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Unformatted text preview: The Concorde 264 supersonic airliner. Flying more than twice as fast as the speed of sound, as
discussed in the present chapter, the Concorde is a milestone in commercial aviation. However,
this great technical achievement is accompanied by high expense for the traveller. (Courtesy of
Don Riepe/Peter Arnold, Inc.)  v v 570  eText Main Menu  Textbook Table of Contents  Study Guide Chapter 9
Compressible Flow Motivation. All eight of our previous chapters have been concerned with “lowspeed’’
or “incompressible’’ flow, i.e., where the fluid velocity is much less than its speed of
sound. In fact, we did not even develop an expression for the speed of sound of a fluid.
That is done in this chapter.
When a fluid moves at speeds comparable to its speed of sound, density changes become significant and the flow is termed compressible. Such flows are difficult to obtain
in liquids, since high pressures of order 1000 atm are needed to generate sonic velocities. In gases, however, a pressure ratio of only 2 1 will likely cause sonic flow. Thus
compressible gas flow is quite common, and this subject is often called gas dynamics.
Probably the two most important and distinctive effects of compressibility on flow
are (1) choking, wherein the duct flow rate is sharply limited by the sonic condition,
and (2) shock waves, which are nearly discontinuous property changes in a supersonic
flow. The purpose of this chapter is to explain such striking phenomena and to familiarize the reader with engineering calculations of compressible flow.
Speaking of calculations, the present chapter is made to order for the Engineering
Equation Solver (EES) in App. E. Compressibleflow analysis is filled with scores of
complicated algebraic equations, most of which are very difficult to manipulate or invert. Consequently, for nearly a century, compressibleflow textbooks have relied upon
extensive tables of Mach number relations (see App. B) for numerical work. With EES,
however, any set of equations in this chapter can be typed out and solved for any variable—see part (b) of Example 9.13 for an especially intricate example. With such a
tool, App. B serves only as a backup and indeed may soon vanish from textbooks. 9.1 Introduction We took a brief look in Chap. 4 [Eqs. (4.13) to (4.17)] to see when we might safely
neglect the compressibility inherent in every real fluid. We found that the proper criterion for a nearly incompressible flow was a small Mach number
Ma V
a 1 where V is the flow velocity and a is the speed of sound of the fluid. Under smallMachnumber conditions, changes in fluid density are everywhere small in the flow field. The
energy equation becomes uncoupled, and temperature effects can be either ignored or  v v 571  eText Main Menu  Textbook Table of Contents  Study Guide 572 Chapter 9 Compressible Flow put aside for later study. The equation of state degenerates into the simple statement that
density is nearly constant. This means that an incompressible flow requires only a momentum and continuity analysis, as we showed with many examples in Chaps. 7 and 8.
This chapter treats compressible flows, which have Mach numbers greater than about
0.3 and thus exhibit nonnegligible density changes. If the density change is significant,
it follows from the equation of state that the temperature and pressure changes are also
substantial. Large temperature changes imply that the energy equation can no longer
be neglected. Therefore the work is doubled from two basic equations to four
1.
2.
3.
4. Continuity equation
Momentum equation
Energy equation
Equation of state to be solved simultaneously for four unknowns: pressure, density, temperature, and
flow velocity ( p, , T, V). Thus the general theory of compressible flow is quite complicated, and we try here to make further simplifications, especially by assuming a reversible adiabatic or isentropic flow. The Mach Number The Mach number is the dominant parameter in compressibleflow analysis, with different effects depending upon its magnitude. Aerodynamicists especially make a distinction between the various ranges of Mach number, and the following rough classifications are commonly used:
0.3 Ma
Ma 0.8 Ma 1.2 Ma
3.0 0.3: incompressible flow, where density effects are negligible.
0.8: subsonic flow, where density effects are important but no
shock waves appear.
1.2: transonic flow, where shock waves first appear, dividing subsonic and supersonic regions of the flow. Powered flight in the
transonic region is difficult because of the mixed character of
the flow field.
3.0: supersonic flow, where shock waves are present but there are
no subsonic regions.
Ma: hypersonic flow [13], where shock waves and other flow
changes are especially strong. The numerical values listed above are only rough guides. These five categories of flow
are appropriate to external highspeed aerodynamics. For internal (duct) flows, the most
important question is simply whether the flow is subsonic (Ma 1) or supersonic (Ma
1), because the effect of area changes reverses, as we show in Sec. 9.4. Since supersonicflow effects may go against intuition, you should study these differences carefully. The SpecificHeat Ratio In addition to geometry and Mach number, compressibleflow calculations also depend
upon a second dimensionless parameter, the specificheat ratio of the gas:  v v k  eText Main Menu  cp
c Textbook Table of Contents (9.1)  Study Guide 9.1 Introduction 573 Earlier, in Chaps. 1 and 4, we used the same symbol k to denote the thermal conductivity of a fluid. We apologize for the duplication; thermal conductivity does not appear in these later chapters of the text.
Recall from Fig. 1.3 that k for the common gases decreases slowly with temperature and
lies between 1.0 and 1.7. Variations in k have only a slight effect upon compressibleflow computations, and air, k 1.40, is the dominant fluid of interest. Therefore, although
we assign some problems involving, e.g., steam and CO2 and helium, the compressibleflow tables in App. B are based solely upon the single value k 1.40 for air.
This text contains only a single chapter on compressible flow, but, as usual, whole
books have been written on the subject. References 1 to 6, 26, 29, and 33 are introductory, fairly elementary treatments, while Refs. 7 to 14, 27 to 28, 31 to 32, and 35
are advanced. From time to time we shall defer some specialized topic to these texts.
We note in passing that there are at least two flow patterns which depend strongly upon
very small density differences, acoustics, and natural convection. Acoustics [9, 14] is the
study of soundwave propagation, which is accompanied by extremely small changes in
density, pressure, and temperature. Natural convection is the gentle circulating pattern set
up by buoyancy forces in a fluid stratified by uneven heating or uneven concentration of
dissolved materials. Here we are concerned only with steady compressible flow where the
fluid velocity is of magnitude comparable to that of the speed of sound. The Perfect Gas In principle, compressibleflow calculations can be made for any fluid equation of state,
and we shall assign problems involving the steam tables [15], the gas tables [16], and
liquids [Eq. (1.19)]. But in fact most elementary treatments are confined to the perfect
gas with constant specific heats
cp
p
RT
R cp c
const
k
const
(9.2)
c
For all real gases, cp, c , and k vary with temperature but only moderately; for example, cp of air increases 30 percent as temperature increases from 0 to 5000°F. Since we
rarely deal with such large temperature changes, it is quite reasonable to assume constant specific heats.
Recall from Sec. 1.6 that the gas constant is related to a universal constant divided by the gas molecular weight
Rgas 49,720 ft2/(s2 °R) where (9.3) Mgas
8314 m2/(s2 K) For air, M 28.97, and we shall adopt the following property values for air throughout this chapter:
1717 ft2/(s2 °R) R
c  v v cp  eText Main Menu  R
k 1 kR
k 1 287 m2/(s2 K) k 1.400 4293 ft2/(s2 °R) 718 m2/(s2 K) 6010 ft2/(s2 R) 1005 m2/(s2 K) Textbook Table of Contents  Study Guide (9.4) 574 Chapter 9 Compressible Flow Experimental values of k for eight common gases were shown in Fig. 1.3. From this
figure and the molecular weight, the other properties can be computed, as in Eqs. (9.4).
The changes in the internal energy û and enthalpy h of a perfect gas are computed
for constant specific heats as
û2 û1 c (T2 T1) h2 h1 cp(T2 T1) (9.5) For variable specific heats one must integrate û
c dT and h
cp dT or use the
gas tables [16]. Most modern thermodynamics texts now contain software for evaluating properties of nonideal gases [17]. Isentropic Process The isentropic approximation is common in compressibleflow theory. We compute the
entropy change from the first and second laws of thermodynamics for a pure substance
[17 or 18]
T ds dh dp (9.6) Introducing dh cp dT for a perfect gas and solving for ds, we substitute T
from the perfectgas law and obtain
2 2 ds
1 cp 1 dT
T 2 R
1 dp
p p/R (9.7) If cp is variable, the gas tables will be needed, but for constant cp we obtain the analytic results
s2 s1 cp ln T2
T1 R ln p2
p1 c ln T2
T1 R ln 2 (9.8) 1 Equations (9.8) are used to compute the entropy change across a shock wave (Sec. 9.5),
which is an irreversible process.
For isentropic flow, we set s2 s1 and obtain the interesting powerlaw relations
for an isentropic perfect gas
p2
p1 T2
T1 k /(k 1) 2 k (9.9) 1 These relations are used in Sec. 9.3. EXAMPLE 9.1
Argon flows through a tube such that its initial condition is p1 1.7 MPa and 1 18 kg/m3
and its final condition is p2 248 kPa and T2 400 K. Estimate (a) the initial temperature, (b)
the final density, (c) the change in enthalpy, and (d) the change in entropy of the gas. Solution  v v From Table A.4 for argon, R 208 m2/(s2 K) and k
at constant pressure from Eq. (9.4):  eText Main Menu  1.67. Therefore estimate its specific heat Textbook Table of Contents  Study Guide 9.2 The Speed of Sound
1.67(208)
1.67 1 kR cp k 1 575 519 m2/(s2 K) The initial temperature and final density are estimated from the ideal gas law, Eq. (9.2):
1.7 E6 N/m2
(18 kg/m3)[208 m2/(s2 K)] p1
1R T1 p2
T2R 2 248 E3 N/m2
(400 K)[208 m2/(s2 K)] 454 K 2.98 kg/m3 Ans. (a)
Ans. (b) From Eq. (9.5) the enthalpy change is
h2 h1 cp(T2 T1) 519(400 454) 28,000 J/kg (or m2/s2) Ans. (c) The argon temperature and enthalpy decrease as we move down the tube. Actually, there may
not be any external cooling; i.e., the fluid enthalpy may be converted by friction to increased kinetic energy (Sec. 9.7).
Finally, the entropy change is computed from Eq. (9.8):
s2 s1 T2
T1 cp ln R ln p2
p1 519 ln 400
454 208 ln 0.248 E6
1.7 E6 66 400 334 m2/(s2 K) Ans. (d) The fluid entropy has increased. If there is no heat transfer, this indicates an irreversible process.
Note that entropy has the same units as the gas constant and specific heat.
This problem is not just arbitrary numbers. It correctly simulates the behavior of argon moving subsonically through a tube with large frictional effects (Sec. 9.7). 9.2 The Speed of Sound The socalled speed of sound is the rate of propagation of a pressure pulse of infinitesimal strength through a still fluid. It is a thermodynamic property of a fluid. Let us
analyze it by first considering a pulse of finite strength, as in Fig. 9.1. In Fig. 9.1a the
pulse, or pressure wave, moves at speed C toward the still fluid (p, , T, V 0) at the
left, leaving behind at the right a fluid of increased properties (p
p,
,T
T) and a fluid velocity V toward the left following the wave but much slower. We
can determine these effects by making a controlvolume analysis across the wave. To
avoid the unsteady terms which would be necessary in Fig. 9.1a, we adopt instead the
control volume of Fig. 9.1b, which moves at wave speed C to the left. The wave appears fixed from this viewpoint, and the fluid appears to have velocity C on the left
and C
V on the right. The thermodynamic properties p, , and T are not affected
by this change of viewpoint.
The flow in Fig. 9.1b is steady and onedimensional across the wave. The continuity equation is thus, from Eq. (3.24),
AC  v v or  eText Main Menu (
V  )(A)(C V) C Textbook Table of Contents (9.10)  Study Guide 576 Chapter 9 Compressible Flow
C p + ∆p
ρ + ∆ρ
T + ∆T p
ρ
T
V=0 ∆V
Moving
wave of
frontal
area A
(a)
Friction and heat
transfer effects are
confined to wave interior p
ρ
T
V=C Fig. 9.1 Controlvolume analysis of
a finitestrength pressure wave:
(a) control volume fixed to still
fluid at left; (b) control volume
moving left at wave speed C. p + ∆p
ρ + ∆ρ
T + ∆T
V = C – ∆V
Fixed
wave
(b) This proves our contention that the induced fluid velocity on the right is much smaller
than the wave speed C. In the limit of infinitesimal wave strength (sound wave) this
speed is itself infinitesimal.
Notice that there are no velocity gradients on either side of the wave. Therefore,
even if fluid viscosity is large, frictional effects are confined to the interior of the wave.
Advanced texts [for example, 14] show that the thickness of pressure waves in gases
is of order 10 6 ft at atmospheric pressure. Thus we can safely neglect friction and apply the onedimensional momentum equation (3.40) across the wave
Fright
or pA (p m (Vout
˙ p)A Vin) ( AC)(C V C) (9.11) Again the area cancels, and we can solve for the pressure change
p CV (9.12) If the wave strength is very small, the pressure change is small.
Finally combine Eqs. (9.10) and (9.12) to give an expression for the wave speed
C2 p 1 (9.13) The larger the strength / of the wave, the faster the wave speed; i.e., powerful explosion waves move much more quickly than sound waves. In the limit of infinitesi→ 0, we have what is defined to be the speed of sound a of a fluid:
mal strength  v v a2  eText Main Menu  p Textbook Table of Contents (9.14)  Study Guide 9.2 The Speed of Sound
Table 9.1 Sound Speed of Various
Materials at 60°F (15.5°C) and 1 atm
Material a, ft/s a, m/s 4,246
3,281
1,117
1,040
873
607
297 1,294
1,000
340
317
266
185
91 6,100
4,890
4,760
3,940 1,860
1,490
1,450
1,200 16,900
16,600
13,200
10,500 Gases:
H2
He
Air
Ar
CO2
CH4
238
UF6
Liquids:
Glycerin
Water
Mercury
Ethyl alcohol
Solids:*
Aluminum
Steel
Hickory
Ice 5,150
5,060
4,020
3,200 *Plane waves. Solids also have a shearwave
speed. 577 But the evaluation of the derivative requires knowledge of the thermodynamic process
undergone by the fluid as the wave passes. Sir Isaac Newton in 1686 made a famous
error by deriving a formula for sound speed which was equivalent to assuming an
isothermal process, the result being 20 percent low for air, for example. He rationalized the discrepancy as being due to the “crassitude’’ (dust particles, etc.) in the air;
the error is certainly understandable when we reflect that it was made 180 years before the proper basis was laid for the second law of thermodynamics.
We now see that the correct process must be adiabatic because there are no temperature gradients except inside the wave itself. For vanishingstrength sound waves
we therefore have an infinitesimal adiabatic or isentropic process. The correct expression for the sound speed is
p a 1/2 k p s 1/2 (9.15) T for any fluid, gas or liquid. Even a solid has a sound speed.
For a perfect gas, From Eq. (9.2) or (9.9), we deduce that the speed of sound is
kp a 1/2 (kRT)1/2 (9.16) The speed of sound increases as the square root of the absolute temperature. For air,
with k 1.4 and R 1717, an easily memorized dimensional formula is
a (ft/s) 49[T (°R)]1/2 a (m/s) 20[T (K)]1/2 (9.17) At sealevel standard temperature, 60°F 520°R, a 1117 ft/s. This decreases in the
upper atmosphere, which is cooler; at 50,000ft standard altitude, T
69.7°F
389.9°R and a 49(389.9)1/2 968 ft/s, or 13 percent less.
Some representative values of sound speed in various materials are given in Table
9.1. For liquids and solids it is common to define the bulk modulus K of the material
p K p s (9.18) s For example, at standard conditions, the bulk modulus of carbon tetrachloride is
163,000 lbf/in2 absolute, and its density is 3.09 slugs/ft3. Its speed of sound is therefore [163,000(144)/3.09]1/2 2756 ft/s, or 840 m/s. Steel has a bulk modulus of
about 29 106 lbf/in2 absolute and water about 320 103 lbf/in2 absolute, or 90
times less.
For solids, it is sometimes assumed that the bulk modulus is approximately equivalent to Young’s modulus of elasticity E, but in fact their ratio depends upon Poisson’s
ratio
E
K 3(1 2) (9.19)  v v 1
The two are equal for
3 , which is approximately the case for many common metals such as steel and aluminum.  eText Main Menu  Textbook Table of Contents  Study Guide 578 Chapter 9 Compressible Flow EXAMPLE 9.2
Estimate the speed of sound of carbon monoxide at 200kPa pressure and 300°C in m/s. Solution
From Table A.4, for CO, the molecular weight is 28.01 and k 1.40. Thus from Eq. (9.3) RCO
8314/28.01 297 m2/(s2 K), and the given temperature is 300°C 273 573 K. Thus from
Eq. (9.16) we estimate
(kRT)1/2 aCO 9.3 Adiabatic and Isentropic
Steady Flow [1.40(297)(573)]1/2 488 m/s Ans. As mentioned in Sec. 9.1, the isentropic approximation greatly simplifies a compressibleflow calculation. So does the assumption of adiabatic flow, even if nonisentropic.
Consider highspeed flow of a gas past an insulated wall, as in Fig. 9.2. There is no
shaft work delivered to any part of the fluid. Therefore every streamtube in the flow
satisfies the steadyflow energy equation in the form of Eq. (3.66)
h1 1
2 V2
1 gz1 h2 1
2 V2
2 gz2 q w (9.20) where point 1 is upstream of point 2. You may wish to review the details of Eq. (3.66)
and its development. We saw in Example 3.16 that potentialenergy changes of a gas
are extremely small compared with kineticenergy and enthalpy terms. We shall neglect the terms gz1 and gz2 in all gasdynamic analyses.
Inside the thermal and velocity boundary layers in Fig. 9.2 the heattransfer and
viscouswork terms q and w are not zero. But outside the boundary layer q and w are
zero by definition, so that the outer flow satisfies the simple relation
1
2 h1 V2
1 1
2 h2 V2
2 const (9.21) The constant in Eq. (9.21) is equal to the maximum enthalpy which the fluid would
achieve if brought to rest adiabatically. We call this value h0, the stagnation enthalpy
of the flow. Thus we rewrite Eq. (9.21) in the form
h 1
2 V2 h0 const (9.22) h0
V δ T > δV if Pr < 1
δV  v v Fig. 9.2 Velocity and stagnationenthalpy distributions near an insulated wall in a typical highspeed
gas flow. Insulated wall  eText Main Menu  Textbook Table of Contents  Study Guide 9.3 Adiabatic and Isentropic Steady Flow 579 This should hold for steady adiabatic flow of any compressible fluid outside the boundary layer. The wall in Fig. 9.2 could be either the surface of an immersed body or the
wall of a duct. We have shown the details of Fig. 9.2; typically the thermallayer thickness T is greater than the velocitylayer thickness V because most gases have a dimensionless Prandtl number Pr less than unity (see, e.g., Ref. 19, sec. 43.2). Note that
the stagnation enthalpy varies inside the thermal boundary layer, but its average value
is the same as that at the outer layer due to the insulated wall.
For nonperfect gases we may have to use the steam tables [15] or the gas tables [16]
to implement Eq. (9.22). But for a perfect gas h cpT, and Eq. (9.22) becomes
1
2 cpT V2 cpT0 (9.23) This establishes the stagnation temperature T0 of an adiabatic perfectgas flow, i.e., the
temperature it achieves when decelerated to rest adiabatically.
An alternate interpretation of Eq. (9.22) occurs when the enthalpy and temperature
drop to (absolute) zero, so that the velocity achieves a maximum value
(2h0)1/2 Vmax (2cpT0)1/2 (9.24) No higher flow velocity can occur unless additional energy is added to the fluid through
shaft work or heat transfer (Sec. 9.8). MachNumber Relations The dimensionless form of Eq. (9.23) brings in the Mach number Ma as a parameter,
by using Eq. (9.16) for the speed of sound of a perfect gas. Divide through by cpT to
obtain
V2
2cpT 1
But, from the perfectgas law, cpT
comes [kR/(k
(k 1
T0
T or 1 k T0
T 1 a2/(k 1)]T 1)V2
2a2 2 (9.25)
1), so that Eq. (9.25) be T0
T Ma2 Ma V
a This relation is plotted in Fig. 9.3 versus the Mach number for k
temperature has dropped to 1 T0.
6
Since a T1/2, the ratio a0 /a is the square root of (9.26)
a0
a T0
T 1/2 1 1
(k
2 Equation (9.27) is also plotted in Fig. 9.3. At Ma
to 41 percent of the stagnation value.  v v Isentropic Pressure and Density
Relations  1)Ma2 (9.26)
1.4. At Ma 5 the 1/2 (9.27) 5 the speed of sound has dropped Note that Eqs. (9.26) and (9.27) require only adiabatic flow and hold even in the presence of irreversibilities such as friction losses or shock waves. eText Main Menu  Textbook Table of Contents  Study Guide 580 Chapter 9 Compressible Flow 1.0
a
a0 T
T0 ρ
ρ 0 p
p0 0.5 Fig. 9.3 Adiabatic (T/T0 and a/a0)
and isentropic (p/p0 and / 0) properties versus Mach number for
k 1.4. 0 1 2
3
Mach number 4 5 If the flow is also isentropic, then for a perfect gas the pressure and density ratios
can be computed from Eq. (9.9) as a power of the temperature ratio
T0
T 0 k/(k 1) T0
T p0
p 1/(k 1) 1 1
(k
2 1) Ma2 1 1
(k
2 1) Ma2 k/(k 1) (9.28a)
1/(k 1) (9.28b) These relations are also plotted in Fig. 9.3; at Ma 5 the density is 1.13 percent of
its stagnation value, and the pressure is only 0.19 percent of stagnation pressure.
The quantities p0 and 0 are the isentropic stagnation pressure and density, respectively, i.e., the pressure and density which the flow would achieve if brought isentropically to rest. In an adiabatic nonisentropic flow p0 and 0 retain their local meaning,
but they vary throughout the flow as the entropy changes due to friction or shock waves.
The quantities h0, T0, and a0 are constant in an adiabatic nonisentropic flow (see Sec.
9.7 for further details). Relationship to Bernoulli’s
Equation The isentropic assumptions (9.28) are effective, but are they realistic? Yes. To see why,
differentiate Eq. (9.22)
Adiabatic: dh Meanwhile, from Eq. (9.6), if ds V dV 0 (9.29) 0 (isentropic process),
dh dp (9.30) Combining (9.29) and (9.30), we find that an isentropic streamtube flow must be  v v dp  eText Main Menu  V dV 0 Textbook Table of Contents (9.31)  Study Guide 9.3 Adiabatic and Isentropic Steady Flow 581 But this is exactly the Bernoulli relation, Eq. (3.75), for steady frictionless flow with
negligible gravity terms. Thus we see that the isentropicflow assumption is equivalent
to use of the Bernoulli or streamline form of the frictionless momentum equation. Critical Values at the Sonic Point The stagnation values (a0, T0, p0, 0) are useful reference conditions in a compressible
flow, but of comparable usefulness are the conditions where the flow is sonic, Ma
1.0. These sonic, or critical, properties are denoted by asterisks: p*, *, a*, and T*.
They are certain ratios of the stagnation properties as given by Eqs. (9.26) to (9.28)
when Ma 1.0; for k 1.4
p*
p0 k/(k 1) 2
k
T*
T0 * 0.5283 1
k 1 1 2 a*
a0 0.8333 0.6339 k 0 2 1/(k 1) 2 1/2 k (9.32)
0.9129 1 In all isentropic flow, all critical properties are constant; in adiabatic nonisentropic flow,
a* and T* are constant, but p* and * may vary.
The critical velocity V* equals the sonic sound speed a* by definition and is often
used as a reference velocity in isentropic or adiabatic flow
V* 1/2 2k (kRT*)1/2 a* k 1 RT0 (9.33) The usefulness of these critical values will become clearer as we study compressible
duct flow with friction or heat transfer later in this chapter. Some Useful Numbers for Air Since the great bulk of our practical calculations are for air, k 1.4, the stagnationproperty ratios p/p0, etc., from Eqs. (9.26) to (9.28), are tabulated for this value in Table
B.1. The increments in Mach number are rather coarse in this table because the values
are only meant as a guide; these equations are now a trivial matter to manipulate on a
hand calculator. Thirty years ago every text had extensive compressibleflow tables with
Machnumber spacings of about 0.01, so that accurate values could be interpolated.
For k 1.4, the following numerical versions of the isentropic and adiabatic flow
formulas are obtained:
T0
T 1 0 0.2 Ma2
p0
p (1 0.2 Ma2)2.5
(9.34) 0.2 Ma2)3.5 (1 Or, if we are given the properties, it is equally easy to solve for the Mach number
(again with k 1.4)
Ma2 5 T0
T 1 5 0 2/5 1 5 p0
p 2/7 1 (9.35)  v v Note that these isentropicflow formulas serve as the equivalent of the frictionless adiabatic momentum and energy equations. They relate velocity to physical properties for
a perfect gas, but they are not the “solution’’ to a gasdynamics problem. The complete  eText Main Menu  Textbook Table of Contents  Study Guide Chapter 9 Compressible Flow solution is not obtained until the continuity equation has also been satisfied, for either
onedimensional (Sec. 9.4) or multidimensional (Sec. 9.9) flow.
One final note: These isentropicratio – versus – Machnumber formulas are seductive, tempting one to solve all problems by jumping right into the tables. Actually, many
problems involving (dimensional) velocity and temperature can be solved more easily
from the original raw dimensional energy equation (9.23) plus the perfectgas law (9.2),
as the next example will illustrate.
EXAMPLE 9.3
Air flows adiabatically through a duct. At point 1 the velocity is 240 m/s, with T1 320 K and
p1 170 kPa. Compute (a) T0, (b) p01, (c) 0, (d) Ma, (e) Vmax, and (f ) V*. At point 2 further
downstream V2 290 m/s and p2 135 kPa. (g) What is the stagnation pressure p02? Solution
For air take k 1.4, cp 1005 m2/(s2 K), and R 287 m2/(s2 K). With V1 and T1 known,
we can compute T01 from Eq. (9.23) without using the Mach number:
T01 V2
1
2cp T1 (240 m/s)2
2[1005 m2/(s2 K)] 320 320 29 349 K Ans. (a) Then compute Ma1 from the known temperature ratio, using Eq. (9.35):
2
Ma1 349
320 Ma1 0.67 Alternately compute a1
kRT1 359 m/s, whence Ma1
stagnation pressure at section 1 follows from Eq. (9.34): V1/a1 240/359 0.67. The 0.2(0.67)2]3.5 230 kPa Ans. (b) p01 5 0.2 Ma2)3.5
1 p1(1 1 0.453 (170 kPa)[1 Ans. (d) We need the density from the perfectgas law before we can compute the stagnation density:
1 p1
RT1 170,000
(287)(320) 0.2 Ma2)2.5
1 1.85 kg/m3
2.29 kg/m3 Ans. (c) Alternately, we could have gone directly to 0 p0/(RT0) (230 E3)/[(287)(349)]
Meanwhile, the maximum velocity follows from Eq. (9.24): 2.29 kg/m3. whence 1(1 01 (2cpT0)1/2 Vmax (1.85)[1 0.2(0.67)2]2.5 [2(1005)(349)]1/2 838 m/s 2(1.4)
(287)(349)
1.4 1 1/2 Ans. (e) and the sonic velocity from Eq. (9.33) is
V* 1/2 2k
k 1 RT0 342 m/s Ans. (f) At point 2, the temperature is not given, but since we know the flow is adiabatic, the stagnation
temperature is constant: T02 T01 349 K. Thus, from Eq. (9.23),
T2 T02 V2
2
2cp 349 (290)2
2(1005) 307 K Then, although the flow itself is not isentropic, the local stagnation pressure is computed by the
local isentropic condition  v v 582  eText Main Menu  Textbook Table of Contents  Study Guide 9.4 Isentropic Flow with Area Changes p02 p2 k/(k 1) T02
T2 (135) 349
307 583 3.5 211 kPa Ans. (g) This is 8 percent less than the upstream stagnation pressure p01. Notice that, in this last part, we
took advantage of the given information (T02, p2, V2) to obtain p02 in an efficient manner. You
may verify by comparison that approaching this part through the (unknown) Mach number Ma2
is more laborious. 9.4 Isentropic Flow with Area
Changes By combining the isentropic and/or adiabaticflow relations with the equation of continuity we can study practical compressibleflow problems. This section treats the onedimensional flow approximation.
Figure 9.4 illustrates the onedimensional flow assumption. A real flow, Fig. 9.4a,
has no slip at the walls and a velocity profile V(x, y) which varies across the duct section (compare with Fig. 7.8). If, however, the area change is small and the wall radius
of curvature large
dh
dx 1 h(x) R(x) (9.36) then the flow is approximately onedimensional, as in Fig. 9.4b, with V V(x) reacting to area change A(x). Compressibleflow nozzles and diffusers do not always satisfy conditions (9.36), but we use the onedimensional theory anyway because of its
simplicity.
For steady onedimensional flow the equation of continuity is, from Eq. (3.24),
(x)V(x)A(x) m
˙ const (9.37) Before applying this to duct theory, we can learn a lot from the differential form of
Eq. (9.37)
d dV
V dA
A 0 (9.38) The differential forms of the frictionless momentum equation (9.31) and the soundspeed relation (9.15) are recalled here for convenience: y Area
A(x) y
V(x, y)
x h(x)  v v Fig. 9.4 Compressible flow through
a duct: (a) realfluid velocity profile; (b) onedimensional approximation.  eText Main Menu V(x) x Wall radius of
curvature R(x)
(a)  Textbook Table of Contents (b)  Study Guide 584 Chapter 9 Compressible Flow Subsonic Ma < 1 Duct geometry dA > 0 dA < 0 Supersonic Ma > 1 dV < 0
dp > 0
Subsonic diffuser dV > 0
dp < 0
Supersonic nozzle dV > 0
dp < 0
Subsonic nozzle dV < 0
dp > 0
Supersonic diffuser Fig. 9.5 Effect of Mach number on
property changes with area change
in duct flow. dp Momentum V dV 0
(9.39) Sound speed: dp 2 ad Now eliminate dp and d between Eqs. (9.38) and (9.39) to obtain the following relation between velocity change and area change in isentropic duct flow:
dV
V dA
1
A Ma2 1 dp
V2 (9.40)  v v Inspection of this equation, without actually solving it, reveals a fascinating aspect of
compressible flow: Property changes are of opposite sign for subsonic and supersonic
flow because of the term Ma2 1. There are four combinations of area change and
Mach number, summarized in Fig. 9.5.
From earlier chapters we are used to subsonic behavior (Ma 1): When area increases, velocity decreases and pressure increases, which is denoted a subsonic diffuser. But in supersonic flow (Ma 1), the velocity actually increases when the area
increases, a supersonic nozzle. The same opposing behavior occurs for an area decrease, which speeds up a subsonic flow and slows down a supersonic flow.
What about the sonic point Ma 1? Since infinite acceleration is physically impossible, Eq. (9.40) indicates that dV can be finite only when dA 0, that is, a minimum area (throat) or a maximum area (bulge). In Fig. 9.6 we patch together a throat
section and a bulge section, using the rules from Fig. 9.5. The throat or convergingdiverging section can smoothly accelerate a subsonic flow through sonic to supersonic
flow, as in Fig. 9.6a. This is the only way a supersonic flow can be created by expanding the gas from a stagnant reservoir. The bulge section fails; the bulge Mach number moves away from a sonic condition rather than toward it.
Although supersonic flow downstream of a nozzle requires a sonic throat, the op  eText Main Menu  Textbook Table of Contents  Study Guide 9.4 Isentropic Flow with Area Changes 585 A max
A min Fig. 9.6 From Eq. (9.40), in flow
through a throat (a) the fluid can
accelerate smoothly through sonic
and supersonic flow. In flow
through the bulge (b) the flow at
the bulge cannot be sonic on physical grounds. Subsonic Subsonic: Ma = 1 Supersonic Ma < 1 (Supersonic:
Ma > 1 (a) Subsonic:
Supersonic) (b) posite is not true: A compressible gas can pass through a throat section without becoming sonic. We can use the perfectgas and isentropicflow relations to convert the continuity relation (9.37) into an algebraic expression involving only area and Mach number, as follows. Equate the mass flow at any section to the mass flow under sonic conditions
(which may not actually occur in the duct) PerfectGas Relations VA
A
A* or *V*A*
* V*
V (9.41) Both the terms on the right are functions only of Mach number for isentropic flow.
From Eqs. (9.28) and (9.32)
* * 2 0 k 0 1 1
(k
2 1 1) Ma2 1/(k 1) (9.42) From Eqs. (9.26) and (9.32) we obtain
V*
V (kRT*)1/2
V (kRT)1/2 T*
V
T0 1
2
1
Ma k 1 1
(k
2 1/2 1) Ma2 Combining Eqs. (9.41) to (9.43), we get the desired result
1 1 (k 1) Ma2 (1/2)(k
2
A
1
A*
For k 1
2 Ma (k T0
T 1/2 1/2 (9.43) 1)/(k 1) 1) (9.44) 1.4, Eq. (9.44) takes the numerical form
A
A* 1 (1
Ma 0.2 Ma2)3
1.728 (9.45)  v v which is plotted in Fig. 9.7. Equations (9.45) and (9.34) enable us to solve any onedimensional isentropicairflow problem given, say, the shape of the duct A(x) and the
stagnation conditions and assuming that there are no shock waves in the duct.  eText Main Menu  Textbook Table of Contents  Study Guide 586 Chapter 9 Compressible Flow 3.0
Curve fit
Eq. (9.48b) Curve fit
Eq. (9.48c) 2.0
A
A*
1.0
Exact Eq. (9.45) Fig. 9.7 Area ratio versus Mach
number for isentropic flow of a
perfect gas with k 1.4. 0 0 0.5 1.0
1.5
Mach number 2.0 2.5 Figure 9.7 shows that the minimum area which can occur in a given isentropic duct
flow is the sonic, or critical, throat area. All other duct sections must have A greater
than A*. In many flows a critical sonic throat is not actually present, and the flow in
the duct is either entirely subsonic or, more rarely, entirely supersonic. Choking From Eq. (9.41) the inverse ratio A*/A equals V/( *V*), the mass flow per unit area
at any section compared with the critical mass flow per unit area. From Fig. 9.7 this
inverse ratio rises from zero at Ma 0 to unity at Ma 1 and back down to zero at
large Ma. Thus, for given stagnation conditions, the maximum possible mass flow
passes through a duct when its throat is at the critical or sonic condition. The duct is
then said to be choked and can carry no additional mass flow unless the throat is
widened. If the throat is constricted further, the mass flow through the duct must decrease.
From Eqs. (9.32) and (9.33) the maximum mass flow is
m max
˙ k1/2
For k 0 k 1 (1/2)(k 1)/(k 1) 2
k 1/(k 1) 2 *A*V* 1 A* 1/2 2k
k 1 RT0 A* 0(RT0)1/2 (9.46a) 1.4 this reduces to
m max
˙ 0.6847A* 0(RT0)1/2 0.6847p0A*
(RT0)1/2 (9.46b) For isentropic flow through a duct, the maximum mass flow possible is proportional
to the throat area and stagnation pressure and inversely proportional to the square root
of the stagnation temperature. These are somewhat abstract facts, so let us illustrate
with some examples.  v v The Local MassFlow Function Equation (9.46) gives the maximum mass flow, which occurs at the choking condition
(sonic exit). It can be modified to predict the actual (nonmaximum) mass flow at any  eText Main Menu  Textbook Table of Contents  Study Guide 9.4 Isentropic Flow with Area Changes 587 section where local area A and pressure p are known.1 The algebra is convoluted, so
here we give only the final result, expressed in dimensionless form:
Massflow function m
˙
A RT0
p0 2k
k p
1 p0 2/k 1 p
p0 (k 1)/k (9.47) We stress that p and A in this relation are the local values at position x. As p/p0 falls,
this function rises rapidly and then levels out at the maximum of Eq. (9.46). A few values may be tabulated here for k 1.4:
p/p0 1.0 0.98 0.95 0.9 0.8 0.7 0.6 Function 0.0 0.1978 0.3076 0.4226 0.5607 0.6383 0.6769 0.5283
0.6847 Equation (9.47) is handy if stagnation conditions are known and the flow is not choked.
The only cumbersome algebra in these problems is the inversion of Eq. (9.45) to
compute the Mach number when A/A* is known. If available, EES is ideal for this situation and will yield Ma in a flash. In the absence of EES, the following curvefitted
formulas are suggested; given A/A*, they estimate the Mach number within 2 percent for k 1.4 if you stay within the ranges listed for each formula: 1 0.27(A/A*) 2 A
1.34
(9.48a) 1.728A/A*
A* subsonic flow A 0.45
A
1.0
1.34 (9.48b) 1 0.88 ln
A*
A* Ma 1/2 A 1 1.2 A
1
1.0
2.9 (9.48c) A*
A* supersonic flow A 2/3 1/5
A 216 A
254
(9.48d)
2.9 A*
A*
A* Formulas (9.48a) and (9.48d) are asymptotically correct as A/A* → , while (9.48b)
and (9.48c) are just curve fits. However, formulas (9.48b) and (9.48c) are seen in Fig.
9.7 to be accurate within their recommended ranges.
Note that two solutions are possible for a given A/A*, one subsonic and one supersonic. The proper solution cannot be selected without further information, e.g., known
pressure or temperature at the given duct section.
EXAMPLE 9.4
Air flows isentropically through a duct. At section 1 the area is 0.05 m2 and V1 180 m/s, p1 500
kPa, and T1 470 K. Compute (a) T0, (b) Ma1, (c) p0, and (d) both A* and m. If at section 2 the
˙
area is 0.036 m2, compute Ma2 and p2 if the flow is (e) subsonic or ( f) supersonic. Assume k 1.4. Solution
Part (a) A general sketch of the problem is shown in Fig. E9.4. With V1 and T1 known, the energy equation (9.23) gives
1  v v The author is indebted to Georges Aigret, of Chimay, Belgium, for suggesting this useful function.  eText Main Menu  Textbook Table of Contents  Study Guide Chapter 9 Compressible Flow Possibly
supersonic Subsonic
Throat V1 = 180 m /s Assume
isentropic
flow p1 = 500 k P a
T1 = 470 K
2E
A2 = 0.036 m 2 2F
A2 = 0.036 m 2 1 A1 = 0.05 m 2 E9.4 The local sound speed a1 Part (b) V2
1
2cp T1 T0 kRT1 (180)2
2(1005) 470 [(1.4)(287)(470)]1/2
V1
a1 Ma1 180
435 486 K Ans. (a) 435 m/s. Hence 0.414 Ans. (b) With Ma1 known, the stagnation pressure follows from Eq. (9.34): Part (c) 0.2 Ma2)3.5
1 p1(1 p0 (500 kPa)[1 0.2(0.414)2]3.5 563 kPa Ans. (c) Similarly, from Eq. (9.45), the critical sonicthroat area is Part (d) A1
A* (1 or 0.2 Ma2)3
1
1.728 Ma1 A* A1
1.547 [1 0.2(0.414)2]3
1.728(0.414) 0.05 m2
1.547 1.547 0.0323 m2 Ans. (d) This throat must actually be present in the duct if the flow is to become supersonic.
We now know A*. So to compute the mass flow we can use Eq. (9.46), which remains valid,
based on the numerical value of A*, whether or not a throat actually exists:
m
˙ 0.6847 p0A*
RT0 0.6847 (563,000)(0.0323)
(287)(486) 33.4 kg/s Ans. (d) Or we could fare equally well with our new “local mass flow” formula, Eq. (9.47), using, say,
the pressure and area at section 1. Given p1/p0 500/563 0.889, Eq. (9.47) yields
m
˙ 287(486)
563,000(0.05) 2(1.4)
(0.889)2/1.4[1
0.4 (0.889)0.4/1.4] 0.447 m
˙ 33.4 kg
Ans. (d)
s Assume subsonic flow corresponds to section 2E in Fig. E9.4. The duct contracts to an area ratio A2/A* 0.036/0.0323 1.115, which we find on the left side of Fig. 9.7 or the subsonic
part of Table B.1. Neither the figure nor the table is that accurate. There are two accurate options. First, Eq. (9.48b) gives the estimate Ma2 1 0.88 ln (1.115)0.45 0.676 (error less than
0.5 percent). Second, EES (App. E) will give an arbitrarily accurate solution with only three
statements (in SI units): Part (e) EES A2 0.036
Astar 0.0323
A2/Astar (1+0.2*Ma2^2)^3/1.2^3/Ma2  v v 588  eText Main Menu  Textbook Table of Contents  Study Guide 9.4 Isentropic Flow with Area Changes Specify that you want a subsonic solution (e.g., limit Ma2
Ma2 [1 1), and EES reports
Ans. (e) 0.6758 1.4001, which is the answer to part (f).] [Ask for a supersonic solution and you receive Ma2
The pressure is given by the isentropic relation
p2 p0
0.2(0.676)2]3.5 589 563 kPa
1.358 415 kPa Ans. (e) Part (e) does not require a throat, sonic or otherwise; the flow could simply be contracting subsonically from A1 to A2. Part (f) EES This time assume supersonic flow, corresponding to section 2F in Fig. E9.4. Again the area ratio is A2/A* 0.036/0.0323 1.115, and we look on the right side of Fig. 9.7 or the supersonic
part of Table B.1 — the latter can be read quite accurately as Ma2 1.40. Again there are two
other accurate options. First, Eq. (9.48c) gives the curvefit estimate Ma2 1 1.2(1.115
1)11/2 1.407, only 0.5 percent high. Second, EES will give a very accurate solution with the
same three statements from part (e). Specify that you want a supersonic solution (e.g., limit
Ma2 1), and EES reports
Ma2 Ans. (f) 1.4001 Again the pressure is given by the isentropic relation at the new Mach number:
p2 [1 p0
0.2(1.4001)2]3.5 563 kPa
3.183 177 kPa Ans. (f) Note that the supersonicflow pressure level is much less than p2 in part (e), and a sonic throat
must have occurred between sections 1 and 2F. EXAMPLE 9.5
It is desired to expand air from p0 200 kPa and T0 500 K through a throat to an exit Mach
number of 2.5. If the desired mass flow is 3 kg/s, compute (a) the throat area and the exit (b)
pressure, (c) temperature, (d) velocity, and (e) area, assuming isentropic flow, with k 1.4. Solution
The throat area follows from Eq. (9.47), because the throat flow must be sonic to produce a supersonic exit:
A* m (RT0)1/2
˙
0.6847p0 3.0[287(500)]1/2
0.6847(200,000) or Dthroat 0.00830 m2 1
4 10.3 cm D*2
Ans. (a) With the exit Mach number known, the isentropicflow relations give the pressure and temperature:
pe [1  v v Te  eText Main Menu  p0
0.2(2.5)2]3.5
1 T0
0.2(2.5)2 Textbook Table of Contents 200,000
17.08
500
2.25  11,700 Pa
222 K Study Guide Ans. (b)
Ans. (c) 590 Chapter 9 Compressible Flow
The exit velocity follows from the known Mach number and temperature
Ve Mae (kRTe)1/2 2.5[1.4(287)(222)]1/2 2.5(299 m/s) 747 m/s Ans. (d) The exit area follows from the known throat area and exit Mach number and Eq. (9.45):
[1 Ae
A*
or Ae 0.2(2.5)2]3
1.728(2.5) 2.64(0.0083 m2) 2.64A* or De 2.64
0.0219 m2 1
4 D2
e 16.7 cm Ans. (e) One point might be noted: The computation of the throat area A* did not depend in any way on
the numerical value of the exit Mach number. The exit was supersonic; therefore the throat is
sonic and choked, and no further information is needed. 9.5 The NormalShock Wave A common irreversibility occurring in supersonic internal or external flows is the
normalshock wave sketched in Fig. 9.8. Except at nearvacuum pressures such shock waves
are very thin (a few micrometers thick) and approximate a discontinuous change in flow
properties. We select a control volume just before and after the wave, as in Fig. 9.8.
The analysis is identical to that of Fig. 9.1; i.e., a shock wave is a fixed strong pressure wave. To compute all property changes rather than just the wave speed, we use
all our basic onedimensional steadyflow relations, letting section 1 be upstream and
section 2 be downstream:
1V1 p1
Energy: h1 1
2 V2
1 2V2 p2
h2 G 1
2 Constant cp: h cpT (9.49a) 2
1V 1 V2
2 p1
1T1 Perfect gas: const 2
2V 2 (9.49b) h0 const (9.49c) p2
2T2
k (9.49d) const (9.49e) Note that we have canceled out the areas A1 A2, which is justified even in a variable
duct section because of the thinness of the wave. The first successful analyses of these
normalshock relations are credited to W. J. M. Rankine (1870) and A. Hugoniot (1887),
hence the modern term RankineHugoniot relations. If we assume that the upstream
conditions ( p1, V1, 1, h1, T1) are known, Eqs. (9.49) are five algebraic relations in the
five unknowns ( p2, V2, 2, h2, T2). Because of the velocitysquared term, two solutions
are found, and the correct one is determined from the second law of thermodynamics,
which requires that s2 s1.
The velocities V1 and V2 can be eliminated from Eqs. (9.49a) to (9.49c) to obtain
the RankineHugoniot relation  v v h2  eText Main Menu  h1 1
(p
22 p1) 1 1 2 1 Textbook Table of Contents  (9.50) Study Guide 9.5 The NormalShock Wave 591 Fixed
normal
shock
Isoenergetic
T01 = T02
1
Isentropic
upstream
s = s1 2 Ma 1 > 1 Ma 2 < 1 Isentropic
downstream
s = s2 > s1
A* > A*
1
2
p02 < p01 Thin
control
volume
A1≈ A2 Fig. 9.8 Flow through a fixed
normalshock wave. This contains only thermodynamic properties and is independent of the equation
of state. Introducing the perfectgas law h cpT kp/[(k 1) ], we can rewrite
this as
k
k p2/p1
p2/p1 1 2
1 1
1 (9.51) We can compare this with the isentropicflow relation for a very weak pressure wave
in a perfect gas
p2
p1 2
1 1/k (9.52) Also, the actual change in entropy across the shock can be computed from the perfectgas relation
s2 s1
c ln p2
p1 k 1 (9.53) 2 Assuming a given wave strength p2/p1, we can compute the density ratio and the entropy change and list them as follows for k 1.4:
2/ 1 v   Eq. (9.51) 0.5
0.9
1.0
1.1
1.5
2.0 v p2
p1 0.6154
0.9275
1.0
1.00704
1.3333
1.6250 eText Main Menu  s2 0.6095
0.9275
1.0
1.00705
1.3359
1.6407 Textbook Table of Contents s1
c Isentropic 0.0134
0.00005
0.0
0.00004
0.0027
0.0134  Study Guide 592 Chapter 9 Compressible Flow We see that the entropy change is negative if the pressure decreases across the shock,
which violates the second law. Thus a rarefaction shock is impossible in a perfect gas.2
We see also that weakshock waves ( p2/p1 2.0) are very nearly isentropic. MachNumber Relations For a perfect gas all the property ratios across the normal shock are unique functions
of k and the upstream Mach number Ma1. For example, if we eliminate 2 and V2 from
Eqs. (9.49a) to (9.49c) and introduce h kp/[(k 1) ], we obtain
1 p2
p1
But for a perfect gas k 2
1V1/p1 1 2
2 1V 1
p1 kV2/(kRT1)
1 p2
p1 1
k 1 (k 1) (9.54) k Ma2, so that Eq. (9.54) is equivalent to
1 [2k Ma2
1 (k 1)] (9.55) From this equation we see that, for any k, p2 p1 only if Ma1 1.0. Thus for flow
through a normalshock wave, the upstream Mach number must be supersonic to satisfy the second law of thermodynamics.
What about the downstream Mach number? From the perfectgas identity V2
kp Ma2, we can rewrite Eq. (9.49b) as
p2
p1 1
1 k Ma2
1
k Ma2
2 (9.56) which relates the pressure ratio to both Mach numbers. By equating Eqs. (9.55) and
(9.56) we can solve for
(k 1) Ma2
1
2k Ma2 (k
1 Ma2
2 2
1) (9.57) Since Ma1 must be supersonic, this equation predicts for all k 1 that Ma2 must be
subsonic. Thus a normalshock wave decelerates a flow almost discontinuously from
supersonic to subsonic conditions.
Further manipulation of the basic relations (9.49) for a perfect gas gives additional
equations relating the change in properties across a normalshock wave in a perfect gas
(k 2
1 T2
T1 [2 (k
(k 1) Ma2
1
1) Ma2 2
1 1) Ma2]
1
T02 p02
p01 (k 02 2 01 1) Ma2
1
(k 1) Ma2
1 V1
V2 2k Ma2 (k 1)
1
(k 1)2 Ma2
1 (9.58) T01
k/(k 1) k
2k Ma2
1 1
(k 1/(k 1) 1) Of additional interest is the fact that the critical, or sonic, throat area A* in a duct increases across a normal shock  v v 2  This is true also for most real gases; see Ref. 14, sec. 7.3. eText Main Menu  Textbook Table of Contents  Study Guide 9.5 The NormalShock Wave A*
2
A*
1 Ma2
Ma1 2
2 (k
(k 1) Ma2
1
1) Ma2
2 593 (1/2)(k 1)/(k 1) (9.59) All these relations are given in Table B.2 and plotted versus upstream Mach number
Ma1 in Fig. 9.9 for k 1.4. We see that pressure increases greatly while temperature
and density increase moderately. The effective throat area A* increases slowly at first
and then rapidly. The failure of students to account for this change in A* is a common
source of error in shock calculations.
The stagnation temperature remains the same, but the stagnation pressure and density decrease in the same ratio; i.e., the flow across the shock is adiabatic but nonisentropic. Other basic principles governing the behavior of shock waves can be summarized as follows:
1. The upstream flow is supersonic, and the downstream flow is subsonic.
2. For perfect gases (and also for real fluids except under bizarre thermodynamic
conditions) rarefaction shocks are impossible, and only a compression shock can
exist.
3. The entropy increases across a shock with consequent decreases in stagnation
pressure and stagnation density and an increase in the effective sonicthroat area.
4. Weak shock waves are very nearly isentropic.
Normalshock waves form in ducts under transient conditions, e.g., shock tubes, and
in steady flow for certain ranges of the downstream pressure. Figure 9.10a shows a
normal shock in a supersonic nozzle. Flow is from left to right. The oblique wave pattern to the left is formed by roughness elements on the nozzle walls and indicates that
the upstream flow is supersonic. Note the absence of these Mach waves (see Sec. 9.10)
in the subsonic flow downstream. 6 A*
2 5 A*
1 p2
p1 V1 ρ2
=
V2 ρ1 4 T2
T1 3 2 p02 ρ02
p01 = ρ
01 1
Ma 2  v v Fig. 9.9 Change in flow properties
across a normalshock wave for
k 1.4.  0 1 1.5 eText Main Menu 2  2.5
Ma1 3 3.5 Textbook Table of Contents 4  Study Guide 594 Chapter 9 Compressible Flow (a ) Fig. 9.10 Normal shocks form in
both internal and external flows:
(a) Normal shock in a duct; note
the Machwave pattern to the left
(upstream), indicating supersonic
flow. (Courtesy of U.S. Air Force
Arnold Engineering Development
Center.) (b) Supersonic flow past a
blunt body creates a normal shock
at the nose; the apparent shock
thickness and bodycorner curvature are optical distortions. (Courtesy of U.S. Army Ballistic Research Laboratory, Aberdeen
Proving Ground.) (b )  v v Normalshock waves occur not only in supersonic duct flows but also in a variety
of supersonic external flows. An example is the supersonic flow past a blunt body
shown in Fig. 9.10b. The bow shock is curved, with a portion in front of the body
which is essentially normal to the oncoming flow. This normal portion of the bow shock
satisfies the propertychange conditions just as outlined in this section. The flow inside the shock near the body nose is thus subsonic and at relatively high temperature
T2 T1, and convective heat transfer is especially high in this region.
Each nonnormal portion of the bow shock in Fig. 9.10b satisfies the obliqueshock
relations to be outlined in Sec. 9.9. Note also the oblique recompression shock on the
sides of the body. What has happened is that the subsonic nose flow has accelerated
around the corners back to supersonic flow at low pressure, which must then pass
through the second shock to match the higher downstream pressure conditions.  eText Main Menu  Textbook Table of Contents  Study Guide 9.5 The NormalShock Wave 595 Note the finegrained turbulent wake structure in the rear of the body in Fig. 9.10b.
The turbulent boundary layer along the sides of the body is also clearly visible.
The analysis of a complex multidimensional supersonic flow such as in Fig. 9.10 is
beyond the scope of this book. For further information see, e.g., Ref. 14, chap. 9, or
Ref. 8, chap. 16. Moving Normal Shocks The preceding analysis of the fixed shock applies equally well to the moving shock if
we reverse the transformation used in Fig. 9.1. To make the upstream conditions simulate a still fluid, we move the shock of Fig. 9.8 to the left at speed V1; that is, we fix
our coordinates to a control volume moving with the shock. The downstream flow then
appears to move to the left at a slower speed V1 V2 following the shock. The thermodynamic properties are not changed by this transformation, so that all our Eqs. (9.50)
to (9.59) are still valid. EXAMPLE 9.6 12 3 Air flows from a reservoir where p 300 kPa and T 500 K through a throat to section 1 in
Fig. E9.6, where there is a normalshock wave. Compute (a) p1, (b) p2, (c) p02, (d ) A*, (e) p03,
2
( f ) A*, (g) p3, (h) T03, and (i) T3.
3 Solution
1 m2
2 m2 3 m2 The reservoir conditions are the stagnation properties, which, for assumed onedimensional adiabatic frictionless flow, hold through the throat up to section 1 E9.6 p01 300 kPa T01 500 K A shock wave cannot exist unless Ma1 is supersonic; therefore the flow must have accelerated
through a throat which is sonic
At A*
1 1 m2 We can now find the Mach number Ma1 from the known isentropic area ratio
2 m2
1 m2 2.0 1.2(2.0 A1
A*
1 1)1/2 From Eq. (9.48c)
Ma1 1 2.20 Further iteration with Eq. (9.45) would give Ma1 2.1972, showing that Eq. (9.48c) gives satisfactory accuracy. The pressure p1 follows from the isentropic relation (9.28) (or Table B.1)
p01
p1  v v or  eText Main Menu p1  [1 0.2(2.20)2]3.5 300 kPa
10.7 Textbook Table of Contents 10.7 28.06 kPa  Study Guide Ans. (a) Chapter 9 Compressible Flow
The pressure p2 is now obtained from Ma1 and the normalshock relation (9.55) or Table B.2
p2
p1 1
[2.8(2.20)2
2.4 or 0.4] 5.48(28.06) p2 5.48 154 kPa Ans. (b) In similar manner, for Ma1 2.20, p02/p01 0.628 from Eq. (9.58) and A*/A*
2
1
Eq. (9.59), or we can read Table B.2 for these values. Thus
p02 0.628(300 kPa)
1.592(1 m2) A*
2 188 kPa 1.592 from
Ans. (c) 1.592 m2 Ans. (d) The flow from section 2 to 3 is isentropic (but at higher entropy than the flow upstream of the
shock). Thus
p03 p02 188 kPa Ans. (e) A*
3 A*
2 1.592 m2 Ans. (f) Knowing A*, we can now compute p3 by finding Ma3 and without bothering to find Ma2 (which
3
happens to equal 0.547). The area ratio at section 3 is
3 m2
1.592 m2 A3
A*
3 1.884 Then, since Ma3 is known to be subsonic because it is downstream of a normal shock, we use
Eq. (9.48a) to estimate
1 Ma3 0.27/(1.884)2
1.728(1.884) 0.330 The pressure p3 then follows from the isentropic relation (9.28) or Table B.1
p03
p3
or [1 0.2(0.330)2]3.5
188 kPa
1.078 p3 1.078 174 kPa Ans. (g) Meanwhile, the flow is adiabatic throughout the duct; thus
T01 T02 T03 500 K Ans. (h) Therefore, finally, from the adiabatic relation (9.26)
T03
T3
or 1
T3 0.2(0.330)2
500 K
1.022 1.022 489 K Ans. (i) Notice that this type of ductflow problem, with or without a shock wave, requires straightforward application of algebraic perfectgas relations coupled with a little thought given to which
formula is appropriate for the particular situation.  v v 596  eText Main Menu  Textbook Table of Contents  Study Guide 9.5 The NormalShock Wave 597 EXAMPLE 9.7
An explosion in air, k 1.4, creates a spherical shock wave propagating radially into still air at
standard conditions. At the instant shown in Fig. E9.7, the pressure just inside the shock is 200
lbf/in2 absolute. Estimate (a) the shock speed C and (b) the air velocity V just inside the shock.
C p = 14.7 1bf / in2 abs
T = 520° R
200 1bf / in2 abs
V
POW! E9.7 Solution
Part (a) In spite of the spherical geometry the flow across the shock moves normal to the spherical wavefront; hence the normalshock relations (9.50) to (9.59) apply. Fixing our control volume to the
moving shock, we find that the proper conditions to use in Fig. 9.8 are
C V1 14.7 lbf/in2 absolute p1 V V1 V2 p2 The speed of sound outside the shock is a1
known pressure ratio across the shock
p2
p1 T1 520°R 200 lbf/in2 absolute
49T 1/2
1 1117 ft/s. We can find Ma1 from the 200 lbf/in2 absolute
14.7 lbf/in2 absolute 13.61 From Eq. (9.55) or Table B.2
1
(2.8 Ma2
1
2.4 13.61 0.4) or Ma1 3.436 Then, by definition of the Mach number,
C Part (b) V1 Ma1 a1 3.436(1117 ft/s) 3840 ft/s Ans. (a) To find V2, we need the temperature or sound speed inside the shock. Since Ma1 is known, from
Eq. (9.58) or Table B.2 for Ma1 3.436 we compute T2/T1 3.228. Then
T2 3.228T1 3.228(520°R) 1679°R At such a high temperature we should account for nonperfectgas effects or at least use the gas
tables [16], but we won’t. Here just estimate from the perfectgas energy equation (9.23) that
V2
2
or 2cp(T1 T2) V2
1 2(6010)(520
V2 1679) (3840)2 815,000 903 ft/s  v v Notice that we did this without bothering to compute Ma2, which equals 0.454, or a2
2000 ft/s.  eText Main Menu  Textbook Table of Contents  Study Guide 1/2
49T2 598 Chapter 9 Compressible Flow
Finally, the air velocity behind the shock is
V V1 V2 3840 903 2940 ft/s Ans. (b) Thus a powerful explosion creates a brief but intense blast wind as it passes. 3 9.6 Operation of Converging
and Diverging Nozzles By combining the isentropicflow and normalshock relations plus the concept of sonic
throat choking, we can outline the characteristics of converging and diverging nozzles. Converging Nozzle First consider the converging nozzle sketched in Fig. 9.11a. There is an upstream reservoir at stagnation pressure p0. The flow is induced by lowering the downstream outside, or back, pressure pb below p0, resulting in the sequence of states a to e shown in
Fig. 9.11b and c.
For a moderate drop in pb to states a and b, the throat pressure is higher than the
critical value p* which would make the throat sonic. The flow in the nozzle is subsonic throughout, and the jet exit pressure pe equals the back pressure pb. The mass
flow is predicted by subsonic isentropic theory and is less than the critical value m max,
˙
as shown in Fig. 9.11c.
For condition c, the back pressure exactly equals the critical pressure p* of the throat.
The throat becomes sonic, the jet exit flow is sonic, pe pb, and the mass flow equals
its maximum value from Eq. (9.46). The flow upstream of the throat is subsonic everywhere and predicted by isentropic theory based on the local area ratio A(x)/A* and
Table B.1.
Finally, if pb is lowered further to conditions d or e below p*, the nozzle cannot respond further because it is choked at its maximum throat mass flow. The throat remains sonic with pe p*, and the nozzlepressure distribution is the same as in state
c, as sketched in Fig. 9.11b. The exit jet expands supersonically so that the jet pressure can be reduced from p* down to pb. The jet structure is complex and multidimensional and is not shown here. Being supersonic, the jet cannot send any signal upstream to influence the choked flow conditions in the nozzle.
If the stagnation plenum chamber is large or supplemented by a compressor, and if
the discharge chamber is larger or supplemented by a vacuum pump, the convergingnozzle flow will be steady or nearly so. Otherwise the nozzle will be blowing down,
with p0 decreasing and pb increasing, and the flow states will be changing from, say,
state e backward to state a. Blowdown calculations are usually made by a quasisteady
analysis based on isentropic steadyflow theory for the instantaneous pressures p0(t)
and pb(t). EXAMPLE 9.8
A converging nozzle has a throat area of 6 cm2 and stagnation air conditions of 120 kPa and
400 K. Compute the exit pressure and mass flow if the back pressure is (a) 90 kPa and (b) 45
kPa. Assume k 1.4.  v v 3
This is the principle of the shocktube wind tunnel, in which a controlled explosion creates a brief
flow at very high Mach number, with data taken by fastresponse instruments. See, e.g., Ref. 5, sec. 4.5.  eText Main Menu  Textbook Table of Contents  Study Guide 9.6 Operation of Converging and Diverging Nozzles 599 pb ⋅pe p0 Jet
boundary
(a)
1.0
a Subsonic
b
jet
p*
p0 c p
p0 d Supersonic
jet
e expansion Sonic
point x 0 (b) e d c 1.0
b ⋅
m
⋅
m max
Fig. 9.11 Operation of a converging
nozzle: (a) nozzle geometry showing characteristic pressures;
(b) pressure distribution caused
by various back pressures;
(c) mass flow versus back pressure. a p*
p0 0 pb
p0 1.0 (c) Solution
From Eq. (9.32) for k
p*
p0 1.4 the critical (sonic) throat pressure is
0.5283 or p* (0.5283)(120 kPa) 63.4 kPa If the back pressure is less than this amount, the nozzle flow is choked. Part (a) For pb 90 kPa p*, the flow is subsonic, not choked. The exit pressure is pe
Mach number is found from the isentropic relation (9.35) or Table B.1:
Ma2
e 5 p0
pe 2/7 1 5 120
90 pb. The throat 2/7 1 0.4283 Mae 0.654  v v To find the mass flow, we could proceed with a serial attack on Mae, Te, ae, Ve, and  eText Main Menu  Textbook Table of Contents  Study Guide e, hence 600 Chapter 9 Compressible Flow
to compute eAeVe. However, since the local pressure is known, this part is ideally suited for the
dimensionless massflow function in Eq. (9.47). With pe /p0 90/120 0.75, compute
m
˙ RT0
Ap0 hence m
˙ 2(1.4)
(0.75)2/1.4[1
0.4
0.6052 (0.0006)(120,000)
287(400) for Part (b) pe For pb 45 kPa
sure is sonic: (0.75)0.4/1.4] pb 0.6052 0.129 kg/s Ans. (a) 90 kPa Ans. (a) p*, the flow is choked, similar to condition d in Fig. 9.11b. The exit prespe p* 63.4 kPa Ans. (b) The (choked) mass flow is a maximum from Eq. (9.46b):
m
˙ 0.6847p0Ae
(RT0)1/2 m max
˙ 0.6847(120,000)(0.0006)
[287(400)]1/2 0.145 kg/s Ans. (b) Any back pressure less than 63.4 kPa would cause this same choked mass flow. Note that the
50 percent increase in exit Mach number, from 0.654 to 1.0, has increased the mass flow only
12 percent, from 0.128 to 0.145 kg/s.  v v ConvergingDiverging Nozzle Now consider the convergingdiverging nozzle sketched in Fig. 9.12a. If the back pressure pb is low enough, there will be supersonic flow in the diverging portion and a variety of shockwave conditions may occur, which are sketched in Fig. 9.12b. Let the
back pressure be gradually decreased.
For curves A and B in Fig. 9.12b the back pressure is not low enough to induce
sonic flow in the throat, and the flow in the nozzle is subsonic throughout. The pressure distribution is computed from subsonic isentropic areachange relations, e.g., Table
B.1. The exit pressure pe pb, and the jet is subsonic.
For curve C the area ratio Ae /At exactly equals the critical ratio Ae /A* for a subsonic
Mae in Table B.1. The throat becomes sonic, and the mass flux reaches a maximum in Fig.
9.12c. The remainder of the nozzle flow is subsonic, including the exit jet, and pe pb.
Now jump for a moment to curve H. Here pb is such that pb/p0 exactly corresponds
to the criticalarea ratio Ae /A* for a supersonic Mae in Table B.1. The diverging flow
is entirely supersonic, including the jet flow, and pe pb. This is called the design
pressure ratio of the nozzle and is the back pressure suitable for operating a supersonic
wind tunnel or an efficient rocket exhaust.
Now back up and suppose that pb lies between curves C and H, which is impossible according to purely isentropicflow calculations. Then back pressures D to F occur in Fig. 9.12b. The throat remains choked at the sonic value, and we can match pe
pb by placing a normal shock at just the right place in the diverging section to cause a
subsonicdiffuser flow back to the backpressure condition. The mass flow remains at
maximum in Fig. 9.12c. At back pressure F the required normal shock stands in the
duct exit. At back pressure G no single normal shock can do the job, and so the flow
compresses outside the exit in a complex series of oblique shocks until it matches pb.  eText Main Menu  Textbook Table of Contents  Study Guide 9.6 Operation of Converging and Diverging Nozzles Possible
normal shock
Throat pb pt p0 601 Possible
complex jet
geometry pe Adverse (a)
pressure
gradient
1.0 A
B p*
p0
p
p0 C
D
E Shock Sonic
throat F Supersonic x 0 G
H
I (b)
1.0
I ⋅
m
⋅
m max
Fig. 9.12 Operation of a convergingdiverging nozzle: (a) nozzle
geometry with possible flow configurations; (b) pressure distribution
caused by various back pressures;
(c) mass flow versus back pressure. H G F E D C Design
pressure
ratio 0 B A p*
p0 pb
p0 1.0
(c)  v v Finally, at back pressure I, pb is lower than the design pressure H, but the nozzle is
choked and cannot respond. The exit flow expands in a complex series of supersonic
wave motions until it matches the low back pressure. See, e.g., Ref. 9, sec. 5.4, for further details of these offdesign jetflow configurations.
Note that for pb less than back pressure C, there is supersonic flow in the nozzle
and the throat can receive no signal from the exit behavior. The flow remains choked,
and the throat has no idea what the exit conditions are.
Note also that the normal shockpatching idea is idealized. Downstream of the shock
the nozzle flow has an adverse pressure gradient, usually leading to wall boundarylayer separation. Blockage by the greatly thickened separated layer interacts strongly
with the core flow (recall Fig. 6.27) and usually induces a series of weak twodimensional compression shocks rather than a single onedimensional normal shock
(see, e.g., Ref. 14, pp. 292 and 293, for further details).  eText Main Menu  Textbook Table of Contents  Study Guide Chapter 9 Compressible Flow EXAMPLE 9.9
A convergingdiverging nozzle (Fig. 9.12a) has a throat area of 0.002 m2 and an exit area
of 0.008 m2. Air stagnation conditions are p0 1000 kPa and T0 500 K. Compute the exit
pressure and mass flow for (a) design condition and the exit pressure and mass flow if (b) pb
300 kPa and (c) pb 900 kPa. Assume k 1.4. Solution
Part (a) The design condition corresponds to supersonic isentropic flow at the given area ratio Ae/At
0.008/0.002 4.0. We can find the design Mach number either by iteration of the arearatio formula (9.45), using EES, or by the curve fit (9.48d)
Mae,design 254(4.0)2/3]1/5 [216(4.0) 2.95 (exact 2.9402) The accuracy of the curve fit is seen to be satisfactory. The design pressure ratio follows from
Eq. (9.34)
p0
pe
or [1 pe,design 0.2(2.95)2]3.5
1000 kPa
34.1 34.1 29.3 kPa Ans. (a) Since the throat is clearly sonic at design conditions, Eq. (9.46b) applies
m max
˙ m design
˙ 0.6847(106 Pa)(0.002 m2)
[287(500)]1/2 0.6847p0At
(RT0)1/2 Ans. (a) 3.61 kg/s Part (b) For pb 300 kPa we are definitely far below the subsonic isentropic condition C in Fig. 9.12b,
but we may even be below condition F with a normal shock in the exit, i.e., in condition G,
where oblique shocks occur outside the exit plane. If it is condition G, then pe pe,design 29.3
kPa because no shock has yet occurred. To find out, compute condition F by assuming an exit
normal shock with Ma1 2.95, that is, the design Mach number just upstream of the shock.
From Eq. (9.55)
p2
p1
or p2 1
[2.8(2.95)2
2.4
9.99p1 0.4] 9.99pe,design 9.99
293 kPa Since this is less than the given pb 300 kPa, there is a normal shock just upstream of the exit
plane (condition E). The exit flow is subsonic and equals the back pressure
pe
Also m
˙ pb
m max
˙ 300 kPa Ans. (b) 3.61 kg/s Ans. (b) The throat is still sonic and choked at its maximum mass flow. Part (c) Finally, for pb 900 kPa, which is up near condition C, we compute Mae and pe for condition
C as a comparison. Again Ae/At 4.0 for this condition, with a subsonic Mae estimated from
the curvefitted Eq. (9.48a):
Mae(C)  v v 602  eText Main Menu  1 0.27/(4.0)2
1.728(4.0) 0.147 (exact Textbook Table of Contents  0.14655) Study Guide 9.7 Compressible Duct Flow with Friction 603 Then the isentropic exitpressure ratio for this condition is
p0
pe 0.2(0.147)2]3.5 pe or [1 1000
1.0152 1.0152 985 kPa The given back pressure of 900 kPa is less than this value, corresponding roughly to condition
D in Fig. 9.12b. Thus for this case there is a normal shock just downstream of the throat, and
the throat is choked
pe pb 900 kPa m
˙ m max
˙ 3.61 kg/s Ans. (c) For this large exitarea ratio, the exit pressure would have to be larger than 985 kPa to cause a
subsonic flow in the throat and a mass flow less than maximum. 9.7 Compressible Duct Flow
with Friction4 Section 9.4 showed the effect of area change on a compressible flow while neglecting
friction and heat transfer. We could now add friction and heat transfer to the area change
and consider coupled effects, which is done in advanced texts [for example, 8, chap.
8]. Instead, as an elementary introduction, this section treats only the effect of friction,
neglecting area change and heat transfer. The basic assumptions are
1.
2.
3.
4.
5. Steady onedimensional adiabatic flow
Perfect gas with constant specific heats
Constantarea straight duct
Negligible shaftwork and potentialenergy changes
Wall shear stress correlated by a Darcy friction factor In effect, we are studying a Moodytype pipefriction problem but with large changes
in kinetic energy, enthalpy, and pressure in the flow.
Consider the elemental duct control volume of area A and length dx in Fig. 9.13.
The area is constant, but other flow properties ( p, , T, h, V ) may vary with x. Appli4 This section may be omitted without loss of continuity. Control volume τwπD d x V V + dV
p p + dp ρ ρ + dρ T Area A
Diameter D h  v v Fig. 9.13 Elemental control volume
for flow in a constantarea duct
with friction.  T + dT
h + dh dx
x + dx x eText Main Menu  Textbook Table of Contents  Study Guide 604 Chapter 9 Compressible Flow cation of the three conservation laws to this control volume gives three differential
equations
Continuity: G d or
x momentum: m
˙
A V pA (p or
h dV
V dp)A 1
2 0 D dx w 4 wdx
D dp Energy: const V2 or h0 (9.60a)
m (V
˙ V dV
cpT0 V) 0 cpT V dV cp dT dV (9.60b)
1
2 V2 0 (9.60c) Since these three equations have five unknowns—p, , T, V, and
ditional relations. One is the perfectgas law
p RT dp
p or d w—we need two ad dT
T (9.61) To eliminate w as an unknown, it is assumed that wall shear is correlated by a local
Darcy friction factor f
w 1
8 f V2 1
8 f kp Ma2 (9.62) where the last form follows from the perfectgas speedofsound expression a2 kp/ .
In practice, f can be related to the local Reynolds number and wall roughness from,
say, the Moody chart, Fig. 6.13.
Equations (9.60) and (9.61) are firstorder differential equations and can be integrated, by using frictionfactor data, from any inlet section 1, where p1, T1, V1, etc.,
are known, to determine p(x), T(x), etc., along the duct. It is practically impossible to
eliminate all but one variable to give, say, a single differential equation for p(x), but
all equations can be written in terms of the Mach number Ma(x) and the friction factor, by using the definition of Mach number
V2
2 dV
V or Adiabatic Flow Ma2 kRT
dT
T 2 d Ma
Ma (9.63) By eliminating variables between Eqs. (9.60) to (9.63), we obtain the working
relations
dp
p  v v d  eText Main Menu  k Ma2 1 (k
2(1 1) Ma2 dx
f
Ma2)
D k Ma2
dx
f
2(1 Ma2) D Textbook Table of Contents dV
V  (9.64a)
(9.64b) Study Guide 9.7 Compressible Duct Flow with Friction dp0
p0 d k(k 1) Ma4 dx
f
2(1 Ma2)
D dT
T
d Ma2
Ma2 1
dx
k Ma2 f
2
D 0
0 (k 1) Ma2
dx
f
1 Ma2
D 605 (9.64c)
(9.64d) 1
2 1
2 k Ma (9.64e) All these except dp0/p0 have the factor 1 Ma2 in the denominator, so that, like the
areachange formulas in Fig. 9.5, subsonic and supersonic flow have opposite effects:
Property Subsonic Supersonic p Decreases
Decreases
Increases
Decreases
Decreases
Increases
Increases Increases
Increases
Decreases
Decreases
Increases
Decreases
Increases V
p0, 0
T
Ma
Entropy We have added to the list above that entropy must increase along the duct for either
subsonic or supersonic flow as a consequence of the second law for adiabatic flow. For
the same reason, stagnation pressure and density must both decrease.
The key parameter above is the Mach number. Whether the inlet flow is subsonic
or supersonic, the duct Mach number always tends downstream toward Ma 1 because this is the path along which the entropy increases. If the pressure and density are
computed from Eqs. (9.64a) and (9.64b) and the entropy from Eq. (9.53), the result can
be plotted in Fig. 9.14 versus Mach number for k 1.4. The maximum entropy occurs
at Ma 1, so that the second law requires that the ductflow properties continually approach the sonic point. Since p0 and 0 continually decrease along the duct due to the
frictional (nonisentropic) losses, they are not useful as reference properties. Instead, the
sonic properties p*, *, T*, p*, and * are the appropriate constant reference quanti0
0
ties in adiabatic duct flow. The theory then computes the ratios p/p*, T/T*, etc., as a
function of local Mach number and the integrated friction effect.
To derive working formulas, we first attack Eq. (9.64e), which relates the Mach number to friction. Separate the variables and integrate:
L* f
0 dx
D 1.0
Ma2 1
k Ma4[1 1
2 Ma2
d Ma2
(k 1) Ma2] (9.65)  v v The upper limit is the sonic point, whether or not it is actually reached in the duct flow.
The lower limit is arbitrarily placed at the position x 0, where the Mach number is
Ma. The result of the integration is (k 1) Ma2
1 Ma2
k1
f L*
ln
(9.66)
2
k Ma
2k
2 (k 1) Ma2
D where f is the average friction factor between 0 and L*. In practice, an average f is always assumed, and no attempt is made to account for the slight changes in Reynolds  eText Main Menu  Textbook Table of Contents  Study Guide 606 Chapter 9 Compressible Flow 4.0 Supersonic
duct flow
3.0 Mach number k = 1.4 2.0 Maximum
entropy
at Ma = 1.0
1.0
Subsonic
duct flow Fig. 9.14 Adiabatic frictional flow
in a constantarea duct always approaches Ma 1 to satisfy the second law of thermodynamics. The
computed curve is independent of
the value of the friction factor. 0 0.2 0.4 0.6
s
cv 0.8 1.0 1.2 number along the duct. For noncircular ducts, D is replaced by the hydraulic diameter
Dh (4 area)/perimeter as in Eq. (6.74).
Equation (9.66) is tabulated versus Mach number in Table B.3. The length L* is the
length of duct required to develop a duct flow from Mach number Ma to the sonic
point. Many problems involve short ducts which never become sonic, for which the solution uses the differences in the tabulated “maximum,’’ or sonic, length. For example,
the length L required to develop from Ma1 to Ma2 is given by f L*
f L*
L
(9.67)
f
D
D1
D2
This avoids the need for separate tabulations for short ducts. It is recommended that the friction factor f be estimated from the Moody chart (Fig.
6.13) for the average Reynolds number and wallroughness ratio of the duct. Available
data [20] on duct friction for compressible flow show good agreement with the Moody
chart for subsonic flow, but the measured data in supersonic duct flow are up to 50
percent less than the equivalent Moody friction factor. EXAMPLE 9.10  v v Air flows subsonically in an adiabatic 2cmdiameter duct. The average friction factor is 0.024.
What length of duct is necessary to accelerate the flow from Ma1 0.1 to Ma2 0.5? What additional length will accelerate it to Ma3 1.0? Assume k 1.4.  eText Main Menu  Textbook Table of Contents  Study Guide 9.7 Compressible Duct Flow with Friction 607 Solution Equation (9.67) applies, with values of f L*/D computed from Eq. (9.66) or read from Table B.3:
L
fD f L* 0.024 L
0.02 m D 66.9216
Thus L to go from Ma
L
fD or L L Ma
* 55 m Ans. (a) 1.0 is taken directly from Table B.2
1.0691 Ma 0.5 1.0691(0.02 m)
0.024 0.5 Ma 0.5 65.8525 0.5 to Ma fL*
D D Ma 0.1 65.8525(0.02 m)
0.024 L The additional length 1.0691 f L* 0.9 m This is typical of these calculations: It takes 55 m to accelerate up to Ma
0.9 m more to get all the way up to the sonic point. Ans. (b)
0.5 and then only Formulas for other flow properties along the duct can be derived from Eqs. (9.64).
Equation (9.64e) can be used to eliminate f dx/D from each of the other relations, giving, for example, dp/p as a function only of Ma and d Ma2/Ma2. For convenience in
tabulating the results, each expression is then integrated all the way from (p, Ma) to
the sonic point ( p*, 1.0). The integrated results are
p
p*
* 1
Ma 2
V*
V T
T*
p0
p*
0 0 *
0 k
(k 12
Ma 1
1) Ma2 1/2 (k
k 1) Ma2
1 (9.68a) 1
1) Ma2 a2
a*2 2 k
(k 12
Ma (k
k 1) Ma2
1 1/2 (9.68b)
(9.68c) (1/2)(k 1)/(k 1) (9.68d ) All these ratios are also tabulated in Table B.3. For finding changes between points
Ma1 and Ma2 which are not sonic, products of these ratios are used. For example,
p2
p1 p2 p*
p* p1 (9.69) since p* is a constant reference value for the flow.
EXAMPLE 9.11  v v For the duct flow of Example 9.10 assume that, at Ma1 0.1, we have p1 600 kPa and T1
450 K. At section 2 farther downstream, Ma2 0.5. Compute (a) p2, (b) T2, (c) V2, and (d ) p02.  eText Main Menu  Textbook Table of Contents  Study Guide 608 Chapter 9 Compressible Flow Solution
As preliminary information we can compute V1 and p01 from the given data:
V1
p01 p1(1 0.1[(1.4)(287)(450)]1/2 Ma1 a1 0.2 Ma2)3.5
1 0.1(425 m/s) 42.5 m/s 0.2(0.1)2]3.5 604 kPa (600 kPa)[1 Now enter Table B.3 or Eqs. (9.68) to find the following property ratios:
Section Ma p/p* T/T* V/V* p0/p*
0 1
2 0.1
0.5 10.9435
2.1381 1.1976
1.1429 0.1094
0.5345 5.8218
1.3399 Use these ratios to compute all properties downstream:
2.1381
10.9435 p2 p1 p2/p*
p1/p* (600 kPa) T2 T1 T2/T*
T1/T* (450 K) V2 V1 V2/V*
V1/V* (42.5 m/s) p02 p01 p02/p*
0
p01/p*
0 1.1429
1.1976 (604 kPa) 117 kPa
429 K 0.5345
0.1094
1.3399
5.8218 Ans. (a)
Ans. (b) m
s Ans. (c) 139 kPa Ans. (d) 208 Note the 77 percent reduction in stagnation pressure due to friction. The formulas are seductive,
so check your work by other means. For example, check p02 p2(1 0.2 Ma2)3.5.
2 The theory here predicts that for adiabatic frictional flow in a constantarea duct, no
matter what the inlet Mach number Ma1 is, the flow downstream tends toward the sonic
point. There is a certain duct length L*(Ma1) for which the exit Mach number will be
exactly unity. The duct is then choked.
But what if the actual length L is greater than the predicted “maximum’’ length L*?
Then the flow conditions must change, and there are two classifications.
Subsonic inlet. If L L*(Ma1), the flow slows down until an inlet Mach number Ma2
is reached such that L L*(Ma2). The exit flow is sonic, and the mass flow has been
reduced by frictional choking. Further increases in duct length will continue to decrease
the inlet Ma and mass flow.
Supersonic inlet. From Table B.3 we see that friction has a very large effect on supersonic duct flow. Even an infinite inlet Mach number will be reduced to sonic con
ditions in only 41 diameters for f
0.02. Some typical numerical values are shown in Fig. 9.15, assuming an inlet Ma 3.0 and f
0.02. For this condition L* 26 diameters. If L is increased beyond 26D, the flow will not choke but a normal shock will
form at just the right place for the subsequent subsonic frictional flow to become sonic
exactly at the exit. Figure 9.15 shows two examples, for L/D 40 and 53. As the length
increases, the required normal shock moves upstream until, for Fig. 9.15, the shock is
at the inlet for L/D 63. Further increase in L causes the shock to move upstream of  v v Choking due to Friction  eText Main Menu  Textbook Table of Contents  Study Guide 9.7 Compressible Duct Flow with Friction 609 3.0 2.5 f = 0.020
k = 1.4 Fig. 9.15 Behavior of duct flow
with a nominal supersonic inlet
condition Ma 3.0: (a) L/D 26,
flow is supersonic throughout duct;
(b) L/D 40 L*/D, normal
shock at Ma 2.0 with subsonic
flow then accelerating to sonic exit
point; (c) L/D 53, shock must
now occur at Ma 2.5; (d) L/D
63, flow must be entirely subsonic
and choked at exit. Mach number 2.0 1.5 c b a 1.0 0.5
d 0 20 10 30
x
D 40 50 60 the inlet into the supersonic nozzle feeding the duct. Yet the mass flow is still the same
as for the very short duct, because presumably the feed nozzle still has a sonic throat.
Eventually, a very long duct will cause the feednozzle throat to become choked, thus
reducing the duct mass flow. Thus supersonic friction changes the flow pattern if L
L* but does not choke the flow until L is much larger than L*. EXAMPLE 9.12
Air enters a 3cmdiameter duct at p0 200 kPa, T0 500 K, and V1 100 m/s. The friction
factor is 0.02. Compute (a) the maximum duct length for these conditions, (b) the mass flow if
the duct length is 15 m, and (c) the reduced mass flow if L 30 m. Solution
Part (a) First compute V2
1
cp 1
2 T1 T0 a1
Thus (100 m/s)2
1005 m2/(s2 K)
1
2 500 (kRT1)1/2
Ma1 V1
a1 20(495)1/2
100
445 500 5 445 m/s 0.225  v v For this Ma1, from Eq. (9.66) or interpolation in Table B.3, f L*
11.0
D  eText Main Menu  Textbook Table of Contents  Study Guide 495 K 610 Chapter 9 Compressible Flow
The maximum duct length possible for these inlet conditions is 11.0(0.03 m)
(f L*/D)D
L*
16.5 m 0.02
f Part (b) Ans. (a) The given L 15 m is less than L*, and so the duct is not choked and the mass flow follows
from inlet conditions
p01
RT0 01 200,000 Pa
287(500 K) 1.394
1.0255 01 whence 0.2(0.225)2]2.5 [1 1 m
˙ 1.394 kg/m3
1.359 kg/m3 (1.359 kg/m3) 4 (0.03 m)2 (100 m/s) 1AV1 0.0961 kg/s Part (c) Ans. (b) Since L 30 m is greater than L*, the duct must choke back until L
lower inlet Ma1:
L* f L*
D L 30 m 0.02(30 m)
0.03 m 20.0 It is difficult to interpolate for fL/D 20 in Table B.3 and impossible to invert Eq. (9.66) for
the Mach number without laborious iteration. But it is a breeze for EES to solve Eq. (9.66) for
the Mach number, using the following three statements: EES k 1.4
fLD 20
fLD (1 Ma^2)/k/Ma^2 Simply specify Ma (k 1)/2/k*LN((k (k 1)*Ma^2)) 0.174 (23 percent less)
T0
0.2(0.174)2 T1,new 1 a1,new 20(497 K)1/2 V1,new Ma1 a1 497 K 446 m/s 0.174(446)
01 1,new m new
˙ [1
1AV1 0.2(0.174)2]2.5
1.373 0.0753 kg/s 77.6 m/s
1.373 kg/m3 π
(0.03)2 (77.6)
4
(22 percent less) Ans. (c) The adiabatic frictionalflow assumption is appropriate to highspeed flow in short
ducts. For flow in long ducts, e.g., naturalgas pipelines, the gas state more closely ap Isothermal Flow with Friction v v 1)*Ma^2/(2 1 in the Variable Information menu and EES cheerfully reports Machoked  L*, corresponding to a  eText Main Menu  Textbook Table of Contents  Study Guide 9.7 Compressible Duct Flow with Friction 611 proximates an isothermal flow. The analysis is the same except that the isoenergetic
energy equation (9.60c) is replaced by the simple relation
T const dT 0 (9.70) Again it is possible to write all property changes in terms of the Mach number. Integration of the Machnumber – friction relation yields 1 k Ma2
f Lmax
ln (k Ma2)
(9.71)
k Ma2
D
which is the isothermal analog of Eq. (9.66) for adiabatic flow.
This friction relation has the interesting result that Lmax becomes zero not at the
sonic point but at Macrit 1/k1/2 0.845 if k 1.4. The inlet flow, whether subsonic
or supersonic, tends downstream toward this limiting Mach number 1/k1/2. If the tube
length L is greater than Lmax from Eq. (9.71), a subsonic flow will choke back to a
smaller Ma1 and mass flow and a supersonic flow will experience a normalshock adjustment similar to Fig. 9.15.
The exit isothermal choked flow is not sonic, and so the use of the asterisk is inappropriate. Let p , , and V represent properties at the choking point L Lmax. Then
the isothermal analysis leads to the following Machnumber relations for the flow properties:
p
p V
V 1
Ma k1/2 Ma k1/2 (9.72) The complete analysis and some examples are given in advanced texts [for example,
8, sec. 6.4]. Mass Flow for a Given
Pressure Drop An interesting byproduct of the isothermal analysis is an explicit relation between the
pressure drop and duct mass flow. This is a common problem which requires numerical iteration for adiabatic flow, as outlined below. In isothermal flow, we may substitute dV/V
dp/p and V2 G2/[p/(RT)]2 in Eq. (9.63) to obtain
2p dp
G2RT f dx
D 2 dp
p 0 Since G2RT is constant for isothermal flow, this may be integrated in closed form between (x, p) (0, p1) and (L, p2):
m
˙
A G2 2 p2 p2
1
2 RT [ f L/D 2 ln (p1/p2)] (9.73) Thus mass flow follows directly from the known end pressures, without any use of
Mach numbers or tables.
The writer does not know of any direct analogy to Eq. (9.73) for adiabatic flow.
However, a useful adiabatic relation, involving velocities instead of pressures, is derived in several textbooks [5, p. 212; 34, p. 418]:  v v V2
1  eText Main Menu  a2[1 (V1/V2)2]
0 k f L/D (k 1) ln (V2/V1) Textbook Table of Contents  Study Guide (9.74) Chapter 9 Compressible Flow where a0 (kRT0)1/2 is the stagnation speed of sound, constant for adiabatic flow. We
assign the proof of this as a problem exercise. This may be combined with continuity
for constant duct area V1/V2
2/ 1, plus the following combination of adiabatic energy and the perfectgas relation:
V1
V2 p2 T1
p1 T2 p2
p1 2a2
0
2a2
0 1)V2
1
1)V2
2 (k
(k (9.75) If we are given the end pressures, neither V1 nor V2 will likely be known in advance.
Here, if EES is not available, we suggest only the following simple procedure. Begin
with a0 a1 and the bracketed term in Eq. (9.75) approximately equal to 1.0. Solve
Eq. (9.75) for a first estimate of V1/V2, and use this value in Eq. (9.74) to get a better
estimate of V1. Use V1 to improve your estimate of a0, and repeat the procedure. The
process should converge in a few iterations.
Equations (9.73) and (9.74) have one flaw: With the Mach number eliminated, the
frictional choking phenomenon is not directly evident. Therefore, assuming a subsonic
inlet flow, one should check the exit Mach number Ma2 to ensure that it is not greater
than 1/k1/2 for isothermal flow or greater than 1.0 for adiabatic flow. We illustrate both
adiabatic and isothermal flow with the following example.
EXAMPLE 9.13 Air enters a pipe of 1cm diameter and 1.2m length at p1 220 kPa and T1 300 K. If f
0.025 and the exit pressure is p2 140 kPa, estimate the mass flow for (a) isothermal flow and
(b) adiabatic flow. Solution
Part (a) For isothermal flow Eq. (9.73) applies without iteration: fL
p
(0.025)(1.2 m)
220
2 ln 1
2 ln
D
0.01 m
140
p2
(220,000 Pa)2 (140,000 Pa)2
[287 m2/(s2 K)](300 K)(3.904) G2
Since A ( /4)(0.01 m)2 85,700 3.904 or G 293 kg/(s m2) 7.85 E5 m2, the isothermal mass flow estimate is
m
˙ GA (293)(7.85 E5) 0.0230 kg/s Ans. (a) Check that the exit Mach number is not choked:
2 or p2
RT 140,000
(287)(300)
Ma2 V2
kRT 1.626 kg/m3 V2 G 293
1.626 2 180
[1.4(287)(300)]1/2 180
347 180 m/s 0.52 This is well below choking, and the isothermal solution is accurate. Part (b) For adiabatic flow, we can iterate by hand, in the timehonored fashion, using Eqs. (9.74) and
(9.75) plus the definition of stagnation speed of sound. A few years ago the author would have
done just that, laboriously. However, EES makes handwork and manipulation of equations un EES  v v 612  eText Main Menu  Textbook Table of Contents  Study Guide 9.8 Frictionless Duct Flow with Heat Transfer 613
necessary, although careful programming and good guesses are required. If we ignore superfluous output such as T2 and V2, 13 statements are appropriate. First, spell out the given physical
properties (in SI units):
k 1.4
P1 220000
P2 140000
T1 300
Next, apply the adiabatic friction relations, Eqs. (9.66) and (9.67), to both points 1 and 2:
fLD1 (1 Ma1^2)/k/Ma1^2 (k 1)/2/k*LN((k 1)*Ma1^2/(2 (k1)*Ma1^2))
fLD2 (1 Ma2^2)/k/Ma2^2 (k 1)/2/k*LN((k 1)*Ma2^2/(2 (k1)*Ma2^2))
DeltafLD 0.025*1.2/0.01
fLD1 fLD2 DeltafLD
Then apply the pressureratio formula (9.68a) to both points 1 and 2:
P1/Pstar
P2/Pstar ((k 1)/(2 (k1)*Ma1^2))^0.5/Ma1
((k 1)/(2 (k1)*Ma2^2))^0.5/Ma2 These are adiabatic relations, so we need not further spell out quantities such as T0 or a0 unless
we want them as additional output.
The above 10 statements are a closed algebraic system, and EES will solve them for Ma1 and
Ma2. However, the problem asks for mass flow, so we complete the system:
V1 Ma1*sqrt(1.4*287*T1)
Rho1 P1/287/T1
Mdot Rho1*(pi/4*0.01^2)*V1
If we apply no constraints, EES reports “cannot solve”, because its default allows all variables
to lie between
and
. So we enter Variable Information and constrain Ma1 and Ma2 to
lie between 0 and 1 (subsonic flow). EES still complains that it “cannot solve” but hints that
“better guesses are needed”. Indeed, the default guesses for EES variables are normally 1.0, too
large for the Mach numbers. Guess the Mach numbers equal to 0.8 or even 0.5, and EES still
complains, for a subtle reason: Since f L/D 0.025(1.2/0.01) 3.0, Ma1 can be no larger than
0.36 (see Table B.3). Finally, then, we guess Ma1 and Ma2 0.3 or 0.4, and EES happily reports the solution:
fL
fL
Ma1 0.3343
Ma2 0.5175
3.935
0.9348
D1
D2
p* 67,892 Pa m
˙ 0.0233 kg/s Ans. (b) Though the programming is complicated, the EES approach is superior to hand iteration and, of
course, we can save this program for use again with new data. 9.8 Frictionless Duct Flow with
Heat Transfer5 Heat addition or removal has an interesting effect on a compressible flow. Advanced
texts [for example, 8, chap. 8] consider the combined effect of heat transfer coupled
with friction and area change in a duct. Here we confine the analysis to heat transfer
with no friction in a constantarea duct.  v v 5  This section may be omitted without loss of continuity. eText Main Menu  Textbook Table of Contents  Study Guide 614 Chapter 9 Compressible Flow
Control
volume q= δQ
δm A2 = A1
V1, p1, T1, T01 Fig. 9.16 Elemental control volume
for frictionless flow in a constantarea duct with heat transfer. The
length of the element is indeterminate in this simplified theory. V2, p2, T2, T02
1 2 τw = 0 Consider the elemental duct control volume in Fig. 9.16. Between sections 1 and 2
an amount of heat Q is added (or removed) to each incremental mass m passing
through. With no friction or area change, the controlvolume conservation relations are
quite simple:
1V1 Continuity:
x momentum: p1
˙
Q Energy:
or 2V2 p2 m (h2
˙
q ˙
Q
m
˙ G const
V1) G(V2
1
2 V2
2 Q
m (9.76a)
(9.76b) h1
h02 1
2 V2)
1 h01 (9.76c) The heat transfer results in a change in stagnation enthalpy of the flow. We shall not specify exactly how the heat is transferred—combustion, nuclear reaction, evaporation, condensation, or wall heat exchange—but simply that it happened in amount q between 1
and 2. We remark, however, that wall heat exchange is not a good candidate for the theory because wall convection is inevitably coupled with wall friction, which we neglected.
To complete the analysis, we use the perfectgas and Machnumber relations
p2
2T2
V2
V1 p1
1T1 h02 h01 Ma2 a2
Ma1 a1 Ma2
Ma1 T2
T1 cp(T02 T01)
(9.77) 1/2  v v For a given heat transfer q
Q/ m or, equivalently, a given change h02 h01, Eqs.
(9.76) and (9.77) can be solved algebraically for the property ratios p2/p1, Ma2/Ma1,
etc., between inlet and outlet. Note that because the heat transfer allows the entropy to
either increase or decrease, the second law imposes no restrictions on these solutions.
Before writing down these propertyratio functions, we illustrate the effect of heat
transfer in Fig. 9.17, which shows T0 and T versus Mach number in the duct. Heating
increases T0, and cooling decreases it. The maximum possible T0 occurs at Ma 1.0,
and we see that heating, whether the inlet is subsonic or supersonic, drives the duct
Mach number toward unity. This is analogous to the effect of friction in the previous
section. The temperature of a perfect gas increases from Ma 0 up to Ma 1/k1/2 and
then decreases. Thus there is a peculiar—or at least unexpected—region where heat  eText Main Menu  Textbook Table of Contents  Study Guide 9.8 Frictionless Duct Flow with Heat Transfer 615 T0 (max) at Ma = 1.0 k = 1.4 T0 Hea ting Coo
ling T (max) at
Ma = Co oli 1
k1/ 2 ng Hea
ting T, T0 Fig. 9.17 Effect of heat transfer on
Mach number. 0 0.5 T 1
1.5
Mach number 2 2.5 ing (increasing T0) actually decreases the gas temperature, the difference being reflected
in a large increase of the gas kinetic energy. For k 1.4 this peculiar area lies between
Ma 0.845 and Ma 1.0 (interesting but not very useful information).
The complete list of the effects of simple T0 change on ductflow properties is as
follows:
Heating Cooling Subsonic V
p0
s
T Subsonic Supersonic Increases
Increases
Decreases
Decreases
Increases
Decreases
Increases
* T0
Ma
p Supersonic
Increases
Decreases
Increases
Increases
Decreases
Decreases
Increases
Increases Decreases
Decreases
Increases
Increases
Decreases
Increases
Decreases
† Decreases
Increases
Decreases
Decreases
Increases
Increases
Decreases
Decreases * Increases up to Ma 1/k1/2 and decreases thereafter.
Decreases up to Ma 1/k1/2 and increases thereafter. † Probably the most significant item on this list is the stagnation pressure p0, which always
decreases during heating whether the flow is subsonic or supersonic. Thus heating does
increase the Mach number of a flow but entails a loss in effective pressure recovery. MachNumber Relations Equations (9.76) and (9.77) can be rearranged in terms of the Mach number and the
results tabulated. For convenience, we specify that the outlet section is sonic, Ma 1,
with reference properties T*, T*, p*, *, V*, and p*. The inlet is assumed to be at ar0
0
bitrary Mach number Ma. Equations (9.76) and (9.77) then take the following form:  v v T0
T*
0  eText Main Menu  (k 1) Ma2 [2 (k 1) Ma2]
(1 k Ma2)2 Textbook Table of Contents  Study Guide (9.78a) Chapter 9 Compressible Flow T
T* k
1 k
1
* V
V*
p0
p*
0 1)2 Ma2
k Ma2)2 (k
(1 p
p* 1
k Ma2 (9.78b) 1
k Ma2 (9.78c) (k
1
2 1) Ma2
k Ma2 (9.78d) 1) Ma2
1 (k
k k/(k 1) (9.78e) These formulas are all tabulated versus Mach number in Table B.4. The tables are
very convenient if inlet properties Ma1, V1, etc., are given but are somewhat cumbersome if the given information centers on T01 and T02. Let us illustrate with an
example. EXAMPLE 9.14
A fuelair mixture, approximated as air with k 1.4, enters a duct combustion chamber at V1
75 m/s, p1 150 kPa, and T1 300 K. The heat addition by combustion is 900 kJ/kg of mixture. Compute (a) the exit properties V2, p2, and T2 and (b) the total heat addition which would
have caused a sonic exit flow. Solution
First compute T01 T1 V 2/(2cp) 300
1
in stagnation temperature of the gas: Part (a) (75)2/[2(1005)] q
or T02 T01 q
cp cp(T02 303 K 303 K. Then compute the change T01) 900,000 J/kg
1005 J/(kg K) 1199 K We have enough information to compute the initial Mach number:
a1 kRT1 [1.4(287)(300)]1/2 347 m/s V1
a1 Ma1 75
347 0.216 For this Mach number, use Eq. (9.78a) or Table B.4 to find the sonic value T*0:
At Ma1 0.216: T01
T*
0 0.1992 or T*
0 303 K
0.1992 1521 K Then the stagnation temperature ratio at section 2 is T02/T* 1199/1521 0.788, which cor0
responds in Table B.4 to a Mach number Ma2 0.573.
Now use Table B.4 at Ma1 and Ma2 to tabulate the desired property ratios.
Section v   Ma V/V* p/p* T/T* 1
2 v 616 0.216
0.573 0.1051
0.5398 2.2528
1.6442 0.2368
0.8876 eText Main Menu  Textbook Table of Contents  Study Guide 9.8 Frictionless Duct Flow with Heat Transfer 617
The exit properties are computed by using these ratios to find state 2 from state 1:
V2 V1 V2/V*
V1/V* p2 p1 p2/p*
p1/p* T1 T2 Part (b) (75 m/s) 385 m/s Ans. (a) (150 kPa) 1.6442
2.2528 109 kPa Ans. (a) (300 K) T2/T*
T1/T* 0.5398
0.1051 0.8876
0.2368 1124 K Ans. (a) The maximum allowable heat addition would drive the exit Mach number to unity:
T02
qmax Choking Effects due
to Simple Heating cp(T*
0 T01) T*
0 1521 K [1005 J/(kg K)](1521 303 K) 1.22 E6 J/kg Ans. (b) Equation (9.78a) and Table B.4 indicate that the maximum possible stagnation temperature in simple heating corresponds to T *, or the sonic exit Mach number. Thus,
0
for given inlet conditions, only a certain maximum amount of heat can be added to the
flow, for example, 1.22 MJ/kg in Example 9.14. For a subsonic inlet there is no theoretical limit on heat addition: The flow chokes more and more as we add more heat,
with the inlet velocity approaching zero. For supersonic flow, even if Ma1 is infinite,
there is a finite ratio T01/ T * 0.4898 for k 1.4. Thus if heat is added without limit
0
to a supersonic flow, a normalshockwave adjustment is required to accommodate the
required property changes.
In subsonic flow there is no theoretical limit to the amount of cooling allowed: The
exit flow just becomes slower and slower, and the temperature approaches zero. In supersonic flow only a finite amount of cooling can be allowed before the exit flow approaches infinite Mach number, with T02/T * 0.4898 and the exit temperature equal
0
to zero. There are very few practical applications for supersonic cooling. EXAMPLE 9.15
What happens to the inlet flow in Example 9.14 if the heat addition is increased to 1400 kJ/kg
and the inlet pressure and stagnation temperature are fixed? What will be the subsequent decrease in mass flow? Solution
For q 1400 kJ/kg, the exit will be choked at the stagnation temperature
T*
0 T01 q
cp 303 1.4 E6 J/kg
1005 J/(kg K) 1696 K  v v This is higher than the value T* 1521 K in Example 9.14, so we know that condition 1 will
0
have to choke down to a lower Mach number. The proper value is found from the ratio
T01/ T* 303/1696 0.1787. From Table B.4 or Eq. (9.78a) for this condition, we read the
0  eText Main Menu  Textbook Table of Contents  Study Guide 618 Chapter 9 Compressible Flow
new, lowered entrance Mach number: Ma1,new
properties follow from this Mach number:
T1
a1
V1
1 T01
0.2 Ma2
1 1 kRT1
Ma1 a1
p1
RT1 1 0.203. With T01 and p1 known, the other inlet
303
0.2(0.203)2 [1.4(287)(301)] 1/2 (0.202)(348 m/s)
150,000
(287)(301) 301 K 348 m/s
70 m/s 1.74 kg/m3 Finally, the new lowered mass flow per unit area is
m new
˙
A 1V1 (1.74 kg/m3)(70 m/s) 122 kg/(s m2) This is 7 percent less than in Example 9.14, due to choking by excess heat addition. The normalshockwave relations of Sec. 9.5 actually lurk within the simple heating relations as a special case. From Table B.4 or Fig. 9.17 we see that for a given stagnation temperature less than T* there are two flow states which satisfy the simple heating relations,
0
one subsonic and the other supersonic. These two states have (1) the same value of T0,
(2) the same mass flow per unit area, and (3) the same value of p
V2. Therefore these
two states are exactly equivalent to the conditions on each side of a normalshock wave.
The second law would again require that the upstream flow Ma1 be supersonic.
To illustrate this point, take Ma1 3.0 and from Table B.4 read T01/T* 0.6540 and
0
p1/p* 0.1765. Now, for the same value T02/T* 0.6540, use Table B.4 or Eq. (9.78a)
0
to compute Ma2 0.4752 and p2/p* 1.8235. The value of Ma2 is exactly what we
read in the shock table, Table B.2, as the downstream Mach number when Ma1 3.0.
The pressure ratio for these two states is p2/p1 (p2/p*)/(p1/p*) 1.8235/0.1765
10.33, which again is just what we read in Table B.2 for Ma1 3.0. This illustration is
meant only to show the physical background of the simple heating relations; it would
be silly to make a practice of computing normalshock waves in this manner. 9.9 TwoDimensional
Supersonic Flow Up to this point we have considered only onedimensional compressibleflow theories.
This illustrated many important effects, but a onedimensional world completely loses
sight of the wave motions which are so characteristic of supersonic flow. The only
“wave motion’’ we could muster in a onedimensional theory was the normalshock
wave, which amounted only to a flow discontinuity in the duct. Mach Waves When we add a second dimension to the flow, wave motions immediately become apparent if the flow is supersonic. Figure 9.18 shows a celebrated graphical construction
which appears in every fluidmechanics textbook and was first presented by Ernst Mach
in 1887. The figure shows the pattern of pressure disturbances (sound waves) sent out
by a small particle moving at speed U through a still fluid whose sound velocity is a.  v v Relationship to the
NormalShock Wave  eText Main Menu  Textbook Table of Contents  Study Guide 9.9 TwoDimensional Suspersonic Flow 619 Limiting
Mach
wave
a δt a δt U<a U=a Typical pressure
disturbance caused
by particle passage U δt U δt (a) (b)
1
µ = sin–1 Ma
Zone of
silence a δt U>a
U δt Fig. 9.18 Wave patterns set up by a
particle moving at speed U into
still fluid of sound velocity a:
(a) subsonic, (b) sonic, and
(c) supersonic motion. Zone of
action Supersonic
Mach wave
(c)  v v As the particle moves, it continually crashes against fluid particles and sends out
spherical sound waves emanating from every point along its path. A few of these spherical disturbance fronts are shown in Fig. 9.18. The behavior of these fronts is quite different according to whether the particle speed is subsonic or supersonic.
In Fig. 9.18a, the particle moves subsonically, U a, Ma U/a 1. The spherical disturbances move out in all directions and do not catch up with one another. They move well
out in front of the particle also, because they travel a distance a t during the time interval
t in which the particle has moved only U t. Therefore a subsonic body motion makes its
presence felt everywhere in the flow field: You can “hear’’ or “feel’’ the pressure rise of an
oncoming body before it reaches you. This is apparently why that pigeon in the road, without turning around to look at you, takes to the air and avoids being hit by your car.
At sonic speed, U a, Fig. 9.18b, the pressure disturbances move at exactly the
speed of the particle and thus pile up on the left at the position of the particle into a
sort of “front locus,’’ which is now called a Mach wave, after Ernst Mach. No disturbance reaches beyond the particle. If you are stationed to the left of the particle, you
cannot “hear’’ the oncoming motion. If the particle blew its horn, you couldn’t hear
that either: A sonic car can sneak up on a pigeon.
In supersonic motion, U a, the lack of advance warning is even more pronounced.
The disturbance spheres cannot catch up with the fastmoving particle which created
them. They all trail behind the particle and are tangent to a conical locus called the
Mach cone. From the geometry of Fig. 9.18c the angle of the Mach cone is seen to be  eText Main Menu  Textbook Table of Contents  Study Guide 620 Chapter 9 Compressible Flow Fig. 9.19 Supersonic wave pattern
emanating from a projectile moving
at Ma 2.0. The heavy lines are
obliqueshock waves and the light
lines Mach waves (Courtesy of U.S.
Army Ballistic Research Laboratory, Aberdeen Proving Ground.) sin 1 a
U t
t sin 1 a
U sin 1 1
Ma (9.79)  v v The higher the particle Mach number, the more slender the Mach cone; for example,
is 30° at Ma 2.0 and 11.5° at Ma 5.0. For the limiting case of sonic flow, Ma
1,
90°; the Mach cone becomes a plane front moving with the particle, in agreement with Fig. 9.18b.
You cannot “hear’’ the disturbance caused by the supersonic particle in Fig. 9.18c
until you are in the zone of action inside the Mach cone. No warning can reach your
ears if you are in the zone of silence outside the cone. Thus an observer on the ground
beneath a supersonic airplane does not hear the sonic boom of the passing cone until
the plane is well past.
The Mach wave need not be a cone: Similar waves are formed by a small disturbance of any shape moving supersonically with respect to the ambient fluid. For example, the “particle’’ in Fig. 9.18c could be the leading edge of a sharp flat plate, which
would form a Mach wedge of exactly the same angle . Mach waves are formed by
small roughnesses or boundarylayer irregularities in a supersonic wind tunnel or at
the surface of a supersonic body. Look again at Fig. 9.10: Mach waves are clearly visible along the body surface downstream of the recompression shock, especially at the
rear corner. Their angle is about 30°, indicating a Mach number of about 2.0 along this
surface. A more complicated system of Mach waves emanates from the supersonic projectile in Fig. 9.19. The Mach angles change, indicating a variable supersonic Mach  eText Main Menu  Textbook Table of Contents  Study Guide 9.9 TwoDimensional Suspersonic Flow 621 number along the body surface. There are also several stronger obliqueshock waves
formed along the surface.
EXAMPLE 9.16
An observer on the ground does not hear the sonic boom caused by an airplane moving at 5km
altitude until it is 9 km past her. What is the approximate Mach number of the plane? Assume
a small disturbance and neglect the variation of sound speed with altitude. Solution
A finite disturbance like an airplane will create a finitestrength obliqueshock wave whose angle will be somewhat larger than the Machwave angle and will curve downward due to the
variation in atmospheric sound speed. If we neglect these effects, the altitude and distance are a
measure of , as seen in Fig. E9.16. Thus Ma = ?
Bow wave
5 km BOOM!
µ 9 km E9.16 tan 5 km
9 km 0.5556 or 29.05° Hence, from Eq. (9.79),
Ma  v v The ObliqueShock Wave  csc 2.06 Ans. Figures 9.10 and 9.19 and our earlier discussion all indicate that a shock wave can form
at an oblique angle to the oncoming supersonic stream. Such a wave will deflect the
stream through an angle , unlike the normalshock wave, for which the downstream
flow is in the same direction. In essence, an oblique shock is caused by the necessity
for a supersonic stream to turn through such an angle. Examples could be a finite wedge
at the leading edge of a body and a ramp in the wall of a supersonic wind tunnel.
The flow geometry of an oblique shock is shown in Fig. 9.20. As for the normal
shock of Fig. 9.8, state 1 denotes the upstream conditions and state 2 is downstream.
The shock angle has an arbitrary value , and the downstream flow V2 turns at an angle which is a function of and state 1 conditions. The upstream flow is always supersonic, but the downstream Mach number Ma2 V2/a2 may be subsonic, sonic, or
supersonic, depending upon the conditions.
It is convenient to analyze the flow by breaking it up into normal and tangential
components with respect to the wave, as shown in Fig. 9.20. For a thin control volume eText Main Menu  Textbook Table of Contents  Study Guide 622 Chapter 9 Compressible Flow
Oblique shock wave
Vt2 = Vt1
Vt1
β V2 V1 > a1 θ Vn1 > a1 Deflection
angle Vn2 < a2
β Fig. 9.20 Geometry of flow
through an obliqueshock wave. just encompassing the wave, we can then derive the following integral relations, canceling out A1 A2 on each side of the wave:
Continuity: 1Vn1 Normal momentum: p1 Tangential momentum:
Energy: p2 h1 V21
n 1
2 V21
t (9.80b) Vt1)
1
2 h2 (9.80a)
2
1V n1 1Vn1(Vt2 0
1
2 2Vn2
2
2V n2 V22
n (9.80c)
1
2 V22
t h0 (9.80d) We see from Eq. (9.80c) that there is no change in tangential velocity across an oblique
shock
Vt2 Vt1 Vt const
(9.81)
Thus tangential velocity has as its only effect the addition of a constant kinetic energy
12
2 V t to each side of the energy equation (9.80d). We conclude that Eqs. (9.80) are identical to the normalshock relations (9.49), with V1 and V2 replaced by the normal components Vn1 and Vn2. All the various relations from Sec. 9.5 can be used to compute
properties of an obliqueshock wave. The trick is to use the “normal’’ Mach numbers
in place of Ma1 and Ma2:
Man1 Vn1
a1 Ma1 sin Man2 Vn2
a2 Ma2 sin ( (9.82)
) Then, for a perfect gas with constant specific heats, the property ratios across the oblique
shock are the analogs of Eqs. (9.55) to (9.58) with Ma1 replaced by Man1:
p2
p1
tan
tan ( 2
1 T2
T1 [2 (k 1
k 1 [2k Ma2 sin2
1
(k ) (k v v   eText Main Menu  1)] 1) Ma2 sin2
1
1) Ma2 sin2
1 1) Ma2 sin2 ]
1
T02 (k 2 (9.83a)
Vn1
Vn2 2k Ma2 sin2
(k
1
(k 1)2 Ma2 sin2
1 (9.83b)
1) T01 Textbook Table of Contents (9.83c)
(9.83d)  Study Guide 9.9 TwoDimensional Suspersonic Flow p02
p01 (k
2 1) Ma2 sin2
1
(k 1) Ma2 sin2
1 k/(k 1) 2k Ma22
n Ma2
1 k1
sin2
(k (k 1) Ma21
n
2k Ma21 (k
n 623 1/(k 1) 1) 2
1) (9.83e)
(9.83f ) All these are tabulated in the normalshock Table B.2. If you wondered why that table
listed the Mach numbers as Man1 and Man2, it should be clear now that the table is
also valid for the obliqueshock wave.
Thinking all this over, we realize by hindsight that an obliqueshock wave is the flow
pattern one would observe by running along a normalshock wave (Fig. 9.8) at a constant tangential speed Vt. Thus the normal and oblique shocks are related by a galilean,
or inertial, velocity transformation and therefore satisfy the same basic equations.
If we continue with this runalongtheshock analogy, we find that the deflection
angle increases with speed Vt up to a maximum and then decreases. From the geometry of Fig. 9.20 the deflection angle is given by
V
V
tan 1 t
tan 1 t
(9.84)
Vn2
Vn1
If we differentiate with respect to Vt and set the result equal to zero, we find that the
maximum deflection occurs when Vt /Vn1 (Vn2/Vn1)1/2. We can substitute this back
into Eq. (9.84) to compute
max tan 1 r1/2 tan 1 r 1/2 r Vn1
Vn2 (9.85)  v v For example, if Man1 3.0, from Table B.2 we find that Vn1/Vn2 3.8571, the square
root of which is 1.9640. Then Eq. (9.85) predicts a maximum deflection of tan 1 1.9640
tan 1 (1/1.9640) 36.03°. The deflection is quite limited even for infinite Man1: From
Table B.2 for this case Vn1/Vn2 6.0, and we compute from Eq. (9.85) that max 45.58°.
This limiteddeflection idea and other facts become more evident if we plot some
of the solutions of Eqs. (9.83). For given values of V1 and a1, assuming as usual that
k 1.4, we can plot all possible solutions for V2 downstream of the shock. Figure 9.21
does this in velocitycomponent coordinates Vx and Vy, with x parallel to V1. Such a
plot is called a hodograph. The heavy dark line which looks like a fat airfoil is the locus, or shock polar, of all physically possible solutions for the given Ma1. The two
dashedline fishtails are solutions which increase V2; they are physically impossible
because they violate the second law.
Examining the shock polar in Fig. 9.21, we see that a given deflection line of small
angle crosses the polar at two possible solutions: the strong shock, which greatly decelerates the flow, and the weak shock, which causes a much milder deceleration. The
flow downstream of the strong shock is always subsonic, while that of the weak shock
is usually supersonic but occasionally subsonic if the deflection is large. Both types of
shock occur in practice. The weak shock is more prevalent, but the strong shock will
occur if there is a blockage or highpressure condition downstream.
Since the shock polar is only of finite size, there is a maximum deflection max,
shown in Fig. 9.21, which just grazes the upper edge of the polar curve. This verifies
the kinematic discussion which led to Eq. (9.85). What happens if a supersonic flow
is forced to deflect through an angle greater than max? The answer is illustrated in Fig.
9.22 for flow past a wedgeshaped body.  eText Main Menu  Textbook Table of Contents  Study Guide 624 Chapter 9 Compressible Flow
Vy
Weak
wave
angle Rarefaction
shock
impossible
by second law θmax
β
Weak
shock θ Fig. 9.21 The obliqueshock polar
hodograph, showing double solutions (strong and weak) for small
deflection angle and no solutions at
all for large deflection. Strong
shock
Vx
V1 Normal
shock Mach
wave
(V2 = V1) In Fig. 9.22a the wedge halfangle is less than max, and thus an oblique shock
forms at the nose of wave angle just sufficient to cause the oncoming supersonic
stream to deflect through the wedge angle . Except for the usually small effect of
boundarylayer growth (see, e.g., Ref. 19, sec. 7 – 5.2), the Mach number Ma2 is constant along the wedge surface and is given by the solution of Eqs. (9.83). The pressure, density, and temperature along the surface are also nearly constant, as predicted
by Eqs. (9.83). When the flow reaches the corner of the wedge, it expands to higher
Mach number and forms a wake (not shown) similar to that in Fig. 9.10.
Weak shock family
above sonic line
Ma > 1
Sonic line
Strong shock family
below sonic line
Ma 2 Ma < 1 θ < θ max
Ma1 > 1 θ > θ max Ma1 > 1
Ma < 1
Ma 2
Sonic line
Ma > 1 (a)
(b)  v v Fig. 9.22 Supersonic flow past a
wedge: (a) small wedge angle, attached oblique shock forms; (b)
large wedge angle, attached shock
not possible, broad curved detached
shock forms. In Fig. 9.22b the wedge halfangle is greater than max, and an attached oblique
shock is impossible. The flow cannot deflect at once through the entire angle max, yet
somehow the flow must get around the wedge. A detached curve shock wave forms in
front of the body, discontinuously deflecting the flow through angles smaller than max.  eText Main Menu  Textbook Table of Contents  Study Guide 9.9 TwoDimensional Suspersonic Flow 625 50°
k = 1.4 Ma1 = ∞
10 40° 6 Deflection angle θ 4
3 30° 2.5 2 20° 1.8 Fig. 9.23 Obliqueshock deflection
versus wave angle for various upstream Mach numbers, k 1.4:
dashdot curve, locus of max, divides strong (right) from weak
(left) shocks; dashed curve, locus
of sonic points, divides subsonic
Ma2 (right) from supersonic Ma2
(left). 1.6
10°
1.4
1.2
30° 0° 60°
Wave angle β 90° The flow then curves, expands, and deflects subsonically around the wedge, becoming
sonic and then supersonic as it passes the corner region. The flow just inside each point
on the curved shock exactly satisfies the obliqueshock relations (9.83) for that particular value of and the given Ma1. Every condition along the curved shock is a point
on the shock polar of Fig. 9.21. Points near the front of the wedge are in the strongshock family, and points aft of the sonic line are in the weakshock family. The analysis of detached shock waves is extremely complex, and experimentation is usually
needed, e.g., the shadowgraph optical technique of Fig. 9.10.
The complete family of obliqueshock solutions can be plotted or computed from
Eqs. (9.83). For a given k, the wave angle varies with Ma1 and , from Eq. (9.83b).
By using a trigonometric identity for tan (
) this can be rewritten in the more convenient form
tan 2 cot (Ma2 sin2
1
Ma2 (k cos 2 )
1 1)
2 (9.86) All possible solutions of Eq. (9.86) for k 1.4 are shown in Fig. 9.23. For deflections
max there are two solutions: a weak shock (small ) and a strong shock (large ),
as expected. All points along the dashdot line for max satisfy Eq. (9.85). A dashed
line has been added to show where Ma2 is exactly sonic. We see that there is a narrow
region near maximum deflection where the weakshock downstream flow is subsonic.
For zero deflections (
0) the weakshock family satisfies the waveangle relation  v v sin  eText Main Menu  Textbook Table of Contents 1  1
Ma1 Study Guide (9.87) 626 Chapter 9 Compressible Flow Thus weak shocks of vanishing deflection are equivalent to Mach waves. Meanwhile
the strong shocks all converge at zero deflection to the normalshock condition
90°.
Two additional obliqueshock charts are given in App. B, where Fig. B.1 gives the
downstream Mach number Ma2 and Fig. B.2 the pressure ratio p2/p1, each plotted as
a function of Ma1 and . Additional graphs, tables, and computer programs are given
in Refs. 24 and 25. Very Weak Shock Waves For any finite the wave angle for a weak shock is greater than the Mach angle .
For small Eq. (9.86) can be expanded in a power series in tan with the following
linearized result for the wave angle:
sin k1
tan
4 cos sin (tan2 ) (9.88) For Ma1 between 1.4 and 20.0 and deflections less than 6° this relation predicts to
within 1° for a weak shock. For larger deflections it can be used as a useful initial
guess for iterative solution of Eq. (9.86).
Other property changes across the oblique shock can also be expanded in a power
series for small deflection angles. Of particular interest is the pressure change from Eq.
(9.83a), for which the linearized result for a weak shock is
p2 p1
p1 k Ma2
1
tan
(Ma2 1)1/2
1 (tan2 ) (9.89) 3.0
k = 1.4 Ma 1 = 10
8
2.0
6
p2 – p1
p1 4
3
1.0
2 Eq. (9.89),
Ma 1 = 2  v v Fig. 9.24 Pressure jump across a
weak obliqueshock wave from Eq.
(9.83a) for k 1.4. For very small
deflections Eq. (9.89) applies. 0 15° 10° 5° 0 Flow deflection θ  eText Main Menu  Textbook Table of Contents  Study Guide 9.9 TwoDimensional Suspersonic Flow 627 The differential form of this relation is used in the next section to develop a theory for
supersonic expansion turns. Figure 9.24 shows the exact weakshock pressure jump
computed from Eq. (9.83a). At very small deflections the curves are linear with slopes
given by Eq. (9.89).
Finally, it is educational to examine the entropy change across a very weak shock.
Using the same powerseries expansion technique, we can obtain the following result
for small flow deflections:
s2 s1
cp (k2 1)Ma6
1
tan3
12(Ma2 1)3/2
1 (tan4 ) (9.90) The entropy change is cubic in the deflection angle . Thus weak shock waves are very
nearly isentropic, a fact which is also used in the next section.
EXAMPLE 9.17
Ma 2 β Ma 1 = 2.0
p1 = 10 lbf / in2 10° Air at Ma 2.0 and p 10 lbf/in2 absolute is forced to turn through 10° by a ramp at the body
surface. A weak oblique shock forms as in Fig. E9.17. For k 1.4 compute from exact obliqueshock theory (a) the wave angle , (b) Ma2, and (c) p2. Also use the linearized theory to estimate (d) and (e) p2. Solution
With Ma1 2.0 and
10° known, we can estimate
40° 2° from Fig. 9.23. For more
(hand calculated) accuracy, we have to solve Eq. (9.86) by iteration. Or we can program Eq.
(9.86) in EES with six statements (in SI units, with angles in degrees): E9.17 EES Ma 2.0
k 1.4
Theta 10
Num 2*(Ma^2*SIN(Beta)^2 1)/TAN(Beta)
Denom Ma^2*(k COS(2*Beta)) 2
Theta ARCTAN(Num/Denom)
Specify that Beta 0 and EES promptly reports an accurate result:
Ans. (a) 39.32°
The normal Mach number upstream is thus
Man1 Ma1 sin 2.0 sin 39.32° 1.267 With Man1 we can use the normalshock relations (Table B.2) or Fig. 9.9 or Eqs. (9.56) to (9.58)
to compute
Man2 0.8031 p2
p1 1.707 Thus the downstream Mach number and pressure are
Man2
0.8031
Ma2
sin (
)
sin (39.32° 10°)
p2 (10 lbf/in2 absolute)(1.707) 1.64 17.07 lbf/in2 absolute  v v Notice that the computed pressure ratio agrees with Figs. 9.24 and B.2.  eText Main Menu  Textbook Table of Contents  Study Guide Ans. (b)
Ans. (c) 628 Chapter 9 Compressible Flow
For the linearized theory the Mach angle is
estimates that
sin sin 30° or sin 1 (1/2.0) 2.4 tan 10°
4 cos 30° 30°. Equation (9.88) then 0.622 38.5° Ans. (d) Equation (9.89) estimates that
p2
p1
or 1 1.4(2)2 tan 10°
(22 1)1/2 1.57(10 lbf/in2 absolute) p2 1.57 15.7 lbf/in2 absolute Ans. (e) These are reasonable estimates in spite of the fact that 10° is really not a “small’’ flow deflection. 9.10 PrandtlMeyer Expansion
Waves The obliqueshock solution of Sec. 9.9 is for a finite compressive deflection which
obstructs a supersonic flow and thus decreases its Mach number and velocity. The present section treats gradual changes in flow angle which are primarily expansive; i.e.,
they widen the flow area and increase the Mach number and velocity. The property
changes accumulate in infinitesimal increments, and the linearized relations (9.88) and
(9.89) are used. The local flow deflections are infinitesimal, so that the flow is nearly
isentropic according to Eq. (9.90).
Figure 9.25 shows four examples, one of which (Fig. 9.25c) fails the test for gradual changes. The gradual compression of Fig. 9.25a is essentially isentropic, with a
Oblique
shock
Mach
waves Slip
line Ma
decreases Ma > 1 Mach
waves
Ma > 1 Ma
increases  v v 9.25 Some examples of supersonic
expansion and compression: (a)
gradual isentropic compression on
a concave surface, Mach waves coalesce farther out to form oblique
shock; (b) gradual isentropic expansion on convex surface, Mach
waves diverge; (c) sudden compression, nonisentropic shock forms;
(d) sudden expansion, centered
isentropic fan of Mach waves
forms. (a) (b) Oblique
shock Mach
waves
Ma 2 < Ma 1 Ma 1 > 1 Ma > 1
Ma
increases (c)  eText Main Menu (d)  Textbook Table of Contents  Study Guide 9.10 PrandtlMeyer Expansion Waves 629 smooth increase in pressure along the surface, but the Mach angle decreases along the
surface and the waves tend to coalesce farther out into an obliqueshock wave. The
gradual expansion of Fig. 9.25b causes a smooth isentropic increase of Mach number
and velocity along the surface, with diverging Mach waves formed.
The sudden compression of Fig. 9.25c cannot be accomplished by Mach waves: An
oblique shock forms, and the flow is nonisentropic. This could be what you would see
if you looked at Fig. 9.25a from far away. Finally, the sudden expansion of Fig. 9.25d
is isentropic and forms a fan of centered Mach waves emanating from the corner. Note
that the flow on any streamline passing through the fan changes smoothly to higher
Mach number and velocity. In the limit as we near the corner the flow expands almost
discontinuously at the surface. The cases in Fig. 9.25a, b, and d can all be handled by
the PrandtlMeyer supersonicwave theory of this section, first formulated by Ludwig
Prandtl and his student Theodor Meyer in 1907 to 1908.
Note that none of this discussion makes sense if the upstream Mach number is subsonic, since Mach wave and shock wave patterns cannot exist in subsonic flow. The PrandtlMeyer PerfectGas
Function Consider a small, nearly infinitesimal flow deflection d such as occurs between the
first two Mach waves in Fig. 9.25a. From Eqs. (9.88) and (9.89) we have, in the limit,
1
Ma (9.91a) k Ma2
d
(Ma2 1)1/2 (9.91b) sin
dp
p 1 Since the flow is nearly isentropic, we have the frictionless differential momentum
equation for a perfect gas
dp kp Ma2 V dV dV
V (9.92) Combining Eqs. (9.91a) and (9.92) to eliminate dp, we obtain a relation between turning angle and velocity change
(Ma2 d 1)1/2 dV
V (9.93) This can be integrated into a functional relation for finite turning angles if we can relate V to Ma. We do this from the definition of Mach number
V
dV
V or Ma a
d Ma
Ma da
a (9.94) Finally, we can eliminate da/a because the flow is isentropic and hence a0 is constant
for a perfect gas
a
da
a v v or   eText Main Menu a0[1  1) Ma2] 1/2
1
1) Ma d Ma
2 (k
1
1 2 (k 1) Ma2
1
2 (k Textbook Table of Contents  Study Guide (9.95) 630 Chapter 9 Compressible Flow Eliminating dV/V and da/a from Eqs. (9.93) to (9.95), we obtain a relation solely between turning angle and Mach number
(Ma2 d 1
2 1 1)1/2 d Ma
1) Ma Ma (9.96) 2 (k Before integrating this expression, we note that the primary application is to expansions, i.e., increasing Ma and decreasing . Therefore, for convenience, we define
the PrandtlMeyer angle (Ma) which increases when decreases and is zero at the
sonic point
d d 0 at Ma (9.97) 1 Thus we integrate Eq. (9.96) from the sonic point to any value of Ma
Ma d
1 0 (Ma2
1 1 (k
2 1)1/2
d Ma
2
1) Ma Ma (9.98) The integrals are evaluated in closed form, with the result, in radians,
K1/2 tan (Ma)
where 1 Ma2 1
K
K k
k 1/2 tan 1 (Ma2 1)1/2 (9.99) 1
1 This is the PrandtlMeyer supersonic expansion function, which is plotted in Fig. 9.26 140°
Ma → ∞:
ω = 130.45° 120° 100° 80° ω
60° 40° 20°
k = 1.4  v v Fig,. 9.26 The PrandtlMeyer supersonic expansion from Eq. (9.99)
for k 1.4. 0°  01 4 eText Main Menu 8
12
Mach number  16 20 Textbook Table of Contents  Study Guide 9.10 PrandtlMeyer Expansion Waves 631 and tabulated in Table B.5 for k 1.4, K 6. The angle changes rapidly at first and
then levels off at high Mach number to a limiting value as Ma → :
max (K1/2 2 1) 130.45° if k 1.4 (9.100) Thus a supersonic flow can expand only through a finite turning angle before it reaches
infinite Mach number, maximum velocity, and zero temperature.
Gradual expansion or compression between finite Mach numbers Ma1 and Ma2, neither of which is unity, is computed by relating the turning angle
to the difference
in PrandtlMeyer angles for the two conditions
(Ma2) 1→2 (Ma1) (9.101) The change
may be either positive (expansion) or negative (compression) as long
as the end conditions lie in the supersonic range. Let us illustrate with an example. EES EXAMPLE 9.18
Air (k 1.4) flows at Ma1 3.0 and p1 200 kPa. Compute the final downstream Mach number and pressure for (a) an expansion turn of 20° and (b) a gradual compression turn of 20°. Solution
Part (a) The isentropic stagnation pressure is
p0 p1[1 0.2(3.0)2]3.5 7347 kPa and this will be the same at the downstream point. For Ma1 3.0 we find from Table B.5 or
Eq. (9.99) that 1 49.757°. The flow expands to a new condition such that
2 EES 1 49.757° 20° 69.757° Linear interpolation in Table B.5 is quite accurate, yielding Ma2 4.32. Inversion of Eq. (9.99),
to find Ma when is given, is impossible without iteration. Once again, our friend EES easily
handles Eq. (9.99) with four statements (angles specified in degrees): k 1.4
C ((k 1)/(k 1))^0.5
Omega 69.757
Omega C*ARCTAN((Ma^2 – 1)^0.5/C) – ARCTAN((Ma^2 – 1)^0.5)
Specify that Ma 1, and EES readily reports an accurate result:6
Ma2 Ans. (a) 4.32 The isentropic pressure at this new condition is
p2 [1 p0
0.2(4.32)2]3.5 7347
230.1 31.9 kPa 6 Ans. (a)  v v The author saves these little programs for further use, giving them names such as PrandtlMeyer.  eText Main Menu  Textbook Table of Contents  Study Guide 632 Chapter 9 Compressible Flow
The flow compresses to a lower PrandtlMeyer angle Part (b) 49.757° 2 20° 29.757° Again from Eq. (9.99), Table B.5, or EES we compute that
Ma2
p2 [1 2.125 p0
0.2(2.125)2]3.5 Ans. (b) 7347
9.51 773 kPa Similarly, density and temperature changes are computed by noticing that T0 and
for isentropic flow. Application to Supersonic Airfoils Ans. (b)
0 are constant The obliqueshock and PrandtlMeyer expansion theories can be used to patch together
a number of interesting and practical supersonic flow fields. This marriage, called shock
expansion theory, is limited by two conditions: (1) Except in rare instances the flow
must be supersonic throughout, and (2) the wave pattern must not suffer interference
from waves formed in other parts of the flow field.
A very successful application of shock expansion theory is to supersonic airfoils.
Figure 9.27 shows two examples, a flat plate and a diamondshaped foil. In contrast to
subsonicflow designs (Fig. 8.21), these airfoils must have sharp leading edges, which
form attached oblique shocks or expansion fans. Rounded supersonic leading edges
would cause detached bow shocks, as in Fig. 9.19 or 9.22b, greatly increasing the drag
and lowering the lift.
In applying shock expansion theory, one examines each surface turning angle to see
whether it is an expansion (“opening up’’) or compression (obstruction) to the surface
flow. Figure 9.27a shows a flatplate foil at an angle of attack. There is a leadingedge
shock on the lower edge with flow deflection
, while the upper edge has an expansion fan with increasing PrandtlMeyer angle
. We compute p3 with expansion theory and p2 with obliqueshock theory. The force on the plate is thus F
(p2 p3)Cb, where C is the chord length and b the span width (assuming no wingtip
effects). This force is normal to the plate, and thus the lift force normal to the stream
is L F cos , and the drag parallel to the stream is D F sin . The dimensionless
coefficients CL and CD have the same definitions as in lowspeed flow, Eq. (7.66), ex1
2
cept that the perfectgaslaw identity 1 V2
2
2 kp Ma is very useful here
L CL 1
2 2 kp Ma bC CD D
1
2 kp Ma2 bC (9.102)  v v The typical supersonic lift
coefficient is much smaller than the subsonic value CL 2 , but the lift can be very
1
large because of the large value of 2 V2 at supersonic speeds.
At the trailing edge in Fig. 9.27a, a shock and fan appear in reversed positions and
bend the two flows back so that they are parallel in the wake and have the same pressure. They do not have quite the same velocity because of the unequal shock strengths
on the upper and lower surfaces; hence a vortex sheet trails behind the wing. This is
very interesting, but in the theory you ignore the trailingedge pattern entirely, since it  eText Main Menu  Textbook Table of Contents  Study Guide 9.10 PrandtlMeyer Expansion Waves
Expansion
fan 633 Oblique
shock Ma 3 > Ma ∞ α p3 < p∞
p03 = p0∞ Ma ∞
p∞
p0∞ Vortex
sheet Ma 2 < Ma ∞
p2 > p∞ Oblique
shock p02 < p0∞
Expansion
fan (a) p3 > p∞ α Fig. 9.27 Supersonic airfoils:
(a) flat plate, higher pressure on
lower surface, drag due to small
downstream component of net pressure force; (b) diamond foil, higher
pressures on both lower surfaces,
additional drag due to body thickness. p5 < p3 Ma ∞
p∞ p2 > p3 p4 > p5
p4 < p2 (b)  v v does not affect the surface pressures: The supersonic surface flow cannot “hear’’ the
wake disturbances.
The diamond foil in Fig. 9.27b adds two more wave patterns to the flow. At this
particular less than the diamond halfangle, there are leadingedge shocks on both
surfaces, the upper shock being much weaker. Then there are expansion fans on each
shoulder of the diamond: The PrandtlMeyer angle change
equals the sum of the
leadingedge and trailingedge diamond halfangles. Finally, the trailingedge pattern
is similar to that of the flat plate (9.27a) and can be ignored in the calculation. Both
lowersurface pressures p2 and p4 are greater than their upper counterparts, and the lift
is nearly that of the flat plate. There is an additional drag due to thickness, because p4
and p5 on the trailing surfaces are lower than their counterparts p2 and p3. The diamond drag is greater than the flatplate drag, but this must be endured in practice to
achieve a wing structure strong enough to hold these forces.
The theory sketched in Fig. 9.27 is in good agreement with measured supersonic
lift and drag as long as the Reynolds number is not too small (thick boundary layers)
and the Mach number not too large (hypersonic flow). It turns out that for large ReC
and moderate supersonic Ma the boundary layers are thin and separation seldom occurs, so that the shock expansion theory, although frictionless, is quite successful. Let
us look now at an example.  eText Main Menu  Textbook Table of Contents  Study Guide Chapter 9 Compressible Flow EXAMPLE 9.19
A flatplate airfoil with C 2 m is immersed at
8° in a stream with Ma
2.5 and p
100 kPa. Compute (a) CL and (b) CD, and compare with lowspeed airfoils. Compute (c) lift and
(d) drag in newtons per unit span width. Solution
Instead of using a lot of space outlining the detailed obliqueshock and PrandtlMeyer expansion computations, we list all pertinent results in Fig. E9.19 on the upper and lower surfaces.
Using the theories of Secs. 9.9 and 9.10, you should verify every single one of the calculations
in Fig. E9.19 to make sure that all details of shock expansion theory are well understood.
∆ ω = 8° = α
ω 3 = 47.124°
Ma 3 = 2.867
p 03 = p0∞ = 1709 k Pa
p03
p3 = 30.05
p 3 = 56.85 k Pa 8°
Ma ∞ = 2.5
p∞ = 100 k Pa
p0∞ = 1709 k Pa
ω ∞ = 39.124° θ = α = 8°
β = 30.01°
Ma 2 = 2.169
p2
p∞ = 1.657
p 2 = 165.7 k Pa Do not
compute E9.19
The important final results are p2 and p3, from which the total force per unit width on the
plate is
F (p2 p3)bC (165.7 56.85)(kPa)(1 m)(2 m) 218 kN The lift and drag per meter width are thus
L
D F cos 8° 216 kN Ans. (c) F sin 8° 30 kN Ans. (d) 2 These are very large forces for only 2 m of wing area.
From Eq. (9.102) the lift coefficient is
CL 216 kN
1
2 (1.4)(100 kPa)(2.5)2(2 m2) The comparable lowspeed coefficient from Eq. (8.64) is CL
times larger.
From Eq. (9.102) the drag coefficient is
CD 30 kN
1
2 (1.4)(100 kPa)(2.5)2(2 m2) From Fig. 7.25 for the NACA 0009 airfoil CD at  v v 634  eText Main Menu  0.246
2 sin 8° Ans. (a)
0.874, which is 3.5 0.035 Ans. (b) 8° is about 0.009, or about 4 times smaller. Textbook Table of Contents  Study Guide 9.10 PrandtlMeyer Expansion Waves 635 Notice that this supersonic theory predicts a finite drag in spite of assuming frictionless flow
with infinite wing aspect ratio. This is called wave drag, and we see that the d’Alembert paradox of zero body drag does not occur in supersonic flow. ThinAirfoil Theory In spite of the simplicity of the flatplate geometry, the calculations in Example 9.19
were laborious. In 1925 Ackeret [21] developed simple yet effective expressions for
the lift, drag, and center of pressure of supersonic airfoils, assuming small thickness
and angle of attack.
surface deThe theory is based on the linearized expression (9.89), where tan
flection relative to the free stream and condition 1 is the free stream, Ma1 Ma . For
the flatplate airfoil, the total force F is based on
p2 p3 p2 p p p3 p p
p k Ma2
[
(Ma2
1)1/2 ( )] (9.103) Substitution into Eq. (9.102) gives the linearized lift coefficient for a supersonic flatplate airfoil
CL 1
2 4 (p2 p3)bC
kp Ma2 bC (Ma2 1)1/2 (9.104) Computations for diamond and other finitethickness airfoils show no firstorder effect
of thickness on lift. Therefore Eq. (9.104) is valid for any sharpedged supersonic thin
airfoil at a small angle of attack.
The flatplate drag coefficient is
CD CL tan 4 CL (Ma2 2 1)1/2 (9.105) However, the thicker airfoils have additional thickness drag. Let the chord line of the
airfoil be the xaxis, and let the uppersurface shape be denoted by yu(x) and the lower
profile by yl (x). Then the complete Ackeret drag theory (see, e.g., Ref. 8, sec. 14.6, for
details) shows that the additional drag depends on the mean square of the slopes of the
upper and lower surfaces, defined by
y 2 1
C C
0 dy
dx 2 dx (9.106) The final expression for drag [8, p. 442] is
CD 4
(Ma2 1)1/2 1
(y 2
2u 2 yl 2) (9.107)  v v These are all in reasonable agreement with more exact computations, and their extreme
simplicity makes them attractive alternatives to the laborious but accurate shock expansion theory. Consider the following example.  eText Main Menu  Textbook Table of Contents  Study Guide Chapter 9 Compressible Flow EXAMPLE 9.20
Repeat parts (a) and (b) of Example 9.19, using the linearized Ackeret theory. Solution
From Eqs. (9.104) and (9.105) we have, for Ma
CL 4(0.1396)
(2.52 1)1/2 0.244 2.5 and
CD 8° 4(0.1396)2
(2.52 1)1/2 0.1396 rad,
0.034 Ans. These are less than 3 percent lower than the more exact computations of Example 9.19. A further result of the Ackeret linearized theory is an expression for the position
xCP of the center of pressure (CP) of the force distribution on the wing:
xCP
C 0.5 Su Sl
2 C2 (9.108) where Su is the crosssectional area between the upper surface and the chord and Sl is
the area between the chord and the lower surface. For a symmetric airfoil (Sl Su) we
obtain xCP at the halfchord point, in contrast with the lowspeed airfoil result of Eq.
(8.66), where xCP is at the quarterchord.
The difference in difficulty between the simple Ackeret theory and shock expansion
theory is even greater for a thick airfoil, as the following example shows. EXAMPLE 9.21
By analogy with Example 9.19 analyze a diamond, or doublewedge, airfoil of 2° halfangle and
C 2 m at
8° and Ma
2.5. Compute CL and CD by (a) shock expansion theory and (b)
Ackeret theory. Pinpoint the difference from Example 9.19. Solution
Again we omit the details of shock expansion theory and simply list the properties computed on
each of the four airfoil surfaces in Fig. E9.21. Assume p
100 kPa. There are both a force F
normal to the chord line and a force P parallel to the chord. For the normal force the pressure
difference on the front half is p2 p3 186.4 65.9 120.5 kPa, and on the rear half it is
p4 p5 146.9 48.1 98.1 kPa. The average pressure difference is 1 (120.5 98.1) 109.3
2
kPa, so that the normal force is Part (a) F (109.3 kPa)(2 m2) 218.6 kN For the chordwise force P the pressure difference on the top half is p3 p5 65.9 48.8
17.1 kPa, and on the bottom half it is p2 p4 186.4 146.9 39.5 kPa. The average difference is 1 (17.1 39.5) 28.3 kPa, which when multiplied by the frontal area (maximum
2
thickness times 1m width) gives
P  v v 636  eText Main Menu  (28.3 kPa)(0.07 m)(1 m) 2.0 kN Textbook Table of Contents  Study Guide 9.10 PrandtlMeyer Expansion Waves 637 Chord length = 2 m ∆ ω = 6°
ω 3 = 45.124°
Ma 3 = 2.770
p 3 = 65.9 kPa 8° Ma ∞ = 2.5
p∞ = 100 k Pa
p0∞ = 1709 k Pa
ω ∞ = 39.124° ∆ ω = 4°
ω 5 = 49.124°
Ma 5 = 2.967
p 5 = 48.8 kPa 4˚ θ = 10°
β = 31.85°
Ma 2 = 2.086
ω 2 = 28.721°
p 02 = 1668 k Pa
p 2 = 186.4 k Pa 0.07 m ∆ ω = 4°
ω 4 = 32.721°
Ma 4 = 2.238
p 4 = 146.9 k Pa E9.21 Both F and P have components in the lift and drag directions. The lift force normal to the free
stream is
L P sin 8° 216.2 kN D and F cos 8°
F sin 8° P cos 8° 32.4 kN For computing the coefficients, the denominator of Eq. (9.102) is the same as in Example 9.19:
1
Ma 2 bC 1 (1.4)(100 kPa)(2.5)2(2 m2) 875 kN. Thus, finally, shock expansion theory
2 kp
2
predicts
CL Part (b) 216.2 kN
875 kN 0.247 32.4 kN
875 kN CD 0.0370 Ans. (a) Meanwhile, by Ackeret theory, CL is the same as in Example 9.20:
CL 4(0.1396)
(2.52 1)1/2 0.244 Ans. (b) This is 1 percent less than the shock expansion result above. For the drag we need the meansquare slopes from Eq. (9.106)
yu2 yl 2 tan2 2° 0.00122 Then Eq. (9.107) predicts the linearized result
CD 4
(2.52 1)1/2 [(0.1396)2 1
2 (0.00122 0.00122)] 0.0362 Ans. (b) This is 2 percent lower than shock expansion theory predicts. We could judge Ackeret theory to
be “satisfactory.’’ Ackeret theory predicts p2 167 kPa ( 11 percent), p3 60 kPa ( 9 percent), p4 140 kPa ( 5 percent), and p5 33 kPa ( 6 percent).  v v ThreeDimensional Supersonic
Flow  We have gone about as far as we can go in an introductory treatment of compressible
flow. Of course, there is much more, and you are invited to study further in the references at the end of the chapter. eText Main Menu  Textbook Table of Contents  Study Guide 638 Chapter 9 Compressible Flow Fig. 9.28 Shadowgraph of flow past
an 8° halfangle cone at Ma
2.0. The turbulent boundary layer is
clearly visible. The Mach lines
curve slightly, and the Mach number varies from 1.98 just inside the
shock to 1.90 at the surface. (Courtesy of U.S. Army Ballistic Research Center, Aberdeen Proving
Ground.)  v v Threedimensional supersonic flows are highly complex, especially if they concern blunt bodies, which therefore contain embedded regions of subsonic and transonic flow, e.g., Fig. 9.10. Some flows, however, yield to accurate theoretical treatment such as flow past a cone at zero incidence, as shown in Fig. 9.28. The exact
theory of cone flow is discussed in advanced texts [for example, 8, chap. 17], and
extensive tables of such solutions have been published [22, 23]. There are similarities between cone flow and the wedge flows illustrated in Fig. 9.22: an attached
oblique shock, a thin turbulent boundary layer, and an expansion fan at the rear corner. However, the conical shock deflects the flow through an angle less than the
cone halfangle, unlike the wedge shock. As in the wedge flow, there is a maximum
cone angle above which the shock must detach, as in Fig. 9.22b. For k 1.4 and
Ma
, the maximum cone halfangle for an attached shock is about 57°, compared with the maximum wedge angle of 45.6° (see Ref. 23).
For more complicated body shapes one usually resorts to experimentation in a
supersonic wind tunnel. Figure 9.29 shows a windtunnel study of supersonic flow
past a model of an interceptor aircraft. The many junctions and wingtips and shape
changes make theoretical analysis very difficult. Here the surfaceflow patterns,
which indicate boundarylayer development and regions of flow separation, have
been visualized by the smearing of oil drops placed on the model surface before the
test.  eText Main Menu  Textbook Table of Contents  Study Guide 9.10 PrandtlMeyer Expansion Waves 639 Fig. 9.29 Windtunnel test of the
Cobra P530 supersonic interceptor.
The surface flow patterns are visualized by the smearing of oil
droplets. (Courtesy of Northrop
Corp.) As we shall see in the next chapter, there is an interesting analogy between gasdynamic shock waves and the surface water waves which form in an openchannel
flow. Chapter 11 of Ref. 14 explains how a water channel can be used in an inexpensive simulation of supersonicflow experiments.  v v Reusable Hypersonic Launch
Vehicles  Having achieved reliable supersonic flight with both military and commercial aircraft, the next step is probably to develop a hypersonic vehicle that can achieve orbit, yet be retrieved. Presently the United States employs the Space Shuttle, where
only the manned vehicle is retrieved, the very expensive giant rocket boosters being lost. In 1996, NASA selected LockheedMartin to develop the X33, the first
smallerscale step toward a retrievable singlestagetoorbit (SSTO) vehicle, to be
called the VentureStar [36].
The X33, shown in an artist’s rendering in Fig. 9.30, will be 20 m long, about
half the size of the VentureStar, and it will be suborbital. It will take off vertically,
rise to a height of 73 km, and coast back to earth at a steep (stressful) angle. Such
a flight will test many new plans for the VentureStar [37], such as metallic tiles, titanium components, graphite composite fuel tanks, highvoltage control actuators,
and Rocketdyne’s novel “aerospike” rocket nozzles. If successful, the VentureStar
is planned as a standard reusable, lowcost orbital vehicle. VentureStar will be
39 m long and weigh 9.7 MN, of which 88 percent (965 tons!) will be propellant eText Main Menu  Textbook Table of Contents  Study Guide 640 Chapter 9 Compressible Flow Fig. 9.30 The X33 is a halfsize
suborbital test version of the VentureStar, which is planned as an orbital, lowcost retrievable space vehicle. It takes off vertically but then
uses its lifting shape to glide back
to earth and land horizontally [36,
37]. (Courtesy of Lockheed Martin
Corp.) and only 2.7 percent (260 kN) will be payload. The dream is that the X33 and VentureStar and their progeny will lead to an era of routine, lowcost space travel appropriate to the new millennium. Summary  v v This chapter is a brief introduction to a very broad subject, compressible flow, sometimes called gas dynamics. The primary parameter is the Mach number Ma V/a,
which is large and causes the fluid density to vary significantly. This means that the
continuity and momentum equations must be coupled to the energy relation and the
equation of state to solve for the four unknowns (p, , T, V).
The chapter reviews the thermodynamic properties of an ideal gas and derives a
formula for the speed of sound of a fluid. The analysis is then simplified to onedimensional steady adiabatic flow without shaft work, for which the stagnation enthalpy of the gas is constant. A further simplification to isentropic flow enables formulas to be derived for highspeed gas flow in a variablearea duct. This reveals the
phenomenon of sonicflow choking (maximum mass flow) in the throat of a nozzle. At
supersonic velocities there is the possibility of a normalshock wave, where the gas
discontinuously reverts to subsonic conditions. The normal shock explains the effect
of back pressure on the performance of convergingdiverging nozzles.
To illustrate nonisentropic flow conditions, there is a brief study of constantarea
duct flow with friction and with heat transfer, both of which lead to choking of the exit
flow.
The chapter ends with a discussion of twodimensional supersonic flow, where
obliqueshock waves and PrandtlMeyer (isentropic) expansion waves appear. With
a proper combination of shocks and expansions one can analyze supersonic airfoils.  eText Main Menu  Textbook Table of Contents  Study Guide Problems 641 Problems
Most of the problems herein are fairly straightforward. More difficult or openended assignments are labeled with an asterisk. Problems labeled with an EES icon will benefit from the use of the Engineering Equations Solver (EES), while problems labeled with a
computer icon may require the use of a computer. The standard
endofchapter problems 9.1 to 9.157 (categorized in the problem
list below) are followed by word problems W9.1 to W9.8, fundamentals of engineering exam problems FE9.1 to FE9.10, comprehensive problems C9.1 to C9.3, and design projects D9.1 and D9.2. P9.7 P9.8 Problem distribution
Section Topic Problems 9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
9.9
9.9
9.10
9.10 Introduction
The speed of sound
Adiabatic and isentropic flow
Isentropic flow with area changes
The normalshock wave
Converging and diverging nozzles
Duct flow with friction
Frictionless duct flow with heat transfer
Mach waves
The obliqueshock wave
PrandtlMeyer expansion waves
Supersonic airfoils 9.1 – 9.9
9.10 – 9.18
9.19 – 9.33
9.34 – 9.53
9.54 – 9.62
9.63 – 9.85
9.86 – 9.107
9.108 – 9.115
9.116 – 9.121
9.122 – 9.139
9.140 – 9.147
9.148 – 9.157 P9.2 P9.3 P9.4 P9.5 P9.6
EES An ideal gas flows adiabatically through a duct. At section
1, p1 140 kPa, T1 260°C, and V1 75 m/s. Farther
downstream, p2 30 kPa and T2 207°C. Calculate V2
in m/s and s2 s1 in J/(kg K) if the gas is (a) air, k
1.4, and (b) argon, k 1.67.
Solve Prob. 9.1 if the gas is steam. Use two approaches:
(a) an ideal gas from Table A.4 and (b) real gas data from
the steam tables [15].
If 8 kg of oxygen in a closed tank at 200°C and 300 kPa
is heated until the pressure rises to 400 kPa, calculate (a)
the new temperature, (b) the total heat transfer, and (c) the
change in entropy.
Compressibility effects become important when the Mach
number exceeds approximately 0.3. How fast can a twodimensional cylinder travel in sealevel standard air before compressibility becomes important somewhere in its
vicinity?
Steam enters a nozzle at 377°C, 1.6 MPa, and a steady
speed of 200 m/s and accelerates isentropically until it exits at saturation conditions. Estimate the exit velocity and
temperature.
Is it possible for the steam in Prob. 9.5 to continue accelerating until it exits with a moisture content of 12
percent? If so, estimate the new exit velocity and temperature.  v v P9.1  eText Main Menu  P9.9 P9.10 P9.11 P9.12 P9.13 P9.14 P9.15 P9.16 Carbon dioxide (k 1.28) enters a constantarea duct at
400°F, 100 lbf/in2 absolute, and 500 ft/s. Farther downstream the properties are V2 1000 ft/s and T2 900°F.
Compute (a) p2, (b) the heat added between sections, (c)
the entropy change between sections, and (d) the mass flow
per unit area. Hint: This problem requires the continuity
equation.
Atmospheric air at 20°C enters and fills an insulated tank
which is initially evacuated. Using a controlvolume analysis from Eq. (3.63), compute the tank air temperature when
it is full.
Liquid hydrogen and oxygen are burned in a combustion
chamber and fed through a rocket nozzle which exhausts
at Vexit 1600 m/s to an ambient pressure of 54 kPa. The
nozzle exit diameter is 45 cm, and the jet exit density is
0.15 kg/m3. If the exhaust gas has a molecular weight of
18, estimate (a) the exit gas temperature, (b) the mass flow,
and (c) the thrust developed by the rocket.
A certain aircraft flies at the same Mach number regardless of its altitude. Compared to its speed at 12,000m standard altitude, it flies 127 km/h faster at sea level. Determine its Mach number.
At 300°C and 1 atm, estimate the speed of sound of (a)
nitrogen, (b) hydrogen, (c) helium, (d) steam, and (e)
238
UF6 (k 1.06).
Assume that water follows Eq. (1.19) with n 7 and B
3000. Compute the bulk modulus (in kPa) and the speed
of sound (in m/s) at (a) 1 atm and (b) 1100 atm (the deepest part of the ocean). (c) Compute the speed of sound at
20°C and 9000 atm and compare with the measured value
of 2650 m/s (A. H. Smith and A. W. Lawson, J. Chem.
Phys., vol. 22, 1954, p. 351).
From Prob. 1.33, mercury data fit Eq. (1.19) with n 6
and B 41,000. Estimate (a) the bulk modulus and (b) the
speed of sound of mercury at 2500 atm and compare with
Table 9.1.
Assume steady adiabatic flow of a perfect gas. Show that
the energy equation (9.21), when plotted as speed of sound
versus velocity, forms an ellipse. Sketch this ellipse; label
the intercepts and the regions of subsonic, sonic, and supersonic flow; and determine the ratio of the major and
minor axes.
A weak pressure wave (sound wave) with a pressure
change p 40 Pa propagates through air at 20°C and 1
atm. Estimate (a) the density change, (b) the temperature
change, and (c) the velocity change across the wave.
A weak pressure pulse p propagates through still air. Discuss the type of reflected pulse which occurs and the
boundary conditions which must be satisfied when the Textbook Table of Contents  Study Guide 642 Chapter 9 Compressible Flow wave strikes normal to, and is reflected from, (a) a solid
wall and (b) a free liquid surface.
P9.17 A submarine at a depth of 800 m sends a sonar signal and
receives the reflected wave back from a similar submerged
object in 15 s. Using Prob. 9.12 as a guide, estimate the
distance to the other object.
*P9.18 The properties of a dense gas (high pressure and low temperature) are often approximated by van der Waals’ equation of state [17, 18]:
p 1 RT
b1 a1 2 the air velocity V, assuming (a) incompressible flow and
(b) compressible flow.
P9.23 A large rocket engine delivers hydrogen at 1500°C and 3
MPa, k 1.41, R 4124 J/(kg K), to a nozzle which exits with gas pressure equal to the ambient pressure of 54
kPa. Assuming isentropic flow, if the rocket thrust is 2 MN,
what is (a) the exit velocity and (b) the mass flow of hydrogen?
P9.24 For lowspeed (nearly incompressible) gas flow, the stagnation pressure can be computed from Bernoulli’s equation where constants a1 and b1 can be found from the critical
temperature and pressure
a1 2
27R2T c
64pc for air, and
RTc
8pc b1 P9.20 P9.21 P9.22 p0 3 0.65 ft /slug for air. Find an analytic expression for the speed of sound
of a van der Waals gas. Assuming k 1.4, compute the
speed of sound of air in ft/s at 100°F and 20 atm for (a)
a perfect gas and (b) a van der Waals gas. What percentage higher density does the van der Waals relation predict?
The Concorde aircraft flies at Ma 2.3 at 11km standard
altitude. Estimate the temperature in °C at the front stagnation point. At what Mach number would it have a front
stagnationpoint temperature of 450°C?
A gas flows at V 200 m/s, p 125 kPa, and T 200°C.
For (a) air and (b) helium, compute the maximum pressure and the maximum velocity attainable by expansion or
compression.
Air expands isentropically through a duct from p1 125
kPa and T1 100°C to p2 80 kPa and V2 325 m/s.
Compute (a) T2, (b) Ma2, (c) T0, (d) p0, (e) V1, and ( f) Ma1.
Given the pitot stagnation temperature and pressure and
the staticpressure measurements in Fig. P9.22, estimate V p 1
V2 1
2 2 1
Ma2
4 k
24 Ma4 (b) Suppose that a pitotstatic tube in air measures the pressure difference p0 p and uses the Bernoulli relation, with
stagnation density, to estimate the gas velocity. At what
Mach number will the error be 4 percent?
P9.25 If it is known that the air velocity in the duct is 750 ft/s,
use the mercurymanometer measurement in Fig. P9.25
to estimate the static pressure in the duct in lbf/in2 absolute. Air at 100°F 8 in Mercury P9.25 V2 80 k Pa 120 k Pa P9.22 v  eText Main Menu 2cpT0 1 p
p0 (k 1)/k What would be a source of error if a shock wave were
formed in front of the probe?
P9.27 In many problems the sonic (*) properties are more useful reference values than the stagnation properties. For
isentropic flow of a perfect gas, derive relations for p/p*, 100°C  1
V2
2 P9.26 Show that for isentropic flow of a perfect gas if a pitotstatic probe measures p0, p, and T0, the gas velocity can
be calculated from Air v P9.19 p (a) For higher subsonic speeds, show that the isentropic
relation (9.28a) can be expanded in a power series as follows: 105 lbf ft4/slug2 9.0 p0  Textbook Table of Contents  Study Guide Problems 643 T/T*, and / * as functions of the Mach number. Let us
help by giving the densityratio formula:
2 * k
(k 1
1) Ma.2 1/(k 1) P9.28 A large vacuum tank, held at 60 kPa absolute, sucks sealevel standard air through a converging nozzle whose
throat diameter is 3 cm. Estimate (a) the massflow rate
through the nozzle and (b) the Mach number at the throat.
P9.29 Steam from a large tank, where T 400°C and p 1 MPa,
expands isentropically through a nozzle until, at a section
of 2cm diameter, the pressure is 500 kPa. Using the steam
tables [15], estimate (a) the temperature, (b) the velocity,
and (c) the mass flow at this section. Is the flow subsonic?
P9.30 Air flows in a duct of diameter 5 cm. At one section, T0
300°C, p 120 kPa, and m 0.4 kg/s. Estimate, at this
˙
section, (a) V, (b) Ma, and (c) 0.
P9.31 Air flows adiabatically through a duct. At one section V1
400 ft/s, T1 200°F, and p1 35 lbf/in2 absolute, while
farther downstream V2 1100 ft/s and p2 18 lbf/in2 absolute. Compute (a) Ma2, (b) Umax, and (c) p02/p01.
P9.32 The large compressedair tank in Fig. P9.32 exhausts from
a nozzle at an exit velocity of 235 m/s. The mercury
manometer reads h 30 cm. Assuming isentropic flow,
compute the pressure (a) in the tank and (b) in the atmosphere. (c) What is the exit Mach number?
30°C
pa ?
235 m /s Air h
ptank? P9.32 Mercury  v v P9.33 Air flows isentropically from a reservoir, where p 300
kPa and T 500 K, to section 1 in a duct, where A1 0.2
m2 and V1 550 m/s. Compute (a) Ma1, (b) T1, (c) p1, (d)
m, and (e) A*. Is the flow choked?
˙
P9.34 Steam in a tank at 450°F and 100 lbf/in2 absolute exhausts
through a converging nozzle of 0.1in2 throat area to a 1atm environment. Compute the initial mass flow (a) for an
ideal gas and (b) from the steam tables [15].
P9.35 Helium, at T0 400 K, enters a nozzle isentropically. At
section 1, where A1 0.1 m2, a pitotstatic arrangement
(see Fig. P9.25) measures stagnation pressure of 150 kPa
and static pressure of 123 kPa. Estimate (a) Ma1, (b) mass
flow m, (c) T1, and (d) A*.
˙  eText Main Menu  P9.36 An air tank of volume 1.5 m3 is initially at 800 kPa and
20°C. At t 0, it begins exhausting through a converging
nozzle to sealevel conditions. The throat area is 0.75 cm2.
Estimate (a) the initial mass flow in kg/s, (b) the time required to blow down to 500 kPa, and (c) the time at which
the nozzle ceases being choked.
P9.37 Make an exact controlvolume analysis of the blowdown
process in Fig. P9.37, assuming an insulated tank with negligible kinetic and potential energy within. Assume critical flow at the exit, and show that both p0 and T0 decrease
during blowdown. Set up firstorder differential equations
for p0(t) and T0(t), and reduce and solve as far as you can.
Insulated tank p0 (t)
T0 (t)
Volume V ⋅
Ae, Ve, me
Measurements
of tank
pressure and
temperature P9.37
P9.38 Prob. 9.37 makes an ideal senior project or combined laboratory and computer problem, as described in Ref. 30,
sec. 8.6. In Bober and Kenyon’s lab experiment, the tank
had a volume of 0.0352 ft3 and was initially filled with air
at 50 lb/in2 gage and 72°F. Atmospheric pressure was 14.5
lb/in2 absolute, and the nozzle exit diameter was 0.05 in.
After 2 s of blowdown, the measured tank pressure was 20
lb/in2 gage and the tank temperature was 5°F. Compare
these values with the theoretical analysis of Prob. 9.37.
P9.39 Consider isentropic flow in a channel of varying area, from
section 1 to section 2. We know that Ma1 2.0 and desire
that the velocity ratio V2/V1 be 1.2. Estimate (a) Ma2 and
(b) A2/A1. (c) Sketch what this channel looks like. For example, does it converge or diverge? Is there a throat?
P9.40 Air, with stagnation conditions of 800 kPa and 100°C, expands isentropically to a section of a duct where A1 20
cm2 and p1 47 kPa. Compute (a) Ma1, (b) the throat area,
and (c) m. At section 2 between the throat and section 1,
˙
the area is 9 cm2. (d) Estimate the Mach number at section 2.
P9.41 Air, with a stagnation pressure of 100 kPa, flows through
the nozzle in Fig. P9.41, which is 2 m long and has an area
variation approximated by Textbook Table of Contents  Study Guide 644 Chapter 9 Compressible Flow A(x) P9.45
p0
p (x)?
p P9.46
0
0 1m P9.41 2m x A 20 20x P9.47 10x2  v v with A in cm2 and x in m. It is desired to plot the complete family of isentropic pressures p(x) in this nozzle, for
the range of inlet pressures 1 p(0) 100 kPa. Indicate
those inlet pressures which are not physically possible and
discuss briefly. If your computer has an online graphics
routine, plot at least 15 pressure profiles; otherwise just
hit the highlights and explain.
P9.42 A bicycle tire is filled with air at an absolute pressure of
169.12 kPa, and the temperature inside is 30.0°C. Suppose
the valve breaks, and air starts to exhaust out of the tire
into the atmosphere (pa 100 kPa absolute and Ta
20.0°C). The valve exit is 2.00 mm in diameter and is the
smallest crosssectional area of the entire system. Frictional losses can be ignored here, i.e., onedimensional
isentropic flow is a reasonable assumption. (a) Find the
Mach number, velocity, and temperature at the exit plane
of the valve (initially). (b) Find the initial massflow rate
out of the tire. (c) Estimate the velocity at the exit plane
using the incompressible Bernoulli equation. How well
does this estimate agree with the “exact” answer of part
(a)? Explain.
P9.43 Air flows isentropically through a duct with T0 300°C.
At two sections with identical areas of 25 cm2, the pressures are p1 120 kPa and p2 60 kPa. Determine (a)
the mass flow, (b) the throat area, and (c) Ma2.
P9.44 In Prob. 3.34 we knew nothing about compressible flow
at the time, so we merely assumed exit conditions p2 and
T2 and computed V2 as an application of the continuity
equation. Suppose that the throat diameter is 3 in. For the
given stagnation conditions in the rocket chamber in Fig.
P3.34 and assuming k 1.4 and a molecular weight of 26,  eText Main Menu  P9.48 compute the actual exit velocity, pressure, and temperature
according to onedimensional theory. If pa 14.7 lbf/in2
absolute, compute the thrust from the analysis of Prob.
3.68. This thrust is entirely independent of the stagnation
temperature (check this by changing T0 to 2000°R if you
like). Why?
At a point upstream of the throat of a convergingdiverging nozzle the properties are V1 200 m/s, T1 300 K,
and p1 125 kPa. If the exit flow is supersonic, compute,
from isentropic theory, (a) m and (b) A1. The throat area
˙
is 35 cm2.
If the author did not falter, the results of Prob. 9.43 are (a)
0.671 kg/s, (b) 23.3 cm2, and (c) 1.32. Do not tell your
friends who are still working on Prob. 9.43. Consider a
control volume which encloses the nozzle between these
two 25cm2 sections. If the pressure outside the duct is
1 atm, determine the total force acting on this section of
nozzle.
In windtunnel testing near Mach 1, a small area decrease
caused by model blockage can be important. Suppose the
test section area is 1 m2, with unblocked test conditions
Ma 1.10 and T 20°C. What model area will first cause
the test section to choke? If the model cross section is
0.004 m2 (0.4 percent blockage), what percentage change
in testsection velocity results?
A force F 1100 N pushes a piston of diameter 12 cm
through an insulated cylinder containing air at 20°C, as in
Fig. P9.48. The exit diameter is 3 mm, and pa 1 atm.
Estimate (a) Ve, (b) Vp, and (c) me.
˙
Insulated
Air
at
20°C Vp F ⋅
Ve, me
De = 3 mm P9.48 Dp = 12 cm pa = 1 atm P9.49 Air expands through a nozzle and exits supersonically. The
throat area is 10 cm2, and the throat pressure is 100 kPa.
Find the pressure on either side of the throat where the
duct area is 24 cm2.
P9.50 Argon expands isentropically in a converging nozzle whose
entrance conditions are D1 10 cm, p1 150 kPa, T1
EES
100°C, and m 1 kg/s. The flow discharges smoothly to
˙
an ambient pressure of 101 kPa. (a) What is the exit diameter of the nozzle? (b) How much further can the ambient pressure be reduced before it affects the inlet mass
flow?
P9.51 Air, at stagnation conditions of 500 K and 200 kPa, flows
through a nozzle. At section 1, where the area is 12 cm2, Textbook Table of Contents  Study Guide Problems 645
the density is 0.32 kg/m3. Assuming isentropic flow, (a)
find the mass flow. (b) Is the flow choked? If so, estimate
A*. Also estimate (c) p1 and (d) Ma1.
P9.52 A convergingdiverging nozzle exits smoothly to sealevel
standard atmosphere. It is supplied by a 40m3 tank initially at 800 kPa and 100°C. Assuming isentropic flow in
the nozzle, estimate (a) the throat area and (b) the tank
pressure after 10 s of operation. The exit area is 10 cm2.
P9.53 Air flows steadily from a reservoir at 20°C through a nozzle of exit area 20 cm2 and strikes a vertical plate as in
Fig. P9.53. The flow is subsonic throughout. A force of
135 N is required to hold the plate stationary. Compute (a)
Ve, (b) Mae, and (c) p0 if pa 101 kPa.
Ae = 20 cm2 Plate Air
20°C 135 N P9.53
P9.54 For flow of air through a normal shock the upstream conditions are V1 600 m/s, T01 500 K, and p01 700
kPa. Compute the downstream conditions Ma2, V2, T2, p2,
and p02.
P9.55 Air, supplied by a reservoir at 450 kPa, flows through a convergingdiverging nozzle whose throat area is 12 cm2. A normal shock stands where A1 20 cm2. (a) Compute the pressure just downstream of this shock. Still farther downstream,
at A3 30 cm2, estimate (b) p3, (c) A*, and (d) Ma3.
3
P9.56 Air from a reservoir at 20°C and 500 kPa flows through a
duct and forms a normal shock downstream of a throat of
area 10 cm2. By an odd coincidence it is found that the
stagnation pressure downstream of this shock exactly
equals the throat pressure. What is the area where the shock
wave stands?
P9.57 Air flows from a tank through a nozzle into the standard
atmosphere, as in Fig. P9.57. A normal shock stands in the
exit of the nozzle, as shown. Estimate (a) the pressure in
the tank and (b) the mass flow.
14 cm2
10 cm2 Air at
100°C Shock Sealevel air  v v P9.57  eText Main Menu  P9.58 Argon (Table A.4) approaches a normal shock with V1
700 m/s, p1 125 kPa, and T1 350 K. Estimate (a) V2
and (b) p2. (c) What pressure p2 would result if the same
velocity change V1 to V2 were accomplished isentropically?
P9.59 Air, at stagnation conditions of 450 K and 250 kPa, flows
through a nozzle. At section 1, where the area is 15 cm2,
there is a normal shock wave. If the mass flow is 0.4 kg/s,
estimate (a) the Mach number and (b) the stagnation pressure just downstream of the shock.
P9.60 When a pitot tube such as in Fig. 6.30 is placed in a supersonic flow, a normal shock will stand in front of the
probe. Suppose the probe reads p0 190 kPa and p 150
kPa. If the stagnation temperature is 400 K, estimate the
(supersonic) Mach number and velocity upstream of the
shock.
P9.61 Repeat Prob. 9.56 except this time let the odd coincidence
be that the static pressure downstream of the shock exactly
equals the throat pressure. What is the area where the shock
stands?
P9.62 An atomic explosion propagates into still air at 14.7 lbf/in2
absolute and 520°R. The pressure just inside the shock is
5000 lbf/in2 absolute. Assuming k 1.4, what are the speed
C of the shock and the velocity V just inside the shock?
P9.63 Sealevel standard air is sucked into a vacuum tank through
a nozzle, as in Fig. P9.63. A normal shock stands where
the nozzle area is 2 cm2, as shown. Estimate (a) the pressure in the tank and (b) the mass flow. 2 cm2
1 cm2
Sealevel air Vacuum tank
3 cm2 P9.63
P9.64 Air in a large tank at 100°C and 150 kPa exhausts to the
atmosphere through a converging nozzle with a 5cm2
throat area. Compute the exit mass flow if the atmospheric
pressure is (a) 100 kPa, (b) 60 kPa, and (c) 30 kPa.
P9.65 Air flows through a convergingdiverging nozzle between
two large reservoirs, as shown in Fig. P9.65. A mercury
manometer between the throat and the downstream reservoir reads h 15 cm. Estimate the downstream reservoir
pressure. Is there a normal shock in the flow? If so, does
it stand in the exit plane or farther upstream?
P9.66 In Prob. 9.65 what would be the mercurymanometer reading h if the nozzle were operating exactly at supersonic design conditions? Textbook Table of Contents  Study Guide 646 Chapter 9 Compressible Flow A t = 10 cm2
100°C
300 k Pa Ae =30 cm2 h
Mercury P9.65
P9.67 In Prob. 9.65 estimate the complete range of manometer
readings h for which the flow through the nozzle is entirely isentropic, except possibly in the exit plane.
P9.68 Air in a tank at 120 kPa and 300 K exhausts to the atmosphere through a 5cm2throat converging nozzle at a
rate of 0.12 kg/s. What is the atmospheric pressure? What
is the maximum mass flow possible at low atmospheric
pressure?
P9.69 With reference to Prob. 3.68, show that the thrust of a
rocket engine exhausting into a vacuum is given by k 1 1
2 Ma2
e P9.75 v v Note that stagnation temperature does not enter into the thrust.
P9.70 Air, at stagnation temperature 100°C, expands isentropically through a nozzle of 6cm2 throat area and 18cm2
exit area. The mass flow is at its maximum value of 0.5
kg/s. Estimate the exit pressure for (a) subsonic and (b)
supersonic exit flow.
P9.71 For the nozzle of Prob. 9.70, allowing for nonisentropic
flow, what is the range of exit tank pressures pb for which
(a) the diverging nozzle flow is fully supersonic, (b) the
exit flow is subsonic, (c) the mass flow is independent of
pb, (d) the exit plane pressure pe is independent of pb, and
(e) pe pb?
P9.72 Suppose the nozzle flow of Prob. 9.70 is not isentropic but
instead has a normal shock at the position where area is
15 cm2. Compute the resulting mass flow, exit pressure,
and exit Mach number.
P9.73 Air flows isentropically in a convergingdiverging nozzle
with a throat area of 3 cm2. At section 1, the pressure is
101 kPa, the temperature is 300 K, and the velocity is 868
m/s. (a) Is the nozzle choked? Determine (b) A1 and (c)
the mass flow. Suppose, without changing stagnation conditions or A1, the (flexible) throat is reduced to 2 cm2. As  eText Main Menu 2 3 100 lbf / in2 abs k/(k 1) where Ae exit area
Mae exit Mach number
p0 stagnation pressure in combustion chamber  1 Air k Ma2)
e p0 Ae(1 F suming shockfree flow, will there be any change in the
gas properties at section 1? If so, compute new p1, V1, and
T1 and explain.
P9.74 The perfectgas assumption leads smoothly to Machnumber relations which are very convenient (and tabulated).
This is not so for a real gas such as steam. To illustrate,
let steam at T0 500°C and p0 2 MPa expand isentropically through a converging nozzle whose exit area is
10 cm2. Using the steam tables, find (a) the exit pressure
and (b) the mass flow when the flow is sonic, or choked.
What complicates the analysis?
*P9.75 A doubletank system in Fig. P9.75 has two identical converging nozzles of 1in2 throat area. Tank 1 is very large,
and tank 2 is small enough to be in steadyflow equilibrium with the jet from tank 1. Nozzle flow is isentropic,
but entropy changes between 1 and 3 due to jet dissipation in tank 2. Compute the mass flow. (If you give up,
Ref. 14, pp. 288 – 290, has a good discussion.)  10 lbf / in2 abs 520° R P9.76 A large reservoir at 20°C and 800 kPa is used to fill a small
insulated tank through a convergingdiverging nozzle with
1cm2 throat area and 1.66cm2 exit area. The small tank
has a volume of 1 m3 and is initially at 20°C and 100 kPa.
Estimate the elapsed time when (a) shock waves begin to
appear inside the nozzle and (b) the mass flow begins to
drop below its maximum value.
P9.77 A perfect gas (not air) expands isentropically through a supersonic nozzle with an exit area 5 times its throat area.
EES
The exit Mach number is 3.8. What is the specificheat ratio of the gas? What might this gas be? If p0 300 kPa,
what is the exit pressure of the gas?
P9.78 The orientation of a hole can make a difference. Consider
holes A and B in Fig. P9.78, which are identical but reversed. For the given air properties on either side, compute
the mass flow through each hole and explain why they are
different.
p1 = 150 k Pa, T1 = 20°C 0.2 cm2 B
A
0.3 cm2 ⋅
mA? P9.78 Textbook Table of Contents p2 = 100 k Pa  ⋅
m B? Study Guide Problems 647
P9.79 Air with p0 300 kPa and T0 500 K flows through a
convergingdiverging nozzle with throat area of 1 cm2 and
exit area of 3 cm2 into a receiver tank. The mass flow is
195.2 kg/h. For what range of receiver pressure is this mass
flow possible?
P9.80 A sealevel automobile tire is initially at 32 lbf/in2 gage
pressure and 75°F. When it is punctured with a hole which
resembles a converging nozzle, its pressure drops to 15
lbf/in2 gage in 12 min. Estimate the size of the hole, in
thousandths of an inch. The tire volume is 2.5 ft2.
P9.81 Helium, in a large tank at 100°C and 400 kPa, discharges
to a receiver through a convergingdiverging nozzle designed to exit at Ma 2.5 with exit area 1.2 cm2. Compute (a) the receiver pressure and (b) the mass flow at design conditions. (c) Also estimate the range of receiver
pressures for which mass flow will be a maximum.
P9.82 Air at 500 K flows through a convergingdiverging nozzle
with throat area of 1 cm2 and exit area of 2.7 cm2. When
the mass flow is 182.2 kg/h, a pitotstatic probe placed in
the exit plane reads p0 250.6 kPa and p 240.1 kPa.
Estimate the exit velocity. Is there a normal shock wave
in the duct? If so, compute the Mach number just downstream of this shock.
P9.83 When operating at design conditions (smooth exit to sealevel pressure), a rocket engine has a thrust of 1 million
lbf. The chamber pressure and temperature are 600 lbf/in2
absolute and 4000°R, respectively. The exhaust gases approximate k 1.38 with a molecular weight of 26. Estimate (a) the exit Mach number and (b) the throat diameter.
P9.84 Air flows through a duct as in Fig. P9.84, where A1 24
cm2, A2 18 cm2, and A3 32 cm2. A normal shock
stands at section 2. Compute (a) the mass flow, (b) the
Mach number, and (c) the stagnation pressure at section 3.
1 P9.86 P9.87 P9.88 P9.89 P9.90 P9.91 (a) the throat pressure and (b) the stagnation pressure in
the upstream supply tank.
Air enters a 3cmdiameter pipe 15 m long at V1 73 m/s,
p1 550 kPa, and T1 60°C. The friction factor is 0.018.
Compute V2, p2, T2, and p02 at the end of the pipe. How
much additional pipe length would cause the exit flow to
be sonic?
Air enters a duct of L/D 40 at V1 170 m/s and T1
300 K. The flow at the exit is choked. What is the average friction factor in the duct for adiabatic flow?
Air enters a 5 by 5cm square duct at V1 900 m/s and
T1 300 K. The friction factor is 0.02. For what length
duct will the flow exactly decelerate to Ma 1.0? If the
duct length is 2 m, will there be a normal shock in the
duct? If so, at what Mach number will it occur?
Air flows adiabatically in a 5cmdiameter tube with f
0.025. At section 1, V1 75 m/s, T1 350 K, and p1
300 kPa. How much further down the tube will (a) the
pressure be 156 kPa, (b) the temperature be 343 K, and
(c) the flow reach the choking point?
Air, supplied at p0 700 kPa and T0 330 K, flows
through a converging nozzle into a pipe of 2.5cm diame
ter which exits to a near vacuum. If f 0.022, what will
be the mass flow through the pipe if its length is (a) 0 m,
(b) 1 m, and (c) 10 m?
Air flows steadily from a tank through the pipe in Fig.
P9.91. There is a converging nozzle on the end. If the mass
flow is 3 kg/s and the nozzle is choked, estimate (a) the
Mach number at section 1 and (b) the pressure inside the
tank.
Air at
100°C L = 9 m, D = 6 cm 3
2
1 Ma1 = 2.5
p1 = 40 k Pa Normal
shock P9.84
P9.85 A large tank delivers air through a nozzle of 1cm2 throat
area and 2.7cm2 exit area. When the receiver pressure is
125 kPa, a normal shock stands in the exit plane. Estimate v v 2 Nozzle P9.91 T1 = 30°C  f = 0.025 Pa = 100 k Pa Air  De = 5 cm eText Main Menu  P9.92 Modify Prob. 9.91 as follows. Let the pressure in the tank
be 700 kPa, and let the nozzle be choked. Determine (a)
Ma2 and (b) the mass flow.
P9.93 Air flows adiabatically in a 3cmdiameter duct. The average friction factor is 0.015. If, at the entrance, V 950
m/s and T 250 K, how far down the tube will (a) the
Mach number be 1.8 or (b) the flow be choked?
P9.94 Compressible pipe flow with friction, Sec. 9.7, assumes
constant stagnation enthalpy and mass flow but variable Textbook Table of Contents  Study Guide 648 Chapter 9 Compressible Flow momentum. Such a flow is often called Fanno flow, and a
line representing all possible property changes on a temperatureentropy chart is called a Fanno line. Assuming a
perfect gas with k 1.4 and the data of Prob. 9.86, draw
a Fanno curve of the flow for a range of velocities from
very low (Ma 1) to very high (Ma 1). Comment on
the meaning of the maximumentropy point on this curve.
P9.95 Helium (Table A.4) enters a 5cmdiameter pipe at p1
550 kPa, V1 312 m/s, and T1 40°C. The friction factor is 0.025. If the flow is choked, determine (a) the length
of the duct and (b) the exit pressure.
P9.96 Derive and verify the adiabaticpipeflow velocity relation
of Eq. (9.74), which is usually written in the form 2
k1
a0 1
1
V
fL
ln 2
k
k V2
V1
V2
D
1
2
P9.97 By making a few algebraic substitutions, show that Eq.
(9.74), or the relation in Prob. 9.96, may be written in the
density form 2k f L
1
2
2
*2
2 ln
1
2
k 1D
2
Why is this formula awkward if one is trying to solve for
the mass flow when the pressures are given at sections 1
and 2?
P9.98 Compressible laminar flow, f 64/Re, may occur in capillary tubes. Consider air, at stagnation conditions of 100°C
and 200 kPa, entering a tube 3 cm long and 0.1 mm in diameter. If the receiver pressure is near vacuum, estimate
(a) the average Reynolds number, (b) the Mach number at
the entrance, and (c) the mass flow in kg/h.
P9.99 A compressor forces air through a smooth pipe 20 m long
and 4 cm in diameter, as in Fig. P9.99. The air leaves at
EES
101 kPa and 200°C. The compressor data for pressure rise
versus mass flow are shown in the figure. Using the Moody chart to estimate f , compute the resulting mass flow.
D = 4 cm
L = 20 m Pe = 101 k Pa ⋅
m Te = 200°C
250 kPa
Parabola ∆p ⋅
m
P9.99 0.4 kg / s  v v P9.100 Modify Prob. 9.99 as follows. Find the length of 4cmdiameter pipe for which the pump pressure rise will be exactly 200 kPa.  eText Main Menu  P9.101 How do the compressiblepipeflow formulas behave for
small pressure drops? Let air at 20°C enter a tube of di
ameter 1 cm and length 3 m. If f 0.028 with p1 102
kPa and p2 100 kPa, estimate the mass flow in kg/h for
(a) isothermal flow, (b) adiabatic flow, and (c) incompressible flow (Chap. 6) at the entrance density.
P9.102 Air at 550 kPa and 100°C enters a smooth 1mlong pipe
and then passes through a second smooth pipe to a 30kPa
reservoir, as in Fig. P9.102. Using the Moody chart to com
pute f , estimate the mass flow through this system. Is the
flow choked? 550
kPa L=1m
D = 5 cm L = 1.2 m
D = 3 cm 100°C Pe = 30 kPa Converging
nozzle P9.102
P9.103 Natural gas, with k 1.3 and a molecular weight of 16, is
to be pumped through 100 km of 81cmdiameter pipeline.
The downstream pressure is 150 kPa. If the gas enters at 60°C, the mass flow is 20 kg/s, and f 0.024, estimate
the required entrance pressure for (a) isothermal flow and
(b) adiabatic flow.
P9.104 A tank of oxygen (Table A.4) at 20°C is to supply an astronaut through an umbilical tube 12 m long and 2 cm in
diameter. The exit pressure in the tube is 40 kPa. If the de
sired mass flow is 90 kg/h and f 0.025, what should be
the pressure in the tank?
P9.105 Air enters a 5cmdiameter pipe at p1 200 kPa and T1
350 K. The downstream receiver pressure is 74 kPa. The
friction factor is 0.02. If the exit is choked, what is (a) the
length of the pipe and (b) the mass flow? (c) If p1, T1, and
preceiver stay the same, what pipe length will cause the mass
flow to increase by 50 percent over (b)? Hint: In part (c)
the exit pressure does not equal the receiver pressure. P9.106 Air at 300 K flows through a duct 50 m long with f
0.019. What is the minimum duct diameter which can carry
the flow without choking if the entrance velocity is (a) 50
m/s, (b) 150 m/s, and (c) 420 m/s?
P9.107 A fuelair mixture, assumed equivalent to air, enters a duct
combustion chamber at V1 104 m/s and T1 300 K.
What amount of heat addition in kJ/kg will cause the exit
flow to be choked? What will be the exit Mach number
and temperature if 504 kJ/kg is added during combustion? Textbook Table of Contents  Study Guide Problems 649
P9.108 What happens to the inlet flow of Prob. 9.107 if the combustion yields 1500 kJ/kg heat addition and p01 and T01
remain the same? How much is the mass flow reduced?
P9.109 A jet engine at 7000m altitude takes in 45 kg/s of air and
adds 550 kJ/kg in the combustion chamber. The chamber
cross section is 0.5 m2, and the air enters the chamber at
80 kPa and 5°C. After combustion the air expands through
an isentropic converging nozzle to exit at atmospheric pressure. Estimate (a) the nozzle throat diameter, (b) the nozzle exit velocity, and (c) the thrust produced by the engine.
P9.110 Compressible pipe flow with heat addition, Sec. 9.8, assumes constant momentum ( p
V2) and constant mass
flow but variable stagnation enthalpy. Such a flow is often called Rayleigh flow, and a line representing all possible property changes on a temperatureentropy chart is
called a Rayleigh line. Assuming air passing through the
flow state p1 548 kPa, T1 588 K, V1 266 m/s, and
A 1 m2, draw a Rayleigh curve of the flow for a range
of velocities from very low (Ma 1) to very high (Ma
1). Comment on the meaning of the maximumentropy
point on this curve.
P9.111 Add to your Rayleigh line of Prob. 9.110 a Fanno line (see
Prob. 9.94) for stagnation enthalpy equal to the value associated with state 1 in Prob. 9.110. The two curves will
intersect at state 1, which is subsonic, and at a certain state
2, which is supersonic. Interpret these two states visàvis
Table B.2.
P9.112 Air enters a duct subsonically at section 1 at 1.2 kg/s. When
650 kW of heat is added, the flow chokes at the exit at
p2 95 kPa and T2 700 K. Assuming frictionless heat
addition, estimate (a) the velocity and (b) the stagnation
pressure at section 1.
P9.113 Air enters a constantarea duct at p1 90 kPa, V1 520
m/s, and T1 558°C. It is then cooled with negligible friction until it exits at p2 160 kPa. Estimate (a) V2, (b) T2,
and (c) the total amount of cooling in kJ/kg.
P9.114 We have simplified things here by separating friction (Sec.
9.7) from heat addition (Sec. 9.8). Actually, they often occur together, and their effects must be evaluated simultaneously. Show that, for flow with friction and heat transfer in a constantdiameter pipe, the continuity, momentum,
and energy equations may be combined into the following
differential equation for Machnumber changes: 8m
3m
V Particle P9.118 P9.119 The particle in Fig. P9.119 is moving supersonically in
sealevel standard air. From the two given disturbance
spheres, compute the particle Mach number, velocity, and
Mach angle.
Particle 8m
3m V
8m P9.119 P9.120 The particle in Fig. P9.120 is moving in sealevel standard
air. From the two disturbance spheres shown, estimate (a)
the position of the particle at this instant and (b) the temperature in °C at the front stagnation point of the particle.
6m
3m 1 k Ma2 dQ
1 Ma2 cpT d Ma2
Ma2 k Ma2 [2 (k 1) Ma2] f dx
D
2(1 Ma2)
where dQ is the heat added. A complete derivation, including many additional combined effects such as area
change and mass addition, is given in chap. 8 of Ref. 8.  v v P9.115 Air flows subsonically in a duct with negligible friction.
When heat is added in the amount of 948 kJ/kg, the pressure drops from p1 200 to p2 106 kPa. Estimate
(a) Ma1, (b) T1, and (c) V1, assuming T01 305 K.
P9.116 An observer at sea level does not hear an aircraft flying at
12,000ft standard altitude until it is 5 (statute) mi past her.
Estimate the aircraft speed in ft/s.
P9.117 An observer at sea level does not hear an aircraft flying at
6000m standard altitude until 15 s after it has passed overhead. Estimate the aircraft speed in m/s.
P9.118 A particle moving at uniform velocity in sealevel standard air creates the two disturbance spheres shown in Fig.
P9.118. Compute the particle velocity and Mach number.  eText Main Menu  P9.120
P9.121 A thermistor probe, in the shape of a needle parallel to the
flow, reads a static temperature of 25°C when inserted
into a supersonic airstream. A conical disturbance cone of
halfangle 17° is created. Estimate (a) the Mach number, Textbook Table of Contents  Study Guide 650 Chapter 9 Compressible Flow (b) the velocity, and (c) the stagnation temperature of the
stream.
P9.122 Supersonic air takes a 5° compression turn, as in Fig.
P9.122. Compute the downstream pressure and Mach number and the wave angle, and compare with smalldisturbance theory. Ma = 3.0
p = 100 k Pa δ 12 cm P9.128
P9.129 Air flows at supersonic speed toward a compression ramp,
as in Fig. P9.129. A scratch on the wall at point a creates
a wave of 30° angle, while the oblique shock created has
a 50° angle. What is (a) the ramp angle and (b) the wave
angle caused by a scratch at b? Ma2, p2
Ma 1 = 3
p1 = 100 k Pa a 5° b 30° P9.122
P9.123 Modify Prob. 9.122 as follows. Let the 5° total turn be in
the form of five separate compression turns of 1° each.
Compute the final Mach number and pressure, and compare the pressure with an isentropic expansion to the same
final Mach number.
P9.124 Determine the validity of the following alternate relation
for the pressure ratio across an oblique shock wave:
p2
p1 cot
cot sin 2
sin 2 cos 2
cos 2 k
k If necessary, your proof (or disproof) may be somewhat
tentative and heuristic.
P9.125 Show that, as the upstream Mach number approaches infinity, the Mach number downstream of an attached
obliqueshock wave will have the value
k1
2k sin2 ( Ma2 v v  50° eText Main Menu θ φ P9.129
P9.130 Modify Prob. 9.129 as follows. If the wave angle is 42°,
determine (a) the shockwave angle and (b) the deflection
angle.
P9.131 In Fig. P9.128, assume that the approach stream temperature is 20°C. For what wedge halfangle will the stream
temperature along the wedge surface be 200°C?
P9.132 Air flows at Ma 3 and p 10 lbf/in2 absolute toward a
wedge of 16° angle at zero incidence in Fig. P9.132. If the
pointed edge is forward, what will be the pressure at point
A? If the blunt edge is forward, what will be the pressure
at point B? ) A P9.126 Consider airflow at Ma1 2.2. Calculate, to two decimal
places, (a) the deflection angle for which the downstream
flow is sonic and (b) the maximum deflection angle.
P9.127 Do the Mach waves upstream of an obliqueshock wave
intersect with the shock? Assuming supersonic downstream flow, do the downstream Mach waves intersect the
shock? Show that for small deflections the shockwave angle lies halfway between 1 and 2
for any Mach
number.
P9.128 Air flows past a twodimensional wedgenosed body as in
Fig. P9.128. Determine the wedge halfangle for which
the horizontal component of the total pressure force on the
nose is 35 kN/m of depth into the paper.  Ma > 1  16°
Ma = 3 p = 10 lbf / in2 abs
B 16° P9.132
P9.133 Air flows supersonically toward the doublewedge system
in Fig. P9.133. The (x, y) coordinates of the tips are given. Textbook Table of Contents  Study Guide Problems 651
(1 m, 1 m)
Shock Ma = 3.5 Shocks Ma h
1m Shock (0, 0) φ P9.133
P9.136
The shock wave of the forward wedge strikes the tip of
the aft wedge. Both wedges have 15° deflection angles.
What is the freestream Mach number?
P9.134 When an oblique shock strikes a solid wall, it reflects as
a shock of sufficient strength to cause the exit flow Ma3
EES
to be parallel to the wall, as in Fig. P9.134. For airflow
with Ma1 2.5 and p1 100 kPa, compute Ma3, p3, and
the angle . P9.137 Generalize Prob. 9.136 into a computer study as follows.
Assuming weak shocks, find and plot all combinations of
and h in Fig. P9.136 for which the canceled or “swallowed’’ reflected shock is possible.
P9.138 The supersonic nozzle of Fig. P9.138 is overexpanded
(case G of Fig. 9.12) with Ae/At 3.0 and a stagnation
pressure of 350 kPa. If the jet edge makes a 4° angle with
the nozzle centerline, what is the back pressure pr in kPa? Ma 2 Ma 1 = 2.5
40° pr ? 4° Ma 3 φ
Air P9.134
P9.135 A bend in the bottom of a supersonic duct flow induces
a shock wave which reflects from the upper wall, as in
Fig. P9.135. Compute the Mach number and pressure in
region 3. 3
2
Air:
p1 = 100 k Pa Jet
edge P9.138
P9.139 Airflow at Ma 2.2 takes a compression turn of 12° and
then another turn of angle in Fig. P9.139. What is the
maximum value of for the second shock to be attached?
Will the two shocks intersect for any less than max?
3 Ma 1 = 3.0 θ max? Ma 1 = 2.2 10 ° 2 P9.135 12°  v v P9.136 Figure P9.136 is a special application of Prob. 9.135. With
careful design, one can orient the bend on the lower wall
so that the reflected wave is exactly canceled by the return
bend, as shown. This is a method of reducing the Mach
number in a channel (a supersonic diffuser). If the bend
angle is
10°, find (a) the downstream width h and (b)
the downstream Mach number. Assume a weak shock
wave.  eText Main Menu  P9.139
P9.140 The solution to Prob. 9.122 is Ma2 2.750, and p2
145.5 kPa. Compare these results with an isentropic compression turn of 5°, using PrandtlMeyer theory.
P9.141 Supersonic airflow takes a 5° expansion turn, as in Fig.
P9.141. Compute the downstream Mach number and pressure, and compare with smalldisturbance theory. Textbook Table of Contents  Study Guide 652 Chapter 9 Compressible Flow
pa = 10 kPa Ma1 = 3
p1 = 100 k Pa Jet
edge φ
5° Ma 2 Ma 2, p2 Ma 2 P9.141
P9.142 A supersonic airflow at Ma1 3.2 and p1 50 kPa undergoes a compression shock followed by an isentropic expansion turn. The flow deflection is 30° for each turn.
Compute Ma2 and p2 if (a) the shock is followed by the
expansion and (b) the expansion is followed by the shock.
P9.143 Airflow at Ma1 3.2 passes through a 25° obliqueshock
deflection. What isentropic expansion turn is required to
bring the flow back to (a) Ma1 and (b) p1?
P9.144 Consider a smooth isentropic compression turn of 20°, as
shown in Fig. P9.144. The Mach waves thus generated will
form a converging fan. Sketch this fan as accurately as possible, using at least five equally spaced waves, and demonstrate how the fan indicates the probable formation of an
obliqueshock wave. Jet
edge φ P9.147 8°? Why does the drag coefficient not have the simple
parabolic form CD K 2 in this range?
P9.150 A flatplate airfoil with C 1.2 m is to have a lift of 30
kN/m when flying at 5000m standard altitude with U
641 m/s. Using Ackeret theory, estimate (a) the angle of
attack and (b) the drag force in N/m.
P9.151 Air flows at Ma 2.5 past a halfwedge airfoil whose angles are 4°, as in Fig. P9.151. Compute the lift and drag
coefficient at equal to (a) 0° and (b) 6°.
4° 4° Ma ∞ = 2.5 P9.151
20° Mach waves P9.152 A supersonic airfoil has a parabolic symmetric shape for
upper and lower surfaces Ma = 3.0 yu,l Finish
Circulararc turn P9.144 Start  v v P9.145 Air at Ma1 2.0 and p1 100 kPa undergoes an isentropic expansion to a downstream pressure of 50 kPa. What
EES
is the desired turn angle in degrees?
P9.146 Helium, at 20°C and V1 2010 m/s, undergoes a PrandtlMeyer expansion until the temperature is 50°C. Estimate
the turn angle in degrees.
P9.147 A convergingdiverging nozzle with a 4:1 exitarea ratio
and p0 500 kPa operates in an underexpanded condition
(case I of Fig. 9.12b) as in Fig. P9.147. The receiver pressure is pa 10 kPa, which is less than the exit pressure,
so that expansion waves form outside the exit. For the
given conditions, what will the Mach number Ma2 and the
angle of the edge of the jet be? Assume k 1.4 as usual.
P9.148 Repeat Example 9.19 for an angle of attack of 6°. Is the
lift coefficient linear with angle in this range of 0°
8°? Is the drag coefficient parabolic with in this range?
P9.149 Repeat Example 9.21 for an angle of attack of 2°. Is the
lift coefficient linear with angle in this range of 0°  eText Main Menu  2t x
C x2
C2 such that the maximum thickness is t at x 1 C. Compute
2
the drag coefficient at zero incidence by Ackeret theory,
and compare with a symmetric double wedge of the same
thickness.
P9.153 A supersonic transport has a mass of 65 Mg and cruises at
11km standard altitude at a Mach number of 2.25. If the
angle of attack is 2° and its wings can be approximated by
flat plates, estimate (a) the required wing area in m2 and
(b) the thrust required in N.
P9.154 A symmetric supersonic airfoil has its upper and lower surfaces defined by a sinewave shape:
x
t
y
sin
C
2
where t is the maximum thickness, which occurs at x
C/2. Use Ackeret theory to derive an expression for the
drag coefficient at zero angle of attack. Compare your result with Ackeret theory for a symmetric doublewedge airfoil of the same thickness.
P9.155 For the sinewave airfoil shape of Prob. 9.154, with Ma
2.5, k 1.4, t/C 0.1, and
0°, plot (without com Textbook Table of Contents  Study Guide Fundamentals of Engineering Exam Problems
puting the overall forces) the pressure distribution p(x)/p
along the upper surface for (a) Ackeret theory and (b) an
oblique shock plus a continuous PrandtlMeyer expansion.
P9.156 A thin circulararc airfoil is shown in Fig. P9.156. The
leading edge is parallel to the free stream. Using linearized
(smallturningangle) supersonicflow theory, derive a formula for the lift and drag coefficient for this orientation,
and compare with Ackerettheory results for an angle of
attack
tan 1 (h/L). Ma > 1 LE 653 Circulararc foil h P9.156 L TE P9.157 Prove from Ackeret theory that for a given supersonic airfoil shape with sharp leading and trailing edges and a given
thickness, the minimumthickness drag occurs for a symmetric doublewedge shape. Word Problems
W9.1 Notice from Table 9.1 that (a) water and mercury and (b)
aluminum and steel have nearly the same speeds of sound,
yet the second of the two materials is much denser. Can you
account for this oddity? Can molecular theory explain it?
W9.2 When an object approaches you at Ma 0.8, you can hear
it, according to Fig. 9.18a. But would there be a Doppler
shift? For example, would a musical tone seem to you to
have a higher or a lower pitch?
W9.3 The subject of this chapter is commonly called gas dynamics. But can liquids not perform in this manner? Using water as an example, make a ruleofthumb estimate of the
pressure level needed to drive a water flow at velocities
comparable to the sound speed.
W9.4 Suppose a gas is driven at compressible subsonic speeds by
a large pressure drop, p1 to p2. Describe its behavior on an
appropriately labeled Mollier chart for (a) frictionless flow W9.5 W9.6 W9.7 W9.8 in a converging nozzle and (b) flow with friction in a long
duct.
Describe physically what the “speed of sound’’ represents.
What kind of pressure changes occur in air sound waves during ordinary conversation?
Give a physical description of the phenomenon of choking
in a convergingnozzle gas flow. Could choking happen even
if wall friction were not negligible?
Shock waves are treated as discontinuities here, but they actually have a very small finite thickness. After giving it some
thought, sketch your idea of the distribution of gas velocity,
pressure, temperature, and entropy through the inside of a
shock wave.
Describe how an observer, running along a normalshock
wave at finite speed V, will see what appears to be an
obliqueshock wave. Is there any limit to the running speed? Fundamentals of Engineering Exam Problems
Onedimensional compressibleflow problems have become quite
popular on the FE Exam, especially in the afternoon sessions. In
the following problems, assume onedimensional flow of ideal air,
R 287 J/(kg K) and k 1.4.  v v FE9.1 For steady isentropic flow, if the absolute temperature increases 50 percent, by what ratio does the static pressure
increase?
(a) 1.12, (b) 1.22, (c) 2.25, (d) 2.76, (e) 4.13
FE9.2 For steady isentropic flow, if the density doubles, by what
ratio does the static pressure increase?
(a) 1.22, (b) 1.32, (c) 1.44, (d) 2.64, (e) 5.66
FE9.3 A large tank, at 500 K and 200 kPa, supplies isentropic
airflow to a nozzle. At section 1, the pressure is only 120
kPa. What is the Mach number at this section?
(a) 0.63, (b) 0.78, (c) 0.89, (d) 1.00, (e) 1.83
FE9.4 In Prob. FE9.3 what is the temperature at section 1?
(a) 300 K, (b) 408 K, (c) 417 K, (d) 432 K, (e) 500 K  eText Main Menu  FE9.5 In Prob. FE9.3, if the area at section 1 is 0.15 m2, what is
the mass flow?
(a) 38.1 kg/s, (b) 53.6 kg/s, (c) 57.8 kg/s, (d) 67.8 kg/s,
(e) 77.2 kg/s
FE9.6 For steady isentropic flow, what is the maximum possible
mass flow through the duct in Fig. FE9.6?
(a) 9.5 kg/s, (b) 15.1 kg/s, (c) 26.2 kg/s, (d) 30.3 kg/s,
(e) 52.4 kg/s Throat area 0.05 m2 Tank:
400 K, 300 kPa Exit FE9.6 Textbook Table of Contents  Study Guide 654 Chapter 9 Compressible Flow FE9.7 If the exit Mach number in Fig. FE9.6 is 2.2, what is the
exit area?
(a) 0.10 m2, (b) 0.12 m2, (c) 0.15 m2 , (d) 0.18 m2,
(e) 0.22 m2
FE9.8 If there are no shock waves and the pressure at one duct
section in Fig. FE9.6 is 55.5 kPa, what is the velocity at
that section?
(a) 166 m/s, (b) 232 m/s, (c) 554 m/s, (d) 706 m/s,
(e) 774 m/s FE9.9 If, in Fig. FE9.6, there is a normal shock wave at a section where the area is 0.07 m2, what is the air density just
upstream of that shock?
(a) 0.48 kg/m3, (b) 0.78 kg/m3, (c) 1.35 kg/m3,
(d) 1.61 kg/m3, (e) 2.61 kg/m3
FE9.10 In Prob. FE9.9, what is the Mach number just downstream
of the shock wave?
(a) 0.42, (b) 0.55, (c) 0.63, (d) 1.00, (e) 1.76 Comprehensive Problems
C9.1 The convergingdiverging nozzle sketched in Fig. C9.1 is
designed to have a Mach number of 2.00 at the exit plane
(assuming the flow remains nearly isentropic). The flow C9.2
travels from tank a to tank b, where tank a is much larger
than tank b. (a) Find the area at the exit Ae and the back
pressure pb which will allow the system to operate at design conditions. (b) As time goes on, the back pressure will
grow, since the second tank slowly fills up with more air. *C9.3
Since tank a is huge, the flow in the nozzle will remain the
same, however, until a normal shock wave appears at the
exit plane. At what back pressure will this occur? (c) If tank
b is held at constant temperature, T 20°C, estimate how
long it will take for the flow to go from design conditions
T 500 K
p 1.00 MPa
Air (k 1.4)
Volume huge to the condition of part (b), i.e., with a shock wave at the
exit plane.
Two large air tanks, one at 400 K and 300 kPa and the other
at 300 K and 100 kPa, are connected by a straight tube 6 m
long and 5 cm in diameter. The average friction factor is
0.0225. Assuming adiabatic flow, estimate the mass flow
through the tube.
Figure C9.3 shows the exit of a convergingdiverging nozzle, where an obliqueshock pattern is formed. In the exit
plane, which has an area of 15 cm2, the air pressure is 16
kPa and the temperature is 250 K. Just outside the exit shock,
which makes an angle of 50° with the exit plane, the temperature is 430 K. Estimate (a) the mass flow, (b) the throat
area, (c) the turning angle of the exit flow, and, in the tank
supplying the air, (d) the pressure and (e) the temperature. Volume 100,000 L
T 20.0 C
430 K
Ae , Ve , Mae
50
Tank b Throat area Tank a Shock waves 0.07 m2 C9.3 C9.1 Design Projects  v v D9.1 It is desired to select a rectangular wing for a fighter aircraft. The plane must be able (a) to take off and land on a
4500ftlong sealevel runway and (b) to cruise supersonically at Ma 2.3 at 28,000ft altitude. For simplicity, assume a wing with zero sweepback. Let the aircraft maximum weight equal (30 n)(1000) lbf, where n is the
number of letters in your surname. Let the available sealevel maximum thrust be onethird of the maximum weight,
decreasing at altitude proportional to ambient density. Mak  eText Main Menu  ing suitable assumptions about the effect of finite aspect ratio on wing lift and drag for both subsonic and supersonic
flight, select a wing of minimum area sufficient to perform
these takeoff/landing and cruise requirements. Some
thought should be given to analyzing the wingtips and wing
roots in supersonic flight, where Mach cones form and the
flow is not twodimensional. If no satisfactory solution is
possible, gradually increase the available thrust to converge
to an acceptable design. Textbook Table of Contents  Study Guide References 655 h p , Ma D9.2 D9.2 Consider supersonic flow of air at sealevel conditions past
a wedge of halfangle , as shown in Fig. D9.2. Assume
that the pressure on the back of the wedge equals the fluid
pressure as it exits the PrandtlMeyer fan. (a) Suppose
Ma
3.0. For what angle will the supersonic wavedrag
coefficient CD, based on frontal area, be exactly 0.5? (b)
Suppose that
20°. Is there a freestream Mach number
for which the wavedrag coefficient CD, based on frontal
area, will be exactly 0.5? (c) Investigate the percent increase
in CD from (a) and (b) due to including boundarylayer friction drag in the calculation. References 2.
3.
4.
5.
6.
7. 8.
9.
10.
11.
12.
13.
14.
15.
16. A. Y. Pope, Aerodynamics of Supersonic Flight, 2d ed., Pitman, New York, 1958.
A. B. Cambel and B. H. Jennings, Gas Dynamics, McGrawHill, New York, 1958.
F. Cheers, Elements of Compressible Flow, Wiley, New York,
1963.
J. E. John, Gas Dynamics, 2d ed., Allyn & Bacon, Boston,
1984.
A. J. Chapman and W. F. Walker, Introductory Gas Dynamics, Holt, New York, 1971.
B. W. Imrie, Compressible Fluid Flow, Halstead, New York,
1974.
R. Courant and K. O. Friedrichs, Supersonic Flow and Shock
Waves, Interscience, New York, 1948; reprinted by SpringerVerlag, New York, 1977.
A. H. Shapiro, The Dynamics and Thermodynamics of Compressible Fluid Flow, Ronald, New York, 1953.
H. W. Liepmann and A. Roshko, Elements of Gas Dynamics,
Wiley, New York, 1957.
R. von Mises, Mathematical Theory of Compressible Fluid
Flow, Academic, New York, 1958.
J. A. Owczarek, Gas Dynamics, International Textbook,
Scranton, PA, 1964.
W. G. Vincenti and C. Kruger, Introduction to Physical Gas
Dynamics, Wiley, New York, 1965.
J. D. Anderson, Hypersonic and HighTemperature Gas Dynamics, McGrawHill, New York, 1989.
P. A. Thompson, Compressible Fluid Dynamics, McGrawHill, New York, 1972.
J. H. Keenan et al., Steam Tables: SI Version, 2 vols., Wiley,
New York, 1985.
J. H. Keenan et al., Gas Tables, 2d ed., WileyInterscience,
New York, 1983.  v v 1.  eText Main Menu  17.
18.
19.
20. 21.
22. 23. 24. 25.
26.
27.
28.
29.
30.
31. Y. A. Cengel and M. A. Boles, Thermodynamics: An Engineering Approach, 3d ed., McGrawHill, New York, 1998.
K. Wark, Thermodynamics, 6th ed., McGrawHill, New York,
1999.
F. M. White, Viscous Fluid Flow, 2d ed., McGrawHill, New
York, 1991.
J. H. Keenan and E. P. Neumann, “Measurements of Friction
in a Pipe for Subsonic and Supersonic Flow of Air,’’ J. Appl.
Mech., vol. 13, no. 2, 1946, p. A91.
J. Ackeret, “Air Forces on Airfoils Moving Faster than Sound
Velocity,’’ NACA Tech. Mem. 317, 1925.
Z. Kopal, “Tables of Supersonic Flow around Cones,’’ M.I.T.
Center Anal. Tech. Rep. 1, 1947 (see also Tech. Rep. 3 and
5, 1947).
J. L. Sims, Tables for Supersonic Flow around Right Circular Cones at Zero Angle of Attack, NASA SP3004, 1964 (see
also NASA SP3007).
J. Palmer et al., Compressible Flow Tables for Engineers:
With Appropriate Computer Programs, Scholium Intl., Port
Washington, NY, 1989.
S. M. Yahya (ed.), Gas Tables for Compressible Flow Calculations, Wiley Eastern, New Delhi, India, 1985.
S. M. Yahya, Fundamentals of Compressible Flow, Wiley
Eastern, New Delhi, India, 1982.
S. Schreier, Compressible Flow, Wiley, New York, 1982.
M. A. Saad, Compressible Fluid Flow, 2d ed., PrenticeHall,
Englewood Cliffs, NJ, 1992.
A. Y. Pope and K. L. Goin, High Speed Wind Tunnel Testing,
Wiley, New York, 1965.
W. Bober and R. A. Kenyon, Fluid Mechanics, Wiley, New
York, 1980.
J. D. Anderson, Modern Compressible Flow: With Historical
Perspective, 2d ed., McGrawHill, New York, 1990. Textbook Table of Contents  Study Guide 656 Chapter 9 Compressible Flow M. J. Zucrow and J. D. Hoffman, Gas Dynamics, Wiley, New
York, 1976.
33. Z. Husain, Gas Dynamics through Problems, Halsted Press,
New York, 1989.
34. J. E. Plapp, Engineering Fluid Mechanics, PrenticeHall, Englewood Cliffs, NJ, 1968.  v v 32.  eText Main Menu  35. P. H. Oosthuizen and W. E. Carscallen, Compressible Fluid
Flow, McGrawHill, New York, 1997.
36. M. H. Kaplan, “The Reusable Launch Vehicle: Is the Stage
Set?” Launchspace Magazine, March 1997, pp. 26 – 30.
37. T. K. Mattingly, “A Simpler Ride into Space,” Scientific
American, October 1997, pp. 120 – 125. Textbook Table of Contents  Study Guide ...
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