 v

(12.36)

Note that the work done by the superconductor on the surroundings to keep the magnetic ﬁeld out as the external ﬁeld is raised from zero to H is
H H

W =−
0

Hdm = +µ0 v
0

HdH =

µ0 v 2 H . 2

(12.37)

So the increase in Gibbs free energy of the superconductor is due to the work it has to do to keep the ﬁeld out; eventually, (at Hc ) it is no longer worth it energetically and the substance reverts to the normal phase and lets the ﬁeld in. Now consider the latent heat associated with the normal to superconducting transition. Doing entropy accounting for a substance which undergoes a phase change from the normal to the superconducting state, ss (T ) = sn (T ) + Q , T (12.38)

where Q is the heat added to bring about the normal → superconducting transition. Since s = −(∂g/∂T )H , Q = T (ss (T ) − sn (T )) ∂(gs (T, H) − gn (T )) = −T ∂T ∂(gs (T, 0) + µ0 vH 2 − gn (T )) = −T ∂T d(gs (T, 0) − gn (T )) = −T dT

CHAPTER 12. THERMODYNAMICS OF MAGNETIC SYSTEMS = T
2 d(µ0 vHc (T )/2) dT

299 (12.39) (12.40)

Therefore, Q = µ0 vT Hc (T ) dHc(T ) dT (12.41)

From the phase diagram, we see that Hc = 0 at Tc , and the transition occurs at T < Tc if a ﬁeld is applied. Therefore, from Eq. (12.41), at zero applied ﬁeld the normal → superconducting transition has a latent heat of zero. Also, as T → 0, Q → 0, so |Q| must have a maximum at a temperature between 0 and Tc . Since dHc(T )/dT < 0, Q < 0 for non-zero H. Thus, when a substance is cooled and becomes superconducting, heat is given oﬀ, similar to when a gas condenses to form a liquid. Finally, we can derive an expression for the diﬀerence between the speciﬁc heats of the normal and superconducting phases. Since for both phases the entropy is a function only of temperature, the speciﬁc heat is simply c = T (ds/dT ), and is the same whether H or m is held constant. Since ss (T ) − sn (T ) = Q/T , we have from Eq. (12.41) ss (T ) − sn (T ) = µ0 vHc (T ) dHc(T ) dT (12.42)

Diﬀerentiating with respect to T and multiplying by T , cs (T ) − cn (T ) = µ0 vT dHc(T ) dT
2

+ Hc(T )

d2 Hc . dT 2

(12.43)

At zero ﬁeld, the transition occurs at T = Tc . Since Hc = 0 in this case, cs (Tc ) − cn (Tc ) = µ0 vTc dHc(T ) dT
2

.
T =Tc

(12.44)

This important formula, known as Rutger’s formula, shows that there will be a jump in speciﬁc heat as the substance is cooled below Tc and becomes superconducting. The magnitude of the jump depends only on v, Tc , and the slope of the coexistence curve at T = Tc . The speciﬁc heat of tin is shown in Figure 12.2, which illustrates the jump in speciﬁc heat when the tin becomes superconducting. This relationship provides a good way to determine how much of a sample is superconducting. In many cases, a sample may have impurities or other phases

CHAPTER 12. THERMODYNAMICS OF MAGNETIC SYSTEMS

300

Figure 12.2: The speciﬁc heat of tin as a function of temperature. Open circles: no magnetic ﬁeld; solid circles: H > Hc (normal phase); dotted line: electron contribution to normal speciﬁc heat; dashed line: lattice contribution to normal speciﬁc heat. From W. Buckel, Superconductivity: Fundamentals and Applications, VCH, 1991.

CHAPTER 12. THERMODYNAMICS OF MAGNETIC SYSTEMS

301

present which do not become superconducting. As long as the superconducting portion forms a continuous network, the electrical resistance will be zero, even if the superconduction portion is very small. But the jump in speciﬁc heat is dependent on the amount of the sample which undergoes the phase transition (through v) and so can be used to determine the superconducting fraction. 12.3.1 Microscopic Interpretation The results derived above were based on a few experimental observations, such as the Meissner eﬀect (M = −H). They did not require us to know anything about what causes superconductivity, or why M = −H. In fact, these thermodynamic results were known long before a satisfactory microscopic theory of superconductivity had been developed. The theory which ﬁnally gave a clear microscopic picture of superconductivity was developed in the 1950’s by Bardeen, Cooper, and Schrieﬀer, and their theory is known as the BCS theory. The BCS theory shows that the reason some materials become superconducting at low temperature is that the electrons pair up in quantum-mechanical bound states called “Cooper pairs.” The pairing is brought about by an interaction between the electrons and the positivey-charged lattice. Since an electron has negative charge, the lattice atoms near an electron are attracted to it, leading to excess positive charge around an electron. This positive charge in turn attracts another electron, which creates an “eﬀective” attractive interaction between the two electrons. Of course, the electrons also repel one another, but in some cases the attractive interaction can be stronger than the repulsion and the two electrons become bound together. The situation is actually very similar to how two protons and two electrons bind together to form an H2 molecule. Even though the electrons repel one another and so do the protons, by keeping the electrons most of the time between the protons, the electron-proton attraction is greater than the proton-proton and electron-electron repulsion and a stable molecule forms. The electrons and postively-charged lattice distortions (“phonons”) arrange themselves similarly to form a stable Cooper pair. Just like a molecule, a Cooper pair requires energy to break it up. It turns out that the number of Cooper pairs nc (T ) is a function of temperature, but hardly depends on H. The number of pairs goes continuously to zero at T = Tc . This is why the speciﬁc heat of a superconductor is greater than in the normal phase – it requires energy to break up some Cooper pairs if the temperature is raised. A Cooper pair has zero total angular momentum, and so are bosons obeying

CHAPTER 12. THERMODYNAMICS OF MAGNETIC SYSTEMS

302

Bose-Einstein statistics (the individual electrons, in contrast are fermions and obey Fermi-Dirac statistics). There is no restriction on the number of bosons which can occupy a single quantum level. In fact, the BCS theory shows the most-stable state of the system (the ground state) is one in which all Cooper pairs occupy the same quantum state. Thus, a single quantum-mechanical wavefunction which extends throughout the superconductor describes all of the Cooper pairs, which are rigidly locked together in phase (much like photons in laser light). It is this macroscopic quantum-mechanical aspect which is responsible for all of the strange properties such as zero resistance.

APPENDIX A
STATISTICAL MECHANICS OF THE MONATOMIC IDEAL GAS
NOTE: this appendix is taken unmodiﬁed from some older lecture notes I wrote for a diﬀerent class. The notation may be a little diﬀerent. Consider a container of volume V which contains N atoms, which we will assume are ”structureless” – that is, they contain no internal energy modes (vibration, rotation, or electronic excitation). Let us also assume that the potential energy due to interatomic forces is negligible compared to the kinetic energy of the atoms. These assumptions deﬁne the ”monatomic ideal gas.” For simplicity, let the container be a cube of side length L, such that V = L3 . (The results we obtain will be independent of the assumed container geometry.) The microstate of the system is speciﬁed by the positions and momenta of all atoms. If there are N atoms, then 3N coordinates (3 for each atom) must be speciﬁed, and 3N momenta, for a total of 6N degrees of freedom. The microstate of the gas is then fully speciﬁed by the 6N numbers {x1 , . . . , x3N , p1 , . . . , p3N }. We may represent the microstate geometrically as a point in a 6N -dimensional space, where the space is spanned by 3N orthogonal spatial coordinate axes xi and 3N orthogonal momentum axes pi. We call this 6N -dimensional space ”phase space.” It is also useful to consider 2 subspaces of phase space. We will deﬁne “coordinate space” to be that space spanned by the 3N coordinate axes, and “momentum space” to be that space spanned by the 3N momentum axes. In order keep the number of microstates ﬁnite, we divide up each coordinate axis into segments of length ∆x, and each momentum axis into segments of length ∆p. We can then construct a grid, or lattice, which divides phase space into small hypercubes of volume ∆W = (∆x∆p)3N . We will take all points within a given cube to correspond to one microstate. The number of microstates in a region of phase space with volume W is then simply Ω = W/∆W. (A.1)

Therefore, rather than trying to count microstates directly (counting becomes 157

APPENDIX A. STATISTICAL MECHANICS OF THE MONATOMIC IDEAL GAS158 diﬃcult when the numbers get very large!), we can ﬁnd the volume of the region of phase space which satisﬁes some criteria, and then get Ω from Eq. (A.1). Of course, the number of microstates will depend upon our choices of ∆x and ∆p. We’ll see later that the only implication of this is that the entropy can only be determined to within some additive constant. But this was also true in thermodynamics – for the ideal gas, we integrated the Gibbs equation to get the entropy, which introduced an arbitrary integration constant s0 . Recall that we introduced a postulate (the 3rd law of thermodynamics) which ﬁxed the entropy at one temperature (T = 0) and removed the ambiguity. In statistical mechanics, no such postulate is needed – the problem is taken care of by a proper quantum-mechanical treatment of the problem. Our objective is to count how many microstates (i.e., cubes in phase space) there are which are consistent with our macroscopic knowledge of the state of the system, for example the volume V and the internal energy U . Some of the microstates are consistent with our macroscopic speciﬁcations, but many are not (for example, those for which one or more atom lies outside the container). There are two constraints on the system, which determine the region of phase space we are interested in. The ﬁrst is that all atoms are located within the container: (A.2) 0 < xi < L, i = 1, . . . , 3N. The second constraint is the speciﬁcation of the total internal energy of the system. The total kinetic energy E of the gas is given by
3N

E= i=1 1 2 mv = 2 i

3N

i=1

p2 i , 2m

(A.3)

where m is the mass of one atom. This equation shows that microstates with √ a given kinetic energy all lie on a sphere of radius R = 2mE in momentum space. Since we are assuming that translational kinetic energy is the only contribution to the total internal energy U of the gas (no potential energy or internal energy modes), the speciﬁcation of the energy of the system is that E = U . Actually, we will assume that there is a very small, but non-zero, uncertainty in the energy. That is, the constraint on energy is U − δU < E < U, (A.4)

where δU U . We’ll assume that ∆x and ∆p are chosen to be small enough that there are very many microstates with kinetic energy in this range.

APPENDIX A. STATISTICAL MECHANICS OF THE MONATOMIC IDEAL GAS159 Let us now ﬁnd the volume W of the region of phase space satisfying eqs. (A.2) and (A.4). This volume may be expressed as a multidimensional integral:
L L L

W =
0

dx1
0

dx2 . . .
0

dx3N Wshell

(A.5)

where Wshell is the volume of the spherical shell in momentum space deﬁned by Eq. (A.4). Since the kinetic energy does not depend on the spatial coordinates (it only depends on the momentum coordinates), the integral in momentum space deﬁning Wshell is independent of the coordinate values, and we may reverse the order of integration and do the coordinate integrals ﬁrst. Each of these contributes a factor L, and therefore W = L3N Wshell = V N Wshell (A.6)

where V = L3 is the container volume (in real 3D space). To determine Wshell we may use a result from mathematics for the volume Wn (R) of a sphere of radius R is n-dimensional space. It may be shown that Wn (R) = πR2 n 2 n/2

!

.

(A.7)

Note that if n is odd we can evaluate (n/2)! in terms of the Gamma function: z! = Γ(z + 1) for any z. The Gamma function is deﬁned by
∞

Γ(z) =
0

tz−1 e−t dt

(A.8)

and satisﬁes the recurrence formula Γ(z + 1) = zΓ(z). Therefore, Γ(n/2) may √ be determined from this recurrence formula, given that Γ(3/2) = π/2. For the familiar cases of n = 2 and n = 3, Eq. (A.7) reduces to W2 = πR2 and W3 = (4/3)πR3 , respectively, as expected. Wshell may be determined by the diﬀerence of two sphere volumes: √ Wshell = W3N ( 2mU ) − W3N 2m(U − δU ) (A.9) Or, using Eq. (A.7), √ δU Wshell = W3N ( 2mU ) 1 + 1 − U
3N/2

.

(A.10)

Note that the term (1 − δU/U ) is very slightly less than one. Therefore, for any ﬁnite (but very small) δU , we have that lim 1− δU U
3N/2

N→∞

=0

(A.11)

APPENDIX A. STATISTICAL MECHANICS OF THE MONATOMIC IDEAL GAS160 and therefore √ lim Wshell = W3N ( 2mU ).

N→∞

(A.12)

This seemingly-paradoxical result states that in the limit of large N , the shell volume equals the volume of the entire sphere! This is in fact the case: as the dimensionality of the space increases, the distribution of the volume of the sphere with radius becomes highly peaked near the surface of the sphere. For any small but ﬁnite shell width, if n is large enough then virtually all of the sphere volume is located within the shell. This means that for large N , the number of microstates available to a gas which have kinetic energy very close to U is overwhelmingly larger than the number of microstates which have kinetic energy less than U . In the large N limit (which is what we are interested in), we have then that √ Ω = W3N ( 2mU )/∆W or Ω= (2πmU )3N/2 . ∆W 3N ! 2 (A.13)

(A.14)

Actually, there is one correction we need to make to Eq. (A.14). In counting microstates by integrating over phase space, we have implicitly assumed that the atoms are distinguishable. That is, suppose we interchange the position and momentum coordinates of particles 1 and 2. Then if the original microstate was {x1 , x2 , x3 , x4, x5 , x6 , x7 , . . . , x3N , p1 , p2 , p3 , p4 , p5 , p6 , p7, . . . , p3N }, atom 1 atom 2 atom 1 atom 2

the microstate with atoms 1 and 2 interchanged will be {x4 , x5 , x6 , x1, x2 , x3 , x7 , . . . , x3N , p4 , p5 , p6 , p1 , p2 , p3 , p7, . . . , p3N }. atom 2 atoms 1 atom 2 atom 1

The second microstate is diﬀerent from the ﬁrst unless atoms 1 and 2 happen to have the same position and momentum vectors, such that x1 = x4 , etc. For a low-density gas, this is extremely unlikely, so in general exchanging atoms 1 and 2 produces a diﬀerent microstate of the gas, which corresponds to a diﬀerent point in phase space. Since this microstate satisﬁes the macroscopic constraints [eqs. Eq. (A.2) and Eq. (A.4)], both of the above microstates are in the accessible region of phase space, and therefore we have counted both of them in Ω. This is correct if atoms 1 and 2 are distinguishable, so that exchanging them really does produce a new microstate of the gas. However, it turns out

APPENDIX A. STATISTICAL MECHANICS OF THE MONATOMIC IDEAL GAS161 that atoms which are identical (i.e., the same isotope of the same element) are actually indistinguishable in nature. That is, exchanging them does not produce a new microstate of the system. Since we have in eﬀect treated the atoms as distinguishable when counting microstates, we have overcounted the number of microstates, and need to correct our expression for Ω. Let us assume that the gas density is low enough that we can neglect the possibility that any 2 atoms are at the same position (to within ∆x) and have the same momentum (to within ∆p). Then we simply need to divide Ω by the number of ways to permute the atom labels among the N atoms, which is N !. Therefore, the correct expression for Ω is Ω= (2πmU )3N/2 . ∆W N ! 3N ! 2 (A.15)

Using Stirling’s formula to approximate the factorial terms (n! ≈ nn e−n ), we have N 3N/2 1 eV 4πemU Ω= (A.16) N 3N (∆x∆p)3N Let us take the product ∆x∆p to be some constant a. Then the entropy is S = k ln Ω = N k ln v + ˆ 3 ln 2 4πmˆ u 3a2 + 5 2 (A.17) (A.18)

where v = V /N and u = U/N . Since this expresses S as a function of U , V , ˆ ˆ and N , we have succeeded in deriving a fundamental relation for the ideal gas. The only uncertainty remaining is the additive constant resulting from the unknown value of a. If we had analyzed this problem from the point of view of quantum mechanics (rather than classical mechanics), we would have found the same result for S as in Eq. (A.18), but with the constant a replaced by Planck’s constant h, which is a fundamental constant, given by h = 6.626 × 10−34 With the substitution a = h, we then have s = k ln v + ˆ ˆ 3 ln 2 4πmˆ u 2 3h + 5 . 2 (A.20) J-s. (A.19)

(Note that the units of s are J/molecule/K. If we would rather work with moles, ˆ ˆ we only need replace k by R.)

APPENDIX A. STATISTICAL MECHANICS OF THE MONATOMIC IDEAL GAS162 Evaluating the temperature from 1/T = (∂ˆ/∂ u)v we ﬁnd s ˆ ˆ 3 u = kT. ˆ 2 Therefore, Eq. (A.20) may also be expressed as s = k ln v + ˆ ˆ 3 ln 2 2πmkT h2 + 5 . 2 (A.22) (A.21)

Index adiabatic, 35 air-standard cycle, 250 Avagadro’s Number, 149 boiler, 126 Boltzmann’s constant, 87 Brayton cycle, 250 cavitation, 108 center of mass, 20 chemical potential, 54 choked, 281 coexistence lines, 77 compressibility factor, 88 condenser, 126 conjugate variables, 218 conservative force, 19 deﬁnition, 20 control mass, 4 control volume, 4 COP, 168 Corresponding States Principle of, 100 critical point, 73 Curie substance, 43 datum state, 71 degrees of freedom, 63 dielectric, 42 diﬀerential equations of state, 192 diﬀuser, 119 electric dipole moment, 40 electric susceptibility, 41 endoreversible, 174 309 endothermic, 200 energy, 3 accounting, 9 balance, 9 conservation of, 24 enthalpy deﬁnition, 70 stagnation, 119 enthalpy of vaporization, 126 entropy, 3 ideal gas, 142 environment, 3 equations of state, 65 equilibrium, 50 chemical, 55 diﬀusive, 53 mechanical, 52 phase, 54 restricted, 55 thermal, 52 equilibrium properties, 108 equipartition of energy, 91 evaporator, 126 exact diﬀerential, 189 exothermic, 200 extensive, 62 First Law of Thermodynamics, 34 force centrifugal, 17 Lorentz, 17, 21 traction, 17 fusion, 206 gas constant, 86

INDEX gas turbine cycle, 249 generalized displacements, 44 generalized equations of state, 98 generalized forces, 44 Gibbs free energy, 196 heat heat heat heat capacity, 86 exchangers, 124 of vaporization, 126 transfer, 32 rate, 35 Helmholtz free energy, 195 hydrostatics, 57 ideal cycles, 229 ideal gas, 87 incompressible substance, 102 inertial mass, 7 intensive, 62 intercooling, 259 internal energy, 28 irreversibility rate, 181 irreversible expansion, 135 isentropic, 159 isobar, 78 isothermal compressibility, 84 kinetic energy, 8 latent heat, 200 Lee-Kesler generalized compressibility, 100 lever rule, 80 liquid compressed, 79 subcooled, 79 local thermodynamic equilibrium, 107 Mach Number, 276 macrostate, 130 magnetization, 43 manometer, 57 mass density, 20 microstate, 130 normal shock waves, 285 nozzle, 117 oblique, 285 one-dimensional ﬂow, 108 phase, 54 phase diagrams, 72 Planck’s constant, 142 polar, 40 polarization, 41 potential energy deﬁnition, 23 gravitational, 19 power, 12 pressure, 33 critical, 72 reduced, 98 quality, 79 quasi-static, 37 reference state, 71 regenerator eﬀectiveness, 258

310

saturation, 78 saturation pressure, 77 saturation temperature, 77 Second Law Clausius statement, 167 Kelvin-Planck statement, 164 speciﬁc, 62 speciﬁc heat cp , 85 cv , 85

INDEX stagnation state, 276 steady ﬂow, 109 steady ﬂow availability function, 181 steady state, 116 Stirling’s approximation, 142 sublimation, 73 subsonic, 276 supercooled vapor, 98 superheated liquid, 98 supersonic, 276 surface tension, 39 system, 3 classiﬁcation, 4 temperature, 53 critical, 73 reduced, 98 thermodynamic, 61 thermal conductivity, 161 thermal eﬃciency, 165 thermal energy, 1 thermal expansion coeﬃcient, 83 thermal reservoir, 163 thermodynamic pressure, 147 thermodynamic state, 63 thermodynamic temperature, 146 triple line, 75 triple point, 75 unrestrained expansion, 135 valve, 120 van der Waals equation of state, 94 vapor pressure, 77 work, 8, 10 deﬁnition of, 10

311

Physical Chemistry

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