MIT Course 16 Fall 2002

Thermal Energy

16.050

Prof. Z. S. Spakovszky

Notes by E.M. Greitzer Z. S. Spakovszky

Table of Contents

PART 0 - PRELUDE: REVIEW OF “UNIFIED ENGINEERING THERMODYNAMICS”

0.1 What it's all about 0-1 0.2 Definitions and fundamentals ideas of thermodynamics 0-1 0.3 Review of thermodynamics concepts 0-2

PART 1 - THE SECOND LAW OF THERMODYNAMICS

1.A- Background to the Second Law of Thermodynamics 1.A.1 Some Properties of Engineering Cycles: Work and Efficiency 1A-1 1.A.2 Carnot Cycles 1A-3 1.A.3 Brayton Cycles (or Joule Cycles): The Power Cycle for a Gas Turbine Jet 1A-5 Engine 1.A.4 Gas Turbine Technology and Thermodynamics 1A-8 1.A.5 Refrigerators and heat pumps 1A-11 1.A.6 Reversibility and Irreversibility in Natural Processes 1A-12 1.A.7 Difference between Free Expansion of a Gas and Reversible Isothermal 1A-14 Expansion 1.A.8 Features of reversible Processes 1A-16

1.B The Second Law of Thermodynamics 1.B.1 Concept and Statements of the Second Law 1B-1 1.B.2 Axiomatic statements of the Laws of Thermodynamics 1B-3 1.B.3 Combined First and Second Law Expressions 1B-5 1.B.4 Entropy Changes in an Ideal Gas 1B-6 1.B.5 Calculation of Entropy Change in Some Basic Processes 1B-7

1.C Applications of the Second Law 1.C.1 Limitations on the Work that Can be Supplied by a Heat Engine 1C-1 1.C.2 The thermodynamic Temperature Scale 1C-3 1.C.3 Representation of Thermodynamic Processes in T-s coordinates 1C-4 1.C.4 Brayton Cycle in T-s coordinates 1C-5 1.C.5 Irreversibility, Entropy Changes, and Lost Work 1C-8 1.C.6 Entropy and Unavailable Energy 1C-11 1.C.7 Examples of Lost Work in Engineering Processes 1C-14 1.C.8 Some Overall Comments on Entropy, Reversible and Irreversible Processes 1C-23

1.D Interpretation of Entropy on the Microscopic Scale - The Connection between Randomness and Entropy

1.D.1 Entropy Change in Mixing of Two ideal Gases 1D-1 1.D.2 Microscopic and Macroscopic Descriptions of a System 1D-2 1.D.3 A Statistical Definition of Entropy 1D-3 1.D.4 Connection between the Statistical Definition of Entropy and Randomness 1D-5

1.D.5 Numerical Example of the Approach to the Equilibrium Distribution 1D-6 1.D.6 Summary and Conclusions 1D-11

PART 2 - POWER AND PROPULSION CYCLES

2.A Gas Power and Propulsion Cycles 2.A.1 The Internal Combustion Engine (Otto Cycle) 2A-1 2.A.2 Diesel Cycle 2A-4 2.A.3 Brayton Cycle 2A-5 2.A.4 Brayton Cycle for Jet Propulsion: the Ideal Ramjet 2A-6 2.A.5 The Breguet Range Equation 2A-8 2.A.6 Performance of the Ideal Ramjet 2A-11 2.A.7 Effect of departures from Ideal Behavior 2A-14

2.B Power Cycles with two-Phase media 2.B.1Behavior of Two-Phase Systems 2B-1 2.B.2 Work and Heat Transfer with Two-Phase Media 2B-5 2.B.3 The Carnot Cycle as a Two-Phase Power Cycle 2B-8 2.B.4 Rankine Power Cycles 2B-13 2.B.5 Enhancements of, and Effect of Design Parameters on Rankine Cycles 2B-15 2.B.6 Combined Cycles in Stationary Gas Turbine for Power Production 2B-19 2.B.7 Some Overall Comments on Thermodynamic Cycles 2B-21

2.C Introduction to Thermochemistry 2.C.1 Fuels 2C-1 2.C.2 Fuel-Air Ratio 2C-2 2.C.3 Enthalpy of Formation 2C-2 2.C.4 First Law analysis of Reacting systems 2C-4 2.C.5 Adiabatic Flame Temperature 2C-7

PART 3 - INTRODUCTION TO ENGINEERING HEAT TRANSFER

1.0 Heat Transfer Modes.......................................................................................................... HT-52.0 Conduction Heat Transfer .................................................................................................. HT-52.1 Steady-State One-Dimensional Conduction.................................................................... HT-82.2 Thermal Resistance Circuits ..........................................................................................HT-102.3 Steady Quasi-One-Dimensional Heat Flow in Non-Planar Geometry ........................... HT-143.0 Convective Heat Transfer..................................................................................................HT-183.1 The Reynolds Analogy..................................................................................................HT-193.2 Combined Conduction and Convection .........................................................................HT-243.3 Dimensionless Numbers and Analysis of Results ..........................................................HT-294.0 Temperature Distributions in the Presence of Heat Sources...............................................HT-325.0 Heat Transfer From a Fin ..................................................................................................HT-356.0 Transient Heat Transfer (Convective Cooling or Heating) .................................................HT-407.0 Some Considerations in Modeling Complex Physical Processes........................................HT-428.0 Heat Exchangers ...............................................................................................................HT-43

8.1 Efficiency of a Counterflow Heat Exchanger.................................................................HT-509.0 Radiation Heat Transfer (Heat transfer by thermal radiation).............................................HT-529.1 Ideal Radiators ..............................................................................................................HT-539.2 Kirchhoff's Law and "Real Bodies" ...............................................................................HT-559.3 Radiation Heat Transfer Between Planar Surfaces .........................................................HT-559.4 Radiation Heat Transfer Between Black Surfaces of Arbitrary Geometry ......................HT-60

ACKNOWLEDGEMENT

Preparation of these notes has benefited greatly from the expertise of a number of individuals, and we are pleased to acknowledge this help. Jessica Townsend, Vincent Blateau, Isabel Pauwels, and David Milanes, the successive Teaching Assistants in this core department course, provided ideas, corrected errors, inserted “Muddy Points”, supplied the index, and in general, created a much more readable document. Any errors that remain, or lack of readability, are thus the sole responsibility of the authors. We also appreciate the work of Diana Park and Robin Palazzolo, who contributed greatly to the editing and graphics. Finally, we are grateful to have had the opportunity to discuss some of the material with Professor Frank Marble of Caltech, whose understanding, insight, and ability to describe thermofluids concepts provide a model of how to address important technical problems.

PART 0

PRELUDE: REVIEW OF "UNIFIED ENGINEERINGTHERMODYNAMICS"

0-1

PART 0 - PRELUDE: REVIEW OF “UNIFIED ENGINEERING THERMODYNAMICS”[IAW pp 2-22, 32-41 (see IAW for detailed SB&VW references); VN Chapter 1]

0.1 What it’s All About

The focus of thermodynamics in 16.050 is on the production of work, often in the form ofkinetic energy (for example in the exhaust of a jet engine) or shaft power, from different sources ofheat. For the most part the heat will be the result of combustion processes, but this is not always thecase. The course content can be viewed in terms of a “propulsion chain” as shown below, where wesee a progression from an energy source to useful propulsive work (thrust power of a jet engine). Interms of the different blocks, the thermodynamics in Unified Engineering and in this course aremainly about how to progress from the second block to the third, but there is some examination ofthe processes represented by the other arrows as well. The course content, objectives, and lectureoutline are described in detail in Handout #1.

0.2 Definitions and Fundamental Ideas of Thermodynamics

As with all sciences, thermodynamics is concerned with the mathematical modeling of thereal world. In order that the mathematical deductions are consistent, we need some precisedefinitions of the basic concepts.

A continuum is a smoothed-out model of matter, neglecting the fact that real substances arecomposed of discrete molecules. Classical thermodynamics is concerned only with continua. Ifwe wish to describe the properties of matter at a molecular level, we must use the techniques ofstatistical mechanics and kinetic theory.

A closed system is a fixed quantity of matter around which we can draw a boundary. Everythingoutside the boundary is the surroundings. Matter cannot cross the boundary of a closed systemand hence the principle of the conservation of mass is automatically satisfied whenever we employa closed system analysis.

The thermodynamic state of a system is defined by the value of certain properties of that system.For fluid systems, typical properties are pressure, volume and temperature. More complex systemsmay require the specification of more unusual properties. As an example, the state of an electricbattery requires the specification of the amount of electric charge it contains.

Properties may be extensive or intensive. Extensive properties are additive. Thus, if the system isdivided into a number of sub-systems, the value of the property for the whole system is equal tothe sum of the values for the parts. Volume is an extensive property. Intensive properties do notdepend on the quantity of matter present. Temperature and pressure are intensive properties.

Specific properties are extensive properties per unit mass and are denoted by lower case letters.For example:

specific volume = V/m = v.

Energy sourcechemicalnuclear, etc.

Heat(combustionprocess)

Mechanicalwork,electric power...

Useful propulsivework (thrustpower)

0-2

Specific properties are intensive because they do not depend on the mass of the system,

A simple system is a system having uniform properties throughout. In general, however,properties can vary from point to point in a system. We can usually analyze a general system bysub-dividing it (either conceptually or in practice) into a number of simple systems in each ofwhich the properties are assumed to be uniform.

If the state of a system changes, then it is undergoing a process. The succession of states throughwhich the system passes defines the path of the process. If, at the end of the process, theproperties have returned to their original values, the system has undergone a cyclic process. Notethat although the system has returned to its original state, the state of the surroundings may havechanged.

Muddy points Specific properties (MP 0.1)What is the difference between extensive and intensive properties? (MP 0.2)

0.3 Review of Thermodynamic Concepts

The following is a brief discussion of some of the concepts introduced in UnifiedEngineering, which we will need in 16.050. Several of these will be further amplified in thelectures and in other handouts. If you need additional information or examples concerning thesetopics, they are described clearly and in-depth in the Unified Notes of Professor Waitz, wheredetailed references to the relevant sections of the text (SB&VW) are given. They are also covered,although in a less detailed manner, in Chapters 1 and 2 of the book by Van Ness.

1) Thermodynamics can be regarded as a generalization of an enormous body of empiricalevidence. It is extremely general, and there are no hypotheses made concerning the structureand type of matter that we deal with.

2) Thermodynamic system :A quantity of matter of fixed identity. Work or heat (see below) can be transferred across

the system boundary, but mass cannot.

Gas, FluidSystem

Boundary

3) Thermodynamic properties :For engineering purposes, we want "averaged" information, i.e., macroscopic not

microscopic (molecular) description. (Knowing the position and velocity of each of 1020+

molecules that we meet in typical engineering applications is generally not useful.)

4) State of a system :The thermodynamic state is defined by specifying values of a (small) set of measured

properties which are sufficient to determine all the remaining properties.

0-3

5) Equilibrium :The state of a system in which properties have definite (unchanged) values as long as

external conditions are unchanged is called an equilibrium state. Properties (P, pressure, T,temperature, ρ, density) describe states only when the system is in equilibrium.

Mg + PoA = PA

Gas atPressure, P

Mass

MechanicalEquilibrium

Po

Insulation

Copper Partition

Thermal Equilibrium

GasT1

Over time, T1 → T2

GasT2

6) Equations of state: For a simple compressible substance (e.g., air, water) we need to know two properties to set

the state. Thus:

P = P(v,T), or v = v(P, T), or T = T(P,v)

where v is the volume per unit mass, 1/ρ.

Any of these is equivalent to an equation f(P,v,T) = 0 which is known as an equation of state. Theequation of state for an ideal gas, which is a very good approximation to real gases at conditionsthat are typically of interest for aerospace applications is:

Pv– = RT,

where v– is the volume per mol of gas and R is the "Universal Gas Constant", 8.31 kJ/kmol-K.

A form of this equation which is more useful in fluid flow problems is obtained if wedivide by the molecular weight, M:

Pv = RT, or P = ρRT

where R is R/M, which has a different value for different gases. For air at room conditions, R is0.287 kJ/kg-K.

7) Quasi-equilibrium processes: A system in thermodynamic equilibrium satisfies:

a) mechanical equilibrium (no unbalanced forces)b) thermal equilibrium (no temperature differences)c) chemical equilibrium.

For a finite, unbalanced force, the system can pass through non-equilibrium states. We wish todescribe processes using thermodynamic coordinates, so we cannot treat situations in which suchimbalances exist. An extremely useful idealization, however, is that only "infinitesimal"unbalanced forces exist, so that the process can be viewed as taking place in a series of "quasi-equilibrium" states. (The term quasi can be taken to mean "as if"; you will see it used in a numberof contexts such as quasi-one-dimensional, quasi-steady, etc.) For this to be true the process mustbe slow in relation to the time needed for the system to come to equilibrium internally. For a gas

0-4

at conditions of interest to us, a given molecule can undergo roughly 1010 molecular collisions per

second, so that, if ten collisions are needed to come to equilibrium, the equilibration time is on theorder of 10-9 seconds. This is generally much shorter than the time scales associated with the bulkproperties of the flow (say the time needed for a fluid particle to move some significant fraction ofthe lighten of the device of interest). Over a large range of parameters, therefore, it is a very goodapproximation to view the thermodynamic processes as consisting of such a succession ofequilibrium states.

8) Reversible process For a simple compressible substance,

Work = ∫PdV.

If we look at a simple system, for example a cylinder of gas and a piston, we see that there can betwo pressures, Ps, the system pressure and Px, the external pressure.

Ps

Px

The work done by the system on the environment is

Work = ∫PxdV.

This can only be related to the system properties if Px ≈ Ps. For this to occur, there cannot be anyfriction, and the process must also be slow enough so that pressure differences due to accelerationsare not significant.

Work during anirreversible process ≠ ∫Ps dV

∫PxdV ≠ 0 but ∫Ps dV = 0

Ps (V)

Px with friction

P

Vs

➀

➀ ➀➁

➁

Under these conditions, we say that the process is reversible. The conditions for reversibility arethat:

a) If the process is reversed, the system and the surroundings will be returned to theoriginal states.

b) To reverse the process we need to apply only an infinitesimal dP. A reversible processcan be altered in direction by infinitesimal changes in the external conditions (see VanNess, Chapter 2).

9) Work: For simple compressible substances in reversible processes , the work done by the system

on the environment is ∫PdV. This can be represented as the area under a curve in a Pressure-volume diagram:

0-5

P

Volume

Work is area under curve of P(V)

V1 V2

W1-2 ≠P

Work depends on the path

V

I

IIIW1-2

II

a) Work is path dependent;b) Properties only depend on states;c) Work is not a property, not a state variable;d) When we say W1-2, the work between states 1 and 2, we need to specify the path;e) For irreversible (non-reversible) processes, we cannot use ∫PdV; either the work must

be given or it must be found by another method.

Muddy points How do we know when work is done? (MP 0.3)

10) Heat Heat is energy transferred due to temperature differences.

a) Heat transfer can alter system states;b) Bodies don't "contain" heat; heat is identified as it comes across system boundaries;c) The amount of heat needed to go from one state to another is path dependent;d) Heat and work are different modes of energy transfer;e) Adiabatic processes are ones in which no heat is transferred.

11) First Law of Thermodynamics For a system,

∆E Q W= −

E is the energy of the system,Q is the heat input to the system, andW is the work done by the system.E = U (thermal energy) + Ekinetic + Epotential + ....

If changes in kinetic and potential energy are not important,WQU −=∆ .

a) U arises from molecular motion.b) U is a function of state, and thus ∆U is a function of state (as is ∆E ).c) Q and W are not functions of state.

Comparing (b) and (c) we have the striking result that:

d) ∆U is independent of path even though Q and W are not!

0-6

Muddy points What are the conventions for work and heat in the first law? (MP 0.4)When does E->U? (MP 0.5)

12) Enthalpy:A useful thermodynamic property, especially for flow processes, is the enthalpy. Enthalpy

is usually denoted by H, or h for enthalpy per unit mass, and is defined by:

H = U + PV.

In terms of the specific quantities, the enthalpy per unit mass is

h = u + Pv = u P+ / ρ .

13) Specific heats - relation between temperature change and heat inputFor a change in state between two temperatures, the “specific heat” is:

Specific heat = Q/(Tfinal - Tinitial)

We must, however, specify the process, i.e., the path, for the heat transfer. Two useful processesare constant pressure and constant volume. The specific heat at constant pressure is denoted as Cpand that at constant volume as Cv, or cp and cv per unit mass.

ch

Tc

u

Tpp

vv

=

=

∂∂

∂∂

and

For an ideal gas

dh = cpdT and du = cvdT.

The ratio of specific heats, cp/cv is denoted by γ. This ratio is 1.4 for air at room conditions.

The specific heats cv. and cp have a basic definition as derivatives of the energy and enthalpy.Suppose we view the internal energy per unit mass, u, as being fixed by specification ofT, the temperature and v, the specific volume, i.e., the volume per unit mass. (For a simplecompressible substance, these two variables specify the state of the system.) Thus,

u = u(T,v).

The difference in energy between any two states separated by small temperature and specific volumedifferences, dT and dv is

du = ∂u∂T v

dT + ∂u∂v T

dv

The derivative ∂u ∂T∂u ∂T v represents the slope of a line of constant v on a u-T plane. The derivative isalso a function of state, i. e., a thermodynamic property, and is called the specific heat at constantvolume, cv.

The name specific heat is perhaps unfortunate in that only for special circumstances is thederivative related to energy transfer as heat. If a process is carried out slowly at constant volume, nowork will be done and any energy increase will be due only to energy transfer as heat. For such a

0-7

process, cv does represent the energy increase per unit of temperature (per unit of mass) andconsequently has been called the "specific heat at constant volume". However, it is more useful tothink of cv in terms of its definition as a certain partial derivative, which is a thermodynamicproperty, rather than a quantity related to energy transfer as heat in the special constant volumeprocess.

The enthalpy is also a function of state. For a simple compressible substance we can regardthe enthalpy as a function of T and P, that is view the temperature and pressure as the two variablesthat define the state. Thus,

h = h(T,P).

Taking the differential,

dh = ∂h∂T P

dT + ∂h∂P T

dP

The derivative ∂h ∂T∂h ∂T P is called the specific heat at constant pressure, denoted by cp.

The derivatives cv and cp constitute two of the most important thermodynamic derivativefunctions. Values of these properties have been experimentally determined as a function of thethermodynamic state for an enormous number of simple compressible substances.

14) Ideal GasesThe equation of state for an ideal gas is

PV = NRT,

where N is the number of moles of gas in the volume V. Ideal gas behavior furnishes an extremelygood approximation to the behavior of real gases for a wide variety of aerospace applications. Itshould be remembered, however, that describing a substance as an ideal gas constitutes a model ofthe actual physical situation , and the limits of model validity must always be kept in mind.

One of the other important features of an ideal gas is that its internal energy depends onlyupon its temperature. (For now, this can be regarded as another aspect of the model of actualsystems that the perfect gas represents, but it can be shown that this is a consequence of the form ofthe equation of state.) Since u depends only on T,

du = cv (T)dT

In the above equation we have indicated that cv can depend on T.

Like the internal energy, the enthalpy is also only dependent on T for an ideal gas. (If u is afunction of T, then, using the perfect gas equation of state, u + Pv is also.) Therefore,

dh = cP(T)dT.

Further, dh = du + d(Pv) = cv dT + R dT. Hence, for an ideal gas,

cv = cP - R.

In general, for other substances, u and h depend on pressure as well as on temperature. In thisrespect, the ideal gas is a very special model.

0-8

The specific heats do not vary greatly over wide ranges in temperature, as shown in VWB&SFigure 5.11. It is thus often useful to treat them as constant. If so

u2 - u1 = cv (T2 - T1)

h2 - h1 = cp (T2 - T1)

These equations are useful in calculating internal energy or enthalpy differences, but it should beremembered that they hold only for an ideal gas with constant specific heats.

In summary, the specific heats are thermodynamic properties and can be used even if theprocesses are not constant pressure or constant volume. The simple relations between changes inenergy (or enthalpy) and temperature are a consequence of the behavior of an ideal gas, specificallythe dependence of the energy and enthalpy on temperature only, and are not true for more complexsubstances.

Adapted from "Engineering Thermodynamics", Reynolds, W. C and Perkins, H. C,McGraw-Hill Publishers

15) Specific Heats of an Ideal Gas

1. All ideal gases:

(a) The specific heat at constant volume (cv for a unit mass or CV for one kmol) is a function of Tonly.

(b) The specific heat at constant pressure (cp for a unit mass or CP for one kmol) is a function of Tonly.

(c) A relation that connects the specific heats cp, cv, and the gas constant is

cp - cv = R

where the units depend on the mass considered. For a unit mass of gas, e. g., a kilogram, cpand cv would be the specific heats for one kilogram of gas and R is as defined above. For onekmol of gas, the expression takes the form:

CP - CV = R,

where CP and CV have been used to denote the specific heats for one kmol of gas and R is theuniversal gas constant.

(d) The specific heat ratio, γ, = cp/cv (or CP/CV), is a function of T only and is greater than unity.

2. Monatomic gases, such as He, Ne, Ar, and most metallic vapors:

(a) cv (or CV) is constant over a wide temperature range and is very nearly equal to (3/2)R [or(3/2)R , for one kmol].

(b) cp (or CP) is constant over a wide temperature range and is very nearly equal to (5/2)R [or(5/2)R , for one kmol].

(c) γ is constant over a wide temperature range and is very nearly equal to 5/3 [γ = 1.67].

0-9

3. So-called permanent diatomic gases, namely H2, O2, N2, Air, NO, and CO:

(a) cv (or CV ) is nearly constant at ordinary temperatures, being approximately (5/2)R [(5/2)R ,for one kmol], and increases slowly at higher temperatures.

(b) cp (or CP ) is nearly constant at ordinary temperatures, being approximately (7/2)R [(7/2)R ,for one kmol], and increases slowly at higher temperatures.

(c) γ is constant over a temperature range of roughly 150 to 600K and is very nearly equal to 7/5

[γ = 1.4]. It decreases with temperature above this.

4. Polyatomic gases and gases that are chemically active, such as CO2, NH3, CH4, and Freons:The specific heats, cv and cp, and γ vary with the temperature, the variation being different for eachgas. The general trend is that heavy molecular weight gases (i.e., more complex gas molecules thanthose listed in 2 or 3), have values of γ closer to unity than diatomic gases, which, as can be seen

above, are closer to unity than monatomic gases. For example, values of γ below 1.2 are typical ofFreons which have molecular weights of over one hundred.

Adapted from Zemansky, M. W. and Dittman, R. H., "Heat and Thermodynamics", SixthEdition, McGraw-Hill book company, 1981

16) Reversible adiabatic processes for an ideal gasFrom the first law, with Q = 0, du = cvdT, and Work = Pdv

du + Pdv = 0 (i)

Also, using the definition of enthalpy

dh = du + Pdv + vdP. (ii)

The underlined terms are zero for an adiabatic process. Re-writing (i) and (ii),

γ cvdT = - γpdT = vdP.

Pdvc

Combining the above two equations we obtain

-γ Pdv = vdP or -γ dv/v = dP/P (iii)

Equation (iii) can be integrated between states 1 and 2 to give

γln(v2/v1) = ln(P2/P1), or, equivalently,( )( ) 1/ 1122 =γγ vPvP

For an ideal gas undergoing a reversible, adiabatic process, the relation between pressure andvolume is thus:

Pvγ = constant, orP = constant ×ργ .

17) Examples of flow problems and the use of enthalpya) Adiabatic, steady, throttling of a gas (flow through a valve or other restriction)Figure 0-1 shows the configuration of interest. We wish to know the relation between propertiesupstream of the valve, denoted by “1” and those downstream, denoted by “2”.

0-10

Figure 0-1: Adiabatic flow through a valve, a generic throttling process

To analyze this situation, we can define the system (choosing the appropriate system is often acritical element in effective problem solving) as a unit mass of gas in the following two states.Initially the gas is upstream of the valve and just through the valve as indicated. In the final statethe gas is downstream of the valve plus just through the valve. The figures on the left show theactual configuration just described. In terms of the system behavior, however, we could replacethe fluid external to the system by pistons which exert the same pressure that the external fluidexerts, as indicated schematically on the right side of Figure 0-2 below.

Figure 0-2: Equivalence of actual system and piston model

The process is adiabatic, with changes in potential energy and kinetic energy assumed to benegligible. The first law for the system is therefore

WU −=∆ .

The work done by the system is

1122 VPVPW −= .

Use of the first law leads to

111222 VPUVPU +=+ .

In words, the initial and final states of the system have the same value of the quantity U+PV. Forthe case examined, since we are dealing with a unit mass, the initial final states of the system havethe same value of u+Pv.

P1, v1, u1... P2, v2, u2...

system

system

=

=

pistons

pistons

Initialstate

Finalstate

Valve

0-11

Muddy points When is enthalpy the same in initial and final states? (MP 0.6)

b) Another example of a flow process, this time for an unsteady flow, is the transient process offilling a tank, initially evacuated, from a surrounding atmosphere, which is at a pressure P0 and atemperature T0 . The configuration is shown in Figure 0-3.

Figure 0-3: A transient problem—filling of a tank from the atmosphere

At a given time, the valve at the tank inlet is opened and the outside air rushes in. Theinflow stops when the pressure inside is equal to the pressure outside. The tank is insulated, sothere is no heat transfer to the atmosphere. What is the final temperature of the gas in the tank?

This time we take the system to be all the gas that enters the tank. The initial state has thesystem completely outside the tank, and the final state has the system completely inside the tank.The kinetic energy initially and in the final state is negligible, as is the change in potential energyso the first law again takes the form,

WU −=∆ .

Work is done on the system, of magnitude 00VP , where V0 is the initial volume of the system, so

00VPU =∆ .

In terms of quantities per unit mass ( umU ∆=∆ , 00 mvV = , where m is the mass of the system),

00vPuuu ifinal =−=∆ .

The final value of the internal energy is

. 0

00

hh

vPuu

i

ifinal

==

+=

For a perfect gas with constant specific heats,

TchTcu pv == ; ,

0TcTc pfinalv = ,

V0

System(all the gas that goesinto the tank)

P0

VacuumP0 ,T0

Valve

0-12

00 TTc

cT

v

pfinal γ== .

The final temperature is thus roughly 200oF hotter than the outside air!

It may be helpful to recap what we used to solve this problem. There were basically four steps:1 Definition of the system2 Use of the first law3 Equating the work to a “PdV” term4 Assuming the fluid to be a perfect gas with constant specific heats.

A message that can be taken from both of these examples (as well as from a large numberof other more complex situations, is that the quantity h = u + Pv occurs naturally in problems offluid flow. Because the combination appears so frequently, it is not only defined but also tabulatedas a function of temperature and pressure for a number of working fluids.

Muddy points In the filling of a tank, why (physically) is the final temperature in the tank higher thanthe initial temperature? (MP 0.7)

18) Control volume form of the system laws (Waitz pp 32-34, VWB&S, 6.1, 6.2)The thermodynamic laws (as well as Newton’s laws) are for a system, a specific quantity of

matter. More often, in propulsion and power problems, we are interested in what happens in afixed volume, for example a rocket motor or a jet engine through which mass is flowing. For thisreason, the control volume form of the system laws is of great importance. A schematic of thedifference is shown below. Rather than focus on a particle of mass which moves through theengine, it is more convenient to focus on the volume occupied by the engine. This requires us touse the control volume form of the thermodynamic laws.

Figure 0-4: Control volume and system for flow through a propulsion device

The first of these is conservation of mass. For the control volume shown, the rate ofchange of mass inside the volume is given by the difference between the mass flow rate in and themass flow rate out. For a single flow coming in and a single flow coming out this is

outinCV mmdt

dm ••

−= .

If the mass inside the control volume changes with time it is because some mass is added or someis taken out.

The first law of thermodynamics can be written as a rate equation:

••

−= WQdt

dE.

To derive the first law as a rate equation for a control volume we proceed as with the massconservation equation. The physical idea is that any rate of change of energy in the control

System at time tf

Engine

System at ti

0-13

volume must be caused by the rates of energy flow into or out of the volume. The heat transferand the work are already included and the only other contribution must be associated with the massflow in and out, which carries energy with it. The figure below shows a schematic of this idea.

dEcvdt

Wboundary⋅

Wshaft⋅

me

Pe Teve ee

Pi Tivi ei

⋅Q⋅

mi⋅

Figure 0-5: Schematic diagram illustrating terms in the energy equation for a control volume

The fluid that enters or leaves has an amount of energy per unit mass given by

gzcue ++= 2/2,

where c is the fluid velocity. In addition, whenever fluid enters or leaves a control volume there isa work term associated with the entry or exit. We saw this in example 16a, and the presentderivation is essentially an application of the ideas presented there. Flow exiting at station “e”must push back the surrounding fluid, doing work on it. Flow entering the volume at station “i” ispushed on by, and receives work from the surrounding air. The rate of flow work at exit is givenby the product of the pressure times the exit area times the rate at which the external flow is“pushed back”. The last of these, however, is equal to the volume per unit mass times the rate ofmass flow. Put another way, in a time dt, the work done on the surroundings by the flow at the exitstation is

eflow PvdmdW = .

The net rate of flow work is

iiieeeflow mvPmvPW && −= .

Including all possible energy flows (heat, shaft work, shear work, piston work etc.), thefirst law can then be written as:

∑ ∑∑∑∑∑∑ +++++++= )( 221 gzcumWWWWQE

dt

dflowpistonshearshaftCV &&&&&&

0-14

where Σ includes the sign associated with the energy flow. If heat is added or work is done on thesystem then the sign is positive, if work or heat are extracted from the system then the sign isnegative. NOTE: this is consistent with ∆E = Q – W, where W is the work done by the system onthe environment, thus work is flowing out of the system.

We can then collect the specific energy term e included in Ecv and the specific flow term Pv tomake the enthalpy appear:

Total energy associated with mass flow: thgzchPvgzcuPve =++=+++=+ 2/2/ 22,

where ht is the stagnation enthalpy (IAW, p.36).

Thus, the first law can be written as:

∑∑∑∑∑∑ ++++++= )( 221 gzchmWWWQE

dt

dpistonshearshaftCV &&&&& .

For most of the applications done in this course, there will be no shear work and no piston work.Hence, the first law for a control volume will be most often used as:

( ) ( )eeeeiiiishaftCVCV gzchmgzchmWQdt

dE++−+++−=

••••

2/2/ 22.

The rate of work term is the sum of the shaft work and the flow work. In writing the controlvolume form of the equation we have assumed only one entering and one leaving stream, but thiscould be generalized to any number of inlet and exit streams.

Muddy pointsWhat distinguishes shaft work from other works? (MP 0.8)

For problems of interest in aerospace applications the velocities are high and the term thatis associated with changes in the elevation is small. From now on, we will neglect this term unlessexplicitly stated. The control volume form of the first law is thus

( ) ( )2/2/ 22eeeiiishaftCV

CV chmchmWQdt

dE+−++−=

••••

teetiishaftCVhmhmWQ

••••

−+−= .

For steady flow (d/dt = 0) the inlet and exit mass flow rates are the same and the control volumeform of the first law becomes the “Steady Flow Energy Equation” (SFEE)

Steady Flow Energy Equation: ( ) shaftCVtt WQhhmie

•••

−=− .

The steady flow energy equation finds much use in the analysis of power and propulsion devicesand other fluid machinery. Note the prominent role of enthalpy.

0-15

Using what we have just learned we can attack the tank filling problem solved in (16b) from analternate point of view using the control volume form of the first law. In this problem the shaftwork is zero, and the heat transfer, kinetic energy changes, and potential energy changes areneglected. In addition there is no exit mass flow.

Figure 0-6: A control volume approach to the tank filling problem

The control volume form of the first law is therefore

ii hmdt

dU •

= .

The equation of mass conservation is

imdt

dm •

= .

Combining we have

ihdt

dm

dt

dU= .

Integrating from the initial time to the final time (the incoming enthalpy is constant) and using U =mu gives the result u h hfinal i= = 0 as before.

Muddy pointsDefinition of a control volume (MP 0.9)What is the difference between enthalpy and stagnation enthalpy? (MP 0.10)

(mass flow)

controlvolume

control surface

•

m

Muddiest Points on Part 0

0.1 Specific properties

Energy, volume, enthalpy are all extensive properties. Their value depends not only onthe temperature and pressure but also on “how much”, i.e., what the mass of the systemis. The internal energy of two kilograms of air is twice as much as the internal energy ofone kilogram of air. It is very often useful to work in terms of properties that do notdepend on the mass of the system, and for this purpose we use the specific volume,specific energy, specific enthalpy, etc., which are the values of volume, energy, andenthalpy for a unit mass (kilogram) of the substance. For a system of mass m, therelations between the two quantities are:V mv U mu H mh= = = ; ;

0.2 What is the difference between extensive and intensive properties?

Intensive properties are properties that do not depend on the quantity of matter. Forexample, pressure and temperature are intensive properties. Energy, volume and enthalpyare all extensive properties. Their value depends on the mass of the system. For example,the enthalpy of a certain mass of a gas is doubled if the mass is doubled; the enthalpy of asystem that consists of several parts is equal to the sum of the enthalpies of the parts.

0.3 How do we know when work is done?

A rigorous test for whether work is done or not is whether a weight could have beenraised in the process under consideration. I will hand out some additional material tosupplement the notes on this point, which seems simple, but can be quite subtle tounravel in some situations.

0.4 What are the conventions for work and heat in the first law?

Heat is positive if it is given to the system. Work is positive if it is done by the system.

0.5 When does E->U?

We deal with changes in energy. When the changes in the other types of energy (kinetic,potential, strain, etc) can be neglected compared to the changes in thermal energy, then itis a good approximation to use ∆U as representing the total energy change.

0.6 When is enthalpy the same in initial and final states?

Initial and final stagnation enthalpy is the same if the flow is steady and if there is no netshaft work plus heat transfer. If the change in kinetic energy is negligible, the initial andfinal enthalpy is the same. The “tank problem” is unsteady so the initial and final

enthalpies are not the same. See the discussion of steady flow energy equation in notes[(17) in Section 0].

0.7 In the filling of a tank, why (physically) is the final temperature in the tank higherthan the initial temperature?

Work is done on the system, which in this problem is the mass of gas that is pushed intothe tank.

0.8 What distinguishes shaft work from other works?

The term shaft work arises in using a control volume approach. As we have defined it,“shaft work” is all work over and above work associated with the “flow work” (the workdone by pressure forces). Generally this means work done by rotating machinery, whichis carried by a shaft from the control volume to the outside world. There could also bework over and above the pressure force work done by shear stresses at the boundaries ofthe control volume, but this is seldom important if the control boundary is normal to theflow direction.

If we consider a system (a mass of fixed identity, say a blob of gas) flowing through somedevice, neglecting the effects of raising or lowering the blob the only mode of workwould be the work to compress the blob. This would be true even if the blob wereflowing through a turbine or compressor. (In doing this we are focusing on the samematerial as it undergoes the unsteady compression or expansion processes in the device,rather than looking at a control volume, through which mass passes.)

The question about shaft work and non shaft work has been asked several times. I am notsure how best to answer, but it appears that the difficulty people are having might beassociated with being able to know when one can say that shaft work occurs. There areseveral features of a process that produces ( or absorbs) shaft work. First of all, the viewtaken of the process is one of control volume, rather than control mass (see the discussionof control volumes in section 0 or in IAW). Second, there need to be a shaft or equivalentdevice ( a moving belt, a row of blades) that can be identified as the work carrier. Third,the shaft work is work over and above the flow work that is done by (or received by) thestreams that exit and enter the control volume.

0.9 Definition of a control volume.

A control volume is an enclosure that separates a quantity of matter from thesurroundings or environment. The enclosure does not necessarily have to consist of asolid boundary like the walls of a vessel. It is only necessary that the enclosure forms aclosed surface and that its properties are defined everywhere. An enclosure may transmitheat or be a heat insulator. It may be deformable and thus capable of transmitting work tothe system. It may also be capable of transmitting mass.

PART 1

THE SECOND LAW OF THERMODYNAMICS

1A-1

PART 1 - THE SECOND LAW OF THERMODYNAMICS

1.A. Background to the Second Law of Thermodynamics [IAW 23-31 (see IAW for detailed VWB&S references); VN Chapters 2, 3, 4]

1.A.1 Some Properties of Engineering Cycles; Work and EfficiencyAs motivation for the development of the second law, we examine two types of processes thatconcern interactions between heat and work. The first of these represents the conversion of workinto heat. The second, which is much more useful, concerns the conversion of heat into work. Thequestion we will pose is how efficient can this conversion be in the two cases.

Figure A-1: Examples of the conversion of work into heat

Three examples of the first process are given above. The first is the pulling of a block on a roughhorizontal surface by a force which moves through some distance. Friction resists the pulling.After the force has moved through the distance, it is removed. The block then has no kineticenergy and the same potential energy it had when the force started to act. If we measured thetemperature of the block and the surface we would find that it was higher than when we started.(High temperatures can be reached if the velocities of pulling are high; this is the basis of inertiawelding.) The work done to move the block has been converted totally to heat.

The second example concerns the stirring of a viscous liquid. There is work associated with thetorque exerted on the shaft turning through an angle. When the stirring stops, the fluid comes torest and there is (again) no change in kinetic or potential energy from the initial state. The fluidand the paddle wheels will be found to be hotter than when we started, however.

The final example is the passage of a current through a resistance. This is a case of electrical workbeing converted to heat, indeed it models operation of an electrical heater.

All the examples in Figure A-1 have 100% conversion of work into heat. This 100% conversioncould go on without limit as long as work were supplied. Is this true for the conversion of heatinto work?

To answer the last question, we need to have some basis for judging whether work is done in agiven process. One way to do this is to ask whether we can construct a way that the process couldresult in the raising of a weight in a gravitational field. If so, we can say “Work has been done”. Itmay sometimes be difficult to make the link between a complicated thermodynamic process andthe simple raising of a weight, but this is a rigorous test for the existence of work.

One example of a process in which heat is converted to work is the isothermal (constanttemperature) expansion of an ideal gas, as sketched in the figure. The system is the gas inside thechamber. As the gas expands, the piston does work on some external device. For an ideal gas, theinternal energy is a function of temperature only, so that if the temperature is constant for someprocess the internal energy change is zero. To keep the temperature constant during the expansion,

i

R+

-

Viscous liquid

Block on rough surface

Resistive heating

F

1A-2

P, T

Q

Work received, W

Patm

heat must be supplied. Because ∆U = 0, the firstlaw takes the form Q=W. This is a process thathas 100% conversion of heat into work.

The work exerted by the system is given by

Work = PdV1

2

∫

where 2 and 1 denote the two states at thebeginning and end of the process. The equation ofstate for an ideal gas is

P = NRT/V,

with N the number of moles of the gas contained in the chamber. Using the equation of state, theexpression for work can be written as

Work during an isothermal expansion = NRT dV V/1

2

∫ = NRTlnV

V2

1

. (A.1.1)

For an isothermal process, PV = constant, so that P P V V1 2 2 1/ /= . The work can be written in termsof the pressures at the beginning and end as

Work during an isothermal expansion = NRTlnP

P1

2

. (A.1.2)

The lowest pressure to which we can expand and still receive work from the system is atmosphericpressure. Below this, we would have to do work on the system to pull the piston out further.There is thus a bound on the amount of work that can be obtained in the isothermal expansion; wecannot continue indefinitely. For a power or propulsion system, however, we would like a sourceof continuous power, in other words a device that would give power or propulsion as long as fuelwas added to it. To do this, we need a series of processes where the system does not progressthrough a one-way transition from an initial state to a different final state, but rather cycles back tothe initial state. What is looked for is in fact a thermodynamic cycle for the system.

We define several quantities for a cycle:QA is the heat absorbed by the systemQR is the heat rejected by the systemW is the net work done by the system.

The cycle returns to its initial state, so the overall energy change, ∆U , is zero. The net work doneby the system is related to the magnitudes of the heat absorbed and the heat rejected by

W Q QA R= = − Net work .

The thermal efficiency of the cycle is the ratio of the work done to the heat absorbed. (Efficienciesare often usefully portrayed as “What you get” versus “What you pay for”. Here what we get iswork and what we pay for is heat, or rather the fuel that generates the heat.) In terms of the heatabsorbed and rejected, the thermal efficiency is:

η = thermal efficiency = Work done

Heat absorbed

=−

= −Q Q

Q

Q

QA R

A

R

A

1 . (A.1.3)

1A-3

The thermal efficiency can only be 100% (complete conversion of heat into work) if QR = 0 , and abasic question is what is the maximum thermal efficiency for any arbitrary cycle? We examinethis for two cases, the Carnot cycle and the Brayton (or Joule) cycle which is a model for thepower cycle in a jet engine.

1.A.2 Carnot CyclesA Carnot cycle is shown below. It has four processes. There are two adiabatic reversible legs andtwo isothermal reversible legs. We can construct a Carnot cycle with many different systems, butthe concepts can be shown using a familiar working fluid, the ideal gas. The system can beregarded as a chamber filled with this ideal gas and with a piston.

P

V

Q1T1

T2

T2

Q2

Q2 T1 Q1

a

b

d

c

12 4

3

Reservoir Insulating stand Reservoir

Figure A-2: Carnot cycle – thermodynamic diagram on left and schematic of the different stages inthe cycle for a system composed of an ideal gas on the right

The four processes in the Carnot cycle are:1) The system is at temperature T2 at state (a). It is brought in contact with a heat reservoir,

which is just a liquid or solid mass of large enough extent such that its temperature does notchange appreciably when some amount of heat is transferred to the system. In other words, theheat reservoir is a constant temperature source (or receiver) of heat. The system thenundergoes an isothermal expansion from a to b, with heat absorbed Q2 .

2) At state b, the system is thermally insulated (removed from contact with the heat reservoir) andthen let expand to c. During this expansion the temperature decreases to T1. The heatexchanged during this part of the cycle, Qbc = 0.

3) At state c the system is brought in contact with a heat reservoir at temperature T1. It is thencompressed to state d, rejecting heat Q1 in the process.

4) Finally, the system is compressed adiabatically back to the initial state a. The heat exchangeQda = 0 .

The thermal efficiency of the cycle is given by the definition

η = − = +1 1 1

2

Q

Q

Q

QR

A

. (A.2.1)

In this equation, there is a sign convention implied. The quantities Q ,QA R as defined are themagnitudes of the heat absorbed and rejected. The quantities Q ,Q1 2 on the other hand are definedwith reference to heat received by the system. In this example, the former is negative and the latteris positive. The heat absorbed and rejected by the system takes place during isothermal processesand we already know what their values are from Eq. (A.1.1):

1A-4

Q Wab2 = =NRT2 ln V Vb a/( )[ ]Q Wcd1 = = NRT1 ln V Vd c/( )[ ] = - ln V Vc d/( )[ ]. ( Q1 is negative.)

The efficiency can now be written in terms of the volumes at the different states as:

η = +( )[ ]

( )[ ]1 1

2

T V V

T V Vd c

b a

ln

ln

/

/. (A.2.2)

The path from states b to c and from a to d are both adiabatic and reversible. For a reversibleadiabatic process we know that PV γ = constant. Using the ideal gas equation of state, wehave TV γ −1= constant. Along curve b-c, therefore T V T Vb c2

11

1γ γ− −= . Along the curve d-a,

T V T Va d21

11γ γ− −= .

Thus,

V

V

T T

T T

V

Vd

c

a

b

=

( )( )

− −γ γ12 1

2 1

1/

/, which means that

V

V

V

Vd

c

a

b

= , or V V V Vd c a b/ /= .

Comparing the expression for thermal efficiency Eq. (A.2.1) with Eq. (A.2.2) shows twoconsequences. First, the heats received and rejected are related to the temperatures of theisothermal parts of the cycle by

Q

T

Q

T1

1

2

2

0+ = . (A.2.3))

Second, the efficiency of a Carnot cycle is given compactly by

ηcT

T= −1 1

2

. Carnot cycle efficiency (A.2.4)

The efficiency can be 100% only if the temperature at which the heat is rejected is zero. The heatand work transfers to and from the system are shown schematically in Figure A-3.

Q2

Q1

T1

W (net work)

T2

System

Figure A-3: Work and heat transfers in a Carnot cycle between two heat reservoirs

1A-5

Muddy points

Since η = −1

TT

1

2

, looking at the P-V graph, does that mean the farther apart the T1, T2

isotherms are, the greater efficiency? And that if they were very close, it would be veryinefficient? (MP 1A.1)In the Carnot cycle, why are we only dealing with volume changes and not pressurechanges on the adiabats and isotherms? (MP 1A.2)Is there a physical application for the Carnot cycle? Can we design a Carnot engine for apropulsion device? (MP 1A.3)How do we know which cycles to use as models for real processes? (MP 1A.4)

1.A.3 Brayton Cycles (or Joule Cycles): The Power Cycle for a Gas Turbine Jet EngineFor a Brayton cycle there are two adiabatic legs and two constant pressure legs. Sketches of anengine and the corresponding cycle are given in Figure A-4.

Turbine and nozzle

Heat rejectionto atmosphere

Inlet andcompressor

Combustor

q2PCompressor exit

Patmq1

da

P

V

b c

Combustor

Compressor Turbine

NozzleInlet

Figure A-4: Sketch of the jet engine components and corresponding thermodynamic states

Gas turbines are also used for power generation and for closed cycle operation (for example forspace power generation). A depiction of the cycle in this case is shown in Figure A-5.

1A-6

Wnet

32

1 4

Equivalent heat transferat constant pressure

Equivalent heat transferat constant pressure

Turbine

Compressor⋅

Wcomp⋅

Q⋅

Q⋅

Figure A-5: Thermodynamic model of gas turbine engine cycle for power generation

The objective now is to find the work done, the heat absorbed, and the thermal efficiency of thecycle. Tracing the path shown around the cycle from a-b-c-d and back to a, the first law gives(writing the equation in terms of a unit mass),

∆u q q wa b c d a− − − − = = + −0 2 1 .

The net work done isw q q= +2 1,

where q q1 2, are defined as heat received by the system ( q1 is negative). We thus need to evaluatethe heat transferred in processes b-c and d-a.

For a constant pressure process the heat exchange per unit mass is

dh c dT dqp= = , or dq dhcons t P[ ] =tan .

The heat exchange can be expressed in terms of enthalpy differences between the relevant states.Treating the working fluid as an ideal gas, for the heat addition from the combustor,

q h h c T Tc b p c b2 = − = −( ).

The heat rejected is, similarly, q h h c T Ta d p a d1 = − = −( ).

The net work per unit mass is given by

Net work per unit mass = q q c T T T Tp c b a d1 2+ = −( ) + −( )[ ].

The thermal efficiency of the Brayton cycle can now be expressed in terms of the temperatures:

1A-7

η = =−( ) − −( )[ ]

−[ ]Net work

Heat in

c T T T T

c T Tp c b d a

p c b

= −−( )−( )

= −−( )−( )

1 11

1

T T

T T

T T T

T T Td a

c b

a d a

b c b

/

/. (A.3.1)

To proceed further, we need to examine the relationships between the different temperatures. Weknow that points a and d are on a constant pressure process as are points b and c,and P P P Pa d b c= =; . The other two legs of the cycle are adiabatic and reversible, so

P

P

P

P

T

T

T

Td

c

a

b

d

c

a

b

= ==

=

−( ) −( )

> γ γ γ γ/ /1 1

.

Therefore T

T

T

Td

c

a

b

= , or, finally, T

T

T

Td

a

c

b

= . Using this relation in the expression for thermal

efficiency, Eq. (A.1.3) yields an expression for the thermal efficiency of a Brayton cycle:

Ideal Brayton cycle efficiency:ηBa

b

T

T= −1 (A.3.2)

= −1T

Tatmospheric

compressor exit

.

The temperature ratio across the compressor, T T TRb a/ = . In terms of compressor temperatureratio, and using the relation for an adiabatic reversible process we can write the efficiency in termsof the compressor (and cycle) pressure ratio, which is the parameter commonly used:

ηγ γB TR PR

= − = −( ) −( )1

11

11 /

. (A.3.3)

Figure A-6 shows pressures and temperatures through a gas turbine engine (the afterburning J57, which powers the F-8 and the F-101).

Figure A-6: Gas turbine engine pressures and temperatures

1A-8

Equation (A.3.3) says that for a high cycle efficiency, the pressure ratio of the cycle should beincreased. Figure A-7 shows the history of aircraft engine pressure ratio versus entry into service,and it can be seen that there has been a large increase in cycle pressure ratio. The thermodynamicconcepts apply to the behavior of real aerospace devices!

Figure A-7: Gas turbine engine pressure ratio trends (Jane’s Aeroengines, 1998)

Muddy pointsWhen flow is accelerated in a nozzle, doesn’t that reduce the internal energy of the flowand therefore the enthalpy? (MP 1A.5)Why do we say the combustion in a gas turbine engine is constant pressure? (MP 1A.6) Why is the Brayton cycle less efficient than the Carnot cycle? (MP 1A.7) If the gas undergoes constant pressure cooling in the exhaust outside the engine, is thatstill within the system boundary ? (MP 1A.8) Does it matter what labels we put on the corners of the cycle or not? (MP 1A.9)Is the work done in the compressor always equal to the work done in the turbine pluswork out (for a Brayton cyle) ? (MP 1A.10)

1.A.4 Gas Turbine Technology and ThermodynamicsThe turbine entry temperature, Tc , is fixed by materials technology and cost. (If the temperature istoo high, the blades fail.) Figures A-8 and A-9 show the progression of the turbine entrytemperatures in aeroengines. Figure A-8 is from Rolls Royce and Figure A-9 is fromPratt&Whitney. Note the relation between the gas temperature coming into the turbine blades andthe blade melting temperature.

19600

10

20

JT3D

Conway 508

Conway 550

JT8D-17JT8D-219

CFM56-3C

PW4084

PW4168

CFM56-5B

CFM56-5C4

CFM56-2

JT8D-1

Tay 611Spey 555

Spey 512

JT9D-3A

JT9D-70TF39-1RB211-22

RB211-524D4CF6-50E

PW4052

CF6-80C2A8

CF6-50A

CF6-6

JT9D-7R4G

Spey 512-14

Spey 505

Tay 651

GE90Trent 890

Trent 775

30

40

1970 1980Year of Certification

Ove

rall

Pre

ssu

re R

atio

(O

PR

), S

ea L

evel

, T-O

1990 2000

CF6-80C2A8CF6-80E1A4

1A-9

Figure A-8: Rolls-Royce high temperature Figure A-9: Turbine blade coolingtechnology technology [Pratt & Whitney]

For a given level of turbine technology (in other words given maximum temperature) a designquestion is what should the compressor TR be? What criterion should be used to decide this?Maximum thermal efficiency? Maximum work? We examine this issue below.

Figure A-10: Efficiency and work of two Brayton cycle engines

The problem is posed in Figure A-10, which shows two Brayton cycles. For maximum efficiencywe would like TR as high as possible. This means that the compressor exit temperature approachesthe turbine entry temperature. The net work will be less than the heat received; as T Tb c→ theheat received approaches zero and so does the net work.

The net work in the cycle can also be expressed as Pdv∫ , evaluated in traversing the cycle. This isthe area enclosed by the curves, which is seen to approach zero as T Tb c→ .

V

T = Tatm = Ta

T = Tc

Cycle with Tb → Tc

Tb2

Tb1

Patm

PCycle with lower Tb

1A-10

The conclusion from either of these arguments is that a cycle designed for maximum thermalefficiency is not very useful in that the work (power) we get out of it is zero.

A more useful criterion is that of maximum work per unit mass (maximum power per unit massflow). This leads to compact propulsion devices. The work per unit mass is given by:

Work unit mass c T T T Tp c b d a/ = −( ) − −( )[ ]Max. turbine temp. Atmospheric temperature

(Design constraint)

The design variable is the compressor exit temperature, Tb , and to find the maximum as this isvaried, we differentiate the expression for work with respect to Tb :

dWork

dTc

dT

dT

dT

dT

dT

dTbp

c

b

d

b

a

b

= − − +

1 .

The first and the fourth term on the right hand side of the above equation are both zero (the turbineentry temperature is fixed, as is the atmospheric temperature). The maximum work occurs wherethe derivative of work with respect to Tb is zero:

dWork

dT

dT

dTb

d

b

= = − −0 1 . (A.4.1)

To use Eq. (A.4.1), we need to relate T and Td b. We know thatT

T

T

TT

T T

Td

a

c

bd

a c

b

= = or .

Hence,dT

dT

T T

Td

b

a c

b

=−

2 .

Plugging this expression for the derivative into Eq. (A.4.1) gives the compressor exit temperaturefor maximum work as T T Tb a c= . In terms of temperature ratio,

Compressor temperature ratio for maximum work: T

T

T

Tb

a

c

a

= .

The condition for maximum work in a Brayton cycle is different than that for maximum efficiency.The role of the temperature ratio can be seen if we examine the work per unit mass which isdelivered at this condition:

Work unit mass c T T TT T

T TTp c a c

a c

a ca/ = − − +

.

Ratioing all temperatures to the engine inlet temperature,

Work unit mass c TT

T

T

Tp ac

a

c

a

/ = − +

2 1 .

To find the power the engine can produce, we need to multiply the work per unit mass by the massflow rate:

1A-11

Power mc TT

T

T

Tp ac

a

c

a

= − +

•2 1 ; Maximum power for an ideal Brayton cycle (A.4.2)

(The units are kg

s

J

kg - KK =

J

s = Watts .)

a) Gas turbine engine core b) Core power vs. turbine entry temperature

Figure A-11: Aeroengine core power [Koff/Meese, 1995]

Figure A-11 shows the expression for power of an ideal cycle compared with data from actual jetengines. Figure 11a shows the gas turbine engine layout including the core (compressor, burner,and turbine). Figure 11b shows the core power for a number of different engines as a function ofthe turbine rotor entry temperature. The equation in the figure for horsepower (HP) is the same asthat we just derived, except for the conversion factors. The analysis not only shows the qualitativetrend very well but captures much of the quantitative behavior too.

A final comment (for now) on Brayton cycles concerns the value of the thermal efficiency. TheBrayton cycle thermal efficiency contains the ratio of the compressor exit temperature toatmospheric temperature, so that the ratio is not based on the highest temperature in the cycle, asthe Carnot efficiency is. For a given maximum cycle temperature, the Brayton cycle is thereforeless efficient than a Carnot cycle.

Muddy points

What are the units of w in power m w=•

? (MP 1A.11)Precision about the assumptions made in the Brayton cycle for maximum efficiency andmaximum work (MP 1A.12)You said that for a gas turbine engine modeled as a Brayton cycle the work done isw=q 1 +q 2 , where q 2 is the heat added and q 1 is the heat rejected. Does this suggest that thework that you get out of the engine doesn't depend on how good your compressor andturbine are?…since the compression and expansion were modeled as adiabatic. (MP1A.13)

1.A.5 Refrigerators and Heat PumpsThe Carnot cycle has been used for power, but we can also run it in reverse. If so, there is now network into the system and net heat out of the system. There will be a quantity of heatQ2 rejected atthe higher temperature and a quantity of heat Q1 absorbed at the lower temperature. The former of

1A-12

these is negative according to our convention and the latter is positive. The result is that work isdone on the system, heat is extracted from a low temperature source and rejected to a hightemperature source. The words “low” and “high” are relative and the low temperature sourcemight be a crowded classroom on a hot day, with the heat extraction being used to cool the room.The cycle and the heat and work transfers are indicated in Figure A-12. In this mode of operation

P

d

a

b

c

V

Q2

Q1

T1

W

T2

System

Figure A-12: Operation of a Carnot refrigerator

the cycle works as a refrigerator or heat pump. “What we pay for” is the work, and “what we get”is the amount of heat extracted. A metric for devices of this type is the coefficient of performance,defined as

Coefficient of performance = Q

W

Q

Q Q1 1

1 2

=+

.

For a Carnot cycle we know the ratios of heat in to heat out when the cycle is run forward and,since the cycle is reversible, these ratios are the same when the cycle is run in reverse. Thecoefficient of performance is thus given in terms of the absolute temperatures as

Coefficient of performance =−

T

T T1

2 1

.

This can be much larger than unity.

The Carnot cycles that have been drawn are based on ideal gas behavior. For different workingmedia, however, they will look different. We will see an example when we discuss two-phasesituations. What is the same whatever the medium is the efficiency for all Carnot cycles operatingbetween the same two temperatures.

Muddy pointsWould it be practical to run a Brayton cycle in reverse and use it as rerigerator? (MP1A.14)

1.A.6 Reversibility and Irreversibility in Natural ProcessesWe wish to characterize the “direction” of natural processes; there is a basic

“directionality” in nature. We start by examining a flywheel in a fluid filled insulated enclosure asshown in Figure A-13.

1A-13

State A: flywheel spinning,system cool

State B: flywheel stationary

Figure A-13: Flywheel in insulated enclosure at initial and final states

A question to be asked is whether we could start with state B and then let events proceed to stateA? Why or why not? The first law does not prohibit this.

The characteristics of state A are that the energy is in an organized form, the molecules inthe flywheel have some circular motion, and we could extract some work by using the flywheelkinetic energy to lift a weight. In state B, in contrast, the energy is associated with disorganizedmotion on a molecular scale. The temperature of the fluid and flywheel are higher than in state A,so we could probably get some work out by using a Carnot cycle, but it would be much less thanthe work we could extract in state A. There is a qualitative difference between these states, whichwe need to be able to describe more precisely.

Muddy pointsWhy is the ability to do work decreased in B? How do we know? (MP 1A.15)

Another example is a system composed of many bricks, half at a high temperature TH and half at alow temperature TL (see IAW p. 42). With the bricks separated thermally, we have the ability toobtain work by running a cycle between the two temperatures. Suppose we put two brickstogether. Using the first law we can write

CT CT CTH L M+ = 2 .

T T TH L M+( ) =/2

where C is the “heat capacity” = ∆ ∆Q T/ . (For solids the heat capacities (specific heats) atconstant pressure and constant volume are essentially the same.) We have lost the ability to getwork out of these two bricks.

Can we restore the system to the original state without contact with the outside? The answer is no.Can we restore the system to the original state with contact with the outside? The answer is yes.We could run a refrigerator to take heat out of one brick and put it into the other, but we wouldhave to do work.

We can think of the overall process involving the system (the two bricks in an insulated setting)and the surroundings (the rest of the universe) as:

System is changedSurroundings are unchanged.

The composite system (system and the surroundings) is changed by putting the bricks together.The process is not reversible—there is no way to undo the change and leave no mark on thesurroundings.

1A-14

V1, T1 Vacuum

What is the measure of change in the surroundings?a) Energy? This is conserved.b) Ability to do work? This is decreased.The measurement and characterization of this type of changes is the subject of the second law ofthermodynamics.

1.A.7 Difference between Free Expansion of a Gas and Reversible Isothermal ExpansionThe difference between reversible and irreversible processes is brought out through

examination of the isothermal expansion of an ideal gas. The question to be asked is what is thedifference between the “free expansion” of a gas and the isothermal expansion against a piston?To answer this, we address the steps that we would have to take to reverse, in other words, to undothe process.

By free expansion, we mean the unrestrainedexpansion of a gas into a volume as shown at the right.Initially all the gas is in the volume designated as V1 with therest of the insulated enclosure a vacuum. The total volume( V1 plus the evacuated volume) is V2.At a given time a hole is opened in the partitionand the gas rushes through to fill the rest of the enclosure.

During the expansion there is no work exchanged with the surroundings because there is nomotion of the boundaries. The enclosure is insulated so there is no heat exchange. The first lawtells us therefore that the internal energy is constant (∆U = 0). For an ideal gas, the internalenergy is a function of temperature only so that the temperature of the gas before the freeexpansion and after the expansion has been completed is the same. Characterizing the before andafter states;

Before: State 1, V V T T= 1 1, =After: State 2, V V T T= 2 1, = .

Q=W=0, so there is no change in the surroundings.

To restore the original state, i.e., to go back to the original volume at the same temperatureV V T T2 1 1→( ) at constant = we can compress the gas isothermally (using work from an external

agency). We can do this in a quasi-equilibrium manner, with P Psystem external≈ . If so the work that

we need to do is W PdV= ∫1

2

. We have evaluated the work in a reversible isothermal expansion

(Eq. A.1.1), and we can apply the arguments to the case of a reversible isothermal compression.The work done on the system to go from state “2” to state “1” is

W = Work done on system = NR TV

V12

1 ln

.

From the first law, this amount of heat must also be rejected from the gas to the surroundings if thetemperature of the gas is to remain constant. A schematic of the compression process, in terms ofheat and work exchanged is shown in Figure A-14.

1A-15

WeightBlock

g

System W

Q Q

Work received, WP, T

System W (work in)

Q (heat out)

Figure A-14: Work and heat exchange in the reversible isothermal compression process

At the end of the combined process (free expansion plus reversible compression):a) The system has been returned to its initial state (no change in system state).b) The surroundings (us!) did work on the system of magnitude W.c) The surroundings received an amount of heat, Q, which is equal to W.d) The sum of all of these events is that we have converted anamount of work, W, into an amount of heat, Q, with W and Q numerically equal in Joules.

The net effect is the same as if we let a weight falland pull a block along a rough surface, as at right.There is 100% conversion of work into heat.

The results of the free expansion can be contrasted against a process of isothermal expansionagainst a pressure dP which is slightly different than that of the system, as shown below.

Figure A-15: Work and heat transfer in reversible isothermal expansion

1A-16

During the expansion, work is done on the surroundings of magnitudeW PdV= ∫ , where P can betaken as the system pressure. As evaluated in Eq. (A.1.1), the magnitude of the work done by the

system is W=NR TV

V12

1 ln

. At the end of the isothermal expansion, therefore:

a) The surroundings have received work Wb) The surroundings have given up heat, Q, numerically equal to W.

We now wish to restore the system to its initial state, just as we did in the free expansion. To dothis we need to do work on the system and extract heat from the system, just as in the freeexpansion. In fact, because we are doing a transition between the same states along the same path,the work and heat exchange are the same as those for the compression process examined justabove.

The overall result when we have restored the system to the initial state, however, is quite differentfor the reversible expansion and for the free expansion. For the reversible expansion, the work weneed to do on the system to compress it has the same magnitude as the work we received duringthe expansion process. Indeed, we could raise a weight during the expansion and then allow it tobe lowered during the compression process. Similarly the heat put into the system by us (thesurroundings) during the expansion process has the same magnitude as the heat received by usduring the compression process. The result is that when the system has been restored back to itsinitial state, so have the surroundings. There is no trace of the overall process on either the systemor the surroundings. That is another meaning of the word “reversible”.

Muddy points

With the isothermal reversible expansion is Pexternal constant? If so, how can we have

P Psystem external≅ ? (MP 1A.16)Why is the work done equal to zero in the free expansion? (MP 1A.17) Is irreversibility defined by whether or not a mark is left on the outside environment?(MP 1A.18)

1.A.8 Features of reversible processes Reversible processes are idealizations or models of real processes. One familiar, and widely used,example is Bernoulli’s equation, which you saw last year. They are extremely useful for defininglimits to system or device behavior, for enabling identification of areas in which inefficienciesoccur, and in giving targets for design.

An important feature of a reversible process is that, depending on the process, it represents themaximum work that can be extracted in going from one state to another, or the minimum work thatis needed to create the state change. This is shown clearly in the discussion on page 46 of theWaitz notes. Said differently:

The work done by a system during a reversible process is the maximum work we can get.The work done on a system in a reversible process is the minimum work we need to do toachieve that state change.

A process must be quasi-static (quasi-equilibrium) to be reversible. This means that thefollowing effects must be absent or negligible.

1) FrictionIf P Pexternal system≠ , as shown in (8) of Part 0, we would have to do work to bring the systemfrom one volume to another and return it to the initial condition. [Review (8) of Part 0]

1A-17

2) Free (unrestrained) expansion3) Heat transfer through a finite temperature difference.

Q2

Q1

Q1 = Q2

T1

T2

System

Figure A-16: Heat transfer across a finite temperature difference

Suppose we have heat transfer from a high temperature to a lower temperature as shownabove. How do we restore the situation to the initial conditions? One thought would be to run aCarnot refrigerator to get an amount of heat, Q, from the lower temperature reservoir to the highertemperature reservoir. We could do this but the surroundings, again us, would need to providesome amount of work (which we could find using our analysis of the Carnot refrigerator). The net(and only) result at the end of the combined process would be a conversion of an amount of workinto heat. For reversible heat transfer from a heat reservoir to a system, the temperatures of thesystem and the reservoir must be T T dTheat reservoir system = ± . In other words the difference betweenthe temperatures of the two entities involved in the heat transfer process can only differ by aninfinitesimal amount, dT.

All natural processes are irreversible to some extent, and reversible processes are idealized models.In natural processes, the conditions for mechanical, thermal, and chemical equilibrium may not besatisfied. In addition, dissipative effects (viscosity, friction) exist. Reversible processes are quasi-equilibrium, with no dissipative effects. It cannot be emphasized too strongly that there are anumber of engineering situations where the effect of irreversibility can be neglected and thereversible process furnishes an excellent approximation to reality.

The second law, which is the next topic we address, allows us to make a quantitative statementconcerning the irreversibility of a given physical process.

Figure A-17: this is Nicolas Sadi Carnot (1796-1832), an engineer and an officer in the French army.Carnot’s work is all the more remarkable because it was made without benefit of the first law, whichwas not discovered until 30 years later. [Atkins, The Second Law].

1A-18

Muddy pointsIs heat transfer across a finite temperature difference only irreversible if no device ispresent between the two to harvest the potential difference ? (MP 1A.19)

Muddiest Points on Part 1A

1A.1 Since h = 1 – T1/T2, looking at the P-V graph, does that mean the farther apart theT1, T2 isotherms are, the greater the efficiency? And that if they were very close, itwould be very inefficient?

This is correct. However, there is a limit on the maximum achievable efficiency. Wecannot convert the absorbed heat into 100% work, that is, we always must reject someamount of heat. The amount of heat we must reject is

QR = - T1/T2*QA (see notes for derivation).

Thus for given values of T2 and QA, QR depends only on the temperature of the coldreservoir T1, which is limited by the temperatures naturally available to us. Thesetemperatures are all well above absolute zero, and there are no means to reduce QR tonegligible values. The consequence of this is that the Carnot cycle efficiency cannotapproach one (η = 1 only if QR = 0, which is not possible).

1A.2 In the Carnot cycle, why are we only dealing with volume changes and not pressurechanges on the adiabats and isotherms?

We are not neglecting the pressure terms and we are also dealing with pressure changes.On the adiabats we know that dq = 0 (adiabatic process), so that for reversible processeswe can write the first law as du = -Pdv and, using enthalpy, also as dh = vdP. With dh =cp dT and du = cv dT for an ideal gas, we can write the ratio of dh/du as

dh/du = cp/cv = γ = -(v dP) / ( Pdv).

By arranging terms we obtain

dP/P = -γ dv/v.

For a process we can integrate from 1 to 2 and get P2v2γ = P1v1

γ, or Pvγ = const. Thisrelation shows how pressure and volume changes are related to one another during anadiabatic reversible process.

During an isothermal process, the temperature stays constant. Using the equation of statefor an ideal gas Pv = RT, we find that Pv = const on an isotherm. Again, this relation tellsus how pressure changes are related to volume changes during an isothermal process.Note that in the P-V diagram, adiabats (Pvγ = constant) are steeper curves than isotherms(Pv = constant).

1A.3 Is there a physical application for the Carnot cycle? Can we design a Carnot enginefor a propulsion device?

We will see that Carnot cycles are the best we can do in terms of efficiency. A constanttemperature heat transfer process is, however, difficult to attain in practice for devices inwhich high rates of power are required. The main role of the Carnot engine is therefore asa standard against which all other cycles are compared and which shows us the directionin which design of efficient cycles should go.

1A.4 How do we know which cycles to use as models for real processes?

We have discussed this briefly for the Brayton cycle, in that we looked at theapproximation that was made in saying heat addition occurred at constant pressure. Youcan also see that the Carnot cycle is not a good descriptor of a gas turbine engine! Wewill look further at this general point, not only for the Brayton cycle, but also for theRankine cycle and for some internal combustion engine cycles. I will try to make clearwhat are the approximations and why the cycle under study is being used as a model.

1A.5 When flow is accelerated in a nozzle, doesn't that reduce the internal energy of theflow and therefore the enthalpy?

Indeed both enthalpy and internal energy are reduced. The stagnation enthalpy is thequantity that is constant.

1A.6 Why do we say that the combustion in a gas turbine engine is at constantpressure?

This is an approximation, and a key question is indeed how accurate it is and what thejustification is. The pressure change in the combustor can be analyzed using the 1-dimensional compressible flow equations. The momentum equation isdP cdc= −ρ , where c is the velocity.If we divide both sides by P, we obtain:dPP

cP/

dcc

ca

dcc

Mdcc

2 2

22= − = − = −

γγ ρ

γγ ,

where a is the speed of sound and M is the Mach number.

Changes in velocity are due to changes in density and in flow-through area, as given bythe 1-dimensional continuity equation

Hence

Differentiating,

ρca = constant

ln ln lnρ + + =c a constant

d dcc

dAA

0, ordcc

dAA

dρρ

ρρ

+ + = = − −

Velocity changes are therefore related to area changes (geometry) and density changes(basically heat input). For a gas turbine combustion process the change in density iscomparable with (a significant fraction of) the initial density and the area change isseveral times the initial area. This means that the change in velocity divided by the initialvelocity is roughly of the order of magnitude of unity. The momentum equation thus tellsus that for small Mach number (say 0.1) the ratio dP/P will be much less than one, so thatthe pressure can be approximated as constant. In reality the pressure does drop in thecombustor, but the overall drop from inlet to exit is about 3-4%, small compared to theinitial level of pressure, so that the approximation of constant pressure is a useful one.

The rapidity of the combustion process does not really have anything to do with thisapproximation. We could have a process, such as a nozzle, in which there wascombustion at the same time that the pressure was dropping. As seen from themomentum equation, the heat addition does not "directly" affect the pressure - changes inpressure are associated with changes in velocity.

1A.7 Why is the Brayton cycle less efficient than the Carnot cycle?

Consider the Brayton cycle and the corresponding work done as being approximated by anumber of elementary Carnot cycles, as shown by the dashed lines in Figure 1. All ofthese Carnot cycles have the same pressure ratio, thus the same temperature ratio, andthus the same efficiency. The temperature ratio that figures into the efficiency of theelementary Carnot cycles is the inlet temperature divided by the compressor exittemperature, not the maximum cycle temperature, which is at the combustor exit. Thebasic reason for the lower efficiency is that heat is absorbed at an average temperaturethat is lower than the maximum temperature and rejected at an average temperaturehigher than the minimum temperature. We will come back to this important point (whichhas implications for all cycles), but if you cannot wait, see Section 1-C of the notes.

Figure 1 - Brayton cycle considered as a number of elementary Carnot cycles, all havingthe same pressure ratio and therefore the same temperature ratio, which is lower than theoverall cycle temperature ratio,

P

v

Brayton cycle (solid line)

Elementary Carnot Cycle(dashed lines)

Tmax / .minT

1A.8 If the gas undergoes constant pressure cooling in the exhaust outside the engine,is that still within the system boundary?

When we analyze the state changes as we trace them around the cycle, we are viewingthe changes in a system, a mass of fixed identity. Thus we follow the mass as it movesthrough the device and the cooling of the gas outside the engine is happening to oursystem.

1A.9 Does it matter what labels we put on the corners of the cycle or not?

It does not matter what labels we use on the corners of the cycle. A cycle is a series ofprocesses. Independent of where you start in the cycle, it always brings you back to thestate where you started.

1A.10 Is the work done in the compressor always equal to the work done in the turbineplus work out (for a Brayton cyle)?

NO. The work done in the compressor plus net work out equals the total turbine work.Using the 1st law, the net work we get out of the Brayton cycle is

w = q2 + q1 = cp [(Ta – Td) + (Tc – Tb)]

(see notes for details). Rearranging the temperatures we can also write

w = cp [(Tc – Td) – (Tb – Ta)] = Dhturbine – Dhcompressor.

Thus the net work is the difference between the enthalpy drop across the turbine (we getwork from the turbine) and the enthalpy rise through the compressor (we have to supplywork to the compressor).

1A.11 What are the units of w in power mw=•

?

The units of power are J/s (kJ/s, MJ/s) or Watts (kW, MW). The mass flow is kg/s. Theunits of w, work per unit mass, are thus J/kg. For the aeroengine, we can think of a givendiameter (frontal area) as implying a given mass flow (think of a given Mach number andhence a given ratio of flow to choked flow). If so, for a given fan diameter power scalesdirectly as work per unit mass.

1A.12 Precision about the assumptions made in the Brayton cycle for maximumefficiency and maximum work

We have first derived a general expression for the thermal efficiency of an ideal Braytoncycle (see Equation A.3.3 in your notes). The assumptions we made for the cycle werethat both the compressor and turbine are ideal, such that they can be modeled adiabaticand reversible. We then looked at possible ideal Brayton cycles that would yield (A)maximum efficiency and (B) maximum work, keeping the assumptions of an ideal cycle

(the assumptions of adiabatic and reversible compression and expansion stem from thechoice of an ideal cycle). One way to construct an ideal Brayton cycle in the P-Vdiagram is to choose the inlet temperature Ta and inlet pressure Pa, the compressorpressure ratio Pb/Pa or temperature ratio Tb/Ta, and the turbine inlet temperature Tc. Apartfrom setting the inlet conditions (these mainly depend on the flight altitude and Machnumber and the day), we decided to fix the turbine inlet temperature (fixed by materialtechnology or cost). So the only two "floating" cycle parameters that remain to be definedare the compressor exit temperature Tb and the turbine exit temperature Td. Looking atequation A.3.3 we know that the higher the compressor temperature ratio Tb/Ta the higherthe thermal efficiency. So, for (A) maximum efficiency we would chose the compressorexit temperature as high as possible, that is in the limit Tb = Tc. Constructing this cycle inthe P-V diagram and letting Tb approach Tc shows that the area enclosed by the cycle, orin other words the net work, becomes zero. Thus a cycle constructed under the given inletconditions and constraints on Tc is not very useful because we don't get any work out ofit.

For the derivation of Tb for maximum work (keeping Tc fixed as above), see notes fordetails.

1A.13 You said that for a gas turbine engine modeled as a Brayton cycle the work doneis w=q1+q2, where q2 is the heat added and q1 is the heat rejected. Does thissuggest that the work that you get out of the engine doesn't depend on how goodyour compressor and turbine are?…since the compression and expansion weremodeled as adiabatic.

Using the 1st law, the net work we get out of the cycle is

w = q2 + q1 = cp [(Ta – Td) + (Tc – Tb)]

(see notes for details). Rearranging the temperatures we can also write

w = cp [(Tc – Td) – (Tb – Ta)] = ∆hturbine – ∆hcompressor.

Thus the net work is the difference between the enthalpy drop across the turbine (we getwork from the turbine) and the enthalpy rise through the compressor (we have to supplywork to the compressor, this is done through the drive shaft that connects turbine andcompressor).

In class we analyzed an ideal Brayton cycle with the assumptions of adiabatic reversiblecompression and expansion processes, meaning that the work done by the turbine is themaximum work we can get from the given turbine (operating between Tc and Td), andthe work needed to drive the given compressor is the minimum work required. In theassumptions the emphasis is put on reversible rather than adiabatic. For real engines theassumption of adiabatic flow through the compressor and turbine still holds. This is anapproximation – the surface inside the compressor or turbine where heat can betransferred is much smaller than the mass flow of the fluid moving through the machine

so that the heat transfer is negligible – we will discuss the different concepts of heattransfer later in class. However the compression and expansion processes in real enginesare irreversible due to non-ideal behavior and loss mechanisms occurring in theturbomachinery flow. Thus the thermal efficiency and work for a real jet engine withlosses depend on the component efficiencies of turbine and compressor and are less thanfor an ideal jet engine. We will discuss these component efficiencies in more detail inclass.

1A.14 Would it be practical to run a Brayton cycle in reverse and use it as rerigerator?

Yes. In fact people in Cryogenics use reversed Brayton cycles to cool down systemswhere very low temperatures are required (e.g. space applications, liquefaction ofpropellants). One major difference between a regular Brayton cycle (such as a jet-engineor a gas-turbine) and a reversed Brayton cycle is the working fluid. In order to make areversed Brayton cycle practical we have to choose a working fluid that is appropriate forthe application.

Extremely low temperatures can be achieved when using a regenerator – a heatexchanger that preheats the fluid before it enters the compressor and cools the fluidfurther down before it enters the turbine. In this configuration the fluid is expanded tomuch lower temperatures, and more heat can be absorbed from the cooling compartment.

1A.15 Why is the ability to do work decreased in B? How do we know?

In state A, the energy is in organized form and the molecules move along circular pathsaround the spinning flywheel. We could get work out this system by using all of thekinetic energy of the flywheel and for example lift a weight with it. The energy of thesystem in state B (flywheel not spinning) is associated with disorganized motion (on themolecular scale). The temperature in state B is higher than in state A. We could alsoextract work from state B by running for example an ideal Carnot cycle between TB andsome heat reservoir at lower temperature. However the work we would get from thisideal Carnot cycle is less than the work we get from state A (all of the kinetic energy),because we must reject some heat when we convert heat into work (we cannot convertheat into 100% work). Although the energy of the system in state A is the same as in stateB (we know this from 1st law) the "organization" of the energy is different, and thus theability to do work is different.

1A.16 With the isothermal reversible expansion, is Pexternal constant? If so, how can wehave Psystem≅Pexternal?

For a reversible process, if the external pressure were constant, there would need to be aforce that pushed on the piston so the process could be considered quasi-equilibrium.This force could be us, it could be a system of weights, or it could be any other workreceiver. Under these conditions the system pressure would not necessarily be near the

external pressure but we would haveP PF

Asystem externalwork receiver

piston

≅ + . We can of course

think of a situation in which the external pressure was varied so it was always close to thesystem pressure, but that is not necessary.

1A.17 Why is the work done equal to zero in the free expansion?

In this problem, the system is everything inside the rigid container. There is no change involume, no “dV”, so no work done on the surroundings. Pieces of the gas might beexpanding, pushing on other parts of the gas, and doing work locally inside the container(and other pieces might be compressed and thus receive work) during the free expansionprocess, but we are considering the system as a whole, and there is no net work done.

1A.18 Is irreversibility defined by whether or not a mark is left on the outsideenvironment?

A process is irreversible when there is no way to undo the change without leaving a markon the surroundings or "the rest of the universe". In the example with the bricks, we couldundo the change by putting a Carnot refrigerator between the bricks (both at TM afterputting them together) and cooling one brick down to TL and heating the other brick to TH

to restore the initial state. To do this we have to supply work to the refrigerator and wewill also reject some heat to the surroundings. Thus we leave a mark on the environmentand the process is irreversible.

1A.19 Is heat transfer across a finite temperature difference only irreversible if no deviceis present between the two to harvest the potential difference?

If we have two heat reservoirs at different temperatures, the irreversibility associated withthe transfer of heat from one to the other is indeed dependent on what is between them.If there is a copper bar between them, all the heat that comes out of the high temperaturereservoir goes into the low temperature reservoir, with the result given in Section 1.B.5.If there were a Carnot cycle between them, some (not all ) heat from the high temperaturereservoir would be passed on to the low temperature reservoir, the process would bereversible, and work would be done. The extent to which the process is irreversible forany device can be assessed by computing the total entropy change (device plussurroundings) associated with the heat transfer.

1.B: The Second Law of Thermodynamics [IAW 42-50; VN Chapter 5; VWB&S-6.3, 6.4, Chapter 7]

1.B.1 Concept and Statements of the Second Law (Why do we need a second law?) The unrestrained expansion, or the temperature equilibration of the two bricks, are familiar

processes. Suppose you are asked whether you have ever seen the reverse of these processes take place? Do two bricks at a medium temperature ever go to a state where one is hot and one is cold? Will the gas in the unrestrained expansion ever spontaneously return to occupying only the left side of the volume? Experience hints that the answer is no. However, both these processes, unfamiliar though they may be, are compatible with the first law. In other words the first law does not prohibit their occurrence. There thus must be some other “great principle” that describes the direction of natural processes, that tells us which first law compatible processes will not be observed. This is contained in the second law. Like the first law, it is a generalization from an enormous amount of observation.

There are several ways in which the second law of thermodynamics can be stated. Listed below are three that are often encountered. As described in class (and as derived in almost every thermodynamics textbook), although the three may not appear to have much connection with each other, they are equivalent.

1) No process is possible whose sole result is the absorption of heat from a reservoir and the conversion of this heat into work. [Kelvin-Planck statement of the second law]

Q

System

T2

W This is not possible

T1

2) No process is possible whose sole result is the transfer of heat from a cooler to a hotter body. [Clausius statement of the second law]

Q

T2

T1

For T1 < T2 , this is not possible

Q

1B-1

3) There exists a property called entropy, S, which is a thermodynamic property of a system. For a reversible process, changes in this property are given by

dS = (dQreversible)/T

The entropy change of any system and its surroundings, considered together, is positive and approaches zero for any process which approaches reversibility.

∆ Stotal > 0

For an isolated system, i.e., a system that has no interaction with the surroundings, changes in the system have no effect on the surroundings. In this case, we need to consider the system only, and the first and second laws become:

∆ E system = 0 ∆ S system > 0

For an isolated system the total energy (E = U + Kinetic Energy + Potential Energy + ....) is constant. The entropy can only increase or, in the limit of a reversible process, remain constant.

All of these statements are equivalent, but (3) gives a direct, quantitative measure of the departure from reversibility.

Entropy is not a familiar concept and it may be helpful to provide some additional rationale for its appearance. If we look at the first law,

dU = dQ − dW

the term on the left is a function of state, while the two terms on the right are not. For a simple compressible substance, however, we can write the work done in a reversible process as dW = PdV , so that

dU = dQ − PdV ; First law for a simple compressible substance, reversible process.

Two out of the three terms in this equation are expressed in terms of state variables. It seems plausible that we ought to be able to express the third term using state variables as well, but what are the appropriate variables? If so, the term dQ = ( ) [ ] should perhaps be viewed as analogous to dW = PdV where the parenthesis denotes an intensive state variable and the square bracket denotes an extensive state variable. The second law tells us that the intensive variable is the temperature, T, and the extensive state variable is the entropy, S.

The first law for a simple compressible substance in terms of state variables is thus

dU = TdS − PdV . (B.1.1)

Because Eq. (B.1.1) includes the second law, it is referred to as the combined first and second law. Because it is written in terms of state variables, it is true for all processes, not just reversible ones.

We list below some attributes of entropy: a) S is an extensive variable. The entropy per unit mass, or specific entropy, is s.

1B-2

b) The units of entropy are Joules per degree Kelvin (J/K). The units for specific entropy are J/K-kg.

dQc) For a system, dS = rev , where the numerator is the heat given to the system and the

T denominator is the temperature of the system at the location where the heat is received.

d) dS = 0 for pure work transfer.

Muddy pointsWhy is dU = TdS − PdV always true? (MP 1B.1) What makes dQrev different than dQ? (MP 1B.2)

1.B.2 Axiomatic Statements of the Laws of Thermodynamics1

(i.) Introduction As a further aid in familiarization with the second law of thermodynamics and the idea of

entropy, we draw an analogy with statements made previously concerning quantities that are closer to experience. In particular, we wish to (re-) present the Zeroth and First Laws of thermodynamics in the same framework as we have used for the Second Law. In this so-called "axiomatic formulation", the Zeroth, First and Second Laws are all introduced in a similar fashion.

(ii.) Zeroth Law

We start with a statement which is based on two observations: a) If two bodies are in contact through a thermally-conducting boundary for a sufficiently long

time, no further observable changes take place; thermal equilibrium is said to prevail. b) Two systems which are individually in thermal equilibrium with a third are in thermal

equilibrium with each other; all three systems have the same value of the property called temperature.

The closely connected ideas of temperature and thermal equilibrium are formally expressed in the “Zeroth Law of Thermodynamics”:

Zeroth Law There exists for every thermodynamic system in equilibrium a property called temperature. Equality of temperature is a necessary and sufficient condition for thermal equilibrium.

The Zeroth law thus defines a property (temperature) and describes its behavior.

(iii.) First Law Observations also show that for any system there is a property called the energy. The First

Law asserts that one must associate such a property with every system.

First Law There exists for every thermodynamic system a property called the energy. The change of energy of a system is equal to the mechanical work done on the system in an adiabatic process. In a non-adiabatic process, the change in energy is equal to the heat added to the system minus the mechanical work done by the system.

1 From notes of Professor F. E. C. Culick, California Institute of Technology (with minor changes)

1B-3

On the basis of experimental results, therefore, one is led to assert the existence of two new properties, the temperature and internal energy, which do not arise in ordinary mechanics. In a similar way, a further remarkable relationship between heat and temperature will be established, and a new property, the entropy, defined. Although this is a much less familiar property, it is to be stressed that the general approach is quite like that used to establish the Zeroth and First Laws. A general principle and a property associated with any system are extracted from experimental results. Viewed in this way, the entropy should appear no more mystical than the internal energy. The increase of entropy in a naturally occurring process is no less real than the conservation of energy.

(iv.) Second Law Although all natural processes must take place in accordance with the First Law, the

principle of conservation of energy is, by itself, inadequate for an unambiguous description of the behavior of a system. Specifically, there is no mention of the familiar observation that every natural process has in some sense a preferred direction of action. For example, the flow of heat occurs naturally from hotter to colder bodies, in the absence of other influences, but the reverse flow certainly is not in violation of the First Law. So far as that law is concerned, the initial and final states are symmetrical in a very important respect.

The Second Law is essentially different from the First Law; the two principles are independent and cannot in any sense be deduced from one another. Thus, the concept of energy is not sufficient, and a new property must appear. This property can be developed, and the Second Law introduced, in much the same way as the Zeroth and First Laws were presented. By examination of certain observational results, one attempts to extract from experience a law which is supposed to be general; it is elevated to the position of a fundamental axiom to be proved or disproved by subsequent experiments. Within the structure of classical thermodynamics, there is no proof more fundamental than observations. A statement which can be adopted as the Second Law of thermodynamics is:

Second Law There exists for every thermodynamic system in equilibrium an extensive scalar property called the entropy, S, such that in an infinitesimal reversible change of state of the system, dS = dQ/T, where T is the absolute temperature and dQ is the amount of heat received by the system. The entropy of a thermally insulated system cannot decrease and is constant if and only if all processes are reversible.

As with the Zeroth and First Laws, the existence of a new property is asserted and its behavior is described.

(v.) Reversible Processes In the course of this development, the idea of a completely reversible process is central, and

we can recall the definition, “a process is called completely reversible if, after the process has occurred, both the system and its surroundings can be wholly restored by any means to their respective initial states”. Especially, it is to be noted that the definition does not, in this form, specify that the reverse path must be identical with the forward path. If the initial states can be restored by any means whatever, the process is by definition completely reversible. If the paths are identical, then one usually calls the process (of the system) reversible, or one may say that the state of the system follows a reversible path. In this path (between two equilibrium states 1 and 2), (i) the system passes through the path followed by the equilibrium states only, and (ii) the system will take the reversed path 2 to 1 by a simple reversal of the work done and heat added.

Reversible processes are idealizations not actually encountered. However, they are clearly

1B-4

h u

ρ

useful idealizations. For a process to be completely reversible, it is necessary that it be quasi-static and that there be no dissipative influences such as friction and diffusion. The precise (necessary and sufficient) condition to be satisfied if a process is to be reversible is the second part of the Second Law.

The criterion as to whether a process is completely reversible must be based on the initial and final states. In the form presented above, the Second Law furnishes a relation between the properties defining the two states, and thereby shows whether a natural process connecting the states is possible.

Muddy points What happens when all the energy in the universe is uniformly spread, ie, entropy at a maximum? (MP 1B.3)

1.B.3 Combined First and Second Law Expressions First Law:

dU = dQ − dW - Always true

Work and heat exchange in terms of state variables:

dQ = TdS; dW = PdV - Only true for reversible processes.

dU = dQ − PdV ; Simple compressible substance, reversible process dU = dQ − PdV − XdY ; Substance with other work modes (e.g., stress-strain), X is a pressure-like quantity, Y is a volume like quantity dU = TdS − dW ; Only true for a reversible process

First law in terms of state variables:

dU = TdS − PdV ; This is a relation between properties and is always true

In terms of specific quantities (per unit mass):

du = Tds − Pdv Combined first and second law (a) or Gibbs equation (a)

The combined first and second law expressions are often more usefully written in terms of the enthalpy, or specific enthalpy, = + Pv:

dh = du + Pdv + vdP = Tds − Pdv + Pdv + vdP , using the first law.

dh = Tds + vdP

Or, since v = 1/ ρ dP

dh = Tds + . Combined first and second law (b) or Gibbs equation (b)

In terms of enthalpy (rather than specific enthalpy) the relation is dH = TdS + VdP .

1B-5

=

=

)

+ =

1.B.4 Entropy Changes in an Ideal Gas Many aerospace applications involve flow of gases (e.g., air) and we thus examine the

entropy relations for ideal gas behavior. The starting point is form (a) of the combined first and second law,

du = Tds − Pdv .

For an ideal gas, du = cvdT . Thus dT P

Tds = c dT + Pdv or ds = c + dv .v v T T Using the equation of state for an ideal gas ( Pv = RT ), we can write the entropy change as an expression with only exact differentials:

ds = cv

dT + R

dv . (B.4.1)

T v

Integrating between two states “1” and “2”:

T2 dT v2 dv∆s s2 − s1 = ∫T1

cv + R∫v1

. T v

For constant specific heat

∆s s2 − s1 = cvln T2

+ Rln v2

. T1 v1

In non-dimensional form (using R

= (γ − 1 ) cv

∆s = ln

T2

+ (γ − 1)ln v2

. Entropy change of an ideal gas (B.4.2)

cv T1 v1

Equation (B.4.2) is in terms of specific quantities. For N moles of gas

∆S = N

ln

T2

+ (γ − 1)ln V2

.

Cv T1 V1

Rather than temperature and volume, we can develop an alternative form of the expression, in terms of pressure and volume, for entropy change, which allows us to examine an assumption we have used over the past year. The ideal gas equation of state can be written as

lnP + lnv = lnR + lnT.

Taking differentials of both sides yields

dP dv dT

P v T

Using the above equation in Eq. (B.4.1), and making use of the relations cp = cv + R; cp / cv = γ , we find

1B-6

ds = cv dP

+ dv + R

dv , P v v

or ds dP dv

= + γ . c P vv

Integrating between two states 1 and 2

γ ∆s

= ln P2

+ γln

v2 = ln

P2 v2 . (B.4.3)

cv P1 v1 P1 v1

Using both sides of (B.4.3) as exponents we obtain

P vγ 2 s cv2 2 = [Pvγ ]1 = e ∆ / . (B.4.4)

Pvγ 1 1

Equation (B.4.4) describes a general process. For the specific situation in which∆s = 0, i.e., the entropy is constant, we recover the expression Pvγ = constant. It was stated that this expression applied to a reversible, adiabatic process. We now see, through use of the second law, a deeper meaning to the expression, and to the concept of a reversible adiabatic process, in that both are characteristics of a constant entropy, or isentropic, process.

Muddy points

Why do you rewrite the entropy change in terms of Pvγ? (MP 1B.4) What is the difference between isentropic and adiabatic? (MP 1B.5)

1.B.5 Calculation of Entropy Change in Some Basic Processes

a) Heat transfer from, or to, a heat reservoir. A heat reservoir is a constant temperature heat source

or sink. Because the temperature is uniform, there is no heat transfer across a finite temperature difference and the heat exchange is reversible. From the definition of entropy (dS = dQrev /T ) ,

∆S = Q

,T

where Q is the heat into the reservoir (defined here as positive if heat flows into the reservoir)

b) Heat transfer between two heat reservoirs

The entropy changes of the two reservoirs are the sum of the entropy change of each. If the high

and the lowtemperature reservoir is at THtemperature reservoir is at TL , the total entropy change is

TH

QH QH

Heat transfer from/to a heat reservoir

Q TH

TL

Device (block of copper) no work no change in state

Heat transfer between two reservoirs

1B-7

H

∆S =

−Q +

Q =

Q (TH − TL ) TH TL T TL

The second law says that the entropy change must be equal to or greater than zero. This corresponds to the statement that heat must flow from the higher temperature source to the lower temperature source. This is one of the statements of the second law given in Section 1.B.1.

Muddy pointsIn the single reservoir example, why can the entropy decrease? (MP 1B.6)9Why does the entropy of a heat reservoir change if the temperature stays the same? (MP91B.7)9How can the heat transfer from or to a heat reservoir be reversible? (MP 1B.8)9How can ∆S be less than zero in any process? Doesn't entropy always increase? (MP 1B.9)9

If Q

= ∆S for a reservoir, could you add Q to any size reservoir and still get the same ∆S? T

(MP 1B.10)

c) Possibility of obtaining work from a single heat reservoir We can regard the proposed process as the

absorption of heat, Q, by a device or system, operating in a cycle, rejecting no heat, and producing work. The total entropy change is the sum of the change in the reservoir, the system or device, and the surroundings. The entropy change of the reservoir is ∆S = −Q/TH . The entropy change of the device is zero, because we are considering a complete cycle (return to initial state) and entropy is a function of state. The surroundings receive work only so the entropy change of the surroundings is zero. The total entropy change is

Work from a single heat reservoir

∆Stotal = ∆Sreservoir + ∆Sdevice + ∆Ssurroundings

/ += −Q TH + 0 0

The total entropy change in the proposed process is thus less than zero, ∆Stotal < 0

which is not possible. The second law thus tells us that we cannot get work from a single reservoir only. The “only” is important; it means without any other changes occurring. This is the other statement of the second law we saw in Section 1.B.1.

Muddy pointsWhat is the difference between the isothermal expansion of a piston and the (forbidden) production of work using a single reservoir? (MP 1B.11) For the "work from a single heat reservoir" example, how do we know there is no ∆Ssurr? (MP 1B.12)

1B-8

How does a cycle produce zero ∆S? I thought that the whole thing about cycles was an entropy that the designers try to minimize. (MP 1B.13)

d) Entropy changes in the “hot brick problem” We can examine in a more quantitative manner the changes that occurred when we put the two

bricks together, as depicted on the left-hand side of the figure below. The process by which the two bricks come to the same temperature is not a reversible one, so we need to devise a reversible path. To do this imagine a large number of heat reservoirs at varying temperatures spanning the range TH − dT,............,TL + dT , as in the right hand side of the figures. The bricks are put in contact with them sequentially to raise the temperature of one and lower the temperature of the

TH

TL

TM

TM

TH ........ TL

TH - dT TL + dT

Temperature equalization of two bricks Reservoirs used in reversible state transformation

other in a reversible manner. The heat exchange at any of these steps is dQ = CdT . For the high temperature brick, the entropy change is:

TM CdT = C ln

TM ∆Shot brick = ∫TH T TH

where C is the heat capacity of the brick (J/kg). This quantity is less than zero. For the cold brick,

TM CdT = C ln

TM .∆Scold brick = ∫TL T TL

The entropy change of the two bricks is

2

∆Sbricks = C ln

TM

+ ln

TM

= C ln TM > 0 .

TH TL T TH L

The process is not reversible.

e) Difference between the free expansion and the reversible isothermal expansion of an ideal gas The essential difference between the free expansion in an insulated enclosure and the

reversible isothermal expansion of an ideal gas can also be captured clearly in terms of entropy changes. For a state change from initial volume and temperature V T1 to final volume and (the1, same) temperature V T1 the entropy change is2 ,

2 2 dU 2 PdV∆S = ∫1

dS = ∫1 T + ∫1

,T

1B-9

T S

T S

V

V −

or, making use of the equation of state and the fact that dU = 0 for an isothermal process,

∆S = NR ln 2 .

V1 This is the entropy change that occurs for the free expansion as well as for the isothermal reversible expansion processes—entropy changes are state changes and the two system final and end states are the same for both processes.

For the free expansion:

∆Ssystem = NR ln

V2

; ∆Ssurroundings = 0 V1

There is no change in the entropy of the surroundings because there is no interaction between the system and the surroundings. The total entropy change is therefore,

∆Stotal = ∆Ssystem + ∆Ssurroundings = NR ln

V2 > 0.

V1 There are several points to note from this result.

i) ∆Stotal > 0 so the process is not reversible 2 dQ

ii) ∆Ssystem > ∫1 T = 0; the equality between ∆S and dQ is only for a reversibleT

process iii) There is a direct connection between the work needed to restore the system to the original state and the entropy change:

W = NRT ln 2 = ∆ 2 1 V1

The quantity ∆ has a physical meaning as “lost work” in the sense of work which we lost the opportunity to utilize. We will make this connection stronger in Section 1.C.

For the reversible isothermal expansion:The entropy is a state variable so the entropy change of the system is the same as before. In thiscase, however, heat is transferred to the system from the surroundings ( Qsurroundings < 0) so that

∆Ssurroundings = Qsurroundings

< 0. T

The heat transferred from the surroundings, however, is equal to the heat received by the system: Qsurroundings = Qsystem = W .

∆Ssurroundings = Qsurroundings

=−W

= - NR ln

V2 . T T V1

The total change in entropy (system plus surroundings) is therefore

∆Stotal = ∆Ssystem + ∆Ssurroundings = Q

− Q

= 0. T T

The reversible process has zero total change in entropy.

1B-109

Muddy pointsOn the example of free expansion versus isothermal expansion, how do we know that the9pressure and volume ratios are the same? We know for each that P91B.14)9

2>P1 and V2>V1. (MP

Where did ∆Ssystem = NRln

V2 come from? (MP 1B.15)

V1

1B-119

Muddiest Points on Part 1B

1B.1 Why is dU = TdS − PdV always true?

This is a relation between state variables. As such it is not path dependent, only depends on the initial and final states, and thus must hold no matter how we transition from initial state to final state. What is not always true, and what holds only for reversible processes are the relations Tds = dq and Pdv = dw. One example of this is the free expansion where dq = dw = 0, but where the quantities Tds and Pdv (and the integrals of these quantities) are not zero.

1B.2 What makes dQrev different than dQ?

The term dQrev denotes the heat exchange during a reversible process. We use the notation dQ to denote heat exchange during any process, not necessarily reversible. The distinction between the two is important for the reason given above in (3).

1B.3 What happens when all the energy in the universe is uniformly spread, ie, entropy at a maximum?

I quote from The Refrigerator and the Universe, by Goldstein and Goldstein:“The entropy of the universe is not yet at its maximum possible value and it seems to beincreasing all the time. Looking forward to the future, Kelvin and Clausius foresaw atime when the maximum possible entropy would be reached and the universe would be atequilibrium forever afterward; at this point, a state called the “heat death” of the universe,nothing would happen forever after”. The book also gives comments on the inevitabilityof this fate.

1B.4 Why do you rewrite the entropy change in terms of Pvγ?

We have discussed the representation of thermodynamic changes in P-v coordinates a number of times and it is familiar, as is the idea of the “ Pvγ = constant ” process. I want to relate this to the more general expression involving the entropy change (Equation B.4.4) to show (i) when the simple form applied and (ii) how valid an approximation it was. Using the entropy change, we now have a quantitative metric for doing just that.

1B.5 What is the difference between isentropic and adiabatic?

Isentropic means no change in entropy (dS = 0). An adiabatic process is a process with no heat transfer (dQ = 0). We defined for reversible processes TdS = dQ. So generally an adiabatic process is not necessarily isentropic – only if the process is reversible and adiabatic we can call it isentropic. For example a real compressor can be assumed adiabatic but is operating with losses. Due to the losses the compression is irreversible. Thus the compression is not isentropic.

1B.6 In the single reservoir example, why can the entropy decrease?

When we looked at the single reservoir, our “system” was the reservoir itself. The example I did in class had heat leaving the reservoir, so that Q was negative. Thus the entropy change of the reservoir is also negative. The second law, however, guarantees that there is a positive change in entropy somewhere else in the surroundings that will be as large, or larger, than this decrease.

1B.7 Why does the entropy of a heat reservoir change if the temperature stays the same?

A heat reservoir is an idealization (like an ideal gas, a rigid body, an inviscid fluid, a discrete element mass-spring-damper system). The basic idea is that the heat capacity of the heat reservoir is large enough so that the transfer of heat in whatever problem we address does not apprecibly alter the temperature of the reservoir. In grappling with approximations such as this it is useful to think about extreme cases. Therefore, suppose the thermal reservoir is the atmosphere. The mass of the atmosphere is roughly 1019 kg (give or take an order of magnitude). Let us calculate the temperature rise due to the heat dumped into the atmosphere by a jet engine during a transcontinental flight. A large gas turbine engine might produce on the order of 100 MW of heat, so that the rise in atmospheric temperature, δTatm , for the heat transfer Q associated with a 6 hour flight is given by

M cpδTatm = × 3600 × 108 J .atm 6

Substituting for the atmospheric mass and the specific heat gives a value for temperature change of roughly 10-10 K. To a very good approximation, we can say that the temperature of this heat reservoir is constant and we can evaluate the entropy change of the reservoir as Q/T.

1B.8 How can the heat transfer from or to a heat reservoir be reversible?

We made the assumption that the heat reservoir is very large, and therefore it is a constant temperature heat source or sink. Since the temperature is uniform there is no heat transfer across a finite temperature difference and this heat exchange is reversible. We discussed this in the second example "Heat transfer between two heat reservoirs".

1B.9 How can ∆S be less than zero in any process? Doesn't entropy always increase?

The second law says that the total entropy (system plus surroundings) always increases.(See Section 1.B.1). This means that either the system or the surroundings can have itentropy decrease if there is heat transfer between the two, although the sum of all entropychanges must be positive.For an isolated system, with no heat transfer to the surroundings, the entropy must alwaysincrease.

1B.10 If Q T = ∆S for a reservoir, could you add Q to any size reservoir and still get the

same ∆S?

Yes, as long as the system you were adding heat to fulfilled the conditions for being a reservoir.

1B.11 What is the difference between the isothermal expansion of a piston and the (forbidden) production of work using a single reservoir?

The difference is contained in the word sole in the Kelvin-Planck statement of the second law given in Section 1.B.1 of the notes.

For the isothermal expansion the changes are: a) The reservoir loses heat Q b) The system does work W (equal in magnitude to Q) c) The system changes its volume and pressure. d) The system changes its entropy (the entropy increases by Q/T).

For the “forbidden” process, a) The reservoir loses heat Q b) The system does work W (= Q) and that’s all the changes that there are.

leave it to you to calculate the total entropy changes (system plus surroundings) that occur in the two processes.

1B.12 For the "work from a single heat reservoir" example, how do we know there is no ∆Ssurr?

Our system was the heat reservoir itself. In the example we had heat leaving the reservoir, thus Q was negative and the entropy change of the reservoir was also negative. Using the second law, it is guaranteed that somewhere else in the surroundings a positive entropy change will occur that is as large or larger than the decrease of the entropy of the reservoir.

1B 13 How does a cycle produce zero ∆S? I thought that the whole thing about cycles was an entropy that the designers try to minimize.

The change in entropy during a cycle is zero because we are considering a complete cycle (returning to initial state) and entropy is a function of state (holds for ideal and real cycles!).

The entropy you are referring to is entropy that is generated in the components of a non-ideal cycle. For example in a real jet engine we have a non-ideal compressor, a non-ideal combustor and also a non-ideal turbine. All these components operate with some loss and generate entropy – this is the entropy that the designers try to minimize. Although the change in entropy during a non-ideal cycle is zero, the total entropy change (cycle and

I

heat reservoirs!) is ∆Stotal > 0. Basically the entropy generated due to irreversibilities in the engine is additional heat rejected to the environment (to the lower heat reservoir). We will discuss this in detail in Section 1.C.1.

1B.14 On the example of free expansion versus isothermal expansion, how do we know that the pressure and volume ratios are the same? We know for each that P2>P1

and V2>V1.

During the free-expansion no work is done and no heat is transferred (insulated system). Thus the internal energy stays constant and so does the temperature. This means that P1V1 = P2V2 holds also for the free-expansion and that the pressure and volume ratios are the same when comparing free-expansion to reversible isothermal expansion.

1B.15 Where did ∆Ssystem = NRln

V2 come from?

V1

We were using the 1st and 2nd law combined (Gibbs) and in the example discussed there was no change in internal energy (dU=0). If we then integrate dS = P/TdV using P/T = NR/V (with N being the number of moles of gas in volume V and R is the universal gas constant) we obtain ∆Ssystem = NR ln(V2/V1).

1C-1

TH

QH

We

QL

TL

Carnot cycle

1.C Applications of the Second Law[VN-Chapter 6; VWB&S-8.1, 8.2, 8.5, 8.6, 8.7, 8.8, 9.6]

1.C.1 Limitations on the Work that Can be Supplied by a Heat EngineThe second law enables us to make powerful and general statementsconcerning the maximum work that can bederived from any heat engine which operates ina cycle. To illustrate these ideas, we use aCarnot cycle which is shown schematically atthe right. The engine operates between two heatreservoirs, exchanging heatQH with the high temperature reservoir at TH

and QL with the reservoir at TL.. The entropychanges of the two reservoirs are:

∆SQ

TQH

H

HH= <; 0

∆SQ

TQL

L

LL= >; 0

The same heat exchanges apply to the system, but with opposite signs; the heat received from thehigh temperature source is positive, and conversely. Denoting the heat transferred to the enginesby subscript “e”,

Q Q Q QHe H Le L= − = − ; .

The total entropy change during any operation of the engine is,

∆ ∆ ∆ ∆S S S StotalH

servoirat TH

L

servoirat TL

e

Engine

= + +Re Re

For a cyclic process, the third of these ∆Se( ) is zero, and thus (remembering that QH < 0 ),

∆ ∆ ∆S S SQ

T

Q

Ttotal

H LH

H

L

L

= + = + (C.1.1)

For the engine we can write the first law as

∆U Q Q We He Le e= = + −0 (cyclic process) .

Or, W Q Qe He Le

= +

= − −Q QH L .

Hence, using (C.1.1)

W Q T S QT

Te H Ltotal

HL

H

= − − +

∆

1C-2

= −( ) −

−Q

T

TT SH

L

HL

total1 ∆ .

The work of the engine can be expressed in terms of the heat received by the engine as

W QT

TT Se He

L

HL

total= ( ) −

−1 ∆ .

The upper limit of work that can be done occurs during a reversible cycle, for which the totalentropy change ( ∆Stotal ) is zero. In this situation:

Maximum work for an engine working between T TH L and : W QT

Te HeL

H

= ( ) −

1

Also, for a reversible cycle of the engine,

Q

T

Q

TH

H

L

L

+ = 0.

These constraints apply to all reversible heat engines operating between fixed temperatures. Thethermal efficiency of the engine is

η = =Work done

Heat received

W

QHe

= − =1T

TL

HCarnotη .

The Carnot efficiency is thus the maximum efficiency that can occur in an engine workingbetween two given temperatures.

We can approach this last point in another way. The engine work is given by

W Q T S Q T Te H Ltotal

H L H= − − + ( )∆ /

or, T S Q Q T T WLtotal

H H L H e∆ = − + ( ) −/

The total entropy change can be written in terms of the Carnot cycle efficiency and the ratio of thework done to the heat absorbed by the engine. The latter is the efficiency of any cycle we candevise:

∆SQ

T

T

T

W

Q

Q

Ttotal He

L

L

H

e

He

He

LCarnot Any other

cycle

= − −

= −

1 η η .

The second law says that the total entropy change is equal to or greater than zero. This means thatthe Carnot cycle efficiency is equal to or greater than the efficiency for any other cycle, with theequality only occurring if ∆Stotal = 0 .

1C-3

Muddy pointsSo, do we lose the capability to do work when we have an irreversible process andentropy increases? (MP 1C.1)Why do we study cycles starting with the Carnot cycle? Is it because I is easier to workwith? (MP 1C.2)

1.C.2 The Thermodynamic Temperature ScaleThe considerations of Carnot cycles in this section have not mentioned the working

medium. They are thus not limited to an ideal gas and hold for Carnot cycles with any medium.Because we derived the Carnot efficiency with an ideal gas as a medium, the temperaturedefinition used in the ideal gas equation is not essential to the thermodynamic arguments. Morespecifically, we can define a thermodynamic temperature scale that is independent of the workingmedium. To see this, consider the situation shown below in Figure C-1, which has three reversiblecycles. There is a high temperature heat reservoir at T3 and a low temperature heat reservoir at T1.For any two temperatures T T1 2, , the ratio of the magnitudes of the heat absorbed and rejected in aCarnot cycle has the same value for all systems.

WA

WC

A

B

C

Q1

Q3

Q3

Q1

Q2T2

T1

T3

Q2

WB

Figure C-1: Arrangement of heat engines to demonstrate the thermodynamic temperature scale

We choose the cycles so Q1 is the same for A and C. Also Q3 is the same for B and C. For aCarnot cycle

η = + = ( )1Q

QF T TL

HL H, ; η is only a function of temperature.

AlsoQ Q F T T1 2 1 2= ( ),

Q Q F T T2 3 2 3= ( ),

Q Q F T T1 3 1 3= ( ), .

But

1C-4

Q

Q

Q

Q

Q

Q1

3

1

2

2

3

= .

HenceF T T F T T F T T

TT

1 3

2

1 2 2 3

2

, , ,( ) = ( ) × ( )Not a functionof

Cannot be a function of 1 24 34 1 2444 3444

.

We thus conclude that F T T1 2,( ) has the form f T f T1 2( ) ( )/ , and similarly

F T T f T f T2 3 2 3, /( ) = ( ) ( ). The ratio of the heat exchanged is therefore

Q

QF T T

f T

f T1

31 3

1

3

= ( ) =( )( )

, .

In general,Q

Q

f T

f TH

L

H

L

=( )( )

,

so that the ratio of the heat exchanged is a function of the temperature. We could choose anyfunction that is monotonic, and one choice is the simplest: f T T( ) = . This is the thermodynamicscale of temperature, Q Q T TH L H L= . The temperature defined in this manner is the same as thatfor the ideal gas; the thermodynamic temperature scale and the ideal gas scale are equivalent

1.C.3 Representation of Thermodynamic Processes in T-s coordinates.It is often useful to plot the thermodynamic state transitions and the cycles in terms of

temperature (or enthalpy) and entropy, T,S, rather than P,V. The maximum temperature is oftenthe constraint on the process and the enthalpy changes show the work done or heat receiveddirectly, so that plotting in terms of these variables provides insight into the process. A Carnotcycle is shown below in these coordinates, in which it is a rectangle, with two horizontal, constanttemperature legs. The other two legs are reversible and adiabatic, hence isentropic( dS dQ Trev= / = 0), and therefore vertical in T-s coordinates.

T

TH

TL

Isothermal

Adiabatic

s

Carnot cycle in T,s coordinates

If the cycle is traversed clockwise, the heat added is

a

c

b

d

1C-5

Heat added: Q TdS T S S T SH ab

H b a H= ∫ = −( ) = ∆ .

The heat rejected (from c to d) has magnitude Q T SL L= ∆ .The work done by the cycle can be found using the first law for a reversible process:

dU dQ dW= − . = −TdS dW (This form is only true for a reversible process).

We can integrate this last expression around the closed path traced out by the cycle:

dU TdS dW= − ∫∫∫

However dU is an exact differential and its integral around a closed contour is zero:

0 = − ∫∫ TdS dW .

The work done by the cycle, which is represented by the term dW∫ , is equal to Tds∫ , the areaenclosed by the closed contour in the T-S plane. This area represents the difference between theheat absorbed ( TdS∫ at the high temperature) and the heat rejected ( TdS∫ at the low temperature).Finding the work done through evaluation of TdS∫ is an alternative to computation of the work in areversible cycle from PdV∫ . Finally, although we have carried out the discussion in terms of theentropy, S, all of the arguments carry over to the specific entropy, s; the work of the reversiblecycle per unit mass is given by Tds.∫

Muddy pointsHow does one interpret h-s diagrams? (MP 1C.3)Is it always OK to "switch" T-s and h-s diagram? (MP 1C.4)What is the best way to become comfortable with T-s diagrams? (MP 1C.5)What is a reversible adiabat physically? (MP 1C.6)

1.C.4 Brayton Cycle in T-s CoordinatesThe Brayton cycle has two reversible adiabatic (i.e., isentropic) legs and two reversible,

constant pressure heat exchange legs. The former are vertical, but we need to define the shape ofthe latter. For an ideal gas, changes in specific enthalpy are related to changes in temperature bydh c dTp= , so the shape of the cycle in an h-s plane is the same as in a T-s plane, with a scale

factor of cp between the two. This suggests that a place to start is with the combined first and

second law, which relates changes in enthalpy, entropy, and pressure:

dh Tdsdp

= +ρ

.

On constant pressure curves dP=0 and dh Tds= . The quantity desired is the derivative oftemperature, T, with respect to entropy, s, at constant pressure: ∂ ∂T s

p( ) . From the combined first

and second law, and the relation between dh and dT, this is

1C-6

∂∂T

s

T

cp p

= (C.4.1)

The derivative is the slope of the constant pressure legs of the Brayton cycle on a T-s plane. For agiven ideal gas (specific cp) the slope is positive and increases as T.

We can also plot the Brayton cycle in an h-s plane. This has advantages because changesin enthalpy directly show the work of the compressor and turbine and the heat added and rejected.The slope of the constant pressure legs in the h-s plane is ∂ ∂h s T

p( ) = .

Note that the similarity in the shapes of the cycles in T-s and h-s planes is true for idealgases only. As we will see when we examine two-phase cycles, the shapes look quite different inthese two planes when the medium is not an ideal gas.

Plotting the cycle in T-s coordinates also allows another way to address the evaluation ofthe Brayton cycle efficiency which gives insight into the relations between Carnot cycle efficiencyand efficiency of other cycles. As shown in Figure C-2, we can break up the Brayton cycle into

many small Carnot cycles. The " "ith Carnot cycle has an efficiency of ηci lowi highiT T= − ( )[ ]1 ,

where the indicated lower temperature is the heat rejection temperature for that elementary cycleand the higher temperature is the heat absorption temperature for that cycle. The upper and lowercurves of the Brayton cycle, however, have constant pressure. All of the elementary Carnot cyclestherefore have the same pressure ratio:

P T

P TPR

high

low

( )( )

= = constant (the same for all the cycles).

From the isentropic relations for an ideal gas, we know that pressure ratio, PR, and temperatureratio, TR, are related by : PR TRγ γ−( ) =1 / .

1C-7

Tmax

Tmin

T

s

Figure C-2: Ideal Brayton cycle as composed of many elementary Carnot cycles [Kerrebrock]

The temperature ratios T Tlowi highi( ) of any elementary cycle “i” are therefore the same and each of

the elementary cycles has the same thermal efficiency. We only need to find the temperature ratioacross any one of the cycles to find what the efficiency is. We know that the temperature ratio ofthe first elementary cycle is the ratio of compressor exit temperature to engine entry (atmosphericfor an aircraft engine) temperature, T T2 0/ in Figure C-2. If the efficiency of all the elementarycycles has this value, the efficiency of the overall Brayton cycle (which is composed of theelementary cycles) must also have this value. Thus, as previously,

ηBraytoninlet

compressor exit

T

T= −

1

.

A benefit of this view of efficiency is that it allows us a way to comment on the efficiencyof any thermodynamic cycle. Consider the cycle shown on the right, which operates betweensome maximum and minimumtemperatures. We can break it up into small Carnot cyclesand evaluate the efficiency of each. It can be seen that theefficiency of any of the small cycles drawn will be less than theefficiency of a Carnot cycle between Tmax and Tmin . Thisgraphical argument shows that the efficiency of any otherthermodynamic cycle operating between these maximum andminimum temperatures has an efficiency less than that of aCarnot cycle.

Muddy pointsIf there is an ideal efficiency for all cycles, is there a maximum work or maximum powerfor all cycles? (MP 1C.7)

Arbitrary cycle operatingbetween T Tmin, max

1C-8

1.C.5 Irreversibility, Entropy Changes, and “Lost Work”Consider a system in contact with a heat reservoir during a reversible process. If there is

heat Q absorbed by the reservoir at temperature T, the change in entropy of the reservoir is∆S Q T= / . In general, reversible processes are accompanied by heat exchanges that occur atdifferent temperatures. To analyze these, we can visualize a sequence of heat reservoirs atdifferent temperatures so that during any infinitesimal portion of the cycle there will not be anyheat transferred over a finite temperature difference.

During any infinitesimal portion, heat dQrev will be transferred between the system and one of thereservoirs which is at T. If dQrev is absorbed by the system, the entropy change of the system is

dSdQ

Tsystem rev= .

The entropy change of the reservoir is

dSdQ

Treservoir rev= − .

The total entropy change of system plus surroundings is

dS dS dStotal system reservoir= + = 0 .

This is also true if there is a quantity of heat rejected by the system.

The conclusion is that for a reversible process, no change occurs in the total entropy produced, i.e.,the entropy of the system plus the entropy of the surroundings: ∆Stotal = 0.

We now carry out the same type of analysis for an irreversible process, which takes the systembetween the same specified states as in the reversible process. This is shown schematicallyat the right, with I and R denoting the irreversible and reversible processes.In the irreversible process, the system receives heat dQ and does work dW.The change in internal energy for the irreversible process is

dU dQ dW= − (Always true - first law).

For the reversible process

dU TdS dWrev= − .

Because the state change is the same in the two processes(we specified that it was), the change in internal energy is thesame. Equating the changes in internal energy in the above two expressions yields

dQ dW TdS dWactual actual rev− = − .

Irreversible and reversiblestate changes

1C-9

The subscript “actual” refers to the actual process (which is irreversible). The entropy changeassociated with the state change is

dSdQ

T TdW dWactual

rev actual= + −[ ]1. (C.5.1)

If the process is not reversible, we obtain less work (see IAW notes) than in a reversible process,dW dWactual rev< , so that for the irreversible process,

dSdQ

Tactual> . (C.5.2)

There is no equality between the entropy change dS and the quantity dQ/T for an irreversibleprocess. The equality is only applicable for a reversible process.

The change in entropy for any process that leads to a transformation between an initial state “a”and a final state “b” is therefore

∆S S SdQ

Tb aactual

ab= − ≥ ∫

where dQactual is the heat exchanged in the actual process. The equality only applies to areversible process.

The difference dW dWrev actual− represents work we could have obtained, but did not. It is referredto as lost work and denoted by Wlost . In terms of this quantity we can write,

dSdQ

T

dW

Tactual lost= + . (C.5.3)

The content of Equation (C.5.3) is that the entropy of a system can be altered in two ways: (i)through heat exchange and (ii) through irreversibilities. The lost work ( dWlost in Equation C.5.3) isalways greater than zero, so the only way to decrease the entropy of a system is through heattransfer.

To apply the second law we consider the total entropy change (system plus surroundings). If thesurroundings are a reservoir at temperature T, with which the system exchanges heat,

dS dSdQ

Treservoir surroundings actual =( ) = − .

The total entropy change is

dS dS dSdQ

T

dW

T

dQ

Ttotal system surroundings actual lost actual= + = +

−

1C-10

dSdW

Ttotal lost= ≥ 0 .

The quantity ( dW Tlost / ) is the entropy generated due to irreversibility.

Yet another way to state the distinction we are making is

dS dS dS dS dSsystemfromheattransfer

generated due toirreversibleprocesses

heat transfer Gen= + = + . (C.5.4)

The lost work is also called dissipation and noted dφ. Using this notation, the infinitesimal entropychange of the system becomes:

dS dSdT

systemheat transfer= +

φ

or TdS dQ dsystemr= + φ

Equation (C.5.4) can also be written as a rate equation,

dS

dtS S Sheat transfer Gen= = +˙ ˙ ˙

. (C.5.5)

Either of equation (C.5.4) or (C.5.5) can be interpreted to mean that the entropy ofthe system, S, is affected by two factors: the flow of heat Q and the appearance ofadditional entropy, denoted by dSGen, due to irreversibility1. This additional entropy iszero when the process is reversible and always positive when the process is irreversible.Thus, one can say that the system develops sources which create entropy during anirreversible process. The second law asserts that sinks of entropy are impossible innature, which is a more graphic way of saying that dSGen and SGen are positive definite,

or zero, for reversible processes.

The term ˙ ,˙

ST

dQ

dt

Q

Theat transfer or =

1, which is associated with heat transfer to

the system, can be interpreted as a flux of entropy. The boundary is crossed by heat andthe ratio of this heat flux to temperature can be defined as a flux of entropy. There are norestrictions on the sign of this quantity, and we can say that this flux either contributestowards, or drains away, the system's entropy. During a reversible process, only this fluxcan affect the entropy of the system. This terminology suggests that we interpret entropyas a kind of weightless fluid, whose quantity is conserved (like that of matter) during areversible process. During an irreversible process, however, this fluid is not conserved; itcannot disappear, but rather is created by sources throughout the system. While thisinterpretation should not be taken too literally, it provides an easy mode of expressionand is in the same category of concepts such as those associated with the phrases "flux of 1 This and the following paragraph are excerpted with minor modifications from A Course inThermodynamics, Volume I, by J. Kestin, Hemisphere Press (1979)

1C-11

energy" or "sources of heat". In fluid mechanics, for example, this graphic language isvery effective and there should be no objections to copying it in thermodynamics.

Muddy pointsDo we ever see an absolute variable for entropy? So far, we have worked withdeltas only (MP 1C.8)

I am confused as to dSdQrev

T= as opposed to dS

dQrev

T≥ .(MP 1C.9)

For irreversible processes, how can we calculate dS if not equal to dQ

T(MP

1C.10)

1.C.6 Entropy and Unavailable Energy (Lost Work by Another Name)Consider a system consisting of a heat reservoir at T2 in surroundings (the atmosphere) at

T0 . The surroundings are equivalent to a second reservoir at T0 . For an amount of heat, Q,transferred from the reservoir, the maximum work we could derive is Q times the thermalefficiency of a Carnot cycle operated between these two temperatures:

Maximum work we could obtain = W Q T Tmax /= −( )1 0 2 . (C.6.1)

Only part of the heat transferred can be turned into work, in other words only part of the heatenergy is available to be used as work.

Suppose we transferred the same amount of heat from the reservoir directly to another reservoir ata temperature T T1 2< . The maximum work available from the quantity of heat, Q , before thetransfer to the reservoir at T1 is,

W Q T TT Tmax

,/

2 00 21= −( ); [Maximum work between T T2 0, ].

The maximum amount of work available after the transfer to the reservoir at T1 is,W Q T T

T Tmax

,/

1 00 11= −( ); [Maximum work between T T1 0, ].

There is an amount of energy that could have been converted to work prior to the irreversible heattransfer process of magnitude ′E ,

′ = −

− −

= −

E Q

T

T

T

TQ

T

T

T

T1 10

2

0

1

0

1

0

2

,

or

′ = −

E T

Q

T

Q

T01 2

.

1C-12

However, Q T/ 1 is the entropy gain of the reservoir at T1 and (-Q T/ 2) is the entropy decrease of thereservoir at T2 . The amount of energy, ′E , that could have been converted to work (but nowcannot be) can therefore be written in terms of entropy changes and the temperature of thesurroundings as

′ = +

=

E T S S

T S

reservoirat T

reservoirat T

irreversible heat transfer process

01 2

0

∆ ∆

∆

′E = “Lost work”, or energy which is no longer available as work.

The situation just described is a special case of an important principle concerning entropy changes,irreversibility and the loss of capability to do work. We thus now develop it in a more generalfashion, considering an arbitrary system undergoing an irreversible state change, which transfersheat to the surroundings (for example the atmosphere), which can be assumed to be at constanttemperature, T0 . The change in internal energy of the system during the state change is∆U Q W= − . The change in entropy of the surroundings is (with Q the heat transfer to the system)

∆SQ

Tsurroundings = −

0

.

Now consider restoring the system to the initial state by a reversible process. To do this we needto do work, Wrev on the system and extract from the system a quantity of heat Qrev . (We did this,for example, in “undoing” the free expansion process.) The change in internal energy is (with thequantities Qrev and Wrev both regarded, in this example, as positive for work done by thesurroundings and heat given to the surroundings)2

∆U Q Wrev rev rev= − + .

In this reversible process, the entropy of the surroundings is changed by

∆SQ

Tsurroundings rev= .

For the combined changes (the irreversible state change and the reversible state change back to theinitial state), the energy change is zero because the energy is a function of state,

∆ ∆U U Q W Q Wrev rev rev+ = = − + − +( )0 .

Thus,Q Q W Wrev rev− = − .

2 In the above equation, and in the arguments that follow, the quantities Qrev and Wrev are both regarded

as positive for work done by the surroundings and heat given to the surroundings. Although this is not inaccord with the convention we have been using, it seems to me, after writing the notes in both ways, thatdoing this gives easier access to the ideas. I would be interested in your comments on whether thisperception is correct.

1C-13

For the system, the overall entropy change for the combined process is zero, because the entropy isa function of state,

∆ ∆ ∆S S Ssystem combined process irreversible process reversible process; = + = 0.

The total entropy change is thus only reflected in the entropy change of the surroundings:

∆ ∆S Stotalsurroundings= .

The surroundings can be considered a constant temperature heat reservoir and their entropy changeis given by

∆SQ Q

Ttotal rev=

−( )0

.

We also know that the total entropy change, for system plus surroundings is, 0

∆ ∆ ∆S S Stotalirreversibleprocess

reversibleprocess system surroundings

= +

+

The total entropy change is associated only with the irreversible process and is related to the workin the two processes by

∆SW W

Ttotal rev=

−( )0

.

The quantity W Wrev − represents the extra work required to restore the system to the originalstate. If the process were reversible, we would not have needed any extra work to do this. Itrepresents a quantity of work that is now unavailable because of the irreversibility. The quantityWrev can also be interpreted as the work that the system would have done if the original processwere reversible. From either of these perspectives we can identify ( W Wrev − ) as the quantity wedenoted previously as ′E , representing lost work. The lost work in any irreversible process cantherefore be related to the total entropy change (system plus surroundings) and the temperature ofthe surroundings by

Lost work = W W T Srevtotal− = 0∆ .

To summarize the results of the above arguments for processes where heat can be exchanged withthe surroundings at T0 :

1) W Wrev − represents the difference between work we actually obtained and work thatwould be done during a reversible state change. It is the extra work that would be needed torestore the system to its initial state.2) For a reversible process, W W Srev

total= =; ∆ 0

1C-14

1 2

c1P1T1

c2P2T2

3) For an irreversible process, W W Srevtotal> > ; ∆ 0

4) W W E T Srevtotal−( ) = ′ = 0∆ is the energy that becomes unavailable for work during an

irreversible process.

Muddy pointsIs ∆S path dependent? (MP 1C.11)Are Q rev and Wrev the Q and W going from the final state back to the initial state?(MP 1C.12)

1.C.7 Examples of Lost Work in Engineering Processes

a) Lost work in Adiabatic Throttling: Entropy and Stagnation Pressure ChangesA process we have encountered before is adiabatic throttling of a gas, by a valve or other

device as shown in the figure at the right. Thevelocity is denoted by c. There is no shaftwork and no heat transfer and the flow issteady. Under these conditions we can use thefirst law for a control volume (the Steady FlowEnergy Equation) to make a statement about theconditions upstream and downstream of the valve:

h c h c ht1 12

2 222 2+ = + =/ / ,

where ht is the stagnation enthalpy, corresponding toa (possibly fictitious) state with zero velocity.The stagnation enthalpy is the same at stations 1 and 2 if Q=W=0, even if the flow processes arenot reversible.

For an ideal gas with constant specific heats, h c T h c Tp t p t= = and . The relation between the static

and stagnation temperatures is:

T

T

c

c T

c

RT

c

at

p

a

= + = +−( )

= +−( )

12

11

21

1

2

2 2

2

2

2

γ

γ

γ

,

T

TMt = +

−

11

22γ,

where a is the speed of sound and M is the Mach number, M = c/a. In deriving this result, use hasonly been made of the first law, the equation of state, the speed of sound, and the definition of theMach number. Nothing has yet been specified about whether the process of stagnating the fluid isreversible or irreversible.

When we define the stagnation pressure, however, we do it with respect to isentropicdeceleration to the zero velocity state. For an isentropic process

Adiabatic throttling

1C-15

P

P

T

T2

1

2

1

1

=

−( )γ γ/

.

The relation between static and stagnation pressures is

P

P

T

Tt t=

−( )γ γ/ 1

.

The stagnation state is defined by P Tt t, . In addition, s sstagnation state static state = . The static and

stagnation states are shown below in T-s coordinates.

T

s

Pt

PTt

T

c2

2

Figure C-1: Static and stagnation pressures and temperatures

Stagnation pressure is a key variable in propulsion and power systems. To see why, weexamine the relation between stagnation pressure, stagnation temperature, and entropy. The formof the combined first and second law that uses enthalpy is

Tds dh dP= −1

ρ. (C.7.1)

This holds for small changes between any thermodynamic states and we can apply it to a situationin which we consider differences between stagnation states,say one state having properties T Pt t,( )and the other having properties T dT P dPt t t t+ +( ), (see at

right). The corresponding static states arealso indicated. Because the entropy is the same at static andstagnation conditions, ds needs no subscript. Writing (1.C.8)in terms of stagnation

conditions yields dsc dT

T TdP

c dT

T

R

PdPp t

t t tt

p t

t tt= − = −

1

ρ.

Both sides of the above are perfect differentials and can beintegrated as

Stagnation and static states

T

dsA-B

s

At

Bt

A

B

1C-16

T∆s1-2

Pt1 Pt2

Tt

s

∆s

R

T

T

P

Pt

t

t

t

=−

−

γγ 1

2

1

2

1

ln ln .

For a process with Q = W = 0, the stagnation enthalpy, and hence the stagnation temperature, isconstant. In this situation, the stagnation pressure is related directly to the entropy as,

∆s

R

P

Pt

t

= −

ln 2

1

. (C.7.2)

The figure on the right shows this relation on a T-s diagram.We have seen that the entropy is related to the loss, orirreversibility. The stagnation pressure plays the role of anindicator of loss if the stagnation temperature is constant.The utility is that it is the stagnation pressure (andtemperature) which are directly measured, not the entropy.The throttling process is a representation of flow throughinlets, nozzles, stationary turbomachinery blades, and the useof stagnation pressure as a measure of loss is a practice that has widespread application.Eq. (C.7.2) can be put in several useful approximate forms. First, we note that for aerospaceapplications we are (hopefully!) concerned with low loss devices, so that the stagnation pressure

change is small compared to the inlet level of stagnation pressure ∆P P P P Pt t t t t/ /1 1 2 1 1= −( ) << .

Expanding the logarithm [using ln (1-x) ≅ -x + ….],

lnPP

ln 1P

PP

P,t2

t1

t

t1

t

t1

= −

≈

∆ ∆

or, ∆ ∆sR

PP

t

t1

≈ .

Another useful form is obtained by dividing both sides by c2/2 and taking the limiting forms ofthe expression for stagnation pressure in the limit of low Mach number (M<<1). Doing this, wefind:

T s

c

P

ct∆ ∆

2 22 2/ /( )≅

( )ρ

The quantity on the right can be interpreted as the change in the “Bernoulli constant” forincompressible (low Mach number) flow. The quantity on the left is a non-dimensional entropychange parameter, with the term T ∆s now representing the loss of mechanical energy associatedwith the change in stagnation pressure.

To summarize:1) for many applications the stagnation temperature is constant and the change in stagnation

pressure is a direct measure of the entropy increase2) stagnation pressure is the quantity that is actually measured so that linking it to entropy (which

is not measured) is useful

1C-17

3) we can regard the throttling process as a “free expansion” at constant temperature Tt1 from the

initial stagnation pressure to the final stagnation pressure. We thus know that, for the process,the work we need to do to bring the gas back to the initial state is T st ∆ , which is the ”lostwork” per unit mass.

Muddy pointsWhy do we find stagnation enthalpy if the velocity never equals zero in the flow?(MP 1C.13)Why does Tt remain constant for throttling? (MP 1C.14)

b) Adiabatic Efficiency of a Propulsion System Component (Turbine)A schematic of a turbine and the accompanying thermodynamic diagram are given in

Figure C-2. There is a pressure and temperature drop through the turbine and it produces work.

1P1

2

2s

P2

∆h

∆s

hor T

Work

2

1m.

s

Figure C-2: Schematic of turbine and associated thermodynamic representation in h-s coordinates

There is no heat transfer so the expressions that describe the overall shaft work and the shaft workper unit mass are:

m h h W

h h w

t t shaft

t t shaft

• •−( ) =

−( ) =

2 1

2 1

(C.7.3)

If the difference in the kinetic energy at inlet and outlet can be neglected, Equation (C.7.3) reducesto

h h wshaft2 1−( ) = .

The adiabatic efficiency of the turbine is defined as

ηad =( )

actual work

ideal work s = 0For a given pressure ratio

∆.

The performance of the turbine can be represented in an h-s plane (similar to a T-s plane for anideal gas) as shown in Figure C-2. From the figure the adiabatic efficiency is

ηads

s s

s

h h

h h

h h h h

h h=

−

−=

− − −( )−

1 2

1 2

1 2 2 2

1 2

1C-18

P, T Work receiver

Friction

Isothermal expansion with friction

The adiabatic efficiency can therefore be written as

ηadh

= −

1∆

Ideal work.

The non-dimensional term ( ∆h /Ideal work) represents the departure from isentropic (reversible)processes and hence a loss. The quantity ∆h is the enthalpy difference for two states along aconstant pressure line (see diagram). From the combined first and second laws, for a constantpressure process, small changes in enthalpy are related to the entropy change by Tds dh= . , orapproximately,

T s h2∆ ∆= .

The adiabatic efficiency can thus be approximated as

ηads

T s

h h= −

−= −

1 12

1 2

∆ Lost work

Ideal work.

The quantity T s∆ represents a useful figure of merit for fluid machinery inefficiency due toirreversibility.

Muddy pointsHow do you tell the difference between shaft work/power and flow work in aturbine, both conceptually and mathematically? (MP 1C.15)

c) Isothermal Expansion with FrictionIn a more general look at

the isothermal expansion, we nowdrop the restriction to frictionlessprocesses. As seen in the diagramat the right, work is done toovercome friction. If the kineticenergy of the piston is negligible, abalance of forces tells us that

W W Wsystemon piston

done byfriction

received

= + .

During the expansion, the piston and the walls of the container will heat up because of the friction.The heat will be (eventually) transferred to the atmosphere; all frictional work ends up as heattransferred to the surrounding atmosphere.

W Qfriction friction=

1C-19

T

Q

Work

Qo

To

The amount of heat transferred to the atmosphere due to the frictional work only is thus,

Q W Wfriction systemon piston

received= −

Work producedWorkreceived

124 34124 34

.

The entropy change of the atmosphere (considered as a heat reservoir) due to the frictional work is

∆SQ

T

W W

Tatm

only

friction

atm

system received

atm

( ) = =−

due to frictionalwork

The difference between the work that the system did (the work we could have received if therewere no friction) and the work that we actually received can be put in a (by now familiar) form as

W W T Ssystem received atm atm− = ∆ = Lost or unavailable work

Muddy pointsIs the entropy change in the equations two lines above the total entropy change?If so, why does it say ∆Satm? (MP 1C.16)

d) Entropy Generation, Irreversibility, and Cycle EfficiencyAs another example, we show the links between entropy changes and cycle efficiency for anirreversible cycle. The conditions are:i) A source of heat at temperature, Tii) A sink of heat (rejection of heat) at T0

iii) An engine operating in a cycle irreversiblyDuring the cycle the engine extracts heat Q, rejectsheat Q0 and produces work,W:

W Q Q= − 0 .∆ ∆ ∆S S Sengine surroundings= + .

The engine operates in a cycle and the entropychange for the complete cycle is zero.Therefore,

∆ ∆ ∆

∆

S S Sheatsource

heatk

Ssurroundings

= + +0sin1 244 344

.

The total entropy change is,

∆ ∆ ∆S S SQ

T

Q

Ttotal

heatsource

heatk

= + =−

+sin

0

0

.

1C-20

Suppose we had an ideal reversible engine working between these same two temperatures, whichextracted the same amount of heat, Q, from the high temperature reservoir, and rejected heat ofmagnitude Q

rev0 to the low temperature reservoir. The work done by this reversible engine is

W Q Qrev rev= − 0 .

For the reversible engine the total entropy change over a cycle is

∆ ∆ ∆S S SQ

T

Q

Ttotal

heatsource

heatk

rev= + =−

+ =sin

0

0

0 .

Combining the expressions for work and for the entropy changes,

Q Q W Wrev rev0 0= + −

The entropy change for the irreversible cycle can therefore be written as

∆SQ

T

Q

T

W W

Ttotal rev rev=

−+ +

−

=

0

0

0

01 24 34

.

The difference in work that the two cycles produce is equal to the entropy that is generated duringthe cycle:

T S W Wtotalrev0∆ = − .

The second law states that the total entropy generated is greater than zero for an irreversibleprocess, so that the reversible work is greater than the actual work of the irreversible cycle.

An “engine effectiveness”, Eengine , can be defined as the ratio of the actual work obtained divided

by the work that would have been delivered by a reversible engine operating between the twotemperatures T T and 0 .

EW

W would be delivered

T T

engineengine

reversibleengine

rev

= = =η

ηActual work obtained

Work that

by a reversible cycle between , 0

EW T S

W

T S

Wenginerev

total

rev

total

rev

=−

= −0 01∆ ∆

The departure from a reversible process is directly reflected in the entropy change and the decreasein engine effectiveness.

1C-21

Muddy pointsWhy does ∆Sirrev=∆Stotal in this example? (MP 1C.17)In discussing the terms "closed system" and "isolated system", can you assumethat you are discussing a cycle or not? (MP 1C.18)Does a cycle process have to have ∆S=0? (MP 1C.19)In a real heat engine, with friction and losses, why is ∆S still 0 if TdS=dQ+dφ?(MP 1C.20)

e) Propulsive Power and Entropy FluxThe final example in this section combines a number of ideas presented in this subject and

in Unified in the development of a relation between entropy generation and power needed topropel a vehicle. Figure C-3 shows an aerodynamic shape (airfoil) moving through the atmosphereat a constant velocity. A coordinate system fixed to the vehicle has been adopted so that we seethe airfoil as fixed and the air far away from the airfoil moving at a velocity c0 . Streamlines of theflow have been sketched, as has the velocity distribution at station “0” far upstream and station “d”far downstream. The airfoil has a wake, which mixes with the surrounding air and grows in the

A0

Streamlines (control surface)Actual wake profile

Wake

A2

A1

∆c

c0

Figure C-3: Airfoil with wake and control volume for analysis of propulsive power requirement

downstream direction. The extent of the wake is also indicated. Because of the lower velocity inthe wake the area between the stream surfaces is larger downstream than upstream.

We use a control volume description and take the control surface to be defined by the two streamsurfaces and two planes at station 0 and station d. This is useful in simplifying the analysisbecause there is no flow across the stream surfaces. The area of the downstream plane controlsurface is broken into A1, which is area outside the wake and A2 , which is the area occupied bywake fluid, i.e., fluid that has suffered viscous losses. The control surface is also taken far enoughaway from the vehicle so that the static pressure can be considered uniform. For fluid which is notin the wake (no viscous forces), the momentum equation is cdc dP= − / ρ . Uniform static pressuretherefore implies uniform velocity, so that on A1 the velocity is equal to the upstream value, c0 .The downstream velocity profile is actually continuous, as indicated. It is approximated in theanalysis as a step change to make the algebra a bit simpler. (The conclusions apply to the more

1C-22

general velocity profile as well and we would just need to use integrals over the wake instead ofthe algebraic expressions below.)

The equation expressing mass conservation for the control volume is

ρ ρ ρ0 0 0 0 1 0 2 2 2A c A c A c= + . (C.7.5)

The vertical face of the control surface is far downstream of the body. By this station, the wakefluid has had much time to mix and the velocity in the wake is close to the free stream value, c0 .We can thus write,

wake velocity = = −( )c c c2 0 ∆ ; ∆c c/ .0 1<< (C.7.6)

(We chose our control surface so the condition ∆c c/ 0 1<< was upheld.)

The integral momentum equation (control volume form of the momentum equation) can beused to find the drag on the vehicle.

ρ ρ ρ0 0 02

0 1 02

2 2 22A c Drag A c A c= − + + . (C.7.7)

There is no pressure contribution in Eq. (C.7.7) because the static pressure on the control surface isuniform. Using the form given for the wake velocity, and expanding the terms in the momentumequation out we obtain,

ρ ρ ρ0 0 02

0 1 02

2 2 02

0 222A c Drag A c A c c c c= − + + − + ( )[ ]∆ (C.7.8)

The last term in the right hand side of the momentum equation, ρ2 22A c∆( ) , is small by virtue of

the choice of control surface and we can neglect it. Doing this and grouping the terms on the righthand side of Eq. (C.7.8) in a different manner, we have

c A c c A c A c c Drag A c c0 0 1 0 0 0 1 0 2 2 0 2 2 0ρ ρ ρ ρ[ ] = + −( )[ ] + − − ∆ ∆

The terms in the square brackets on both sides of this equation are the continuity equationmultiplied by c0 . They thus sum to zero leaving the curly bracketed terms as

Drag A c c= −ρ2 2 0∆ . (C.7.9)

The wake mass flow is ρ2 A2 c2 = ρ2 A2 (c0 − ∆ c). All this flow has a velocity defect (compared tothe free stream) of ∆c , so that the defect in flux of momentum (the mass flow in the wake timesthe velocity defect) is, to first order in ∆c ,

Momentum defect in wake = −ρ2 2 0A c c∆ , = Drag.

1C-23

The combined first and second law gives us a means of relating the entropy and velocity:

Tds dh dP= − / ρ .

The pressure is uniform (dP=0) at the downstream station. There is no net shaft work or heattransfer to the wake so that the mass flux of stagnation enthalpy is constant. We can alsoapproximate that the condition of constant stagnation enthalpy holds locally on all streamlines.Applying both of these to the combined first and second law yields

Tds dh cdct= − .

For the present situation, dh cdc c ct = 0 0; = ∆ , so that

T s c c0 0∆ ∆= − (C.7.10)

In Equation (C.7.10) the upstream temperature is used because differences between wakequantities and upstream quantities are small at the downstream control station. The entropy can berelated to the drag as

Drag = ρ2 2 0A T s∆ (C.7.11)

The quantity ρ2 2 0A c s∆ is the entropy flux (mass flux times the entropy increase per unit mass; inthe general case we would express this by an integral over the locally varying wake velocity anddensity).

The power needed to propel the vehicle is the product of drag x flight speed, Drag co× .From Eq. (C.7.11), this can be related to the entropy flux in the wake to yield a compactexpression for the propulsive power needed in terms of the wake entropy flux:

Propulsive power needed = T A c s T Entropy flux in wake0 2 2 0 0ρ ∆( ) = × (C.7.12)

This amount of work is dissipated per unit time in connection with sustaining the vehicle motion.Equation (C.7.12) is another demonstration of the relation between lost work and entropygeneration, in this case manifested as power that needs to be supplied because of dissipation in thewake.

Muddy pointsIs it safe to say that entropy is the tendency for a system to go into disorder? (MP 1C.21)

1.C.8 Some Overall Comments on Entropy, Reversible and Irreversible Processess[Mainly excerpted (with some alterations) from: Engineering Thermodynamics, WilliamC. Reynolds and Henry C. Perkins, McGraw-Hill Book Company, 1977]

Muddy pointsIsn't it possible for the mixing of two gases to go from the final state to the initialstate? If you have two gases in a box, they should eventually separate by density,right? (MP 1C.22)

Muddiest Points on Part 1C

1C.1 So, do we lose the capability to do work when we have an irreversible process andentropy increases?

Absolutely. We will see this in a more general fashion very soon. The idea of lost workis one way to view what “entropy is all about”!

1C.2 Why do we study cycles starting with the Carnot cycle? Is it because it is easier towork with?

Carnot cycles are the best we can do in terms of efficiency. We use the Carnot cycle as astandard against which all other cycles are compared. We will see in class that we canbreak down a general cycle into many small Carnot cycles. Doing this we can gaininsight in which direction the design of efficient cycles should go.

1C.3 How does one interpret h-s diagrams?

I find h-s diagrams useful, especially in dealing with propulsion systems, because thedifference in stagnation h can be related (from the Steady Flow Energy Equation) to shaftwork and heat input. For processes that just have shaft work (compressors or turbines)the change in stagnation enthalpy is the shaft work. For processes that just have heataddition or rejection at constant pressure, the change in stagnation enthalpy is the heataddition or rejection.

1C.4 Is it always OK to "switch" T-s and h-s diagram?

No! This is only permissible for perfect gases with constant specific heats. We will see,when we examine cycles with liquid-vapor mixtures, that the h-s diagrams and the T-sdiagrams look different.

1C.5 What is the best way to become comfortable with T-s diagrams?

I think working with these diagrams may be the most useful way to achieve thisobjective. In doing this, the utility of using these coordinates (or h-s coordinates) shouldalso become clearer. I find that I am more comfortable with T-s or h-s diagrams thanwith P-v diagrams, especially the latter because it conveys several aspects of interest topropulsion engineers: work produced or absorbed, heat produced or absorbed, and loss.

1C.6 What is a reversible adiabat physically?

Let's pick an example process involving a chamber filled with a compressible gas and apiston. We assume that the chamber is insulated (so no heat-transfer to or from thechamber) and the process is thus adiabatic. Let us also assume that the piston is ideal,such that there is no friction between the walls of the chamber and the piston. The gas is

at some Temperature T1. We now push the piston in and compress the gas. The internalenergy of the system will then increase by the amount of work we put in and the gas willheat up and be at higher pressure. If we now let the piston expand again, it will return toits original position (no friction, ideal piston) and the work we took from the environmentwill be returned (we get the exact same amount of work back and leave no mark on theenvironment). Also, the temperature and the pressure of the gas return to the initialvalues. We thus have an adiabatic reversible process. For both compression andexpansion we have no change in entropy of the system because there is no heat transferand also no irreversibility. If we now draw this process in the h-s or T-s diagram we get avertical line since the entropy stays constant: S = constant or ∆S = 0 and we can also callthis process an isentropic process.

1C.7 If there is an ideal efficiency for all cycles, is there a maximum work or maximumpower for all cycles?

Yes. As with the Brayton cycle example, we could find the maximum as a function ofthe appropriate design parameters.

1C.8 Do we ever see an absolute variable for entropy? So far, we have worked withdeltas only.

It is probably too strong a statement to say that for “us” the changes in entropy are whatmatters, but this has been my experience for the type of problems aerospace engineerswork on. Some values of absolute entropy are given in Table A.8 in SB&VW. We willalso see, in the lectures on Rankine cycles, that the entropy of liquid water at atemperature of 0.01 C and a pressure of 0.6113 kPa has been specified as zero forproblems involving two-phase (steam and water) behavior.

1C.9 I am confused as to dSdQrev

T= as opposed to dS

dQrev

T≥ .

Both of these are true and both can always be used. The first is the definition of entropy.The second is a statement of how the entropy behaves. Section 1C.5 attempts to make

the relationship clearer through the development of the equality dSdQ

T

dW

Tlost= + .

1C.10 For irreversible processes, how can we calculate dS if not equal to dQ

T?

We need to define a reversible process between the two states in order to calculate theentropy (see muddy point 3, above). See VN Chapter 5 (especially) for discussion ofentropy or section 1C.5. If you are still in difficulty, come and see me.

1C.11 Is ∆S path dependent?

No. Entropy is a function of state (see Gibbs) and thus ∆S is path independent. Forexample we could have three different paths connecting the same two states and thereforehave the same change in entropy

∆S path I = ∆S path II = ∆S path III .

1C.12 Are Q rev and Wrev the Q and W going from the final state back to the initial state?

Yes. We have an irreversible process from state 1 to state 2. We then used a reversibleprocess to restore the initial state 1 (we had to do work on the system and extract heatfrom the system).

1C.13 Why do we find stagnation enthalpy if the velocity never equals zero in the flow?

The stagnation enthalpy (or temperature) is a useful reference quantity. Unlike the statictemperature it does not vary along a streamline in an adiabatic flow, even if irreversible.It was thus the natural reference temperature to use in describing the throttling process.In addition, changes in stagnation pressure are direct representations of the shaft work orheat associated with a fluid component. The enthalpy is not, unless we assume thatchanges in KE are small. Measurement of stagnation temperature thus allows directassessment of shaft work in a turbine or compressor, for example.

1C.14 Why does Tt remain constant for throttling?

Because for a steady adiabatic flow with no shaft work done the Steady Flow EnergyEquation yields constant stagnation enthalpy even though the flow processes might not bereversible (see notes). For a perfect gas h = cpT, thus the stagnation temperature remainsconstant.

1C.15 How do you tell the difference between shaft work/power and flow work in aturbine, both conceptually and mathematically?

Let us look at the expansion of a flow through a turbine using both the control massapproach and the control volume approach. Using the control mass approach we canmodel the situation by tracking 1kg of air as follows: state 1 – before the expansion wehave 1kg of air upstream of the turbine. We then push the gas into the turbine and expandit through the blade rows. After the expansion we take 1kg of air out of the turbine andthe mass of air is downstream of the turbine – state 2. The work done by the gas is workdone by the turbine (blades moved around by the gas) plus the work done by pressures(flow work). Using the first law we can then write for the change of internal energy of1kg of air:

u2 – u1 = - wshaft + p1v1 – p2v2 (adiabatic turbine: dq = 0)

When entering the turbine, the fluid has to push the surroundings out of the way to makeroom for itself (it has volume v1 and is at p1) – the work to do this is +p1v1. When leavingthe turbine the fluid is giving up room and the work to keep that volume v2 at pressure p2

is freed; thus –p2v2. We can then write for the shaft work

wshaft = u1 + p1v1 – (u2 + p2v2).The right hand side of the above equation is the change in enthalpy (h1 - h2). This isanother example to show how useful enthalpy is (enthalpy is the total energy of a fluid:the internal energy plus the extra energy that it is credited with by having a volume v atpressure p). The shaft work outputted by the turbine is equal to the change in enthalpy(enthalpy contains the flow work!).

wshaft = h1 - h2.

We can also solve this problem by using the 1st law in general form (control volumeapproach).

. . . . . d/dtΣ ECV = Σ Q + Σ Wshaft + Σ Wshear + Σ Wpiston + Σ m ( h + _ c2 + gz).

Note that in this form the flow work is buried in h already! For this turbine, we can dropthe unsteady term on the left and neglect heat fluxes (adiabatic turbine), shear work andpiston work (no pistons but blades, so we keep the shaft work). Further we assume thatchanges in potential energy and kinetic energy are negligible and we obtain for 1kg/s air

0 = - wshaft + h1 - h2.

We obtain the same result as before: wshaft = h1 - h2.

1C.16 Is the entropy change in the equations two lines above the total entropy change? Ifso, why does it say ∆Satm?

The entropy change in question is the entropy change due to the heat produced by frictiononly.

∆SQ

T

W W

Tatmonly

friction

atm

system received

atm

( ) = =−( )

due to frictionalwork

.

1C.17 Why does ∆Sirrev=∆Stotal in this example?

When we wrote this equality, we were considering a system that was returned to itsoriginal state, so that there were no changes in any of the system properties. All evidenceof irreversibility thus resides in the surroundings.

1C.18 In discussing the terms "closed system" and "isolated system", can you assume thatyou are discussing a cycle or not?

The terms closed system and isolated system have no connection to whether we arediscussing a cycle or not. They are attributes of a system (any system), whetherundergoing cyclic behavior, one-way transitions, or just sitting there.

1C.19 Does a cycle process have to have ∆S=0?

Entropy is a state function. If the process is cyclic, initial and final states are the same.So, for a cyclic process, ∆S = 0 .

1C.20 In a real heat engine, with friction and losses, why is ∆S still 0 if TdS=dQ+dφ?

The change in entropy during a real cycle is zero because we are considering a completecycle (returning to initial state) and entropy is a function of state (holds for both ideal andreal cycles!). Thus if we integrate dS = dQ/T + dΦ/T around the real cycle we will obtain∆Scycle = 0. What essentially happens is that all irreversibilities (dΦ's) are turned intoadditional heat that is rejected to the environment. The amount of heat rejected in the realcycle QR

real is going to be larger than the amount of heat rejected in an ideal cycle QRideal

QRideal = QA TR/TA (from ∆Scycle = 0)

QRreal = QA TR/TA + TR∆SΦ (from ∆Scycle = 0)

We will see this better using the T-s diagram. The change of entropy of the surroundings(heat reservoirs) is ∆Ssurr = -QA/TA + QR

real/TR = ∆SΦ > 0. So ∆Scycle = 0 even for realcycles, but ∆Stotal = ∆Scycle + ∆Ssurr = ∆SΦ > 0.

1C.21 Is it safe to say that entropy is the tendency for a system to go into disorder?

Entropy can be given this interpretation from a statistical perspective, and this provides adifferent, and insightful view of this property. At the level in which we have engaged theconcept, however, we focus on the macroscopic properties of systems, and there is noneed to address the idea of order and disorder ; as we will see, entropy is connected to theloss of our ability to do work, and that is sufficient to make it a concept of great utility forthe evaluation and design of engineering systems. We will look at this in a later lecture. Ifyou are interested in pursuing this, places to start might be the book by Goldstein andGoldstein referred to above, Great Ideas in Physics by Lightman ( paperback book by anMIT professor), or Modern Thermodynamics, by Kondepudi and Prigogine.

1C.22 Isn't it possible for the mixing of two gases to go from the final state to the initialstate? If you have two gases in a box, they should eventually separate by density,right?

Let us assume that gas X is oxygen and gas Y is nitrogen. When the membrane breaks theentire volume will be filled with a mixture of oxygen and nitrogen. This may beconsidered as a special case of an unrestrained expansion, for each gas undergoes anunrestrained expansion as it fills the entire volume. It is impossible for these twouniformly mixed gases to separate without help from the surroundings or environment. Acertain amount of work is necessary to separate the gases and to bring them back into theleft and right chambers.

1D-1

1.D: Interpretation of Entropy on the Microscopic Scale - The Connection betweenRandomness and Entropy

1.D.1 Entropy Change in Mixing of Two Ideal Gases

Consider an insulated rigid container of gas separated into two halves by a heat conductingpartition so the temperature of the gas in each part is the same. One side contains air, the otherside another gas, say argon, both regarded as ideal gases. The mass of gas in each side is suchthat the pressure is also the same.

The entropy of this system is the sum of the entropies of the two parts: S S Ssystem air on= + arg .

Suppose the partition is taken away so the gases are free to diffuse throughout the volume. Foran ideal gas, the energy is not a function of volume, and, for each gas, there is no change intemperature. (The energy of the overall system is unchanged, the two gases were at the sametemperature initially, so the final temperature is the same as the initial temperature.) Theentropy change of each gas is thus the same as that for a reversible isothermal expansion fromthe initial specific volume vi to the final specific volume, v f . For a mass m of ideal gas, the

entropy change is ∆S mR v vf i= ( )ln . The entropy change of the system is

∆ ∆ ∆S S S m R v v m R v vsystem air on air air f air i air on on f on i on= + = [ ] + [ ]arg arg arg arg argln ( ) ( ) ln ( ) ( ) . (D.1.1)

Equation (D.1.1) states that there is an entropy increase due to the increased volume that eachgas is able to access.

Examining the mixing process on a molecular level gives additional insight. Suppose we wereable to see the gas molecules in different colors, say the air molecules as white and the argonmolecules as red. After we took the partition away, we would see white molecules start to moveinto the red region and, similarly, red molecules start to come into the white volume. As wewatched, as the gases mixed, there would be more and more of the different color molecules inthe regions that were initially all white and all red. If we moved further away so we could nolonger pick out individual molecules, we would see the growth of pink regions spreading into theinitially red and white areas. In the final state, we would expect a uniform pink gas to existthroughout the volume. There might be occasional small regions which were slightly more redor slightly more white, but these fluctuations would only last for a time on the order of severalmolecular collisions.

In terms of the overall spatial distribution of the molecules, we would say this final state wasmore random, more mixed, than the initial state in which the red and white molecules wereconfined to specific regions. Another way to say this is in terms of “disorder”; there is moredisorder in the final state than in the initial state. One view of entropy is thus that increases inentropy are connected with increases in randomness or disorder. This link can be made rigorousand is extremely useful in describing systems on a microscopic basis. While we do not havescope to examine this topic in depth, the purpose of Section 1.D is to make plausible the linkbetween disorder and entropy through a statistical definition of entropy.

1D-2

1.D.2 Microscopic and Macroscopic Descriptions of a System

The microscopic description of a system is the complete description of each particle in thissystem. In the above example, the microscopic description of the gas would be the list of thestate of each molecule: position and velocity in this problem. It would require a great deal ofdata for this description; there are roughly 1019 molecules in a cube of air one centimeter on aside at room temperature and pressure. The macroscopic description, which is in terms of a few(two!) properties is thus far more accessible and useable for engineering applications, although itis restricted to equilibrium states.

To address the description of entropy on a microscopic level, we need to state some resultsconcerning microscopic systems. These, and the computations and arguments below are takenalmost entirely from the excellent discussion in Chapter 6 of Engineering Thermodynamics byReynolds and Perkins (1977)*.

For a given macroscopic system, there are many microscopic states. A key idea from quantummechanics is that the states of atoms, molecules, and entire systems are discretely quantized.This means that a system of particles under certain constraints, like being in a box of a specifiedsize, or having a fixed total energy, can exist in a finite number of allowed microscopic states.This number can be very big, but it is finite.

The microstates of the system keep changing with time from one quantum state to another asmolecules move and collide with one another. The probability for the system to be in a particularquantum state is defined by its quantum-state probability pi . The set of the pi is called thedistribution of probability. The sum of the probabilities of all the allowed quantum states must beunity, hence for any time t,

pii

=∑ 1 (D.2.1)

When the system reaches equilibrium, the individual molecules still change from one quantumstate to another. In equilibrium, however, the system state does not change with time; so theprobabilities for the different quantum states are independent of time. This distribution is thencalled the equilibrium distribution, and the probability pi can be viewed as the fraction of time asystem spends in the ithquantum state. In what follows, we limit consideration to equilibriumstates.

We can get back to macroscopic quantities from the microscopic description using theprobability distribution. For instance, the macroscopic energy of the system would be theweighted average of the successive energies of the system (the energies of the quantum states);weighted by the relative time the system spends in the corresponding microstates. In terms ofprobabilities, the average energy, E , is

E pi ii

= ∑ ε , where iε is the energy of a quantum state. (D.2.2)

* Reynolds, W.C., and Perkins, H.C., Engineering Thermodynamics, McGraw-Hill Book Co., 1977.

1D-3

The probability distribution provides information on the randomness of the equilibrium quantumstates. For example, suppose the system can only exist in three states (1, 2 and 3). If thedistribution probability is

p1=1, p2=0, p3=0the system is in quantum state 1 and there is no randomness. If we were asked what quantumstate the system is in, we would be able to say it is always in state 1. If the distribution were

p1=0.3, p2=0.2, p3=0.5or

p1=0.5, p2=0.2, p3=0.3

the randomness would not be zero and would be equal in both cases. We would be moreuncertain about the instantaneous quantum state than in the first situation.

Maximum randomness corresponds to the case where the three states are equally probable:

p1=1/3, p2=1/3, p3=1/3

In this case, we can only guess the instantaneous state with 33 per cent probability.

1.D.3 A Statistical Definition of Entropy

The list of the pi is a precise description of the randomness in the system, but the number ofquantum states in almost any industrial system is so high this list is not useable. We thus lookfor a single quantity, which is a function of the pi that gives an appropriate measure of therandomness of a system. As shown below, the entropy provides this measure.

There are several attributes that the sought for function should have. The first is that the averageof the function over all of the microstates should have an extensive behavior, in other words themicroscopic description of the entropy of a system C, composed of parts A and B is given by

S S SC A B= + . (D.3.1)

Second is that entropy should increase with randomness and should be largest for a given energywhen all the quantum states are equiprobable.

The average of the function over all the microstates is defined by

S f p f pii

i= = ∑ ( ) (D.3.2)

where the function f pi( ) is to be found. Suppose that system A has n microstates and system B

has m microstates. The entropies of systems A, B, and C, are defined by

1D-4

S p f pA i ii

n= ( )∑

=1 (D.3.3a)

S p f pB j jj

m= ( )∑

=1 (D.3.3b)

S p f p p f p p f pC ij ijj

m

i

n

i ii

n

j jj

m= ( ) =∑∑ ( )∑ ( )∑

== = =11 1 1 (D.3.3c)

In Equations (D.3.2 and D.3.3), the term pij means the probability of a microstate in which

system A is in state i and system B is in state j.

For Equation (D.3.1) to hold,

S p f p p f p p f p p f p S SC i ii

n

j jj

m

i ii

n

j jj

m

A B= ( )∑ ( )∑ = ( )∑ + ( )∑ = += = = =1 1 1 1

. (D.3.4)

The function f must be such that this is true regardless of the values of the probabilitiesp pi j and . This will occur if f ( ) = ( )ln because ln ln lna b a b⋅( ) = ( ) + ( ) . Making this

substitution, the expression for Sc in Equation (D.3.4) becomes

p p p p p pi j ij

m

i

n

i j jj

m

i

nln ln( ) +∑∑ ( )∑∑

== ==11 11. (D.3.5a)

Rearranging the sums, D.3.5a becomes

p p p p p pi i jj

m

i

n

j j ii

n

j

mln ln( ) ∑

∑ + ( ) ∑

∑== ==11 11

. (D.3.5b)

Because

p pii

n

jj

m

= =∑ = ∑ =

1 11, (D.3.6)

the left hand side of Equation (D.3.4) can be written as

S p p p pC i ii

n

j jj

m= ( ) +∑ ( )∑

= =ln ln

1 1. (D.3.7)

This means that Equation (D.3.4) is satisfied for any p p n mi j, , , . Reynolds and Perkins show

that the most general f pi( ) is f C pi= ( )ln , where C is an arbitrary constant. Because the pi are

less than unity, the constant is chosen to be negative to make the entropy positive.

Based on the above a the statistical definition of entropy can be given as:

S k p pii

i= − ∑ ln . (D.3.8)

1D-5

The constant k is known as the Boltzmann constant,

kJ

K= × −1 380 10 23. . (D.3.9)

The value of k is (another wonderful result!) given by k = R/ NAvogadro , where R is the universal

gas constant, 8.3143 J/(mol-K) and NAvogadro is Avogadro's number, 6 02 1023. × molecules per

mol. Sometimes k is called the gas constant per molecule. With this value for k, the statisticaldefinition of entropy is identical with the macroscopic definition of entropy.

1.D.4 Connection between the Statistical Definition of Entropy and Randomness

We need now to examine the behavior of the statistical definition of entropy as regardsrandomness. Because a uniform probability distribution reflects the largest randomness, asystem with n allowed states will have the greatest entropy when each state is equally likely. Inthis situation, the probabilities become

p pi = =1Ω

, (D.4.1)

where Ω is the total number of microstates. The entropy is thus

S k k ki

= − ∑

= −

= −

=

1 1 1 1 1

1Ω ΩΩ

Ω Ω Ω

Ωln ln ln

S k= lnΩ . (D.4.2)

Equation (D.4.1) states that the larger the number of possible states the larger the entropy.

The behavior of the entropy stated in Equation (D.4.2) can be summarized as follows:a) S is maximum when Ω is maximum, which means many permitted quantum states,hence much randomness,b) S is minimum when Ω is minimum. In particular, for Ω=1, there is no randomnessand S=0.

These trends are in accord with our qualitative ideas concerning randomness. Equation (D.4.2)is carved on Boltzmann's tombstone (he died about a hundred years ago) in Vienna.

We can also examine the additive property of entropy with respect to probabilities. If we havetwo systems, A and B, which are viewed as a combined system, C, the quantum states for thecombined system are the combinations of the quantum states from A and B. The quantum statewhere A is in its state x and B in its state y would have a probability p pAx By

⋅ because the two

probabilities are independent. The number of probabilities for the combined system, ΩC , is thusdefined by Ω Ω ΩC A B= ⋅ . The entropy of the combined system is

1D-6

S k k k S SC A B A B A B= ⋅( ) = + = +ln ln lnΩ Ω Ω Ω (D.4.3)

Equation (D.4.2) is sometimes taken as the basic definition of entropy, but it should beremembered that it is only appropriate when each quantum state is equally likely. Equation(D.3.8) is more general and applies equally for equilibrium and non-equilibrium situations.

A simple numerical example shows trends in entropy changes and randomness for a systemwhich can exist in three states. Consider the five probability distributions

i) p p p1 2 31 0 0 0= = =. , , ; S k= − + +( ) =1 1 0 0 0 0 0ln ln ln

ii) p p p1 2 30 8 0 2 0= = =. , . , ; S k k= − ( ) + ( ) + ( )[ ] =0 8 0 8 0 2 0 2 0 0 0 500. ln . . ln . ln .

iii) p p p1 2 30 8 0 1 0 1= = =. , . , . ; S k k= − ( ) + ( ) + ( )[ ] =0 8 0 8 0 1 0 1 0 1 0 1 0 639. ln . . ln . . ln . .

iv) p p p1 2 30 5 0 3 0 2= = =. , . , . ; S k k= − ( ) + ( ) + ( )[ ] =0 5 0 5 0 3 0 3 0 2 0 2 1 03. ln . . ln . . ln . .

v) p p p1 2 31 3 1 3 1 3= = =/ , / , / ; S k k= −

=3

13

13

1 099ln . .

The first distribution has no randomness. For the second, we know that state 3 is never found.States (iii) and (iv) have progressively greater uncertainty about the distribution of states andthus higher randomness. State (v) has the greatest randomness and uncertainty and also thelargest entropy.

1.D.5. Numerical Example of the Approach to the Equilibrium Distribution

Reynolds and Perkins give a numerical example which illustrates the above concepts and also thetendency of a closed isolated system to tend to equilibrium.

Reynolds and Perkins,Engineering Thermodynamics, McGraw-Hill, 1977. Sec. 6.7. pp.177-183.

1D-7

1.D.6 Summary and Conclusions

a) Entropy as defined from a microscopic point of view is a measure of randomness in a system.

b) The entropy is related to the probabilities pi of the individual quantum states of the system by

S k p pii

i= − ∑ ln

where k, the Boltzmann constant is given by R/ NAvogadro .

c) For a system in which there are Ω quantum states, all of which are equally probable (forwhich the probability is pi = 1/Ω ), the entropy is given by

S k= lnΩ .

The more quantum states, the more the randomness and uncertainty that a system is in aparticular quantum state.

d) From the statistical point of view there is a finite, but exceedingly small possibility that asystem that is well mixed could suddenly "unmix" and that all the air molecules in the roomcould suddenly come to the front half of the room. The unlikelihood of this is well described byDenbigh [Principles of Chemical Equilibrium, 1981] in a discussion of the behavior of anisolated system:

"In the case of systems containing an appreciable number of atoms, it becomes increasinglyimprobably that we shall ever observe the system in a non-uniform condition. For example,it is calculated that the probability of a relative change of density, ∆ρ ρ , of only 0.001% in 1

cm3 of air is smaller than 10 108− and would not be observed in trillions of years. Thus,according to the statistical interpretation the discovery of an appreciable and spontaneousdecrease in the entropy of an isolated system, if it is separated into two parts, is notimpossible, but exceedingly improbable. We repeat , however, that it is an absoluteimpossibility to know when it will take place."

1D-8

e) The definition of entropy in the form S k p pii

i= − ∑ ln arises in other aerospace fields, notably

that of information theory. In this context, the constant k is taken as unity and the entropybecomes a dimensionless measure of the uncertainty represented by a particular message. Thereis no underlying physical connection with thermodynamic entropy, but the underlyinguncertainty concepts are the same.

f) The presentation of entropy in this subject is focused on the connection to macroscopicvariables and behavior. These involve the definition of entropy given in Section 1.B of the notesand the physical link with lost work, neither of which makes any mention of molecular(microscopic) behavior. The approach in other sections of the notes is only connected to thesemacroscopic processes and does not rely at all upon the microscopic viewpoint. Exposure to thestatistical definition of entropy, however, is helpful as another way not only to answer thequestion of "What is entropy?" but also to see the depth of this fundamental concept and theconnection with other areas of technology.

PART 2

POWER AND PROPULSION CYCLES

2A-1

PART 2 – POWER AND PROPULSION CYCLES

2A – Gas Power and Propulsion Cycles[SB&VW - 11.8, 11.9, 11.10, 11.11, 11.12, 11.13, 11.14]

In this section we analyze several gas cycles used in practical applications for propulsionand power generation, using the air standard cycle. The air standard cycle is an approximation tothe actual cycle behavior, and the term specifically refers to analysis using the followingassumptions:

• Air is the working fluid (the presence of combustion products is neglected)

• Combustion is represented by heat transfer from an external heat source

• The cycle is ‘completed’ by heat transfer to the surroundings

• All processes are internally reversible

• Air is a perfect gas with constant specific heats

2.A.1 The Internal combustion engine (Otto Cycle)The different processes of an idealized Otto cycle (internal combustion engine) are shown

in Figure 2A-1:

P

V

P0

V2 = V3 V1 = V4

5

2

3

4

1

QH

QL

Adiabatic reversible

Figure 2A-1: Ideal Otto cycle

i. Intake stroke, gasoline vapor and air drawn into engine (5 -> 1)ii. Compression stroke, P, T increase (1->2)iii. Combustion (spark), short time, essentially constant volume (2->3)

Model: heat absorbed from a series of reservoir at temperatures T2 to T3

iv. Power stroke: expansion (3 ->4)v. Valve exhaust: valve opens, gas escapesvi. (4->1) Model: rejection of heat to series of reservoirs at temperatures T4 to T1

vii. Exhaust stroke, piston pushes remaining combustion products out of chamber 1->5

2A-2

The actual cycle does not have these sharp transitions between the different processes and might beas sketched in Figure 2A-2

Spark Exhaustvalveopens

Notisentropic

Exhaust valvecloses

P

P0

V

Figure 2A-2: Sketch of actual Otto cycle

Efficiency of an ideal Otto cycleThe starting point is the general expression for the thermal efficiency of a cycle:

η = =+

= +work

heat input

Q Q

Q

Q

QH L

H

L

H

1 .

The convention, as previously, is that heat exchange is positive if heat is flowing into the system orengine, so QL is negative. The heat absorbed occurs during combustion when the spark occurs,roughly at constant volume. The heat absorbed can be related to the temperature change from state2 to state 3 as:

Q Q U W

C dT C T T

H

vTT

v

= = =( )= ∫ = −( )

23 23 23

23

3 2

0∆

The heat rejected is given by (for a perfect gas with constant specific heats)

Q Q U C T TL v= = = −( )41 41 1 4∆

Substituting the expressions for the heat absorbed and rejected in the expression for thermalefficiency yields

η = −−

−1 4 1

3 2

T T

T T

2A-3

We can simplify the above expression using the fact that the processes from 1 to 2 and from 3 to4 are isentropic:

T V T V T V T V

T T V T T V

T T

T T

V

V

4 11

3 21

1 11

2 21

4 1 11

3 2 21

4 1

3 2

2

1

1

γ γ γ γ

γ γ

γ

− − − −

− −

−

= =

−( ) = −( )

−

−=

,

The quantity V

Vr1

2

= is called the compression ratio. In terms of compression ratio, the

efficiency of an ideal Otto cycle is:

η γ γOtto VV

r= − = −

− −11

11

12

1 1.

The ideal Otto cycle efficiency is shown atthe right, as a function of the compression ratio.As the compression ratio, r, increases,ηOtto increases, but so does T2 . If T2 istoo high, the mixture will ignite without aspark (at the wrong location in the cycle).

Ideal Otto cycle thermal efficiency

Engine work, rate of work per unit enthalpy flux:The non-dimensional ratio of work done (the power) to the enthalpy flux through the

engine is given by

PowerEnthalpy flux

= =˙

˙

˙

˙W

mc T

Q

mc Tp

Otto

p1

23

1

η

There is often a desire to increase this quantity, because it means a smaller engine for the samepower. The heat input is given by

˙ ˙Q m hfuel fuel23 = ( )∆ ,where

• ∆hfuel is the heat of reaction, ie the chemical energy liberated per unit mass of fuel

• m fuel is the fuel mass flow rate .

The non-dimensional power is

˙

˙

˙

˙W

mc T

m

m

h

c T rp

fuel fuel

p1 11

11

= −

−

∆γ

.

00

10

20

30

40

50

60

70

2 4 6 8Compression ratio, rv

The

rmal

eff

icie

ncy,

ηth

10 12 14 16

2A-4

The quantities in this equation, evaluated at stoichiometric conditions are:˙

˙,

m

m

1

15

h

c T

4 10

10 288fuel fuel

p 1

7

3≈ ≈

×

×

∆

so,˙

˙W

mc T rp 11

9 11

≈ −

−γ .

Muddy points

How is ∆ h fuel calculated? (MP 2A.1)What are "stoichiometric conditions"? (MP 2A.2)

2.A.2. Diesel CycleThe Diesel cycle is a compression ignition (rather than spark ignition) engine. Fuel is sprayedinto the cylinder at P2 (high pressure) when the compression is complete, and there is ignitionwithout a spark. An idealized Diesel engine cycle is shown in Figure 2A-3.

QHP

V2

VV3 V4 = V1

QL

Adiabaticreversible

2 3

4

1

Figure 2A-3 Ideal Diesel cycle

The thermal efficiency is given by:

η

ηγ

DieselL

H

v

p

Diesel

TT

TT

Q

Q

C T T

C T T

T

T

= + = +−( )−( )

= −−( )−( )

1 1

11

1

1 4

3 2

14

1

23

2

This cycle can operate with a higher compression ratio than Otto cycle because only air iscompressed and there is no risk of auto-ignition of the fuel. Although for a given compressionratio the Otto cycle has higher efficiency, because the Diesel engine can be operated to highercompression ratio, the engine can actually have higher efficiency than an Otto cycle when bothare operated at compression ratios that might be achieved in practice.

2A-5

Muddy pointsWhen and where do we use c v and c p ? Some definitions use dU=c v dT. Is it everdU=c p dT? (MP 2A.3)Explanation of the above comparison between Diesel and Otto. (MP 2A.4)

2.A.3 Brayton CycleThe Brayton cycle is the cycle that represents the operation of a gas turbine engine. The

“simple gas turbine” can be operated in open cycle or closed cycle (recirculating working fluid)modes, as shown below.

FuelQH

wnetwnet

ProductsAir

QL

Combustionchamber

Compressor Turbine

Heatexchanger

Heatexchanger

Compressor Turbine

(a) (b)

Figure 2A-4: Gas turbine engine operating on the Brayton cycle – (a) open cycle operation, (b)closed cycle operation

Efficiency of the Brayton cycle:We derived the ideal Brayton cycle efficiency in Section 1.A:

η γ γBraytoninlet

compressorexit

T

T PR= − = − −1 1

11( ) /

.

Net work per unit mass flow in a Brayton cycle:The net mechanical work of the cycle is given by:

Net mechanical work/unit mass = −w wturbine compressor ,where

w h h

w h h

compressor comp

turbine turb

= − = −

= − = −

∆ ∆

∆ ∆12

34

If kinetic energy changes across the compressor and turbine are neglected, the temperature ratio,TR, across the compressor and turbine is related to the enthalpy changes:

TRh

h

h

hcomp turb− = =11 4

∆ ∆,

2A-6

∆ ∆h hh

hturb comp= − 4

1

The net work is thus

net work = −

∆h

h

hcomp4

1

1

The turbine work is greater than the work needed to drive the compressor. The thermodynamicstates in an enthalpy-entropy (h,s) diagram, and the work of the compressor and turbine, areshown below for an ideal Brayton cycle.

h4 T4=Tmax

wturb

qR

s0 T0=Tinlet

5

P0

wcomp

qA

P3

3

Figure 2A-5: Brayton cycle in enthalpy-entropy (h-s) representation showing compressorand turbine work

Muddy pointsWhat is shaft work? (MP 2A.5)

2.A.4 Brayton Cycle for Jet Propulsion: the Ideal RamjetA schematic of a ramjet is given in Figure 2A-6 below.

0 1 3 4 5inletstreamtube

cθp0 T0

c3p3 T3

diffuserπd

burnerτd πd

nozzle

Station Numbers

fuel, mf. exhaust

streamtube

T4c5p5 T5

Figure 2A-6: Ideal ramjet [adapted from J. L. Kerrebrock, Aircraft Engines and Gas Turbines]

2A-7

In the ramjet there are “no moving parts”. The processes that occur in this propulsion device are: 0->3 isentropic diffusion (slowing down) and compression, with a decrease in Machnumber, M M0 3 1→ <<

3->4 Constant pressure combustion, 4->5 Isentropic expansion through the nozzle.

Thrust of an ideal engine ramjetThe coordinate system and control volume are chosen to be fixed to the ramjet. The

thrust, F, is given by:

F c c= −( )m 5 0 ,

where c5 and co are the inlet and exit flow velocities. The thrust can be put in terms of non-dimensional parameters as follows:

F

a

c

a

a

a

c

am 0

5

5

5

0

0

0= − , where a RT= γ is the speed of sound.

F

aM

a

aM M

T

TM

m 05

5

00 5

5

00= − = −

Using M M32

42 1, << in the expression for stagnation pressure, P

PMT = +

−

−1

1

22 1γ

γγ

,P P P P P P P PT T T T3 3 0 4 4 5 4 3≈ = ≈ = ≈; ;

The ratios of stagnation pressure to static pressure at inlet and exit of the ramjet are:

P

P

P

P

P

P

P

PT

M

Te

e

Me

0

0

3

0

0

4

0

= = =

determines determines1 24 34 1 24 34

The ratios of stagnation to static pressure at exit and at inlet are the same, with the consequencethat the inlet and exit Mach numbers are also the same.

M M5 0= .

To find the thrust we need to find the ratio of the temperature at exit and the temperature at inlet.This is given by:

TT

T

M

M

TTT

TT

T T

T

T

Tb

5

0

5

52

02

0

5

0

4

311

2

11

2=+

−+

−

= = =γ

γ

τ,

2A-8

where τ b is the stagnation temperature ratio across the combustor (burner). The thrust is thus:

F

aM bm 0

0 1= −( )τ

Cycle efficiency in terms of aerodynamic parameters:

ηBraytoncompressor exit T

T

T

T

T

T

T= − = − = −1 1 10 0

3

0

0

, and T

T MT

0

0 02

1

11

2

=+

−γ, so:

η

γ

γBrayton

M

M=

−

+−

1

2

11

2

02

02

: Ramjet thermodynamic cycle efficiency in terms of flight

Mach number, M0 .

For propulsion engines, the figures of merit includes more than thrust and ηBrayton .

The specific impulse, Isp measures how effectively fuel is used:

IF

g

F

f gspf

= =˙ ˙m m

; Specific Impulse,

where ˙ ˙m mf f= is the fuel mass flow rate.

To find the fuel-air ratio, f, we employ a control volume around the combustor and carry out anenergy balance. Before doing this, however, it is useful to examine the way in which Isp appearsin expressions for range.

Muddy pointsWhat exactly is the specific impulse, Isp, a measure of? (MP 2A.6)How is Isp found for rockets in space where g ~ 0? (MP 2A.7)Why does industry use TSCP rather than Isp? Is there an advantage to this? (MP2A.8)Why isn't mechanical efficiency an issue with ramjets? (MP 2A.9)How is thrust created in a ramjet? (MP 2A.10)Why don't we like the numbers 1 and 2 for the stations? Why do we go 0-3? (MP2A.11)For the Brayton cycle efficiency, why does T 3 =T t0 ? (MP 2A.12)

2.A.5 The Breguet Range Equation[See Waitz Unified Propulsion Notes, No. IV (see the 16.050 Web site)]

Consider an aircraft in level flight, with weight W. The rate of change of the gross weightof the vehicle is equal to the fuel weight flow:

2A-9

L

D

F

W

gdW

dt

F

Ispfm = − = −

W L D LD F L

D= = ( ) = ( )The rate of change of aircraft gross weight is thus

dW

dt

WL

D Isp

= −( ) .

Suppose L/D and Isp remain constant along flight path:

dW

W

dtL

D Isp

= −( ) .

We can integrate this equation for the change in aircraft weight to yield a relation between theweight change and the time of flight:

lnW

W

tL

D Ii sp

= −( )

, where Wi is the initial weight.

If Wf is the final weight of vehicle and tinitial=0, the relation between vehicle parameters and flighttime, t f , is

L

DI

W

Wtsp

i

ffln =.

The range is the flight time multiplied by the flight speed, or,

Range c tL

Dc I

W

Wf spi

f

= =

× × ( ) ×

0 0

aircraft designerpropulsion systemdesigner structural designer

ln

123

2A-10

The above equation is known as the Breguet range equation. It shows the influence of aircraft,propulsion system, and structural design parameters.

Relation of overall efficiency, I sp , and thermal efficiency

Suppose ∆hfuel is the heating value (‘heat of combustion’) of fuel (i.e., the energy per unitof fuel mass), in J/kg. The rate of energy release is m f fuelh∆ , so

c I cF

g

h

h

Fc

h

h

gspf

fuel

fuel f fuel

fuel0 0

0= =˙ ˙m m

∆

∆ ∆

∆

and Fc

hf fueloverall

0m ∆

=( )

=Thrust power usefulwork

Ideal available energyη (overall propulsion system efficiency)

ηoverallfuel

spg

hc I=

∆ 0

and Range =

∆h

g

L

D

W

Wfuel

overalli

f

η ln

η η η ηoverall thermal propulsive combustion=

The Propulsion Energy Conversion ChainThe above concepts can be depicted as parts of the propulsion energy conversion train

mentioned in Part 0, which shows the process from chemical energy contained in the fuel toenergy useful to the vehicle.

Figure 2A-7: The propulsion energy conversion chain.

The combustion efficiency is near unity unless conditions are far off design and we can regard thetwo main drivers as the thermal and propulsive1 efficiencies. The evolution of the overallefficiency of aircraft engines in terms of these quantities is shown below in Figure 2A-8.

1 The transmission efficiency represents the ratio between compressor and turbine power, which is less than unity due toparasitic frictional effects. As with the combustion efficiency, however, this is very close to one and the horizontal axiscan thus be regarded essentially as propulsive efficiency.

Chemical1 energy

m hf fuel

⋅∆

Heat

Mechanicalwork

Useful work:Thrust power

ηcombustion 1≅ ηthermalηpropulsive

exitc c= +2

1 0/

mc cexit•

−2

02

2 2

2A-11

0.1 0.2

0.2

0.3

0.3

0.3

0.4

0.4

0.4

0.5

0.5

0.5

0.6

0.6

0.6

0.8

0.8

0.7

0.7

0.6 0.5 0.4 0.30.8

0.7

SFCFuturetrend

AdvancedUDF

'777'Engines

CF6-80C2

Low BPR

Turbojets CurrentHigh BPR

UDFEngine

Cor

e T

herm

al E

ffic

ienc

y

Propulsive x Transmission Efficiency

Whittle

Overall Efficiency

Figure 2A-8: Overall efficiency of aircraft engines as functions of thermal and propulsiveefficiencies [data from Koff].

Muddy pointsHow can we idealize fuel addition as heat addition? (MP 2A.13)

2.A.6 Performance of the Ideal RamjetWe now return to finding the ramjet fuel-air ratio, f. Using a control volume around the

burner, we get:

Heat given to the fluid: ˙ ˙ ˙Q h f hf fuel fuel= =m m∆ ∆

From the steady flow energy equation:

˙ ˙ ˙m m m4 3 34 3h h f ht t fuel− = ∆

The exit mass flow is not greatly different from theinlet mass flow, ˙ ˙ ˙m m f m4 3 31= +( ) ≈ , because the

fuel-air ratio is much less than unity (generally several percent). We thus neglect the differencebetween the mass flows and obtain

h h c T T f ht t p t t fuel4 3 4 3− = −( ) = ∆

T c f ht p b fuel31τ −( ) = ∆ , with T T T Mt t3 0 0

12 0

2

0

1= = +( )−γ

Θ1 24 34

Q3 4

.

2A-12

Fuel-air ratio, f:

fh

c T

b

fuel

p

=−τ 1

0 0

∆Θ

,

The fuel-air ratio, f, depends on the fuel properties ( ∆hfuel ), the desired flight parameters (Θ0 ),

the ramjet performance (τ b ), and the temperature of the atmosphere ( T0 ).

Specific impulse, I sp :The specific impulse for the ramjet is given by

IF

f g g

c hc T

sp

fuel

pb

b= =

−( )

−( )m

11

1

0

0

0

∆

Θ

τ

τ

The specific impulse can be written in terms of fuel properties and flight and vehiclecharacteristics as,

Ia h

gc T

Msp

fuel

p b

= ×+( )

0

0

0

0 1

∆

Θfuel properties flight characteristics, ramjet temp increase124 34 1 244 344

τ.

We wish to explore the parameter dependency of the above expression, which is a complicatedformula. How can we do this? What are the important effects of the different parameters? Howdo we best capture the ramjet performance behavior?

To make effective comparisons, we need to add some additional information concerning theoperational behavior. An important case to examine is when f is such that all the fuel burns, i.e.when we have stoichiometric conditions. What happens in this situation as the flight Machnumber, M0, increases? T0 is fixed so Tt3

increases, but the maximum temperature does not increase

much because of dissociation: the reaction does not go to completion at high temperature. A usefulapproximation is therefore to take Tt4

constant for stoichiometric operation. In the stratosphere,

from 10 to 30 km, T K0 212≈ ≈constant . The maximum temperature ratio is

τ maxmax= = =

T

T

T

TT

0

4

0

const ,

ττ

bT

T

TTT

TTT

T

T= = =4

3

40

30 0

max

Θ

For the stoichiometric ramjet:

IF

f g

F

a

a

f gM

a

f gspstoich

bstoich

= = = −( )˙ ˙m m 0

00

01τ

2A-13

Using the expression forτ b , the specific impulse is

I Ma

f gspstoich

= −

0

0

01τ max

Θ

Representative performance values:A plot of the performance of the stoichiometric ramjet is shown in Figure 2A-9.

1 2 3 4 5 6 7

0.5

1

1.5

2

2.5F

m a

I g fa

f = f0

stoic

0

stoic

q = 10b.F

m a0.

Isp gfstoicha0

M0

τmax = 10

f = fstoich

Figure 2A-9: Thrust per unit mass flow and specific impulse for ideal ramjet withstoichiometric combustion [adapted from Kerrebrock]

The figure shows that for the parameters used, the best operating range of a hydrocarbon-fueledramjet is 2 40≤ ≤M . The parameters used areτ max = 10 , a ms0

1300≈ − in the stratosphere,

fstoich = 0 067. for hydrocarbonsa

g fs

stoich

0 450≈ .

Recapitulation:In this section we have:

Examined the Brayton Cycle for propulsionFound ηBrayton as a function of M0

Found ηoverall and the relation between ηoverall and ηBrayton

Examined F

am 0 and Isp as a function of M0 for the ideal ramjet.

Muddy pointsWhat is the relation between h h f ht4 t3 f− = ∆ and the existence of the maximum value

of T t4 ? (MP 2A.14a) Why didn’t we have a 2s point for the Brayton cycle with non-ideal components ? (MP2A.14b)What is the variable f stoich ? (MP 2A.15)

2A-14

2.A.7 Effect of Departures from Ideal Behavior - Real Cycle behavior[See also charts 69-82 in 16.050: Gas Turbine Engine Cycles]

What are the sources of non-ideal performance and departures from reversibility?- Losses (entropy production) in the compressor and the turbine- Stagnation pressure decrease in the combustor- Heat transfer

We take into account here only irreversibility in the compressor and in the turbine. Because ofthese irreversibilities, we need more work, ∆hcomp (the changes in kinetic energy from inlet to exit

of the compressor are neglected), to drive the compressor than in the ideal situation. We also getless work, ∆hturb , back from the turbine. The consequence, as can be inferred from Figure 2A-10

below, is that the net work from the engine is less than in the cycle with ideal components.

h4

5

5s

3

0

3s

s

qR

qAwturb

T4 = Tmax

T0 = Tinlet

P3

P0

wcomp

Figure 2A-10: Gas turbine engine (Brayton) cycle showing effect of departure from idealbehavior in compressor and turbine

To develop a quantitative description of the effect of these departures from reversiblebehavior, consider a perfect gas with constant specific heats and neglect kinetic energy at theinlet and exit of the turbine and compressor. We define the turbine adiabatic efficiency as:

ηturbturbactual

turbideal

S

w

w

h h

h h= =

−

−4 5

4 5

where wturbactual is specified to be at the same pressure ratio as wturb

ideal . (See charts 69-76 in 16.050

Gas Turbine Cycles.)

There is a similar metric for the compressor, the compressor adiabatic efficiency:

2A-15

ηcompcompideal

compactual

Sw

w

h h

h h= =

−

−3 0

3 0

again for the same pressure ratio. Note that the ratio is the actual work delivered divided by theideal work for the turbine, whereas the ratio is the ideal work needed divided by the actual workrequired for the compressor. These are not thermal efficiencies, but rather measures of thedegree to which the compression and expansion approach the ideal processes.

We now wish to find the net work done in the cycle and the efficiency. The net work is giveneither by the difference between the heat received and rejected or the work of the compressorand turbine, where the convention is that heat received is positive and heat rejected is negativeand work done is positive and work absorbed is negative.

Net work =H

heat in heat out

turb comp

q q h h h h

w w h h h h

L + = −( ) − −( )

+ = −( ) − −( )

4 3 5 0

4 5 3 0

The thermal efficiency is:

ηthermal =Net work

Heat input

We need to calculate T3 , T5

From the definition ofηcomp , T TT T

TS

c

T ST

c3 0

3 00

30

1− =

−( )=

−( )η η

With T ST

comp

comp3

0

1 1

( ) =

=

− −

isentropic temperature ratio =P

Pexit

inlet

γγ

γγΠ

T T T

comp

comp3 0

1

0

1

=

−

+

−

Πγγ

η

Similarly, by the definition of ηturbext

inlet

P

P

=actual work received

ideal work for same

, we can find T5 :

T T T T TT

TTturb S turb

Sturb Turb4 5 4 5 4

5

44

1

1 1− = −( ) = −

= −

−

η η ηγγΠ

T T Tturb turb5 4 4

1

1= − −

−

ηγγΠ

2A-16

The thermal efficiency can now be found:

ηthermalL

H

Q

Q

T T

T T= + = −

−

−1 1 5 0

4 3

with ΠΠ

Πct

= =1 , and τ

γγ

S =

−

Π1

the isentropic cycle temperature ratio,

η

ητ

ητ

thermal

turbS

compS

T T

T T

= −

− −

−

− −( ) +

1

1 11

11 1

4 0

4 0

or,

ητ

η η τ

η τthermal

Scomp turb S

c S

T

T

T

T

=

−

−

+ −

−

11

1 1

4

0

4

0

There are several non-dimensional parameters that appear in this expression for thermalefficiency. We list these just below and show their effects in subsequent figures:

Parameters reflecting design choices

τγγ

S =

−

Π1

: cycle pressure ratioT

T4

0

: maximum turbine inlet temperature

Parameters reflecting the ability to design and execute efficient componentsηcomp : compressor efficiency

ηturb : turbine efficiency

In addition to efficiency, net rate of work is a quantity we need to examine,˙ ˙ ˙W W Wnet turbine compressor= −

Putting this in a non-dimensional form:˙

˙W

m

work to drive compressor work extracted from flow by turbine

net

p compS turb

Sc T

T

T0

4

0

11 1

1= − −( ) + −

ητ η

τ1 244 344 1 244 344

˙

˙W

c T

TTnet

pS

T

S compm 0

401

1= −( ) −

τη

τ η

2A-17

Trends in net power and efficiency are shown in Figure 2A-11 for parameters typical ofadvanced civil engines. Some points to note in the figure:• For anyη ηcomp turb, ≠ 1 , the optimum pressure ratio ( Π ) for maximum ηth is not the

highest that can be achieved, as it is for the ideal Brayton cycle. The ideal analysis is tooidealized in this regard. The highest efficiency also occurs closer to the pressure ratio formaximum power than in the case of an ideal cycle. Choosing this as a design criterion willtherefore not lead to the efficiency penalty inferred from ideal cycle analysis.

• There is a strong sensitivity to the component efficiencies. For example, forη ηturb comp= = 0 85. , the cycle efficiency is roughly two-thirds of the ideal value.

• The maximum power occurs at a value of τ S or pressure ratio ( Π ) less than that for

maxη . (this trend is captured by ideal analysis).

• The maximum power and maximum ηthermal are strongly dependent on the maximum

temperature, TT

4

0.

Muddy points

How can

TT

4

0

be the maximum turbine inlet temperature? (MP 2A.16)

When there are losses in the turbine that shift the expansion in T-s diagram to theright, does this mean there is more work than ideal since the area is greater? (MP2A.17)

2A-18

Figure 2A-11: Non-dimensional power and efficiency for a non-ideal gas turbine engine - (a)Non-dimensional work as a function of cycle pressure ratio for different values of turbine entrytemperature divided by compressor entry temperature, (b) Overall cycle efficiency as a functionof pressure ratio for different values of turbine entry temperature divided by compressor entrytemperature, (c) Overall cycle efficiency as a function of cycle pressure ratio for differentcomponent efficiencies. [from Cumpsty, Jet Propulsion]

2A-19

A scale diagram of a Brayton cycle with non-ideal compressor and turbine behaviors, in terms oftemperature-entropy (h-s) and pressure-volume (P-v) coordinates is given below as Figure 2A-12.

Figure 2A-12: Scale diagram of non-ideal gas turbine cycle. Nomenclature is shown in the figure.Pressure ratio 40, T0 = 288, T4 =1700, compressor and turbine efficiencies = 0.9 [adapted fromCumpsty, Jet Propulsion]

Muddy pointsFor an afterburning engine, why must the nozzle throat area increase if thetemperature of the fluid is increased? (MP 2A.18a)Why doesn’t the pressure in the afterburner go up if heat is added? (MP 2A.18b)Why is the flow in the nozzle choked? (MP 2A.18c)What’s the point of having a throat if it creates a retarding force? (MP 2A.18d)Why isn’t the stagnation temperature conserved in this steady flow? (MP 2A.18e)

Muddiest points on part 2A

2A.1 How is ∆hfuel calculated?

For now, we rely on tabulated values. In the lectures accompanying Section 2.C of thenotes, we will see how one can calculate the heat , fuelh∆ , liberated in a given reaction.

2A.2 What are "stoichiometric conditions"?

Stoechiometric conditions are those in which the proportions of fuel and air are such thatthere is not an excess of each one--all the fuel is burned, and all the air (oxidizer) is usedup doing it. See Notes Sections 2.C.

2A.3 When and where do we use cv and cp? Some definitions use dU=cvdT. Is it everdU=cpdT?

The answer is no. The definitions of cp and cv are derived in the notes on page 0-6. cp isthe specific heat at constant pressure and for an ideal gas dh = cp dT always holds.Similarly cv is the specific heat at constant volume and for an ideal gas du = cv dT alwaysholds. A discussion on this is also given in the notes on pages 0-6 and 0-7. If you thinkabout how you would measure the specific heat c = q/(Tfinal – Tinitial) for a certain knownchange of state you could do the following experiments.For a process during which heat ∆q is transferred (reversibly) and the volume staysconstant (e.g. a rigid, closed container filled with a substance, or the heat transfer in anOtto engine during combustion – the piston is near the top-dead-center and the volume isapproximately constant for the heat transfer) the first law is du = dq since v = const.Using the definition du = cv dT we obtain for the specific heat at constant volume

cv = ∆q/ ∆T ,

where both the heat transferred ∆q and the temperature difference ∆T can be measured.

Similarly we can do an experiment involving a process where the pressure is keptconstant during the reversible heat transfer ∆q (e.g. a rigid container filled with asubstance that is closed by a lid with a certain weight, or the heat transfer in a jet enginecombustor where the pressure is approximately constant during heat addition). The firstlaw can be written in terms of enthalpy as dh – vdp = dq, and since p = const we obtaindh = dq. Using the definition dh = cpdT we obtain for the specific heat at constantpressure

cp = ∆q/ ∆T .

2A.4 Explanation of the above comparison between Diesel and Otto.

Basically we can operate the diesel cycle at much higher compression ratio than the Ottocycle because only air is compressed and we don't run into the auto-ignition problem(knocking problem). Because of the higher compression ratios in the diesel engine we gethigher efficiencies.

2A.5 What is shaft work?

I am not sure how best to answer, but it appears that the difficulty people are havingmight be associated with being able to know when one can say that shaft work occurs.There are several features of a process that produces (or absorbs) shaft work. First of allthe view taken of the process is one of control volume, rather than control mass (see thediscussion of control volumes in Section 0 or in IAW). Second, there needs to be a shaftor equivalent device (a moving belt, a row of blades) that can be identified as the workcarrier. Third, the shaft work is work over and above the “flow work” that is done by (orreceived by) the streams that exit and enter the control volume.

2A.6 What exactly is the specific impulse, Isp, a measure of?

The specific impulse is a measure of how well the fuel is used in creating thrust. For arocket engine, the specific impulse is the effective exit velocity divided by theacceleration of gravity, g. In terms of relating the specific impulse to some characteristictime, we can write the definition of Isp as

FgmIsp =& .

From this, one can regard the specific impulse as the time that it would take to flow aquantity of fuel that has a weight equal to the thrust force

2A.7 How is Isp found for rockets in space where g ~ 0?

The impulse I given to a rocket is the thrust force integrated over the burn time.Traditionally, for the case of constant exhaust velocity cex, the specific impulse has beenused Isp = I /(mp g) = cex / g0, where mp is the propellant mass and g0 is the Earth's surfacegravity. Thus Isp is measured in seconds and is a force per weight flow. Often today,however, specific impulse is measured in units meters/second [m/s], recognizing thatforce per mass flow is more logical. The specific impulse is then simply equal to theexhaust velocity Isp = cex.

2A.8 Why does industry use TSCP rather than Isp? Is there an advantage to this?

I am not sure why Thrust*Specific Fuel Consumption was originally used. The gasturbine industry uses TSFC; the rocket propulsion industry uses essentially its inverse,Specific Impulse. Perhaps an advantage is that TSFC is a number of magnitude unity,whereas specific impulse is not.

2A.9 Why isn't mechanical efficiency an issue with ramjets?

As defined, the mechanical efficiency represents bearing friction, and other parasitictorques on the rotating shaft in a gas turbine. The work associated with this needs to beprovided by the turbine, but does not go into driving the compressor. The ramjet has noshaft, and hence does not encounter this.

2A.10 How is thrust created in a ramjet?

You can look at thrust in several ways. One is through the integral form of themomentum equation, which relates thrust to the difference between exit and inletvelocities, multiplied by the mass flow. Another way, however, is to look at the forces onthe ramjet structure, basically the summation of pressure forces on all the surfaces. Iattempted to do this in class using the turbojet with an afterburner. For the ramjet, fromthe same considerations, we would have an exit nozzle that was larger in diameter thanthe inlet so that the structural area on which there is a force in the retarding direction issmaller than the area on which there is a force in the thrust direction.

2A.11 Why don't we like the numbers 1 and 2 for the stations? Why do we go 0-3?

A common convention in the industry is that station 0 is far upstream, station 1 is afterthe shock in the inlet (if there is one), station 2 is at inlet to the compressor (after theinlet/diffuser) and station 3 is after the compressor. In class, when we examined theramjet we considered no changes in stagnation pressure between 0 and 2, so I have used 0as the initial state for the compression process. It would be more precise to differentiatebetween stations 0 and 2, and I will do this where appropriate.

2A.12 For the Brayton cycle efficiency, why does T3=Tt0?

The ramjet is operating as a Brayton cycle where ηb= 1 – Tinlet / T compressor exit. For theramjet discussed in class the inlet temperature is T0 and since there is no compressor (nomoving parts) the only compression we get is from diffusion. We assumed isentropicdiffusion in the diffuser and found for very low Mach numbers that the diffuser exit orcombustor inlet temperature T3 is Tt3. From first law we know that for a steady, adiabaticflow where no work is done the stagnation enthalpy stays constant. Assuming perfect gaswe thus get Tt0 = Tt3 = T3. So we can write for the ramjet thermal efficiency

ηb= 1 – T0 / Tt0 .

2A.13 How can we idealize fuel addition as heat addition?

The validity of an approximation rests on what the answer is going to be used for. Weare seeking basically only one item concerning combustor exit conditions, namely theexit temperature or the exit enthalpy. The final state is independent of how we add theheat, and depends only on whether we add the heat. If it is done from an electrical heateror from combustion, and if we neglect the change in the constitution of the gas due to the

combustion products (most of the gas is nitrogen) the enthalpy rise is the same no matterhow the temperature rise is achieved.

2A.14a What is the relation between h h f ht4 t3 f− = ∆ and the existence of the maximum

value of Tt4?

The two are very different physical statements. The first is the SFEE (steady flow energyequation) plus the approximation that inlet and exit mass flows to the control volume arethe same. The heat received within the volume is represented by the quantity f h f∆ ,

where ∆hf is the heat liberated per kilogram of fuel. The second statement is a

representation of the fact that the degree of completion of the reaction in the combustordepends on temperature, so that even though the inlet temperature increases strongly asthe Mach number increases, the combustor exit temperature does not change greatly.This is an attempt to represent a complex physical process (or set of processes) in anapproximate manner, not a law of nature.

2A.14b Why didn’t we have a 2s point for the Brayton cycle with non-ideal components?

If we didn’t, we should have, or I should at least have marked the point at which thecompressor exit would be if the compression process was isentropic.

2A.15 What is the variable fstoich?

fstoich is the fuel-to-air ratio for stoichiometric combustion, or in other words the fuel-to-air ratio for a chemically correct combustion process during which all fuel is burnt.

2A.16 How can TT

4

0

be the maximum turbine inlet temperature?

I agree that the T4/T0 is a temperature ratio. If we assume constant ambient temperaturethen this ratio reflects the maximum cycle temperature. The main point was to emphasizethat the higher your turbine inlet temperature the higher your power and efficiency levels.

2A.17 When there are losses in the turbine that shift the expansion in T-s diagram tothe right, does this mean there is more work than ideal since the area isgreater?

We have to be careful when looking at the area enclosed by a cycle or underneath a pathin the T-s diagram. Only for a reversible cycle, the area enclosed is the work done by thecycle (see notes page 1C-5). Looking at the Brayton cycle with losses in compressor andturbine the net work is the difference between the heat absorbed and the heat rejected(from 1st law). The heat absorbed can be found by integrating TdS = dQ along the heataddition process. The heat rejected during the cycle with losses in compressor and turbineis larger than in the ideal cycle (look at the area underneath the path where heat isrejected, this area is larger than when there are no losses dsirrev = 0 – see also muddy point

1C.1). So we get less net work if irreversibilities are present. It is sometimes easier tolook at work and heat (especially shaft work for turbines and compressors) in the h-sdiagram because the enthalpy difference between two states directly reflects the shaftwork (remember, enthalpy includes the flow work!) and / or heat transfer.

2A.18a For an afterburning engine, why must the nozzle throat area increase if thetemperature of the fluid is increased?

The Mach number of the flow is unity at the throat with and without the afterburnerlit. The ratio of static pressure to stagnation pressure at the throat is thus the samewith and without the afterburner lit. The ratio of static temperature to stagnationtemperature at the throat is thus the same with and without the afterburner lit.

T

TM

P

PMt t= +

−= +

−

−( )1

1

21

1

22 2

1γ γ γ γ

; /

The flow through the throat is

m cA aAP

RTRTAthroat throat throat= = =ρ ρ γ .

The flow through the throat thus scales as

˙

˙/

/

/

/

m

m

P

TA

P

TA

A B

noA B

throatA B

throatnoA B

=

From what we have said, however, the pressure at the throat is the same in both cases.Also, we wish to have the mass flow the same in both cases in order to have theengine operate at near design conditions. Putting these all together, plus use of theidea that the ratio of stagnation to static temperature at the throat is the same for bothcases gives the relation

A

A

T

TthroatA B

throatnoA B

tA B

tnoA B

/

/

/

/

=

The necessary area to pass the flow is proportional to the square root of the stagnationtemperature.

If too much fuel is put into the afterburner, the increase in area cannot be met and theflow will decrease. This can stall the engine, a serious consequence for a singleengine fighter.

2A.18b Why doesn’t the pressure in the afterburner go up if heat is added?

From discussions after lecture, the main point here seems to be that the process ofheat addition in the afterburner, or the combustor, is not the same as heat addition to agas in a box. In that case the density (mass/volume) would be constant and, fromP RT= ρ , increasing the temperature would increase the pressure. In a combustor,the geometry is such that the pressure is approximately constant; this happens because

the fluid has the freedom to expand so the density decreases. From the equationP RT= ρ if the temperature goes up, the density must go down.

2A.18c Why is the flow in the nozzle choked?

As seen in Unified, choking occurs when the stagnation to static pressureratio P Pt /( )gets to a certain value, 1.89 for gas with γ of 1.4. Almost all jet aircraft

operate at flight conditions such that this is achieved. If you are not comfortable withthe way in which the concepts of choking are laid out in the Unified notes, please seeme and I can give some references.

2A.18d What’s the point of having a throat if it creates a retarding force?

As shown in Unified, to accelerate the flow from subsonic to supersonic, i.e., tocreate the high velocities associated with high thrust, one must have a converging-diverging nozzle, and hence a throat.

2A.18e Why isn’t the stagnation temperature conserved in this steady flow?

Heat is added in the afterburner, so the stagnation temperature increases.

2B-1

2.B Power Cycles with Two-Phase Media (Vapor Power Cycles)[SB&VW – Chapter 3, Chapter 11, Sections 11.1 to 11.7]

In this section, we examine cycles that use two-phase media as the working fluid. Thesecan be combined with gas turbine cycles to provide combined cycles which have higher efficiencythan either alone. They can also be used by themselves to provide power sources for bothterrestrial and space applications. The topics to be covered are:i) Behavior of two-phase systems: equilibrium, pressure temperature relationsii) Carnot cycles with two-phase mediaiii) Rankine cyclesiv) Combined cycles

2.B.1 Behavior of Two-Phase SystemsThe definition of a phase, as given by SB&VW, is “a quantity of matter that is

homogeneous throughout”. Common examples of systems that contain more than one phase are aliquid and its vapor and a glass of ice water. A system which has three phases is a container withice, water, and water vapor.

We wish to find the relations between phases and the relations that describe the change ofphase (from solid to liquid, or from liquid to vapor) of a pure substance, including the work doneand the heat transfer. To start we consider a system consisting of a liquid and its vapor inequilibrium, which are enclosed in a container under a moveable piston, as shown in Figure 2B-1.The system is maintained at constant temperature through contact with a heat reservoir attemperature T, so there can be heat transfer to or from the system.

(a)

Liquid water Liquid water

Water vaporWater vapor

(b) (c)

Figure 2B-1: Two-phase system in contact with constant temperature heat reservoir

For a pure substance, asshown at the right, there isa one-to-one correspondencebetween the temperature at whichvaporization occurs and the pressure.These values are called the saturationpressure and saturation temperature(see Ch. 3 in SB&VW).

P-T relation for liquid-vapor system

2B-2

This means there is an additional constraint for a liquid-vapor mixture, in addition to the equationof state. The consequence is that we only need to specify one variable to determine the state of thesystem. For example, if we specify T then P is set. In summary, for two phases in equilibrium,P P T= ( ). If both phases are present, any quasi-static process at constant T is also at constant P.

Let us examine the pressure-volume behavior of a liquid-vapor system at constant temperature.For a single-phase perfect gas we know that the curve would be Pv = constant. For the two-phasesystem the curve looks quite different, as indicated in Figure 2B-2.

Volume, V

Pres

sure

, P

Liq

uid

phas

e

Mixture ofliquid and

vapor

Liquid saturation curve

Vaporsaturation

curve

Critical point

Vapor phase

Gas phase

Critical isotherm

D

B AC

Figure 2B-2 – P-v diagram for two-phase system showing isotherms

Several features of the figure should be noted. First, there is a region in which liquid and vaporcan coexist. This is roughly dome-shaped and is thus often referred to as the “vapor dome”.Outside of this regime, the equilibrium state will be a single phase. The regions of the diagram inwhich the system will be in the liquid and vapor phases respectively are indicated. Second is thesteepness of the isotherms in the liquid phase, due to the small compressibility of most liquids.Third, the behavior of isotherms at temperatures below the “critical point” (see below) in theregion to the right of the vapor dome approach those of an ideal gas as the pressure decreases andthe ideal gas relation is a good approximation in this region.

The behavior shown is found for all the isotherms that go through the vapor dome. At a highenough temperature, specifically at a temperature corresponding to the pressure at the peak of thevapor dome, there is no transition from liquid to vapor and the fluid goes continuously from aliquid-like behavior to a gas-type behavior. This behavior is unfamiliar, mainly because thetemperatures and pressures are not ones that we typically experience; for water the criticaltemperature is 374oC and the associated critical pressure is 220 atmospheres.

2B-3

There is a distinct nomenclature used for systems with more than one phase. In this, the terms“vapor” and “gas” seem to be used interchangeably. In the zone where both liquid and vapor exist,there are two bounding situations. When the last trace of vapor condenses, the state becomessaturated liquid. When the last trace of liquid evaporates the state becomes saturated vapor (or dryvapor). If we put heat into a saturated vapor it is referred to as superheated vapor. Nitrogen atroom temperature and pressure (at one atmosphere the vaporization temperature of nitrogen is 77K) is a superheated vapor.

Figure 2B-3: Constant pressure curves in T-v coordinates showing vapor dome

Figure 2B-3 shows lines of constant pressure in temperature-volume coordinates. Inside the vapordome the constant pressure lines are also lines of constant temperature.

It is useful to describe the situations encountered as we decrease the pressure or equivalentlyincrease the specific volume, starting from a high pressure-low specific volume state (the upperleft-hand side of the isotherm in Figure 2B-2). The behavior in this region is liquid-like with verylittle compressibility. As the pressure is decreased, the volume changes little until the boundary ofthe vapor dome is reached. Once this occurs, however, the pressure is fixed because thetemperature is constant. As the piston is withdrawn, the specific volume increases through moreliquid evaporating and more vapor being produced. During this process, since the expansion isisothermal (we specified that it was), heat is transferred to the system. The specific volume willincrease at constant pressure until the right hand boundary of the vapor dome is reached. At thispoint, all the liquid will have been transformed into vapor and the system again behaves as asingle-phase fluid. For water at temperatures near room temperature, the behavior would beessentially that of a perfect gas in this region. To the right of the vapor dome, as mentioned above,the behavior is qualitatively like that of a perfect gas.

Referring to Figure 2B-4, we define notation to be used in what follows. The states a and c denotethe conditions at which all the fluid is in the liquid state and the gaseous state respectively.

2B-4

State of liquid in mixture

State of vaporin mixture

Liquid + gas

vf v vg

Mixture"state"Liquid

Standard-liquid line

Critical point

Saturated-vapor line

Gas (vapor)

(constant T line)

T

v

ab

c

Figure 2B-4: Specific volumes at constant temperature and states within the vapor domein a liquid-vapor system

The specific volumes corresponding to these states are:

v f - specific volume of liquid phase

vg - specific volume of gas phase

For conditions corresponding to specific volumes between these two values, i.e., for state b, thesystem would exist with part of the mass in a liquid state and part of the mass in a gaseous (vapor)state. The average specific volume for this condition is

v - average specific volume of two-phase system.

We can relate the average specific volume to the specific volumes for liquid and vapor and themass that exists in the two phases as follows. The total mass of the system is given by

total mass = liquid mass + vapor mass = m m mf g= + .

The volume of the system is

Volume of liquid =V m vf f f=

Volume of vapor =V m vg g g=

Total volume =V m v m vf f g g= + .

The average specific volume, v , is the ratio of the total volume to the total mass of the system

vm v m v

m mf f g g

f g

=+

+ = average specific volume.

2B-5

The fraction of the total mass in the vapor phase is called quality, and denoted by X.

Xm

m mg

f g

=+

= quality of a liquid-vapor system.

In terms of the quality and specific volumes, the average specific volume can be expressed as:

v X v X vg f= ⋅ + − ⋅( )1

In reference to Figure 2B-5, ab v v ac v vf g f= − = − , .

ab

ac

v v

v vXf

g f

=−

−= = quality .

(a)

T

L

L-V

T

T

V

p

a b c

vgv vvf

(b)

Figure 2B-5: Liquid vapor equilibrium in a two-phase medium

2.B.2 - Work and Heat Transfer with Two-Phase MediaWe examine the work and heat transfer in quasi-static processes with two-phase systems.

For definiteness, consider the system to be a liquid-vapor mixture in a container whose volume canbe varied through movement of a piston, as shown in Figure 2B-5. The system is kept at constanttemperature through contact with a heat reservoir at temperature T. The pressure is thus alsoconstant, but the volume, V, can change. For a fixed mass, the volume is proportional to thespecific volume v so that point b in Figure 2B-5 must move to the left or the right as V changes.This implies that the amount of mass in each of the two phases, and hence the quality, also changesbecause mass is transferred from one phase to the other. We wish to find the heat and worktransfer associated with the change in mass in each phase. The change in volume can be related tothe changes in mass in the two phases as,

dV v dm v dmg g f f= + .

2B-6

The system mass is constant (m m mf g= + = constant ) so that for any changes

dm dm dmf g= = +0 .

We can define the quantity dm fg

dm dm dmfg g f= = - = mass transferred from liquid to vapor.

In terms of dm fg the volume change of the system is

dV v v dmg f fg= −( ) .

The work done is given by

dW = PdV = −( )P v v dmg f fg .

The change in internal energy, ∆U , can be found as follows. The internal energy of the systemcan be expressed in terms of the mass in each phase and the specific internal energy (internalenergy per unit mass, u) of the phase as,

U u m u mf f g g= +

dU u dm u dm u u dmf f g g g f fg= + = −( ) .

Note that the specific internal energy can be expressed in a similar way as the specific volume interms of the quality and the specific enthalpy of each phase:

u X u X ug f= ⋅ + − ⋅( )1

Writing the first law for this process:

dQ = dU + dW = −( ) + −( )u u dm P v v dmg f fg g f fg .

= +( ) − +( )[ ]u Pv u Pv dmg g f f fg

= −( )h h dmg f fg.

The heat needed for the transfer of mass is proportional to the difference in specific enthalpybetween vapor and liquid. The pressure and temperature are constant, so that the specific internalenergy and the specific enthalpy for the liquid phase and the gas phase are also constant. For afinite change in mass from liquid to vapor, mfg, therefore, the quantity of heat needed is

Q h h m Hg f fg= −( ) = ∆ (enthalpy change).

2B-7

The heat needed per unit mass, q, for transformation between the two phases is

qQ

mh h h

fgg f fg= = −( ) = .

The notation hfg refers to the specific enthalpy change between the liquid state and the vapor state.

The expression for the amount of heat needed, q, is a particular case of the general result that inany reversible process at constant pressure, the heat flowing into, or out of, the system is equal tothe enthalpy change. Heat is absorbed if the change is from solid to liquid (heat of fusion), liquidto vapor (heat of vaporization), or solid to vapor (heat of sublimation).

A numerical example is furnished by the vaporization of water at 100oC:i) How much heat is needed per unit mass of fluid vaporized?ii) How much work is done per unit mass of fluid vaporized?iii) What is the change in internal energy per unit mass of fluid vaporized?.

In addressing these questions, we make use of the fact that problems involving heat and workexchanges in two-phase media are important enough that the values of the specific thermodynamicproperties that characterize these transformations have been computed for many different workingfluids. The values are given in SB&VW in Tables B.1.1 and B.1.2 for water at saturatedconditions and in Tables B.1.3, B.1.4, and B.1.5 for other conditions, as well as for other workingfluids. From these:

- At 100oC, the vapor pressure is 0.1013 MPa,- The specific enthalpy of the vapor, hg , is 2676 kJ/kg and the specific enthalpy of the

liquid, hf , is 419 kJ/kg

- The difference in enthalpy between liquid and vapor, hfg , occurs often enough so that it is

tabulated also. This is 2257 kJ/kg,- The specific volume of the vapor is 1.6729 m3/kg and the specific volume of the liquid is0.001044.

The heat input to the system is the change in enthalpy between liquid and vapor, hfg , and is equal

to 2.257 x 106 J/kg.

The work done is P v vg f−( ) which has a value of

P v vg f−( )=0.1013 x 106 x [1.629 – 0.001044] =0.169 x 106 J/kg.

The change in internal energy per unit mass (u fg) can be found from ∆u q w= − or from the

tabulated values as 2.088 x 106 J/kg. This is much larger than the work done. Most of the heatinput is used to change the internal energy rather than appearing as work.

Muddy points

2B-8

For the vapor dome, is there vapor and liquid inside the dome and outside is it just liquidor just gas? Is it interchangeable? Is it true for the plasma phase? (MP 2B.1)What is hfg ? How do we find it? (MP 2B.2)Reasoning behind the slopes for T=cst lines in the P-V diagram. (MP 2B.3)For a constant pressure heat addition, why is q=∆h? (MP 2B.4)What is latent heat? (MP 2B.5)Why is U a function of x? (MP 2B.6)

2.B.3 The Carnot Cycle as a Two-Phase Power CycleA Carnot cycle that uses a two-phase fluid as the working medium is shown below in

Figure 2B-6. Figure 2B-6a gives the cycle in P-v coordinates, 2B-6b in T-s coordinates, and 2B-6cin h-s coordinates. The boundary of the region in which there is liquid and vapor both present (thevapor dome) is also indicated. Note that the form of the cycle is different in the T-s and h-srepresentation; it is only for a perfect gas with constant specific heats that cycles in the twocoordinate representations have the same shapes.

a

a

b

b

c

c

d d

e

T

T2

T2

T2

T1

T1

T1

s1 s2s s

h

f

g h

a b

cd

p

p2

p1

v

(a) p-v diagram

(b) T-s diagram (c) h-s diagram

Figure 2B-6: Carnot cycle with two-phase medium. (a) cycle in P-v coordinates, (b) cycle in T-scoordinates, (c) cycle in h-s coordinates

The processes in the cycle are as follows:i) Start at state a with saturated liquid (all of mass in liquid condition). Carry out a reversible

isothermal expansion to b (a b) until all the liquid is vaporized. During this process aquantity of heat qH per unit mass is received from the heat source at temperature T2 .

ii) Reversible adiabatic (i.e., isentropic) expansion (b c) lowers the temperature to T1.Generally state c will be in the region where there is both liquid and vapor.

2B-9

iii) Isothermal compression (c d) at T1 to state d. During this compression, heat qL per unit

mass is rejected to the source at T1.iv) Reversible adiabatic (i.e., isentropic) compression (d a) in which the vapor condenses to

liquid and the state returns to a.

In the T-s diagram the heat received, qH , is abef and the heat rejected, qL , is dcef. The net work isrepresented by abcd. The thermal efficiency is given by

η = = = −w

q

abcd

abef

T

Tnet

H

Area

Area 1 1

2

.

In the h-s diagram, the isentropic processes are vertical lines as in the T-s diagram. The isothermsin the former, however, are not horizontal as they are in the latter. To see their shape we note thatfor these two-phase processes the isotherms are also lines of constant pressure (isobars), since P =P(T). The combined first and second law is

Tds dhdp

= −ρ

.

For a constant pressure reversible process, dq Tds dhrev = = . The slope of a constant pressure linein h-s coordinates is thus,

∂∂h

sT

P

= = constant ; slope of constant pressure line for two-phase medium.

The heat received and rejected per unit mass is given in terms of the enthalpy at the different statesas,

q h hH b a= −q h hL d c= − . (In accord with our convention this is less than zero.)

The thermal efficiency is

η = =+

=−( ) + −( )

−( )w

q

q q

q

h h h h

h hnet

H

H L

H

b a d c

b a,

or, in terms of the work done during the isentropic compression and expansion processes, whichcorrespond to the shaft work done on the fluid and received by the fluid,

η =−( ) − −( )

−( )h h h h

h hb c a d

b a

.

Example: Carnot steam cycle:Heat source temperature = 300oCHeat sink temperature = 20oC

What is the (i) thermal efficiency and (ii) ratio of turbine work to compression (pump) work ifa) all processes are reversible?b) the turbine and the pump have adiabatic efficiencies of 0.8?

2B-10

Neglect the changes in kinetic energy at inlet and outlet of the turbine and pump.

a) For the reversible cycle,

η ηthermal CarnotT

T= = −

= − =

1

1293

5730 489

1

2

.

To find the work in the pump (compression process) or in the turbine, we need to find the enthalpychanges between states b and c, ∆hbc , and the change between a and d, ∆had . To obtain these theapproach is to use the fact that s = constant during the expansion to find the quality at state c andthen, knowing the quality, calculate the enthalpy as h Xh X hg f= + −( )1 . We know the conditions

at state b, where the fluid is all vapor, i.e., we know T h sb b b, , :

h h C h Cb vaporo

go= ( ) = ( ) =300 300 2749 kJ/kg

s s C s Cb vaporo

go= ( ) = ( ) =300 300 5 7045. kJ/kg - K

s sb c= in the isentropic expansion process.

We now need to find the quality at state c, Xc . Using the definition of quality given in Section

2.B.1, and noting that s X s X sc c g c f= + −( )1 , we obtain,

Xs s T

s T s T

s s T

s Tcc f c

g c f c

c f c

fg c

=− ( )

( ) − ( )=

− ( )( )

.

The quantity sc is the mass-weighted entropy at state c, which is at temperature Tc.

The quantity s Tf c( ) is the entropy of the liquid at temperature Tc .

The quantity s Tg c( ) is the entropy of the gas (vapor) at temperature Tc .

The quantity ∆ ∆s T s Tfg c liquid gas c( ) = → at .

We know: s sc b= = 5 7045. kJ/kg - K

s fg = 8 3706. kJ/kg - K

s f = 0 2966. kJ/kg - K .

The quality at state c is thus,

Xc =−

=5 7045 0 2966

8 37060 646

. .

.. .

The enthalpy at state c is,

2B-11

h X h X hc c g c f= + −( )1 at Tc .

Substituting the values,

hc = × + ×0 646 2538 1 0 354 83 96. . . . kJ/kg =1669.4 kJ/kg.

The turbine work/unit mass is the difference between the enthalpy at state b and state c,

h h wb c turbine− = = − =2749 1669 4 1079 6. . kJ/kg.

We can apply a similar process to find the conditions at state d:

Xs s T

s T s T

s s T

s Tdd f d

g d f d

c f d

fg d

=− ( )

( ) − ( )=

− ( )( )

.

We have given that T Tc d= . Also s s sd a f= = at 300 Co . The quality at state d is

X Xd c=−

=3 253 0 2966

8 37060 353

. .

.. <

The enthalpy at state d is

h X h X hd d g d f= + −( )1

= 0.353 x 2538.1 + 0.647 x 83.96 = 950.8 kJ/kg.

The work of compression (pump work) is ∆h h had a d= − . Substituting the numerical values,

∆had= 1344- 950.8 = 393.3 kJ/kg.

The ratio of turbine work to compression work (pump work) is w

wturbine

compression

= 2 75.

We can check the efficiency by computing the ratio of net work (w w wnet turbine compression= − ) to the

heat input (T sa fg ). Doing this gives, not surprisingly, the same value as the Carnot equation.

b) Efficiency and work ratio for a cycle with adiabatic efficiencies of pump and turbine both equalto 0.8 (non-ideal components).

We can find the turbine work using the definition of turbine and compressor adiabatic efficiencies.The relation between the enthalpy changes is

w h h h hturbine b c turbine b c= − = −( )′ η = actual turbine work received.

Substituting the numerical values, the turbine work per unit mass is 863.7 kJ/kg.

2B-12

For the compression process, we use the definition of compressor (or pump) adiabatic efficiency:

w h h h hcompression a dcompression

a d= − =( )

−( )′1

η

= actual work to achieve given pressure difference

= 491.6 kJ/kg.

The value of the enthalpy at state ′a is 1442.4 kJ/kg.

The thermal efficiency is given by

ηthermalnet turbine compressionw w w

= =−

heat input heat input

=−( ) − −( )

−( )′ ′

′

h h h h

h hb c a d

b a

.

Substituting the numerical values, we obtain for the thermal efficiency with non-ideal components,ηthermal = 0 285. .

A question arises as to whether the Carnot cycle can be practically applied for power generation.The heat absorbed and the heat rejected both take place at constant temperature and pressurewithin the two-phase region. These can be closely approximated by a boiler for the heat additionprocess and a condenser for the heat rejection. Further, an efficient turbine can produce areasonable approach to reversible adiabatic expansion, because the steam is expanded with onlysmall losses. The difficulty occurs in the compression part of the cycle. If compression is carriedout slowly, there is equilibrium between the liquid and the vapor, but the rate of power generationmay be lower than desired and there can be appreciable heat transfer to the surroundings. Rapidcompression will result in the two phases coming to very different temperatures (the liquidtemperature rises very little during the compression whereas the vapor phase temperature changesconsiderably). Equilibrium between the two phases cannot be maintained and the approximationof reversibility is not reasonable.

Another circumstance is that in a Carnot cycle all the heat is added at the same temperature. Forhigh efficiency we need to do this at a higher temperature than the critical point, so that the heataddition no longer takes place in the two-phase region. Isothermal heat addition under thiscircumstance is difficult to accomplish. Also, if the heat source and the cycle are consideredtogether, the products of combustion which provide the heat can be cooled only to the highesttemperature of the cycle. The source will thus be at varying temperature while the system requiresconstant temperature heat addition, so there will be irreversible heat transfer. In summary, thepractical application of the Carnot cycle is limited because of the inefficient compression process,the low work per cycle, the upper limit on temperature for operation in the two-phase flow regime,and the irreversibility in the heat transfer from the heat source. In the next section, we examine theRankine cycle, which is much more compatible with the characteristics of two-phase media andavailable machinery for carrying out the processes.

2B-13

Muddy pointsWhat is the reason for studying two-phase cycles? (MP 2B.7)How did you get thermal efficiency? How does a boiler work? (MP 2B.8)

2.B.4 Rankine Power CyclesA schematic of the components of a Rankine cycle is shown in Figure 2B-7. The cycle is

shown on P-v, T-s, and h-s coordinates in Figure 2B-8.

The processes in the Rankine cycle are as follows:

i) d e: Cold liquid at initial temperatureT1 is pressurized reversibly to a high pressureby a pump. In this, the volume changes slightly.

ii) e a: Reversible constant pressure heating in a boiler to temperature T2

iii) a b: Heat added at constant temperature T2 (constant pressure), with transition ofliquid to vapor

iv) b c: Isentropic expansion through a turbine. The quality decreases from unity atpoint b to Xc < 1

v) c d: Liquid-vapor mixture condensed at temperature T1 by extracting heat.

Figure 2B-7: Rankine power cycle with two-phase working fluid [Moran and Shapiro,Fundamentals of Engineering Thermodynamics]

2B-14

a

a

b

b

c c′

c′

c′

c

d

e

d

e

T

T2

T2

T2

T1

T1

T1

s s

h

ae b

cd

p

p2

p1

v

(a) p-v coordinates

(b) T-s coordinates (c) h-s coordinates

Figure 2B-8: Rankine cycle diagram. (a) P-v coordinates, (b) T-s coordinates, (c) h-s coordinates.Stations correspond to those in Figure 2B-7

In the Rankine cycle, the mean temperature at which heat is supplied is less than the maximumtemperature, T2 , so that the efficiency is less than that of a Carnot cycle working between the samemaximum and minimum temperatures. The heat absorption takes place at constant pressure overeab, but only the part ab is isothermal. The heat rejected occurs over cd; this is at both constanttemperature and pressure.

To examine the efficiency of the Rankine cycle, we define a mean effective temperature, Tm interms of the heat exchanged and the entropy differences:

q T sH m=2 2∆

q T sL m=1 1∆ .

The thermal efficiency of the cycle is

ηthermalm b e m c d

m b e

T s s T s s

T s s=

−( ) − −( )−( )

2 1

2

.

The compression and expansion processes are isentropic, so the entropy differences are related by

2B-15

s s s sb e c d− = − .

The thermal efficiency can be written in terms of the mean effective temperatures as,

ηthermalm

m

T

T= −1 1

2

.

For the Rankine cycle, T T T Tm m1 1 2 2≈ <, . From Equation (B.4.1), we see not only the reason that

the cycle efficiency is less than that of a Carnot cycle, but the direction to move in terms of cycledesign (increased Tm2

) if we wish to increase the efficiency.

There are several features that should be noted about Figure 2B-8 and the Rankine cycle ingeneral:

i) The T-s and the h-s diagrams are not similar in shape, as they were with the perfect gas.The slope of a constant pressure reversible heat addition line is, as derived in Section 1.C.4,

∂∂h

sT

p

= . In the two-phase region, constant pressure means also constant temperature,

so the slope of the constant pressure heat addition line is constant and the line is straight.ii) The effect of irreversibilities is represented by the dashed line from b to ′c . Irreversible

behavior during the expansion results in a value of entropy. sc′ .at the end state of theexpansion that is higher than sc. The enthalpy at the end of the expansion (the turbine exit)is thus higher for the irreversible process than for the reversible process, and, as seen forthe Brayton cycle, the turbine work thus lower in the irreversible case.

iii) The Rankine cycle is less efficient than the Carnot cycle for given maximum and minimumtemperatures, but, as said earlier, it is more effective as a practical power productiondevice.

Muddy pointsWhere does degrees Rankine come from? Related to Rankine cycles? (MP 2B 9)

2.B.5: Enhancements of, and Effect of Design Parameters on, Rankine CyclesThe basic Rankine cycle can be enhanced through processes such as superheating and

reheat. Diagrams for a Rankine cycle with superheating are given in Figure 2B-9. The heataddition is continued past the point of vapor saturation, in other words the vapor is heated so thatits temperature is higher than the saturation temperature associated with P P P Pa b c d ( )= = = . This

2B-16

a

f

a

b

bc

e′

e′

e′

c

d

e f

d

e

T

T2

T3

T2T3

T3T2

T1

T1

T1

s s

h

ba c

ef

d

p

p2

p1

v(a)

(b) (c)

Figure 2B-9: Rankine cycle with superheating

does several things. First, it increases the mean temperature at which heat is added, Tm2, thus

increasing the efficiency of the cycle (see Equation B.4.1). Second is that the quality of the two-phase mixture during the expansion is higher with superheating, so that there is less moisturecontent in the mixture as it flows through the turbine. (The moisture content at e is less than that at

′e .) This is an advantage in terms of decreasing the mechanical deterioration of the blading.

The heat exchanges in the superheated cycle are;

Along abcd, which is a constant pressure (isobaric) process: q h hd a2 = − .Along ef: q h hf e1 = − , (< 0).

The thermal efficiency of the ideal Rankine cycle with superheating is

ηthermald a e f

d a

h h h h

h h=

− − −( )−

This can be expressed explicitly in terms of turbine work and compression (pump) work as:

2B-17

ηthermald e a f

d a

h h h h

h h=

− − −( )−

.

Compared to the basic cycle, superheating has increased the turbine work, increased the meantemperature at which heat is received, Tm2

, and increased the cycle efficiency.

A comparison of the Carnot cycle and the Rankine cycle with superheat is given in Figure 2B-10.The maximum and minimum temperatures are the same, but the average temperature at which heatis absorbed is lower for the Rankine cycle.

T

T1

T2

s

f e

g d

cba

fa

bc

d

e

=Rankine

= Carnotd

e

g

f

Isothermal

Isentropic

Figure 2B-10: Comparison of Rankine cycle with superheat and Carnot cycle

To alleviate the problem of having moisture in the turbine, one can heat again after an initialexpansion in a turbine, as shown in Figure 2B-11, which gives a schematic of a Rankine cycle forspace power application. This process is known as reheat. The main practical advantage of reheat(and of superheating) is the decrease in moisture content in the turbine because most of the heataddition in the cycle occurs in the vaporization part of the heat addition process.

We can also examine the effect of variations in design parameters on the Rankine cycle. Considerfirst the changes in cycle output due to a decrease in exit pressure. In terms of the cycle shown inFigure 2B-12, the exit pressure would be decreased from P P dPexit exit exit to −( ). The original cycle

is abcdea, and the modified cycle is abfgha. The consequences are that the cycle work, which is

2B-18

1′

2

34

Nuclearheat

source

Pump

CondensingRadiator

Electricpower

Generator

Turbine

Reheat

1st stage

2nd stage

Win

Qreject

5

4

32

2′

4′

4′

2′

1′

Q′

5

Entropy (s)

Tem

pera

ture

(T

)

Rankine cycle with reheat

Figure 2B-11: Rankine cycle with superheating and reheat for space power application

12

2′

1′

3

4

4′

T

sa′

p4 ′

p4

a b

Figure 2B-12: Effect of exit pressure on Rankine cycle efficiency

the integral of Tds around the cycle, is increased. In addition, as drawn, although the levels of themean temperature at which the heat is absorbed and rejected both decrease, the largest change inthe mean temperature of the heat rejection, so that the thermal efficiency increases.

Another design parameter is the maximum cycle pressure. Figure 2B-13 shows comparison of twocycles with different maximum pressure but the same maximum temperature, which is set bymaterials properties. The average temperature at which the heat is supplied for the cycle with ahigher maximum pressure is increased over the original cycle, so that the efficiency increases.

2B-19

122′

3′ 3

44′

T

sa bb′

Figure 2B-13: Effect of maximum boiler pressure on Rankine cycle efficiency

Muddy pointsWhy do we look at the ratio of pump (compression) work to turbine work? We did not dothat for the Brayton cycle. (MP 2B.10)Shouldn't the efficiency of the super/re-heated Rankine cycle be larger because its area isgreater? (MP 2B.11)Why can't we harness the energy in the warm water after condensing the steam in apower plant? (MP 2B.12)

2.B.6 Combined Cycles in Stationary Gas Turbine for Power ProductionThe turbine entry temperature in a gas turbine (Brayton) cycle is considerably higher than

the peak steam temperature. Depending on the compression ratio of the gas turbine, the turbineexhaust temperature may be high enough to permit efficient generation of steam using the “wasteheat” from the gas turbine. A configuration such as this is known as a gas turbine-steam combinedcycle power plant. The cycle is illustrated in Figure 2B-14.

Figure 2B-14: Gas turbine-steam combined cycle [Kerrebrock, Aircraft Engines and GasTurbines]

The heat input to the combined cycle is the same as that for the gas turbine, but the work output islarger (by the work of the Rankine cycle steam turbine). A schematic of the overall heat engine,

2B-20

which can be thought of as composed of an upper and a lower heat engine in series, is given inFigure 2B-15. The upper engine is the gas turbine (Brayton cycle) which expels heat to the lower

Figure 2B-15: Schematic of combine cycle using gas turbine (Brayton cycle) and steam turbine(Rankine cycle) [Langston]

engine, the steam turbine (Rankine cycle). The overall efficiency of the combined cycle can bederived as follows. We denote the heat received by the gas turbine as Qin and the heat rejected tothe atmosphere as Qout . The heat out of the gas turbine is denoted as Q1 . The hot exhaust gasesfrom the gas turbine pass through a heat exchanger where they are used as the heat source for thetwo-phase Rankine cycle, so that Q1 is also the heat input to the steam cycle. The overallcombined cycle efficiency is

ηCCin

B R

in

W

Q

W W

Q= =

+,

where the subscripts refer to combined cycle (CC), Brayton cycle (B) and Rankine cycle (R)respectively.

From the first law, the overall efficiency can be expressed in terms of the heat inputs and heatrejections of the two cycles as (using the quantity Q1 to denote the magnitude of the heattransferred):

ηCCin out

in in

out

in

Q Q Q Q

Q

Q

Q

Q

Q

Q

Q=

− + −( )= −

+ −

1 1 1

1

11 1 .

2B-21

Tmax

Tmin

T

s

The first square bracket term on the right hand side is the Brayton cycle efficiency, ηB , the secondis the Rankine cycle efficiency, ηR , and the term in parentheses is (1- ηB). The combined cycleefficiency can thus be written as,

η η η η ηCC B R B R= + − ; Combined cycle efficiency. (B.6.1)

Equation (B.6.1) gives insight into why combined cycles are so successful. Suppose that the gasturbine cycle has an efficiency of 40%, which is a representative value for current Brayton cyclegas turbines, and the Rankine cycle has an efficiency of 30%. The combined cycle efficiencywould be 58%, which is a very large increase over either of the two simple cycles. Somerepresentative efficiencies and power outputs for different cycles are shown in Figure 2B-16.

Figure 2B-16: Comparison of efficiency and power output of various power products [Bartol(1997)]

2.B.7 Some Overall Comments on Thermodynamic Cyclesi) There are many different power and propulsion cycles, and we have only looked at a few of

these. Many other cycles have been devised in the search for ways to increase efficiencyand power in practical devices.

ii) We can view a given cycle in term ofelementary Carnot cycles, as sketched inthe figure on the right. This shows thatthe efficiency of any other cycle operatingbetween two given temperatures will beless than that of a Carnot cycle.

2B-22

iii) If we view the thermal efficiency as

ηthermalAverage

Average

T

T= −

( )( )

1heat rejected

heat absorbed

,

(derived in Section 2.B.4), this means that we should accept heat at a high temperature and reject itat a low temperature for high efficiency. This objective must be tempered by considerations ofpractical application.

iv) The cycle diagrams in T-s and h-s coordinates will only be similar if the working mediumis an ideal gas. For other media (for example, a two-phase mixture) they will lookdifferent.

v) Combined cycles make use of the rejected heat from a “topping” cycle as heat source for a“bottoming” cycle. The overall efficiency is higher than the efficiency of either cycle.

Muddiest Points on Part 2B

2B.1 For the vapor dome, is there vapor and liquid inside the dome and outside is it justliquid or just gas? Is it interchangeable? Is it true for the plasma phase?

The vapor dome separates the two-phase region from the single-phase region. Inside, wehave a mixture of liquid and vapor. The peak of the vapor dome is called the criticalpoint. The left-hand side leg of the vapor dome (from the critical point) is called thesaturated liquid line along which the quality x is zero (purely liquid). The right-hand sideleg is denoted the saturated vapor line and the quality x is one (purely vapor). For furtherdetails see the notes.

Heating of a solid or liquid substance leads to phase transition to a liquid or gaseous state,respectively. This takes place at a constant temperature for a given pressure, and requiresan amount of energy known as latent heat. On the other hand, the transition from a gas toan ionized gas, i.e., plasma, is not a phase transition, since it occurs gradually withincreasing temperature. During the process, a molecular gas dissociates first into anatomic gas which, with increasing temperature, is ionized. Resulting plasma consists of amixture of neutral particles, positive ions (atoms or molecules that have lost one or moreelectrons), and negative electrons.

2B.2 What is hfg ? How do we find it?

The quantity hfgrepresents the specific enthalpy change between the liquid and vapor

phases of a substance at constant temperature, and thus constant pressure, and thusconstant temperature. It is therefore the heat input, per unit mass, to vaporize a kilogramof liquid. See the notes, Section 2.B.2.

2B.3 Reasoning behind the slopes for T=cst lines in the P-V diagram.

The slope of an isotherm in the gaseous phase (to the right of the vapor dome) is similarto the slope we found for the isotherm of an ideal gas (PV=const). Inside the vapor domepressure and temperature are directly related to one another (P = P(T), vapor pressurecurve) such that an isotherm is a horizontal line (isobar inside the vapor dome). In theliquid phase the isotherms are very steep lines, because for liquids the volume is aboutconstant (very low compressibility).

2B.4 For a constant pressure heat addition, why is q=∆h?

The combined first and second law is dh Tds vdP= + . For a reversible constant pressureprocess, dP = 0, and the heat input, dq, is TdS. Thus for a reversible constant pressureprocess, the answer is yes.

For an irreversible process we can say from the steady flow energy equation:dq dh dh cdct= = + . For a steady flow, the one-dimensional momentum equation is

cdcdP Fviscous= − +ρ ρ

,

where Fviscous represents the viscous forces in an irreversible flow. Combining these twoexpressions, and using dP = 0 (the condition of constant pressure) gives

dq dh cdc dhFviscous= + = +ρ

.

Without going into any detail concerning the form of the viscous forces, this equationshows that the equality between heat input and enthalpy change does not hold for generalirreversible flow processes at constant pressure.

2B.5 What is latent heat?

Latent heat is a term for the enthalpy change needed for vaporization.

2B.6 Why is U a function of x?

Inside the vapor dome we have a mixture of liquid and vapor. The internal energy U ofthe system (liquid and vapor) can be expressed in terms of the mass in each phase and thespecific internal energy u of each phase as,

U = uf mf + ug mg.

Introducing the quality x as the fraction of the total mass in the vapor phase x =mg/(mg+mf) we can write

U = (1 – x) uf + x ug.

Since the specific energy of the saturated liquid and the saturated vapor are functions oftemperature the internal energy U of the two-phase system is a function of x and T (seealso notes on page 2B-6).

2B.7 What is the reason for studying two-phase cycles?

Their immense practical utility in a number of industrial devices and their intrinsicinterest as applications of the basic principles.

2B.8 How did you get thermal efficiency? How does a boiler work?

The thermal efficiency is, as previously, the net work done divided by the heat input.Using the first law for a control volume we can write both of these quantities in terms ofthe enthalpy at different states of the cycle.

For the steam cycles discussed in class, a boiler is a large (as in the viewgraph of theMitsubishi power plant) structure with a lot of tubes running through it. The water (orwhatever medium is used in the cycle) runs through the tubes. Hot gases wash over the

outside of the tubes. The hot gases could be from a combustor or from the exit of a gasturbine.

2B.9 Where does degrees Rankine come from? Related to Rankine cycles?

I think the answer is yes, although I do not know for sure. If so, this is the same Rankinewho has his name on the Rankine-Hugoniot conditions across a shock wave.

2B.10 Why do we look at the ratio of pump (compression) work to turbine work? We didnot do thatfor the Brayton cycle.

If the ratio of compression work to turbine work were close to unity for an ideal cycle,small changes in component efficiencies would have large effects on cycle efficiency andwork. For the Rankine cycle this is not true. (The effect of pump efficiency on Rankinecycle efficiency is clearly small in the class example.) For the Brayton cycle, where thenet work is the difference of two numbers which are of (relatively) similar sizes, theeffect of compressor and turbine efficiency on cycle efficiency can be much larger.

I used the word “sensitive” and the meaning was that the cycle performance respondedstrongly to changes in the compressor and turbine behavior.

2B.11 Shouldn't the efficiency of the super/re-heated Rankine cycle be larger because itsarea is greater?

The area enclosed by an ideal cycle in a T-s diagram is the net work done, but it does nottell you about efficiency. We saw that for example when we looked at the Brayton cyclefor the condition of maximum work, rather than maximum efficiency (Section 1.A.4 inthe notes).

2B.12 Why can't we harness the energy in the warm water after condensing the steam ina power plant?

Let's assume the temperature of the warm water after condensing the steam is at atemperature of about 30 to 40 degrees C. If we consider running a heat engine betweenthis heat reservoir (say 35 degrees C) and the surroundings at 20 degrees C, we would getan ideal thermal efficiency of about 5%. In other words, the available useful work isrelatively small if we considered the lower heat reservoir to be the surroundings. Ingeneral, there is a property that only depends on state variables called availability. Thechange in availability gives the maximum work between two states, where one state isreferred to the surroundings (dead state).

Part 2.C: Introduction to Thermochemistry [SB&VW-14.1-14.6]

Until now, we have specified the heat given to the devices analyzed, and not concerned ourselves with how this heat might be produced. In this section, we examine the issue of how we obtain the heat needed for work production. For the most part, this is from converting chemical energy into heat, so the discussion will be on reacting mixtures of gas which are involved in chemical combustion processes.

The topic addressed is “thermochemistry”, which is the combining of thermodynamics with chemistry to predict such items as how much heat is released from a chemical reaction. This is the “Q” or “q” that we have used in the cycle analysis. The principal components of the approach are use of a chemical balance plus the steady flow energy equation (SFEE) which equates the sum of shaft work (from) and heat transfer (to) a control volume to the difference in control volume inlet and exit enthalpy fluxes.

2.C.1 FuelsThere are a wide variety of fuels used for aerospace power and propulsion. A primary one

is jet fuel (octane, essentially kerosene) which has the chemical formula C H18

. Other fuels we8

consider are hydrogen (H2 ) and methane ( CH4 ).

The chemical process in which a fuel, for example methane, is burned consists of (on a very basic level—there are many intermediate reactions that need to be accounted for when computations of the combustion process are carried out):

CH4 + 2O2 → CO2 + 2H O .2

(Reactants) (Products)

The reactions we describe are carried out in air, which can be approximated as 21% O2 and 79% N2 . This composition is referred to as “theoretical air”. There are other components of air (for example Argon, which is roughly 1%), but the results given using the theoretical air approximation are more than adequate for our purposes. With this definition, for each mole of O2 , 3.76 (79/21) moles of N2 are involved:

CH4 + 2O2 + 2 3 76 )N → CO2 + 2H O + 7.52N2( . 2 2

Even if the nitrogen is not part of the combustion process, it leaves the combustion chamber at the same temperature as the other products, and this change in state (change in enthalpy) needs to be accounted for in the steady flow energy equation. At the high temperatures achieved in internal combustion engines (aircraft and automobile) reaction does occur between the nitrogen and oxygen, which gives rise to oxides of nitrogen, although we will not consider these reactions.

The condition at which the mixture of fuel and air is such that both completely participate in the reaction is called stoichiometric. In gas turbines, excess air is often used so that the

2C-1

temperatures of the gas exiting the combustor is kept to within desired limits (see Figures A-8, A-9, A-11 in Part 1 for data on these limits.)

Muddy pointsWhy is there 3.76 N2? (MP 2C.1)What is the most effective way to solve for the number of moles in the reactions? (MP2C.2)

2.C.2 Fuel-Air Ratio The reaction for aeroengine fuel at stoichiometric conditions is:

C H +12.5 O2 +12 5 (3 76 ) N → 8 CO2 + 9 H O + 47.0 N . . 28 18 2 2

On a molar basis, the ratio of fuel to air is [1/(12.5+47.0)] = 1/59.5 = 0.0167.

To find the ratio on a mass flow basis, which is the way in which the aeroengine industry discusses it, we need to “weight” the molar proportions by the molecular weight of the components. The fuel molecular weight is 114, the oxygen molecular weight is 32 and the nitrogen molecular weight is (approximately) 28. The fuel/air ratio on a mass flow basis is thus

1 ×114Fuel-air ratio = = 0 0664 .

12 5 × 32 +12 5 × 3 76 × 28 . . .

If we used the actual constituents of air we would get 0.0667, a value about 0.5% different.

Muddy pointsDo we always assume 100% complete combustion? How good an approximation is this? (MP 2C.3)

2.C.3 Enthalpy of formation The systems we have worked with until now have been of fixed chemical composition.

Because of this, we could use thermodynamic properties relative to an arbitrary base, since all comparisons could be made with respect to the chosen base. For example, the specific energy u f ( . o .0 01 C) = 0 0 for steam. If there are no changes in composition, and only changes in

properties of given substances, this is adequate. If there are changes in composition, however, we need to have a reference state so there is consistency for different substances.

The convention used is that the reference state is a temperature of 25oC (298 K) and a pressure of 0.1 MPa. (These are roughly room conditions.) At these reference conditions, the enthalpy of the elements (oxygen, hydrogen, nitrogen, carbon, etc.) is taken as zero.

The results of a combustion process can be diagrammed as below. The reactants enter at standard conditions; the combustion (reaction) takes place in the volume indicated. Downstream of the reaction zone there is an appropriate amount of heat transfer with the surroundings so that the products leave at the standard conditions. For the reaction of carbon and oxygen to produce CO2, the heat that has to be extracted is QCV = −393 522 kJ/kmole ; this is heat that comes out of,the control volume.

2C-2

C02C + 02

1 kmole C 1 kmole C02 25o C, 0.1 MPa

25o C, 0.1 MPa Volume

1 kmole C02

= -393,522 KJ, heat is out of control volumeQcv

Figure C-1: Constant pressure combustion

There is no shaft work done in the control volume and the first law for the control volume (SFEE) reduces to:

mass flow of enthalpy in + rate of heat addition = mass flow of enthalpy out.

We can write this statement in the form

˙∑m h + QCV = ∑m he (C.3.1)˙ i i e R P

In Eq. (C.3.1) the subscripts “R” and “P” on the summations refer to the reactants (R) and products (P) respectively. The subscripts on the mass flow rates and enthalpies refer to all of the components at inlet and at exit.

The relation in terms of mass flows can be written in molar form, which is often more convenient for reacting flow problems, by using the molecular weight, Mi , to define the molar mass flow rate,

ni , and molar enthalpy, hi , for any individual ith (or eth) component as

ni = mi / M ; mass flow rate in terms of kmoles/seci

hi = M hi ; enthalpy per kmolei

The SFEE is, in these terms,

˙∑n h + QCV = ∑n he . (C.3.2)˙i i e R P

The statements that have been made do not necessarily need to be viewed in the context of flow processes. Suppose we have one unit of C and one unit of O2 at the initial conditions and we carry out a constant pressure reaction at ambient pressure, Pamb . If so,

2C-3

= Q −WU final −Uinitial

= QCV − Pamb (Vfinal −Vinitial ) ,

since P = Pf = Pamb . Combining terms,i

U final + PfinalVfinal − (Uinitial + PinitialVinitial ) = QCV ,

or,

= QCV .H final − Hinitial

In terms of the numbers of moles and the specific enthalpy this is

∑n h +QCV = ∑n he (C.3.3)i i eR P

The enthalpy of CO2, at 25oC and 0.1 MPa, with reference to a base where the enthalpy of the oelements is zero, is called the enthalpy of formation and denoted by hf . Values of the heat of

formation for a number of substances are given in Table A.9 in SB&VW.

The enthalpies of the reactants and products for the formation of CO2 are:

= hC = 0

h

hO2 of ,For one kmole: QCV = ∑n he = HP = ( ) = −393 522 kJ/kmole.e

P CO2

The enthalpy of CO2 in any other state (T,P)is given by

hT P = ( ) + (∆h ) . , h0 f 298K, . 0 1 Mpa→T P 0 1 MPa 298K, . ,

These descriptions can be applied to any compound. For elements or compounds that exist in more than one state at the reference conditions (for example, carbon exists as diamond and as graphite), we also need to specify the state.

Note that there is a minus sign for the heat of formation. The heat transfer is out of the control ,volume and is thus negative by our convention. This means that helements > = − 393 522 kJ .hCO2

Muddy pointsIs the enthalpy of formation equal to the heat transfer out of the combustion during theformation reaction? (MP 2C.4)Are the enthalpies of H2 and H (monoatomic hydrogen) both zero at 298K? (MP 2C.5)

2C-4

2.C.4 First Law Analysis of Reacting Systems The form of the first law for the control volume is (there is no shaft work):

˙∑n h + QCV = ∑n h .˙i i e eR P

This is given in terms of the moles of the different constituents, and it reduces to the more familiar form for a single fluid (say air) with no reactions occurring. We need to specify one parameter as the basis of the solution; 1 kmole of fuel, 1 kmole of air, 1 kmole total, etc. We use 1 kmole of fuel as the basic unit and examine the burning of hydrogen.

H2 + 20 → 2H O 2 2

The reactants and the products are both taken to be at 0.1MPa and 25oC, so the inlet and exit P and T are specified. The control volume is the combustion chamber. There is no shaft work done and the SFEE is in the form of Equation (C.1.2). The enthalpy of the entering gas is zero for both the hydrogen and the oxygen (elements have enthalpies defined as zero at the reference state). If the exit products are in the gaseous state, the exit enthalpy is therefore related to the enthalpy of formation of the product by:

ohfneH Ohe H O = ne ( )

2 ( )H O H O g 2 2 2

= 2 x (- 241,827)kJ = - 483,654 kJ; gaseous state at exit.

If the water is in a liquid state at the exit of the process: ohfneH O

he H O = ne ( )2 ( )H O H O l 2 2 2

= 2 x (- 285,783) kJ = - 571, 676.

There is more heat given up if the products emerge as liquid. The difference between the two values is the enthalpy needed to turn the liquid into gas at 25oC: hfg = 2442 kJ/kmole.

A more complex example is provided by the burning of methane (natural gas) in oxygen, producing .

CH4 + 2O2 → CO2 + 2H O ( )l2

The components in this reaction equation are three ideal gases (methane, oxygen, and CO2) and liquid water. We again specify that the inlet and exit states are at the reference conditions so that:

∑n h hfo ,i i = ( ) = −74 873 kJ

R CH4

e e = ( ) + 2( )n h h∑ fo

CO2 hf

o

H O l P 2 ( )

Q =-393,522 + 2(-285,838) = -965,198 kJ

CV = −965 198 kJ – (-74,873) = -890,325 kJ.,

2C-5

Suppose the substances which comprise the reactants and the products are not at 25oC and 0.1MPa. If so, the expression that connects the reactants and products is;

h o f + h

Pi and reference conditions

∆ e

h o f + h

Pe and reference conditions

∆ . (C.4.1)QCV + ∑ = ∑n ni R PBetween Between

Ti Te, ,

i e

Equation (C.4.1) shows that we must compute the enthalpy difference ∆h between the reference conditions and the given state if the inlet or exit conditions are not the reference pressure and temperature.

There are different levels of approximation for the computation: (a) assume the specific heat is constant over the range at some average value, (b) use the polynomial expressions (Table A.6) in the integral, and (c) use tabulated values. The first is the simplest and the crudest. Combustion processes often involve changes of a thousand degrees or more and, as Figure C-2 shows, the specific heat for some gases can change by a factor of two or more over this range, although the changes for air are more modest. This means that, depending on the accuracy desired, one may need to consider the temperature dependence of the specific heat in computing ∆h .

Figure C-2: Specific heat as a function of temperature [from SB&VW]

Muddy pointsWhen doing cycle analysis, do we have to consider combustion products and their effect on specific heat ratio (γ is not 1.4)? (MP 2C.6)

2C-6

2.C.5 Adiabatic Flame Temperature For a combustion process that takes place adiabatically with no shaft work, the temperature

of the products is referred to as the adiabatic flame temperature. This is the maximum temperature that can be achieved for given reactants. Heat transfer, incomplete combustion, and dissociation, all result in lower temperature. The maximum adiabatic flame temperature for a given fuel and oxidizer combination occurs with a stoichiometric mixture (correct proportions such that all fuel and all oxidizer are consumed). The amount of excess air can be tailored as part of the design to control the adiabatic flame temperature. The considerable distance between present temperatures in a gas turbine engine and the maximum adiabatic flame temperature at stoichiometric conditions is shown in Figure A-11 of Part 1, based on a compressor exit temperature of 1200oF (922 K).

An initial view of the concept of adiabatic flame temperature is provided by examining two reacting gases, at a given pressure, and asking what the end temperature is. The process is shown schematically at the right,

1 2

∆h1

Constant P

aActual path

∆h f = Final state where temperature is plotted versus the percentage completion of the reaction. T ∆h2 The initial state is i and the final state is f, with the final state at a higher Constant P temperature than the initial state. The State i 2 solid line in the figure shows a representation of the “actual” process. 0 Percentage 100%To see how we would arrive at the final

completionstate the dashed lines break the state of reaction change into two parts. Process (1) is reaction at constant T and P. To carry out such a process, we would need to extract heat. Suppose the total amount of heat extracted per unit mass is q1. The relation between Schematic of adiabatic flame temperature the enthalpy changes in Process (1) is

h2 − = − q1 = ( )unithi hof

mass

where q1 is the “heat of reaction”.

For Process (2), we put this amount back into the products to raise their temperature to the final level. For this process, h − h2 = q1, or, if we can approximate the specific heat as constant (usingf

some appropriate average value) cpav. (T − T2 ) = q1 . For the overall process there is no work donef

and no heat exchanged so that the difference in enthalpy between initial and final states is zero:

∆h1 + ∆h = 0.2 = ∆hadiabatic

The temperature change during this second process is therefore given by (approximately)

2C-7

ohf( )unit mass1(T − T2 ) =

q = . (C.5.1)f c cpav. pav.

The value of the adiabatic flame temperature given in Equation (C.5.1) is for 100% completion of the reaction. In reality, as the temperature increases, the tendency is for the degree of reaction to be less than 100%. For example, for the combustion of hydrogen and oxygen, at high temperatures the combustion product (water) dissociates back into the simpler elemental reactants. The degree of reaction is thus itself a function of temperature that needs to be computed. We used this idea in discussing the stoichiometric ramjet, when we said that the maximum temperature was independent of flight Mach number and hence of inlet stagnation temperature. It is also to be emphasized that the idea of a constant (average) specific heat, cpav.

, is for illustration

and not inherently part of the definition of adiabatic flame temperature.

An example computation of adiabatic flame temperature is furnished by the combustion of liquid octane at 25oC with 400% theoretical air. The reaction is

C H ( ) +12.5O +12.5 3.76N2 ) + 3 12.5O +12.5 3.76N2 )]→ 8CO + 9H O g 8 18 l 2 ( [ 2 ( 2 2 ( ) + 37.5O2 +188N2 .

For an adiabatic process

o o∑n hf + ∆h ) = ∑ne (hf + ∆h ) . (C.5.2)i ( i eR P

At adiabatic flame temperature

We can again think of the general process in steps:

a) Bring reactants to 25oC [the term (∆h ) ] from the initial temperature, using whateveri

heat transfer, qa , is needed. In this example we do not need step (i) because we are already at the reference temperature.

ob) Reaction at 25oC - the term ( ) . There will be some heat transfer in this step,h f reactants→products qb , out of the combustor.c) Put back heat q + qb into the products of combustion. The resulting temperature is thea

adiabatic flame temperature.

In the present case Equation (C.1.6) is, explicitly:

o o ohf ( ) = 8hf CO2 + 9hf +∆hCO2

+ 9∆hH2O + 37.5∆h +188∆hN2 C8H18 l

H2O O2

We can examine the terms in the SFEE separately, starting with the heat of formation terms,

o o o ohf : 8hf CO2 + 9hf − hf ( )= 8 (-393,522) + 9 (-241,827) – (-249,952)l

H2O C8H18

2C-8

= -5.075 X 106 kJ/kmole. The exit state at the adiabatic flame temperature is specified by:

n h∑ ∆ = 5.075 X 106 kJ/kmolee eP

We find the adiabatic flame temperature in three ways, approximate solution using an average value of cp , a more accurate one using the tabulated evolution of cp with temperature and a more

precise solution using the tabulated values for gas enthalpy in Table A.8 of SB&VW.

a) Approximate solution using “average” values of specific heat: From Figure C-2 we can use the values at 500K as representative. These are:

Gas cp (kJ/kmole)

CO2 45H2O 35 O2 30N2 30.

Using ∆h cp ∆T ,= "ave"

∆ c c c c∑n he = ∆T8( ) + 9( ) + 37.5( ) +188( ) e p CO2 p H2O p O2

pP N2

o =where ∆T Tfinal − 25 C = Tfinal − 298 K

∆∑n he = 7440∆T kJeP

∆T = 682 ⇒ Tfinal = 980 K .

b) Solution for adiabatic flame temperature using evolutions of specific heats with temperature

Tables give the following evolutions of specific heats with temperature: Gas Evolution of cp/ R with T (kJ/kmol) CO2 2.401+8.735.10-3xT-6.607.10-6xT2+2.002.10-9xT3

H2O 4.070-1.108.10-3xT+4.152.10-6xT2-2.964.10-9xT3+0.807.10-12xT4

O2 3.626-1.878.10-3xT+7.055.10-6xT2-6.764.10-9xT3+2.156.10-12xT4

N2 3.675-1.208.10-3xT+2.324.10-6xT2-0.632.10-9xT3-0.226.10-12xT4

Tf in K

Using ∆h = ∫ c ( )T .dT and the same equation as above, we obtain:p298 K

Tf= 899 K

c) Solution for adiabatic flame temperature using tabulated values for gas enthalpy:

∆hCO2 ∆hH2O ∆hO2

∆hN2

T= 900 K 28,041 21,924 19,246 18,221 kJ/kmole T=1000K 33,405 25,978 22,707 21,460 kJ/kmole

2C-9

Plugging in the numbers shows the answer is between these two conditions. Linearly interpolating gives a value of Tfinal = 962 K .

Muddy pointsDoes "adiabatic flame temperature" assume 100% combustion? (MP 2C.7)What part of the computation for adiabatic flame temperature involves iteration? (MP2C.8)

2C-10

Muddiest points on part 2C

2C.1 Why is there 3.76 N2?

This is to represent the components other than oxygen that are in air. From SB&VW, page 525, “The assumption that air is 21% oxygen and 79% nitrogen by volume leads to the conclusion that for each mole of oxygen, 79/21 =3.76 moles of nitrogen are involved.”

2C.2 What is the most effective way to solve for the number of moles in the reactions?

What we are doing is basically counting atoms on both sides of the reactantsproducts statement. See Section 14.2 of SB&VW for a description of the combustion process and for going from ratios in terms of moles to ratios in terms of mass.

2C.3 Do we always assume 100% complete combustion? How good an approximation is this?

In the problems we do, we will only consider 100% combustion. It is a good approximation for the range of problems that we address.

2C.4 Is the enthalpy of formation equal to the heat transfer out of the combustion during the formation reaction?

As defined, the enthalpy of formation relates to a process in which the initial and final states are at the same temperature. If there is combustion in between, heat will have to be removed for this condition to occur. The enthalpy of formation is equal to the negative of the magnitude of the heat outflow. If we consider the combustion as occurring in a control volume, then per kmole hf

o = −Qcv , where Qcv is the heat transfer

out of the control volume per kmole. This is not in accord with our convention, and if you please feel free to transform it back into the notation we have used before. (I find that if I do this there are too many minus signs to keep track of easily.)

2C.5 Are the enthalpies of H2 and H (monoatomic hydrogen) both zero at 298K?

The enthalpies of the elements are taken as zero at 298 and 0.1 MPa. In some cases there are more than one form of the element. In that case the form chosen to have the value of zero is that which is chemically stable at the reference state. The other forms then have an enthalpy which is consistent with the reaction that produces this form of the element. For hydrogen H2 has zero enthalpy at the reference conditions and H has an enthalpy of 217,999 kJ/kmole (see Table A.8 in SB&VW), consistent with the idea that energy has to be supplied to break the molecule apart.

2C.6 When doing a cycle analysis, do we have to consider combustion products and their effect on specific heat ratio (γ is not 1.4)?

The specific heat ratio does depend on combustion products but the effect is not large because the fuel air ratio is small. For example, for conditions of fuel air ratio ).034, the specific heat ratio at room temperature is about 1.38. A larger variation encountered in practice is with temperature; for a temperature of 1750K the specific heat ratio of pure air is 1.3.

2C.7 Does "adiabatic flame temperature" assume 100% combustion?

Yes. This is the maximum temperature that could be produced. Incomplete combustion will lower the temperature, as will heat transfer out of the combustion region.

2C.8 What part of the computation for adiabatic flame temperature involves iteration?

If the specific heat is not a simple analytic function of temperature (i.e., suppose it is known only in tabular form), we cannot get a closed form solution for the adiabatic flame temperature. We can, however, readily solve the enthalpy balance (SFEE) numerically (this is where the iteration comes in) to find at what temperature the products have to come out to have the same enthalpy (including the enthalpy of formation) as the reactants. We did not do this calculation yet, but we will do it to show what the iteration is all about.

Remember that the assumption of constant specific heat is just that, an assumption. While this is an excellent assumption for many practical problems, if the precision of the answer needed is very high, or if the range of temperatures is large (see Figure 5.11 in SB&VW), then we cannot assume constant specific heat.

PART 3

INTRODUCTION TO ENGINEERING HEAT TRANSFER

HT-1

Introduction to Engineering Heat Transfer

These notes provide an introduction to engineering heat transfer. Heat transfer processes set limitsto the performance of aerospace components and systems and the subject is one of an enormousrange of application. The notes are intended to describe the three types of heat transfer and providebasic tools to enable the readers to estimate the magnitude of heat transfer rates in realistic aerospaceapplications. There are also a number of excellent texts on the subject; some accessible referenceswhich expand the discussion in the notes are listen in the bibliography.

HT-2

Table of Tables

Table 2.1: Thermal conductivity at room temperature for some metals and non-metals ............. HT-7Table 2.2: Utility of plane slab approximation..........................................................................HT-17Table 9.1: Total emittances for different surfaces [from: A Heat Transfer Textbook, J. Lienhard ]HT-63

HT-3

Table of Figures

Figure 1.1: Conduction heat transfer ......................................................................................... HT-5Figure 2.1: Heat transfer along a bar ......................................................................................... HT-6Figure 2.2: One-dimensional heat conduction ........................................................................... HT-8Figure 2.3: Temperature boundary conditions for a slab............................................................ HT-9Figure 2.4: Temperature distribution through a slab .................................................................HT-10Figure 2.5: Heat transfer across a composite slab (series thermal resistance) ............................HT-11Figure 2.6: Heat transfer for a wall with dissimilar materials (Parallel thermal resistance)........HT-12Figure 2.7: Heat transfer through an insulated wall ..................................................................HT-11Figure 2.8: Temperature distribution through an insulated wall ................................................HT-13Figure 2.9: Cylindrical shell geometry notation........................................................................HT-14Figure 2.10: Spherical shell......................................................................................................HT-17Figure 3.1: Turbine blade heat transfer configuration ...............................................................HT-18Figure 3.2: Temperature and velocity distributions near a surface. ...........................................HT-19Figure 3.3: Velocity profile near a surface................................................................................HT-20Figure 3.4: Momentum and energy exchange in turbulent flow. ...............................................HT-20Figure 3.5: Heat exchanger configurations ...............................................................................HT-23Figure 3.6: Wall with convective heat transfer .........................................................................HT-25Figure 3.7: Cylinder in a flowing fluid .....................................................................................HT-26Figure 3.8: Critical radius of insulation ....................................................................................HT-29Figure 3.9: Effect of the Biot Number [hL / kbody] on the temperature distributions in the solid and in

the fluid for convective cooling of a body. Note that kbody is the thermal conductivity of thebody, not of the fluid.........................................................................................................HT-31

Figure 3.10: Temperature distribution in a convectively cooled cylinder for different values of Biotnumber, Bi; r2 / r1 = 2 [from: A Heat Transfer Textbook, John H. Lienhard] .....................HT-32

Figure 4.1: Slab with heat sources (a) overall configuration, (b) elementary slice.....................HT-32Figure 4.2: Temperature distribution for slab with distributed heat sources ..............................HT-34Figure 5.1: Geometry of heat transfer fin .................................................................................HT-35Figure 5.2: Element of fin showing heat transfer ......................................................................HT-36Figure 5.3: The temperature distribution, tip temperature, and heat flux in a straight one-

dimensional fin with the tip insulated. [From: Lienhard, A Heat Transfer Textbook, Prentice-Hall publishers].................................................................................................................HT-40

Figure 6.1: Temperature variation in an object cooled by a flowing fluid .................................HT-41Figure 6.2: Voltage change in an R-C circuit............................................................................HT-42Figure 8.1: Concentric tube heat exchangers. (a) Parallel flow. (b) Counterflow.......................HT-44Figure 8.2: Cross-flow heat exchangers. (a) Finned with both fluids unmixed. (b) Unfinned with one

fluid mixed and the other unmixed ....................................................................................HT-45Figure 8.3: Geometry for heat transfer between two fluids .......................................................HT-45Figure 8.4: Counterflow heat exchanger...................................................................................HT-46Figure 8.5: Fluid temperature distribution along the tube with uniform wall temperature .........HT-46Figure 9.1: Radiation Surface Properties ..................................................................................HT-52Figure 9.2: Emissive power of a black body at several temperatures - predicted and observed..HT-53Figure 9.3: A cavity with a small hole (approximates a black body) .........................................HT-54Figure 9.4: A small black body inside a cavity .........................................................................HT-54Figure 9.5: Path of a photon between two gray surfaces ...........................................................HT-55

HT-4

Figure 9.6: Thermocouple used to measure temperature...........................................................HT-59Figure 9.7: Effect of radiation heat transfer on measured temperature. .....................................HT-59Figure 9.8: Shielding a thermocouple to reduce radiation heat transfer error ............................HT-60Figure 9.9: Radiation between two bodies................................................................................HT-60Figure 9.10: Radiation between two arbitrary surfaces .............................................................HT-61Figure 9.11: Radiation heat transfer for concentric cylinders or spheres ...................................HT-62Figure 9.12: View Factors for Three - Dimensional Geometries [from: Fundamentals of Heat

Transfer, F.P. Incropera and D.P. DeWitt, John Wiley and Sons] ......................................HT-64Figure 9.13: Fig. 13.4--View factor for aligned parallel rectangles [from: Fundamentals of Heat

Transfer, F.P. Incropera and D.P. DeWitt, John Wiley and Sons] ......................................HT-65Figure 9.14: Fig 13.5--View factor for coaxial parallel disk [from: Fundamentals of Heat Transfer,

F.P. Incropera and D.P. DeWitt, John Wiley and Sons] .....................................................HT-65Figure 9.15: Fig 13.6--View factor for perpendicular rectangles with a common edge .............HT-66

HT-5

1.0 Heat Transfer Modes

Heat transfer processes are classified into three types. The first is conduction, which is definedas transfer of heat occurring through intervening matter without bulk motion of the matter. Figure1.1 shows the process pictorially. A solid (a block of metal, say) has one surface at a hightemperature and one at a lower temperature. This type of heat conduction can occur, for example,through a turbine blade in a jet engine. The outside surface, which is exposed to gases from thecombustor, is at a higher temperature than the inside surface, which has cooling air next to it. Thelevel of the wall temperature is critical for a turbine blade.

Thigh Tlow

Solid

Heat “flows” to right ( q&)

Figure 1.1: Conduction heat transfer

The second heat transfer process is convection, or heat transfer due to a flowing fluid. Thefluid can be a gas or a liquid; both have applications in aerospace technology. In convection heattransfer, the heat is moved through bulk transfer of a non-uniform temperature fluid.

The third process is radiation or transmission of energy through space without the necessarypresence of matter. Radiation is the only method for heat transfer in space. Radiation can beimportant even in situations in which there is an intervening medium; a familiar example is the heattransfer from a glowing piece of metal or from a fire.

Muddy pointsHow do we quantify the contribution of each mode of heat transfer in a given situation?(MP HT.1)

2.0 Conduction Heat Transfer

We will start by examining conduction heat transfer. We must first determine how to relate theheat transfer to other properties (either mechanical, thermal, or geometrical). The answer to this isrooted in experiment, but it can be motivated by considering heat flow along a "bar" between twoheat reservoirs at TA, TB as shown in Figure 2.1. It is plausible that the heat transfer rate Q& , is a

HT-6

function of the temperature of the two reservoirs, the bar geometry and the bar properties. (Are thereother factors that should be considered? If so, what?). This can be expressed as

Q& = f1 (TA , TB , bar geometry, bar properties) (2.1)

It also seems reasonable to postulate that Q& should depend on the temperature difference TA - TB. IfTA – TB is zero, then the heat transfer should also be zero. The temperature dependence can thereforebe expressed as

Q& = f2 [ (TA - TB), TA, bar geometry, bar properties] (2.2)

L

TBTA

Q&

Figure 2.1: Heat transfer along a bar

An argument for the general form of f2 can be made from physical considerations. Onerequirement, as said, is f2 = 0 if TA = TB. Using a MacLaurin series expansion, as follows:

f( T) f(0)f

( T)T

0

∆∆

∆= +∂

∂+L (2.3)

If we define ∆T = TA – TB and f = f2, we find that (for small TA – TB),

f (T T ) Q f (0)f

(T T )T T .2 A B 2

2

A B T A TB 0A B− = = +

∂∂ −

−( ) +⋅

− =

L (2.4)

We know that f2(0) = 0 . The derivative evaluated at TA = TB (thermal equilibrium) is a measurable

property of the bar. In addition, we know that Q T Tf

T TA B2

A B

⋅

> >∂

∂ −( )>0 0 if or . It also seems

reasonable that if we had two bars of the same area, we would have twice the heat transfer, so thatwe can postulate that Q& is proportional to the area. Finally, although the argument is by no means

rigorous, experience leads us to believe that as L increases Q& should get smaller. All of these leadto the generalization (made by Fourier in 1807) that, for the bar, the derivative in equation (2.4) hasthe form

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∂

∂ −( )=

− =

f

T T

kAL

2

A B T A TB 0

. (2.5)

In equation (2.5), k is a proportionality factor that is a function of the material and thetemperature, A is the cross-sectional area and L is the length of the bar. In the limit for anytemperature difference ∆T across a length ∆x as both L, TA - TB → 0, we can say

( ) ( )dx

dTkA

L

TTkA

L

TTkAQ ABBA −=

−−=

−=& . (2.6)

A more useful quantity to work with is the heat transfer per unit area, defined as

qAQ

&&= . (2.7)

The quantity q& is called the heat flux and its units are Watts/m2. The expression in (2.6) canbe written in terms of heat flux as

dx

dTkq −=& . (2.8)

Equation 2.8 is the one-dimensional form of Fourier's law of heat conduction. Theproportionality constant k is called the thermal conductivity. Its units are W / m-K. Thermalconductivity is a well-tabulated property for a large number of materials. Some values for familiarmaterials are given in Table 1; others can be found in the references. The thermal conductivity is afunction of temperature and the values shown in Table 1 are for room temperature.

Table 2.1: Thermal conductivity at room temperature for some metals and non-metals

Metals Ag Cu Al Fe Steelk [W/m-K] 420 390 200 70 50

Non-metals H20 Air Engine oil H2 Brick Wood Corkk [W/m-K] 0.6 0.026 0.15 0.18 0.4 -0 .5 0.2 0.04

HT-8

2.1 Steady-State One-Dimensional Conduction

( )xQ& ( )dxxQ +&

dx

x

Insulated(no heat transfer)

Figure 2.2: One-dimensional heat conduction

For one-dimensional heat conduction (temperature depending on one variable only), we candevise a basic description of the process. The first law in control volume form (steady flow energyequation) with no shaft work and no mass flow reduces to the statement that Q&Σ for all surfaces = 0(no heat transfer on top or bottom of figure 2.2). From equation (2.8), the heat transfer rate in at theleft (at x) is

Q x k AdT

dx x( ) = − ⎛

⎝⎞⎠

. (2.9)

The heat transfer rate on the right is

˙ ˙˙

Q x dx Q xdQ

dxdx

x

+( ) = ( ) + +L. (2.10)

Using the conditions on the overall heat flow and the expressions in (2.9) and (2.10)

˙ ˙˙

Q QQ

x xddx

x dx 0( ) − ( ) + ( ) +⎛⎝⎜ ⎞

⎠⎟ =L . (2.11)

Taking the limit as dx approaches zero we obtain

dQ x

dx

˙ ( )= 0 , (2.12a)

or

HT-9

ddx

kAdTdx

0⎛

⎝⎜

⎞

⎠⎟ = . (2.12b)

If k is constant (i.e. if the properties of the bar are independent of temperature), this reduces to

ddx

AdTdx

0⎛

⎝⎜

⎞

⎠⎟ = (2.13a)

or (using the chain rule)

d T

dx

1A

dAdx

dTdx

02

2+⎛

⎝⎜

⎞

⎠⎟ = . (2.13b)

Equations (2.13a) or (2.13b) describe the temperature field for quasi-one-dimensional steady state(no time dependence) heat transfer. We now apply this to some examples.

Example 2.1: Heat transfer through a plane slab

x

T = T1 T = T2

Slab

x = 0 x = L

Figure 2.3: Temperature boundary conditions for a slab

For this configuration, the area is not a function of x, i.e. A = constant. Equation (2.13) thus became

02

2

=dx

Td . (2.14)

Equation (2.14) can be integrated immediately to yield

adx

dT= (2.15)

HT-10

and baxT += . (2.16)

Equation (2.16) is an expression for the temperature field where a and b are constants of integration.For a second order equation, such as (2.14), we need two boundary conditions to determine a and b.One such set of boundary conditions can be the specification of the temperatures at both sides of theslab as shown in Figure 2.3, say T (0) = T1; T (L) = T2.

The condition T (0) = T1 implies that b = T1. The condition T2 = T (L) implies that T2 = aL + T1, or

LTT

a 12 −= .

With these expressions for a and b the temperature distribution can be written as

T x TT T

Lx1

2 1( ) = +−⎛

⎝⎜

⎞

⎠⎟ . (2.17)

This linear variation in temperature is shown in Figure 2.4 for a situation in which T1 > T2.

T2

T1

T

x

Figure 2.4: Temperature distribution through a slab

The heat flux q& is also of interest. This is given by

( )constant12 =

−−=−=

L

TTk

dx

dTkq& . (2.18)

Muddy pointsHow specific do we need to be about when the one-dimensional assumption is valid? Is itenough to say that dA/dx is small? (MP HT.2)Why is the thermal conductivity of light gases such as helium (monoatomic) or hydrogen(diatomic) much higher than heavier gases such as argon (monoatomic) or nitrogen(diatomic)? (MP HT.3)

HT-11

2.2 Thermal Resistance Circuits

There is an electrical analogy with conduction heat transfer that can be exploited in problemsolving. The analog of Q& is current, and the analog of the temperature difference, T1 - T2, is voltagedifference. From this perspective the slab is a pure resistance to heat transfer and we can define

R

TTQ 21 −=& (2.19)

where R = L/kA, the thermal resistance. The thermal resistance R increases as L increases, as Adecreases, and as k decreases.

The concept of a thermal resistance circuit allows ready analysis of problems such as a compositeslab (composite planar heat transfer surface). In the composite slab shown in Figure 2.5, the heatflux is constant with x. The resistances are in series and sum to R = R1 + R2. If TL is the temperatureat the left, and TR is the temperature at the right, the heat transfer rate is given by

21 RR

TT

R

TTQ RLRL

+

−=

−=& . (2.20)

1 2

TL

Q&

TR

R1 R2

x

Figure 2.5: Heat transfer across a composite slab (series thermal resistance)

Another example is a wall with a dissimilar material such as a bolt in an insulating layer. Inthis case, the heat transfer resistances are in parallel. Figure 2.6 shows the physical configuration,the heat transfer paths and the thermal resistance circuit.

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R2

Q&

k2

k1

k1

R1

model

Figure 2.6: Heat transfer for a wall with dissimilar materials (Parallel thermal resistance)

For this situation, the total heat flux Q& is made up of the heat flux in the two parallel paths:

21 QQQ &&& += with the total resistance given by:

21

111

RRR+= . (2.21)

More complex configurations can also be examined; for example, a brick wall with insulationon both sides.

T2

Brick0.1 m

T1 = 150 °C T4 = 10 °C

T2 T3

Insulation0.03 m

T3T1 T4

R1 R2 R3

Figure 2.7: Heat transfer through an insulated wall

The overall thermal resistance is given by

33

3

22

2

11

1321 Ak

L

Ak

L

Ak

LRRRR ++=++= . (2.22)

Some representative values for the brick and insulation thermal conductivity are:

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kbrick = k2 = 0.7 W/m-Kkinsulation = k1 = k3 = 0.07 W/m-K

Using these values, and noting that A1 = A2 = A3 = A, we obtain:

K/W m 0.42K W/m0.07

m 0.03 2

1

131 ====

k

LARAR

K/W m 0.14K W/m0.7

m 0.1 2

2

22 ===

k

LAR .

This is a series circuit so

22

41 W/m142K/W m 0.98

K 140 hroughout constant t ==

−===

RA

TT

AQ

q&

&

x

41

4

TTTT

−−

0

1.01 2 3 4

Figure 2.8: Temperature distribution through an insulated wall

The temperature is continuous in the wall and the intermediate temperatures can be foundfrom applying the resistance equation across each slab, since Q& is constant across the slab. Forexample, to find T2:

2

1

21 W/m142=−

=AR

TTq&

This yields T1 – T2 = 60 K or T2 = 90 °C.

The same procedure gives T3 = 70 °C. As sketched in Figure 2.8, the larger drop is across theinsulating layer even though the brick layer is much thicker.

Muddy pointsWhat do you mean by continuous? (MP HT.4)Why is temperature continuous in the composite wall problem? Why is it continuous at theinterface between two materials? (MP HT.5)

HT-14

Why is the temperature gradient dT/dx not continuous? (MP HT.6)Why is ∆T the same for the two elements in a parallel thermal circuit? Doesn't the relativearea of the bolt to the wood matter? (MP HT.7)

2.3 Steady Quasi-One-Dimensional Heat Flow in Non-Planar Geometry

The quasi one-dimensional equation that has been developed can also be applied to non-planargeometries. An important case is a cylindrical shell, a geometry often encountered in situationswhere fluids are pumped and heat is transferred. The configuration is shown in Figure 2.9.

r1

control volumer1

r2

r2

Figure 2.9: Cylindrical shell geometry notation

For a steady axisymmetric configuration, the temperature depends only on a single coordinate (r)and Equation (2.12b) can be written as

kddr

A rdTdr

0( )⎛

⎝⎜

⎞

⎠⎟ = (2.23)

or, since A = 2π r,

ddr

rdTdr

0⎛

⎝⎜

⎞

⎠⎟ = . (2.24)

The steady-flow energy equation (no flow, no work) tells us that outin QQ && = or

0=drQd &

(2.25)

The heat transfer rate per unit length is given by

Q k 2 rdTdr

⋅= − ⋅ π .

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Equation (2.24) is a second order differential equation for T. Integrating this equation once gives

adr

dTr = . (2.26)

where a is a constant of integration. Equation (2.26) can be written as

r

dradT = (2.27)

where both sides of equation (2.27) are exact differentials. It is useful to cast this equation in termsof a dimensionless normalized spatial variable so we can deal with quantities of order unity. To dothis, divide through by the inner radius, r1

( )( )1

1

/

/

rr

rrdadT = (2.28)

Integrating (2.28) yields

T arr

b1

=⎛

⎝⎜

⎞

⎠⎟ +ln . (2.29)

To find the constants of integration a and b, boundary conditions are needed. These will be taken tobe known temperatures T1 and T2 at r1 and r2 respectively. Applying T = T1 at r = r1 gives T1 = b.Applying T = T2 at r = r2 yields

11

22 ln T

r

raT += ,

or

( )12

12

/ln rr

TTa

−= .

The temperature distribution is thus

( ) ( )( ) 1

12

112 /ln

/lnT

rrrr

TTT +−= . (2.30)

As said, it is generally useful to put expressions such as (2.30) into non-dimensional andnormalized form so that we can deal with numbers of order unity (this also helps in checkingwhether results are consistent). If convenient, having an answer that goes to zero at one limit is alsouseful from the perspective of ensuring the answer makes sense. Equation (2.30) can be put in non-dimensional form as

HT-16

( )( )12

1

12

1

/ln/ln

rrrr

TTTT

=−−

. (2.31)

The heat transfer rate, Q& , is given by

( )( )

( )( )12

21

112

121 /ln

21

/ln2

rr

TTk

rrr

TTkr

dr

dTkAQ

−=

−−=−= ππ&

per unit length. The thermal resistance R is given by

( )k

rrR

π2

/ln 12= (2.32)

R

TTQ 21 −=& .

The cylindrical geometry can be viewed as a limiting case of the planar slab problem. To

make the connection, consider the case when 11

12 <<−

r

rr. From the series expansion for ln (1 + x)

we recall that

ln 1+ x x - x2

+x3

+ 2 3

( ) ≈ K (2.33)

(Look it up, try it numerically, or use the binomial theorem on the series below and integrate term by

term. K++−=+

211

1xx

x)

The logarithms in Equation (2.31) can thus be written as

ln ln1r r

rr r

rrr

r rr

1

1

1

1

2

1

2 1

1

+−⎛

⎝⎜

⎞

⎠⎟ ≅

−≅

− and (2.34)

in the limit of (r2 – r1) << r1. Using these expressions in equation (2.30) gives

( ) ( )( ) 1

12

112 T

rrrr

TTT +−−

−= . (2.35)

With the substitution of r – r1 = x, and r2 – r1 = L we obtain

( )L

xTTTT 121 −+= (2.36)

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which is the same as equation (2.17). The plane slab is thus the limiting case of the cylinder if (r -r1) / r << 1, where the heat transfer can be regarded as taking place in (approximately) a planar slab.

To see when this is appropriate, consider the expansion ( )

xx+1ln

, which is the ratio of heat flux for a

cylinder and a plane slab.

Table 2.2: Utility of plane slab approximation

x .1 .2 .3 .4 .5

( )x

x+1ln .95 .91 .87 .84 .81

For < 10% error, the ratio of thickness to inner radius should be less than 0.2, and for 20% error, thethickness to inner radius should be less than 0.5.

A second example is the spherical shell with specified temperatures T (r1) = T1 and T (r2) =T2, as sketched in Figure 2.10.

r1

T2T1

r2

Figure 2.10: Spherical shell

The area is now 24)( rrA π= , so the equation for the temperature field is

ddr

rdTdr

02⎛

⎝⎜

⎞

⎠⎟ = . (2.37)

Integrating equation (2.37) once yields

2/ radr

dT= . (2.38)

Integrating again gives

HT-18

br

aT +−=

or, normalizing the spatial variable

( )b

rra

T +′

=1/

(2.39)

where a′ and b are constants of integration. As before, we specify the temperatures at r = r1 and r =r2. Use of the first boundary condition gives ( ) baTrT +′== 11 . Applying the second boundarycondition gives

( )( )

brr

aTrT +

′==

1222 /

Solving for a′ and b,

./1

/1

21

211

21

21

rr

TTTb

rr

TTa

−

−−=

−

−=′

(2.40)

In non-dimensional form the temperature distribution is thus:

( )( )21

1

21

1

/1

/1

rr

rr

TT

TT

−

−=

−

− (2.41)

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3.0 Convective Heat Transfer

The second type of heat transfer to be examined is convection, where a key problem isdetermining the boundary conditions at a surface exposed to a flowing fluid. An example is the walltemperature in a turbine blade because turbine temperatures are critical as far as creep (and thusblade) life. A view of the problem is given in Figure 3.1, which shows a cross-sectional view of aturbine blade. There are three different types of cooling indicated, all meant to ensure that the metalis kept at a temperature much lower than that of the combustor exit flow in which the turbine bladeoperates. In this case, the turbine wall temperature is not known and must be found as part of thesolution to the problem.

Figure 3.1: Turbine blade heat transfer configuration

To find the turbine wall temperature, we need to analyze convective heat transfer, which means weneed to examine some features of the fluid motion near a surface. The conditions near a surface areillustrated schematically in Figure 3.2.

T

c∞Ve locitydistribution;c = 0 at surface

c (velocity)

δ′

TwT∞

y y

Figure 3.2: Temperature and velocity distributions near a surface.

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In a region of thickness δ′, there is a thin "film" of slowly moving fluid through which most of thetemperature difference occurs. Outside this layer, T is roughly uniform (this defines δ′). The heatflux can thus be expressed as

qQA

k T T

dw

⋅⋅

∞= =−( )′

(3.1)

It cannot be emphasized enough that this is a very crude picture. The general concept, however, iscorrect, in that close to the wall, there is a thin layer in which heat is transferred basically byconduction. Outside of this region is high mixing. The difficulty is that the thickness of the layer isnot a fluid property. It depends on velocity (Reynolds number), structure of the wall surface,pressure gradient and Mach number. Generally δ′ is not known and needs to be found and it iscustomary to calculate the heat transfer using [kfluid / δ′]. This quantity has the symbol h and isknown as the convective heat transfer coefficient. The units of h are W/m2K. The convective heattransfer coefficient is defined by

qQA

h T Tw

⋅⋅

∞= = −( ) (3.2)

Equation 3.2 is often called Newton’s Law of Cooling. For many situations of practical interest, thequantity h is still known mainly through experiments.

Muddy pointsHow do we know that δ' is not a fluid property? (MP HT.8)

3.1 The Reynolds Analogy

We describe the physical mechanism for the heat transfer coefficient in a turbulent boundary layerbecause most aerospace vehicle applications have turbulent boundary layers. The treatment closelyfollows that in Eckert and Drake (1959). Very near the wall, the fluid motion is smooth and laminar,and molecular conduction and shear are important. The shear stress, τ, at a plane is given by

τµ =dy

dc (where µ is the dynamic viscosity), and the heat flux by dy

dTkq −=& . The latter is the same

expression that was used for a solid. The boundary layer is a region in which the velocity is lowerthan the free stream as shown in Figures 3.2 and 3.3. In a turbulent boundary layer, the dominantmechanisms of shear stress and heat transfer change in nature as one moves away from the wall.

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plane

Figure 3.3: Velocity profile near a surface

As one moves away from the wall (but still in the boundary layer), the flow is turbulent. The fluidparticles move in random directions and the transfer of momentum and energy is mainly throughinterchange of fluid particles, shown schematically in Figure 3.4.

2 2

a a

1 1

m′ cp T′

m′ cp T

Figure 3.4: Momentum and energy exchanges in turbulent flow.

With reference to Figure 3.4, because of the turbulent velocity field, a fluid mass m′ penetrates theplane aa per unit time and unit area. In steady flow, the same amount crosses aa from the other side.Fluid moving up transports heat m′ cp T. Fluid moving down transports m′ cp T′ downwards. If T >T′, there is a turbulent downwards heat flow, qt , given by q m c T Tt p= ′ ′−( ) that results.

Fluid moving up also has momentum m′ c and fluid moving down has momentum m′ c′. The netflux of momentum down per unit area and time is therefore m′ (c′ - c). This net flux of momentumper unit area and time is a force per unit area or stress, given by

t m' c ct = ′ −( ) (3.3)

Based on these considerations, the relation between heat flux and shear stress at plane aa is

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qt =′−′−

⎛⎝

⎞⎠

τ t pcT Tc c

(3.4)

or (again approximately)

qt t pcdTdc

= τ (3.5)

since the locations of planes 1-1 and 2-2 are arbitrary. For the laminar region, the heat flux towardsthe wall is

q = τµk dT

dc

The same relationship is applicable in laminar or turbulent flow if pck=

µ or, expressed slightly

differently,

1/

/===

αυ

ρρµ

p

p

ckk

c

where υ is the kinematic viscosity, and α is the thermal diffusivity.

The quantity µcp/k is known as the Prandtl number (Pr), after the man who first presented the idea ofthe boundary layer and was one of the pioneers of modern fluid mechanics. For gases, Prandtlnumbers are in fact close to unity and for air Pr = 0.71 at room temperature. The Prandtl numbervaries little over a wide range of temperatures; approximately 3% from 300-2000 K.

We want a relation between the values at the wall (at which T = Tw and c = 0) and those in the freestream. To get this, we integrate the expression for dT from the wall to the free stream

dT1

cdc

p=

q

τ (3.6)

where the relation between heat transfer and shear stress has been taken as the same for both thelaminar and the turbulent portions of the boundary layer. The assumption being made is that themechanisms of heat and momentum transfer are similar. Equation (3.6) can be integrated from thewall to the freestream (conditions "at ∞"):

dTc

dcpw w

1∞ ∞

∫ ∫= ⎛⎝⎞⎠

q

τ (3.7)

where q

τ and cp are assumed constant.

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Carrying out the integration yields

T Tccw

w

w p∞

∞− =q

τ (3.8)

where c∞ is the velocity and cp is the specific heat. In equation (3.8), wq& is the heat flux to the wall

and τw is the shear stress at the wall. The relation between skin friction (shear stress) at the wall andheat transfer is thus

qw

p w

w2c T T c cρ

τ

ρ∞ ∞ ∞ ∞ ∞−( )= . (3.9)

The quantity τ

ρ

w2

1/2 c∞ ∞

is known as the skin friction coefficient and is denoted by Cf . The skin

friction coefficient has been tabulated (or computed) for a large number of situations. If we define anon-dimensional quantity

qw

p w

w

p w pc T T ch T Tc T T c

hc cρ ρ ρ∞ ∞ ∞

∞

∞ ∞ ∞ ∞ ∞−( )=

−( )−( )

= = St,

known as the Stanton Number, we can write an expression for the heat transfer coefficient, h as

h c cC

2pf≈ ∞ ∞ρ . (3.10)

Equation (3.10) provides a useful estimate of h, or wq& , based on knowing the skin friction, or drag.

The direct relationship between the Stanton Number and the skin friction coefficient is

St =C

2f

The relation between the heat transfer and the skin friction coefficient

qww p wc T T

c≈

−( )∞

∞

τ

is known as the Reynolds analogy between shear stress and heat transfer. The Reynolds analogy isextremely useful in obtaining a first approximation for heat transfer in situations in which the shearstress is "known".

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An example of the use of the Reynolds analogy is in analysis of a heat exchanger. One type of heatexchanger has an array of tubes with one fluid flowing inside and another fluid flowing outside, withthe objective of transferring heat between them. To begin, we need to examine the flow resistanceof a tube. For fully developed flow in a tube, it is more appropriate to use an average velocity c anda bulk temperature TB . Thus, an approximate relation for the heat transfer is

qw w pB wc

T Tc

≈−

τ . (3.11)

The fluid resistance (drag) is all due to shear forces and is given by τw Aw = D, where Aw is the tube“wetted” area (perimeter x length). The total heat transfer, Q& , is Aqw& w, so that

c

TTcDQ WB

p

−=& (3.12)

The power, P, to drive the flow through a resistance is given by the product of the drag and thevelocity, cD , so that

Q

P

c T T

cp B w=

−( )2

(3.13)

The mass flow rate is given by Acm ρ=& where A is the cross sectional area. For given mass flowrate and overall heat transfer rate, the power scales as c2 or as 1/A2, i.e.

Pc T T A2

p B w2∝

−( )

˙ ˙Qm2 1

ρ (3.14)

Equations (3.13) and (3.14) show that to decrease the power dissipated, we need to decrease c ,which can be accomplished by increasing the cross-sectional area. Two possible heat exchangerconfigurations are sketched in Figure 3.5; the one on the right will have a lower loss.

heat exchangerheat exchanger

high loss

lower loss

m⋅ diffuser

Figure 3.5: Heat exchanger configurations

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To recap, there is an approximate relation between skin friction (momentum flux to the wall) andheat transfer called the Reynolds analogy that provides a useful way to estimate heat transfer rates insituations in which the skin friction is known. The relation is expressed by

St =C

2f (3.15a)

or

flux momentum convected

wallflux to momentum

fluxheat convected

wallflux toheat = (3.15b)

orqw

p w

w2c c T T cρ

τ

ρ∞ ∞ ∞ ∞ ∞−( )≈ (3.15c)

The Reynolds analogy can be used to give information about scaling of various effects as well asinitial estimates for heat transfer. It is emphasized that it is a useful tool based on a hypothesis aboutthe mechanism of heat transfer and shear stress and not a physical law.

Muddy pointsWhat is the "analogy" that we are discussing? Is it that the equations are similar? (MP HT.9)In what situations does the Reynolds analogy "not work"? (MP HT.10)

3.2 Combined Conduction and Convection

We can now analyze problems in which both conduction and convection occur, starting with a wallcooled by flowing fluid on each side. As discussed, a description of the convective heat transfer canbe given explicitly as

˙˙

Qq

Ah T Tw= = −( )∞ (3.16)

This could represent a model of a turbine blade with internal cooling. Figure 3.6 shows theconfiguration.

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T1

T2

Tw1

Tw2

L

1δ ′ 2δ ′

T

Figure 3.6: Wall with convective heat transfer

The heat transfer in fluid 1 is given by

Q

Ah T T1 w 11

= −( ) ,which is the heat transfer per unit area to the fluid. The heat transfer in fluid (2) is similarly given by

Q

Ah T T2 2 w2= −( ) .

Across the wall, we have

Q

AkL

T Tw2 w1= −( ).

The quantity Q& /A is the same in all of these expressions. Putting them all together to write theknown overall temperature drop yields a relation between heat transfer and overall temperature drop,T2 – T1 :

T T T T T T T TA

1h

Lk

1h2 1 2 w2 w2 w1 w1 1

1 2− = −( ) + −( ) + −( ) = + +

⎡

⎣⎢⎤

⎦⎥Q

. (3.17)

We can define a thermal resistance, R, as before, such that

Q =−( )T TR

2 1 ,

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where R is given by

R1

h AL

Ak1

h A1 2

= + + . (3.18)

Equation (3.18) is the thermal resistance for a solid wall with convection heat transfer on each side.

For a turbine blade in a gas turbine engine, cooling is a critical consideration. In terms of Figure 3.6,T2 is the combustor exit (turbine inlet) temperature and T1 is the temperature at the compressor exit.We wish to find T w2

because this is the highest metal temperature. From (3.17), the wall

temperature can be written as

Tw2 = T2 - Q

Ah2 = −

−T

T T

R

1

Ah22 1

2 (3.19)

Using the expression for the thermal resistance, the wall temperatures can be expressed in terms ofheat transfer coefficients and wall properties as

12

1

2

1222

++

−−=

k

Lh

h

hTT

TTw (3.20)

Equation (3.20) provides some basic design guidelines. The goal is to have a low value of T w2.

This means h1/h2 should be large, k should be large (but we may not have much flexibility in choiceof material) and L should be small. One way to achieve the first of these is to have h2 low (forexample, to flow cooling air out as in Figure 3.1 to shield the surface).

A second example of combined conduction and convection is given by a cylinder exposed to aflowing fluid. The geometry is shown in Figure 3.7.

c∞

T∞

r1

r2

r1

Figure 3.7: Cylinder in a flowing fluid

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For the cylinder the heat flux at the outer surface is given by

˙˙

qQ

= = −( ) =∞Ah T T r rw 2at

The boundary condition at the inner surface could either be a heat flux condition or a temperaturespecification; we use the latter to simplify the algebra. Thus, 11 at rrTT == . This is a model for theheat transfer in a pipe of radius r1 surrounded by insulation of thickness r2 - r1. The solution for acylindrical region was given in Section 2.3 as

T r ar

rb( ) ln= +

1

Use of the boundary condition T (r1) = T1 yields b = T1.

At the interface between the cylinder and the fluid, r = r2, the temperature and the heat flow arecontinuous. (Question: Why is this? How would you argue the point?)

q = − = − = +⎛⎝⎜ ⎞

⎠⎟ −

⎡

⎣⎢⎤

⎦⎥∞k

dT

drk

a

rh aln

r

rT T

2

2

11 (3.21)

Plugging the form of the temperature distribution in the cylinder into Equation (3.21) yields

− +⎛⎝⎜ ⎞

⎠⎟ = −( )∞a

k

rhln

r

rh T T

2

2

11 .

The constant of integration, a, is

ah T T

kr

hrr

T T

khr

rr

1

2

2

1

1

2

2

1

=− −( )

+= −

−( )+

∞ ∞

ln ln,

and the expression for the temperature is, in normalized non-dimensional form

T TT T

r/r

khr

r /r

1

1

1

22 1

−−

=( )+ ( )∞

ln

ln. (3.22)

heat flux just inside cylinder surface heat flux to fluid

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The heat flow per unit length, Q, is given by

Q =−( )

+ ( )

∞2 T T kk

hrr /r

1

22 1

π

ln (3.23)

The units in Equation (3.23) are W / m-s.

A problem of interest is choosing the thickness of insulation to minimize the heat loss for a fixedtemperature difference T1 - T∞ between the inside of the pipe and the flowing fluid far away fromthe pipe. (T1 - T∞ is the driving temperature distribution for the pipe). To understand the behavior ofthe heat transfer we examine the denominator in Equation (3.23) as r2 varies. The thickness ofinsulation that gives maximum heat transfer is given by

ddr

khr

rr

02 2

2

1

+⎛

⎝⎜

⎞

⎠⎟ =ln (3.24)

(Question: How do we know this is a maximum?)

From Equation (3.24), the value of r2 for maximum Q& is thus

(r2)maximum heat transfer = k/h. (3.25)

If r2 is less than this, we can add insulation and increase heat loss. To understand why this occurs,consider Figure 3.8, which shows a schematic of the thermal resistance and the heat transfer. As r2

increases from a value less than r2 = k/h, two effects take place. First, the thickness of the insulationincreases, tending to drop the heat transfer because the temperature gradient decreases. Secondly,the area of the outside surface of the insulation increases, tending to increase the heat transfer. Thesecond of these is (loosely) associated with the k/hr2 term, the first with the ln(r2/r1) term. There arethus two competing effects which combine to give a maximum Q at r2 = k/h.

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R

Q⋅

r2

ln rr2

1

khr2

kh

Figure 3.8: Critical radius of insulation

Muddy pointsIn the expression

1h.A

, what is A? (MP HT.11)

It seems that we have simplified convection a lot. Is finding the heat transfer coefficient, h,really difficult? (MP HT.12)What does the "K" in the contact resistance formula stand for? (MP HT.13)In the equation for the temperature in a cylinder (3.22), what is "r"? (MP HT.14)

3.3 Dimensionless Numbers and Analysis of Results

Phenomena in fluid flow and heat transfer depend on dimensionless parameters. The Mach numberand the Reynolds number are two you have already seen. These parameters give information as tothe relevant flow regimes of a given solution. Casting equations in dimensionless form helps showthe generality of application to a broad class of situations (rather than just one set of dimensionalparameters). It is generally good practice to use non-dimensional numbers, forms of equations, andresults presentation whenever possible. The results for heat transfer from the cylinder are already indimensionless form but we can carry the idea even further. For the cylinder:

T - TT T

r/r

k/hr r /r1

1

1

2 2 1

=( )+ ( )

ln

ln (3.26)

The parameter hr

k

2 or hL

k, where L is a relevant length for the particular problem of interest, is

called the Biot number denoted by Bi. In terms of this parameter,

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T - TT T

r/r

1Bi

r /r

1

1

1

2 1∞ −

=( )

+ ( )

ln

ln (3.27)

The size of the Biot number gives a key to the regimes in which different features are dominant. ForBi >> 1 the convection heat transfer process offers little resistance to heat transfer. There is thusonly a small ∆T outside (i.e. T (r2) close to T∞) compared to the ∆T through the solid with a limitingbehavior of

T TT T

r/rr /r

1 1

2 1

−−

=∞

lnln

as Bi goes to infinity. This is much like the situation with an external temperature specified.

For Bi << 1 the conduction heat transfer process offers little resistance to heat transfer. Thetemperature difference in the body (i.e. from r1 to r2) is small compared to the external temperaturedifference, T1 - T∞ . In this situation, the limiting case is

T TT T

Bi r/r1

11

−−

= ( )∞

ln

In this regime there is approximately uniform temperature in the cylinder. The size of the Biotnumber thus indicates the regimes where the different effects become important.

Figure 3.9 shows the general effect of Biot number on temperature distribution. Figure 3.10 is a plotof the temperature distribution in the cylinder for values of Bi = 0.1, 1.0 and 10.0.

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Figure 3.9: Effect of the Biot Number [hL / kbody] on the temperature distributions in the solidand in the fluid for convective cooling of a body. Note that kbody is the thermal conductivity of

the body, not of the fluid.

[adapted from: A Heat Transfer Textbook, John H. Lienhard, Prentice-Hall Publishers, 1980]

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Figure 3.10: Temperature distribution in a convectively cooled cylinder for different values ofBiot number, Bi; r2 / r1 = 2 [from: A Heat Transfer Textbook, John H. Lienhard]

4.0 Temperature Distributions in the Presence of Heat Sources

There are a number of situations in which there are sources of heat in the domain of interest.Examples are:

1) Electrical heaters where electrical energy is converted resistively into heat2) Nuclear power supplies3) Propellants where chemical energy is the source

These situations can be analyzed by looking at a model problem of a slab with heat sources α (W/m3)distributed throughout. We take the outside walls to be at temperature Tw. and we will determine themaximum internal temperature.

TwTw

(a)

heatsources

3m

Wα

x x x + dx

q& dxdx

qdq

&& +

Slice at x, x+dx

(b)

Figure 4.1: Slab with heat sources (a) overall configuration, (b) elementary slice

With reference to Figure 4.1(b), a steady-state energy balance yields an equation for the heat flux, q& .

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q adx qd qdx

dx 0⋅ ⋅

⋅

+ − +⎛

⎝

⎜⎜

⎞

⎠

⎟⎟= (4.1)

ord qdx

⋅

= α . (4.2)

There is a change in heat flux with x due to the presence of the heat sources. The equation for thetemperature is

0/2

2

=+ kdx

Tdα (4.3)

Equation (4.3) can be integrated once,

axkdx

dT+−=

α (4.4)

and again to give

baxxk

T ++−= 2

2

α (4.5)

where a and b are constants of integration. The boundary conditions imposed are ( ) ( ) wTLTT ==0 .

Substituting these into Equation (4.5) gives wTb = and k

La

2

α= . The temperature distribution is thus

WTLxkk

xT ++−=

22

2 αα . (4.6)

Writing (4.6) in a normalized, non-dimensional fashion gives a form that exhibits in a more usefulmanner the way in which the different parameters enter the problem:

T T

L /k

12

xL

x

LW

2

2

2

−= −

⎛

⎝⎜

⎞

⎠⎟

α (4.7)

This distribution is sketched in Figure 4.2. It is symmetric about the mid-plane at 2

Lx = , with half

the energy due to the sources exiting the slab on each side.

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T T

aL /kW

2

−⎛⎝⎜

⎞⎠⎟

0.125

X

L0 0.5 1.0

Figure 4.2:Temperature distribution for slab with distributed heat sources

The heat flux at the side of the slab, x = 0, can be found by differentiating the temperature

distribution and evaluating at x = 0 : = ==

kdTdx

kL

2K1L

L/2 x 0

2

.

This is half of the total heat generated within the slab. The magnitude of the heat flux is the same atx = L, although the direction is opposite.

A related problem would be one in which there were heat flux (rather than temperature) boundaryconditions at x = 0 and x = L, so that Tw is not known. We again determine the maximumtemperature. At x = 0 and L, the heat flux and temperature are continuous so

= ( ) =kdTdx

h T T x 0,Lat . (4.8)

Referring to the temperature distribution of Equation (4.6) gives for the two terms in Equation (4.8),

kdTdx

k -x

ka x ka= + = +( ) (4.9)

h T T hxk

x b T2

( ) = + + (4.10)

Evaluating (4.10) at x = 0 and L allows determination of the two constants a and b. This is left as anexercise for the reader.

Muddy pointsFor an electric heated strip embedded between two layers, what would the temperaturedistribution be if the two side temperatures were not equal? (MP HT.15)

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5.0 Heat Transfer From a Fin

Fins are used in a large number of applications to increase the heat transfer from surfaces.Typically, the fin material has a high thermal conductivity. The fin is exposed to a flowing fluid,which cools or heats it, with the high thermal conductivity allowing increased heat being conductedfrom the wall through the fin. The design of cooling fins is encountered in many situations and wethus examine heat transfer in a fin as a way of defining some criteria for design.

A model configuration is shown in Figure 5.1. The fin is of length L. The other parameters of theproblem are indicated. The fluid has velocity c∞ and temperature T∞.

y

x

Fin

x = 0(wall)

x = L

T0 (wall)

T∞ , c∞

Figure 5.1: Geometry of heat transfer fin

We assume (using the Reynolds analogy or other approach) that the heat transfer coefficient for thefin is known and has the value h. The end of the fin can have a different heat transfer coefficient,which we can call hL.

The approach taken will be quasi-one-dimensional, in that the temperature in the fin will be assumedto be a function of x only. This may seem a drastic simplification, and it needs some explanation.With a fin cross-section equal to A and a perimeter P, the characteristic dimension in the transversedirection is A / P (For a circular fin, for example, A / P = r / 2). The regime of interest will be taken

to be that for which the Biot number is much less than unity, ( )

1/

<<=k

PAhBi , which is a realistic

approximation in practice.The physical content of this approximation can be seen from the following. Heat transfer per unitarea out of the fin to the fluid is roughly of magnitude ~ h(Tw - T∞ ) per unit area . The heat transferper unit area within the fin in the transverse direction is (again in the same approximate terms)

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PATT

k w

/)( 1 −

, where T1 is an internal temperature. These two quantities must be of the same

magnitude. If 1/

<<k

PAh , then

T TT T

11 w

w

−−

<<∞

. In other words, if Bi << 1, there is a much larger

capability for heat transfer per unit area across the fin than there is between the fin and the fluid, andthus little variation in temperature inside the fin in the transverse direction. To emphasize the point,consider the limiting case of zero heat transfer to the fluid i.e., an insulated fin. Under theseconditions, the temperature within the fin would be uniform and equal to the wall temperature.

If there is little variation in temperature across the fin, an appropriate model is to say that thetemperature within the fin is a function of x only, T = T(x), and use a quasi-one-dimensionalapproach. To do this, consider an element, dx, of the fin as shown in Figure 5.2. There is heat flow

of magnitude inQ& at the left-hand side and heat flow out of magnitude dxdxQd

QQ inout

&&& += at the

right hand side. There is also heat transfer around the perimeter on the top, bottom, and sides of thefin. From a quasi-one-dimensional point of view, this is a situation similar to that with internal heatsources, but here, for a cooling fin, in each elemental slice of thickness dx there is essentially a heat

sink of magnitude Pdxh T T− ∞( ), where Pdx is the area for heat transfer to the fluid.

inQ&

x x+dx

outQ&

Figure 5.2:Element of fin showing heat transfer

The heat balance for the element in Figure 5.2 can be written in terms of the heat flux using AqQ && = ,for a fin of constant area:

˙ ˙˙

q qq

A Ph T T dx Addx

dxA= −( ) + +⎛⎝

⎞⎠∞ . (5.1)

From Equation (5.1) we obtain

Addx

Ph T T 0q

+ −( ) =∞ . (5.2)

In terms of the temperature distribution, T(x):

d T

dx

PhAk

T T 02

2− −( ) =∞ . (5.3)

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The quantity of interest is the temperature difference (T - T∞ ), and we can change variables to putEquation (5.3) in terms of this quantity using the substitution

ddx

(T T )dTdx

− =∞ . (5.4)

Equation (5.3) can therefore be written as

d

dx(T T )

hPAk

(T T ) 02

2− − − =∞ ∞ . (5.5)

Equation (5.5) describes the temperature variation along the fin. It is a second order equation andneeds two boundary conditions. The first of these is that the temperature at the end of the fin thatjoins the wall is equal to the wall temperature. (Does this sound plausible? Why or why not?)

(T T ) T Tx 0 0− = −∞ = ∞ . (5.6)

The second boundary condition is at the other end of the fin. We will assume that the heat transferfrom this end is negligible1. The boundary condition at x = L is

ddx

T T 0x L

−( ) =∞ =. (5.7)

The last step is to work in terms of non-dimensional variables to obtain a more compact description.

In this we define T - TT T

T0

~∞

∞− as ∆ , where the values of ∆T

~

range from zero to one and Lx /=ξ ,

where ξ also ranges over zero to one. The relation between derivatives that is needed to cast theequation in terms of ξ is

ξξξ

d

d

Ld

d

dx

d

dx

d 1== .

1Equation (5.5) can be written in this dimensionless form as

d Td

hPkA

L T2

~

22

~∆∆

ξ− ⎛

⎝⎞⎠

= 0 . (5.8)

There is one non-dimensional parameter in Equation (5.8), which we will call m and define by 1 Note: We don’t need to make this assumption, and if we were looking at the problem in detail we would solve itnumerically and not worry about whether an analytic solution existed. In the present case, developing the analyticsolution is useful in presenting the structure of the solution as well as the numbers, so we resort to the mild fictionof no heat transfer at the fin end. We need to assess, after all is said and done, whether this is appropriate or not.

HT-39

kALhP

Lm2

22 = . (5.9)

The equation for the temperature distribution we have obtained is

d T

dm L T2 2

2

20

∆∆

~~

ξ− = . (5.10)

This second order equation has the solution

∆T~

ae bemL mL= + −ξ ξ. (5.11)

(Try it and see). The boundary condition at ξ = 0 is

∆T(0) a b 1~

= + = . (5.12a)

The boundary condition at ξ = 1 is that the temperature gradient is zero or

d Td

(L) mLae mLbe

~

m m∆ξ

= − =− 0 . (5.12b)

Solving the two equations given by the boundary conditions for a and b gives an expression for ∆T~

in terms of the hyperbolic cosine or cosh: cosh xe e

2

x x

=+⎛

⎝⎜

⎞

⎠⎟

−

∆TmL

mL

~

=−( )cosh

cosh

1 ξ (5.13)

This is the solution to Equation (5.8) for a fin with no heat transfer at the tip. In terms of the actualheat transfer parameters it is written as

T TT T

1xL

hPkA

L

hPkA

L0

−

−=

−⎛⎝

⎞⎠

⎛

⎝⎜

⎞

⎠⎟

⎛

⎝⎜

⎞

⎠⎟

∞

∞

cosh

cosh

. (5.14)

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The amount of heat removed from wall due to the fin, which is the quantity of interest, can be foundby differentiating the temperature and evaluating the derivative at the wall, x = 0:

Q kAddx

T T x 0

⋅

∞ == − −( ) (5.15)

or

Q L

kA T T

dDTd0

~

0

⋅

∞ =−( )= −

ξ ξ = mLmL

mL

mLmL tanh

cosh

sinh = (5.16)

Q

kAhP T TmL

0

⋅

∞−( )= tanh (5.16a)

The solution is plotted in Figure 5.3, which is taken from the book by Lienhard. Several features ofthe solution should be noted. First, one does not need fins which have a length such that m is muchgreater than 3. Second, the assumption about no heat transfer at the end begins to be inappropriateas m gets smaller than 3, so for very short fins the simple expression above would not be a goodestimate. We will see below how large the error is.

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Dimensionless axial position ξ = x L/

Figure 5.3:The temperature distribution, tip temperature, and heat flux in a straight one-dimensional fin with the tip insulated. [Adapted from: Lienhard, A Heat Transfer Textbook, Prentice-

Hall publishers]

Muddy pointsWhy did you change the variable and write the derivative

d Tdx

2

2 as d T - T

dx

2

2∞( )

in the equation

for heat transfer in the fin? (MP HT.16)What types of devices use heat transfer fins? (MP HT.17)Why did the Stegosaurus have cooling fins? Could the stegosaurus have "heating fins"?(MP HT.18)

Dimensionlesstemperature

(T - T∞ ) / (T0 - T∞ )

Dimensionless heatflow into the fin,

Q

hPkA T T0 −( )∞

Dimensionlesstemperature

at tip(T - T∞ ) / (T0 - T∞ )

mL = 5(a long "overdesigned" fin)

mL = 3

mL = 2

mL - 1 (a very stumpy fin)

tanh mL

The temperatureexcess at the tipis less than 1.4%,beyond L = 5 / m

Heat flow cannot be noticeablyimproved by lengthening thefin beyond L = 3 / m

1

cosh mL

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6.0 Transient Heat Transfer (Convective Cooling or Heating)

All the heat transfer problems we have examined have been steady state, but there are oftencircumstances in which the transient response to heat transfer is critical. An example is the heatingup of gas turbine compressors as they are brought up to speed during take-off. The disks that holdthe blades are large and take a long time to come to temperature, while the casing is thin and in thepath of high velocity compressor flow and thus comes to temperature rapidly. The result is that thecase expands away from the blade tips, sometimes enough to cause serious difficulties withaerodynamic performance.

To introduce the topic as well as to increase familiarity with modeling of heat transfer problems, weexamine a lumped parameter analysis of an object cooled by a stream. This will allow us to seewhat the relevant non-dimensional parameters are and, at least in a quantitative fashion, how morecomplex heat transfer objects will behave. We want to view the object as a "lump" described by asingle parameter. We need to determine when this type of analysis would be appropriate. Toaddress this, consider the temperature difference T1 - Tw between two locations in the object, asshown in Figure 6.1.

c∞

T∞

T1

Tw

object

Figure 6.1:Temperature variation in an object cooled by a flowing fluid

If the heat transfer within the body and from the body to the fluid are of the same magnitude,

h T TkL

T Tw 1 w−( ) ≈ −( )∞ (6.1)

where L is a relevant length scale, say half the thickness of the object. The ratio of the temperaturedifference is

T TT T

hLk

1 w

w

−−

≈∞

(6.2)

If the Biot number is small the ratio of temperature differences described in Equation (6.2) is alsoT T

T T11 w

w

−( )−( )

<<∞

. We can thus say T T T T1 w w−( ) << − ∞ and neglect the temperature non-uniformity

within the object.

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The approximation made is to view the object as having a spatially uniform temperature that is afunction of time only. Explicitly, T = T(t). The first law applied to the object is (using the fact thatfor solids cp = cv = c)

Q VcdTdtin

⋅

= ρ (6.3)

where ρ is the density of the object and V is its volume. In terms of heat transferred to the fluid,

dt

dTVcQout ρ−=& . The rate of heat transfer to the fluid is Ah (T - T∞), so the expression for the time

evolution of the temperature is

Ah T T VcdTdt

−( ) = −∞ ρ . (6.4)

The initial temperature, T(0), is equal to some known value, which we can call Ti . Using this,

Equation (6.4) can be written in terms of a non-dimensional temperature difference T TT Ti

−−

⎛

⎝⎜

⎞

⎠⎟

∞

∞

,

ddt

T TT T

hAVc

T TT Ti i

−−

⎛

⎝⎜

⎞

⎠⎟ +

−−

⎛

⎝⎜

⎞

⎠⎟ =∞

∞

∞

∞ρ0 (6.5)

At time t = 0, this non-dimensional quantity is equal to one. Equation (6.5) is an equation you have

seen before, dxdt

x0+ =

⎛

⎝⎜

⎞

⎠⎟τ

which has the solution x ae t= − / τ . For the present problem the form is

T TT T

aei

hAt/pVc−−

=∞

∞

− . (6.6)

The constant a can be seen to be equal to unity to satisfy the initial condition. This form of equation

implies that the solution has a heat transfer "time constant" given by hA

Vcρτ = .

The time constant,τ, is in accord with our "intuition"; high density, large volume, or high specificheat all tend to increase the time constant, while high heat transfer coefficient and large area willtend to decrease the time constant. This is the same form of equation and the same behavior youhave seen for the R-C circuit, as shown schematically below. The time dependence of the voltage in

the R-C circuit when the switch is opened suddenly is given by the equation 0=+RC

E

dt

dE. There

are, in fact, a number of physical processes which have (or can be modeled as having) this type ofexponentially decaying behavior.

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E0 C

R

Figure 6.2: Voltage change in an R-C circuit

Muddy pointsIn equation Q .V.c.

dTdtin

•= ρ (6.3), what is c? (MP HT.19)

In the lumped parameter transient heat transfer problem, does a high density "slow down"heat transfer? (MP HT.20)

7.0 Some Considerations in Modeling Complex Physical Processes

In Sections 5 and 6, a number of assumptions were made about the processes that we wereattempting to describe. These are all part of the general approach to modeling of physical systems.The main idea is that for engineering systems, one almost always cannot compute the processexactly, especially for fluid flow problems. At some level of detail, one generally needs to model,i.e. to define some plausible behavior for attributes of the system that will not be computed.Modeling can span an enormous range from the level of our assumption of uniform temperaturewithin the solid object to a complex model for the small scale turbulent eddies in the flow past acompressor blade. In carrying out such modeling, it is critical to have a clear idea of just what theassumptions really mean, as well as the fidelity that we ascribe to the descriptions of actual physicalphenomena, and we thus look at the statements we have made in this context.

One assertion made was that because hL

k<< 1 and on the basis of a heat balance,

kL

T T h T Tc w W−( ) ≈ −( )∞

we could assume

(T T )

(T T )1w

w

body interior −

−<<

∞

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Based on this, we said that Tbody is approximately uniform and T Tw ≈ body interior. Another aspect is

that setting 1<<Pk

hA erases any geometrical detail of the fin cross section. The only place where P

and A enter the problem is in a non-dimensional combination. A third assumption, made in the finproblem, was that the heat transfer at the far end can be neglected. The solution including this

effect, where h L

k

L is an axial Biot number is given as Equation (7.1). If the quantity h L

k

L is small,

you can see that Equation (7.1) reduces to the previous result (5.13).

T TT T

mL Bi m mL

mL Bi mL mL0

−−

=−( ) + ( ) −( )

+ ( )∞

∞

cosh sinh

cosh sinh

1 1ξ ξaxial

axal

/

/ (7.1)

and

Q

kAhP T T

Bi /mL mL

1Bi

mL mL0

axial

axial

⋅

∞−( )=

( ) +

+

tanh

tanh (7.2)

8.0 Heat Exchangers

The general function of a heat exchanger is to transfer heat from one fluid to another. The basiccomponent of a heat exchanger can be viewed as a tube with one fluid running through it andanother fluid flowing by on the outside. There are thus three heat transfer operations that need to bedescribed:

1) Convective heat transfer from fluid to the inner wall of the tube2) Conductive heat transfer through the tube wall3) Convective heat transfer from the outer tube wall to the outside fluid

Heat exchangers are typically classified according to flow arrangement and type of construction.The simplest heat exchanger is one for which the hot and cold fluids move in the same or oppositedirections in a concentric tube (or double-pipe) construction. In the parallel-flow arrangement ofFigure 8.1a, the hot and cold fluids enter at the same end, flow in the same direction, and leave at thesame end. In the counterflow arrangement of Figure 8.1b, the fluids enter at opposite ends, flow inopposite directions, and leave at opposite ends.

Alternatively, the fluids may be in cross flow (perpendicular to each other), as shown by thefinned and unfinned tubular heat exchangers of Figure 8.2. The two configurations differ accordingto whether the fluid moving over the tubes is unmixed or mixed. In Figure 8.2a, the fluid is said tobe unmixed because the fins prevent motion in a direction (y) that is transverse to the main-flowdirection (x). In this case the fluid temperature varies with x and y. In contrast, for the unfinned tubebundle of Figure 8.2b, fluid motion, hence mixing, in the transverse direction is possible, andtemperature variations are primarily in the main-flow direction. Since the tube flow is unmixed,

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both fluids are unmixed in the finned exchanger, while one fluid is mixed and the other unmixed inthe unfinned exchanger.

(a) (b)

Figure 8.1: Concentric tubes heat exchangers. (a) Parallel flow. (b) Counterflow

Cross flowT = f(x,y)

Cross flowT = f(x)

Tube flow Tube flow

x

y

(a) (b)

Figure 8.2: Cross-flow heat exchangers. (a) Finned with both fluids unmixed. (b) Unfinnedwith one fluid mixed and the other unmixed

To develop the methodology for heat exchanger analysis and design, we look at the problem of heattransfer from a fluid inside a tube to another fluid outside.

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r2

r1

TA

T1

T2

TB

Figure 8.3: Geometry for heat transfer between two fluids

We examined this problem before in Section 3.2 and found that the heat transfer rate per unit lengthis given by

( )

221

2

11

ln

2

hr

k

r

r

hr

kTTk

Q BA

++

−=

π& (8.1)

It is useful to define an overall heat transfer coefficient h0 per unit length as

( )BA TThrQ −= 022π& . (8.2)

From (8.1) and (8.2) the overall heat transfer coefficient, h0 , is

21

22

11

2

0

1ln

1hr

r

k

r

hr

r

h++= . (8.3)

We will make use of this in what follows.

A schematic of a counterflow heat exchanger is shown in Figure 8.4. We wish to know thetemperature distribution along the tube and the amount of heat transferred.

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Tb2

Ta2

Ta1

Tb1

Figure 8.4: Counterflow heat exchanger

To address this we start by considering the general case of axial variation of temperature in a tubewith wall at uniform temperature T0 and a fluid flowing inside the tube.

T2

x = Ldx

x = 0

T1

cooling

heating

T2

∆T2

T0

T1

∆T1

Figure 8.5: Fluid temperature distribution along the tube with uniform wall temperature

The objective is to find the mean temperature of the fluid at x, T(x), in the case where fluid comes inat x = 0 with temperature T1 and leaves at x = L with temperature T2 . The expected distribution forheating and cooling are sketched in Figure 8.5.

For heating (T0 > T), the heat flow from the pipe wall in a length dx is

q Ddx h D(T T)dx0

⋅

= −π π

where D is the pipe diameter. The heat given to the fluid (the change in enthalpy) is given by

ρπ

u cD

c dTm p

2

p4dT m=

⋅

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where ρ is the density of the fluid, um is the mean velocity of the fluid, cp is the specific heat of the

fluid and m⋅

is the mass flow rate of the fluid.

Setting the last two expressions equal and integrating from the start of the pipe, we find

dTT T

4hu c D

dx0 m p0

x

T1

T

−= ∫∫ ρ

. (8.4)

Carrying out the integration,

4hxu c D

dTT T

d T T

T TT T

m p 0

0

00 T1

T

T1

T

T1

T

ρ=

−= −

−( )−

= − −( )∫∫ ln , (8.5)

i.e.

lnT TT T

4hxu c D

0

0 1 m p

−−

⎛

⎝⎜

⎞

⎠⎟ = −

ρ. (8.6)

Equation (8.6) can be written as

xeTT

TT

10

0 β−=−

− (8.7)

where

βρ

π= =

⋅

4hu c D

hD

mcm pp

. (8.8)

This is temperature distribution along the pipe. The exit temperature at x = L is

pcm

hDL

eTT

TT &

π−

=−

−

10

20 (8.9)

The total heat transfer to the wall all along the pipe is

Q mc T Tp 1 2

⋅ ⋅

= −( ).

From Equation (8.9),

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mch DL

T TT T

p0 1

0 2

⋅

=−−

⎛

⎝⎜

⎞

⎠⎟

π

ln

.

The total rate of heat transfer is therefore

Qh DL T T

T TT T

Q h DL T

1 2

1 0

2 0

LM

⋅

⋅

=−( )

−−

=

π

π

ln,

or ∆

(8.10)

where ∆TLM is the logarithmic mean temperature difference, defined as

∆∆ ∆

∆∆

TT TT TT T

T T

TT

LM2 1

0 2

0 1

1

1

2

=−−−

=−

⎛

⎝⎜

⎞

⎠⎟ln

2

ln

. (8.11)

The concept of a logarithmic mean temperature difference is useful in the analysis of heatexchangers. We can define a logarithmic mean temperature difference for a counterflow heatexchanger as follows. With reference to Figure 8.4, an overall heat balance between the twocounterflowing streams is

Q m c T T m c T Ta pa a1 a2 b pb b2 b1

⋅ ⋅ ⋅

= −( ) = −( ). (8.12)

From a local heat balance, the heat given up by stream a in length dx is −⋅

m c dTa pa a. (There is a

negative sign since Ta decreases). The heat taken up by stream b is −⋅

m c dTb pb b. (There is a

negative sign because Tb decreases as x increases). The local heat balance is:

− = − = =⋅ ⋅ ⋅ ⋅

m c dT m c dT q dA q Ddxa pa a b pb b π (8.13)

Solving (8.13) for dTa and dTb, we find:

dTq dA

m c; dT

q dA

m ca

a pa

b

b pb

= − = −

⋅

⋅

⋅

⋅ (8.14)

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d T T d T1

m c

1

m cq dA

1W

1W

q Ddxa b

a pa b pba b

−( ) = = − −

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

= − −⎛

⎝⎜

⎞

⎠⎟⋅ ⋅

⋅ ⋅

∆ π (8.15)

where pcmW &= . Also, q& = h0 ∆T where h0 is the overall heat transfer coefficient. We can then say:

d TT

h D1

W1

Wdx0

a b

∆∆

= − −⎛

⎝⎜

⎞

⎠⎟π .

Integrating from x = 0 to x = L gives

lnT TT T

h DL1

W1

Wa2 b1

a1 b20

a b

−−

⎛

⎝⎜

⎞

⎠⎟ = − −

⎛

⎝⎜

⎞

⎠⎟π . (8.16)

Equation (8.16) can also be written as:

α−=−

−e

TT

TT

ba

ba

21

12 (8.17)

where

α π= −⎛

⎝⎜

⎞

⎠⎟h DL

1W

1W0

a b

We know that

Ta1 - Ta2 = Q⋅

/ Wa Tb2 - Tb1 = Q⋅

/ Wb . (8.18)

Thus

T T T T1

W1

Wa1 b2 a2 b1a b

−( ) − −( ) = −⎛

⎝⎜

⎞

⎠⎟

⋅

Q . (8.19)

Solving for the total heat transfer:

QT T T T

1W

1W

a1 b2 a2 b1

a b

⋅

=−( ) − −( )

−⎛

⎝⎜

⎞

⎠⎟

(8.20)

Rearranging (8.16) allows us to express 1

W1

Wa b

−⎛

⎝⎜

⎞

⎠⎟ in terms of other parameters as

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1W

1W

T TT T

h DLa b

a2 b2

a1 b2

0

−⎛

⎝⎜

⎞

⎠⎟ = −

−−

⎛

⎝⎜

⎞

⎠⎟ln

π. (8.21)

Substituting (8.21) into (8.20) we obtain a final expression for the total heat transfer for acounterflow heat exchanger:

Q h DLT T T T

T TT T

0a1 b2 a2 b1

a1 b2

a2 b1

⋅=

−( ) − −( )−−

⎛

⎝⎜

⎞

⎠⎟

π

ln

(8.22)

or

Q h DL T0 LM

⋅

= π ∆ (8.23)

8.1 Efficiency of a Counterflow Heat Exchanger

Suppose we know the two inlet temperatures Ta1, Tb1, and we need to find the outlet temperatures.

From (8.17)

Ta2 - Tb1 = (Ta1 - Tb2) e-α

Ta2 - Ta1 = Tb1 - Ta1 + (Ta1 - Tb2 ) e-α

Rearranging (8.18),

( )2112 aab

abb TT

W

WTT −+=

Thus

Ta2 - Ta1 = Tb1 - Ta1 + (Ta1 - Tb1 ) e-α -

WW

T T ea

ba1 a2−( ) −α

T T 1WW

e T T 1 ea1 a2a

ba1 b1−( ) −

⎛

⎝⎜

⎞

⎠⎟ = −( ) −( )− −α α

or

( ) ( )1121 baaa TTTT −=− η (8.24)

where η is the efficiency of a counterflow heat exchanger:

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ηα

α

α

α

=−

−=

−

−

−

−

−

⋅

⋅

−

1

1

eWW

e

e

m

m

c

ce

a

ba

b

pa

pb

1

1

(8.25)

From (8.24) and (8.25) we can find outlet temperatures Ta2 and Tb2:

T Tm c

m cT T

m c

m cT Tb2 b1

a pa

b pb

a1 a2a pa

b pb

a1 b2− = −( ) = −( )⋅

⋅

⋅

⋅η

We examine three examples.

i) m c m cb pb a pa

⋅ ⋅>

∆T can approach zero at cold end

η π→ −

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟⋅ ⋅

1 h DLh1

m c

1

m c0 0

a pa b pb

as , surface area ,

Maximum value of ratio 111

21 =−

−

ba

aa

TT

TT

Maximum value of ratio T TT T

m c

m c

b2 b1

a1 b1

a pa

b pb

−−

=

⋅

⋅

ii) m c m c b pb a pa

⋅ ⋅<

α η is negative , m c

m c as

b pb

a pa

→ [ ] → ∞

⋅

⋅ ( )ab WW <

Maximum value of T TT T

m c

m c

a1 a2

a1 b1

b pb

a pa

−−

=

⋅

⋅

1 of valueMaximum21

12 =−

−

ba

bb

TT

TT

iii) m c m ca pa b pb

⋅ ⋅=

0)( =− ba TTd

temperature difference remains uniform, η = 1

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9.0 Radiation Heat Transfer (Heat transfer by thermal radiation)

All bodies radiate energy in the form of photons moving in a random direction, with random phaseand frequency. When radiated photons reach another surface, they may either be absorbed, reflectedor transmitted. The behavior of a surface with radiation incident upon it can be described by thefollowing quantities:

α = absorptance - fraction of incident radiation absorbedρ = reflectance - fraction of incident radiation reflectedτ = transmittance – fraction of incident radiation transmitted

Figure 9.1 shows these processes graphically.

Incident radiationRadiation absorbed, α

Radiation transmitted, τ

Radiation reflected, ρ

Figure 9.1: Radiation Surface Properties

From energy considerations the three coefficients must sum to unity

α + ρ + τ = 1 (9.1)

Reflective energy may be either diffuse or specular (mirror-like). Diffuse reflections areindependent of the incident radiation angle. For specular reflections, the reflection angle equals theangle of incidence.

9.1 Ideal Radiators

An ideal thermal radiator is called a "black body". It has several properties:

1) It has α = 1, and absorbs all radiation incident on it.

2) The energy radiated per unit area is Eb = σT4 where σ is the Stefan-Boltzmann constant,

σ = 5.67 x 10-8 W

m K2 4 (9.2)

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The units of Eb are therefore W

m2.

The energy of a black body, E b , is distributed over a range of wavelengths of radiation. We can

define e dEd

Eb bλ λ λ

= ≈ ∆∆

, the energy radiated per unit area for a range of wavelengths of width ∆λ .

The behavior of eλ is given in Figure 9.2.

Figure 9.2: Emissive power of a black body at several temperatures - predicted and observed

( )maxλλ eT = 0.2898 cm K [adapted from: A Heat Transfer Textbook by Lienhard, J.]

The distribution of eλ varies with temperature. The quantity λT at the condition where eλ is amaximum is given by ( )

maxλλ eT = 0.2898 cm K. As T increases, the wavelength for maximum energy

emission shifts to shorter values. The frequency of the radiation, f, is given by f = c/λ so high energymeans short wavelengths and high frequency.

Mon

ochr

omat

ic e

mis

sive

pow

er, e

λ kW

/m -

mic

ron

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Figure 9.3: A cavity with a small hole (approximates a black body)

A physical realization of a black body is a cavity with a small hole. There are many reflections andabsorptions. Very few entering photons (light rays) will get out. The inside of the cavity hasradiation which is homogeneous and isotropic (the same in any direction, uniform everywhere).

Suppose we put a small black body inside the cavity as seen in Figure 9.4. The cavity and the blackbody are both at the same temperature.

Black body

Temperature T

Cavity

Figure 9.4: A small black body inside a cavity

The radiant energy absorbed by the black body per second and per m2 is αΒH, where H is theirradiance, the radiant energy falling on any surface inside the cavity. The radiant energy emitted bythe black body is EB. Since αB = 1 for a black body, H = EB. The irradiance within a cavity whosewalls are at temperature T is therefore equal to the radiant emittance of a black body at the sametemperature and irradiance is a function of temperature only.

9.2 Kirchhoff's Law and "Real Bodies"

Real bodies radiate less effectively than black bodies. The measurement of this is the emittance, ε ,defined by

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Emittance: ε = EEb

Values of emittance vary greatly for different materials. They are near unity for rough surfaces suchas ceramics or oxidized metals, and roughly 0.02 for polished metals or silvered reflectors. A tableof emittances for different substances is given at the end of this section as Table 9.1, taken from thebook by Lienhard.

The level of the emittance can be related to the absorptance using the following arguments. Supposewe have a small non-black body in the cavity. The power absorbed per unit area is equal to αH.The power emitted is equal to E. An energy balance gives E = Ebε = α H =α Eb.. Thus

εα ==bE

E (9.3)

Equation (9.3), the relation α = ε , is known as Kirchhoff's Law. It implies that good radiators aregood absorbers. It was derived for the case when Tbody = Tsurroundings (cavity) and is not strictly truefor all circumstances when the temperature of the body and the cavity are different, but it is true if αλ

= α ,ελ = ε, so the absorptance and emittance are not functions of λ . This situation describes a"gray body". Also, since αλ , ελ are properties of the surface, αλ = ελ .

9.3 Radiation Heat Transfer Between Planar Surfaces

E1

ρ2 1E

ρ ρ1 2 1E

ρ ρ1 22

1E

α2 1E

α ρ1 2 1E

ρ ρ α1 2 2E1

ρ ρ12

22

1E

ρ ρ α1 22

2 1E Surface 1

Surface 2

Figure 9.5: Path of a photon between two gray surfaces

Consider the two infinite gray surfaces shown in Figure 9.5. We suppose that the surfaces are thickenough so that, α + ρ = 1 (no radiation transmitted so transmittance = 0). Consider a photon emittedfrom Surface 1 (remembering that the reflectance ρ = 1 - α):

Surface 1 emits E1

Surface 2 absorbs E1 2α

← radiation from real body at T← radiation from black body at T

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Surface 2 reflects ( )21 1 α−E

Surface 1 absorbs ( ) 121 1 αα−E

Surface 1 reflects ( )( )121 11 αα −−E

Surface 2 absorbs ( )( ) 2121 11 ααα −−E

Surface 2 reflects ( )( )( )2121 111 ααα −−−E

Surface 1 absorbs ( )( )( ) 12121 111 αααα −−−E

The same can be said for a photon emitted from Surface 2:

Surface 2 emits 2E

Surface 1 absorbs 12αE

Surface 1 reflects ( )12 1 α−E

Surface 2 absorbs ( ) 212 1 αα−E

Surface 2 reflects ( )( )212 11 αα −−Eetc……

We can add up all the energy E1 absorbed in 1 and all the energy E2 absorbed in 2. In doing thebookkeeping, it is helpful to define ( )( )21 11 ααβ −−= . The energy E1 absorbed in 1 is

( ) ( ) ( )( ) K+−−−+− 12121121 1111 αααααα EE

This is equal to

( ) ( )...11 2121 +++− ββααE .

However

( ) +++=−=−

− 21 111

1βββ

β....

We thus observe that the radiation absorbed by surface 1 can be written as ( )β

αα−

−

1

1 121E .

Likewise ( )β

αα−−

11 212E is the radiation generated at 2 and absorbed there as well.

Putting this all together we find that

EE 1

1E12

2 1 2−−( )−

⎛

⎝⎜

⎞

⎠⎟ =

−

α α

βαβ

2 1

is absorbed by 1. The net heat flux from 1 to 2 is

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q EE E E E E E E

net 1 to 2 11 1 1

⋅

= −−( )−

−−

=− − − +( ) − + −

− − − +( )1

1 1

1

1 12 1 2 1 1 2 1 2 1 1 1 1 2 2 1

1 2 1 2

α α

βαβ

α α α α α α α α

α α α α

or

qE E⋅

=−

+ −net 1 to 21 2 2 1

1 2 1 2

α αα α α α

. (9.4)

If 21 TT = , we would have q& = 0, so from Equation (9.4)

2

2

1

1

αα

EE= = f (T) .

If body 2 is black, α2 = 1, and E2 = σT4.

4

1

1 TE

σα

=

4

1

41 T

Tσ

ασε

=

Therefore, again, 11 αε = for any gray surface (Kirchhoff's Law).

Using Kirchhoff's Law we find,

q⋅

=−

+ −net 1 to 2ε σ ε ε σ ε

ε ε ε ε1 1

42 2 2

41

1 2 1 2

T T

or, as the final expression for heat transfer between gray, planar, surfaces:

qT T1

42

4⋅

=−⎛

⎝⎞⎠

+ −net 1 to 2

σ

ε ε1 1

11 2

. (9.5)

HT-60

Example 1: Use of a thermos bottle to reduce heat transfer

ε1 = ε2 = 0.02 for silvered walls

T1 = 100 °C = 373 K T2 = 20 °C = 293 K

qT T W

m2

⋅

=−⎛

⎝⎞⎠

+ −=

−=1 to 2 1

0.021

0.021

1100 42099

6.9σ 1

42

4

For the same ∆T, if we had cork insulation with k = 0.04, what thickness would be needed?

qk T

L

⋅

=∆

so a thickness Lk T

qm= =

( )( )=

⋅

∆ 0.04 80

6.947 0. would be needed. The thermos is indeed a

good insulator.

Example 2: Temperature measurement error due to radiation heat transfer

Thermocouples (see Figure 9.6) are commonly used to measure temperature. There can be errorsdue to heat transfer by radiation. Consider a black thermocouple in a chamber with black walls.

1 2

Silvered Walls

Outside of thermos(Cold)

Inside of thermos(hot fluid)

HT-61

metal 1 metal 2

T1

T2

Measured Voltage

Note: The measured voltage is related to the difference between T1 and T2

(the latter is a known temperature).

Figure 9.6: Thermocouple used to measure temperature

Suppose the air is at 20 °C, the walls are at 100 °C, and the convective heat transfer coefficient is

h 15W

m K2= .

What temperature does the thermocouple read?

We use a heat (energy) balance on the control surface shown in Figure 9.7. The heat balance statesthat heat convected away is equal to heat radiated into the thermocouple in steady state.(Conduction heat transfer along the thermocouple wires is neglected here, although it would beincluded for accurate measurements.)

Heat in (radiation)

Control volume

Heat out (convection)Heat out (convection)

Twall

Tair

Ttc

Figure 9.7: Effect of radiation heat transfer on measured temperature.

HT-62

The heat balance is

hA T T A T Ttc air wall4

tc4

−( ) = −⎛⎝

⎞⎠σ (9.6)

where A is the area of the thermocouple. Substituting the numerical values gives

15 T 293 5.67 x 10 373 Ttc-8 4

tc4

−( ) = −⎛⎝

⎞⎠

from which we find Ttc = 51 °C = 324 K. The thermocouple thus sees a higher temperature than theair. We could reduce this error by shielding the thermocouple as shown in Figure 9.8.

Radiation shield

< 100 ϒ C

100 ϒ C

Figure 9.8: Shielding a thermocouple to reduce radiation heat transfer error

Muddy pointsWhich bodies does the radiation heat transfer occur between in the thermocouple?(MPHT.21)Still muddy about thermocouples. (MP HT.22)Why does increasing the local flow velocity decrease the temperature error for thethermocouple? (MP HT.23)

9.4 Radiation Heat Transfer Between Black Surfaces of Arbitrary Geometry

In general, for any two objects in space, a given object 1 radiates to object 2, and to other places aswell, as shown in Figure 9.9.

HT-63

Qnet T2

elsewhere

Figure 9.9: Radiation between two bodies

We want a general expression for energy interchange between two surfaces at different temperatures.This is given by the radiation shape factor or view factor, Fi - j. For the situation in Figure 9.10

F1 2- = fraction of energy leaving 1 which reaches 2F2 1- = fraction of energy leaving 2 which reaches 1F F1 2 2 1- - , are functions of geometry only

radiation

A2, T2

A1, T1

Body 1

Body 2

Figure 9.10: Radiation between two arbitrary surfaces

For body 1, we know that Eb is the emissive power of a black body, so the energy leaving body 1 isEb1 A1. The energy leaving body 1 and arriving (and being absorbed) at body 2 is Eb1 A1 F1-2. Theenergy leaving body 2 and being absorbed at body 1 is Eb2 A2 F2-1. The net energy interchange frombody 1 to body 2 is

Eb1 A1 F1-2 - Eb2 A2 F2-1 = Q1 2

⋅

− . (9.7)

Suppose both surfaces are at the same temperature so there is no net heat exchange. If so,

Eb1 A1 F1-2 - Eb2 A2 F2-1 = 0, but also Eb1 = Eb2. Thus

A1 F1-2 = A2 F2-1. (9.8)

HT-64

Equation (9.7) is the shape factor reciprocity relation. The net heat exchange between the twosurfaces is

Q A F E E A F E E1 2 1 1 2 b1 b2 2 2 1 b1 b2

⋅

− − −= −( ) −( )[ ]or

Example: Concentric cylinders or concentric spheres

T1

T2

Figure 9.11: Radiation heat transfer for concentric cylinders or spheres

The net heat transfer from surface 1 to surface 2 is

Q A F E E 1 2 1 1 2 b1 b2

⋅

− −= −( ) .

We know that F1-2 = 1, i.e. that all of the energy emitted by 1 gets to 2. Thus

Q A E E1 2 1 b1 b2

⋅

− = −( )

This can be used to find the net heat transfer from 2 to 1.

Q A F E E A F E E A E E2 1 2 2 1 b2 b1 1 1 2 b2 b1 1 b2 b1

⋅

− − −= −( ) = −( ) = −( )

View factors for other configurations can be found analytically or numerically. Shape factors aregiven in textbooks and reports (they are tabulated somewhat like Laplace transforms), and examplesof the analytical forms and numerical values of shape factors for some basic engineeringconfigurations are given in Figures 9.12 through 9.15, taken from the book by Incropera and DeWitt.

Surface Temperature (o

C) Surface Temperature (o

C)

Aluminum Asbestos 40 0.93 - 0.97

Polished, 98% pure 200 - 600 0.04 - 0.06 Brick

Commercial sheet 90 0.09 Red, Rough 40 0.93

Heavily oxidized 90 - 540 0.2 - 0.33 Silica 980 0.8 - 0.85

Brass Fireclay 9980 0.75

Highly polished 260 0.03 Ordinary refractory 1090 0.59

Dull plate 40 - 260 0.22 Magnesite refractory 980 0.38

Oxidized 40 - 260 0.46 - 0.56 White refractory 1090 0.29

Copper Carbon

Highly polished electrolytic 90 0.02 Filament 1040 - 1430 0.53

Slightly polished, to dull 40 0.12 - 0.15 Lampsoot 40 0.95

Black oxidized 40 0.76 Concrete, rough 40 0.94

Gold: pure, polished 90 - 600 0.02 - 0.035 Glass

Iron and Steel Smooth 40 0.94

Mild steel, polished 150 - 480 0.14 - 0.32 Quartz glass (2mm) 260 - 540 0.96 - 0.66

Steel, polished 40 - 260 0.07 - 0.1 Pyrex 260 - 540 0.94 - 0.74

Sheet steel, rolled 40 0.66 Gypsum 40 0.8 - 0.9

Sheet steel, strong rough oxide 40 0.8 Ice 0 0.97 - 0.98

Cast iron, oxidized 40 - 260 0.57 - 0.66 Limestone 40 - 260 0.95 - 0.83

Iron, rusted 40 0.61 - 0.85 Marble 40 0.93 - 0.95

Wrought iron, smooth 40 0.35 Mica 40 0.75

Wrought iron, dull oxidized 20 - 360 0.94 Paints

Stainless, polished 40 0.07 - 0.17 Black gloss 40 0.9

Stainless, after repeated heating 230 - 900 0.5 - 0.7 White paint 40 0.89 - 0.97

Lead Lacquer 40 0.8 - 0.95

Polished 40 - 260 0.05 - 0.08 Various oil paints 40 0.92 - 0.96

Oxidized 40 - 200 0.63 Red lead 90 0.93

Mercury: pure, clean 40 - 90 0.1 - 0.12 Paper

Platinum White 40 0.95 - 0.98

Pure, polished plate 200 - 590 0.05 - 0.1 Other colors 40 0.92 - 0.94

Oxidized at 590oC 260 - 590 0.07 - 0.11 Roofing 40 0.91

Drawn wire and strips 40 - 1370 0.04 - 0.19 Plaster, rough lime 40 - 260 0.92

Silver 200 0.01 - 0.04 Quartz 100 - 1000 0.89 - 0.58

Tin 40 - 90 0.05 Rubber 40 0.86 - 0.94

Tungsten Snow 10 - 20 0.82

Filament 540 - 1090 0.11 - 0.16 Water, thickness 0.1mm 40 0.96

Filament 2760 0.39 Wood 40 0.8 - 0.9

Oak, planed 20 0.9

Metals Nonmetals

Table 9.1:Total emittances for different surfaces [Adapted from: A Heat Transfer Textbook,

J. Lienhard]

HT-66

Figure 9.12: View Factors for Three - Dimensional Geometries

Figure 9.15: Fig 13.6--View factor for perpendicular rectangles with a common edge

[from: Fundamentals of Heat Transfer, F.P. Incropera and D.P. DeWitt, John Wiley and Sons]

Figure 9.13: Fig. 13.4--View factor for aligned parallel rectangles

Figure 9.14: Fig 13.5--View factor for coaxial parallel disk

HT-67

Heat Transfer References

Eckert, E. R. G. and Drake, R. M., Heat and Mass Transfer, McGraw Hill Book Company, 1959.

Eckert, E. R. G. and Drake, R. M., Analysis of Heat and Mass Transfer, Hemisphere Pub. Corp.,1987.

Incropera, F. P. and Dewitt, D. P., Fundamentals of Heat and Mass Transfer, Wiley, 1990.

Lienhard, J. H., A Heat Transfer Textbook, Prentice-Hall, 1987.

Mills, A. F., Heat Transfer, Irwin, 1992.

Muddiest points on heat transfer

HT.1 How do we quantify the contribution of each mode of heat transfer in a givensituation?

Developing the insight necessary to address the important parts of a complex situation(such as turbine heat transfer) and downplay (neglect or treat approximately) the otheraspects is part of having a solid working knowledge of the fundamentals. This is animportant issue, because otherwise every problem will seem very complex. One way tosort out what is important is to make order of magnitude estimates (similar to what wedid to answer when the one-dimensional heat transfer approximation was appropriate) tosee whether all three modes have to be considered. Sometimes one can rule out one ortwo modes on the basis of the problem statement. For example if there were a vacuumbetween the two surfaces in the thermos bottle, we would not have to considerconvection, but often the situation is more subtle.

HT.2 How specific do we need to be about when the one-dimensional assumption isvalid? Is it enough to say that dA/dx is small?

The answer really is “be specific enough to enable one to have faith in the analysis or atleast some idea of how good the answer is”. This is a challenge that comes up a greatdeal. For now, if we say that A is an area defined per unit depth normal to theblackboard then saying dA/dx is small, which is a statement involving a non-dimensionalparameter, is a reasonable criterion.

HT.3 Why is the thermal conductivity of light gases such as helium (monoatomic) orhydrogen (diatomic) much higher than heavier gases such as argon (monoatomic)or nitrogen (diatomic)?

To answer this, we need some basics of the kinetic theory of gases. A reference for thisis Castellan, Physical Chemistry, Benjamin/Cummings Publishers, 1986. Two factorcontribute, the collision cross section and the average molecular velocity. For the gasesmentioned above the dominant factor appears to be the velocity. The kinetic energy permolecule at a given temperature is the same and so the lower the molecular weight thehigher the average molecular velocity.

HT.4 What do you mean by continuous?

The meaning is similar to the definition you have seen in the math subjects. A way tostate it is that the function at a given location has the same value as we approach thelocation independent of the direction we approach from. To say this in a more physicalmanner, the temperature as a function of x has the same value at x = a, say, whether weapproach location a from the left or from the right. In terms of what the function lookslike, it will have no “jumps” or discontinuous (step) changes in value.

HT.5 Why is temperature continuous in the composite wall problem? Why is it continuousat the interface between two materials?

We can argue this point by supposing T were not continuous, i.e., there was a differenttemperature on the two sides of an interface between two materials, with the interfacetruly a sharp plane. If so, there would be a finite temperature difference across aninfinitesimal distance, leading to a very large (infinite in the limit) heat flux which wouldimmediately erase the temperature difference. This argument could also be applied toany location inside a solid of uniform composition, guaranteeing that the temperature iscontinuous in each material.

HT.6 Why is the temperature gradient dT/dx not continuous?

As derived in class, across an interface the heat flux is continuous. From the first law, fora thin control volume that encloses the interface the net heat flow into the control volumeis zero. As sketched below, the contribution from the heat flux at the upper and lowerends of the control volume is negligible so the heat flux in one side must be the same asthe flux out of the other. The heat flux, however, is related to the temperature gradient byrq k T= − ∇ , or, for one-dimension, q k

dT

dxx = − ⎛⎝

⎞⎠

. If the heat flux is continuous, and if

the thermal conductivity, k, is not the same in the two materials, then dT/dx is notcontinuous.

MaterialA

MaterialB

Interface

Control volume

qinqout

HT.7 Why is ∆T the same for the two elements in a parallel thermal circuit? Doesn't therelative area of the bolt to the wood matter?

In terms of the bolt through the wood wall, the approximation made is that the bolt andthe wood are both exposed to the same conditions at the two sides of the wall. Therelative areas of the bolt and the wood indeed do matter. Suppose we consider a squaremeter area of wood without bolts. It has a certain heat resistance. If we now add bolts to

the wall, the resistance of each bolt is RL

k Aboltbolt

bolt bolt

= . If there are n bolts, they will be

in parallel, and the effective area will be n times the area for one bolt. The thermal

resistance of the wood will be increased by a factor of A Awithoutbolts

withn bolts

/⎛

⎝⎜

⎞

⎠⎟ , which is larger

than unity.

In summary, the amount by which the heat transfer is increased depends on the fractionalarea with low thermal resistance compared to the fractional area with high thermalresistance.

HT.8 How do we know that δ' is not a fluid property?

The term ′⎛

⎝⎜

⎞⎠⎟

δk

represents the resistance to heat transfer for a unit area. The resistance to

heat transfer per unit area (1/heat transfer coefficient) can be computed for cases oflaminar flow, or measured experimentally where we cannot compute it, and it is foundthat for the same state variables, it can have a range of values of several orders ofmagnitude depending on the parameters I described (Reynolds number, surface condition,surface shape…). Put another way, if the value of the resistance is affected by the surfacecondition (smooth, bumpy, corrugated, etc.) how can the resistance be just a property ofthe fluid?

HT.9 What is the "analogy" that we are discussing? Is it that the equations are similar?

While the equations are similar, the concepts are deeper than that. The analogy is drawnbetween the heat transfer process (transfer of heat represented by heat flux) and themomentum transfer process (transfer of momentum represented by shear stress)

HT.10 In what situations does the Reynolds analogy "not work"?

The Reynolds Analogy is just that. It is not a law of nature, but rather a plausiblehypothesis that allows useful estimation of the heat transfer coefficients in manysituations in which little or no explicit heat transfer information exists. In the form wehave derived it, the Reynolds Analogy is appropriate for use in air, but it is not strictly

applicable if there are pressure gradients, or if the Prandtl number ( Pr =µc

kp ) is not

unity. However, the conceptual framework provided by the analogy has been founduseful enough that the analogy has been extended (in a more complex form, as brieflydiscussed in class) for application to these situations.

HT.11 In the expression 1

h.A, what is A?

A is the area normal to the heat flow. For a turbine blade, for example, it would be theouter surface area of the blade.

HT.12 It seems that we have simplified convection a lot. Is finding the heat transfercoefficient, h, really difficult?

We have indeed “simplified convection a lot”. We will look at heat transfer byconvection in more depth in a few lectures, but to answer the question in a few sentences,finding the heat transfer coefficient is a difficult problem, because it necessitatesdetermining the fluid dynamics; the latter is key to predictions of heat transfer. Evenwith present computational power, calculating the flow around aerospace devices withthe accuracy needed to be confident about heat transfer coefficients is not by any means a“standard” calculation. For some circumstances, it is still beyond the state of the art. Wewill concentrate on describing (i) the basic mechanisms of convective heat transfer and(ii) ways of estimating the heat transfer coefficient from known fluid dynamicinformation.

HT.13 What does the "K" in the contact resistance formula stand for?

The definition of the resistance comes out of thermal

driving

R

TQ

∆=& . The units of Q& are

Watts/meter2, so the units of Rthermal are [degrees x meters2/Watts]. The K is thus thesymbol for Kelvin, or degree centigrade.

HT.14 In the equation for the temperature in a cylinder (3.22), what is "r"?

The variable r denotes the radial coordinate, in other words the location of the point atwhich we want to know the temperature.

HT.15 For an electric heated strip embedded between two layers, what would thetemperature distribution be if the two side temperatures were not equal?

If the two temperatures on the outer surfaces of the composite layer were T T1 2 and , theheat flux from the two sides of the electrical heater would be different, but the sum of thetwo heat fluxes would still be equal to the heat generated per unit area and unit time. Infact heat could be coming into one side of the heater if one of the temperatures were highenough. (Sketch this out and prove it to yourself.) The temperature distribution wouldbe linear from the heater temperature to the two surface temperatures. If this example isnot clear, please come and see me.

HT.16 Why did you change the variable and write the derivative d Tdx

2

2 as d T - T

dx

2

2∞( )

in

the equation for heat transfer in the fin?

The heat balance and derivation of the equation for temperature (5.3) is given in Section5.0 of the notes. This is

d T

dx

Ph

AkT T

2

2 0− −( ) =∞ (5.3)

It is not necessary to change variables, and one could solve (5.3) as is

(or in the form d T

dx

Ph

AkT T

2

2 − = ∞ ). However, since: (1) the reference temperature is T∞,

and what is of interest is the difference T T− ∞ , (2) making the substitution results in asimpler form of the equation to be solved, and (3) the derivative of T T− ∞ is the same asthe derivative of T , I found it convenient to put it the equation in the form of Eq. (5.5),

d T T

dx

Ph

AkT T

2

2 0−( )

− −( ) =∞∞ . (5.5)

HT.17 What types of devices use heat transfer fins?

A number of types of heat exchangers use fins. Examples of the use of fins you mayhave seen are cooling fins on motorcycle engine heads, cooling fins on electric powertransformers, or cooling fins on air conditioners.

HT.18 Why did the Stegosaurus have cooling fins? Could the stegosaurus have "heatingfins"?

My knowledge of this issue extends only to reading about it in the text by Lienhard (seereference in notes). The journal article referenced in that text is:Farlow, J. O., Thompson, C. V., and Rosner, D.E., “Plates of the Dinosaur Stegosaurus:Forced Convection Heat Loss Fins?, Science, vol. 192, no. 4244, 1976, pp. 1123-1125and cover.

HT.19 In equation Q .V.c.dTdtin

•= ρ (6.3), what is c?

c is the specific heat for a unit mass. We don’t have to differentiate between the twospecific heats for a solid because the volume changes are very small, unlike a gas.

HT.20 In the lumped parameter transient heat transfer problem, does a high density"slow down" heat transfer?

It doesn’t. The high density slows down the rate at which the object changestemperature; high density means more “heat capacity”.

HT.21 Which bodies does the radiation heat transfer occur between in the thermocouple?

The radiation heat transfer was between the walls (or more generally the boundaries ofthe duct) and the thermocouple. The boundaries may not be at the same temperature ofthe flowing fluid, for example in a turbine. If the boundaries are not at the sametemperature as the fluid, and they radiate to the thermocouple, there can be an error in thetemperature that the thermocouple reads. The discussion was about different possiblesources of this type of error.

HT.22 Still muddy about thermocouples.

I didn’t mean to strew confusion about these. As on page 59 of the notes, if we take apair of dissimilar wires (say copper and constantan (an alloy of tin and several othermetals) or platinum and rhodium) which are joined at both ends and subject the twojunctions to a temperature difference, it is found that a voltage difference will be created.If we know the temperature of one junction (say by use of an ice bath) and we know therelation between voltage and temperature difference (these have been measured in detail)we can find the temperature of the other junction from measurement of the voltage. Theassumption is that the temperature of the junction is the temperature that is of interest; thepossible error in this thinking is the subject of muddy point 1.

HT.23 Why does increasing the local flow velocity decrease the temperature error for thethermocouple?

The heat transferred by convection is ( )fluidlethermocoupconvection TThQ −=& . If we neglect

conduction from the wire junction (in other words assume the thermocouple wires arethin and long), the heat balance is between convection heat transfer and radiation heattransfer. For a fixed convectionQ& if the heat transfer increases the temperature difference

between the thermocouple and the fluid decreases. We have seen, however, that the heattransfer may be estimated using the Reynolds analogy (Section 3.1). For fixed skinfriction coefficient the higher the velocity the higher the heat transfer coefficient.

Index

absorptance HT-54absorption HT-57adiabatic 0-5, 0-9adiabatic efficiency see efficiencyadiabatic flame temperature 2C-7

Biot number HT-30, HT-36black body HT-56, HT-63 see also radiationblade see turbine bladeBrayton cycle 1A-5, 1C-5, 2A-5

efficiency 1A-7, 1C-7, 2A-5 maximum work 1A-10 net work 2A-5

Breguet Range equation 2A-9

Carnot cycle 1A-3, 1C-1, 1C-4 efficiency 1A-4, 1C-2 two-phase 2B-8

coefficients of absorption see absorptance reflection see reflectance transmission see transmittance

coefficient of performance 1A-12combined First and Second Law 1B-5composite slab see slabcompressor adiabatic efficiency 2A-15conduction HT-5, HT-26

heat flux HT-6one-dimensional HT-7, HT-8, HT-14,HT-17,HT- 26

continuum 0-1control volume 0-12 see also first law for acontrol volumeconvection HT-5, HT-19, HT-26

heat transfer coefficient HT-20, HT-23near a wall HT-26

convective heat transfer coefficient seeconvectionCOP see coefficient of performancecritical point 2B-3critical radius of insulation HT-30cycle see thermodynamic cycle

cylindrical geometry see non-planar geometry

Diesel cycle 2A-4diffusivity HT-22dissipation 1C-10drag HT-24

efficiencyadiabatic efficiency 1C-17, 2A-14, 2A-15overall efficiency 2A-10propulsive efficiency 2A-10thermal efficiency 1A-2, 1A-4, 1A-7,1C-2, 1C-7, 2A-2, 2A-5, 2A-8, 2B-14

emittance HT-57energy 0-5

internal 0-5, 0-6, 0-7kinetic 0-5potential 0-5

enthalpy 0-6, 0-7of formation 2C-4

entropy 1B-2, 1C-24 generated 1C-10, 1C-20 microscopic 1D

entropy changes 1B-5, 1B-6, 1C-9equilibrium 0-2, 0-3extensive properties 0-1

fin HT-36First Law of thermodynamics 1B-3

control mass 0-5control volume 0-13, 0-14steady flow 0-14

Fourier's law HT-7free expansion 1A-14, 1B-9fuel-air ratio 2A-12, 2C-2

Gibbs equation 1B-5gray body HT-57, HT-59

heat 0-5two-phase systems 2B-6

heat exchanger HT-24, HT-45efficiency HT-52

heat flux HT-7heat pumps 1A-11heat reservoir

heat transfer from 1B-7 heat transfer between two 1B-7 single 1B-8

heat sources HT-33heat transfer coefficient see convectionh-s diagram 1C-6

ideal gases 0-7enthalpy 0-6, 0-7entropy changes 1B-6equation of state 0-7internal energy 0-6, 0-7specific heats 0-7, 0-8 see also specificheats

insulated wall HT-12intensive properties 0-1internal energy see energyirreversibility 1A-12, 1C-8, 1C-24isentropic efficiency see adiabatic efficiencyisothermal process 1A-1, 1A-14, 1B-9, 1C-18

Kirchoff's law HT-57

lost work 1C-9, 1C-10, 1C-13

maximum work 1A-10, 1C-2microscopic see entropy

Newton's Law of cooling HT-20non-planar geometry HT-14

cylindrical HT-14, HT-27, HT-30, HT-47, HT-65spherical HT-17, HT-65

one-dimensional problem HT-34 see also conduction

Otto cycle 2A-1 efficiency 2A-2 work produced 2A-3

overall efficiency see efficiency

phase 2B-1planar geometry HT-7, HT-26, HT-59Prandtl number HT-22process 0-2

adiabatic 0-5isothermal 1A-1, 1A-14reversible 0-4, 1B-4 see alsoreversibilityreversible and adiabatic 0-9

propulsive efficiency see efficiency

quality 2B-5quasi-equilibrium 0-3quasi-steady heat transfer HT-14

radiation HT-5, HT-54of a black body HT-54

ramjet 2A-6 efficiency 2A-8 fuel-air ratio 2A-12 specific impulse 2A-8, 2A-12 thrust 2A-7

Rankine cycle 2B-13 efficiency 2B-14 superheated 2B-16

reflectance HT-54reflection HT-54, HT-57refrigerators 1A-11reversibility 0-4, 0-9, 1A-12, 1A-15, 1A-16,

1B-4, 1C-2, 1C-7, 1C-24Reynolds analogy HT-20, HT-23, HT-25

saturated liquid 2B-3saturated vapor 2B-3saturation state 2B-1Second Law of thermodynamics 1B-1, 1B-4

Combined with First Law 1B-5shape factor HT-63

reciprocity relation HT-64shear stress HT-20, HT-24skin friction coefficient HT-23slab HT-9, HT-16

composite HT-11with heat sources HT-33

sources see heat sourcesspecific heats 0-6, 0-7, 0-8, 0-9

specific impulse 2A-8, 2A-12specific properties 0-1spherical geometry see non-planar geometrystagnation state 1C-15Stanton number HT-23state see thermodynamic statestate equation 0-3, 0-7steady flow energy equation 0-14steady heat transfer HT-8 see also quasi-steadyStephan-Boltzmann constant HT-54stoichiometric 2C-1superheated Rankine cycle see Rankine cyclesuperheated vapor 2B-3

temperature distributionthrough a cylinder section HT-15through a spherical geometry HT-18

temperature scale 1C-3theoretical air 2C-1thermal conductivity HT-7thermal efficiency see efficiencythermal resistance circuits

series HT-11, HT-12, HT-26parallel HT-12

thermocouples HT-60thermodynamic cycle 1A-2, 1C-19

Brayton cycle see Brayton cycleCarnot cycle see Carnot cyclecombined cycle 2B-19Diesel cycle see Diesel cycleOtto cycle see Otto cyclenon ideal 2A-14Rankine cycle see Rankine cycle

thermodynamic properties 0-1, 0-2extensive properties 0-1intensive properties 0-1specific properties 0-1

thermodynamic state 0-2, 0-3thermodynamic system 0-2

closed system 0-1simple system 0-2

throttle 0-9, 1C-14thrust 2A-7transient heat transfer HT-42transmittance HT-54

T-s diagrams 1C-4turbine adiabatic efficiency 1C-17, 2A-14turbine blade HT-19turbulent boundary layer HT-20two-phase systems 2B

heat transfer 2B-6 internal energy 2B-6 P-T diagram 2B-1 P-v diagram 2B-2 T-v diagram 2B-3 work 2B-6

unavailable energy 1C-11

vapor dome 2B-2, 2B-4 view factor see shape factor viscosity

dynamic HT-20kinematic HT-22

work 0-4, 1A-1 from a single heat reservoir 1B-8 isothermal process 1A-2, 1A-14 net work 2A-5, 2A-17 two-phase system 2B-6

Zeroth Law of Thermodynamics 1B-3