Unformatted text preview: cen84959_ch17.qxd 4/21/05 11:08 AM Page 823 Chapter 17
COMPRESSIBLE FLOW F or the most part, we have limited our consideration so
far to flows for which density variations and thus compressibility effects are negligible. In this chapter we lift
this limitation and consider flows that involve significant
changes in density. Such flows are called compressible flows,
and they are frequently encountered in devices that involve
the flow of gases at very high velocities. Compressible flow
combines fluid dynamics and thermodynamics in that both
are necessary to the development of the required theoretical
background. In this chapter, we develop the general relations
associated with onedimensional compressible flows for an
ideal gas with constant specific heats.
We start this chapter by introducing the concepts of stagnation state, speed of sound, and Mach number for compressible flows. The relationships between the static and
stagnation fluid properties are developed for isentropic flows of
ideal gases, and they are expressed as functions of specificheat ratios and the Mach number. The effects of area
changes for onedimensional isentropic subsonic and supersonic flows are discussed. These effects are illustrated by
considering the isentropic flow through converging and
converging–diverging nozzles. The concept of shock waves
and the variation of flow properties across normal and oblique
shocks are discussed. Finally, we consider the effects of heat
transfer on compressible flows and examine steam nozzles. Objectives
The objectives of Chapter 17 are to:
• Develop the general relations for compressible flows
encountered when gases flow at high speeds.
• Introduce the concepts of stagnation state, speed of sound,
and Mach number for a compressible fluid.
• Develop the relationships between the static and stagnation
fluid properties for isentropic flows of ideal gases.
• Derive the relationships between the static and stagnation
fluid properties as functions of specificheat ratios and
Mach number.
• Derive the effects of area changes for onedimensional
isentropic subsonic and supersonic flows.
• Solve problems of isentropic flow through converging and
converging–diverging nozzles.
• Discuss the shock wave and the variation of flow properties
across the shock wave.
• Develop the concept of duct flow with heat transfer and
negligible friction known as Rayleigh flow.
• Examine the operation of steam nozzles commonly used in
steam turbines.  823 cen84959_ch17.qxd 4/21/05 11:08 AM Page 824 824  Thermodynamics 17–1 ■ STAGNATION PROPERTIES When analyzing control volumes, we find it very convenient to combine the
internal energy and the flow energy of a fluid into a single term, enthalpy,
defined per unit mass as h
u
P v. Whenever the kinetic and potential
energies of the fluid are negligible, as is often the case, the enthalpy represents the total energy of a fluid. For highspeed flows, such as those
encountered in jet engines (Fig. 17–1), the potential energy of the fluid is
still negligible, but the kinetic energy is not. In such cases, it is convenient
to combine the enthalpy and the kinetic energy of the fluid into a single
term called stagnation (or total) enthalpy h0, defined per unit mass as
h h0 FIGURE 17–1
Aircraft and jet engines involve high
speeds, and thus the kinetic energy
term should always be considered
when analyzing them.
(a) Photo courtesy of NASA,
http://lisar.larc.nasa.gov/IMAGES/SMALL/EL199900108.jpeg, and (b) Figure courtesy of Pratt
and Whitney. Used by permission. 1 kJ> kg 2 V2
2 (17–1) When the potential energy of the fluid is negligible, the stagnation enthalpy
represents the total energy of a flowing fluid stream per unit mass. Thus it
simplifies the thermodynamic analysis of highspeed flows.
Throughout this chapter the ordinary enthalpy h is referred to as the static
enthalpy, whenever necessary, to distinguish it from the stagnation
enthalpy. Notice that the stagnation enthalpy is a combination property of a
fluid, just like the static enthalpy, and these two enthalpies become identical
when the kinetic energy of the fluid is negligible.
Consider the steady flow of a fluid through a duct such as a nozzle, diffuser, or some other flow passage where the flow takes place adiabatically
and with no shaft or electrical work, as shown in Fig. 17–2. Assuming the
fluid experiences little or no change in its elevation and its potential energy,
.
.
Eout) for this singlestream steadyflow
the energy balance relation (Ein
system reduces to
V2
1
2 h2 h01 h1 V2
2
2 h02 (17–2) or h1
V1
h01 Control
volume h2
V2
h02 = h 01 FIGURE 17–2
Steady flow of a fluid through an
adiabatic duct. (17–3) That is, in the absence of any heat and work interactions and any changes in
potential energy, the stagnation enthalpy of a fluid remains constant during
a steadyflow process. Flows through nozzles and diffusers usually satisfy
these conditions, and any increase in fluid velocity in these devices creates
an equivalent decrease in the static enthalpy of the fluid.
If the fluid were brought to a complete stop, then the velocity at state 2
would be zero and Eq. 17–2 would become
h1 V2
1
2 h2 h 02 Thus the stagnation enthalpy represents the enthalpy of a fluid when it is
brought to rest adiabatically.
During a stagnation process, the kinetic energy of a fluid is converted to
enthalpy (internal energy flow energy), which results in an increase in the
fluid temperature and pressure (Fig. 17–3). The properties of a fluid at the
stagnation state are called stagnation properties (stagnation temperature, cen84959_ch17.qxd 4/21/05 11:08 AM Page 825 Chapter 17
stagnation pressure, stagnation density, etc.). The stagnation state and the
stagnation properties are indicated by the subscript 0.
The stagnation state is called the isentropic stagnation state when the
stagnation process is reversible as well as adiabatic (i.e., isentropic). The
entropy of a fluid remains constant during an isentropic stagnation process.
The actual (irreversible) and isentropic stagnation processes are shown on
the hs diagram in Fig. 17–4. Notice that the stagnation enthalpy of the fluid
(and the stagnation temperature if the fluid is an ideal gas) is the same for
both cases. However, the actual stagnation pressure is lower than the isentropic stagnation pressure since entropy increases during the actual stagnation process as a result of fluid friction. The stagnation processes are often
approximated to be isentropic, and the isentropic stagnation properties are
simply referred to as stagnation properties.
When the fluid is approximated as an ideal gas with constant specific
heats, its enthalpy can be replaced by cpT and Eq. 17–1 can be expressed as
V2
2 FIGURE 17–3
Kinetic energy is converted to
enthalpy during a stagnation process.
© Reprinted with special permission of King
Features Syndicate. h T0 T V2
2cp (17–4) Here T0 is called the stagnation (or total) temperature, and it represents
the temperature an ideal gas attains when it is brought to rest adiabatically.
The term V 2/2cp corresponds to the temperature rise during such a process
and is called the dynamic temperature. For example, the dynamic temperature of air flowing at 100 m/s is (100 m/s)2/(2 1.005 kJ/kg · K) 5.0 K.
Therefore, when air at 300 K and 100 m/s is brought to rest adiabatically (at
the tip of a temperature probe, for example), its temperature rises to the
stagnation value of 305 K (Fig. 17–5). Note that for lowspeed flows, the
stagnation and static (or ordinary) temperatures are practically the same.
But for highspeed flows, the temperature measured by a stationary probe
placed in the fluid (the stagnation temperature) may be significantly higher
than the static temperature of the fluid.
The pressure a fluid attains when brought to rest isentropically is called
the stagnation pressure P0. For ideal gases with constant specific heats, P0
is related to the static pressure of the fluid by
P0
P a T0 k>1k
b
T 12 (17–5) k
By noting that r
1/v and using the isentropic relation Pv k P0v 0 , the
ratio of the stagnation density to static density can be expressed as r0
r a T0 1>1k
b
T 12 (17–6) When stagnation enthalpies are used, there is no need to refer explicitly to
#
#
kinetic energy. Then the energy balance Ein Eout for a singlestream,
steadyflow device can be expressed as
qin win 1 h01 gz1 2 qout wout 1 h02 gz2 2 (17–7) t 0
ac 0, Isentropic
stagnation
state or P cpT 825 P cpT0  h0 V2
2 Actual
stagnation
state P h
Actual state
s FIGURE 17–4
The actual state, actual stagnation
state, and isentropic stagnation state
of a fluid on an hs diagram. cen84959_ch17.qxd 4/21/05 11:08 AM Page 826 826  Thermodynamics
where h01 and h02 are the stagnation enthalpies at states 1 and 2, respectively.
When the fluid is an ideal gas with constant specific heats, Eq. 17–7 becomes Temperature
rise during
stagnation 1 qin 305 K
300 K qout 2 1 win wout 2 cp 1 T02 T01 2 g 1 z2 z1 2 (17–8) where T01 and T02 are the stagnation temperatures.
Notice that kinetic energy terms do not explicitly appear in Eqs. 17–7 and
17–8, but the stagnation enthalpy terms account for their contribution. AIR
100 m/s EXAMPLE 17–1 FIGURE 17–5
The temperature of an ideal gas
flowing at a velocity V rises by V 2/2cp
when it is brought to a complete stop. Compression of HighSpeed Air in an Aircraft An aircraft is flying at a cruising speed of 250 m/s at an altitude of 5000 m
where the atmospheric pressure is 54.05 kPa and the ambient air temperature is 255.7 K. The ambient air is first decelerated in a diffuser before it
enters the compressor (Fig. 17–6). Assuming both the diffuser and the compressor to be isentropic, determine (a) the stagnation pressure at the compressor inlet and (b) the required compressor work per unit mass if the
stagnation pressure ratio of the compressor is 8. Solution Highspeed air enters the diffuser and the compressor of an air Diffuser Compressor craft. The stagnation pressure of air and the compressor work input are to be
determined.
Assumptions 1 Both the diffuser and the compressor are isentropic. 2 Air is
an ideal gas with constant specific heats at room temperature.
Properties The constantpressure specific heat cp and the specific heat ratio
k of air at room temperature are (Table A–2a) T1 = 255.7 K cp Aircraft
engine P1 = 54.05 kPa
V1 = 250 m/s 1 01 02 FIGURE 17–6
Schematic for Example 17–1. 1.005 kJ> kg # K and k 1.4 Analysis (a) Under isentropic conditions, the stagnation pressure at the
compressor inlet (diffuser exit) can be determined from Eq. 17–5. However,
first we need to find the stagnation temperature T01 at the compressor inlet.
Under the stated assumptions, T01 can be determined from Eq. 17–4 to be T01 T1 V2
1
2cp 255.7 K 1 250 m> s 2 2 1 kJ> kg b
a
1 2 2 1 1.005 kJ> kg # K 2 1000 m2> s2 286.8 K
Then from Eq. 17–5, P01 T01 k>1k
b
T1
80.77 kPa
P1 a 12 1 54.05 kPa 2 a 286.8 K 1.4>11.4
b
255.7 K 12 That is, the temperature of air would increase by 31.1°C and the pressure by
26.72 kPa as air is decelerated from 250 m/s to zero velocity. These
increases in the temperature and pressure of air are due to the conversion of
the kinetic energy into enthalpy.
(b) To determine the compressor work, we need to know the stagnation temperature of air at the compressor exit T02. The stagnation pressure ratio
across the compressor P02/P01 is specified to be 8. Since the compression
process is assumed to be isentropic, T02 can be determined from the idealgas isentropic relation (Eq. 17–5): T02 T01 a P02 1k
b
P01 12>k 1 286.8 K 2 1 8 2 11.4 12>1.4 519.5 K cen84959_ch17.qxd 4/21/05 11:08 AM Page 827 Chapter 17  827 Disregarding potential energy changes and heat transfer, the compressor
work per unit mass of air is determined from Eq. 17–8: win cp 1 T02 T01 2 1 1.005 kJ> kg # K 2 1 519.5 K 286.8 K 2 233.9 kJ/kg
Thus the work supplied to the compressor is 233.9 kJ/kg.
Discussion Notice that using stagnation properties automatically accounts
for any changes in the kinetic energy of a fluid stream. 17–2 ■ SPEED OF SOUND AND MACH NUMBER An important parameter in the study of compressible flow is the speed of
sound (or the sonic speed), which is the speed at which an infinitesimally
small pressure wave travels through a medium. The pressure wave may be
caused by a small disturbance, which creates a slight rise in local pressure.
To obtain a relation for the speed of sound in a medium, consider a pipe
that is filled with a fluid at rest, as shown in Fig. 17–7. A piston fitted in the
pipe is now moved to the right with a constant incremental velocity dV, creating a sonic wave. The wave front moves to the right through the fluid at
the speed of sound c and separates the moving fluid adjacent to the piston
from the fluid still at rest. The fluid to the left of the wave front experiences
an incremental change in its thermodynamic properties, while the fluid on
the right of the wave front maintains its original thermodynamic properties,
as shown in Fig. 17–7.
To simplify the analysis, consider a control volume that encloses the wave
front and moves with it, as shown in Fig. 17–8. To an observer traveling
with the wave front, the fluid to the right will appear to be moving toward
the wave front with a speed of c and the fluid to the left to be moving away
from the wave front with a speed of c
dV. Of course, the observer will
think the control volume that encloses the wave front (and herself or himself) is stationary, and the observer will be witnessing a steadyflow process.
The mass balance for this singlestream, steadyflow process can be
expressed as
#
mright or 1r rAc dr 2 A 1 c r dV 2 c
2 h h Stationary
P
fluid
r c V dV
0 x P P + dP
P
x FIGURE 17–7
Propagation of a small pressure wave
along a duct. Control volume
traveling with
the wave front dV 2 0 (a) No heat or work crosses the boundaries of the control volume during this
steadyflow process, and the potential energy change, if any, can be
neglected. Then the steadyflow energy balance ein eout becomes
h h + dh
P + dP
r + dr dV #
mleft By canceling the crosssectional (or flow) area A and neglecting the higherorder terms, this equation reduces to
c dr Moving
wave front Piston dh 1c dV 2 2 2 h + dh
P + dP
r + dr c – dV c h
P
r FIGURE 17–8
Control volume moving with the small
pressure wave along a duct. cen84959_ch17.qxd 4/21/05 11:08 AM Page 828 828  Thermodynamics
which yields
dh c dV 0 (b) dV 2. where we have neglected the secondorder term
The amplitude of the
ordinary sonic wave is very small and does not cause any appreciable
change in the pressure and temperature of the fluid. Therefore, the propagation of a sonic wave is not only adiabatic but also very nearly isentropic.
Then the second T ds relation developed in Chapter 7 reduces to
0
T ds ¡ dh dP
r or
dP
r dh (c) Combining Eqs. a, b, and c yields the desired expression for the speed of
sound as
c2 dP
dr at s constant or
a c2 0P
b
0r s (17–9) It is left as an exercise for the reader to show, by using thermodynamic
property relations (see Chap. 12) that Eq. 17–9 can also be written as
c2 ka 0P
b
0r T (17–10) where k is the specific heat ratio of the fluid. Note that the speed of sound
in a fluid is a function of the thermodynamic properties of that fluid.
When the fluid is an ideal gas (P rRT ), the differentiation in Eq. 17–10
can easily be performed to yield
c2
AIR 284 m/s 347 m/s HELIUM
200 K 300 K 832 m/s 1019 m/s 1000 K
634 m/s 1861 m/s FIGURE 17–9
The speed of sound changes with
temperature and varies with the fluid. ka 0P
b
0r T kc 0 1 rRT 2 2kRT or
c 0r d kRT
T (17–11) Noting that the gas constant R has a fixed value for a specified ideal gas and
the specific heat ratio k of an ideal gas is, at most, a function of temperature, we see that the speed of sound in a specified ideal gas is a function of
temperature alone (Fig. 17–9).
A second important parameter in the analysis of compressible fluid flow
is the Mach number Ma, named after the Austrian physicist Ernst Mach
(1838–1916). It is the ratio of the actual velocity of the fluid (or an object in
still air) to the speed of sound in the same fluid at the same state:
Ma V
c (17–12) Note that the Mach number depends on the speed of sound, which depends
on the state of the fluid. Therefore, the Mach number of an aircraft cruising cen84959_ch17.qxd 4/21/05 12:19 PM Page 829 Chapter 17
at constant velocity in still air may be different at different locations
(Fig. 17–10).
Fluid flow regimes are often described in terms of the flow Mach number.
The flow is called sonic when Ma 1, subsonic when Ma 1, supersonic
when Ma 1, hypersonic when Ma
1, and transonic when Ma 1.
EXAMPLE 17–2 Mach Number of Air Entering a Diffuser Air enters a diffuser shown in Fig. 17–11 with a velocity of 200 m/s. Determine (a) the speed of sound and (b) the Mach number at the diffuser inlet
when the air temperature is 30°C. AIR
200 K  829 V = 320 m/s
Ma = 1.13 V = 320 m/s AIR
300 K Ma = 0.92 FIGURE 17–10
The Mach number can be different at
different temperatures even if the
velocity is the same. Solution Air enters a diffuser with a high velocity. The speed of sound and
the Mach number are to be determined at the diffuser inlet.
Assumptions Air at specified conditions behaves as an ideal gas.
Properties The gas constant of air is R
0.287 kJ/kg · K, and its specific
heat ratio at 30°C is 1.4 (Table A–2a).
Analysis We note that the speed of sound in a gas varies with temperature,
which is given to be 30°C. 2kRT B (a) The speed of sound in air at 30°C is determined from Eq. 17–11 to be c 1 1.4 2 1 0.287 kJ> kg # K 2 1 303 K 2 a 1000 m2> s2
1 kJ> kg b 349 m/s (b) Then the Mach number becomes Ma
Discussion 17–3 ■ V
c 200 m> s 349 m> s 0.573 The flow at the diffuser inlet is subsonic since Ma AIR Diffuser V = 200 m/s
T = 30°C 1. FIGURE 17–11
Schematic for Example 17–2. ONEDIMENSIONAL ISENTROPIC FLOW During fluid flow through many devices such as nozzles, diffusers, and turbine blade passages, flow quantities vary primarily in the flow direction
only, and the flow can be approximated as onedimensional isentropic flow
with good accuracy. Therefore, it merits special consideration. Before presenting a formal discussion of onedimensional isentropic flow, we illustrate
some important aspects of it with an example.
EXAMPLE 17–3 Gas Flow through a Converging–Diverging Duct Carbon dioxide flows steadily through a varying crosssectionalarea duct
such as a nozzle shown in Fig. 17–12 at a mass flow rate of 3 kg/s. The carbon dioxide enters the duct at a pressure of 1400 kPa and 200°C with a low
velocity, and it expands in the nozzle to a pressure of 200 kPa. The duct is
designed so that the flow can be approximated as isentropic. Determine the
density, velocity, flow area, and Mach number at each location along the
duct that corresponds to a pressure drop of 200 kPa. Solution Carbon dioxide enters a varying crosssectionalarea duct at specified conditions. The flow properties are to be determined along the duct. Stagnation
region:
1400 kPa
200°C
CO2 ⋅
m 3 kg/s 1400 1000
767
P, kPa FIGURE 17–12
Schematic for Example 17–3. 200 cen84959_ch17.qxd 4/21/05 11:08 AM Page 830 830  Thermodynamics
Assumptions 1 Carbon dioxide is an ideal gas with constant specific heats
at room temperature. 2 Flow through the duct is steady, onedimensional,
and isentropic.
Properties For simplicity we use cp
0.846 kJ/kg · K and k
1.289
throughout the calculations, which are the constantpressure specific heat
and specific heat ratio values of carbon dioxide at room temperature. The
gas constant of carbon dioxide is R
0.1889 kJ/kg K (Table A–2a).
Analysis We note that the inlet temperature is nearly equal to the stagnation temperature since the inlet velocity is small. The flow is isentropic, and
thus the stagnation temperature and pressure throughout the duct remain
constant. Therefore, T0 T1 200°C 473 K P0 P1 1400 kPa and To illustrate the solution procedure, we calculate the desired properties at
the location where the pressure is 1200 kPa, the first location that corresponds to a pressure drop of 200 kPa.
From Eq. 17–5, T0 a T P 1k
b
P0 12>k 22cp 1 T0 1 473 K 2 a 1200 kPa 11.289
b
1400 kPa 12>1.289 457 K From Eq. 17–4, V B T2 2 1 0.846 kJ> kg # K 2 1 473 K 457 K 2 a 1000 m2> s3
1 kJ> kg b 164.5 m/s
From the idealgas relation, r P
RT 1200 kPa
1 0.1889 kPa # m3> kg # K 2 1 457 K 2 13.9 kg/m3 From the mass flow rate relation, A #
m
rV 2kRT 3 kg> s 1 13.9 kg> m3 2 1 164.5 m> s 2 13.1 10 4 m2 13.1 cm2 1000 m2> s2
1 1.289 21 0.1889 kJ> kg # K 21 457 K 2 a
b
1 kJ> kg
B From Eqs. 17–11 and 17–12, c
Ma V
c 164.5 m> s 333.6 m> s 333.6 m> s 0.493 The results for the other pressure steps are summarized in Table 17–1 and
are plotted in Fig. 17–13.
Discussion Note that as the pressure decreases, the temperature and speed
of sound decrease while the fluid velocity and Mach number increase in the
flow direction. The density decreases slowly at first and rapidly later as the
fluid velocity increases. cen84959_ch17.qxd 4/21/05 11:08 AM Page 831 Chapter 17  TABLE 17–1
Variation of fluid properties in flow direction in duct described in
.
Example 17–3 for m 3 kg/s constant
P, kPa T, K V, m/s r, kg/m3 c, m/s A, cm2 Ma 1400
1200
1000
800
767*
600
400
200 473
457
439
417
413
391
357
306 0
164.5
240.7
306.6
317.2
371.4
441.9
530.9 15.7
13.9
12.1
10.1
9.82
8.12
5.93
3.46 339.4
333.6
326.9
318.8
317.2
308.7
295.0
272.9 ∞
13.1
10.3
9.64
9.63
10.0
11.5
16.3 0
0.493
0.736
0.962
1.000
1.203
1.498
1.946 * 767 kPa is the critical pressure where the local Mach number is unity. Flow direction
A, Ma, r, T, V A T
r Ma V
1400 1200 1000 800
P, kPa 600 400 200 We note from Example 17–3 that the flow area decreases with decreasing
pressure up to a criticalpressure value where the Mach number is unity, and
then it begins to increase with further reductions in pressure. The Mach
number is unity at the location of smallest flow area, called the throat (Fig.
17–14). Note that the velocity of the fluid keeps increasing after passing the
throat although the flow area increases rapidly in that region. This increase
in velocity past the throat is due to the rapid decrease in the fluid density.
The flow area of the duct considered in this example first decreases and
then increases. Such ducts are called converging–diverging nozzles. These
nozzles are used to accelerate gases to supersonic speeds and should not be
confused with Venturi nozzles, which are used strictly for incompressible
flow. The first use of such a nozzle occurred in 1893 in a steam turbine FIGURE 17–13
Variation of normalized fluid
properties and crosssectional area
along a duct as the pressure drops
from 1400 to 200 kPa. 831 cen84959_ch17.qxd 4/21/05 11:08 AM Page 832 832  Thermodynamics
Throat designed by a Swedish engineer, Carl G. B. de Laval (1845–1913), and
therefore converging–diverging nozzles are often called Laval nozzles. Fluid Variation of Fluid Velocity with Flow Area
Converging nozzle
Throat Fluid Converging–diverging nozzle FIGURE 17–14
The cross section of a nozzle at the
smallest flow area is called the throat. It is clear from Example 17–3 that the couplings among the velocity, density, and flow areas for isentropic duct flow are rather complex. In the
remainder of this section we investigate these couplings more thoroughly,
and we develop relations for the variation of statictostagnation property
ratios with the Mach number for pressure, temperature, and density.
We begin our investigation by seeking relationships among the pressure,
temperature, density, velocity, flow area, and Mach number for onedimensional isentropic flow. Consider the mass balance for a steadyflow
process:
#
m rAV constant Differentiating and dividing the resultant equation by the mass flow rate, we
obtain
dr
r dA
A dV
V 0 (17–13) Neglecting the potential energy, the energy balance for an isentropic flow with
no work interactions can be expressed in the differential form as (Fig. 17–15)
dP
r CONSERVATION OF ENERGY
(steady flow, w = 0, q = 0, ∆pe = 0)
h1 + 2
V1
V2
= h2 + 2
2
2 V dV 0 (17–14) This relation is also the differential form of Bernoulli’s equation when
changes in potential energy are negligible, which is a form of the conservation of momentum principle for steadyflow control volumes. Combining
Eqs. 17–13 and 17–14 gives
dA
A dP 1
a
r V2 dr
b
dP (17–15) or V2
= constant
2
Differentiate,
h+ dh + V dV = 0
Also, 0 (isentropic) T ds = dh – v dP
1
dh = v dP = r dP
Substitute,
dP
r + V dV = 0 FIGURE 17–15
Derivation of the differential form of
the energy equation for steady
isentropic flow. Rearranging Eq. 17–9 as (∂r/∂P)s
dA
A 1/c2 and substituting into Eq. 17–15 yield
dP
11
rV 2 Ma2 2 (17–16) This is an important relation for isentropic flow in ducts since it describes
the variation of pressure with flow area. We note that A, r, and V are positive
quantities. For subsonic flow (Ma
1), the term 1
Ma2 is positive; and
thus dA and dP must have the same sign. That is, the pressure of the fluid
must increase as the flow area of the duct increases and must decrease as the
flow area of the duct decreases. Thus, at subsonic velocities, the pressure
decreases in converging ducts (subsonic nozzles) and increases in diverging
ducts (subsonic diffusers).
In supersonic flow (Ma 1), the term 1 Ma2 is negative, and thus dA
and dP must have opposite signs. That is, the pressure of the fluid must cen84959_ch17.qxd 4/21/05 11:08 AM Page 833 Chapter 17  833 increase as the flow area of the duct decreases and must decrease as the
flow area of the duct increases. Thus, at supersonic velocities, the pressure
decreases in diverging ducts (supersonic nozzles) and increases in converging ducts (supersonic diffusers).
Another important relation for the isentropic flow of a fluid is obtained by
substituting rV
dP/dV from Eq. 17–14 into Eq. 17–16:
dA
A dV
11
V Ma2 2 (17–17) This equation governs the shape of a nozzle or a diffuser in subsonic or
supersonic isentropic flow. Noting that A and V are positive quantities, we
conclude the following:
For subsonic flow 1 Ma 6 1 2 , dA
60
dV For supersonic flow 1 Ma 7 1 2 , dA
70
dV For sonic flow 1 Ma dA
dV 12, 0 Thus the proper shape of a nozzle depends on the highest velocity desired
relative to the sonic velocity. To accelerate a fluid, we must use a converging nozzle at subsonic velocities and a diverging nozzle at supersonic velocities. The velocities encountered in most familiar applications are well
below the sonic velocity, and thus it is natural that we visualize a nozzle as
a converging duct. However, the highest velocity we can achieve by a converging nozzle is the sonic velocity, which occurs at the exit of the nozzle.
If we extend the converging nozzle by further decreasing the flow area, in
hopes of accelerating the fluid to supersonic velocities, as shown in
Fig. 17–16, we are up for disappointment. Now the sonic velocity will occur
at the exit of the converging extension, instead of the exit of the original
nozzle, and the mass flow rate through the nozzle will decrease because of
the reduced exit area.
Based on Eq. 17–16, which is an expression of the conservation of mass
and energy principles, we must add a diverging section to a converging nozzle to accelerate a fluid to supersonic velocities. The result is a converging–
diverging nozzle. The fluid first passes through a subsonic (converging) section, where the Mach number increases as the flow area of the nozzle
decreases, and then reaches the value of unity at the nozzle throat. The fluid
continues to accelerate as it passes through a supersonic (diverging) section.
.
Noting that m rAV for steady flow, we see that the large decrease in density makes acceleration in the diverging section possible. An example of this
type of flow is the flow of hot combustion gases through a nozzle in a gas
turbine.
The opposite process occurs in the engine inlet of a supersonic aircraft.
The fluid is decelerated by passing it first through a supersonic diffuser,
which has a flow area that decreases in the flow direction. Ideally, the flow
reaches a Mach number of unity at the diffuser throat. The fluid is further A
P 0, T0 MaA = 1
(sonic) Converging
nozzle MaA < 1
A
B
P 0, T0 Converging
nozzle Ma B = 1
(sonic)
Attachment FIGURE 17–16
We cannot obtain supersonic velocities
by attaching a converging section to a
converging nozzle. Doing so will only
move the sonic cross section farther
downstream and decrease the mass
flow rate. cen84959_ch17.qxd 4/21/05 11:08 AM Page 834 834  Thermodynamics Ma <1 P decreases
V increases
Ma increases
T decreases
r decreases P increases
V decreases
Ma decreases
T increases
r increases <1 Ma Subsonic diffuser Subsonic nozzle
(a ) Subsonic flow Ma FIGURE 17–17
Variation of flow properties in
subsonic and supersonic nozzles and
diffusers. >1 P decreases
V increases
Ma increases
T decreases
r decreases P increases
V decreases
Ma decreases
T increases
r increases >1 Ma Supersonic nozzle Supersonic diffuser
(b ) Supersonic flow decelerated in a subsonic diffuser, which has a flow area that increases in
the flow direction, as shown in Fig. 17–17. Property Relations for Isentropic Flow
of Ideal Gases
Next we develop relations between the static properties and stagnation properties of an ideal gas in terms of the specific heat ratio k and the Mach number
Ma. We assume the flow is isentropic and the gas has constant specific heats.
The temperature T of an ideal gas anywhere in the flow is related to the
stagnation temperature T0 through Eq. 17–4:
T0 T V2
2cp T0
T 1 V2
2cpT or Noting that cp kR/(k V2
2cpT 1), c2 kRT, and Ma V2
2 3 kR> 1 k 1 2 4 T a k 1 V2
b2
2
c V/c, we see that
a k 1
2 b Ma 2 Substituting yields
T0
T 1 a k 1
2 b Ma2 which is the desired relation between T0 and T. (17–18) cen84959_ch17.qxd 4/26/05 4:17 PM Page 835 Chapter 17
The ratio of the stagnation to static pressure is obtained by substituting
Eq. 17–18 into Eq. 17–5:
P0
P c1 a k 1
2 b Ma2 d k>1k 12 (17–19)  835 Throat
Tt
Pt
rt Subsonic
nozzle T*
P*
r*
(if Ma t = 1) The ratio of the stagnation to static density is obtained by substituting
Eq. 17–18 into Eq. 17–6:
r0
r c1 a k 1
2 b Ma2 d 1>1k 12 Throat
(17–20) Numerical values of T/T0, P/P0, and r/r0 are listed versus the Mach number
in Table A–32 for k 1.4, which are very useful for practical compressible
flow calculations involving air.
The properties of a fluid at a location where the Mach number is unity (the
throat) are called critical properties, and the ratios in Eqs. (17–18) through
(17–20) are called critical ratios (Fig. 17–18). It is common practice in the
analysis of compressible flow to let the superscript asterisk (*) represent the
critical values. Setting Ma 1 in Eqs. 17–18 through 17–20 yields
T*
T0 2
k 1 P*
P0 a 2 r*
r0 a (17–21) 1 k b 1 b 2
k k>1k 12 (17–22) 1>1k 12 (17–23) These ratios are evaluated for various values of k and are listed in Table
17–2. The critical properties of compressible flow should not be confused
with the properties of substances at the critical point (such as the critical
temperature Tc and critical pressure Pc). TABLE 17–2
The criticalpressure, criticaltemperature, and criticaldensity ratios for
isentropic flow of some ideal gases
Superheated
steam,
k 1.3
P*
P0
T*
T0
r*
r0 Hot products
of combustion,
k 1.33 Air,
k 1.4 Monatomic
gases,
k 1.667 0.5457 0.5404 0.5283 0.4871 0.8696 0.8584 0.8333 0.7499 0.6276 0.6295 0.6340 0.6495 To
Po
ro Supersonic
nozzle
T *, P*, r*
(Mat = 1) FIGURE 17–18
When Mat 1, the properties at the
nozzle throat become the critical
properties. cen84959_ch17.qxd 4/21/05 11:08 AM Page 836 836  Thermodynamics
EXAMPLE 17–4 P 0 = 1.4 MPa Calculate the critical pressure and temperature of carbon dioxide for the flow
conditions described in Example 17–3 (Fig. 17–19). CO2 T0 = 473 K Critical Temperature and Pressure in Gas Flow Solution For the flow discussed in Example 17–3, the critical pressure and
temperature are to be calculated.
Assumptions 1 The flow is steady, adiabatic, and onedimensional. 2 Carbon
dioxide is an ideal gas with constant specific heats.
Properties The specific heat ratio of carbon dioxide at room temperature is
k
1.289 (Table A–2a).
Analysis The ratios of critical to stagnation temperature and pressure are
determined to be P*
T* FIGURE 17–19
Schematic for Example 17–4. 2 T*
T0 k P*
P0 a 2
1.289 1
2
k 1 b k>1k 12 1
a 0.8737
2
1.289 1 b 1.289>11.289 12 0.5477 Noting that the stagnation temperature and pressure are, from Example
17–3, T0
473 K and P0
1400 kPa, we see that the critical temperature
and pressure in this case are T* 0.8737T0 P* 0.5477P0 1 0.8737 2 1 473 K 2 1 0.5477 2 1 1400 kPa 2 413 K
767 kPa Discussion Note that these values agree with those listed in Table 17–1, as
expected. Also, property values other than these at the throat would indicate
that the flow is not critical, and the Mach number is not unity.
Reservoir
Pr = P 0
Tr = T0 Pe
Pb
(Back
pressure) Vr = 0 17–4 x
P /P0 1 1
2 P*
P0 0 3
Lowest exit
pressure 4
5 Pb = P0
Pb > P* ■ ISENTROPIC FLOW THROUGH NOZZLES Converging or converging–diverging nozzles are found in many engineering
applications including steam and gas turbines, aircraft and spacecraft
propulsion systems, and even industrial blasting nozzles and torch nozzles.
In this section we consider the effects of back pressure (i.e., the pressure
applied at the nozzle discharge region) on the exit velocity, the mass flow
rate, and the pressure distribution along the nozzle. Pb = P* Converging Nozzles Pb < P* Consider the subsonic flow through a converging nozzle as shown in
Fig. 17–20. The nozzle inlet is attached to a reservoir at pressure Pr and
temperature Tr. The reservoir is sufficiently large so that the nozzle inlet
velocity is negligible. Since the fluid velocity in the reservoir is zero and
the flow through the nozzle is approximated as isentropic, the stagnation
pressure and stagnation temperature of the fluid at any cross section
through the nozzle are equal to the reservoir pressure and temperature,
respectively. Pb = 0 FIGURE 17–20
The effect of back pressure on the
pressure distribution along a
converging nozzle. x cen84959_ch17.qxd 4/21/05 11:08 AM Page 837 Chapter 17  837 Now we begin to reduce the back pressure and observe the resulting
effects on the pressure distribution along the length of the nozzle, as shown
in Fig. 17–20. If the back pressure Pb is equal to P1, which is equal to Pr,
there is no flow and the pressure distribution is uniform along the nozzle.
When the back pressure is reduced to P2, the exit plane pressure Pe also
drops to P2. This causes the pressure along the nozzle to decrease in the
flow direction.
When the back pressure is reduced to P3 ( P*, which is the pressure
required to increase the fluid velocity to the speed of sound at the exit plane
or throat), the mass flow reaches a maximum value and the flow is said to
be choked. Further reduction of the back pressure to level P4 or below does
not result in additional changes in the pressure distribution, or anything else
along the nozzle length.
Under steadyflow conditions, the mass flow rate through the nozzle is
constant and can be expressed as
#
m rAV a P
b A 1 Ma 2kRT 2
RT AMaP0 2k> 1 RT0 2 PAMa k
B RT Solving for T from Eq. 17–18 and for P from Eq. 17–19 and substituting,
#
m 31 1k 1 2 Ma2> 2 4 1k 12> 321k 124 (17–24) Thus the mass flow rate of a particular fluid through a nozzle is a function
of the stagnation properties of the fluid, the flow area, and the Mach num.
ber. Equation 17–24 is valid at any cross section, and thus m can be evaluated at any location along the length of the nozzle.
For a specified flow area A and stagnation properties T0 and P0, the maximum mass flow rate can be determined by differentiating Eq. 17–24 with
respect to Ma and setting the result equal to zero. It yields Ma
1. Since
the only location in a nozzle where the Mach number can be unity is the
location of minimum flow area (the throat), the mass flow rate through a
nozzle is a maximum when Ma 1 at the throat. Denoting this area by A*,
we obtain an expression for the maximum mass flow rate by substituting
Ma 1 in Eq. 17–24:
#
mmax A*P0 1k
k
2
a
b
B RT0 k 1 12> 321k 124 ⋅
m
⋅
mmax 5 4 3 2 1
P*
P0 1.0 Pe / P0 Pb
P0 1 1.0
2 (17–25) Thus, for a particular ideal gas, the maximum mass flow rate through a
nozzle with a given throat area is fixed by the stagnation pressure and temperature of the inlet flow. The flow rate can be controlled by changing
the stagnation pressure or temperature, and thus a converging nozzle can be
used as a flowmeter. The flow rate can also be controlled, of course, by
varying the throat area. This principle is vitally important for chemical
processes, medical devices, flowmeters, and anywhere the mass flux of a
gas must be known and controlled.
.
A plot of m versus Pb /P0 for a converging nozzle is shown in Fig. 17–21.
Notice that the mass flow rate increases with decreasing Pb /P0, reaches a
P*, and remains constant for Pb /P0 values less than this
maximum at Pb P* 5
P0 0 4 3 P*
P0 1.0 Pb
P0 FIGURE 17–21
The effect of back pressure Pb on the
#
mass flow rate m and the exit pressure
Pe of a converging nozzle. cen84959_ch17.qxd 4/21/05 11:08 AM Page 838 838  Thermodynamics
critical ratio. Also illustrated on this figure is the effect of back pressure on
the nozzle exit pressure Pe. We observe that
Pe e Pb
P* for Pb P*
for Pb 6 P* To summarize, for all back pressures lower than the critical pressure P*,
the pressure at the exit plane of the converging nozzle Pe is equal to P*, the
Mach number at the exit plane is unity, and the mass flow rate is the maximum (or choked) flow rate. Because the velocity of the flow is sonic at the
throat for the maximum flow rate, a back pressure lower than the critical
pressure cannot be sensed in the nozzle upstream flow and does not affect
the flow rate.
The effects of the stagnation temperature T0 and stagnation pressure P0 on
the mass flow rate through a converging nozzle are illustrated in Fig. 17–22
where the mass flow rate is plotted against the statictostagnation pressure
ratio at the throat Pt /P0. An increase in P0 (or a decrease in T0) will increase
the mass flow rate through the converging nozzle; a decrease in P0 (or an
increase in T0) will decrease it. We could also conclude this by carefully
observing Eqs. 17–24 and 17–25.
A relation for the variation of flow area A through the nozzle relative to
throat area A* can be obtained by combining Eqs. 17–24 and 17–25 for the
same mass flow rate and stagnation properties of a particular fluid. This
yields
1
2
ca
b a1
Ma k 1 A
A* k 1
2 Ma2 b d 1k 12> 321k 124 (17–26) Table A–32 gives values of A/A* as a function of the Mach number for air
(k
1.4). There is one value of A/A* for each value of the Mach number,
but there are two possible values of the Mach number for each value of
A/A*—one for subsonic flow and another for supersonic flow. Mat = 1 Mat <1 ⋅
m
Increase in P0 ,
decrease in T0 ,
or both
P0, T0
Decrease in P0 ,
increase in T0 ,
or both FIGURE 17–22
The variation of the mass flow rate
through a nozzle with inlet stagnation
properties. 0 P*
P0 1.0 Pt
P0 cen84959_ch17.qxd 4/21/05 11:08 AM Page 839 Chapter 17  839 Another parameter sometimes used in the analysis of onedimensional
isentropic flow of ideal gases is Ma*, which is the ratio of the local velocity
to the speed of sound at the throat:
Ma* Vc
c c* Ma 2kRT (17–27) T
Ma
B T* 2kRT * It can also be expressed as
Ma* V
c* Mac
c* Ma where Ma is the local Mach number, T is the local temperature, and T* is
the critical temperature. Solving for T from Eq. 17–18 and for T* from
Eq. 17–21 and substituting, we get
Ma* Ma B2 k
1k 1
1 2 Ma2 (17–28) Values of Ma* are also listed in Table A–32 versus the Mach number for
k
1.4 (Fig. 17–23). Note that the parameter Ma* differs from the Mach
number Ma in that Ma* is the local velocity nondimensionalized with
respect to the sonic velocity at the throat, whereas Ma is the local velocity
nondimensionalized with respect to the local sonic velocity. (Recall that the
sonic velocity in a nozzle varies with temperature and thus with location.) EXAMPLE 17–5 Solution Air enters a converging nozzle. The mass flow rate of air through
the nozzle is to be determined for different back pressures.
Assumptions 1 Air is an ideal gas with constant specific heats at room
temperature. 2 Flow through the nozzle is steady, onedimensional, and
isentropic.
Properties The constantpressure specific heat and the specific heat ratio of
air are cp
1.005 kJ/kg K and k
1.4, respectively (Table A–2a).
Analysis We use the subscripts i and t to represent the properties at the
nozzle inlet and the throat, respectively. The stagnation temperature and
pressure at the nozzle inlet are determined from Eqs. 17–4 and 17–5: Ti P0i Pi a V2
i
2cp 873 K T0 i k>1k
b
Ti 12 1 150 m> s 2 2 2 1 1.005 kJ> kg # K 2 1 1 MPa 2 a 884 K 1.4>11.4
b
873 K a 1 kJ> kg 1000 m2> s2 12 b 884 K 1.045 MPa These stagnation temperature and pressure values remain constant throughout the nozzle since the flow is assumed to be isentropic. That is, T0 .
.
. .
.
. 0.90
1.00
1.10
.
.
. 0.9146
1.0000
1.0812
.
.
. A
A*
.
.
.
1.0089
1.0000
1.0079
.
.
. r
r0
.
.
.
.
.
.
.
0.5913 .
.
.
0.5283 .
.
.
0.4684 .
.
.
.
.
.
.
.
.
P
P0 T
T0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. FIGURE 17–23
Various property ratios for isentropic
flow through nozzles and diffusers are
listed in Table A–32 for k 1.4 for
convenience. Effect of Back Pressure on Mass Flow Rate Air at 1 MPa and 600°C enters a converging nozzle, shown in Fig. 17–24,
with a velocity of 150 m/s. Determine the mass flow rate through the nozzle
for a nozzle throat area of 50 cm2 when the back pressure is (a) 0.7 MPa
and (b) 0.4 MPa. T0i Ma* T0i 884 K and P0 P0i 1.045 MPa AIR
Pi = 1 MPa
Ti = 600°C
Vi = 150 m/s Converging
nozzle Pb
A t = 50 cm2 FIGURE 17–24
Schematic for Example 17–5. cen84959_ch17.qxd 4/21/05 11:08 AM Page 840 840  Thermodynamics
The criticalpressure ratio is determined from Table 17–2 (or Eq. 17–22) to
be P*/P0
0.5283.
(a) The back pressure ratio for this case is Pb
P0 0.7 MPa
1.045 MPa 0.670 which is greater than the criticalpressure ratio, 0.5283. Thus the exit plane
pressure (or throat pressure Pt) is equal to the back pressure in this case. That
is, Pt
Pb
0.7 MPa, and Pt /P0
0.670. Therefore, the flow is not choked.
From Table A–32 at Pt /P0 0.670, we read Mat 0.778 and Tt /T0 0.892.
The mass flow rate through the nozzle can be calculated from Eq. 17–24.
But it can also be determined in a stepbystep manner as follows: Tt 0.892T0 rt Pt
RTt Vt Mat ct 0.892 1 884 K 2 Mat 2kRTt 788.5 K 700 kPa 1 0.287 kPa # m3> kg # K 2 1 788.5 K 2 1 0.778 2 B 437.9 m> s 3.093 kg> m3 1 1.4 2 1 0.287 kJ> kg # K 2 1 788.5 K 2 a 1000 m2> s2
1 kJ> kg b Thus, #
m r t A tVt 1 3.093 kg> m3 2 1 50 10 4 m2 2 1 437.9 m> s 2 6.77 kg/s (b) The back pressure ratio for this case is Pb
P0 0.4 MPa
1.045 MPa 0.383 which is less than the criticalpressure ratio, 0.5283. Therefore, sonic conditions exist at the exit plane (throat) of the nozzle, and Ma
1. The flow is
choked in this case, and the mass flow rate through the nozzle can be calculated from Eq. 17–25: #
m A*P0
1 50 1k
2
k
a
b
B RT0 k 1
4 12>321k 124 m2 2 1 1045 kPa 2 2.4>0.8
1.4
2
b
a
B 1 0.287 kJ> kg # K 2 1 884 K 2 1.4 1 since kPa # m2> 2kJ> kg
21000 kg> s.
Discussion This is the maximum mass flow rate through the nozzle for the
specified inlet conditions and nozzle throat area. 10 7.10 kg/s EXAMPLE 17–6 Gas Flow through a Converging Nozzle Nitrogen enters a duct with varying flow area at T1
400 K, P1
100 kPa,
and Ma1
0.3. Assuming steady isentropic flow, determine T2, P2, and Ma2
at a location where the flow area has been reduced by 20 percent. cen84959_ch17.qxd 4/21/05 11:08 AM Page 841 Chapter 17
Solution Nitrogen gas enters a converging nozzle. The properties at the
nozzle exit are to be determined.
Assumptions 1 Nitrogen is an ideal gas with k
1.4. 2 Flow through the
nozzle is steady, onedimensional, and isentropic.
Analysis The schematic of the duct is shown in Fig. 17–25. For isentropic
flow through a duct, the area ratio A/A* (the flow area over the area of the
throat where Ma
1) is also listed in Table A–32. At the initial Mach
number of Ma1
0.3, we read A1
A* T1
T0 2.0351 With a 20 percent reduction in flow area, A2 A2
A* P1
P0 0.9823 0.8A1, and 1 0.8 2 1 2.0351 2 A2 A1
A1 A* 0.9395 1.6281 For this value of A2/A* from Table A–32, we read T2
T0 P2
P0 0.9701 0.8993 Ma2 0.391 Here we chose the subsonic Mach number for the calculated A2/A* instead
of the supersonic one because the duct is converging in the flow direction
and the initial flow is subsonic. Since the stagnation properties are constant
for isentropic flow, we can write T2
T1
P2
P1 T2> T0
T1> T0 S T2 P2> P0
P1> P0 S P2 T1 a T2> T0
T1> T0 P1 a b P2> P0 P1> P0 1 400 K 2 a
b 0.9701
b
0.9823 1 100 kPa 2 a 395 K 0.8993
b
0.9395 95.7 kPa which are the temperature and pressure at the desired location.
Discussion Note that the temperature and pressure drop as the fluid accelerates in a converging nozzle. Converging–Diverging Nozzles
When we think of nozzles, we ordinarily think of flow passages whose
crosssectional area decreases in the flow direction. However, the highest
velocity to which a fluid can be accelerated in a converging nozzle is limited
to the sonic velocity (Ma 1), which occurs at the exit plane (throat) of the
nozzle. Accelerating a fluid to supersonic velocities (Ma 1) can be accomplished only by attaching a diverging flow section to the subsonic nozzle at
the throat. The resulting combined flow section is a converging–diverging
nozzle, which is standard equipment in supersonic aircraft and rocket propulsion (Fig. 17–26).
Forcing a fluid through a converging–diverging nozzle is no guarantee
that the fluid will be accelerated to a supersonic velocity. In fact, the fluid
may find itself decelerating in the diverging section instead of accelerating
if the back pressure is not in the right range. The state of the nozzle flow is
determined by the overall pressure ratio Pb /P0. Therefore, for given inlet
conditions, the flow through a converging–diverging nozzle is governed by
the back pressure Pb, as will be explained. T 1 = 400 K
P1 = 100 kPa
Ma 1 = 0.3
A1 N2
Nozzle  841 T2
P2
Ma2
A2 = 0.8A1 FIGURE 17–25
Schematic for Example 17–6 (not to
scale). cen84959_ch17.qxd 4/26/05 4:17 PM Page 842 842  Thermodynamics FIGURE 17–26
Converging–diverging nozzles are commonly used in rocket engines to provide high thrust.
Courtesy of Pratt and Whitney, www.prattwhitney.com/how.htm. Used by permission. Consider the converging–diverging nozzle shown in Fig. 17–27. A fluid
enters the nozzle with a low velocity at stagnation pressure P0. When Pb
P0 (case A), there will be no flow through the nozzle. This is expected since
the flow in a nozzle is driven by the pressure difference between the nozzle
inlet and the exit. Now let us examine what happens as the back pressure is
lowered.
1. When P0 Pb PC, the flow remains subsonic throughout the nozzle,
and the mass flow is less than that for choked flow. The fluid velocity
increases in the first (converging) section and reaches a maximum at
the throat (but Ma
1). However, most of the gain in velocity is lost
in the second (diverging) section of the nozzle, which acts as a diffuser. The pressure decreases in the converging section, reaches a
minimum at the throat, and increases at the expense of velocity in the
diverging section.
2. When Pb
PC, the throat pressure becomes P* and the fluid achieves
sonic velocity at the throat. But the diverging section of the nozzle still
acts as a diffuser, slowing the fluid to subsonic velocities. The mass
flow rate that was increasing with decreasing Pb also reaches its maximum value.
Recall that P* is the lowest pressure that can be obtained at the
throat, and the sonic velocity is the highest velocity that can be
achieved with a converging nozzle. Thus, lowering Pb further has no
influence on the fluid flow in the converging part of the nozzle or the cen84959_ch17.qxd 4/21/05 11:08 AM Page 843 Chapter 17
Throat  843 Pe P0 Vi ≅ 0 Pb x P Pb
AP
BA
PB
C
PC
D
PD P0 P*
Sonic flow
at throat PE
Shock
in nozzle 0
Inlet Throat
Shock
in nozzle Ma
Sonic flow
at throat
1 E, F
,G ,G
E, F D }
} } Throat B
A
Exit }
} Subsonic flow
at nozzle exit
(shock in nozzle)
Supersonic flow
at nozzle exit
(no shock in nozzle)
x Exit C 0
Inlet PF
PG } Subsonic flow
at nozzle exit
(no shock) Supersonic flow
at nozzle exit
(no shock in nozzle) Subsonic flow
at nozzle exit
(shock in nozzle)
Subsonic flow
at nozzle exit
(no shock)
x mass flow rate through the nozzle. However, it does influence the character of the flow in the diverging section.
3. When PC
Pb
PE, the fluid that achieved a sonic velocity at the
throat continues accelerating to supersonic velocities in the diverging
section as the pressure decreases. This acceleration comes to a sudden
stop, however, as a normal shock develops at a section between the
throat and the exit plane, which causes a sudden drop in velocity to subsonic levels and a sudden increase in pressure. The fluid then continues
to decelerate further in the remaining part of the converging–diverging
nozzle. Flow through the shock is highly irreversible, and thus it cannot
be approximated as isentropic. The normal shock moves downstream
away from the throat as Pb is decreased, and it approaches the nozzle
exit plane as Pb approaches PE.
When Pb PE, the normal shock forms at the exit plane of the nozzle. The flow is supersonic through the entire diverging section in this
case, and it can be approximated as isentropic. However, the fluid
velocity drops to subsonic levels just before leaving the nozzle as it FIGURE 17–27
The effects of back pressure on the
flow through a converging–diverging
nozzle. cen84959_ch17.qxd 4/21/05 11:08 AM Page 844 844  Thermodynamics
crosses the normal shock. Normal shock waves are discussed in
Section 17–5.
4. When PE
Pb
0, the flow in the diverging section is supersonic,
and the fluid expands to PF at the nozzle exit with no normal shock
forming within the nozzle. Thus, the flow through the nozzle can be
PF, no shocks occur within
approximated as isentropic. When Pb
PF, irreversible mixing and expanor outside the nozzle. When Pb
sion waves occur downstream of the exit plane of the nozzle. When
PF, however, the pressure of the fluid increases from PF to Pb
Pb
irreversibly in the wake of the nozzle exit, creating what are called
oblique shocks. EXAMPLE 17–7
T0 = 800 K
P0 = 1.0 MPa
Vi ≅ 0
Mae = 2
A t = 20 cm 2 FIGURE 17–28
Schematic for Example 17–7. Airflow through a Converging–Diverging Nozzle Air enters a converging–diverging nozzle, shown in Fig. 17–28, at 1.0 MPa
and 800 K with a negligible velocity. The flow is steady, onedimensional, and
isentropic with k
1.4. For an exit Mach number of Ma
2 and a throat
area of 20 cm2, determine (a) the throat conditions, (b) the exit plane conditions, including the exit area, and (c) the mass flow rate through the nozzle. Solution Air flows through a converging–diverging nozzle. The throat and
the exit conditions and the mass flow rate are to be determined.
Assumptions 1 Air is an ideal gas with constant specific heats at room
temperature. 2 Flow through the nozzle is steady, onedimensional, and
isentropic.
Properties The specific heat ratio of air is given to be k
1.4. The gas
constant of air is 0.287 kJ/kg K.
Analysis The exit Mach number is given to be 2. Therefore, the flow must
be sonic at the throat and supersonic in the diverging section of the
nozzle. Since the inlet velocity is negligible, the stagnation pressure and
stagnation temperature are the same as the inlet temperature and pressure,
P0 1.0 MPa and T0
800 K. The stagnation density is r0 P0
RT0 1000 kPa
1 0.287 kPa # m3> kg # K 2 1 800 K 2 (a) At the throat of the nozzle Ma P*
P0 0.5283 Thus, P* 0.8333T0 r*
Also, V* c* 2kRT * 0.6339r 0 517.5 m/s 1, and from Table A–32 we read 0.8333 1 0.5283 2 1 1.0 MPa 2 0.5283P0 T* T*
T0 1 0.8333 2 1 800 K 2
B 4.355 kg> m3 r*
r0 0.6339 0.5283 MPa
666.6 K 1 0.6339 2 1 4.355 kg> m3 2 2.761 kg/m3 1 1.4 2 1 0.287 kJ> kg # K 2 1 666.6 K 2 a 1000 m2> s2
1 kJ> kg b cen84959_ch17.qxd 4/21/05 11:08 AM Page 845 Chapter 17
(b) Since the flow is isentropic, the properties at the exit plane can also be
calculated by using data from Table A–32. For Ma
2 we read Pe
P0 Te
T0 0.1278 re
r0 0.5556 Thus, Pe 0.1278P0 Te 0.5556T0 re 0.2300r 0 Ae 1.6875A* 1 0.1278 2 1 10 MPa 2 1 0.5556 2 1 800 K 2 Mae 2kRTe 0.1278 MPa 33.75 cm2 1 1.6330 2 1 517.5 m> s 2 B 1.002 kg/m3 845.1 m/s The nozzle exit velocity could also be determined from Ve
is the speed of sound at the exit conditions: Maece Ve 2 1.6875 444.5 K 1 1.6875 2 1 20 cm2 2 Mae*c* Ae
A* 1.6330 1 0.2300 2 1 4.355 kg> m3 2 and Ve Ma*
t 0.2300 Maece, where ce 1 1.4 2 1 0.287 kJ> kg # K 2 1 444.5 K 2 a 845.2 m> s 1000 m2> s2
1 kJ> kg b (c) Since the flow is steady, the mass flow rate of the fluid is the same at all
sections of the nozzle. Thus it may be calculated by using properties at any
cross section of the nozzle. Using the properties at the throat, we find that
the mass flow rate is #
m r*A*V* 1 2.761 kg> m3 2 1 20 10 4 m2 2 1 517.5 m> s 2 2.86 kg/s Discussion Note that this is the highest possible mass flow rate that can
flow through this nozzle for the specified inlet conditions. 17–5 ■ SHOCK WAVES AND EXPANSION WAVES We have seen that sound waves are caused by infinitesimally small pressure
disturbances, and they travel through a medium at the speed of sound. We
have also seen that for some back pressure values, abrupt changes in fluid
properties occur in a very thin section of a converging–diverging nozzle
under supersonic flow conditions, creating a shock wave. It is of interest to
study the conditions under which shock waves develop and how they affect
the flow. Normal Shocks
First we consider shock waves that occur in a plane normal to the direction
of flow, called normal shock waves. The flow process through the shock
wave is highly irreversible and cannot be approximated as being isentropic.
Next we follow the footsteps of Pierre Lapace (1749–1827), G. F. Bernhard Riemann (1826–1866), William Rankine (1820–1872), Pierre Henry
Hugoniot (1851–1887), Lord Rayleigh (1842–1919), and G. I. Taylor  845 cen84959_ch17.qxd 4/21/05 11:08 AM Page 846 846  Thermodynamics
Control
volume Ma1 > 1
Flow V1
P1
h1
r1
s1 V2
P2
h2
r2
s2 Ma2 < 1 Shock wave FIGURE 17–29
Control volume for flow across a
normal shock wave. (1886–1975) and develop relationships for the flow properties before and
after the shock. We do this by applying the conservation of mass, momentum, and energy relations as well as some property relations to a stationary
control volume that contains the shock, as shown in Fig. 17–29. The normal
shock waves are extremely thin, so the entrance and exit flow areas for the
control volume are approximately equal (Fig 17–30).
We assume steady flow with no heat and work interactions and no
potential energy changes. Denoting the properties upstream of the shock
by the subscript 1 and those downstream of the shock by 2, we have the
following:
r1AV1 Conservation of mass: r2AV2 (17–29) or
r1V1 r2V2 h1 V2
1
2 h2 h01 Conservation of energy: V2
2
2 h02 (17–30) or
(17–31) Conservation of momentum: Rearranging Eq. 17–14 and integrating yield
A 1 P1
Increase of entropy: FIGURE 17–30
Schlieren image of a normal shock in
a Laval nozzle. The Mach number in
the nozzle just upstream (to the left) of
the shock wave is about 1.3. Boundary
layers distort the shape of the normal
shock near the walls and lead to flow
separation beneath the shock.
Photo by G. S. Settles, Penn State University. Used
by permission. P2 2
s2 #
m 1 V2
s1 0 V1 2 (1732)
(17–33) We can combine the conservation of mass and energy relations into a single equation and plot it on an hs diagram, using property relations. The
resultant curve is called the Fanno line, and it is the locus of states that
have the same value of stagnation enthalpy and mass flux (mass flow per
unit flow area). Likewise, combining the conservation of mass and momentum equations into a single equation and plotting it on the hs diagram yield
a curve called the Rayleigh line. Both these lines are shown on the hs diagram in Fig. 17–31. As proved later in Example 17–8, the points of maximum entropy on these lines (points a and b) correspond to Ma
1. The
state on the upper part of each curve is subsonic and on the lower part
supersonic.
The Fanno and Rayleigh lines intersect at two points (points 1 and 2),
which represent the two states at which all three conservation equations are
satisfied. One of these (state 1) corresponds to the state before the shock,
and the other (state 2) corresponds to the state after the shock. Note that
the flow is supersonic before the shock and subsonic afterward. Therefore
the flow must change from supersonic to subsonic if a shock is to occur.
The larger the Mach number before the shock, the stronger the shock will
be. In the limiting case of Ma 1, the shock wave simply becomes a sound
s1. This is expected since the flow
wave. Notice from Fig. 17–31 that s2
through the shock is adiabatic but irreversible. cen84959_ch17.qxd 4/21/05 11:08 AM Page 847 Chapter 17  847 2
V2 h
P 01 2
02 Subsonic flow
(Ma < 1) P h 01 h 01 = h 02 h 02
2 b Ma = 1
Ma = 1
a W
AV
E h2
2
V1 SH OC K 2 no n
Fa ine Supersonic flow
(Ma > 1) hl ig
yle h1 e lin 1 Ra 0 s1 FIGURE 17–31
The hs diagram for flow across a
normal shock. s s2 The conservation of energy principle (Eq. 17–31) requires that the stagnah02. For ideal gases
tion enthalpy remain constant across the shock; h01
h h(T ), and thus
T01 T02 (17–34) That is, the stagnation temperature of an ideal gas also remains constant
across the shock. Note, however, that the stagnation pressure decreases
across the shock because of the irreversibilities, while the thermodynamic
temperature rises drastically because of the conversion of kinetic energy
into enthalpy due to a large drop in fluid velocity (see Fig. 17–32).
We now develop relations between various properties before and after the
shock for an ideal gas with constant specific heats. A relation for the ratio of
the thermodynamic temperatures T2/T1 is obtained by applying Eq. 17–18
twice:
T01
T1 1 a k 1
2 b Ma2
1 and T02
T2 a 1 k 1
2 b Ma 2
2 Dividing the first equation by the second one and noting that T01
have
T2
T1 1
1 Ma2 1 k
1 1 2 >2 Ma2 1 k
2 1 2 >2 T02, we (17–35) From the idealgas equation of state, Substituting these into the conservation of mass relation r1V1
1kRT , we have
noting that Ma V/c and c
r1 T2
T1 P2V2
P1V1 P1
RT1 P2Ma2c2
P1Ma1c1 and r2 P2Ma2 2T2
P1Ma1 2T1 P2
RT2 a P2 2 Ma2 2
ba
b
P1
Ma1 r2V2 and (17–36) Normal
shock
P
P0
V
Ma
T
T0
r
s increases
decreases
decreases
decreases
increases
remains constant
increases
increases FIGURE 17–32
Variation of flow properties across a
normal shock. cen84959_ch17.qxd 4/21/05 11:08 AM Page 848 848  Thermodynamics Ma1 21
Ma2 21 Combining Eqs. 17–35 and 17–36 gives the pressure ratio across the shock:
P2
P1 Ma2 1 k
1 1 2 >2 Ma2 1 k
2 (17–37) 1 2 >2 Equation 17–37 is a combination of the conservation of mass and energy
equations; thus, it is also the equation of the Fanno line for an ideal gas with
constant specific heats. A similar relation for the Rayleigh line can be
obtained by combining the conservation of mass and momentum equations.
From Eq. 17–32,
P1 P2 #
m
1V
A2 V1 2 However,
rV 2 Thus, a P
b 1 Mac 2 2
RT
P1 1 1 a r 2V 2
2 r 1V 2
1 P
b 1 Ma 2kRT 2 2
RT kMa2 2
1 P2 1 1 PkMa2 kMa2 2
2 or
P2
P1 1
1 kMa2
1
kMa2
2 Combining Eqs. 17–37 and 17–38 yields
Ma2
1 Ma2
2 2Ma2 k >
1 2> 1 k 1k 12 (17–38) 12 1 (17–39) This represents the intersections of the Fanno and Rayleigh lines and relates
the Mach number upstream of the shock to that downstream of the shock.
The occurrence of shock waves is not limited to supersonic nozzles only.
This phenomenon is also observed at the engine inlet of a supersonic aircraft, where the air passes through a shock and decelerates to subsonic
velocities before entering the diffuser of the engine. Explosions also produce powerful expanding spherical normal shocks, which can be very
destructive (Fig. 17–33).
Various flow property ratios across the shock are listed in Table A–33 for
an ideal gas with k
1.4. Inspection of this table reveals that Ma2 (the
Mach number after the shock) is always less than 1 and that the larger the
supersonic Mach number before the shock, the smaller the subsonic Mach
number after the shock. Also, we see that the static pressure, temperature,
and density all increase after the shock while the stagnation pressure
decreases.
The entropy change across the shock is obtained by applying the entropychange equation for an ideal gas across the shock:
s2 s1 cp ln T2
T1 R ln P2
P1 (17–40) which can be expressed in terms of k, R, and Ma1 by using the relations
developed earlier in this section. A plot of nondimensional entropy change cen84959_ch17.qxd 4/21/05 11:08 AM Page 849 Chapter 17  849 FIGURE 17–33
Schlieren image of the blast wave
(expanding spherical normal shock)
produced by the explosion of a
firecracker detonated inside a metal can
that sat on a stool. The shock expanded
radially outward in all directions at a
supersonic speed that decreased with
radius from the center of the explosion.
The microphone at the lower right
sensed the sudden change in pressure
of the passing shock wave and
triggered the microsecond flashlamp
that exposed the photograph.
Photo by G. S. Settles, Penn State University. Used
by permission. across the normal shock (s2
s1)/R versus Ma1 is shown in Fig. 17–34.
Since the flow across the shock is adiabatic and irreversible, the second law
requires that the entropy increase across the shock wave. Thus, a shock
wave cannot exist for values of Ma1 less than unity where the entropy
change would be negative. For adiabatic flows, shock waves can exist only
for supersonic flows, Ma1 1. (s2 s1)/R s2 – s1 > 0
0
s2 – s1 < 0 EXAMPLE 17–8 The Point of Maximum Entropy
o n the Fanno Line Show that the point of maximum entropy on the Fanno line (point b of Fig.
17–31) for the adiabatic steady flow of a fluid in a duct corresponds to the
sonic velocity, Ma
1. Solution It is to be shown that the point of maximum entropy on the Fanno
line for steady adiabatic flow corresponds to sonic velocity.
Assumptions The flow is steady, adiabatic, and onedimensional.
Analysis In the absence of any heat and work interactions and potential
energy changes, the steadyflow energy equation reduces to V2
2 h constant Differentiating yields dh V dV 0 For a very thin shock with negligible change of duct area across the shock, the
steadyflow continuity (conservation of mass) equation can be expressed as rV constant Impossible
Subsonic flow
before shock Ma1 = 1 Supersonic flow Ma1
before shock FIGURE 17–34
Entropy change across the normal
shock. cen84959_ch17.qxd 4/21/05 11:08 AM Page 850 850  Thermodynamics
Differentiating, we have r dV V dr 0 Solving for dV gives dV V dr
r Combining this with the energy equation, we have dh V2 dr
r 0 which is the equation for the Fanno line in differential form. At point a
(the point of maximum entropy) ds
0. Then from the second T ds relation
(T ds
dh
v dP ) we have dh
v dP
dP/r. Substituting yields dP
r V2 dr
r 0 at s constant Solving for V, we have V a 0 P 1>2
b
0r s which is the relation for the speed of sound, Eq. 17–9. Thus the proof is
complete. Shock wave ·
m = 2.86 kg/s 1 Ma1 = 2
P = 1.0 MPa
2 01
P1 = 0.1278 MPa
T1 = 444.5 K
r1 = 1.002 kg/m3 FIGURE 17–35
Schematic for Example 17–9. EXAMPLE 17–9 Shock Wave in a Converging–Diverging Nozzle If the air flowing through the converging–diverging nozzle of Example 17–7
experiences a normal shock wave at the nozzle exit plane (Fig. 17–35), determine the following after the shock: (a) the stagnation pressure, static pressure, static temperature, and static density; (b) the entropy change across the
shock; (c) the exit velocity; and (d ) the mass flow rate through the nozzle.
Assume steady, onedimensional, and isentropic flow with k
1.4 from the
nozzle inlet to the shock location. Solution Air flowing through a converging–diverging nozzle experiences a
normal shock at the exit. The effect of the shock wave on various properties
is to be determined.
Assumptions 1 Air is an ideal gas with constant specific heats at room
temperature. 2 Flow through the nozzle is steady, onedimensional, and
isentropic before the shock occurs. 3 The shock wave occurs at the exit
plane.
Properties The constantpressure specific heat and the specific heat ratio of
air are cp
1.005 kJ/kg · K and k
1.4. The gas constant of air is 0.287
kJ/kg K (Table A–2a).
Analysis (a) The fluid properties at the exit of the nozzle just before the
shock (denoted by subscript 1) are those evaluated in Example 17–7 at the
nozzle exit to be P01 1.0 MPa P1 0.1278 MPa T1 444.5 K r1 1.002 kg> m3 cen84959_ch17.qxd 4/21/05 11:08 AM Page 851 Chapter 17
The fluid properties after the shock (denoted by subscript 2) are related
to those before the shock through the functions listed in Table A–33. For
Ma1
2.0, we read Ma2 0.5774 P02
P01 P2
P1 0.7209 4.5000 T2
T1 1.6875 r2
r1 2.6667 Then the stagnation pressure P02, static pressure P2, static temperature T2,
and static density r2 after the shock are P02 0.7209P01 P2 4.5000P1 T2 1.6875T1 r2 2.6667r 1 1 0.7209 2 1 1.0 MPa 2 0.721 MPa 1 4.5000 2 1 0.1278 MPa 2 1 1.6875 2 1 444.5 K 2 0.575 MPa 750 K 1 2.6667 2 1 1.002 kg> m3 2 2.67 kg/m3 (b) The entropy change across the shock is s2 s1 cp ln T2
T1 R ln P2
P1 1 1.005 kJ> kg # K 2 ln 1 1.6875 2 0.0942 kJ/kg # K 1 0.287 kJ> kg # K 2 ln 1 4.5000 2 Thus, the entropy of the air increases as it experiences a normal shock,
which is highly irreversible. Ma2 2kRT2 Ma2c2,
(c) The air velocity after the shock can be determined from V2
where c2 is the speed of sound at the exit conditions after the shock: V2 Ma2c2
1 0.5774 2 B 1 1.4 2 1 0.287 kJ> kg # K 2 1 750 K 2 a 1000 m2> s2
1 kJ> kg b 317 m/s
(d ) The mass flow rate through a converging–diverging nozzle with sonic
conditions at the throat is not affected by the presence of shock waves in
the nozzle. Therefore, the mass flow rate in this case is the same as that
determined in Example 17–7: #
m 2.86 kg/s Discussion This result can easily be verified by using property values at the
nozzle exit after the shock at all Mach numbers significantly greater than unity. Example 17–9 illustrates that the stagnation pressure and velocity
decrease while the static pressure, temperature, density, and entropy
increase across the shock. The rise in the temperature of the fluid downstream of a shock wave is of major concern to the aerospace engineer
because it creates heat transfer problems on the leading edges of wings and
nose cones of space reentry vehicles and the recently proposed hypersonic
space planes. Overheating, in fact, led to the tragic loss of the space shuttle
Columbia in February of 2003 as it was reentering earth’s atmosphere.  851 cen84959_ch17.qxd 4/27/05 11:00 AM Page 852 852  Thermodynamics FIGURE 17–36
Schlieren image of a small model of
the space shuttle Orbiter being tested
at Mach 3 in the supersonic wind
tunnel of the Penn State Gas
Dynamics Lab. Several oblique shocks
are seen in the air surrounding the
spacecraft.
Photo by G. S. Settles, Penn State University. Used
by permission. Oblique Shocks Oblique
shock
Ma1
Ma 2
Ma1 u b
d FIGURE 17–37
An oblique shock of shock angle b
formed by a slender, twodimensional
wedge of halfangle d. The flow is
turned by deflection angle u
downstream of the shock, and the
Mach number decreases. Not all shock waves are normal shocks (perpendicular to the flow direction). For example, when the space shuttle travels at supersonic speeds
through the atmosphere, it produces a complicated shock pattern consisting
of inclined shock waves called oblique shocks (Fig. 17–36). As you can
see, some portions of an oblique shock are curved, while other portions are
straight.
First, we consider straight oblique shocks, like that produced when a uni1) impinges on a slender, twodimensional
form supersonic flow (Ma1
wedge of halfangle d (Fig. 17–37). Since information about the wedge
cannot travel upstream in a supersonic flow, the fluid “knows” nothing
about the wedge until it hits the nose. At that point, since the fluid cannot
flow through the wedge, it turns suddenly through an angle called the
turning angle or deflection angle u. The result is a straight oblique shock
wave, aligned at shock angle or wave angle b, measured relative to the
oncoming flow (Fig. 17–38). To conserve mass, b must obviously be
greater than d. Since the Reynolds number of supersonic flows is typically
large, the boundary layer growing along the wedge is very thin, and we
ignore its effects. The flow therefore turns by the same angle as the wedge;
namely, deflection angle u is equal to wedge halfangle d. If we take into
account the displacement thickness effect of the boundary layer, the deflection angle u of the oblique shock turns out to be slightly greater than
wedge halfangle d.
Like normal shocks, the Mach number decreases across an oblique shock,
and oblique shocks are possible only if the upstream flow is supersonic.
However, unlike normal shocks, in which the downstream Mach number is
always subsonic, Ma2 downstream of an oblique shock can be subsonic,
sonic, or supersonic, depending on the upstream Mach number Ma1 and the
turning angle. cen84959_ch17.qxd 4/21/05 11:08 AM Page 853 Chapter 17
We analyze a straight oblique shock in Fig. 17–38 by decomposing the
velocity vectors upstream and downstream of the shock into normal and tangential components, and considering a small control volume around the
shock. Upstream of the shock, all fluid properties (velocity, density, pressure, etc.) along the lower left face of the control volume are identical to
those along the upper right face. The same is true downstream of the shock.
Therefore, the mass flow rates entering and leaving those two faces cancel
each other out, and conservation of mass reduces to
r 1V1,n A r 2V2, n A S r 1V1, n r 2V2, n P2 A rV2, n AV2, n rV1, n AV1, n S P1 P2 853 Oblique
shock
P1 V1,t P2 V1,n → V2 → u V1
Control
volume V2,t
V2,n b (17–41) where A is the area of the control surface that is parallel to the shock. Since
A is the same on either side of the shock, it has dropped out of Eq. 17–41.
As you might expect, the tangential component of velocity (parallel to the
V2,t). This is
oblique shock) does not change across the shock (i.e., V1,t
easily proven by applying the tangential momentum equation to the control
volume.
When we apply conservation of momentum in the direction normal to the
oblique shock, the only forces are pressure forces, and we get
P1 A  r 2V 2 n
2, FIGURE 17–38
Velocity vectors through an oblique
shock of shock angle b and deflection
angle u. r 1V 2 n (17–42)
1, Finally, since there is no work done by the control volume and no heat
transfer into or out of the control volume, stagnation enthalpy does not
change across an oblique shock, and conservation of energy yields
h01 But since V1,t h 02 h0 S h1 12
V
2 1,n 12
V
2 1,t h2 12
V
2 2,n 12
V
2 2,t V2,t, this equation reduces to
h1 12
V
2 1, n h2 12
V
2 2, n (17–43) Careful comparison reveals that the equations for conservation of mass,
momentum, and energy (Eqs. 17–41 through 17–43) across an oblique
shock are identical to those across a normal shock, except that they are written in terms of the normal velocity component only. Therefore, the normal
shock relations derived previously apply to oblique shocks as well, but must
be written in terms of Mach numbers Ma1,n and Ma2,n normal to the oblique
shock. This is most easily visualized by rotating the velocity vectors in
Fig. 17–38 by angle p/2 b, so that the oblique shock appears to be vertical (Fig. 17–39). Trigonometry yields
Ma1,n Ma1 sin b and Ma2,n Ma2 sin 1 b u2 (17–44) where Ma1,n
V1, n /c1 and Ma2,n
V2, n /c2. From the point of view shown
in Fig. 17–40, we see what looks like a normal shock, but with some superposed tangential flow “coming along for the ride.” Thus,
All the equations, shock tables, etc., for normal shocks apply to oblique shocks
as well, provided that we use only the normal components of the Mach number. In fact, you may think of normal shocks as special oblique shocks in
which shock angle b
p/2, or 90°. We recognize immediately that an
oblique shock can exist only if Ma1,n 1, and Ma2,n 1. The normal shock P1
Ma 1,n P2 1 Ma 2,n
b u u Oblique
shock
→ V2
V2,t
V1,n
V2,n V1,t
b → V1
P1 P2 FIGURE 17–39
The same velocity vectors of Fig.
17–38, but rotated by angle p/2 – b,
so that the oblique shock is vertical.
Normal Mach numbers Ma1,n and
Ma2,n are also defined. 1 cen84959_ch17.qxd 4/21/05 11:08 AM Page 854 854  Thermodynamics
h01
01 B 2k Ma 2, n
1
1,
→ h 02
02 P2 r1 P02
02
P01
01 V1,, n
1
V2,, n
2 [2
c (k (k 2 1)Ma 2, n
)Ma 1,
1 d 1 1
)Ma 1,
1)Ma 2, n
1 (k 1)Ma 2, n
)Ma 1,
1 (k 1)Ma2, n ]
)Ma1,
1 )Ma 1,
1)Ma 2, n
1 (k
2 1 k k r2
T2
T1 2 k 2k Ma 2, n
1
1, P1 T02
02 1)Ma 2, n
)Ma 1,
1 (k Ma
Ma 2, n T01
01 k /(k 1)
/( 2k Ma 2, n
1
1, c k 1 1)2Ma 2, n
Ma 1,
1 (k
(k 2k Ma 2, n
1
1, 1)
k d
1 1/
1/ (k 1) FIGURE 17–40
Relationships across an oblique
shock for an ideal gas in terms of the
normal component of upstream Mach
number Ma1,n. equations appropriate for oblique shocks in an ideal gas are summarized in
Fig. 17–40 in terms of Ma1,n.
For known shock angle b and known upstream Mach number Ma1, we use
the first part of Eq. 17–44 to calculate Ma1,n, and then use the normal shock
tables (or their corresponding equations) to obtain Ma2,n. If we also knew the
deflection angle u, we could calculate Ma2 from the second part of Eq. 17–44.
But, in a typical application, we know either b or u, but not both. Fortunately,
a bit more algebra provides us with a relationship between u, b, and Ma1. We
begin by noting that tan b V1,n / V1,t and tan(b u) V2,n / V2,t (Fig. 17–39).
But since V1,t V2,t, we combine these two expressions to yield
V2,n
V1,n tan 1 b tan b u2 2 1k 1k 1 2 Ma2 n
1, 1 2 Ma2 n
1, 2 1k 1k 1 2 Ma2 sin2 b
1 1 2 Ma2 sin2 b
1 (17–45) where we have also used Eq. 17–44 and the fourth equation of Fig. 17–40.
We apply trigonometric identities for cos 2b and tan(b u), namely,
cos 2b cos2 b sin2 b and tan 1 b u2 tan b tan u
1 tan b tan u After some algebra, Eq. 17–45 reduces to
The ubMa relationship: tan u 2 cot b 1 Ma2 sin2 b
1
Ma2 1 k
1 cos 2 b 2 12 2 (17–46) Equation 17–46 provides deflection angle u as a unique function of
shock angle b, specific heat ratio k, and upstream Mach number Ma1. For
air (k
1.4), we plot u versus b for several values of Ma1 in Fig. 17–41.
We note that this plot is often presented with the axes reversed (b versus u)
in compressible flow textbooks, since, physically, shock angle b is determined by deflection angle u.
Much can be learned by studying Fig. 17–41, and we list some observations here:
• Figure 17–41 displays the full range of possible shock waves at a given
freestream Mach number, from the weakest to the strongest. For any value
of Mach number Ma1 greater than 1, the possible values of u range from
u 0° at some value of b between 0 and 90°, to a maximum value u umax
at an intermediate value of b, and then back to u 0° at b 90°. Straight
oblique shocks for u or b outside of this range cannot and do not exist. At
Ma1 1.5, for example, straight oblique shocks cannot exist in air with
shock angle b less than about 42°, nor with deflection angle u greater than
about 12. If the wedge halfangle is greater than umax, the shock becomes
curved and detaches from the nose of the wedge, forming what is called a
detached oblique shock or a bow wave (Fig. 17–42). The shock angle b
of the detached shock is 90° at the nose, but b decreases as the shock curves
downstream. Detached shocks are much more complicated than simple
straight oblique shocks to analyze. In fact, no simple solutions exist, and
prediction of detached shocks requires computational methods.
• Similar oblique shock behavior is observed in axisymmetric flow over
cones, as in Fig. 17–43, although the ubMa relationship for
axisymmetric flows differs from that of Eq. 17–46. cen84959_ch17.qxd 4/21/05 11:08 AM Page 855 Chapter 17
50
Ma2 1 u umax u, degrees 40 30
Ma2 Ma1 → 1 Ma2 1 20
Weak Strong 10
10 5 3 2 1.5
1.2 0 0 10 20 30 40
50
b, degrees 60 70 80 90 • When supersonic flow impinges on a blunt body—a body without a
sharply pointed nose, the wedge halfangle d at the nose is 90°, and an
attached oblique shock cannot exist, regardless of Mach number. In fact,
a detached oblique shock occurs in front of all such bluntnosed bodies,
whether twodimensional, axisymmetric, or fully threedimensional. For
example, a detached oblique shock is seen in front of the space shuttle
model in Fig. 17–36 and in front of a sphere in Fig. 17–44.
• While u is a unique function of Ma1 and b for a given value of k, there are
two possible values of b for u umax. The dashed black line in Fig. 17–41
passes through the locus of umax values, dividing the shocks into weak
oblique shocks (the smaller value of b) and strong oblique shocks (the
larger value of b). At a given value of u, the weak shock is more common
and is “preferred” by the flow unless the downstream pressure conditions
are high enough for the formation of a strong shock.
• For a given upstream Mach number Ma1, there is a unique value of u for
which the downstream Mach number Ma2 is exactly 1. The dashed gray
line in Fig. 17–41 passes through the locus of values where Ma2 1.
To the left of this line, Ma2 1, and to the right of this line, Ma2 1.
Downstream sonic conditions occur on the weak shock side of the plot,
with u very close to umax. Thus, the flow downstream of a strong oblique
shock is always subsonic (Ma2 1). The flow downstream of a weak
oblique shock remains supersonic, except for a narrow range of u just
below umax, where it is subsonic, although it is still called a weak
oblique shock.
• As the upstream Mach number approaches infinity, straight oblique
shocks become possible for any b between 0 and 90°, but the maximum
possible turning angle for k 1.4 (air) is umax 45.6°, which occurs at b
67.8°. Straight oblique shocks with turning angles above this value of
umax are not possible, regardless of the Mach number.
• For a given value of upstream Mach number, there are two shock angles
where there is no turning of the flow (u 0°): the strong case, b 90°,  855 FIGURE 17–41
The dependence of straight oblique
shock deflection angle u on shock
angle b for several values of upstream
Mach number Ma1. Calculations are
for an ideal gas with k 1.4. The
dashed black line connects points of
maximum deflection angle (u umax).
Weak oblique shocks are to the left of
this line, while strong oblique shocks
are to the right of this line. The dashed
gray line connects points where the
downstream Mach number is sonic
(Ma2 1). Supersonic downstream
flow (Ma2 1) is to the left of this
line, while subsonic downstream flow
(Ma2 1) is to the right of this line. Detached
oblique
shock
Ma1 d umax FIGURE 17–42
A detached oblique shock occurs
upstream of a twodimensional wedge
of halfangle d when d is greater than
the maximum possible deflection
angle u. A shock of this kind is called
a bow wave because of its
resemblance to the water wave that
forms at the bow of a ship. cen84959_ch17.qxd 4/21/05 11:08 AM Page 856 856  Thermodynamics FIGURE 17–43
Still frames from schlieren
videography illustrating the
detachment of an oblique shock from a
cone with increasing cone halfangle d
in air at Mach 3. At (a) d 20 and
(b) d 40 , the oblique shock remains
attached, but by (c) d 60 , the
oblique shock has detached, forming
a bow wave.
Photos by G. S. Settles, Penn State University.
Used by permission. Ma1
d (a) (b) (c) corresponds to a normal shock, and the weak case, b bmin, represents
the weakest possible oblique shock at that Mach number, which is called
a Mach wave. Mach waves are caused, for example, by very small
nonuniformities on the walls of a supersonic wind tunnel (several can be
seen in Figs. 17–36 and 17–43). Mach waves have no effect on the flow,
since the shock is vanishingly weak. In fact, in the limit, Mach waves are
isentropic. The shock angle for Mach waves is a unique function of the
Mach number and is given the symbol m, not to be confused with the
coefficient of viscosity. Angle m is called the Mach angle and is found by
setting u equal to zero in Eq. 17–46, solving for b m, and taking the
smaller root. We get
Mach angle: m sin 1 1 1> Ma1 2 (17–47) Since the specific heat ratio appears only in the denominator of Eq.
17–46, m is independent of k. Thus, we can estimate the Mach number of
any supersonic flow simply by measuring the Mach angle and applying
Eq. 17–47. Prandtl–Meyer Expansion Waves FIGURE 17–44
Shadowgram of a onehalfin diameter
sphere in free flight through air at Ma
1.53. The flow is subsonic behind
the part of the bow wave that is ahead
of the sphere and over its surface back
to about 45 . At about 90 the laminar
boundary layer separates through an
oblique shock wave and quickly
becomes turbulent. The fluctuating
wake generates a system of weak
disturbances that merge into the
second “recompression” shock wave.
Photo by A. C. Charters, Army Ballistic Research
Laboratory. We now address situations where supersonic flow is turned in the opposite
direction, such as in the upper portion of a twodimensional wedge at an
angle of attack greater than its halfangle d (Fig. 17–45). We refer to this
type of flow as an expanding flow, whereas a flow that produces an oblique
shock may be called a compressing flow. As previously, the flow changes
direction to conserve mass. However, unlike a compressing flow, an expanding flow does not result in a shock wave. Rather, a continuous expanding
region called an expansion fan appears, composed of an infinite number of
Mach waves called Prandtl–Meyer expansion waves. In other words, the
flow does not turn suddenly, as through a shock, but gradually—each successive Mach wave turns the flow by an infinitesimal amount. Since each
individual expansion wave is isentropic, the flow across the entire expansion
fan is also isentropic. The Mach number downstream of the expansion
Ma1), while pressure, density, and temperature decrease,
increases (Ma2
just as they do in the supersonic (expanding) portion of a converging–
diverging nozzle.
Prandtl–Meyer expansion waves are inclined at the local Mach angle m,
as sketched in Fig. 17–45. The Mach angle of the first expansion wave is
sin 1 (1/Ma1). Similarly, m2
sin 1 (1/Ma2),
easily determined as m1 cen84959_ch17.qxd 4/21/05 11:08 AM Page 857 Chapter 17
where we must be careful to measure the angle relative to the new direction
of flow downstream of the expansion, namely, parallel to the upper wall of
the wedge in Fig. 17–45 if we neglect the influence of the boundary layer
along the wall. But how do we determine Ma2? It turns out that the turning
angle u across the expansion fan can be calculated by integration, making
use of the isentropic flow relationships. For an ideal gas, the result is
(Anderson, 2003),
Turning angle across an expansion fan: u n 1 Ma2 2 n 1 Ma1 2 a 2Ma2 (17–48) where n(Ma) is an angle called the Prandtl–Meyer function (not to be confused with the kinematic viscosity),
n 1 Ma 2 k
Bk 1
tan
1 k
1
c
Bk 1
1 Ma2
1 12 d tan 1 1 b (17–49) Note that n(Ma) is an angle, and can be calculated in either degrees or radians. Physically, n(Ma) is the angle through which the flow must expand,
starting with n 0 at Ma 1, in order to reach a supersonic Mach number,
Ma 1.
To find Ma2 for known values of Ma1, k, and u, we calculate n(Ma1) from
Eq. 17–49, n(Ma2) from Eq. 17–48, and then Ma2 from Eq. 17–49, noting
that the last step involves solving an implicit equation for Ma2. Since there
is no heat transfer or work, and the flow is isentropic through the expansion,
T0 and P0 remain constant, and we use the isentropic flow relations derived
previously to calculate other flow properties downstream of the expansion,
such as T2, r2, and P2.
Prandtl–Meyer expansion fans also occur in axisymmetric supersonic
flows, as in the corners and trailing edges of a conecylinder (Fig. 17–46).
Some very complex and, to some of us, beautiful interactions involving
both shock waves and expansion waves occur in the supersonic jet produced by an “overexpanded” nozzle, as in Fig. 17–47. Analysis of such
flows is beyond the scope of the present text; interested readers are referred
to compressible flow textbooks such as Thompson (1972) and Anderson
(2003).  857 Expansion
waves m1
Ma1 Ma 2
m2 1
u
d
Oblique
shock FIGURE 17–45
An expansion fan in the upper
portion of the flow formed by a twodimensional wedge at the angle of
attack in a supersonic flow. The flow
is turned by angle u, and the Mach
number increases across the expansion
fan. Mach angles upstream and
downstream of the expansion fan are
indicated. Only three expansion waves
are shown for simplicity, but in fact,
there are an infinite number of them.
(An oblique shock is present in the
bottom portion of this flow.) FIGURE 17–46
A conecylinder of 12.5 halfangle in
a Mach number 1.84 flow. The
boundary layer becomes turbulent
shortly downstream of the nose,
generating Mach waves that are visible
in this shadowgraph. Expansion waves
are seen at the corners and at the
trailing edge of the cone.
Photo by A. C. Charters, Army Ballistic Research
Laboratory. cen84959_ch17.qxd 4/27/05 5:57 PM Page 858 858  Thermodynamics FIGURE 17–47
The complex interactions between
shock waves and expansion waves in
an “overexpanded” supersonic jet.
The flow is visualized by a schlierenlike differential interferogram.
Photo by H. Oertel sen. Reproduced by courtesy of
the FrenchGerman Research Institute of SaintLouis, ISL. Used with permission. EXAMPLE 17–10 Estimation of the Mach Number
f rom Mach Lines Estimate the Mach number of the freestream flow upstream of the space
shuttle in Fig. 17–36 from the figure alone. Compare with the known value
of Mach number provided in the figure caption. Solution We are to estimate the Mach number from a figure and compare
it to the known value.
Analysis Using a protractor, we measure the angle of the Mach lines in the
freestream flow: m
19°. The Mach number is obtained from Eq. 17–47, Weak
shock
Ma1
bweak
d 10° m sin 1 a 1
b
Ma1 S Ma1 1
sin 19° S Ma1 3.07 Our estimated Mach number agrees with the experimental value of 3.0
Discussion The result is independent of the fluid properties. 0.1. (a)
Strong
shock EXAMPLE 17–11 Ma1
bstrong
d 10° (b) FIGURE 17–48
Two possible oblique shock angles,
(a) bweak and (b) bstrong, formed by a
twodimensional wedge of halfangle
d 10 . Oblique Shock Calculations Supersonic air at Ma1
2.0 and 75.0 kPa impinges on a twodimensional
wedge of halfangle d
10° (Fig. 17–48). Calculate the two possible
oblique shock angles, bweak and bstrong, that could be formed by this wedge.
For each case, calculate the pressure and Mach number downstream of the
oblique shock, compare, and discuss. Solution We are to calculate the shock angle, Mach number, and pressure
downstream of the weak and strong oblique shocks formed by a twodimensional wedge.
Assumptions 1 The flow is steady. 2 The boundary layer on the wedge is
very thin.
Properties The fluid is air with k
1.4. cen84959_ch17.qxd 4/21/05 11:08 AM Page 859 Chapter 17  859 Analysis Because of assumption 2, we approximate the oblique shock
deflection angle as equal to the wedge halfangle, i.e., u
d
10°. With
Ma1
2.0 and u
10°, we solve Eq. 17–46 for the two possible values of
oblique shock angle b: Bweak
39.3° and Bstrong
83.7°. From these values,
we use the first part of Eq. 17–44 to calculate the upstream normal Mach
number Ma1,n, Ma1 sin b S Ma1,n Ma1,n Weak shock: Ma1,n Strong shock: Ma1 sin b S Ma1,n 2.0 sin 39.3° 1.267 2.0 sin 83.7° 1.988 We substitute these values of Ma1,n into the second equation of Fig. 17–40
to calculate the downstream normal Mach number Ma2,n. For the weak
shock, Ma2,n
0.8032, and for the strong shock, Ma2,n
0.5794. We also
calculate the downstream pressure for each case, using the third equation of
Fig. 17–40, which gives Weak shock:
P2
P1 2kMa2 n
1,
k k 1 1 S P2 1 75.0 kPa 2 S P2 1 75.0 kPa 2 2 1 1.4 2 1 1.267 2 2
1.4 1.4 1 1.4 1 1 128 kPa Strong shock:
P2
P1 2 kMa2 n
1,
k k
1 1 2 1 1.4 2 1 1.988 2 2
1.4 1 333 kPa Finally, we use the second part of Eq. 17–44 to calculate the downstream
Mach number, Weak shock: Strong shock: Ma2
Ma2 Ma2,n sin 1 b Ma2,n sin 1 b u2 0.8032
sin 1 39.3° 10° 2 1.64 u2 0.5794
sin 1 83.7° 10° 2 0.604 The changes in Mach number and pressure across the strong shock are
much greater than the changes across the weak shock, as expected.
Discussion Since Eq. 17–46 is implicit in b, we solve it by an iterative
approach or with an equation solver such as EES. For both the weak and
strong oblique shock cases, Ma1,n is supersonic and Ma2,n is subsonic. However, Ma2 is supersonic across the weak oblique shock, but subsonic across
the strong oblique shock. We could also use the normal shock tables in
place of the equations, but with loss of precision. Ma1 2 .0 u
Ma 2 EXAMPLE 17–12 Prandtl–Meyer Expansion Wave Calculations Supersonic air at Ma1
2.0 and 230 kPa flows parallel to a flat wall that
suddenly expands by d
10° (Fig. 17–49). Ignoring any effects caused by
the boundary layer along the wall, calculate downstream Mach number Ma2
and pressure P2. d 10° FIGURE 17–49
An expansion fan caused by the sudden
expansion of a wall with d 10 . cen84959_ch17.qxd 4/21/05 11:08 AM Page 860 860  Thermodynamics
Solution We are to calculate the Mach number and pressure downstream of
a sudden expansion along a wall.
Assumptions 1 The flow is steady. 2 The boundary layer on the wall is
very thin.
Properties The fluid is air with k
1.4.
Analysis Because of assumption 2, we approximate the total deflection
angle as equal to the wall expansion angle (i.e., u
d
10°). With Ma1
2.0, we solve Eq. 17–49 for the upstream Prandtl–Meyer function,
k
Bk n 1 Ma 2 1.4
B 1.4 1
tan
1 1 1
tan
1 c k
Bk 1 c 12 d 1
1 Ma2
1 1.4
B 1.4 1
1 2.0 2
1 1 tan
12 d a 2 Ma2 1 b tan 1 a 22.0 2 1b 26.38° Next, we use Eq. 17–48 to calculate the downstream Prandtl–Meyer function, u n 1 Ma2 2 n 1 Ma1 2 S n 1 Ma2 2 u n 1 Ma1 2 10° 26.38° 36.38° Ma2 is found by solving Eq. 17–49, which is implicit—an equation solver is
helpful. We get Ma2
2.385. There are also compressible flow calculators
on the Internet that solve these implicit equations, along with both normal
and oblique shock equations; e.g., see www.aoe.vt.edu/~devenpor/aoe3114/calc
.html.
We use the isentropic relations to calculate the downstream pressure, P2> P0
P1> P0 P2 P1 c1 a k c1 a k 1 b Ma2 d
2 1 b Ma2 d
1 2
2 k>1k 12
k>1k 12 1 230 kPa 2 126 kPa Since this is an expansion, Mach number increases and pressure decreases,
as expected.
Discussion We could also solve for downstream temperature, density, etc.,
using the appropriate isentropic relations. Fuel nozzles or spray bars Air inlet Flame holders FIGURE 17–50
Many practical compressible flow
problems involve combustion, which
may be modeled as heat gain through
the duct wall. 17–6 ■ DUCT FLOW WITH HEAT TRANSFER AND
NEGLIGIBLE FRICTION (RAYLEIGH FLOW) So far we have limited our consideration mostly to isentropic flow, also
called reversible adiabatic flow since it involves no heat transfer and no
irreversibilities such as friction. Many compressible flow problems encountered in practice involve chemical reactions such as combustion, nuclear
reactions, evaporation, and condensation as well as heat gain or heat loss
through the duct wall. Such problems are difficult to analyze exactly since
they may involve significant changes in chemical composition during flow,
and the conversion of latent, chemical, and nuclear energies to thermal
energy (Fig. 17–50).
The essential features of such complex flows can still be captured by a
simple analysis by modeling the generation or absorption of thermal energy cen84959_ch17.qxd 4/21/05 11:08 AM Page 861 Chapter 17
as heat transfer through the duct wall at the same rate and disregarding any
changes in chemical composition. This simplified problem is still too complicated for an elementary treatment of the topic since the flow may involve
friction, variations in duct area, and multidimensional effects. In this section, we limit our consideration to onedimensional flow in a duct of constant crosssectional area with negligible frictional effects.
Consider steady onedimensional flow of an ideal gas with constant specific heats through a constantarea duct with heat transfer, but with negligible
friction. Such flows are referred to as Rayleigh flows after Lord Rayleigh
(1842–1919). The conservation of mass, momentum, and energy equations
for the control volume shown in Fig. 17–51 can be written as follows:
Mass equation Noting that the duct crosssectional area A is constant, the
.
.
relation m1 m2 or r1A1V1 r2A2V2 reduces to
r1V1 r2V2 (17–50) xMomentum equation Noting that the frictional effects are negligible
and thus there are no shear forces, and assuming there are no external
!
#!
#!
and body forces, the momentum equation a F a bmV a bmV
out in in the flow (or x) direction becomes a balance between static pressure
forces and momentum transfer. Noting that the flows are high speed
and turbulent, the momentum flux correction factor is approximately 1
(b 1) and thus can be neglected. Then,
P1A 1 P2A 2 #
mV2 #
mV1 S P1 1 r 2V2 2 V2 P2 1 r 1V1 2 V1 or
r 1V 2
1 P1 P2 r 2V 2
2 (17–51) Energy equation The control volume involves no shear, shaft, or other
forms of work, and.the potential energy change is negligible. If the rate
of heat transfer is Q and the heat transfer per unit mass of fluid is q
.
.
..
Q/m, the steadyflow energy balance Ein Eout becomes
#
Q #
m a h1 V2
1
b
2 #
m a h2 V2
2
bSq
2 h1 For an ideal gas with constant specific heats, h
q cp 1 T2 T1 2 or
q h02 h01 V2
2 V2
1
2 cp 1 T02 T01 2 2
V1
2 h2 V2
2
(17–52)
2 cp T, and thus
(17–53) (17–54) Therefore, the stagnation enthalpy h0 and stagnation temperature T0
change during Rayleigh flow (both increase when heat is transferred to
the fluid and thus q is positive, and both decrease when heat is transferred from the fluid and thus q is negative).
Entropy change In the absence of any irreversibilities such as friction,
the entropy of a system changes by heat transfer only: it increases with
heat gain, and decreases with heat loss. Entropy is a property and thus  861 . Q
P1, T 1, r 1 P2 , T 2 , r 2 V1 V2 Control
volume FIGURE 17–51
Control volume for flow in a constantarea duct with heat transfer and
negligible friction. cen84959_ch17.qxd 4/21/05 11:08 AM Page 862 862  Thermodynamics
a state function, and the entropy change of an ideal gas with constant
specific heats during a change of state from 1 to 2 is given by
s2 s1 cp ln T2
T1 R ln P2
P1 (17–55) The entropy of a fluid may increase or decrease during Rayleigh flow,
depending on the direction of heat transfer.
Equation of state Noting that P rRT, the properties P, r, and T of an
ideal gas at states 1 and 2 are related to each other by
P1
r1T1 T Mab = 1/ k Tmax
Cooling
(Ma S 0) b Ma < 1 Heating
(Ma S 1)
Ma > 1
Heating
(Ma S 1) a Maa = 1 a
smax Cooling
(Ma S )
s FIGURE 17–52
Ts diagram for flow in a constantarea
duct with heat transfer and negligible
friction (Rayleigh flow). P2
r2T2 (17–56) Consider a gas with known properties R, k, and cp. For a specified inlet
state 1, the inlet properties P1, T1, r1, V1, and s1 are known. The five exit
properties P2, T2, r2, V2, and s2 can be determined from the five equations
17–50, 17–51, 17–53, 17–55, and 17–56 for any specified value of heat
transfer q. When the velocity and temperature are known, the Mach number
can be determined from Ma V> c V> 1kRT .
Obviously there is an infinite number of possible downstream states 2
corresponding to a given upstream state 1. A practical way of determining
these downstream states is to assume various values of T2, and calculate all
other properties as well as the heat transfer q for each assumed T2 from the
Eqs. 17–50 through 17–56. Plotting the results on a Ts diagram gives a
curve passing through the specified inlet state, as shown in Fig. 17–52. The
plot of Rayleigh flow on a Ts diagram is called the Rayleigh line, and several important observations can be made from this plot and the results of the
calculations:
1. All the states that satisfy the conservation of mass, momentum, and
energy equations as well as the property relations are on the Rayleigh
line. Therefore, for a given initial state, the fluid cannot exist at any
downstream state outside the Rayleigh line on a Ts diagram. In fact,
the Rayleigh line is the locus of all physically attainable downstream
states corresponding to an initial state.
2. Entropy increases with heat gain, and thus we proceed to the right on
the Rayleigh line as heat is transferred to the fluid. The Mach number is
Ma 1 at point a, which is the point of maximum entropy (see Example 17–13 for proof). The states on the upper arm of the Rayleigh line
above point a are subsonic, and the states on the lower arm below point
a are supersonic. Therefore, a process proceeds to the right on the Rayleigh line with heat addition and to the left with heat rejection regardless of the initial value of the Mach number.
3. Heating increases the Mach number for subsonic flow, but decreases it
for supersonic flow. The flow Mach number approaches Ma 1 in both
cases (from 0 in subsonic flow and from ∞ in supersonic flow) during
heating.
4. It is clear from the energy balance q
cp(T02
T01) that heating
increases the stagnation temperature T0 for both subsonic and supersonic flows, and cooling decreases it. (The maximum value of T0 occurs
at Ma 1.) This is also the case for the thermodynamic temperature T cen84959_ch17.qxd 4/21/05 11:08 AM Page 863 except for the narrow Mach number range of 1> 1k
Ma 1 in subsonic flow (see Example 17–13). Both temperature and the Mach number increase with heating in subsonic flow, but T reaches a maximum
Tmax at Ma 1> 1k (which is 0.845 for air), and then decreases. It may
seem peculiar that the temperature of a fluid drops as heat is transferred
to it. But this is no more peculiar than the fluid velocity increasing in
the diverging section of a converging–diverging nozzle. The cooling
effect in this region is due to the large increase in the fluid velocity and
the accompanying drop in temperature in accordance with the relation T0
T V 2/2cp. Note also that heat rejection in the region 1> 1k Ma
1 causes the fluid temperature to increase (Fig. 17–53).
5. The momentum equation P
KV
constant, where K
rV
constant (from the conservation of mass equation), reveals that velocity and
static pressure have opposite trends. Therefore, static pressure
decreases with heat gain in subsonic flow (since velocity and the
Mach number increase), but increases with heat gain in supersonic flow
(since velocity and the Mach number decrease).
6. The continuity equation rV constant indicates that density and velocity are inversely proportional. Therefore, density decreases with heat
transfer to the fluid in subsonic flow (since velocity and the Mach number increase), but increases with heat gain in supersonic flow (since
velocity and the Mach number decrease).
7. On the left half of Fig. 17–52, the lower arm of the Rayleigh line is
steeper (in terms of s as a function of T), which indicates that the
entropy change corresponding to a specified temperature change (and
thus a given amount of heat transfer) is larger in supersonic flow. Chapter 17  863 Heating Subsonic
flow T01 T2
T2 T1 or
T1 T02 T1 T 01 Heating T1 Supersonic
flow T01 T2
T02 T1
T 01 FIGURE 17–53
During heating, fluid temperature
always increases if the Rayleigh flow
is supersonic, but the temperature may
actually drop if the flow is subsonic. The effects of heating and cooling on the properties of Rayleigh flow are
listed in Table 17–3. Note that heating or cooling has opposite effects on most
properties. Also, the stagnation pressure decreases during heating and increases
during cooling regardless of whether the flow is subsonic or supersonic. TABLE 17–3
The effects of heating and cooling on the properties of Rayleigh flow
Heating Cooling Property Subsonic Supersonic Subsonic Supersonic Velocity, V
Mach number, Ma
Stagnation temperature, T0
Temperature, T Increase
Increase
Increase
Increase for Ma
Decrease for Ma
Decrease
Decrease
Decrease
Increase Decrease
Decrease
Increase
Increase Decrease
Decrease
Decrease
Decrease for Ma
Increase for Ma
Increase
Increase
Increase
Decrease Increase
Increase
Decrease
Decrease Density, r
Stagnation pressure, P0
Pressure, P
Entropy, s 1/k1/2
1/k1/2 Increase
Decrease
Increase
Increase 1/k1/2
1/k1/2 Decrease
Increase
Decrease
Decrease cen84959_ch17.qxd 4/21/05 11:08 AM Page 864 864  T Thermodynamics
dT
ds Tmax 0
b Ma a 1 Ma 1 EXAMPLE 17–13 b ds
dT 0
a smax s FIGURE 17–54
The Ts diagram of Rayleigh flow
considered in Example 17–13. Extrema of Rayleigh Line Consider the Ts diagram of Rayleigh flow, as shown in Fig. 17–54. Using
the differential forms of the conservation equations and property relations,
show that the Mach number is Maa
1 at the point of maximum entropy
(point a), and Mab
1/ 1k at the point of maximum temperature (point b). Solution It is to be shown that Maa 1 at the point of maximum entropy
and Mab
1/ 1k at the point of maximum temperature on the Rayleigh line.
Assumptions The assumptions associated with Rayleigh flow (i.e., steady
onedimensional flow of an ideal gas with constant properties through a constant crosssectionalarea duct with negligible frictional effects) are valid.
Analysis The differential forms of the mass (rV
constant), momentum
[rearranged as P + (rV)V
constant], ideal gas (P
rRT), and enthalpy
change ( h
cp T) equations can be expressed as
rV constant S r dV 1 rV 2 V P V dr
1 rV 2 dV constant S dP P rRT S dP rR dT dr
r 0S dP
dV 0S RT dr S dV
V (1) rV dP
P dT
T (2) dr
r (3) The differential form of the entropy change relation (Eq. 17–40) of an
ideal gas with constant specific heats is ds cp dT
T dP
P R (4) Substituting Eq. 3 into Eq. 4 gives ds cp dT
T Ra dr
b
r dT
T 1 cp R2 dT
T R dr
r R
k dT
1T R/(k dr
(5)
r 1) R since cp R cv S kcv R cv S cv Dividing both sides of Eq. 5 by dT and combining with Eq. 1, ds
dT T 1k R R dV
V dT 12 (6) Dividing Eq. 3 by dV and combining it with Eqs. 1 and 2 give, after rearranging, dT
dV T
V V
R (7) Substituting Eq. 7 into Eq. 6 and rearranging, ds
dT T 1k R 12 2kRTa T R
V 2> R R2 1 kRT T 1k 1 2 1 RT 2kRTa
2kRTa V 22 V 22 Setting ds /dT
0 and solving the resulting equation R2(kRT
V give the velocity at point a to be Va and Maa Va
ca 1 (8) V 2) 0 for (9) cen84959_ch17.qxd 4/21/05 11:08 AM Page 865 Chapter 17
2RTb  865 Therefore, sonic conditions exist at point a, and thus the Mach number is 1.
Setting dT/ds
(ds/dT ) 1
0 and solving the resulting equation
2)
T(k
1)(RT
V
0 for velocity at point b give 2RTb 2kRTb 2k Therefore, the Mach number at point b is Mab
1/ 1k . For air, k
1.4 and
thus Mab
0.845.
Discussion Note that in Rayleigh flow, sonic conditions are reached as the
entropy reaches its maximum value, and maximum temperature occurs during subsonic flow. Vb EXAMPLE 17–14 and Mab Vb
cb 1 (10) Effect of Heat Transfer on Flow Velocity Starting with the differential form of the energy equation, show that the flow
velocity increases with heat addition in subsonic Rayleigh flow, but decreases
in supersonic Rayleigh flow. Solution It is to be shown that flow velocity increases with heat addition in
subsonic Rayleigh flow and that the opposite occurs in supersonic flow.
Assumptions 1 The assumptions associated with Rayleigh flow are valid.
2 There are no work interactions and potential energy changes are negligible.
Analysis Consider heat transfer to the fluid in the differential amount of dq.
The differential form of the energy equations can be expressed as dq dh 0 dah V2
b
2 cp dT Dividing by cpT and factoring out dV/V give dq
cpT dT
T V dV
cpT 1k dV V dT
c
V dV T V dV
12V 2 kRT where we also used cp
kR/(k
1). Noting that Ma2
using Eq. 7 for dT/dV from Example 17–13 give dq
cpT dV V T
ca
VTV V
b
R 1k 1 2 Ma2 d dV
a1
V dq
cpT 1 1 1
Ma2 2 d (2) V 2/c2 V2
TR Canceling the two middle terms in Eq. 3 since V 2/TR
ing give the desired relation, dV
V (1) kMa2 V 2/kRT and Ma2 b (3) dq V1 kMa2 and rearrang Subsonic
flow V2 V1 V2 V1 dq
(4) In subsonic flow, 1
Ma2
0 and thus heat transfer and velocity change
have the same sign. As a result, heating the fluid (dq
0) increases the
flow velocity while cooling decreases it. In supersonic flow, however, 1
Ma2
0 and heat transfer and velocity change have opposite signs. As a
result, heating the fluid (dq
0) decreases the flow velocity while cooling
increases it (Fig. 17–55).
Discussion Note that heating the fluid has the opposite effect on flow velocity in subsonic and supersonic Rayleigh flows. V1 Supersonic
flow FIGURE 17–55
Heating increases the flow velocity in
subsonic flow, but decreases it in
supersonic flow. cen84959_ch17.qxd 4/21/05 11:08 AM Page 866 866  Thermodynamics It is often desirable to express the variations in properties in terms of the Mach
number Ma. Noting that Ma V> c V> 1kRT and thus V Ma 1kRT , Property Relations for Rayleigh Flow
rV 2 rkRT Ma2 kPMa2 (17–57) since P rRT. Substituting into the momentum equation (Eq. 17–51) gives
2
P2 kP2Ma 2, which can be rearranged as
P1 kP1Ma 1
2
Ma 1kRT , the continuity equation r1V1
P2
P1 Again utilizing V
expressed as r1
r2 kMa2
1
kMa2
2 1
1 Ma2 2kRT2
Ma1 2kRT1 V2
V1 (17–58) Ma2 2T2 r2V2 can be Ma1 2T1 kMa2
Ma2 1T2
1
ba
b
kMa2
Ma1 1T1
2 (17–59) Then the idealgas relation (Eq. 17–56) becomes
T2
T1 a P2 r 1
P1 r 2 1
1 (17–60) Solving Eq. 17–60 for the temperature ratio T2/T1 gives
c T2
T1
T0
T0* (k 1)Ma2[2
)Ma [2 k kMa )
Ma 12
a
kMa2
Ma (k
k *
P0 1 T
T*
P
P* 1 k
kMa2
Ma r* (1 k)Ma2
)M r 1 b Ma(1 k) 2
Ma(
a
b
Ma
1 kMa2 kMa2
Ma V
V* 1 Ma1 1 1 kMa2 2
2 d 2 (17–61) Ma2 1 1
1 kMa2 2
2 Ma2 1 1
2 kMa2 2
1 (17–62) Flow properties at sonic conditions are usually easy to determine, and thus
the critical state corresponding to Ma
1 serves as a convenient reference
point in compressible flow. Taking state 2 to be the sonic state (Ma2 1, and
superscript * is used) and state 1 to be any state (no subscript), the property
relations in Eqs. 17–58, 17–61, and 17–62 reduce to (Fig. 17–56)
P
P* FIGURE 17–56
Summary of relations for Rayleigh
flow. V1
V2 r2
r1 1)Ma2 k/(k 1)
)Ma /(
1 kMa2 2
1 Substituting this relation into Eq. 17–59 gives the density or velocity ratio as 22 (1
(1 P0 1)Ma2]
)Ma (k Ma2 1 1 1
1 k
kMa2 T
T* c Ma 1 1
1 k2
2 kMa d 2 and 11 r*
r V
V* k 2 Ma2
kMa2 1 (17–63) Similar relations can be obtained for dimensionless stagnation temperature and stagnation pressure as follows:
T0
T*
0 T0 T T *
T T* T*
0 a1 k 1
2 Ma2 b c Ma 1 1
1 k2 kMa2 d a1
2 k 1
2 b 1 (17–64) which simplifies to
T0
T*
0 1k 1 2 Ma2 3 2
11 1k kMa2 2 2 1 2 Ma2 4 (17–65) cen84959_ch17.qxd 4/21/05 11:08 AM Page 867 Chapter 17  867 Also,
P0
P*
0 P0 P P*
P P* P*
0 a1 k 1
2 Ma2 b k>1k 12 a 1
1 k
b a1
kMa2 k 1
2 b k>1k 12 (17–66) which simplifies to
P0
P*
0 k
1 2
1
c
2
kMa 1k 1 2 Ma2 k 1 d k>1k 12 (17–67) The five relations in Eqs. 17–63, 17–65, and 17–67 enable us to calculate
the dimensionless pressure, temperature, density, velocity, stagnation temperature, and stagnation pressure for Rayleigh flow of an ideal gas with a
specified k for any given Mach number. Representative results are given in
tabular form in Table A–34 for k 1.4. Choked Rayleigh Flow
It is clear from the earlier discussions that subsonic Rayleigh flow in a duct
may accelerate to sonic velocity (Ma
1) with heating. What happens if
we continue to heat the fluid? Does the fluid continue to accelerate to supersonic velocities? An examination of the Rayleigh line indicates that the fluid
at the critical state of Ma 1 cannot be accelerated to supersonic velocities
by heating. Therefore, the flow is choked. This is analogous to not being
able to accelerate a fluid to supersonic velocities in a converging nozzle by
simply extending the converging flow section. If we keep heating the fluid,
we will simply move the critical state further downstream and reduce the
flow rate since fluid density at the critical state will now be lower. Therefore, for a given inlet state, the corresponding critical state fixes the maximum possible heat transfer for steady flow (Fig. 17–57). That is,
qmax h*
0 h 01 cp 1 T *
0 T01 2 (17–68) Further heat transfer causes choking and thus the inlet state to change (e.g.,
inlet velocity will decrease), and the flow no longer follows the same Rayleigh
line. Cooling the subsonic Rayleigh flow reduces the velocity, and the Mach
number approaches zero as the temperature approaches absolute zero. Note
that the stagnation temperature T0 is maximum at the critical state of Ma 1.
In supersonic Rayleigh flow, heating decreases the flow velocity. Further
heating simply increases the temperature and moves the critical state further
downstream, resulting in a reduction in the mass flow rate of the fluid.
It may seem like supersonic Rayleigh flow can be cooled indefinitely, but it
turns out that there is a limit. Taking the limit of Eq. 17–65 as the Mach
number approaches infinity gives
LimMaS T0
T*
0 1 1
k2 (17–69) which yields T0/T * 0.49 for k
1.4. Therefore, if the critical stagnation
0
temperature is 1000 K, air cannot be cooled below 490 K in Rayleigh flow.
Physically this means that the flow velocity reaches infinity by the time the
temperature reaches 490 K—a physical impossibility. When supersonic flow
cannot be sustained, the flow undergoes a normal shock wave and becomes
subsonic. qmax T1
T01 Rayleigh
flow T2
T02 T*
*
T 01 Choked
flow FIGURE 17–57
For a given inlet state, the maximum
possible heat transfer occurs when
sonic conditions are reached at the exit
state. cen84959_ch17.qxd 4/21/05 11:08 AM Page 868 868  Thermodynamics
. EXAMPLE 17–15 Q
P1
T1 4 80 kPa
5 50 K V1 8 0 m/s Combustor
tube P2, T 2, V 2 FIGURE 17–58
Schematic of the combustor tube
analyzed in Example 17–15. Rayleigh Flow in a Tubular Combustor A combustion chamber consists of tubular combustors of 15cm diameter.
Compressed air enters the tubes at 550 K, 480 kPa, and 80 m/s (Fig.
17–58). Fuel with a heating value of 42,000 kJ/kg is injected into the air
and is burned with an air–fuel mass ratio of 40. Approximating combustion
as a heat transfer process to air, determine the temperature, pressure, velocity, and Mach number at the exit of the combustion chamber. Solution Fuel is burned in a tubular combustion chamber with compressed
air. The exit temperature, pressure, velocity, and Mach number are to be
determined.
Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steady
onedimensional flow of an ideal gas with constant properties through a constant crosssectionalarea duct with negligible frictional effects) are valid.
2 Combustion is complete, and it is treated as a heat transfer process, with
no change in the chemical composition of the flow. 3 The increase in mass
flow rate due to fuel injection is disregarded.
Properties We take the properties of air to be k
1.4, cp
1.005 kJ/kg · K,
and R
0.287 kJ/kg · K (Table A–2a).
Analysis The inlet density and mass flow rate of air are P1
RT1 r1
#
mair 480 kPa
1 0.287 kJ> kg # K 2 1 550 K 2 3.041 kg> m3 1 3.041 kg> m3 2 3 p 1 0.15 m 2 2> 4 4 1 80 m> s 2 r 1A 1V1 4.299 kg> s The mass flow rate of fuel and the rate of heat transfer are #
4.299 kg> s
mair
0.1075 kg> s
AF
40
#
1 0.1075 kg> s 2 1 42,000 kJ> kg 2
m fuel HV
#
4515 kJ> s
Q
1050 kJ> kg
#
mair
4.299 kg> s #
m fuel
#
Q
q 4515 kW The stagnation temperature and Mach number at the inlet are T01 c1 Ma1 2kRT1 T1 2
V1
2cp V1
c1 B 550 K 1 80 m> s 2 2 2 1 1.005 kJ> kg # K 2 a 1 1.4 2 1 0.287 kJ> kg # K 2 1 550 K 2 a 80 m/s
470.1 m/s 1 kJ> kg 1000 m2> s2
1000 m2> s2
1 kJ> kg T01 553.2 K b 470.1 m> s 0.1702 The exit stagnation temperature is, from the energy equation q T02 b q
cp 553.2 K 1050 kJ/ kg
1.005 kJ/ kg # K cp(T02 1598 K T01), cen84959_ch17.qxd 4/21/05 11:08 AM Page 869 Chapter 17
The maximum value of stagnation temperature T 0 occurs at Ma
*
1, and its
value can be determined from Table A–34 or from Eq. 17–65. At Ma1
0.1702 we read T0/T *
0.1291. Therefore,
0 T01
0.1291 T*
0 553.2 K
0.1291 4285 K The stagnation temperature ratio at the exit state and the Mach number corresponding to it are, from Table A–34, T02
T*
0 1598 K
4285 K 0.3729 S Ma2 0.3142 The Rayleigh flow relations corresponding to the inlet and exit Mach numbers are (Table A–34): Ma1 0.1702: T1
T* 0.1541 P1
P* 2.3065 V1
V* 0.0668 Ma2 0.3142: T2
T* 0.4389 P2
P* 2.1086 V2
V* 0.2082 Then the exit temperature, pressure, and velocity are determined to be T2
T1
P2
P1
V2
V1 T2> T*
T1> T* P2/P*
P1/P* V2> V*
V1> V* 0.4389
0.1541 2.848 S T2 2.1086
2.3065 0.9142 S P2 0.2082
0.0668 3.117 S V2 2.848T1
0.9142P1
3.117V1 2.848 1 550 K 2 1566 K 0.9142 1 480 kPa 2
3.117 1 80 m> s 2 439 kPa 249 m/s Discussion Note that the temperature and velocity increase and pressure
decreases during this subsonic Rayleigh flow with heating, as expected. This
problem can also be solved using appropriate relations instead of tabulated
values, which can likewise be coded for convenient computer solutions. 17–7 ■ STEAM NOZZLES We have seen in Chapter 3 that water vapor at moderate or high pressures
deviates considerably from idealgas behavior, and thus most of the relations developed in this chapter are not applicable to the flow of steam
through the nozzles or blade passages encountered in steam turbines. Given
that the steam properties such as enthalpy are functions of pressure as well
as temperature and that no simple property relations exist, an accurate analysis of steam flow through the nozzles is no easy matter. Often it becomes
necessary to use steam tables, an hs diagram, or a computer program for
the properties of steam.
A further complication in the expansion of steam through nozzles occurs
as the steam expands into the saturation region, as shown in Fig. 17–59. As
the steam expands in the nozzle, its pressure and temperature drop, and  869 cen84959_ch17.qxd 4/21/05 11:08 AM Page 870 870  Thermodynamics
P 1 h
1 P2
Saturation
line 2
Wilson line (x = 0.96)
s FIGURE 17–59
The hs diagram for the isentropic
expansion of steam in a nozzle. ordinarily one would expect the steam to start condensing when it strikes
the saturation line. However, this is not always the case. Owing to the high
speeds, the residence time of the steam in the nozzle is small, and there may
not be sufficient time for the necessary heat transfer and the formation of
liquid droplets. Consequently, the condensation of the steam may be
delayed for a little while. This phenomenon is known as supersaturation,
and the steam that exists in the wet region without containing any liquid is
called supersaturated steam. Supersaturation states are nonequilibrium (or
metastable) states.
During the expansion process, the steam reaches a temperature lower than
that normally required for the condensation process to begin. Once the temperature drops a sufficient amount below the saturation temperature corresponding to the local pressure, groups of steam moisture droplets of
sufficient size are formed, and condensation occurs rapidly. The locus of
points where condensation takes place regardless of the initial temperature
and pressure at the nozzle entrance is called the Wilson line. The Wilson
line lies between the 4 and 5 percent moisture curves in the saturation
region on the hs diagram for steam, and it is often approximated by the
4 percent moisture line. Therefore, steam flowing through a highvelocity
nozzle is assumed to begin condensation when the 4 percent moisture line is
crossed.
The criticalpressure ratio P*/P0 for steam depends on the nozzle inlet state
as well as on whether the steam is superheated or saturated at the nozzle inlet.
However, the idealgas relation for the criticalpressure ratio, Eq. 17–22, gives
reasonably good results over a wide range of inlet states. As indicated in
Table 17–2, the specific heat ratio of superheated steam is approximated as
k 1.3. Then the criticalpressure ratio becomes
P*
P0 a 2
k 1 b k>1k 12 0.546 When steam enters the nozzle as a saturated vapor instead of superheated
vapor (a common occurrence in the lower stages of a steam turbine), the
criticalpressure ratio is taken to be 0.576, which corresponds to a specific
heat ratio of k 1.14. EXAMPLE 17–16 Steam Flow through a
C onverging–Diverging Nozzle Steam enters a converging–diverging nozzle at 2 MPa and 400°C with a negligible velocity and a mass flow rate of 2.5 kg/s, and it exits at a pressure of
300 kPa. The flow is isentropic between the nozzle entrance and throat, and
the overall nozzle efficiency is 93 percent. Determine (a) the throat and exit
areas and (b) the Mach number at the throat and the nozzle exit. Solution Steam enters a converging–diverging nozzle with a low velocity.
The throat and exit areas and the Mach number are to be determined.
Assumptions 1 Flow through the nozzle is onedimensional. 2 The flow is
isentropic between the inlet and the throat, and is adiabatic and irreversible
between the throat and the exit. 3 The inlet velocity is negligible. cen84959_ch17.qxd 4/21/05 11:08 AM Page 871 Chapter 17
Analysis We denote the entrance, throat, and exit states by 1, t, and 2,
respectively, as shown in Fig. 17–60.
(a) Since the inlet velocity is negligible, the inlet stagnation and static states
are identical. The ratio of the exittoinlet stagnation pressure is P2
P01 300 kPa
2000 kPa 0.546P01 1 0.546 2 1 2 MPa 2 2 MPa
f
400°C P01
T01 h P1 1 1.09 MPa =P0 1 P2 2s 22 1 h01 1.09 MPa
f
7.1292 kJ> kg # K
B ht
vt ht 2 1000 m2> s2
3076.8 2 kJ/kg 4 a
b
1 kJ> kg 3 2 1 3248.4 FIGURE 17–60
Schematic and hs diagram for
Example 17–16.
585.8 m/s The flow area at the throat is determined from the mass flow rate relation: At 1 2.5 kg/s 2 1 0.2420 m3/kg 2 #
mvt
Vt 585.8 m/s 10.33 4 10 m2 10.33 cm2 At state 2s, P2s
s2s 300 kPa
f
7.1292 kJ> kg # K P2
s1 h 2s 2783.6 kJ> kg The enthalpy of the steam at the actual exit state is (see Chap. 7) hN
0.93 h01
h01 h2
h2 s 3248.4 h2
3248.4 2783.6 ¡ h2 2816.1 kJ> kg Therefore, 22 1 h01 P2
h2 300 kPa
f
2816.1 kJ> kg B v2
s2 0.67723 m3> kg
7.2019 kJ> kg # K Then the exit velocity and the exit area become V2 A2 #
mv2
V2 h2 2 3 2 1 3248.4 2816.1 2 kJ> kg 4 a 1 2.5 kg> s 2 1 0.67723 m3> kg 2
929.8 m> s 18.21 1000 m2> s2 10 1 kJ> kg
4 m2 =3 00 a kP a s 3076.8 kJ> kg
0.24196 m3> kg Then the throat velocity is determined from Eq. 17–3 to be Vt MP 2 Also, at the throat, Pt
st =2 Pt t h 01 3248.4 kJ> kg
st s2s 7.1292 kJ> kg # K h1
s1 ·
m = 2.5 kg/s V1 ≅ 0
STEAM Throat At the inlet, P1
T1 871 hN = 93% 0.15 It is much smaller than the criticalpressure ratio, which is taken to be
P */P01
0.546 since the steam is superheated at the nozzle inlet. Therefore, the flow surely is supersonic at the exit. Then the velocity at the throat
is the sonic velocity, and the throat pressure is Pt P1 = 2 MPa
T1 = 400°C  b 929.8 m> s 18.21 cm2 cen84959_ch17.qxd 4/21/05 11:08 AM Page 872 872  Thermodynamics
(b) The velocity of sound and the Mach numbers at the throat and the exit
of the nozzle are determined by replacing differential quantities with
differences, a c 0 P 1>2
b
0r s ¢ P 1>2
d
¢ 1 1> v 2 s c The velocity of sound at the throat is determined by evaluating the specific
volume at st
7.1292 kJ/kg · K and at pressures of 1.115 and 1.065 MPa
(Pt
25 kPa): c B 1 1> 0.23776 1 1115 1065 2 kPa 1> 0.24633 2 kg> m 3 1000 m2> s2 a 1 kPa # m3> kg 584.6 m> s b The Mach number at the throat is determined from Eq. 17–12 to be Ma V
c 585.8 m> s 584.6 m> s 1.002 Thus, the flow at the throat is sonic, as expected. The slight deviation of
the Mach number from unity is due to replacing the derivatives by
differences.
The velocity of sound and the Mach number at the nozzle exit are determined by evaluating the specific volume at s2
7.2019 kJ/kg · K and at
pressures of 325 and 275 kPa (P2
25 kPa): c B 1 1> 0.63596 1 325 275 2 kPa 1000 m2> s2
a
b
1> 0.72245 2 kg> m3 1 kPa # m3> kg and Ma V
c 929.8 m> s 515.4 m> s 515.4 m> s 1.804 Thus the flow of steam at the nozzle exit is supersonic. SUMMARY
In this chapter the effects of compressibility on gas flow are
examined. When dealing with compressible flow, it is convenient to combine the enthalpy and the kinetic energy of the
fluid into a single term called stagnation (or total) enthalpy
h0, defined as
V2
h0 h
2
The properties of a fluid at the stagnation state are called
stagnation properties and are indicated by the subscript zero.
The stagnation temperature of an ideal gas with constant specific heats is
V2
T0 T
2cp
which represents the temperature an ideal gas would attain if
it is brought to rest adiabatically. The stagnation properties of
an ideal gas are related to the static properties of the fluid by P0
P a T0 k>1k
b
T 12 r0
r and a T0 1>1k
b
T 12 2kRT The speed at which an infinitesimally small pressure wave
travels through a medium is the speed of sound. For an ideal
gas it is expressed as
c B a 0P
b
0r s The Mach number is the ratio of the actual velocity of the
fluid to the speed of sound at the same state:
Ma V
c The flow is called sonic when Ma 1, subsonic when Ma 1,
supersonic when Ma
1, hypersonic when Ma
1, and
transonic when Ma 1. cen84959_ch17.qxd 4/21/05 11:08 AM Page 873 Chapter 17
Nozzles whose flow area decreases in the flow direction
are called converging nozzles. Nozzles whose flow area first
decreases and then increases are called converging–diverging
nozzles. The location of the smallest flow area of a nozzle is
called the throat. The highest velocity to which a fluid can be
accelerated in a converging nozzle is the sonic velocity.
Accelerating a fluid to supersonic velocities is possible only
in converging–diverging nozzles. In all supersonic converging–
diverging nozzles, the flow velocity at the throat is the speed
of sound.
The ratios of the stagnation to static properties for ideal
gases with constant specific heats can be expressed in terms
of the Mach number as
a T0
T
P0
P c1 a k r0
r and 1 c1 a k k 1
2 b Ma T01 b Ma d 1 b Ma d 2 2 2 2 When Ma
1, the resulting statictostagnation property
ratios for the temperature, pressure, and density are called
critical ratios and are denoted by the superscript asterisk:
T*
T0
and k 1 r*
r0 a 2 2 k P*
P0 1 b a 2
k 1 b k>1k 12 1>1k 12 The pressure outside the exit plane of a nozzle is called the
back pressure. For all back pressures lower than P*, the pres T02 T2
T1
and 2 P2
P1 k>1k 12 1>1k 12 873 sure at the exit plane of the converging nozzle is equal to P*,
the Mach number at the exit plane is unity, and the mass flow
rate is the maximum (or choked) flow rate.
In some range of back pressure, the fluid that achieved a
sonic velocity at the throat of a converging–diverging nozzle
and is accelerating to supersonic velocities in the diverging
section experiences a normal shock, which causes a sudden
rise in pressure and temperature and a sudden drop in velocity to subsonic levels. Flow through the shock is highly
irreversible, and thus it cannot be approximated as isentropic. The properties of an ideal gas with constant specific
heats before (subscript 1) and after (subscript 2) a shock are
related by 2 1  1
1 2 Ma2
Ma2 1 k
1
Ma2 1 k
2
kMa2
1
kMa2
2 B 1k 1 2 Ma2
1 2kMa2
1 2 k 1 12
12 2kMa2
1
k k
1 1 These equations also hold across an oblique shock, provided
that the component of the Mach number normal to the
oblique shock is used in place of the Mach number.
Steady onedimensional flow of an ideal gas with constant
specific heats through a constantarea duct with heat transfer
and negligible friction is referred to as Rayleigh flow. The
property relations and curves for Rayleigh flow are given in
Table A–34. Heat transfer during Rayleigh flow can be determined from
q cp 1 T02 T01 2 cp 1 T2 T1 2 V2
2 V2
1
2 REFERENCES AND SUGGESTED READINGS
1. J. D. Anderson. Modern Compressible Flow with Historical
Perspective. 3rd ed. New York: McGrawHill, 2003. NACA Report 1135, http://naca.larc.nasa.gov/reports/
1953/nacareport1135/. 2. Y. A. Çengel and J. M. Cimbala. Fluid Mechanics:
Fundamentals and Applications. New York: McGrawHill, 2006. 8. A. H. Shapiro. The Dynamics and Thermodynamics of
Compressible Fluid Flow. vol. 1. New York: Ronald Press
Company, 1953. 3. H. Cohen, G. F. C. Rogers, and H. I. H. Saravanamuttoo.
Gas Turbine Theory. 3rd ed. New York: Wiley, 1987. 9. P. A. Thompson. CompressibleFluid Dynamics. New
York: McGrawHill, 1972. 4. W. J. Devenport. Compressible Aerodynamic Calculator,
http://www.aoe.vt.edu/~devenpor/aoe3114/calc.html. 10. United Technologies Corporation. The Aircraft Gas
Turbine and Its Operation. 1982. 5. R. W. Fox and A. T. McDonald. Introduction to Fluid
Mechanics. 5th ed. New York: Wiley, 1999. 11. Van Dyke, 1982. 6. H. Liepmann and A. Roshko. Elements of Gas Dynamics.
Dover Publications, Mineola, NY, 2001.
7. C. E. Mackey, responsible NACA officer and curator.
Equations, Tables, and Charts for Compressible Flow. 12. F. M. White. Fluid Mechanics. 5th ed. New York:
McGrawHill, 2003. cen84959_ch17.qxd 4/21/05 11:08 AM Page 874 874  Thermodynamics PROBLEMS*
Stagnation Properties
17–1C A highspeed aircraft is cruising in still air. How
will the temperature of air at the nose of the aircraft differ
from the temperature of air at some distance from the
aircraft?
17–2C How and why is the stagnation enthalpy h0 defined?
How does it differ from ordinary (static) enthalpy?
17–3C What is dynamic temperature?
17–4C In airconditioning applications, the temperature of
air is measured by inserting a probe into the flow stream.
Thus, the probe actually measures the stagnation temperature.
Does this cause any significant error?
17–5 Determine the stagnation temperature and stagnation
pressure of air that is flowing at 44 kPa, 245.9 K, and 470
m/s. Answers: 355.8 K, 160.3 kPa
17–6 Air at 300 K is flowing in a duct at a velocity of (a) 1,
(b) 10, (c) 100, and (d) 1000 m/s. Determine the temperature
that a stationary probe inserted into the duct will read for
each case. be isentropic, determine the power output of the turbine per
unit mass flow.
17–11 Air flows through a device such that the stagnation
pressure is 0.6 MPa, the stagnation temperature is 400°C, and
the velocity is 570 m/s. Determine the static pressure and temperature of the air at this state. Answers: 518.6 K, 0.23 MPa Speed of Sound and Mach Number
17–12C What is sound? How is it generated? How does it
travel? Can sound waves travel in a vacuum?
17–13C Is it realistic to assume that the propagation of
sound waves is an isentropic process? Explain.
17–14C Is the sonic velocity in a specified medium a fixed
quantity, or does it change as the properties of the medium
change? Explain.
17–15C In which medium does a sound wave travel faster:
in cool air or in warm air?
17–16C In which medium will sound travel fastest for a
given temperature: air, helium, or argon? 17–7 Calculate the stagnation temperature and pressure for
the following substances flowing through a duct: (a) helium at
0.25 MPa, 50°C, and 240 m/s; (b) nitrogen at 0.15 MPa, 50°C,
and 300 m/s; and (c) steam at 0.1 MPa, 350°C, and 480 m/s. 17–17C In which medium does a sound wave travel faster:
in air at 20°C and 1 atm or in air at 20°C and 5 atm? 17–8 Air enters a compressor with a stagnation pressure of
100 kPa and a stagnation temperature of 27°C, and it is compressed to a stagnation pressure of 900 kPa. Assuming the
compression process to be isentropic, determine the power
input to the compressor for a mass flow rate of 0.02 kg/s. 17–19 Determine the speed of sound in air at (a) 300 K and
(b) 1000 K. Also determine the Mach number of an aircraft
moving in air at a velocity of 280 m/s for both cases. Answer: 5.27 kW 17–9E Steam flows through a device with a stagnation
pressure of 120 psia, a stagnation temperature of 700°F, and a
velocity of 900 ft/s. Assuming idealgas behavior, determine
the static pressure and temperature of the steam at this state.
17–10 Products of combustion enter a gas turbine with a
stagnation pressure of 1.0 MPa and a stagnation temperature
of 750°C, and they expand to a stagnation pressure of 100
kPa. Taking k 1.33 and R 0.287 kJ/kg · K for the products of combustion, and assuming the expansion process to 17–18C Does the Mach number of a gas flowing at a constant velocity remain constant? Explain. 17–20 Carbon dioxide enters an adiabatic nozzle at 1200 K
with a velocity of 50 m/s and leaves at 400 K. Assuming constant specific heats at room temperature, determine the Mach
number (a) at the inlet and (b) at the exit of the nozzle.
Assess the accuracy of the constant specific heat assumption.
Answers: (a) 0.0925, (b) 3.73 17–21 Nitrogen enters a steadyflow heat exchanger at 150
kPa, 10°C, and 100 m/s, and it receives heat in the amount of
120 kJ/kg as it flows through it. Nitrogen leaves the heat
exchanger at 100 kPa with a velocity of 200 m/s. Determine
the Mach number of the nitrogen at the inlet and the exit of
the heat exchanger.
17–22 Assuming idealgas behavior, determine the speed of
sound in refrigerant134a at 0.1 MPa and 60°C. *Problems designated by a “C” are concept questions, and students
are encouraged to answer them all. Problems designated by an “E”
are in English units, and the SI users can ignore them. Problems
with a CDEES icon
are solved using EES, and complete solutions
together with parametric studies are included on the enclosed DVD.
Problems with a computerEES icon
are comprehensive in nature,
and are intended to be solved with a computer, preferably using the
EES software that accompanies this text. 17–23 The Airbus A340 passenger plane has a maximum
takeoff weight of about 260,000 kg, a length of 64 m, a wing
span of 60 m, a maximum cruising speed of 945 km/h, a
seating capacity of 271 passengers, maximum cruising altitude of 14,000 m, and a maximum range of 12,000 km. The
air temperature at the crusing altitude is about 60°C. Determine the Mach number of this plane for the stated limiting
conditions. cen84959_ch17.qxd 4/21/05 11:08 AM Page 875 Chapter 17
17–24E Steam flows through a device with a pressure of
120 psia, a temperature of 700°F, and a velocity of 900 ft/s.
Determine the Mach number of the steam at this state by
assuming idealgas behavior with k 1.3. Answer: 0.441
17–25E Reconsider Prob. 17–24E. Using EES (or
other) software, compare the Mach number of
steam flow over the temperature range 350 to 700°F. Plot the
Mach number as a function of temperature.  875 pressure that can be obtained at the throat of the nozzle?
Answer: 634 kPa 17–38 Helium enters a converging–diverging nozzle at 0.7
MPa, 800 K, and 100 m/s. What are the lowest temperature
and pressure that can be obtained at the throat of the nozzle?
17–39 Calculate the critical temperature, pressure, and density of (a) air at 200 kPa, 100°C, and 250 m/s, and (b) helium
at 200 kPa, 40°C, and 300 m/s. 17–26 The isentropic process for an ideal gas is expressed
as Pv k constant. Using this process equation and the definition of the speed of sound (Eq. 17–9), obtain the expression
for the speed of sound for an ideal gas (Eq. 17–11). 17–40 Quiescent carbon dioxide at 600 kPa and 400 K is
accelerated isentropically to a Mach number of 0.5. Determine the temperature and pressure of the carbon dioxide after
acceleration. Answers: 388 K, 514 kPa 17–27 Air expands isentropically from 1.5 MPa and 60°C
to 0.4 MPa. Calculate the ratio of the initial to final speed of
sound. Answer: 1.21 17–41 Air at 200 kPa, 100°C, and Mach number Ma 0.8
flows through a duct. Find the velocity and the stagnation
pressure, temperature, and density of the air. 17–28 17–42 Repeat Prob. 17–27 for helium gas. 17–29E Air expands isentropically from 170 psia and
200°F to 60 psia. Calculate the ratio of the initial to final
speed of sound. Answer: 1.16 OneDimensional Isentropic Flow
17–30C Consider a converging nozzle with sonic velocity
at the exit plane. Now the nozzle exit area is reduced while
the nozzle inlet conditions are maintained constant. What will
happen to (a) the exit velocity and (b) the mass flow rate
through the nozzle?
17–31C A gas initially at a supersonic velocity enters an
adiabatic converging duct. Discuss how this affects (a) the
velocity, (b) the temperature, (c) the pressure, and (d) the
density of the fluid.
17–32C A gas initially at a supersonic velocity enters an
adiabatic diverging duct. Discuss how this affects (a) the
velocity, (b) the temperature, (c) the pressure, and (d) the
density of the fluid.
17–33C A gas initially at a supersonic velocity enters an
adiabatic converging duct. Discuss how this affects (a) the
velocity, (b) the temperature, (c) the pressure, and (d) the
density of the fluid.
17–34C A gas initially at a subsonic velocity enters an adiabatic diverging duct. Discuss how this affects (a) the velocity, (b) the temperature, (c) the pressure, and (d) the density
of the fluid.
17–35C A gas at a specified stagnation temperature and
pressure is accelerated to Ma
2 in a converging–diverging
nozzle and to Ma
3 in another nozzle. What can you say
about the pressures at the throats of these two nozzles?
17–36C Is it possible to accelerate a gas to a supersonic
velocity in a converging nozzle?
17–37 Air enters a converging–diverging nozzle at a pressure of 1.2 MPa with negligible velocity. What is the lowest Reconsider Prob. 17–41. Using EES (or other)
software, study the effect of Mach numbers in
the range 0.1 to 2 on the velocity, stagnation pressure, temperature, and density of air. Plot each parameter as a function
of the Mach number. 17–43E Air at 30 psia, 212°F, and Mach number Ma 0.8
flows through a duct. Calculate the velocity and the stagnation pressure, temperature, and density of air.
Answers: 1016 ft/s, 45.7 psia, 758 R, 0.163 lbm/ft3 17–44 An aircraft is designed to cruise at Mach number
Ma
1.2 at 8000 m where the atmospheric temperature is
236.15 K. Determine the stagnation temperature on the leading edge of the wing. Isentropic Flow through Nozzles
17–45C Consider subsonic flow in a converging nozzle
with fixed inlet conditions. What is the effect of dropping the
back pressure to the critical pressure on (a) the exit velocity,
(b) the exit pressure, and (c) the mass flow rate through the
nozzle?
17–46C Consider subsonic flow in a converging nozzle
with specified conditions at the nozzle inlet and critical pressure at the nozzle exit. What is the effect of dropping the
back pressure well below the critical pressure on (a) the exit
velocity, (b) the exit pressure, and (c) the mass flow rate
through the nozzle?
17–47C Consider a converging nozzle and a converging–
diverging nozzle having the same throat areas. For the same
inlet conditions, how would you compare the mass flow rates
through these two nozzles?
17–48C Consider gas flow through a converging nozzle
with specified inlet conditions. We know that the highest
velocity the fluid can have at the nozzle exit is the sonic
velocity, at which point the mass flow rate through the nozzle
is a maximum. If it were possible to achieve hypersonic cen84959_ch17.qxd 4/21/05 11:08 AM Page 876 876  Thermodynamics velocities at the nozzle exit, how would it affect the mass
flow rate through the nozzle?
17–49C How does the parameter Ma* differ from the Mach
number Ma?
17–50C What would happen if we attempted to decelerate
a supersonic fluid with a diverging diffuser?
17–51C What would happen if we tried to further accelerate a supersonic fluid with a diverging diffuser?
17–52C Consider the isentropic flow of a fluid through a
converging–diverging nozzle with a subsonic velocity at the
throat. How does the diverging section affect (a) the velocity,
(b) the pressure, and (c) the mass flow rate of the fluid?
17–53C Is it possible to accelerate a fluid to supersonic
velocities with a velocity other than the sonic velocity at the
throat? Explain. 17–54 Explain why the maximum flow rate per unit area
for a given gas depends only on P0/ 1T0. For an ideal gas
with k
1.4 and R
0.287 kJ/kg · K, find the constant a
.
such that m /A* aP0/ 1T0.
17–55 For an ideal gas obtain an expression for the ratio of
the velocity of sound where Ma
1 to the speed of sound
based on the stagnation temperature, c*/c0.
17–56 An ideal gas flows through a passage that first converges and then diverges during an adiabatic, reversible,
steadyflow process. For subsonic flow at the inlet, sketch the
variation of pressure, velocity, and Mach number along the
length of the nozzle when the Mach number at the minimum
flow area is equal to unity.
17–57 Repeat Prob. 17–56 for supersonic flow at the inlet. 17–58 Air enters a nozzle at 0.2 MPa, 350 K, and a velocity of 150 m/s. Assuming isentropic flow, determine the pressure and temperature of air at a location where the air
velocity equals the speed of sound. What is the ratio of the
area at this location to the entrance area?
Answers: 0.118 MPa, 301 K, 0.629 17–63 An ideal gas with k 1.4 is flowing through a nozzle such that the Mach number is 2.4 where the flow area is
25 cm2. Assuming the flow to be isentropic, determine the
flow area at the location where the Mach number is 1.2.
17–64 Repeat Prob. 17–63 for an ideal gas with k 1.33. 17–65 Air at 900 kPa and 400 K enters a converging
nozzle with a negligible velocity. The throat area
of the nozzle is 10 cm2. Assuming isentropic flow, calculate
and plot the exit pressure, the exit velocity, and the mass flow
rate versus the back pressure Pb for 0.9 Pb 0.1 MPa.
17–66 Reconsider Prob. 17–65. Using EES (or other)
software, solve the problem for the inlet conditions of 1 MPa and 1000 K. 17–67E Air enters a converging–diverging nozzle of a
supersonic wind tunnel at 150 psia and 100°F with a low
velocity. The flow area of the test section is equal to the exit
area of the nozzle, which is 5 ft2. Calculate the pressure, temperature, velocity, and mass flow rate in the test section for a
Mach number Ma 2. Explain why the air must be very dry
for this application. Answers: 19.1 psia, 311 R, 1729 ft/s,
1435 lbm/s Shock Waves and Expansion Waves
17–68C Can a shock wave develop in the converging section of a converging–diverging nozzle? Explain.
17–69C What do the states on the Fanno line and the
Rayleigh line represent? What do the intersection points of
these two curves represent?
17–70C Can the Mach number of a fluid be greater than 1
after a shock wave? Explain.
17–71C How does the normal shock affect (a) the fluid
velocity, (b) the static temperature, (c) the stagnation temperature, (d) the static pressure, and (e) the stagnation pressure?
17–72C How do oblique shocks occur? How do oblique
shocks differ from normal shocks? 17–59 Repeat Prob. 17–58 assuming the entrance velocity
is negligible. 17–73C For an oblique shock to occur, does the upstream
flow have to be supersonic? Does the flow downstream of an
oblique shock have to be subsonic? 17–60E Air enters a nozzle at 30 psia, 630 R, and a velocity of 450 ft/s. Assuming isentropic flow, determine the pressure and temperature of air at a location where the air
velocity equals the speed of sound. What is the ratio of the
area at this location to the entrance area? 17–74C It is claimed that an oblique shock can be analyzed
like a normal shock provided that the normal component of
velocity (normal to the shock surface) is used in the analysis.
Do you agree with this claim? 17–61 Air enters a converging–diverging nozzle at 0.5 MPa
with a negligible velocity. Assuming the flow to be isentropic, determine the back pressure that will result in an exit
Mach number of 1.8. Answer: 0.087 MPa 17–75C Consider supersonic airflow approaching the nose
of a twodimensional wedge and experiencing an oblique
shock. Under what conditions does an oblique shock detach
from the nose of the wedge and form a bow wave? What is
the numerical value of the shock angle of the detached shock
at the nose? 17–62 Nitrogen enters a converging–diverging nozzle at 700
kPa and 450 K with a negligible velocity. Determine the critical velocity, pressure, temperature, and density in the nozzle. 17–76C Consider supersonic flow impinging on the rounded
nose of an aircraft. Will the oblique shock that forms in front
of the nose be an attached or detached shock? Explain. Answers: 17.4 psia, 539 R, 0.574 cen84959_ch17.qxd 4/21/05 11:08 AM Page 877 Chapter 17  877 17–77C Are the isentropic relations of ideal gases applicable for flows across (a) normal shock waves, (b) oblique
shock waves, and (c) Prandtl–Meyer expansion waves? mal shock for upstream Mach numbers between 0.5 and 1.5
in increments of 0.1. Explain why normal shock waves can
occur only for upstream Mach numbers greater than Ma 1. 17–78 For an ideal gas flowing through a normal shock,
develop a relation for V2/V1 in terms of k, Ma1, and Ma2. 17–89 Consider supersonic airflow approaching the nose of
a twodimensional wedge at a Mach number of 5. Using Fig.
17–41, determine the minimum shock angle and the maximum deflection angle a straight oblique shock can have. 17–79 Air enters a converging–diverging nozzle of a supersonic wind tunnel at 1.5 MPa and 350 K with a low velocity.
If a normal shock wave occurs at the exit plane of the nozzle
at Ma
2, determine the pressure, temperature, Mach number, velocity, and stagnation pressure after the shock wave.
Answers: 0.863 MPa, 328 K, 0.577, 210 m/s, 1.081 MPa 17–80 Air enters a converging–diverging nozzle with low
velocity at 2.0 MPa and 100°C. If the exit area of the nozzle
is 3.5 times the throat area, what must the back pressure be to
produce a normal shock at the exit plane of the nozzle?
Answer: 0.661 MPa 17–81 What must the back pressure be in Prob. 17–80 for a
normal shock to occur at a location where the crosssectional
area is twice the throat area?
17–82 Air flowing steadily in a nozzle experiences a normal
shock at a Mach number of Ma
2.5. If the pressure and
temperature of air are 61.64 kPa and 262.15 K, respectively,
upstream of the shock, calculate the pressure, temperature,
velocity, Mach number, and stagnation pressure downstream
of the shock. Compare these results to those for helium undergoing a normal shock under the same conditions. 17–90 Air flowing at 60 kPa, 240 K, and a Mach number of
3.4 impinges on a twodimensional wedge of halfangle 12°.
Determine the two possible oblique shock angles, bweak and
bstrong, that could be formed by this wedge. For each case,
calculate the pressure, temperature, and Mach number downstream of the oblique shock.
17–91 Consider the supersonic flow of air at upstream conditions of 70 kPa and 260 K and a Mach number of 2.4 over
a twodimensional wedge of halfangle 10°. If the axis of the
wedge is tilted 25° with respect to the upstream airflow,
determine the downstream Mach number, pressure, and temperature above the wedge. Answers: 3.105, 23.8 kPa, 191 K Ma 2
Ma1 2 .4
25° 10° 17–83 Calculate the entropy change of air across the normal shock wave in Prob. 17–82.
17–84E Air flowing steadily in a nozzle experiences a
normal shock at a Mach number of Ma
2.5. If the pressure and temperature of air are 10.0 psia and
440.5 R, respectively, upstream of the shock, calculate the
pressure, temperature, velocity, Mach number, and stagnation
pressure downstream of the shock. Compare these results to
those for helium undergoing a normal shock under the same
conditions.
17–85E Reconsider Prob. 17–84E. Using EES (or
other) software, study the effects of both air
and helium flowing steadily in a nozzle when there is a normal shock at a Mach number in the range 2 Ma1 3.5. In
addition to the required information, calculate the entropy
change of the air and helium across the normal shock. Tabulate the results in a parametric table.
17–86 Air enters a normal shock at 22.6 kPa, 217 K, and
680 m/s. Calculate the stagnation pressure and Mach number
upstream of the shock, as well as pressure, temperature,
velocity, Mach number, and stagnation pressure downstream
of the shock.
17–87 Calculate the entropy change of air across the normal shock wave in Prob. 17–86. Answer: 0.155 kJ/kg · K
17–88 Using EES (or other) software, calculate and
plot the entropy change of air across the nor FIGURE P17–91
17–92 Reconsider Prob. 17–91. Determine the downstream
Mach number, pressure, and temperature below the wedge for
a strong oblique shock for an upstream Mach number of 5.
17–93E Air at 8 psia, 20°F, and a Mach number of 2.0 is
forced to turn upward by a ramp that makes an 8° angle off
the flow direction. As a result, a weak oblique shock forms.
Determine the wave angle, Mach number, pressure, and temperature after the shock.
17–94 Air flowing at P1 40 kPa, T1 280 K, and Ma1
3.6 is forced to undergo an expansion turn of 15°. Determine
the Mach number, pressure, and temperature of air after the
expansion. Answers: 4.81, 8.31 kPa, 179 K
17–95E Air flowing at P1
6 psia, T1
480 R, and
2.0 is forced to undergo a compression turn of 15°.
Ma1
Determine the Mach number, pressure, and temperature of air
after the compression. Duct Flow with Heat Transfer and Negligible Friction
(Rayleigh Flow)
17–96C What is the characteristic aspect of Rayleigh flow?
What are the main assumptions associated with Rayleigh
flow? cen84959_ch17.qxd 4/21/05 11:08 AM Page 878 878  Thermodynamics 17–97C On a Ts diagram of Rayleigh flow, what do the
points on the Rayleigh line represent?
17–98C What is the effect of heat gain and heat loss on the
entropy of the fluid during Rayleigh flow?
17–99C Consider subsonic Rayleigh flow of air with a
Mach number of 0.92. Heat is now transferred to the fluid
and the Mach number increases to 0.95. Will the temperature
T of the fluid increase, decrease, or remain constant during
this process? How about the stagnation temperature T0?
17–100C What is the effect of heating the fluid on the flow
velocity in subsonic Rayleigh flow? Answer the same questions for supersonic Rayleigh flow.
17–101C Consider subsonic Rayleigh flow that is accelerated to sonic velocity (Ma 1) at the duct exit by heating. If
the fluid continues to be heated, will the flow at duct exit be
supersonic, subsonic, or remain sonic?
17–102 Consider a 12cmdiameter tubular combustion
chamber. Air enters the tube at 500 K, 400 kPa, and 70 m/s.
Fuel with a heating value of 39,000 kJ/kg is burned by spraying it into the air. If the exit Mach number is 0.8, determine
the rate at which the fuel is burned and the exit temperature.
Assume complete combustion and disregard the increase in
the mass flow rate due to the fuel mass.
Fuel
P1
T1 4 00 kPa
5 00 K V1 7 0 m/s Ma2 0 .8 Combustor
tube 17–103 Air enters a rectangular duct at T1
300 K, P1
2. Heat is transferred to the air in the
420 kPa, and Ma1
amount of 55 kJ/kg as it flows through the duct. Disregarding
frictional losses, determine the temperature and Mach number at the duct exit. Answers: 386 K, 1.64
55 kJ/kg 3 00 K 17–107 Air enters a frictionless duct with V1
70
600 K, and P1
350 kPa. Letting
m/s, T1
the exit temperature T2 vary from 600 to 5000 K, evaluate the
entropy change at intervals of 200 K, and plot the Rayleigh
line on a Ts diagram.
17–108E Air is heated as it flows through a 4 in
4 in
square duct with negligible friction. At the inlet, air is at T1
700 R, P1 80 psia, and V1 260 ft/s. Determine the rate at
which heat must be transferred to the air to choke the flow at
the duct exit, and the entropy change of air during this
process.
17–109 Compressed air from the compressor of a gas turbine enters the combustion chamber at T1 550 K, P1 600
0.2 at a rate of 0.3 kg/s. Via combustion,
kPa, and Ma1
heat is transferred to the air at a rate of 200 kJ/s as it flows
through the duct with negligible friction. Determine the Mach
number at the duct exit and the drop in stagnation pressure
P01 P02 during this process. Answers: 0.319, 21.8 kPa
Repeat Prob. 17–109 for a heat transfer rate of 300 17–111 Argon gas enters a constant crosssectionalarea
0.2, P1
320 kPa, and T1
400 K at a rate
duct at Ma1
of 0.8 kg/s. Disregarding frictional losses, determine the
highest rate of heat transfer to the argon without reducing the
mass flow rate.
17–112 Consider supersonic flow of air through a 6cmdiameter duct with negligible friction. Air enters the duct at
Ma1 1.8, P01 210 kPa, and T01 600 K, and it is decelerated by heating. Determine the highest temperature that air
can be heated by heat addition while the mass flow rate
remains constant. 4 20 kPa T1 17–106E Air flows with negligible friction through a 4indiameter duct at a rate of 5 lbm/s. The temperature and pres800 R and P1
30 psia, and the
sure at the inlet are T1
1. Determine the rate of
Mach number at the exit is Ma2
heat transfer and the pressure drop for this section of the
duct. 17–110
kJ/s. FIGURE P17–102 P1 flow is observed to be choked, and the velocity and the static
pressure are measured to be 620 m/s and 270 kPa. Disregarding frictional losses, determine the velocity, static temperature, and static pressure at the duct inlet. Ma1 Air 2 Steam Nozzles
17–113C What is supersaturation? Under what conditions
does it occur? FIGURE P17–103
17–104 Repeat Prob. 17–103 assuming air is cooled in the
amount of 55 kJ/kg.
17–105 Air is heated as it flows subsonically through a
duct. When the amount of heat transfer reaches 60 kJ/kg, the 17–114 Steam enters a converging nozzle at 3.0 MPa and
500°C with a negligible velocity, and it exits at 1.8 MPa. For
a nozzle exit area of 32 cm2, determine the exit velocity,
mass flow rate, and exit Mach number if the nozzle (a) is
isentropic and (b) has an efficiency of 90 percent. Answers:
(a) 580 m/s, 10.7 kg/s, 0.918, (b) 551 m/s, 10.1 kg/s, 0.865 cen84959_ch17.qxd 4/21/05 11:08 AM Page 879 Chapter 17
17–115E Steam enters a converging nozzle at 450 psia and
900°F with a negligible velocity, and it exits at 275 psia. For
a nozzle exit area of 3.75 in2, determine the exit velocity,
mass flow rate, and exit Mach number if the nozzle (a) is
isentropic and (b) has an efficiency of 90 percent. Answers:
(a) 1847 ft/s, 18.7 lbm/s, 0.900, (b) 1752 ft/s, 17.5 lbm/s, 0.849 17–116 Steam enters a converging–diverging nozzle at 1 MPa
and 500°C with a negligible velocity at a mass flow rate of 2.5
kg/s, and it exits at a pressure of 200 kPa. Assuming the flow
through the nozzle to be isentropic, determine the exit area and
the exit Mach number. Answers: 31.5 cm2, 1.738
17–117 Repeat Prob. 17–116 for a nozzle efficiency of 95
percent. Review Problems
17–118 Air in an automobile tire is maintained at a pressure of 220 kPa (gauge) in an environment where the atmospheric pressure is 94 kPa. The air in the tire is at the
ambient temperature of 25°C. Now a 4mmdiameter leak
develops in the tire as a result of an accident. Assuming isentropic flow, determine the initial mass flow rate of air through
the leak. Answer: 0.554 kg/min
17–119 The thrust developed by the engine of a Boeing 777
is about 380 kN. Assuming choked flow in the nozzles, determine the mass flow rate of air through the nozzle. Take the
ambient conditions to be 265 K and 85 kPa.
17–120 A stationary temperature probe inserted into a duct
where air is flowing at 250 m/s reads 85°C. What is the
actual temperature of air? Answer: 53.9°C
17–121 Nitrogen enters a steadyflow heat exchanger at
150 kPa, 10°C, and 100 m/s, and it receives heat in the
amount of 125 kJ/kg as it flows through it. The nitrogen
leaves the heat exchanger at 100 kPa with a velocity of 180
m/s. Determine the stagnation pressure and temperature of
the nitrogen at the inlet and exit states.
17–122 Derive an expression for the speed of sound based
on van der Waals’ equation of state P
RT(v
b)
a/v 2.
Using this relation, determine the speed of sound in carbon
dioxide at 50°C and 200 kPa, and compare your result to that
obtained by assuming idealgas behavior. The van der Waals
constants for carbon dioxide are a
364.3 kPa · m6/kmol2
and b 0.0427 m3/kmol.
17–123 Obtain Eq. 17–10 by starting with Eq. 17–9 and
using the cyclic rule and the thermodynamic property relations
cp
T a 0s
b
and
0T P cv
T a 0s
b.
0T v 17–124 For ideal gases undergoing isentropic flows, obtain
expressions for P/P*, T/ T *, and r/r* as functions of k and Ma.
17–125 Using Eqs. 17–4, 17–13, and 17–14, verify that for
dA/A
(1
Ma2)
the steady flow of ideal gases dT0 / T  879 dV/V. Explain the effect of heating and area changes on the
velocity of an ideal gas in steady flow for (a) subsonic flow
and (b) supersonic flow.
17–126 A subsonic airplane is flying at a 3000m altitude
where the atmospheric conditions are 70.109 kPa and 268.65 K.
A Pitot static probe measures the difference between the static
and stagnation pressures to be 35 kPa. Calculate the speed of
the airplane and the flight Mach number. Answers: 257 m/s, .
17–127 Plot the mass flow parameter m 1RT0 /(AP0) versus
the Mach number for k 1.2, 1.4, and 1.6 in the range of 0
Ma 1.
0.783 17–128 Helium enters a nozzle at 0.8 MPa, 500 K, and
a velocity of 120 m/s. Assuming isentropic flow, determine the pressure and temperature of helium at a location
where the velocity equals the speed of sound. What is the
ratio of the area at this location to the entrance area?
17–129 Repeat Prob. 17–128 assuming the entrance velocity is negligible.
17–130 Air at 0.9 MPa and 400 K enters a converging
nozzle with a velocity of 180 m/s. The throat
area is 10 cm2. Assuming isentropic flow, calculate and plot
the mass flow rate through the nozzle, the exit velocity, the
exit Mach number, and the exit pressure–stagnation pressure
ratio versus the back pressure–stagnation pressure ratio for a
back pressure range of 0.9 Pb 0.1 MPa. 17–131 Steam at 6.0 MPa and 700 K enters a converging nozzle with a negligible velocity. The
nozzle throat area is 8 cm2. Assuming isentropic flow, plot
the exit pressure, the exit velocity, and the mass flow rate
Pb
through the nozzle versus the back pressure Pb for 6.0
3.0 MPa. Treat the steam as an ideal gas with k
1.3, cp
1.872 kJ/kg · K, and R 0.462 kJ/kg · K.
17–132 Find the expression for the ratio of the stagnation
pressure after a shock wave to the static pressure before the
shock wave as a function of k and the Mach number upstream
of the shock wave Ma1.
17–133 Nitrogen enters a converging–diverging nozzle at 700
kPa and 300 K with a negligible velocity, and it experiences a
normal shock at a location where the Mach number is Ma
3.0. Calculate the pressure, temperature, velocity, Mach number, and stagnation pressure downstream of the shock. Compare these results to those of air undergoing a normal shock
at the same conditions.
17–134 An aircraft flies with a Mach number Ma1 0.8 at
an altitude of 7000 m where the pressure is 41.1 kPa and the
temperature is 242.7 K. The diffuser at the engine inlet has
0.3. For a mass flow rate of
an exit Mach number of Ma2
65 kg/s, determine the static pressure rise across the diffuser
and the exit area. cen84959_ch17.qxd 5/10/05 12:27 PM Page 880 880  Thermodynamics 17–135 Helium expands in a nozzle from 1 MPa, 500 K,
and negligible velocity to 0.1 MPa. Calculate the throat and
exit areas for a mass flow rate of 0.25 kg/s, assuming the
nozzle is isentropic. Why must this nozzle be converging–
diverging? Answers: 3.51 cm2, 5.84 cm2 at all times. Disregarding frictional losses, determine the
highest rate of heat transfer to the air in the duct without
affecting the inlet conditions. Answer: 1958 kW Qmax 17–136E Helium expands in a nozzle from 150 psia, 900
R, and negligible velocity to 15 psia. Calculate the throat and
exit areas for a mass flow rate of 0.2 lbm/s, assuming the
nozzle is isentropic. Why must this nozzle be converging–
diverging? P1
T1
Ma1 400 kPa
3 60 K
0 .4 17–137 Using the EES software and the relations in
Table A–32, calculate the onedimensional
compressible flow functions for an ideal gas with k 1.667,
and present your results by duplicating Table A–32.
17–138 Using the EES software and the relations in
Table A–33, calculate the onedimensional
normal shock functions for an ideal gas with k
1.667, and
present your results by duplicating Table A–33.
17–139 Consider an equimolar mixture of oxygen and
nitrogen. Determine the critical temperature, pressure, and
density for stagnation temperature and pressure of 800 K
and 500 kPa.
17–140 Using EES (or other) software, determine the
shape of a converging–diverging nozzle for air
for a mass flow rate of 3 kg/s and inlet stagnation conditions
of 1400 kPa and 200°C. Assume the flow is isentropic.
Repeat the calculations for 50kPa increments of pressure
drops to an exit pressure of 100 kPa. Plot the nozzle to scale.
Also, calculate and plot the Mach number along the nozzle. 17–141 Using EES (or other) software and the relations given in Table A–32, calculate the onedimensional isentropic compressibleflow functions by
varying the upstream Mach number from 1 to 10 in increments of 0.5 for air with k 1.4.
17–142 Repeat Prob. 17–141 for methane with k
1.3. 17–143 Using EES (or other) software and the relations given in Table A–33, generate the onedimensional normal shock functions by varying the upstream
Mach number from 1 to 10 in increments of 0.5 for air with
k 1.4. 17–144 Repeat Prob. 17–143 for methane with k
1.3. 17–145 Air is cooled as it flows through a 20cmdiameter
duct. The inlet conditions are Ma1 1.2, T01 350 K, and P01
240 kPa and the exit Mach number is Ma2 2.0. Disregarding frictional effects, determine the rate of cooling of air.
17–146 Air is heated as it flows subsonically through a
10 cm 10 cm square duct. The properties of air at the inlet
are maintained at Ma1 0.4, P1 400 kPa, and T1 360 K FIGURE P17–146 17–147 Repeat Prob. 17–146 for helium. 17–148 Air is accelerated as it is heated in a duct with negligible friction. Air enters at V1 100 m/s, T1 400 K, and
P1 35 kPa and then exits at a Mach number of Ma2 0.8.
Determine the heat transfer to the air, in kJ/kg. Also determine the maximum amount of heat transfer without reducing
the mass flow rate of air.
17–149 Air at sonic conditions and static temperature and
pressure of 500 K and 420 kPa, respectively, is to be accelerated to a Mach number of 1.6 by cooling it as it flows through
a channel with constant crosssectional area. Disregarding
frictional effects, determine the required heat transfer from the
air, in kJ/kg. Answer: 69.8 kJ/kg
17–150 Saturated steam enters a converging–diverging nozzle at 3.0 MPa, 5 percent moisture, and negligible velocity,
and it exits at 1.2 MPa. For a nozzle exit area of 16 cm2,
determine the throat area, exit velocity, mass flow rate, and
exit Mach number if the nozzle (a) is isentropic and (b) has
an efficiency of 90 percent. Fundamentals of Engineering (FE) Exam Problems
17–151 An aircraft is cruising in still air at 5°C at a velocity
of 400 m/s. The air temperature at the nose of the aircraft
where stagnation occurs is
(a) 5°C (b) 25°C (c) 55°C (d) 80°C (e) 85°C 17–152 Air is flowing in a wind tunnel at 15°C, 80 kPa,
and 200 m/s. The stagnation pressure at the probe inserted
into the flow section is
(a) 82 kPa
(d) 101 kPa (b) 91 kPa
(e) 114 kPa (c) 96 kPa 17–153 An aircraft is reported to be cruising in still air at
20°C and 40 kPa at a Mach number of 0.86. The velocity
of the aircraft is
(a) 91 m/s
(d) 280 m/s (b) 220 m/s
(e) 378 m/s (c) 186 m/s cen84959_ch17.qxd 4/21/05 11:08 AM Page 881 Chapter 17  881 17–154 Air is flowing in a wind tunnel at 12°C and 66 kPa
at a velocity of 230 m/s. The Mach number of the flow is (d) There will be no flow through the nozzle if the back
pressure equals the stagnation pressure. (a) 0.54 m/s
(d) 0.36 m/s (e) The fluid velocity decreases, the entropy increases, and
stagnation enthalpy remains constant during flow
through a normal shock. (b) 0.87 m/s
(e) 0.68 m/s (c) 3.3 m/s 17–155 Consider a converging nozzle with a low velocity at
the inlet and sonic velocity at the exit plane. Now the nozzle
exit diameter is reduced by half while the nozzle inlet temperature and pressure are maintained the same. The nozzle
exit velocity will
(a) remain the same (b) double
(c) quadruple
(d) go down by half (e) go down to onefourth
17–156 Air is approaching a converging–diverging nozzle
with a low velocity at 20°C and 300 kPa, and it leaves the
nozzle at a supersonic velocity. The velocity of air at the
throat of the nozzle is
(a) 290 m/s
(d) 343 m/s (b) 98 m/s
(e) 412 m/s (c) 313 m/s 17–157 Argon gas is approaching a converging–diverging
nozzle with a low velocity at 20°C and 120 kPa, and it leaves
the nozzle at a supersonic velocity. If the crosssectional area
of the throat is 0.015 m2, the mass flow rate of argon through
the nozzle is
(a) 0.41 kg/s
(d) 17 kg/s (b) 3.4 kg/s
(e) 22 kg/s (c) 5.3 kg/s 17–158 Carbon dioxide enters a converging–diverging nozzle at 60 m/s, 310°C, and 300 kPa, and it leaves the nozzle at
a supersonic velocity. The velocity of carbon dioxide at the
throat of the nozzle is
(a) 125 m/s
(d) 353 m/s (b) 225 m/s
(e) 377 m/s (c) 312 m/s 17–159 Consider gas flow through a converging–diverging
nozzle. Of the five following statements, select the one that is
incorrect: 17–160 Combustion gases with k 1.33 enter a converging
nozzle at stagnation temperature and pressure of 400°C and
800 kPa, and are discharged into the atmospheric air at 20°C
and 100 kPa. The lowest pressure that will occur within the
nozzle is
(a) 26 kPa
(d) 432 kPa (b) 100 kPa
(e) 272 kPa Design and Essay Problems
17–161 Find out if there is a supersonic wind tunnel on
your campus. If there is, obtain the dimensions of the wind
tunnel and the temperatures and pressures as well as the
Mach number at several locations during operation. For what
typical experiments is the wind tunnel used?
17–162 Assuming you have a thermometer and a device to
measure the speed of sound in a gas, explain how you can
determine the mole fraction of helium in a mixture of helium
gas and air.
17–163 Design a 1mlong cylindrical wind tunnel whose
diameter is 25 cm operating at a Mach number of 1.8.
Atmospheric air enters the wind tunnel through a converging–
diverging nozzle where it is accelerated to supersonic velocities. Air leaves the tunnel through a converging–diverging
diffuser where it is decelerated to a very low velocity before
entering the fan section. Disregard any irreversibilities. Specify
the temperatures and pressures at several locations as well as
the mass flow rate of air at steadyflow conditions. Why is it
often necessary to dehumidify the air before it enters the wind
tunnel? (a) The fluid velocity at the throat can never exceed the
speed of sound.
(b) If the fluid velocity at the throat is below the speed of
sound, the diversion section will act like a diffuser.
(c) If the fluid enters the diverging section with a Mach
number greater than one, the flow at the nozzle exit will
be supersonic. (c) 321 kPa P0
T0 Ma 1.8 D 25 cm FIGURE P17–163 cen84959_ch17.qxd 4/21/05 11:08 AM Page 882 ...
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This note was uploaded on 06/15/2009 for the course MAE 3311 taught by Professor Hajisheik during the Summer '08 term at UT Arlington.
 Summer '08
 HAJISHEIK

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