Review - Comp Flow - Quick Review of Compressible Flow Topics

8/20/2019 Review - Comp Flow - Quick Review of Compressible Flow Topics

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MAE 3050 November 26, 2008

Review of Compressible Flow Topics

D. A. Caughey Sibley School of Mechanical & Aerospace Engineering

Cornell University

Ithaca, New York 14853-7501

These notes provide a brief background to the thermodynamics of perfect gases and to compressible flowtheory. The notes are meant to complement the presentations in the text books used in M&AE 3050,Introduction to Aeronautics, and M&AE 5060, Aerospace Propulsion Systems.


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Chapter 1

Thermodynamic Review

1.1 Thermodynamic Equilibrium and Properties

There are many important concepts and results from the engineering science of thermodynamics that areuseful in fluid mechanics. Most of these results are based on the assumption that the substances withwhich we deal are in thermodynamic equilibrium, which implies that all local mechanical, physical, andthermal properties are independent of both position and time. Less is known about the thermodynamicsof nonequilibrium states, but numerous experiments have verified that the results for equilibrium statesapproximate the behavior of the nonequilibrium states common in most fluid mechanical problems. Thus,although the departures in the properties of moving fluids from equilibrium may seem large, their effect onthermodynamic relations is, in fact, quite small.

Knowledge of thermodynamics is not required to understand the fluid mechanical behavior of liquids inmost cases. Important thermodynamic results relevant to gas flows are summarized here. A brief review of thermodynamics is provided in Section 1.1.1.

Fluid properties such as pressure, specific volume, and temperature are thermodynamic properties, or statevariables. For a pure substance, any two state variables determine the thermodynamic state of the material,so we can always, at least in principle, write an equation of state in the form

f ( p, v, T ) = 0 , (1.1)

which relates the pressure p, specific volume v, and the temperature T . This equation of state can be writtenequivalently as

p = f p(v, T ) . (1.2)

The most common, and useful for aerodynamics, example of an equation of state is the perfect gas law, whichdescribes the behavior of gases at relatively low pressures and densities (relative to their critical points). Theperfect gas law takes the form

pv = RT , (1.3)

where R is the gas constant. In fluid mechanics, it is more common to work with the fluid density, which isthe inverse of the specific volume (ρ = 1/v), and so the perfect gas law can be written

p = ρRT . (1.4)

See Section 1.2 for more details about the perfect gas law.

Pressure and density are defined more carefully in subsequent sections. The temperature is a state variablethat governs the tendency for heat to flow between two masses in thermal contact: heat always flows from

3


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4 CHAPTER 1. THERMODYNAMIC REVIEW

the mass at higher temperature to the mass at lower temperature. If no heat is transferred when two massesare brought into thermal contact, their temperatures must be equal.

The amount of heat that must be added to a substance to raise its temperature by a given amount is animportant property, called the specific heat. The specific heat can be determined either at constant volume,in which case it can be identified as

cv = ∂e∂T

v

, (1.5)

where e is the specific1 internal energy, and the subscript v indicates that the derivative is taken at constantvolume. The specific heat at constant pressure can be identified as

c p =

∂h

∂T

p

, (1.6)

where

h ≡ e + pv = e + pρ

(1.7)

is the specific enthalpy, and the subscript p indicates that the derivative is taken at constant pressure.

The change in specific entropy s of a substance is defined as

∆s = s2 − s1 ≡ 21

dq

T

rev

, (1.8)

where q is the heat added per unit mass during a reversible process between states 1 and 2. If the entropy isintroduced into an expression of the first law of thermodynamics, the following differential equation of stateresults:

T ds = de + p dv = dh− v d p . (1.9)Since only state variables appear in Eq. (1.9), this equation relates changes in these variables for a substancewhenever it is in thermodynamic equilibrium; i.e., it is valid even for processes that are neither reversiblenor adiabatic.

1.1.1 Review of Thermodynamics

The study of most problems in fluid mechanics involving liquids, and even those involving gases when thepressure changes are not too large, does not require a knowledge of thermodynamics. This section is intendedto be a very brief review of the thermodynamic concepts required in the study of gas flows at high speeds,which are treated in more detail in Chapter 2.

The thermodynamic state of any pure substance is determined uniquely by any two state variables. Themost common state variables used to describe fluids are the pressure p, the temperature T , and the specificvolume v. In fluid dynamics, it often is more convenient to work with the inverse of the specific volume,

which is called the fluid density:ρ =

1

v . (1.10)

The first law of thermodynamics introduces the specific internal energy e. The first law states that workand heat are distinguishable, but equivalent, forms of energy transfer and that the rate of change of internalenergy of a system consisting of a mass of fluid at rest is equal to the difference between the rates at which

1The adjective “specific” is used to indicate the amount of a variable per unit mass. Thus, for example, while the internal

energy of a fluid is proportional to its mass, the specific internal energy, or internal energy per unit mass, is independent of the

total amount of mass in the system.


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1.1. THERMODYNAMIC EQUILIBRIUM AND PROPERTIES 5

heat is added to, and work is done by, the system. If we represent these rates, per unit mass of fluid in thesystem, as δq and δw, respectively, then the first law can be written

δe = δq − δw . (1.11)

The most important way in which a fluid system does work on its surroundings is by the action of the fluidpressure on the boundary of the system. If this work is done by a reversible process, then δw = p dv, and

the first law can be writtenδe = δq − p dv . (1.12)

The specific heat represents the amount of heat that must be added to raise the temperature of a unit massof the substance by a unit temperature

c = δq

δT . (1.13)

Two particular versions of this quantity are of interest, depending on whether the process is carried out atconstant volume or at constant pressure. That is, we define the specific heat at constant volume

cv ≡

δq

δT

v

, (1.14)

and the specific heat at constant pressure

c p ≡

δq

δT

p

. (1.15)

As seen directly from Eq. (1.12), the specific heat at constant volume is equivalent to

cv =

∂e

∂T

v

. (1.16)

Also, since from Eq. (1.12)δq = δe + p dv , (1.17)

we see that if we define the specific enthalpy h as

h ≡ e + pv , (1.18)then Eq. (1.17) can be written in the form

δq = dh− v d p . (1.19)From this it is seen that the specific heat at constant pressure is equivalent to

c p =

∂h

∂T

p

. (1.20)

The second law of thermodynamics implies the existence of another state variable, the specific entropy s,defined such that in a reversible process the change in entropy is proportional to the heat added, with the

constant of proportionality being the inverse of the temperature. That is, for a reversible process the changein specific entropy is given by

δs ≡

δq

T

rev

. (1.21)

A direct consequence of this definition is that a reversible process that also is adiabatic takes place at constantentropy; such a process is said to be isentropic.

Substitution of the definition Eq. (1.21) into the first law in the form of Eq. (1.17) gives

T ds = de + p dv . (1.22)


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Since Eq. (1.22) contains only state variables, it must be independent of the path along which the processtakes place. Thus, it can be interpreted as a (differential) equation of state, describing how changes ininternal energy, entropy, and specific volume must be related. This equation of state also can be written interms of the density ρ as

T ds = de− pρ2

dρ . (1.23)

Using Eq. (1.18) to replace the internal energy with the enthalpy also allows Eqs. (1.22) and (1.23) to be

written asT ds = dh− v d p , (1.24)

or

T ds = dh− d pρ

. (1.25)

Equations (1.22), (1.23), (1.24), and (1.25) describe the behavior of any pure substance. In particular,these equations can be used to relate changes in internal energy and density (or enthalpy and pressure) forisentropic processes by setting ds = 0.

The implications of these equations for the behavior of gases that obey the perfect gas law are described inSection 1.2.

1.2 Perfect gases

Over a fairly broad range of conditions, the equations of state for many gases are well approximated by theequation of state for a perfect gas. The perfect gas law states that the pressure p, density ρ, and temperatureT are related according to

p = ρRT , (1.26)

where R is called the gas constant. The gas constant for a particular gas can be determined from theuniversal gas constant Ru by

R = RuM m

, (1.27)

where M m is the molecular weight of the gas. The value of the universal gas constant is

Ru = 8314.3 kg m2

s2 K mole . (1.28)

Thus, air, which has an average molecular weight of M m = 28.964 kg/mole, has a gas constant of

Rair = 287.05 m2

s2 K . (1.29)


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1.2. PERFECT GASES 7

1.2.1 Example: Perfect Gas Law – Air

In the 1962 U. S. Standard Atmosphere [1], the temperature and pressure at 11,000 m above sea level areT = 216.65 K and p = 22, 632 Pa, respectively. What is the corresponding density?

Solution: Solving the perfect gas law for density gives

ρ = p

RT =

22, 632 Pa

(287.05 m2/(s2 K))(216.65 K) = 0.36392 kg/m3 . (1.30)

1.2.2 Example: Perfect Gas Law – Uranium Hexafluoride

One of the heaviest gases of technical importance is uranium hexafluoride, for which the chemical formulais U F6. The molecular weight of this gas is 352 kg/mole. What are the values of the density and specificgravity of this gas at standard sea-level conditions?

Solution: The gas constant for uranium hexafluoride is

RUF6 = RuM m

= 8314.3 kg m2/(s2 K mole)

352 kg/mole = 23.62 m2/(s2 K) . (1.31)

Thus, the perfect gas law gives for the density

ρ = p

RT =

101, 325 Pa

[23.62 m2/(s2 K)](288.15 K) = 14.89 kg/m3 . (1.32)

The specific gravity is then

SG = ρUF6

ρair=

14.89 kg/m3

1.204 kg/m3 = 12.37 . (1.33)

It is an interesting coincidence that the specific gravities of the heaviest known gas and the heaviest liquid,mercury, (SGHg = 13.6) are approximately the same, when referenced to the most common gas (air) andliquid (water), respectively.

1.2.3 Caloric Properties of Perfect Gases

The specific heat at constant volume cv for a perfect gas is a function only of the temperature T

cv = cv(T ) . (1.34)

See Section 1.5 for proof that this follows directly from the equation of state for a perfect gas. Moreover, if the gas is calorically perfect , i.e., if the constant-volume specific heat is constant, then the specific internal

energy e is given by

e =

cv dT = cvT . (1.35)

The corresponding specific enthalpy h, defined as

h ≡ e + pρ

, (1.36)

can be written

h =

c p dT = c pT . (1.37)


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The relationship between the specific heats at constant volume and at constant pressure for a caloricallyperfect gas can then be seen from

h = e + p

ρ = cvT + RT = (cv + R)T = c pT . (1.38)

Thus, for a calorically perfect gas, the difference in specific heats at constant pressure and constant volumeis given by

c p − cv = R . (1.39)

The Ratio of Specific Heats

The ratio γ of the specific heat at constant pressure to that at constant volume

γ ≡ c pcv

(1.40)

plays a very important role in the behavior of compressible fluids. Introduction of this parameter intoEq. (1.39) allows us to write

cv =

R

γ − 1 (1.41)and

c p = γR

γ − 1 . (1.42)

The ratio of specific heats γ is related to the number of modes in which energy can be stored by the moleculesof the gas. It is thus affected by both, the structure of the molecule and quantum effects. If the temperatureof the gas is such that m modes of energy storage are excited, then it can be shown that

γ m = m + 2

m (1.43)

For a fairly broad range of common temperatures, the rotational energy modes of common molecules are

fully excited, but there is no significant energy in vibrational modes. In this case the value of m is the sumof the three translational modes and the number of non-degenerate rotational modes:

m = 3 for monoatomic gases (such as helium), thus γ 3 = 5/3;

m = 5 for diatomic gases (such as oxygen and nitrogen), thus γ 5 = 7/5;

m = 6 for (non-collinear) triatomic gases (such as carbon-dioxide and water), thus γ 6 = 8/6.

For further information about the values of γ for various gases, see any common thermodynamics textbook(e.g., [3]).

1.3 Compressibility

All fluids experience changes in density with changes in pressure. For liquids the changes usually are quitesmall, but for gases the changes can be very large. These changes in density are equivalent to changes inspecific volume, since the density ρ is the inverse of the specific volume v :

ρ ≡ 1v

. (1.44)


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1.3. COMPRESSIBILITY 9

The compressibility of a substance represents the fractional amount by which its volume changes in responseto changes in pressure. Thus, the compressibility β is defined as

β ≡ − lim∆ p→0

∆v/v

∆ p = −1

v

dv

d p , (1.45)

where v = 1/ρ is the specific volume. The minus sign is introduced to make β positive, since an increase inpressure generally results in a decrease in specific volume. By virtue of Eq (1.44), an equivalent definition

of the compressibility is

β ≡ 1ρ

dρ

d p . (1.46)

Note that the minus sign is absent from Eq. (1.46), since a decrease in specific volume corresponds to anincrease in density.

The definitions in Eqs. (1.45) and (1.46) are not complete until we specify how the temperature (or someother state variable) changes during the compression process. Two specifications are commonly used. Theisothermal compressibility β T is defined for a process in which the temperature is held fixed

β T ≡ −1v

∂v

∂p

T

, (1.47)

while the isentropic compressibility β s is defined for a process in which the entropy is held fixed

β s ≡ −1v

∂v

∂p

s

. (1.48)

A thermodynamic analysis shows that the ratio of the isothermal compressibility to the isentropic compress-ibility is equal to the ratio of the specific heat at constant pressure c p to the specific heat at constant volumecv:

β T β s

= c pcv

= γ . (1.49)

For liquids, there is only a very small change in temperature with pressure in isentropic processes, and theisothermal value of the compressibility usually is used. The isothermal compressibility of liquids is a weak

function of temperature that usually must be determined empirically. It also is common to work with thereciprocal of the compressibility, which is called the bulk modulus of elasticity E , defined as

E ≡ −v

∂p

∂v

T

= ρ

∂p

∂ρ

T

= 1

β T . (1.50)

From Eq. (1.50), it is obvious that the dimensions of the bulk modulus are the same as those of pressure.

The dependence of liquid densities on pressure for an isentropic process is reasonably well approximated bythe empirical relationship

ρ

ρref =

p/pref + B

B + 1

1/n, (1.51)

where B and n are constants. The values B = 3, 000 and n = 7 provide a good fit for both fresh and seawater when atmospheric reference conditions are chosen [2]. Differentiation of Eq. (1.51) gives the isentropiccompressibility at the reference conditions p = pref and ρ = ρref as

β s = 1

n(B + 1) pref . (1.52)

The bulk modulus of water at one atmosphere pressure is a weak function of temperature and passes througha maximum value of E = 2.292 × 109 Pa at a temperature T = 49oC. The bulk modulus of water at oneatmosphere pressure and T = 20oC is E = 2.182× 109 Pa.


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1.3.1 Example: Compressibility of Water

To what value must the pressure be increased from one atmosphere to increase the density of water atT = 20oC by 1%?

Solution: Equation (1.51) can be solved for the pressure ratio to give

p

pref = (B + 1)

ρ

ρref

n−B . (1.53)

Thus, a 1% increase in density requires

p

pref = (B + 1) (1.01)

n −B = 3001(1.01)7 − 3000 = 217.5 (1.54)

whence p = 217.5 pref = 217.5 atm . (1.55)

Alternatively, since the density change is so small, we can approximate Eq. (1.46) as

1ρ∂ρ∂p ≈ ∆ρ/ρ

∆ p = β s , (1.56)

which can be solved for the pressure change ∆ p

∆ p = 1

β s

∆ρ

ρ , (1.57)

using the value of isentropic compressibility given by 1/β s = n(B + 1) pref , corresponding to the referenceconditions. Thus, for a 1% change in density

∆ p = 0.01

β s= 0.01n(B + 1) pref = 0.01(7)(3, 001) pref = 210 atm . (1.58)

This approximation gives a slightly smaller value for the required pressure increase because it does not

account for the increase in compressibility with pressure. Either result indicates that the pressure must beincreased to more than 200 atm to produce a 1% change in density of water. Thus, the compressibility of water must be taken into account only for extremely large pressure changes; for most flows, water can betreated as a fluid whose density is independent of the pressure.

As is seen in Example 1.3.1, changes in the density of water due to changes in pressure are very smallunless the pressure changes are very large. Thus, it is usually a good approximation to treat water as anincompressible fluid. It is important to emphasize, however, that does not necessarily mean that the densityis constant. Even though the compressibility of water usually can be neglected, changes in the density of water can become important in certain stratified flows. Such density changes usually are the result of changesin temperature and/or the amount of dissolved solids. This results in incompressible flows in which it isnecessary to account for changes in fluid density.

For gases, the compressibility can be computed from the equation of state. For perfect gases, which obeythe equation of state

p = ρRT , (1.59)

the isothermal compressibility is

β T = 1

p . (1.60)

Equation (1.49) then shows that the isentropic compressibility for a perfect gas is

β s = 1

γp . (1.61)


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1.3. COMPRESSIBILITY 11

The compressibility is related to the speed of sound in the fluid. In Chapter 2 it is shown that the speed of sound a is given by

a =

∂p

∂ρ

s

=

1

ρβ s. (1.62)

Since the isothermal and isentropic compressibilities of liquids are nearly the same, we can write

a = 1

ρβ s≈ E/ρ (1.63)

for liquids, while for a perfect gas,

a =

γRT =

γp

ρ . (1.64)

Thus, the speed of sound in a perfect gas is proportional to the square root of the temperature.

1.3.2 Isentropic Processes involving Perfect Gases

Flows in which the effects of heat transfer and of fluid friction are small are well approximated as isentropic

processes. The differential equation of state for a pure substance relates the changes in entropy s to thosein enthalpy h and those in density ρ according to

T ds = dh− d pρ

(1.65)

Thus, for a constant entropy (or isentropic) process ( ds = 0) involving a perfect gas, we can write

dh− d pρ

= c p dT − d pρ

= 0 ,

γR

γ − 1 d

p

ρR

− d p

ρ = 0 .

(1.66)

Since R is constant, this can be simplified to

d p

p − γ dρ

ρ = 0 (1.67)

which can be integrated to give p

ργ = Const. (1.68)

This important equation relates the pressure and density for any isentropic process involving a perfect gas.For any such isentropic process, Eq. (1.68) and the equation of state Eq. (1.26) can be used to relate thechanges in any one state variable to those of only one other state variable:

p

p1=

ρ

ρ1γ

(1.69)

p

p1=

T

T 1

γ/(γ −1)

(1.70)

ρ

ρ1=

T

T 1

1/(γ −1)(1.71)


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1.3.3 Example: Isentropic Processes

What percentage pressure change is required to produce an increase in density of 5 per cent for an isentropicprocess in air?

Solution: Assuming the process is isentropic, and that for air γ = 1.4, we have

p + ∆ p

p =

ρ + ∆ρ

ρ

γ = (1.05)

1.4= 1.0707 (1.72)

Thus, an increase of approximately 7 per cent is required. Under standard sea level conditions, this corre-sponds to a pressure change of approximately 7,165 Pa (or 1.04 psi). Note that since the pressure and densitychanges are small, these could have been predicted approximately by Eq. (1.67), which can be written as

∆ p

p ≈ γ ∆ρ

ρ = 1.4(0.05) = 0.07 (1.73)

1.4 The Speed of Sound

The speed at which infinitesimal pressure disturbances propagate in a fluid, i.e., the speed of sound, playsa very important role in the flow of compressible fluids. The conservation laws of mass and momentum areused in Chapter 2 to show that the speed of sound is equal to

a2 = d p

dρ . (1.74)

In addition, the processes occurring sound waves of all but the highest frequencies are slow enough to benearly reversible, and the thermal conductivities of common gases, such as air for a broad range of densities,

are sufficiently small that the process of sound propagation also is very nearly adiabatic, so that sound wavesare very nearly isentropic. Thus we can write the equation determining the sound speed as

a2 =

∂p

∂ρ

s

, (1.75)

where the subscript ()s denotes the isentropic derivative of pressure with respect to density. Finally, for acalorically perfect gas undergoing an isentropic process it is seen from Eq. (1.67) that

d p

p − γ dρ

ρ = 0 , (1.76)

so we can write

a2 =∂p

∂ρs

= γp

ρ = γRT , (1.77)

where the final form was obtained using the equation of state for a perfect gas, Eq (1.26). Thus, the soundspeed for a calorically perfect gas is proportional to the square root of its temperature T .

That the speed of sound should be proportional to the square root of the temperature of a gas is consistentwith a simple kinetic theory picture of the gas at the molecular level. This is because the the temperatureis a measure of the kinetic energy of the random molecular motion of the molecules, whence the mean speedof this motion is proportional to the square root of the temperature, and it is reasonable to assume that thespeed at which disturbances are propagated through the gas should be proportional to this mean speed.


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1.5. APPENDIX: SPECIFIC HEATS OF CALORICALLY PERFECT GASES 13

1.5 Appendix: Specific Heats of Calorically Perfect Gases

The Helmholtz function is defined asa ≡ u− T s (1.78)

whence, using Eq. (1.22), we can writeda =

− p dv

−s dT (1.79)

This equation is equivalent to stating that∂a

∂v

T

= − p (1.80)

and∂a

∂T

v

= −s (1.81)

Since, for sufficiently differentiable functions the mixed cross derivative is independent of the order in whichthe derivatives are taken, these two equations lead to

∂p

∂T

v

= ∂s

∂v

T

(1.82)

This one of the four Maxwell relations ; the other three come from equivalent operations on the differentialequations of state, Eqs. (1.22) and (1.24), and the corresponding formula for the differential of the Gibbsfunction.

Now, if we assume that the internal energy is a function of specific volume v and temperature T , we have

du = ∂u

∂v

T

dv + ∂u

∂T

v

dT (1.83)

and, the differential equation of state, Eq. (1.22), tells us that

∂u

∂v

T

= T ∂s

∂v

T

Substituting from Eq. (1.82) gives∂u

∂v

T

= T ∂p

∂T

v

and, since the equation of state for a perfect gas gives

∂p

∂T

v

= ρR = p

T

we have∂u

∂v

T

= T p

T − p = 0 (1.84)

Thus, the internal energy of a perfect gas must be a function only of its temperature.


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Bibliography

[1] Anon., [1962]. U. S. Standard Atmosphere, 1962, U. S. Government Printing Office, Washington.

[2] R. H. Cole, [1948]. Underwater Explosions, Princeton University Press.

[3] John R. Howell & Richard O. Buckius, Fundamentals of Engineering Thermodynamics, SecondEdition, McGraw-Hill, 1992.

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16 BIBLIOGRAPHY


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Chapter 2

Summary of Results from

Compressible Flow Theory

This chapter provides a summary of some useful results from compressible flow theory, concentrating onisentropic flows and flows with shock waves. Further details can be found in the classic text books [4] and[6].

2.1 Steady, Adiabatic Flows

The energy equation for steady, inviscid, adiabatic flow can be derived either from the Bernoulli equationor from the First Law of thermodynamics. When it is derived from the Bernoulli equation, we must includethe assumption that the flow is isentropic while, when it is derived from the First Law, we require only theassumption that the flow be adiabatic – i.e., that there is no heat transfer to the fluid.

2.1.1 Compressible Bernoulli Equation

We first consider the compressible form of the Bernoulli equation. The stream wise component of theMomentum Theorem, for the control volume illustrated in Fig. 2.1, gives the differential form of the Bernoulliequation

d p + ρV dV = 0 , (2.1)

for steady flow along a streamline. In this equation, p and ρ are the fluid pressure and density, respectively,and V is the velocity; the gravitational term has been neglected, as is usual in aerodynamics.

If the flow is compressible, changes in density ρ must be taken into account when integrating this equation.If we assume that changes in density are related to those in pressure by the equation describing an isentropicprocess – i.e., Eq. (1.69) – then the integral of this equation becomes

γ

γ − 1 p

ρ +

V 2

2 = Const. , (2.2)

where γ = c p/cv is the ratio of specific heats. This is the compressible form of the Bernoulli equation,relating pressure (and density!) to velocity.

17


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18 CHAPTER 2. SUMMARY OF RESULTS FROM COMPRESSIBLE FLOW THEORY

p, e, V, A

p + dp

e + de

V + dV

A + dAρ

ρ + ρd

ds

Figure 2.1: Differential element of a stream tube for deriving the compressible Bernoulli and energy equations.

The perfect gas law can be used to replace the ratio of pressure to density in the compressible Bernoulliequation with the product RT of the gas constant and temperature and, since

γR

γ − 1 = c p ,

Eq. (2.2) can be written

c pT + V 2

2 = Const.

Finally, for a calorically perfect gas, for which c pT = h, the enthalpy, we have

h + V 2

2 = hT , (2.3)

where hT is the total, or stagnation, enthalpy. As will be seen in the next section, this is the energy equationfor a steady, adiabatic, but not necessarily reversible, flow.

2.1.2 Steady Energy Equation

The First Law of thermodynamics for steady flow through a control volume can be written

Q−W =

ρ

e +

V 2

2

V · n̂ dS , (2.4)

where e is the specific internal energy, Q is the rate at which heat is being added to the fluid in the controlvolume, and W is the rate at which the fluid in the control volume is doing work on the surroundings. 1

Consistent with the treatment of the preceding section, we neglect the work done by the gravitational force.

For the control volume consisting of the differential element of stream tube illustrated in Fig. 2.1,

W =

p V · n̂

and we will assume there is no heat transfer, i.e., that Q = 0. For the given control volume, Eq. (2.4) then

becomes

pAV − ( p + d p)(A + dA)(V + dV ) = ρAV

e + de + 1

2(V + dV )2 − (e + 1

2V 2)

. (2.5)

The Continuity Equation applied to the same control volume gives

(A + dA)(V + dV ) = ρAV

ρ + dρ = AV

1− dρ

ρ + · · ·

, (2.6)

1Here, unlike Section 1.1.1, the total energy is taken to include the kinetic energy of the flowing gas.


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2.1. STEADY, ADIABATIC FLOWS 19

aV = a V = a + dV

T

ρ

p

T + dT

ρ +

p + dp

d ρ

(a) (b)

Figure 2.2: Determination of the speed of sound. (a) Planar sound wave traveling at speed a throughstationary atmosphere; (b) Sound wave and control volume enclosing it in reference frame traveling withwave.

which can be used to simplify Eq. (2.5) to

d

e + pρ

+ V 2

2

= d

h + V

2

2

= 0 , (2.7)

or

h + V 2

2 = hT , (2.8)

where the stagnation enthalpy hT is constant for a given flow. This equation is seen to be identical to theenergy form of the compressible Bernoulli equation, Eq. (2.3). Our assumption that the flow is inviscid meansthat work done by frictional forces acting on the sides of the stream tube is negligible, but no assumptionshave been made about whether there are irreversible processes inside the control volume. In particular,this equation will hold if the control volume contains shock waves which, as will be seen in Section 2.3, areirreversible (but adiabatic).

2.1.3 Sound Speed in a Continuous Medium

We consider the one-dimensional process corresponding to an infinitesimal pressure wave propagating througha fluid that is initially at rest. The situation is as shown in Fig. 2.2(a). This flow pattern is unsteady, sincethe pressure wave is moving with velocity a relative to our reference frame (which is attached to the fluid atrest). The problem can be viewed as a steady one, provided the problem is transformed to a reference framemoving with the pressure wave; in this case the flow becomes that shown in Fig. 2.2(b).

We apply the laws of conservation of mass and momentum to the control volume shown by the broken linesin Fig. 2.2(b). The sides of the control volume are aligned with the flow, so the only fluxes are across thefaces having area A. Conservation of mass thus requires

ρaA = (ρ + dρ)(a + dV )A (2.9)

or0 = ρ dV + a dρ (2.10)

where the higher-order differential dρ dV has been neglected, since it will vanish in the limit of infinitesimaldisturbances.

The momentum theorem, applied to the same control volume, gives for the x-component

( p + d p)A− pA = −ρAa2 + (ρ + dρ)A(a + dV )2 . (2.11)


20/49


Using the continuity equation, this expression can be simplified to

( p + d p)A− pA = −ρAa dV , (2.12)

ord p + ρa dV = 0 . (2.13)

Note, since a = V , the fluid velocity entering the control volume, this is simply equivalent to the Bernoulli

equation, in differential form, for inviscid flow along a streamline.

Now, if the results from the continuity and momentum equations both are solved for dV (to eliminate thisquantity from the final expression) we find

− dV = d pρa

= a dρ

ρ (2.14)

which, in turn, can be solved for a2 to give

a2 = d p

dρ . (2.15)

Our mechanical analysis here does not specify what is held constant in taking this derivative, but theviscosities and thermal conductivities of common gases, including air, are sufficiently small that the process

is nearly adiabatic and reversible, hence isentropic. Thus, as seen earlier in Chapter 1,

a2 = d p

dρ

s

= γp

ρ = γRT . (2.16)

2.1.4 Example: Speed of Sound

Determine the sound speed for air at standard sea level conditions. What is the speed of sound in hydrogenat the same temperature?

Solution: The temperature of air under standard sea level conditions [1] is T = 288.15 K, so the speed of sound under these conditions is

a =

γRT =

(1.4)(287.05)(288.15) m2/s2 = 340.3 m/s . (2.17)

The speed of sound of gases depends on their molecular weights. The molecular weight of hydrogen isM H2 = 2, so its gas constant is

RH2 = Ru/M H2 = (8314/2) m2/(K s2) = 4157 m2/(K s2) . (2.18)

Thus, the speed of sound in hydrogen at the temperature corresponding to standard sea level conditions is

a = γRH2T = (1.4)(4157)(288.15) m2/s2

= 1295 m/s . (2.19)

2.1.5 Stagnation Temperature, Pressure, and Density

For a calorically perfect gas, the enthalpy can be expressed as h = c pT , and Eq. (2.8) can be rearranged togive

T T T

= 1 + γ − 1

2 M2 , (2.20)


21/49

2.1. STEADY, ADIABATIC FLOWS 21

where T T is the stagnation temperature of the flow and T is the static temperature at a point in the flowwhere the local Mach number is M = V /a, where a is the speed of sound. In developing Eq. (2.20) we haveused the fact that the square of the speed of sound is given by a2 = γRT (see Sections 1.4 and 2.1.3).

The isentropic stagnation pressure pT and the isentropic stagnation density ρT are defined as the values of pressure and density, respectively, achieved when the flow is brought to rest (i.e., to stagnation conditions)by a reversible and adiabatic process. For such an isentropic process, we can combine the equations relating

pressure and density to temperature for isentropic processes with the energy equation in the form of Eq. (2.20)to give

pT p

=

T T T

γ/(γ −1)=

1 +

γ − 12

M2γ/(γ −1)

, (2.21)

andρT ρ

=

T T T

1/(γ −1)=

1 +

γ − 12

M21/(γ −1)

. (2.22)

Note that the stagnation temperature is achieved when the flow is stagnated by any adiabatic process, whilethe stagnation pressure and density are achieved only when the stagnation process also is reversible . Anyirreversibility in the process will cause a loss in stagnation pressure (and stagnation density).

2.1.6 Example: When is a Flow Incompressible?

We might reasonably expect that a flow can be approximated as incompressible so long as the ratio of stagnation density to density in the free stream is less than 1.10; to what maximum Mach number does thiscorrespond?

Solution: We can solve Eq. (2.22) for the Mach number to give

M =

2

γ − 1

ρT ρ

γ −1− 1

1/2.

Thus, for the specified ratio of stagnation to static density, and assuming γ = 1.4 for air,

Mmax =

2

0.4

(1.10)

0.4 − 11/2

= 0.44

Airspeed measurements sometimes interpret the pressure difference pT − p as the dynamic pressure ρV 2/2,using the incompressible Bernoulli equation. The ratio

pT − p12

ρV 2 =

2

γ M2

1 +

γ − 12

M2γ/(γ −1)

− 1

(2.23)

is plotted as a function of free stream Mach number in Fig. 2.3. As expected, the ratio approaches unity inthe limit as M → 0. Furthermore, it is within 5 per cent of unity for Mach numbers less than M = 0.44. Thatis, using the incompressible Bernoulli equation to infer the dynamic pressure from the Pitot-static pressuredifference incurs less than 5 per cent error at Mach numbers up to M = 0.44. The binomial theorem can beused to expand the first term in the square brackets of Eq. (2.23), giving the asymptotic expansion

pT − p12ρV

2 = 1 +

M2

4 +

2 − γ 24

M4 +O(M6) . (2.24)

This result also is plotted in Fig. (2.3), and is seen to closely approximate the exact result for Mach numbersas large as M = 2.0.


22/49


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 11

1.05

1.1

1.15

1.2

1.25

1.3

1.35

Mach number, M∞

D y n a m i c p r e s

s u r e r a t i o ,

( p T − p ) / q

Exact

Asymptotic

0 0.5 1 1.5 2 2.5 31

1.5

2

2.5

3

3.5

4

4.5

5

5.5

6

Mach number, M∞

D y n a m i c p r e s

s u r e r a t i o ,

( p T − p ) / q

Exact

Asymptotic

Figure 2.3: Ratio of the pressure difference pT − p to the dynamic pressure 12ρV 2 as a function of free streamMach number.

Often it is convenient to use sonic conditions as a reference, rather than stagnation values. We use the

superscript asterisk ∗ to denote sonic conditions, so V ∗ = a∗, by definition, where a is the speed of sound.Equations (2.20), (2.21), and (2.22) show that the ratio of any thermodynamic variable at the sonic point toits stagnation value is a function only of the ratio of specific heats γ . That is, setting M = 1 in Eqs. (2.20),(2.21), and (2.22) yields

T T T ∗

= γ + 1

2 , (2.25)

pT p∗

=

γ + 1

2

γ/(γ −1), (2.26)

andρT ρ∗

=

γ + 1

2

1/(γ −1). (2.27)

2.2 Isentropic Flow in Ducts of Varying Area

The properties of steady, isentropic flows can be expressed as functions of the Mach number and the ratioof the stream tube (or duct) cross-sectional area to the area, real or virtual, at which the flow is sonic. Forsteady flow, the continuity equation can be written as

ρAV = ρ∗A∗V ∗ , (2.28)

where the asterisk ∗ again refers to sonic conditions. Since, by definition, V ∗ = a∗ this equation can bere-written as

ρ∗

ρ · a∗

a · 1

M = A

A∗ , (2.29)

orρ∗

ρT · ρT

ρ · a

∗

aT · aT

a · 1

M =

A

A∗. (2.30)

Finally, recalling that aT /a = (T T /T )1/2 and using Eqs (2.20), (2.21), (2.22) and Eqs. (2.25), (2.26),

and (2.27), this relationship can be written as

1

M

2

γ + 1 +

γ − 1γ + 1

M2 γ+1

2(γ−1)

= A

A∗. (2.31)


23/49

2.2. ISENTROPIC FLOW IN DUCTS OF VARYING AREA 23

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Mach number, M

R a t i o

T/TT

p/pT

ρ / ρT

A* /A

Figure 2.4: Area ratio and flow variables as functions of Mach number for isentropic flow; calorically perfect

air (γ = 1.4).

This relationship and the isentropic ratios of the flow variables to their stagnation values given by Eqs (2.20),(2.22), and (2.22) completely determine the properties of isentropic flows in ducts (or stream tubes) of slowlyvarying cross-sectional area; these results are presented in Fig. 2.4.

A consequence of these equations is that the mass flow rate ṁ per unit area for given stagnation conditionsis a function only of the Mach number M (and, parametrically, the ratio of specific heats γ ). That is,

ṁ

A = ρV =

ρ

ρT

V

a

a

aT ρT aT

= pT √

T T γ

R

M

1 +

γ

−1

2 M

2(γ +1)/(2[γ −1]) =

pT √ RT T

√ γ M

1 + γ −12 M2(γ +1)/(2[γ −1])

(2.32)

The quantity in square brackets in the next to last line of the above equation is called the mass flow parameter MFP by Mattingly [5]

MFP =

γ

R

M1 + γ −12 M

2(γ +1)/(2[γ −1]) (2.33)

I prefer to define the related quantity MFP, which is independent of the gas constant R

MFP =

√ γ M

1 + γ −12 M2(γ +1)/(2[γ −1]) (2.34)

Thus, we can express the mass flow rate as

ṁ = pT A√

T T MFP(M, R , γ ) =

pT A√ RT T

MFP(M, γ ) (2.35)

The mass flow parameter MFP and the ratio A/A∗ are inverse measures of the same quantity. In fact,comparison of Eqs. (2.31) and (2.34) shows that

A

A∗=√

γ

2

γ + 1

(γ +1)/(2[γ −1])1

MFP(2.36)


24/49


0 0.5 1 1.5 2 2.50

0.5

1

1.5

2

2.5

V/a*

a/a*

V/a*

ρ / ρ*

Mach

0 0.5 1 1.5 2 2.5 3 3.5 40

0.5

1

1.5

2

2.5

3

Mach

ρ V A

V A

ρ A

(a) (b)

Figure 2.5: Flow variables as functions of flow velocity or Mach number for isentropic flow; calorically perfectair (γ = 1.4). (a) Speed of sound and other flow variables as functions of flow velocity; (b) Continuity equationand it’s low-Mach and high-Mach approximations; all quantities normalized by their sonic values.

The isentropic flow properties are given in tabular form as functions of Mach number for both, subsonicand supersonic flows, in Tables 2.1 and 2.2 in Section 2.8 for a calorically perfect gas having ratio of specificheats γ = 1.4 (representative of air).

2.2.1 The Adiabatic Ellipse

A useful summary of how flow properties change with Mach number, illustrating the qualitative differencesbetween subsonic and supersonic flow, can be developed from the energy equation (see Eq. (2.8)). Since theenthalpy of a calorically perfect gas can be expressed as

h = c pT = γRT

γ − 1 = a2

γ − 1

where, as usual, a represents the speed of sound, Eq. (2.8) can be written as

aa∗

2+

γ − 12

V

a∗

2=

γ + 1

2 , (2.37)

where a∗ again represents the speed of sound at the sonic point. This result is called the adiabatic ellipse ,

since it shows that a plot of the speed of sound a as a function of the fluid velocity V is an ellipse.

The adiabatic ellipse is shown in Fig. 2.5(a) for calorically perfect air (γ = 1.4). Note that a steady, adiabaticflow has a maximum achievable velocity, given by

V

a∗

max

=

γ + 1

γ − 1 . (2.38)

This corresponds to the limit when T → 0 (and, correspondingly a → 0), i.e., when all the thermal energyhas been converted to directed, kinetic energy. Note that, by virtue of the shape of the ellipse, the speed


25/49

2.3. NORMAL SHOCK WAVES 25

of sound approaches a constant as the velocity goes to zero, while the velocity approaches a constant as thespeed of sound goes to zero.

The quantity

M∗ = V

a∗(2.39)

is simply a normalized velocity, although it has some properties in common with the Mach number M = V /a.

In particular, M∗ < 1 when the Mach number is subsonic, and M∗ > 1 when the Mach number is supersonic.Note, however, that M∗ has a finite maximum, corresponding to the maximum achievable velocity. The Machnumber, however, approaches infinity in this limit, since the speed of sound goes to zero there. The Machnumber is also plotted as a function of V /A∗ in Fig. 2.5(a). Note that for subsonic Mach numbers, M∗ isnot very different from the Mach number M – since the speed of sound changes by less than 10% betweenstagnation conditions and the sonic point – but, of course, it deviates significantly for supersonic flows.

Finally, the variable ρ/ρ∗ is also plotted in Fig. 2.5(a). As expected, the density approaches a constant,consistent with the incompressible approximation, in the limit as the velocity goes to zero. Recall that thesonic point is the critical point where changes in density become more important than changes in velocity inthe continuity equation (as the flow transitions from subsonic to supersonic Mach numbers). This is reflectedin the fact that the slope of the density curve is equal, but of opposite sign to the slope of the velocity curveat the sonic point in Fig. 2.5.

The various factors involved in the steady continuity equation are plotted in Fig. 2.5(b); for all three curves,the variables are normalized by their sonic values. The product ρV A is seen to be constant for all Machnumbers, as required by the continuity equation. In the limit of small Mach number, the product V A is seento approach a constant, since the density ρ is nearly constant, at it’s value

ρT /ρ∗ =

γ + 1

2

1/(γ −1).

And, in the limit of large Mach number, the product ρA is seen to approach a constant, since the velocityV /a∗ is nearly constant, at it’s value

V max/a∗ =

(γ + 1)/(γ − 1) .These two observations confirm that, in order to satisfy the steady-flow continuity equation ( ρV A = const.),at small Mach numbers changes in cross-sectional area are balanced primarily by changes in velocity (i.e.,V A ≈ const.); while at large Mach numbers changes in cross-sectional area are balanced primarily bychanges in density (i.e., ρA ≈ const.).

2.3 Normal Shock Waves

The governing laws of conservation of mass, momentum and energy, applied to a control volume containing

a shock, requireρ1V 1 = ρ2V 2 = ṁ , (2.40)

p1 + ρ1V 21 = p2 + ρ2V

22 , (2.41)

h1 + V 21

2 = h2 +

V 222

= hT . (2.42)

Note that the stagnation enthalpy is constant, since we are assuming the process is adiabatic – i.e., that noheat is added – but the flow will not necessarily be isentropic, since we have not assumed the flow to bereversible.


26/49


1 1.5 2 2.5 3 3.5 4 4.5 50

5

10

15

20

25

30

Mach number, M1

R a t i o

p2 /p

1

ρ2 / ρ

1

M2

T2 /T

1

pT

2

/pT

1

1 1.5 2 2.5 3 3.5 4 4.5 50

1

2

3

4

5

6

Mach number, M1

R a t i o

p2 /p

1

ρ2 / ρ

1

M2

T2 /T

1

pT

2

/pT

1

Figure 2.6: Shock relations for calorically perfect air (with γ = 1.4). The same data are plotted in bothfigures; the vertical scale of the figure on the right is chosen to clarify the behavior for small M1.

The ratios of the thermodynamic variables on either side of a normal shock depend only upon the normalMach number M1 of the flow upstream of the shock (and the ratio of specific heats γ ) for the case of acalorically perfect gas. These can be written in the form

p2 p1

= 1 + 2γ

γ + 1

M21 − 1

, (2.43)

ρ2ρ1

= M21

1 + γ −1γ +1 (M21 − 1)

, (2.44)

M2

2

=1 + γ −1γ +1 M

21 − 1

1 + 2γ γ +1 (M21 − 1), (2.45)

T 2T 1

=

1 + 2γ γ +1

M21 − 1

1 + γ −1γ +1

M21 − 1

M21

. (2.46)

Equation (2.45) gives values of M2 < 1 when M1 > 1, and values of M2 > 1 when M < 1; i.e., a normal shockwave must convert a supersonic flow to a subsonic one, and vice versa. Equation (2.43) can be combinedwith Eq. (2.21) to give an expression for the ratio of stagnation pressures across the shock, namely

pT 2 pT 1

=

(γ + 1)M21

2 + (γ − 1)M21

γ/(γ −1) γ + 1

2γ M21 − (γ − 1)

1/(γ −1). (2.47)

The Second Law of Thermodynamics requires that the stagnation pressure not increase across a shock, andinspection of Eq. (2.47) shows that this requires that M1 not be less than unity. Thus, the Second Law limitspossible shock waves to only those having M1 ≥ 1. For these shock waves, as we have seen, we will haveM2 ≤ 1 – thus, only shock waves across which the flow changes from supersonic to subsonic velocities areallowed. For these shock waves, the pressure, density and temperature all must increase across the shock.

The relationships for the behavior of normal shocks in Eqs. (2.43) – (2.47) are plotted in Fig. 2.6 for caloricallyperfect air (with γ = 1.4); consistent with the discussion in the preceding paragraph, the curves are plottedonly for supersonic upstream Mach numbers M1 ≥ 1.


27/49

2.3. NORMAL SHOCK WAVES 27

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

5

10

15

20

25

30

Mach number, M1

R a t i o

pt2

/p1

pt1

/p1

Figure 2.7: Rayleigh Pitot formula compared with isentropic compression as a function of flight Mach number

M1; calorically perfect air with γ = 1.4.

A properly designed Pitot-static probe, when immersed in a supersonic stream, measures the ratio of stag-nation pressure downstream of a normal shock to the upstream static pressure; this is given by

pT 2 p1

=

γ + 1

2 M21

γ/(γ −1) γ + 1

2γ M21 − (γ − 1)

1/(γ −1). (2.48)

This equation is known as the Rayleigh Supersonic Pitot formula, and is the basis for calibrating Machmeters for supersonic flight. Figure 2.7 compares the ratio pT 2/p1 with the isentropic ratio pT 1/p1; it is seenthat there is very little loss in stagnation pressure for flight Mach numbers less than about M1 = 1.3, butthe differences become very significant at larger Mach numbers.

In addition to these general results, several other features of shock waves are evident from Fig. 2.6. Note,for weak shock waves, the stagnation pressure ratio is very nearly equal to unity. In fact, it can be shownfrom Eq. (2.47) that

pT 2 pT 1

= 1 − (γ + 1)2

12γ 2

∆ p

p1

3+O

∆ p

p1

4, (2.49)

where ∆ p = p2 − p1 is the pressure change across the shock. That is, the loss in stagnation pressure isof third order in the strength of the shock. This means that for small values of the pressure change (i.e.,∆ p/p1


28/49


β

V1

V 2 θ

Figure 2.8: Flow through an oblique shock wave, oriented at the shock angle β relative to the upstreamvelocity vector. Since the tangential component of velocity is continuous across the shock, while the normalcomponent decreases, the flow is turned through the angle θ .

The normal shock relations of Eqs. (2.43), (2.44), (2.45), (2.46), and (2.47), are included in Tables 2.1 and 2.2in Section 2.8 for a calorically perfect gas having ratio of specific heats γ = 1.4 (representative of air).

2.4 Oblique Shock Waves

Oblique shock waves are not normal to the upstream flow direction; rather the angle between the upstreamflow direction and the shock surface is the shock angle β < π/2, as illustrated in Fig. 2.8. Application of the Momentum Theorem to a control volume of infinitesimal thickness spanning the shock shows that thetangential component of velocity is the same on both sides of the shock, and the velocity component normalto the shock surface changes in exactly the same way as that for a normal shock. In other words, if we viewthe properties of the shock in a reference frame moving along the shock surface at a velocity equal to thetangential component of velocity, the situation reduces to that of a normal shock. For the shock to exist,the normal component of velocity upstream of the shock must be supersonic, and the normal component

downstream of the shock will be subsonic. Since the tangential component of velocity is the same on bothsides, the flow will be turned through some angle θ , as illustrated in the figure.

From the geometry illustrated in Fig. 2.8 it is easy to see that

β − θ = tan−1

V n2V t2

= tan−1

V n2V n1

tan β

,

since V t2 = V t1 . This equation can be recast as

β − θ = tan−1

Mn2Mn1

T 2T 1

tan β

,

where we have expressed the velocities in terms of Mach numbers and used the fact that the speed of sound isproportional to the square root of the temperature. Using the normal shock relations Eqs. (2.45) and (2.46)allows us to write this equation as

θ = β − tan−1

1 + γ −1γ +1

M2n1 − 1

M2n1tan β

.

Finally, since Mn1 = M1 sin β , we can write

θ = β − tan−1

2 + (γ − 1)M21 sin2 β (γ + 1)M21 sin β cos β

. (2.52)


29/49

2.4. OBLIQUE SHOCK WAVES 29

0 5 10 15 20 25 30 35 40 4510

20

30

40

50

60

70

80

90

Turning angle, θ; degrees

S h o c k a n g l e ,

β ; d e g r e e s

M =1.2

M =1.3

M =1.5

M =1.7

M =2

M =2.5

M =3

M =3.5

M =4

M =5

Figure 2.9: Shock angle as a function of turning angle for oblique shocks in air (approximated as a caloricallyperfect gas with γ = 1.4). Broken blue line shows locus of maximum turning angle for a given Mach number.Broken red line represents dividing line separating flows for which the downstream flow is subsonic from thosefor which downstream flow remains supersonic; points above the broken red line have subsonic downstreamflow, while those below it have supersonic downstream flow.

This equation allows us to determine the turning angle θ for given values of the free stream Mach numberM1 and the shock angle β .

The relation between shock angle and turning angle given by Eq. (2.52) is shown in Fig. 2.9 for various free

stream Mach numbers; the ratio of specific heats is taken to be γ = 1.4. It is seen that there is a maximumturning angle possible for a given upstream Mach number. For any turning angle less than this maximum,there are two different shock angles β that produce the same turning angle θ . Zero turning is achieved forboth, the case of a normal shock β = π/2 and for the limiting case of an infinitesimally weak shock inclinedat the Mach angle β = sin−1(1/M).

The flow downstream of normal shock waves must be subsonic, while the flow downstream of infinitesimallyweak oblique shocks must remain supersonic. (If the tangential component of velocity, which does not changeacross the shock, is large enough, the Mach number downstream of the shock can be supersonic even thoughthe normal component must be subsonic.) For a given Mach number, shock angles greater than that formaximum turning angle are said to correspond to the strong branch of flow solutions, while shock anglesless than that for the maximum turning angle are said to correspond to the weak branch of flow solutions.More simply, shocks on the strong branch are more like normal shocks while those on the weak branch are

more like Mach waves. As seen in Fig. 2.9, all strong shocks have subsonic downstream Mach numbers, andmost weak shocks have supersonic downstream Mach numbers – but shocks on the weak branch very nearthe maximum turning angle can have subsonic downstream flow.

The flow upstream of shock waves must be supersonic, but such supersonic flows can be achieved even whenthe flight Mach number is subsonic (since the flow accelerates around the body, and especially above thewing surface). Figure 2.10 dramatically visualizes the extensive shock patterns in the vicinity of an aircraftin a low-altitude pass over San Francisco Bay at nearly the speed of sound. In particular, compare the shockpatterns at the downstream end of the supersonic zones surrounding the wing with the shock waves shownin the higher Mach number flows past the airfoils shown in Fig. 2.14.


30/49


Figure 2.10: Photograph of an aircraft in flight at a Mach number only slightly less than unity. Shock wavesare made visible by diffraction of light through the sharp density gradients associated with them. They arevisible above the canopy, upstream of the engine inlets, and at the downstream end of the supersonic zonesabove and below the wings. The background is blurred because the photographer was following the aircraft

with his camera; we might be able to estimate the speed of the aircraft if we knew the shutter speed and thedistances from the photographer to the aircraft and to the sailboats in the background.

2.5 Heat Addition

The flow in a gas turbine combustor is very complicated. Liquid fuel is injected as a spray that is thenatomized and evaporated, mixed with air, and burnt. The mixing of fuel and air, and the resulting exothermicreaction take place in a turbulent flow, on length and time scales that make simulation extremely difficult.Nevertheless, for simple parametric studies of engine performance, the net effect of the combustion can beapproximated as a simple addition of heat to the gas – and it is this heating process of a compressible fluidthat we will characterize in this section.

2.5.1 Simple Heating – the Rayleigh Process

We consider the addition of heat to the steady flow of a calorically perfect gas in a duct of constant cross-sectional area, in the absence of viscous stresses; such a process is called simple heating or a Rayleigh process .The differential forms of the continuity equation

dρ

ρ +

dV

V = 0 , (2.53)

and the momentum equationd p + ρV dV = 0 , (2.54)

can be combined to eliminate the velocity V , giving

d p− ṁ2 dρρ2

= 0 , (2.55)

where ṁ = ρV is the (constant) mass flow rate per unit cross-sectional area. This equation can be integratedto give the purely thermodynamic relation

p + ṁ2

ρ = Const. . (2.56)


31/49

2.5. HEAT ADDITION 31

Entropy, s

T e m p e r a t u r e ,

T

Figure 2.11: Rayleigh line, illustrating points accessible by simple heating of a compressible flow in a duct

of constant cross-sectional area. Blue square indicates point of maximum static temperature, correspondingto M = 1/√ γ ; red circle indicates point of maximum entropy, corresponding to sonic flow (M = 1).

A plot of this equation in the T -s (or h-s) plane is called a Rayleigh line ; a typical Rayleigh line is shown inFig. 2.11.

Sinced p

dρ =

ṁ

ρ2 = V 2 , (2.57)

at the point on the curve corresponding to maximum entropy,

d p

dρ = V 2 =

∂p

∂ρs

= a2 , (2.58)

so the flow is sonic at this point. The upper branch of the curve in Fig. 2.11 corresponds to subsonic flow,and the lower branch corresponds to supersonic flow Since adding heat increases the entropy of the gas, thismust always drive the flow toward sonic conditions. In other words, adding heat to a subsonic flow willincrease its Mach number, while adding heat to a supersonic flow will decrease its Mach number.

At the point on the curve corresponding to maximum (static) temperature,

d p

dρ = V 2 =

∂p

∂ρ

T

= RT = a2

γ , (2.59)

so the Mach number at this point is M = 1/√

γ . Note that, for Mach numbers in the range 1/√

γ ≤ M ≤ 1,the temperature decreases as heat is added. Remember, however, that the flow is accelerating toward M = 1,

so this simply means that the kinetic energy is increasing more rapidly than the rate at which heat is beingadded.

In view of the preceding, the sonic point is again a convenient reference point for processes involving simpleheating. Equation 2.56 can be written as

p + ρV 2 = Const. = p∗ + ρ∗V ∗2 , (2.60)

which can be recast into the form p

p∗=

γ + 1

1 + γ M2 . (2.61)


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1

1.5

2

2.5

Mach number, M_1

R a t i o t o

s o n i c v a l u e

Pressure

Temperature

Total Temperature

Total Pressure

0 0.5 1 1.5 2 2.5 3 3.5 40

0.5

1

1.5

2

2.5

3

Mach number, M_1

R a t i o t o

s o n i c v a l u e

Pressure

Temperature

Total Temperature

Total Pressure

Figure 2.12: Variation of flow properties for Rayleigh process; calorically perfect gas with γ = 1.4.

The density appearing in the continuity equation

ρV = ρ∗a∗

can be eliminated in favor of the pressure p and temperature T using the equation of state of a perfect gas;then, substitution of Eq. (2.61) results in an equation for the temperature ratio

T

T ∗=

(γ + 1)2M2

(1 + γ M2)2 . (2.62)

The ratio of stagnation temperatures can be written

T T

T ∗T

= T T

T

T

T ∗

T ∗

T ∗T

=

1 +

γ − 12

M2· (γ + 1)

2M2

(1 + γ M2)2 ·

2

γ + 1

or

T T T ∗T

= (γ + 1)M2

2 + (γ − 1)M2

(1 + γ M2)2 (2.63)

Finally, the ratio of stagnation pressures can be written

pT p∗T

= pT

p

p

p∗ p∗

p∗T

=

1 + γ − 12

M2γ/(γ −1) · (γ + 1)

1 + γ M2 · 2

γ + 1

γ/(γ −1)

or

pT p∗T

= γ + 1

1 + γ M2

2 + (γ − 1)M2

γ + 1

γ/(γ −1)(2.64)

Equations (2.61) – (2.64) are plotted for a gas having γ = 1.4 in Figs. 2.12. Data for the Rayleigh processare also given in Tables 2.3 and 2.4 for γ = 1.4 and in Tables 2.5 and 2.6 for γ = 1.325 in Section 2.8.


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2.6. CRITICAL MACH NUMBER 33

2.5.2 Heating at Constant Mach Number

As an alternative to heating in a constant-area duct, we also consider the addition of heat to a steady flowat constant Mach number. Since we can write the mass flow rate ṁ as

ṁ = ρV A = γ

R

p

pT

T T T

M pT √

T T

A , (2.65)

if the Mach number is held constant we must have

pT √ T T

A = Const. . (2.66)

Thus, the stagnation pressure ratio for constant Mach number heating between stations () 1 and ()2 can bewritten as

pT 2 pT 1

=

T T 2T T 1

A1A2

. (2.67)

Thus, in order to predict the loss in stagnation pressure we need to know how the cross-sectional area changes

as heat is added.

The continuity equation can be expressed for constant Mach number as

dρ

ρ +

dV

V +

dA

A =

dρ

ρ +

dT

2T +

dA

A = 0 , (2.68)

or, using the perfect gas law to eliminate the change in density in favor of pressure and temperature, wehave

d p

p − dT

2T +

dA

A = 0 . (2.69)

The Bernoulli equation for constant Mach number can be written as

d p p

+ ρV dV

p =

d p p

+ γ M2 dT 2T

= 0 , (2.70)

and eliminating the change in pressure between this and the continuity equations gives

dA

A =

1 + γ M2

2

dT

T . (2.71)

Since the Mach number is constant, this can be integrated to give

A1A2

=

T T 1T T 2

(1+γ M2)/2. (2.72)

Finally, combining Eqs. (2.67) and (2.72) gives

pT 2 pT 1

=

T T 2T T 1

−γ M2/2. (2.73)

A comparison of the stagnation pressure loss for Rayleigh and constant-Mach heating is shown in Fig. 2.13.For the Mach numbers shown, constant-Mach heating is seen to result in less total pressure loss for subsonicflows, and greater total pressure loss for supersonic flows.


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Total temperature ratio, ∆ Tt /T

t1

S t a g n a t i o n p r e s s u r e r a t i o , p t 2

/ p t 1

M1 =0.2

M1 =0.4

M1 =0.7

M1 =1.5

M1 =2M

1 =4

Figure 2.13: Stagnation pressure ratio for heat addition in compressible flow. Calorically perfect gas withspecific heat ratio γ = 1.4. Solid lines represent heating at constant Mach number; dashed lines representheating at constant area (and their termination corresponds to thermal choking).

2.6 Critical Mach Number

The nature of the flow pattern past an airfoil as the free stream Mach number M∞ is increased is illustratedin Fig. 2.14. The flow here is that past the RAE 2822 airfoil at an angle of attack of α = 3◦, and the datapresented are based on numerical solutions of the inviscid Euler equations [3].

From the contours of constant Mach number shown in this figure, it is seen that the maximum Mach numberoccurs on the airfoil surface, usually near the leading edge. As the free stream Mach number increases,so does the maximum Mach number in the flow, and, for sufficiently large free stream Mach number themaximum Mach number in the flow is supersonic. The free stream Mach number for which the maximumMach number in the flow becomes sonic is called the critical Mach number ; the critical Mach number forthis airfoil at this angle of attack is approximately Mcrit = 0.52.

Figure 2.15 shows the airfoil surface pressure distributions for several of these cases. The Mach numberM = 0.05 is sufficiently small that this case can be taken to correspond to essentially incompressible flow.As long as the flow remains entirely subsonic (i.e., as long as the free stream Mach number is less thanthe critical Mach number) the pressure distribution has a shape similar to that for incompressible flow. Assupersonic flow begins to develop, however, the pressure distribution changes significantly, and includes ashock wave that terminates the supersonic pocket.

As long as the flow remains subsonic, the properties of flow past thin, two-dimensional airfoils can be related

to those of the incompressible flow past the same body. The Prandtl-Glauert similarity law states that thepressure coefficient

C p ≡ p− p∞12ρ∞V

2∞

, (2.74)

at any point on a thin airfoil for the compressible flow at a free stream Mach number M∞ is related to thepressure coefficient C p0 at the same point for incompressible flow according to

C p = C p0

1−M2∞. (2.75)


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M∞ = 0.20 M∞ = 0.40

M∞ = 0.50 M∞ = 0.60

M∞ = 0.70 M∞ = 0.75

Figure 2.14: Contours of constant Mach number for inviscid flow past RAE 2822 airfoil at α = 3.0◦ angleof attack. Contour spacing is ∆M = 0.05. Solutions to the Euler equations of inviscid flow are computednumerically on 320× 64 cell grids using Jameson-Caughey LU -SGS scheme [3].


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-2

-1.5

-1

-0.5

0

0.5

1

1.5 0 0.2 0.4 0.6 0.8 1

P r e s s u r e C o e f f i c i e n t

Chordwise position, x/c

Surface Pressure, Total Enthalpy, and Entropy Change Distributions

Pressure coefficientTotal enthalpy (x100)Total pressure (x10)

-2

-1.5

-1

-0.5

0

0.5

1

1.5 0 0.2 0.4 0.6 0.8 1





M∞ = 0.05 M∞ = 0.50

-2

-1.5

-1

-0.5

0

0.5

1

1.5 0 0.2 0.4 0.6 0.8 1





-2

-1.5

-1

-0.5

0

0.5

1

1.5 0 0.2 0.4 0.6 0.8 1





M∞ = 0.60 M∞ = 0.70

Figure 2.15: Airfoil surface pressure coefficients for inviscid flow past RAE 2822 airfoil at α = 3.0◦ angleof attack. Solutions to the Euler equations of inviscid flow are computed numerically on 320 × 64 cell gridsusing Jameson-Caughey LU -SGS scheme [3].


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0

0.2

0.4

0.6

0.8

1

1.2

1.4

0 0.2 0.4 0.6 0.8 1-0.02

0

0.02

0.04

0.06

0.08

0.1

0.12

L i f t c o e f f i c

i e n t

D r a g c o e f f i

c i e n t

Free stream Mach number

Prandtl-Glauert Scaling for RAE 2822 Airfoil

Lift, numericalLift, Prandtl-Glauert

Drag, numerical

Figure 2.16: Lift and drag coefficients as functions of Mach number for inviscid flow past RAE 2822 airfoilat α = 3.0◦ angle of attack. Solutions to the Euler equations of inviscid flow are computed numerically on320

×64 cell grids using Jameson-Caughey LU -SGS scheme [3]; lift coefficient is compared with results of

the Prandtl-Glauert similarity law.

Since the lift coefficient Cl is equal to the integral of the difference in pressure coefficients on the lower andupper surfaces of the airfoil, a similar relation holds for it

Cl = Cl0

1−M2∞, (2.76)

where Cl0 is the lift coefficient in the corresponding incompressible flow. A comparison of the lift coeffi-cient computed using the Prandtl-Glauert rule with that computed numerically for the airfoil of the earlierFigs. 2.14 and 2.15 is shown in Fig. 2.16. The Prandtl-Glauert rule is seen to predict the increase in lift coeffi-cient with increasing Mach number quite well up to approximately the critical Mach number (of Mcrit = 0.52

for this case). The computed drag coefficient is also plotted in the figure, showing a rapid rise as the size of the supersonic zone and the strength and extent of the shock wave increase for Mach numbers larger thanabout M∞ = 0.65.

A similar relation also holds in supersonic flow, giving the pressure coefficient at any point on the airfoil as

C p =C p√ 2 M2∞ − 1

, (2.77)

where C p√ 2 represents the pressure coefficient at the same point for the flow at a free stream Mach number

M∞ =√

2. Both the subsonic and supersonic Prandtl-Glauert rules can be derived by studying the way inwhich the equations describing small perturbations from the free stream conditions must scale with changesin the Mach number. For a description of this procedure, and for the more general scaling laws that apply

to three-dimensional flows, see, e.g., the aerodynamics texts [1] or [2].

As was seen above, the flow past a body begins to behave in a qualitatively different fashion from anincompressible one when the maximum local Mach number in the flow exceeds unity, and a pocket of supersonic flow begins to form. This supersonic pocket is almost always terminated by a shock wave, andthe losses associated with the shock wave appear as drag on the airfoil. As the strength and extent of theshock wave grow with increasing free stream Mach number, the drag begins to increase dramatically. Thefree stream Mach number for which the derivative of the drag coefficient with respect to Mach numberreaches

∂ CD∂ M

= 0.10 (2.78)


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is called the drag divergence Mach number .

The critical Mach number is necessarily a conservative estimate for the drag divergence Mach number. Andthe critical Mach number is easily determined by finding the Mach number at which the Prandtl-Glauertrule predicts the minimum pressure coefficient will become equal to its sonic value.

For an isentropic flow, the pressure coefficient can be expressed in terms of the free stream and local Mach

numbers using

C p = p/p∞ − 1

ρ∞2 p∞

V 2∞=

2

γ M2∞

p

pT · pT

p∞− 1

, (2.79)

which, using the isentropic relation, Eq. (2.21), can be written

C p = 2

γ M2∞

1 + γ −12 M2∞

1 + γ −12 M2

γ/(γ −1)− 1

, (2.80)

where M is the local Mach number, corresponding to the point in the flow where the pressure is p, and M∞is the Mach number in the free stream where the pressure is p∞.

The value of the critical pressure coefficient C p∗ at a point where the local Mach number is unity is thengiven by

C p∗ =

2

γ M2∞

2

γ + 1 +

γ − 1γ + 1

M2∞

γ/(γ −1)− 1

. (2.81)

A plot of the critical pressure coefficient as a function of free stream Mach number is shown in Fig. 2.17, alongwith plots of the Prandtl-Glauert scaling given by Eq. (2.75) for several values of incompressible pressurecoefficient of C p0. Since the maximum Mach number in isentropic flow will occur at the point where thepressure is the lowest and, according to the Prandtl-Glauert rule the minimum pressure will always occur atthe same point, the point on the airfoil at which sonic flow will first occur is the point corresponding to theminimum pressure point in the incompressible flow past the body. Thus, the Mach number at which thesetwo curves intersect (seen in the figure to be approximately M∞ = 0.575, .643, and .747, respectively for the

several values of C p0 plotted) will be the critical Mach number for an airfoil having the minimum pressurecoefficient specified in incompressible flow. The critical Mach numbers for other values of C p0 can be foundby a similar procedure.

2.7 Supersonic Airfoil Theory

Linearized supersonic flow theory relates the pressure coefficient at any point on a two-dimensional airfoil tothe local slope θ of the surface relative to the free stream direction according to

C p =

− 2θ M2∞ − 1

, (2.82)

where the angle θ is positive when the surface normal has a downstream component. This simple result canbe used to investigate the properties of airfoils in supersonic flow.

We represent the coordinates of the upper surface yu and the lower surface yl of the airfoil as

yuc

= −x

c

α + f c(x/c) + f t(x/c)

ylc

= −x

c

α + f c(x/c)− f t(x/c) ,

(2.83)


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2.7. SUPERSONIC AIRFOIL THEORY 39

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

−3.5

−3

−2.5

−2

−1.5

−1

−0.5

0

Mach number, M∞

P r e s s u r e C o e f f i

c i e n t , C

p

Cp

*

Cp

0

= −1.2

Cp

0

= −.80

Cp

0

= −.40

Figure 2.17: Critical pressure coefficient and Prandtl-Glauert scaled pressure coefficients (for C p0 =

−0.4,−.8,−1.2) as functions of free stream Mach number M∞.

where the functions f c and f t represent the contributions of camber and thickness, respectively. Since theangle of attack α is represented separately, we have

f c(0) = f c(1) = 0

f t(0) = f t(1) = 0 . (2.84)

The lift coefficient is given by the integral

Cl =

10

C pl −C pu

dx/c, (2.85)

which, using the linearized supersonic flow result of Eq. (2.82), can be written

Cl = − 2 M2∞ − 1

10

dy

dx

l

+ dy

dx

u

dx/c. (2.86)

By virtue of the conditions imposed by Eqs. (2.84), the above integral reduces to

Cl = − 2 M2∞ − 1

[−α− α] = 4α M2∞ − 1

. (2.87)

It is seen that, for supersonic flow the lift coefficient is completely independent of the camber and thicknessdistributions of the airfoil, and depends only on the angle of attack.

The drag coefficient is given by the integral

Cd =

10

−C pl

dy

dx

l

+ C pu

dy

dx

u

dx/c

= 2 M2∞ − 1

10

dy

dx

2l

+

dy

dx

2u

dx/c.

(2.88)

Substituting for the airfoil geometry using Eqs. (2.83), the drag coefficient can be expressed as

Cd = 4 M2∞ − 1

α2 +

10

f ′c

2+ f ′t

2

dx/c

. (2.89)


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Thus, while thickness and camber contribute nothing to the lift coefficient, they do contribute to the dragcoefficient. Thickness is, of course, required for structural reasons, so optimal supersonic wings require acareful compromise between thicker wings that will generally weigh less and thinner wings that will haveless drag. Camber serves no useful purpose in supersonic flow, so wings optimized for supersonic flight willhave little, or no, camber.

The integrals in Eq. (2.89) can be evaluated for particular camber line shapes and thickness distributions.

For example, the circular-arc airfoil having thickness ratio τ is defined by

f t = 2τ x

c

1− x

c

so

f ′t2

= 4τ 2

1− 2 xc

2and the drag coefficient for this thickness distribution is given by

Cd = 16τ 2

3

M2∞ − 1. (2.90)

It can be shown that the minimum drag coefficient for an airfoil of given thickness ratio is

Cd = 4τ 2

M2∞ − 1(2.91)

for the double-diamond shape defined as

f t = τ minx

c,

1− xc

Finally, note that since the lift coefficient for any airfoil shape is the same as that for a flat plate at the sameangle of attack, the lift will be distributed uniformly along the chord line, so the centroid of the incrementallift due to an infinitesimal change in angle of attack will be at mid-chord. This means that the aerodynamic

center of a supersonic airfoil will be at the mid-chord station (rather than at the quarter-chord station, asin low-speed flow).

2.7.1 Prandtl-Meyer Flow

Equation (2.82) can be developed in a variety of ways. The most direct way is from a linearization of theequations of inviscid, supersonic flow, but the result can also be developed from a description of Prandtl-Meyer flow. Inviscid, supersonic flow of a uniform stream past a sharp corner joining two strai

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