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1
Compressible Flow
Chapter 7
Fluid Mechanics
(MEng 2113)
Mechanical Engineering
Department
Prepared by: Addisu Dagne May 2017
Contents of the Chapter
 Introduction
 Thermodynamic Relations
 Stagnation Properties
 Speed of Sound and Mach number
 One Dimensional Isentropic flow: Effect of area
changes of flow parameters
 Converging Nozzle: Effect of back pressure on flow
parameters
 Converging-Diverging Nozzles
 Normal shock waves; Fanno and Rayleigh lines
 Oblique shock waves
2
Introduction
 Flows that involve significant changes in density are called
compressible flows.
 Therefore, ρ(x, y, z) must now be treated as a field variable
rather than simply a constant.
 Typically, significant density variations start to appear when the
flow Mach number exceeds 0.3 or so. The effects become
especially large when the Mach number approaches and
exceeds unity.
 In this chapter we will 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 speeds such as flows in gas turbine
engine components . Many aircraft fly fast enough to involve
compressible flow.
3
 Gas has large compressibility but when its velocity is low
compared with the sonic velocity the change in density is
small and it is then treated as an incompressible fluid.
 When a fluid moves at speeds comparable to its speed of
sound, density changes become significant and the flow is
termed compressible.
 Such flows are difficult to obtain in liquids, since high
pressures of order 1000 atm are needed to generate sonic
velocities. In gases, however, a pressure ratio of only 2:1
will likely cause sonic flow. Thus compressible gas flow is
quite common, and this subject is often called gas
dynamics.
4
Introduction
Thermodynamic Relations
Perfect gas
 A perfect gas is one whose individual molecules interact
only via direct collisions, with no other intermolecular
forces present.
 For such a perfect gas, p, ρ, and the temperature T are
related by the following equation of state
p = ρRT
 where R is the specific gas constant. For air, R =287J/kg-K◦ .
 It is convenient at this point to define the specific volume as
the limiting volume per unit mass,
 which is merely the reciprocal of the density.5
 The equation of state can now be written as
pυ = RT
 which is the more familiar thermodynamic form.
 Here R is the gas constant, and
 where Ro is the universal gas constant (Ro = 8314J/(kg K))
and M is the molecular weight. For example, for air
assuming M = 28.96, the gas constant is
6
Thermodynamic Relations
 Then, assuming internal energy and enthalpy per unit mass
e and h respectively,
 the specific enthalpy, denoted by h, and related to the other
variables by
 For a calorically perfect gas, which is an excellent model
for air at moderate temperatures both e and h are directly
proportional to the temperature.
7
Thermodynamic Relations
 Therefore we have
 where cv and cp are specific heats at constant volume and
constant pressure, respectively.
 and comparing to the equation of state, we see that
 Defining the ratio of specific heats, γ ≡ cp/cv, we can with a
bit of algebra write
8
Thermodynamic Relations
 so that cv and cp can be replaced with the equivalent
variables γ and R. For air, it is handy to remember that
First Law of Thermodynamics
 Consider a thermodynamic system consisting of a small
Lagrangian control volume (CV) moving with the flow.
 Over the short time interval dt, the CV undergoes a process
where it receives work δw and heat δq from its
surroundings, both per unit mass. This process results in
changes in the state of the CV, described by the increments
de, dh, dp . . .
9
Thermodynamic Relations
Thermodynamic Relations
 The first law of thermodynamics
for the process is
 This states that whatever energy
is added to the system, whether
by heat or by work, it must
appear as an increase in the
internal energy of the system.
10
Isentropic relations
 Aerodynamic flows are effectively inviscid outside of
boundary layers. This implies they have negligible heat
conduction and friction forces, and hence are isentropic.
 Therefore, along the pathline followed by the CV in the
figure above, the isentropic version of the first law applies
11
Thermodynamic Relations
 This relation can be integrated after a few substitutions.
First we note that
 and with the perfect gas relation
 the isentropic first law becomes
12
Thermodynamic Relations
→
 The final form can now be integrated from any state 1 to
any state 2 along the pathline.
 From the equation of state we also have
 Which gives the alternative isentropic relation
13
Thermodynamic Relations
Stagnation Properties
 Consider a fluid flowing into a diffuser at a velocity ,
temperature T, pressure P, and enthalpy h, etc. Here the
ordinary properties T, P, h, etc. are called the static
properties; that is, they are measured relative to the flow at
the flow velocity.
 If the diffuser is sufficiently long and the exit area is
sufficiently large that the fluid is brought to rest (zero
velocity) at the diffuser exit while no work or heat transfer
is done. The resulting state is called the stagnation state.
14

V
 We apply the first law per unit mass for one entrance, one
exit, and neglect the potential energies. Let the inlet state
be unsubscripted and the exit or stagnation state have the
subscript o.
 Since the exit velocity, work, and heat transfer are zero,
 The term ho is called the stagnation enthalpy (some
authors call this the total enthalpy).
 It is the enthalpy the fluid attains when brought to rest
adiabatically while no work is done.
15
q h
V
w h
V
net net o
o
    
 2 2
2 2
Stagnation Properties
h h
V
o  
2
2
Stagnation Properties
 If, in addition, the process is
also reversible, the process is
isentropic, and the inlet and
exit entropies are equal.
 The stagnation enthalpy and
entropy define the stagnation
state and the isentropic
stagnation pressure, Po.
 The actual stagnation pressure
for irreversible flows will be
somewhat less than the
isentropic stagnation pressure
as shown in the fig.
16
s so 
Example 1
 Steam at 400o
C, 1.0 MPa, and 300 m/s flows through a
pipe. Find the properties of the steam at the stagnation
state.
Solution
 At T = 400oC and P = 1.0 MPa,
h = 3264.5 kJ/kg s = 7.4670 kJ/kgK
Then
17
2
2
2
2
2
300
3264.5
2
1000
3309.5
o
V
h h
kJm
kJ kgs
mkg
s
kJ
kg
 
 
 
  

and
 We can find Po by trial and error. The resulting stagnation
properties are
18
Example 1…. solution
= = 7.4670o
kJ
s s
kg K
h h P so o o ( , )
3
1.16
422.2
1
3.640
o
o
o
o
o
P MPa
T C
kg
v m



 
19
Ideal Gas Result
Rewrite the equation defining the stagnation enthalpy as
h h
V
o  
2
2
For ideal gases with constant specific heats, the enthalpy
difference becomes
C T T
V
P o( ) 
2
2
where To is defined as the stagnation temperature.
T T
V
C
o
P
 
2
2
20
For the isentropic process, the stagnation pressure can be
determined from
or
The ratio of the stagnation density to static density can be
expressed as
Example 2
 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.
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.
21
Solution
 High-speed air enters the diffuser and the compressor of an
aircraft. The stagnation pressure of the 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 constant-pressure specific heat cp and the
specific heat ratio k of air at room temperature are
cp = 1.005 kJ/kg . K and k = 1.4
 Analysis
(a) the stagnation temperature T01 at the compressor inlet can
be determined from
22
 Then
 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.
23
Solution
 To determine the compressor work, we need to know the
stagnation temperature of air at the compressor exit T02.
 Disregarding potential energy changes and heat transfer, the
compressor work per unit mass of air is determined from
24
Solution
25
Conservation of Energy for Control Volumes Using
Stagnation Properties
The steady-flow conservation of energy for the above figure is
Since
h h
V
o  
2
2
26
For no heat transfer, one entrance, one exit, this reduces to
If we neglect the change in potential energy, this becomes
For ideal gases with constant specific heats we write this as
Conservation of Energy for a Nozzle
We assume steady-flow, no heat transfer, no work, one entrance,
and one exit and neglect elevation changes; then the conservation
of energy becomes
1 1 2 2
in out
o o
E E
m h m h


27
But  m m1 2 thus h ho o1 2
Thus the stagnation enthalpy remains constant throughout the
nozzle. At any cross section in the nozzle, the stagnation
enthalpy is the same as that at the entrance. For ideal gases this
last result becomes
T To o1 2
Thus the stagnation temperature remains constant through out
the nozzle. At any cross section in the nozzle, the stagnation
temperature is the same as that at the entrance.
Assuming an isentropic process for flow through the nozzle,
we can write for the entrance and exit states
So we see that the stagnation pressure is also constant through
out the nozzle for isentropic flow.
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 duct that is filled with a fluid at rest,
28
 A piston fitted in the duct 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.
 To simplify the analysis, consider a control volume that encloses the
wave front and moves with it, as shown in the fig. below.
 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.
29
Speed of Sound and Mach number
30
Speed of Sound and Mach number
31
Speed of Sound and Mach number
Cancel terms and neglect ; we havedV
2
dh CdV 

0
32
Now, apply the conservation of mass or continuity equation
to the control volume.
m AV 

  
    
AC d A C dV
AC A C dV Cd d dV
  
   
( ) ( )
( )

 
Cancel terms and neglect the higher-order terms like .
We have
d dV

Cd dV  

0
Also, we consider the property relation dh T ds vdP
dh T ds dP
 
 
1

33
Let's assume the process to be isentropic; then ds = 0 and
dh dP
1

Using the results of the first law
dh dP CdV 
1


From the continuity equation
dV
Cd



Now we have
34
Thus dP
d
C

 2
Since the process is assumed to be isentropic, the above becomes
By using thermodynamic property relations this can be written
as
where k is the ratio of specific heats, k = CP/CV.
35
Example -3
Find the speed of sound in air at an altitude of 5000 m.
At 5000 m, T = 255.7 K.
C
kJ
kg K
K
m
s
kJ
kg
m
s



14 0287 2557
1000
3205
2
2
. ( . )( . )
.
Ideal Gas Result
For ideal gases →
36
Notice that the temperature used for the speed of sound is the
static (normal) temperature.
Example -4
Find the speed of sound in steam where the pressure is 1 MPa
and the temperature is 350o
C.
At P = 1 MPa, T = 350oC,
1s
s
P P
C
v

 
   
   
       
  
37
Here, we approximate the partial derivative by perturbating
the pressure about 1 MPa. Consider using P±0.025 MPa at the
entropy value s = 7.3011 kJ/kg K, to find the corresponding
specific volumes.
38
What is the speed of sound for steam at 350o
C assuming
ideal-gas behavior?
Assume k = 1.3, then
C
kJ
kg K
K
m
s
kJ
kg
m
s




13 04615 350 273
1000
6114
2
2
. ( . )( )
.
Mach Number
The Mach number M is defined as M
V
C


M <1 flow is subsonic
M =1 flow is sonic
M >1 flow is supersonic
39
Example -5
In the air and steam examples above, find the Mach number if the
air velocity is 250 m/s and the steam velocity is 300 m/s.
M
m
s
m
s
M
m
s
m
s
air
steam
 
 
250
3205
0 780
300
6055
0 495
.
.
.
.
The flow parameters To/T, Po/P, o/, etc. are related to the flow Mach
number. Let's consider ideal gases, then
T T
V
C
T
T
V
C T
o
P
o
P
 
 


2
2
2
1
2
40
but C
k
k
R or
C
k
kR
P
P




1
1 1
T
T
V
T
k
kR
o
 

1
2
12

( )
and
C kRT2

so T
T
k V
C
k
M
o
 

 

1
1
2
1
1
2
2
2
2
( )
( )

The pressure ratio is given by
41
We can show the density ratio to be
For the Mach number equal to 1, the sonic location, the static
properties are denoted with a superscript “*”. This condition,
when M = 1, is called the sonic condition. When M = 1 and k =
1.4, the static-to-stagnation ratios are
Example. Mach Number of Air Entering a Diffuser
 Air enters a diffuser shown in Fig. 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.
42
43
One-Dimensional 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 one-dimensional isentropic flow with good
accuracy.
Effect of Area Changes on Flow Parameters
 Consider the isentropic steady flow of an ideal gas through
the nozzle shown below.
44
45
Air flows steadily through a varying-cross-sectional-area duct
such as a nozzle at a flow rate of 3 kg/s. The air enters the
duct at a low velocity at a pressure of 1500 kPa and a
temperature of 1200 K and it expands in the duct to a pressure
of 100 kPa. The duct is designed so that the flow process is
isentropic. Determine the pressure, temperature, velocity,
flow area, speed of sound, and Mach number at each point
along the duct axis that corresponds to a pressure drop of 200
kPa.
Since the inlet velocity is low, the stagnation properties equal
the static properties.
T T K P P kPao o   1 11200 1500,
46
After the first 200 kPa pressure drop, we have
  


P
RT
kPa
kJ
kg K
K
kJ
m kPa
kg
m
( )
( . )( . )
.
1300
0 287 11519
3932
3
3
47
A
m
V
kg
s
kg
m
m
s
cm
m
cm
 


( . )( . )
.


3
39322 31077
10
24 55
3
4 2
2
2
C kRT
kJ
kg K
K
m
s
kJ
kg
m
s
 


14 0287 11519
1000
68033
2
2
. ( . )( . )
.
M
V
C
m
s
m
s
  
 31077
68033
0457
.
.
.
Now we tabulate the results for the other 200 kPa increments
in the pressure until we reach 100 kPa.
48
Summary of Results for Nozzle Problem

V

Step P
kPa
T
K m/s

kg/m3
C
m/s
A
cm2
M
0 1500 1200 0 4.3554 694.38 0
1 1300 1151.9 310.77 3.9322 680.33 24.55 0.457
2 1100 1098.2 452.15 3.4899 664.28 19.01 0.681
3 900 1037.0 572.18 3.0239 645.51 17.34 0.886
4 792.4 1000.0 633.88 2.7611 633.88 17.14 1.000
5 700 965.2 786.83 2.5270 622.75 17.28 1.103
6 500 876.7 805.90 1.9871 593.52 18.73 1.358
7 300 757.7 942.69 1.3796 551.75 23.07 1.709
8 100 553.6 1139.62 0.6294 471.61 41.82 2.416
Note that at P = 797.42 kPa, M = 1.000, and this state is the
critical state.
 We note from the Nozzle Example
that the flow area decreases with
decreasing pressure down to a
critical-pressure 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 .
 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.
49
 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.
50
51
Now let's see why these relations work this way. Consider the
nozzle and control volume shown below.
The first law for the control volume is
dh VdV 
 
0
The continuity equation for the control volume yields
d dA
A
dV
V


  

 0
Also, we consider the property relation for an isentropic process
Tds dh
dP
  

0
52
and the Mach Number relation dP
d
C
V
M
 2
2
2

Putting these four relations together yields
dA
A
dP
V
M 

2
2
1( )
Let’s consider the implications of this equation for both nozzles
and diffusers. A nozzle is a device that increases fluid velocity
while causing its pressure to drop; thus, d > 0, dP < 0.

V
Nozzle Results dA
A
dP
V
M 

2
2
1( )
Subsonic
Sonic
Supersonic
: ( )
: ( )
: ( )
M dP M dA
M dP M dA
M dP M dA
   
   
   
1 1 0 0
1 1 0 0
1 1 0 0
2
2
2
53
To accelerate subsonic flow, the nozzle flow area must first
decrease in the flow direction. The flow area reaches a
minimum at the point where the Mach number is unity. To
continue to accelerate the flow to supersonic conditions, the
flow area must increase.
The minimum flow area is called the throat of the nozzle.
We are most familiar with the shape of a subsonic nozzle.
That is, the flow area in a subsonic nozzle decreases in the
flow direction.
54
A diffuser is a device that decreases fluid velocity while
causing its pressure to rise; thus, d < 0, dP > 0.

V
Diffuser Results dA
A
dP
V
M 

2
2
1( )
Subsonic
Sonic
Supersonic
: ( )
: ( )
: ( )
M dP M dA
M dP M dA
M dP M dA
   
   
   
1 1 0 0
1 1 0 0
1 1 0 0
2
2
2
To diffuse supersonic flow, the diffuser flow area must first
decrease in the flow direction. The flow area reaches a minimum
at the point where the Mach number is unity. To continue to
diffuse the flow to subsonic conditions, the flow area must
increase. We are most familiar with the shape of a subsonic
diffuser. That is the flow area in a subsonic diffuser increases in
the flow direction.
55
56
Equation of Mass Flow Rate through a Nozzle
Let's obtain an expression for the flow rate through a converging
nozzle at any location as a function of the pressure at that
location. The mass flow rate is given by
m AV 

The velocity of the flow is related to the static and stagnation
enthalpies.

V h h C T T C T
T
T
o P P
o
     2 2 2 10 0( ) ( ) ( )
and
57
Write the mass flow rate as
m AV o
o





We note from the ideal-gas relations that
o
o
o
P
RT

58
What pressure ratios make the mass flow rate zero?
Do these values make sense?
Now let's make a plot of mass flow rate versus the static-to-
stagnation pressure ratio.
0.00 0.20 0.40 0.60 0.80 1.00
0.00
0.02
0.04
0.06
0.08
0.10
0.12
0.14
0.160.16
P*/Po
m[kg/s]
P/Po
Dia.=1 cm
To=1200 K
Po=1500 kPa
59
This plot shows there is a value of P/Po that makes the mass flow rate
a maximum. To find that mass flow rate, we note
The result is
So the pressure ratio that makes the mass flow rate a maximum is the
same pressure ratio at which the Mach number is unity at the flow cross-
sectional area. This value of the pressure ratio is called the critical
pressure ratio for nozzle flow. For pressure ratios less than the critical
value, the nozzle is said to be choked. When the nozzle is choked, the
mass flow rate is the maximum possible for the flow area, stagnation
pressure, and stagnation temperature. Reducing the pressure ratio below
the critical value will not increase the mass flow rate.
60
Using
The mass flow rate becomes
When the Mach number is unity, M = 1, A = A*
Taking the ratio of the last two results gives the ratio of the area
of the flow A at a given Mach number to the area where the
Mach number is unity, A*.
What is the expression for mass flow rate when the nozzle is choked?
61
Then
0.0 0.5 1.0 1.5 2.0 2.5 3.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
M
A/A*
From the above plot we note that for each A/A* there are two
values of M: one for subsonic flow at that area ratio and one for
supersonic flow at that area ratio. The area ratio is unity when
the Mach number is equal to one.
62
Consider the converging nozzle
shown below. The flow is supplied
by a reservoir at pressure Pr and
temperature Tr. The reservoir is
large enough that the velocity in the
reservoir is zero.
Let's plot the ratio P/Po along the
length of the nozzle, the mass flow
rate through the nozzle, and the exit
plane pressure Pe as the back
pressure Pb is varied. Let's consider
isentropic flow so that Po is constant
throughout the nozzle.
Effect of Back Pressure on Flow through a Converging Nozzle
 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 the Fig. above.
 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.
63
Effect of Back Pressure on Flow through a Converging Nozzle
 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 steady-flow conditions, the mass flow rate through the
nozzle is constant and can be expressed as
 Solving for T from and for P from
and substituting
64
Effect of Back Pressure on Flow through a Converging Nozzle
 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 number.
 The maximum mass flow rate can be determined by
differentiating the above equation 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
65
Effect of Back Pressure on Flow through a Converging Nozzle
 A relation for the variation of flow area A through the nozzle
relative to throat area A* can be obtained by combining
equations f or and for the same mass flow rate and
stagnation properties of a particular fluid. This yields
 Another parameter sometimes used in the analysis of one-
dimensional isentropic flow of ideal gases is Ma*, which is
the ratio of the local velocity to the speed of sound at the
throat:
66
Effect of Back Pressure on Flow through a Converging Nozzle
 where Ma is the local Mach number, T is the local
temperature, and T* is the critical temperature.
 Solving for T and for T* and substituting, we get
 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.
67
Effect of Back Pressure on Flow through a Converging Nozzle
68
Table 1
 A plot of versus Pb /P0 for a
converging nozzle is shown in
Fig. below.
 Notice that the mass flow rate
increases with decreasing Pb /P0,
reaches a maximum at Pb = P*,
and remains constant for Pb /P0
values less than this critical
ratio. Also illustrated on this
figure is the effect of back
pressure on the nozzle exit
pressure Pe. We observe that
69
Effect of Back Pressure on Flow through a Converging Nozzle
 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.
70
Effect of Back Pressure on Flow through a Converging Nozzle
71
1. Pb = Po, Pb /Po = 1. No flow occurs. Pe = Pb, Me=0.
2. Pb > P* or P*/Po < Pb /Po < 1. Flow begins to increase as
the back pressure is lowered. Pe = Pb, Me < 1.
3. Pb = P* or P*/Po = Pb /Po < 1. Flow increases to the choked
flow limit as the back pressure is lowered to the critical
pressure. Pe = Pb, Me=1.
4. Pb < P* or Pb /Po < P*/Po < 1. Flow is still choked and does
not increase as the back pressure is lowered below the
critical pressure, pressure drop from Pe to Pb occurs outside
the nozzle. Pe = P*, Me=1.
5. Pb = 0. Results are the same as for item 4.
Consider the converging-diverging nozzle shown below.
Effect of Back Pressure on Flow through a Converging Nozzle
Example. Effect of Back Pressure on Mass Flow Rate
 Air at 1 MPa and 600°C enters a converging nozzle, shown
in Fig., 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.
72
73
74
The critical-pressure ratio is determined from Table 1 (or Eq.
below)
to be P*/P0 = 0.5283.
That is, Pt = Pb = 0.7 MPa, and Pt /P0 = 0.670. Therefore, the
flow is not choked. From Table 1 at Pt /P0 = 0.670, we read Mat =
0.778 and Tt /T0 = 0.892.
75
Converging–Diverging Nozzles
 When we think of nozzles, we ordinarily think of flow
passages whose cross-sectional 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.
76
 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.
77
Converging–Diverging Nozzles
 Consider the converging–
diverging nozzle shown in Fig.
A fluid enters the nozzle with a
low velocity at stagnation
pressure P0. When Pb =P0 (case
A), there is 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.78
Converging–Diverging
Nozzles
79
Converging–Diverging Nozzles
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.
80
Converging–Diverging Nozzles
 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 crosses the normal shock.
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 approximated as isentropic.
 When Pb = PF, no shocks occur within or outside the nozzle.
When Pb < PF, irreversible mixing and expansion waves occur
downstream of the exit plane of the nozzle. When Pb > PF,
however, the pressure of the fluid increases from PF to Pb
irreversibly in the wake of the nozzle exit, creating what are
called oblique shocks.81
Converging–Diverging Nozzles
82
Example -7
Air leaves the turbine of a turbojet engine and enters a
convergent nozzle at 400 K, 871 kPa, with a velocity of 180 m/s.
The nozzle has an exit area of 730 cm2. Determine the mass
flow rate through the nozzle for back pressures of 700 kPa, 528
kPa, and 100 kPa, assuming isentropic flow.
The stagnation temperature and stagnation pressure are
T T
V
C
o
P
 
2
2
83
For air k = 1.4 and from Table or using the equation below the critical
pressure ratio is P*/Po = 0.528. The critical pressure for this nozzle is
P P
kPa kPa
o
*
.
. ( )

 
0528
0528 1000 528
Therefore, for a back pressure of 528 kPa, M = 1 at the nozzle exit and
the flow is choked. For a back pressure of 700 kPa, the nozzle is not
choked. The flow rate will not increase for back pressures below 528
kPa.
84
For the back pressure of 700 kPa,
P
P
kPa
kPa
P
P
B
o o
  
700
1000
0700.
*
Thus, PE = PB = 700 kPa. For this pressure ratio Table 1 gives
ME  0 7324.
T
T
T T K K
E
o
E o

  
09031
09031 09031 4161 3758
.
. . ( . ) .
C kRT
kJ
kg K
K
m
s
kJ
kg
m
s
V M C
m
s
m
s
E E
E E E




 

14 0287 3758
1000
3886
07324 3886
284 6
2
2
. ( . )( . )
.
( . )( . )
.

85
E
E
E
P
RT
kPa
kJ
kg K
K
kJ
m kPa
kg
m
 


( )
( . )( . )
.
700
0 287 3758
6 4902
3
3
Then

. ( )( . )
( )
.
m A V
kg
m
cm
m
s
m
cm
kg
s
E E E




6 4902 730 284 6
100
134 8
3
2
2
2
For the back pressure of 528 kPa,
P
P
kPa
kPa
P
P
E
o o
  
528
1000
0528.
*
86
This is the critical pressure ratio and ME = 1 and PE = PB = P* = 528 kPa.
T
T
T
T
T T K K
E
o o
E o
 
  
*
.
. . ( . ) .
08333
08333 08333 4161 3467
And since ME = 1,

V C kRT
kJ
kg K
K
m
s
kJ
kg
m
s
E E E 



14 0 287 346 7
1000
3732
2
2
. ( . )( . )
.
 E
P
RT
kPa
kJ
kg K
K
kJ
m kPa
kg
m
  


*
*
*
(528 )
( . )( . )
.
0 287 346 7
53064
3
3
87

. ( )( . )
( )
.
m A V
kg
m
cm
m
s
m
cm
kg
s
E E E




53064 730 3732
100
144 6
3
2
2
2
For a back pressure less than the critical pressure, 528 kPa in this case,
the nozzle is choked and the mass flow rate will be the same as that for
the critical pressure. Therefore, at a back pressure of 100 kPa the mass
flow rate will be 144.6 kg/s.
Example -8
Air enters a converging–diverging nozzle, shown in Fig., at 1.0 MPa and
800 K with a negligible velocity. The flow is steady, one-dimensional,
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.
88
89
90
91
92
Normal Shocks
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.
The normal shock 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
93
We assume steady-flow with no heat and work interactions and no
potential energy changes. We have the following
Conservation of mass
1 1 2 2
2 2 2 2
AV AV
V V
 
 


94
Conservation of energy
2 2
1 2
1 2
1 2
1 2
2 2
:
o o
o o
V V
h h
h h
for ideal gases T T
  


Conservation of momentum
Rearranging and integrating yield
1 2 2 1( ) ( )A P P m V V  
Increase of entropy
2 1 0s s 
Thus, we see that from the conservation of energy, the stagnation
temperature is constant across the shock. However, the stagnation
pressure decreases across the shock because of irreversibilities. The
ordinary (static) temperature rises drastically because of the conversion
of kinetic energy into enthalpy due to a large drop in fluid velocity.
95
We can show that the following relations apply across the shock.
2
2 1
2
1 2
2
1 12
2
1 2 2
2
2 1
2 2
1
1 ( 1)/ 2
1 ( 1)/ 2
1 ( 1)/ 2
1 ( 1)/ 2
2/( 1)
2 /( 1) 1
T M k
T M k
M M kP
P M M k
M k
M
M k k
 

 
 

 
 

 
The entropy change across the shock is obtained by applying the
entropy-change equation for an ideal gas, constant properties, across
the shock:
2 2
2 1
1 1
ln lnp
T P
s s C R
T P
  
96
We can combine the conservation of
mass and energy relations into a
single equation and plot it on an h-s
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 h-s
diagram yield a curve called the
Rayleigh line.
97
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 wave. Notice from Fig.
that entropy increases, s2 >s1. This is expected since the flow through
the shock is adiabatic but irreversible.
98
Table 2
99
100
Example -9
Air flowing with a velocity of 600 m/s, a pressure of 60 kPa, and a
temperature of 260 K undergoes a normal shock. Determine the
velocity and static and stagnation conditions after the shock and the
entropy change across the shock.
The Mach number before the shock is
1 1
1
1 1
2
2
600
1000
1.4(0.287 )(260 )
1.856
V V
M
C kRT
m
s
m
kJ sK
kJkg K
kg
 



101
For M1 = 1.856, Table 1 gives
1 1
1 1
0.1597, 0.5921
o o
P T
P T
 
For Mx = 1.856, Table 2 gives the following results.
2 2
2
1 1
2 22
1 1 1
0.6045, 3.852, 2.4473
1.574, 0.7875, 4.931o o
o
P
M
P
P PT
T P P


  
  
102
From the conservation of mass with A2 = A1.
2 2 1 1
1
2
2
1
600
245.2
2.4473
V V
m
V msV
s
 



  
2
2 1
1
2
2 1
1
60 (3.852) 231.1
260 (1.574) 409.2
P
P P kPa kPa
P
T
T T K K
T
  
  
1
1 2
1
1
260
439.1
0.5921
o o
o
T K
T K T
T
T
   
 
 
 
1
1
1
1
60
375.6
0.1597
o
o
P kPa
P kPa
P
P
  
 
 
 
103
The entropy change across the shock is
2 2
2 1
1 1
ln lnP
T P
s s C R
T P
   
     
   
   2 1 1.005 ln 1.574 0.287 ln 3.852
0.0688
kJ kJ
s s
kg K kg K
kJ
kg K
  
 


You are encouraged to read about the following topics in the text:
•Oblique shocks
2
2 1
1
375.6 (0.7875) 295.8o
o o
o
P
P P kPa kPa
P
  
Example 10. Shock Wave in a Converging–Diverging Nozzle
 If the air flowing through the converging–diverging nozzle of
Example 8 experiences a normal shock wave at the nozzle exit plane
(see fig below), 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, one-dimensional, and isentropic flow with k = 1.4 from the
nozzle inlet to the shock location.
104
105
Analysis (a) The fluid properties at the exit of the nozzle just before the
shock (denoted by subscript 1) are those evaluated in Example 8 at the
nozzle exit to be
The fluid properties after the shock (denoted by subscript 2) are
related to those before the shock through the functions listed in Table
2. For Ma1= 2.0, we read
106
107
Oblique Shocks
 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.
Reading Assignment
 Read about oblique shock waves: characteristic features,
governing equations, calculation o properties;
 Textbook to Read
 [YUNUS A. CENGEL] Fluid Mechanics. Fundamentals
and Applications
108
End of Course
109

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Fluid Mechanics Chapter 7. Compressible flow

  • 1. 1 Compressible Flow Chapter 7 Fluid Mechanics (MEng 2113) Mechanical Engineering Department Prepared by: Addisu Dagne May 2017
  • 2. Contents of the Chapter  Introduction  Thermodynamic Relations  Stagnation Properties  Speed of Sound and Mach number  One Dimensional Isentropic flow: Effect of area changes of flow parameters  Converging Nozzle: Effect of back pressure on flow parameters  Converging-Diverging Nozzles  Normal shock waves; Fanno and Rayleigh lines  Oblique shock waves 2
  • 3. Introduction  Flows that involve significant changes in density are called compressible flows.  Therefore, ρ(x, y, z) must now be treated as a field variable rather than simply a constant.  Typically, significant density variations start to appear when the flow Mach number exceeds 0.3 or so. The effects become especially large when the Mach number approaches and exceeds unity.  In this chapter we will 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 speeds such as flows in gas turbine engine components . Many aircraft fly fast enough to involve compressible flow. 3
  • 4.  Gas has large compressibility but when its velocity is low compared with the sonic velocity the change in density is small and it is then treated as an incompressible fluid.  When a fluid moves at speeds comparable to its speed of sound, density changes become significant and the flow is termed compressible.  Such flows are difficult to obtain in liquids, since high pressures of order 1000 atm are needed to generate sonic velocities. In gases, however, a pressure ratio of only 2:1 will likely cause sonic flow. Thus compressible gas flow is quite common, and this subject is often called gas dynamics. 4 Introduction
  • 5. Thermodynamic Relations Perfect gas  A perfect gas is one whose individual molecules interact only via direct collisions, with no other intermolecular forces present.  For such a perfect gas, p, ρ, and the temperature T are related by the following equation of state p = ρRT  where R is the specific gas constant. For air, R =287J/kg-K◦ .  It is convenient at this point to define the specific volume as the limiting volume per unit mass,  which is merely the reciprocal of the density.5
  • 6.  The equation of state can now be written as pυ = RT  which is the more familiar thermodynamic form.  Here R is the gas constant, and  where Ro is the universal gas constant (Ro = 8314J/(kg K)) and M is the molecular weight. For example, for air assuming M = 28.96, the gas constant is 6 Thermodynamic Relations
  • 7.  Then, assuming internal energy and enthalpy per unit mass e and h respectively,  the specific enthalpy, denoted by h, and related to the other variables by  For a calorically perfect gas, which is an excellent model for air at moderate temperatures both e and h are directly proportional to the temperature. 7 Thermodynamic Relations
  • 8.  Therefore we have  where cv and cp are specific heats at constant volume and constant pressure, respectively.  and comparing to the equation of state, we see that  Defining the ratio of specific heats, γ ≡ cp/cv, we can with a bit of algebra write 8 Thermodynamic Relations
  • 9.  so that cv and cp can be replaced with the equivalent variables γ and R. For air, it is handy to remember that First Law of Thermodynamics  Consider a thermodynamic system consisting of a small Lagrangian control volume (CV) moving with the flow.  Over the short time interval dt, the CV undergoes a process where it receives work δw and heat δq from its surroundings, both per unit mass. This process results in changes in the state of the CV, described by the increments de, dh, dp . . . 9 Thermodynamic Relations
  • 10. Thermodynamic Relations  The first law of thermodynamics for the process is  This states that whatever energy is added to the system, whether by heat or by work, it must appear as an increase in the internal energy of the system. 10
  • 11. Isentropic relations  Aerodynamic flows are effectively inviscid outside of boundary layers. This implies they have negligible heat conduction and friction forces, and hence are isentropic.  Therefore, along the pathline followed by the CV in the figure above, the isentropic version of the first law applies 11 Thermodynamic Relations
  • 12.  This relation can be integrated after a few substitutions. First we note that  and with the perfect gas relation  the isentropic first law becomes 12 Thermodynamic Relations →
  • 13.  The final form can now be integrated from any state 1 to any state 2 along the pathline.  From the equation of state we also have  Which gives the alternative isentropic relation 13 Thermodynamic Relations
  • 14. Stagnation Properties  Consider a fluid flowing into a diffuser at a velocity , temperature T, pressure P, and enthalpy h, etc. Here the ordinary properties T, P, h, etc. are called the static properties; that is, they are measured relative to the flow at the flow velocity.  If the diffuser is sufficiently long and the exit area is sufficiently large that the fluid is brought to rest (zero velocity) at the diffuser exit while no work or heat transfer is done. The resulting state is called the stagnation state. 14  V
  • 15.  We apply the first law per unit mass for one entrance, one exit, and neglect the potential energies. Let the inlet state be unsubscripted and the exit or stagnation state have the subscript o.  Since the exit velocity, work, and heat transfer are zero,  The term ho is called the stagnation enthalpy (some authors call this the total enthalpy).  It is the enthalpy the fluid attains when brought to rest adiabatically while no work is done. 15 q h V w h V net net o o       2 2 2 2 Stagnation Properties h h V o   2 2
  • 16. Stagnation Properties  If, in addition, the process is also reversible, the process is isentropic, and the inlet and exit entropies are equal.  The stagnation enthalpy and entropy define the stagnation state and the isentropic stagnation pressure, Po.  The actual stagnation pressure for irreversible flows will be somewhat less than the isentropic stagnation pressure as shown in the fig. 16 s so 
  • 17. Example 1  Steam at 400o C, 1.0 MPa, and 300 m/s flows through a pipe. Find the properties of the steam at the stagnation state. Solution  At T = 400oC and P = 1.0 MPa, h = 3264.5 kJ/kg s = 7.4670 kJ/kgK Then 17 2 2 2 2 2 300 3264.5 2 1000 3309.5 o V h h kJm kJ kgs mkg s kJ kg          
  • 18. and  We can find Po by trial and error. The resulting stagnation properties are 18 Example 1…. solution = = 7.4670o kJ s s kg K h h P so o o ( , ) 3 1.16 422.2 1 3.640 o o o o o P MPa T C kg v m     
  • 19. 19 Ideal Gas Result Rewrite the equation defining the stagnation enthalpy as h h V o   2 2 For ideal gases with constant specific heats, the enthalpy difference becomes C T T V P o( )  2 2 where To is defined as the stagnation temperature. T T V C o P   2 2
  • 20. 20 For the isentropic process, the stagnation pressure can be determined from or The ratio of the stagnation density to static density can be expressed as
  • 21. Example 2  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. 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. 21
  • 22. Solution  High-speed air enters the diffuser and the compressor of an aircraft. The stagnation pressure of the 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 constant-pressure specific heat cp and the specific heat ratio k of air at room temperature are cp = 1.005 kJ/kg . K and k = 1.4  Analysis (a) the stagnation temperature T01 at the compressor inlet can be determined from 22
  • 23.  Then  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. 23 Solution
  • 24.  To determine the compressor work, we need to know the stagnation temperature of air at the compressor exit T02.  Disregarding potential energy changes and heat transfer, the compressor work per unit mass of air is determined from 24 Solution
  • 25. 25 Conservation of Energy for Control Volumes Using Stagnation Properties The steady-flow conservation of energy for the above figure is Since h h V o   2 2
  • 26. 26 For no heat transfer, one entrance, one exit, this reduces to If we neglect the change in potential energy, this becomes For ideal gases with constant specific heats we write this as Conservation of Energy for a Nozzle We assume steady-flow, no heat transfer, no work, one entrance, and one exit and neglect elevation changes; then the conservation of energy becomes 1 1 2 2 in out o o E E m h m h  
  • 27. 27 But  m m1 2 thus h ho o1 2 Thus the stagnation enthalpy remains constant throughout the nozzle. At any cross section in the nozzle, the stagnation enthalpy is the same as that at the entrance. For ideal gases this last result becomes T To o1 2 Thus the stagnation temperature remains constant through out the nozzle. At any cross section in the nozzle, the stagnation temperature is the same as that at the entrance. Assuming an isentropic process for flow through the nozzle, we can write for the entrance and exit states So we see that the stagnation pressure is also constant through out the nozzle for isentropic flow.
  • 28. 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 duct that is filled with a fluid at rest, 28
  • 29.  A piston fitted in the duct 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.  To simplify the analysis, consider a control volume that encloses the wave front and moves with it, as shown in the fig. below.  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. 29 Speed of Sound and Mach number
  • 30. 30 Speed of Sound and Mach number
  • 31. 31 Speed of Sound and Mach number Cancel terms and neglect ; we havedV 2 dh CdV   0
  • 32. 32 Now, apply the conservation of mass or continuity equation to the control volume. m AV           AC d A C dV AC A C dV Cd d dV        ( ) ( ) ( )    Cancel terms and neglect the higher-order terms like . We have d dV  Cd dV    0 Also, we consider the property relation dh T ds vdP dh T ds dP     1 
  • 33. 33 Let's assume the process to be isentropic; then ds = 0 and dh dP 1  Using the results of the first law dh dP CdV  1   From the continuity equation dV Cd    Now we have
  • 34. 34 Thus dP d C   2 Since the process is assumed to be isentropic, the above becomes By using thermodynamic property relations this can be written as where k is the ratio of specific heats, k = CP/CV.
  • 35. 35 Example -3 Find the speed of sound in air at an altitude of 5000 m. At 5000 m, T = 255.7 K. C kJ kg K K m s kJ kg m s    14 0287 2557 1000 3205 2 2 . ( . )( . ) . Ideal Gas Result For ideal gases →
  • 36. 36 Notice that the temperature used for the speed of sound is the static (normal) temperature. Example -4 Find the speed of sound in steam where the pressure is 1 MPa and the temperature is 350o C. At P = 1 MPa, T = 350oC, 1s s P P C v                      
  • 37. 37 Here, we approximate the partial derivative by perturbating the pressure about 1 MPa. Consider using P±0.025 MPa at the entropy value s = 7.3011 kJ/kg K, to find the corresponding specific volumes.
  • 38. 38 What is the speed of sound for steam at 350o C assuming ideal-gas behavior? Assume k = 1.3, then C kJ kg K K m s kJ kg m s     13 04615 350 273 1000 6114 2 2 . ( . )( ) . Mach Number The Mach number M is defined as M V C   M <1 flow is subsonic M =1 flow is sonic M >1 flow is supersonic
  • 39. 39 Example -5 In the air and steam examples above, find the Mach number if the air velocity is 250 m/s and the steam velocity is 300 m/s. M m s m s M m s m s air steam     250 3205 0 780 300 6055 0 495 . . . . The flow parameters To/T, Po/P, o/, etc. are related to the flow Mach number. Let's consider ideal gases, then T T V C T T V C T o P o P       2 2 2 1 2
  • 40. 40 but C k k R or C k kR P P     1 1 1 T T V T k kR o    1 2 12  ( ) and C kRT2  so T T k V C k M o       1 1 2 1 1 2 2 2 2 ( ) ( )  The pressure ratio is given by
  • 41. 41 We can show the density ratio to be For the Mach number equal to 1, the sonic location, the static properties are denoted with a superscript “*”. This condition, when M = 1, is called the sonic condition. When M = 1 and k = 1.4, the static-to-stagnation ratios are
  • 42. Example. Mach Number of Air Entering a Diffuser  Air enters a diffuser shown in Fig. 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. 42
  • 43. 43
  • 44. One-Dimensional 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 one-dimensional isentropic flow with good accuracy. Effect of Area Changes on Flow Parameters  Consider the isentropic steady flow of an ideal gas through the nozzle shown below. 44
  • 45. 45 Air flows steadily through a varying-cross-sectional-area duct such as a nozzle at a flow rate of 3 kg/s. The air enters the duct at a low velocity at a pressure of 1500 kPa and a temperature of 1200 K and it expands in the duct to a pressure of 100 kPa. The duct is designed so that the flow process is isentropic. Determine the pressure, temperature, velocity, flow area, speed of sound, and Mach number at each point along the duct axis that corresponds to a pressure drop of 200 kPa. Since the inlet velocity is low, the stagnation properties equal the static properties. T T K P P kPao o   1 11200 1500,
  • 46. 46 After the first 200 kPa pressure drop, we have      P RT kPa kJ kg K K kJ m kPa kg m ( ) ( . )( . ) . 1300 0 287 11519 3932 3 3
  • 47. 47 A m V kg s kg m m s cm m cm     ( . )( . ) .   3 39322 31077 10 24 55 3 4 2 2 2 C kRT kJ kg K K m s kJ kg m s     14 0287 11519 1000 68033 2 2 . ( . )( . ) . M V C m s m s     31077 68033 0457 . . . Now we tabulate the results for the other 200 kPa increments in the pressure until we reach 100 kPa.
  • 48. 48 Summary of Results for Nozzle Problem  V  Step P kPa T K m/s  kg/m3 C m/s A cm2 M 0 1500 1200 0 4.3554 694.38 0 1 1300 1151.9 310.77 3.9322 680.33 24.55 0.457 2 1100 1098.2 452.15 3.4899 664.28 19.01 0.681 3 900 1037.0 572.18 3.0239 645.51 17.34 0.886 4 792.4 1000.0 633.88 2.7611 633.88 17.14 1.000 5 700 965.2 786.83 2.5270 622.75 17.28 1.103 6 500 876.7 805.90 1.9871 593.52 18.73 1.358 7 300 757.7 942.69 1.3796 551.75 23.07 1.709 8 100 553.6 1139.62 0.6294 471.61 41.82 2.416 Note that at P = 797.42 kPa, M = 1.000, and this state is the critical state.
  • 49.  We note from the Nozzle Example that the flow area decreases with decreasing pressure down to a critical-pressure 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 .  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. 49
  • 50.  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. 50
  • 51. 51 Now let's see why these relations work this way. Consider the nozzle and control volume shown below. The first law for the control volume is dh VdV    0 The continuity equation for the control volume yields d dA A dV V        0 Also, we consider the property relation for an isentropic process Tds dh dP     0
  • 52. 52 and the Mach Number relation dP d C V M  2 2 2  Putting these four relations together yields dA A dP V M   2 2 1( ) Let’s consider the implications of this equation for both nozzles and diffusers. A nozzle is a device that increases fluid velocity while causing its pressure to drop; thus, d > 0, dP < 0.  V Nozzle Results dA A dP V M   2 2 1( ) Subsonic Sonic Supersonic : ( ) : ( ) : ( ) M dP M dA M dP M dA M dP M dA             1 1 0 0 1 1 0 0 1 1 0 0 2 2 2
  • 53. 53 To accelerate subsonic flow, the nozzle flow area must first decrease in the flow direction. The flow area reaches a minimum at the point where the Mach number is unity. To continue to accelerate the flow to supersonic conditions, the flow area must increase. The minimum flow area is called the throat of the nozzle. We are most familiar with the shape of a subsonic nozzle. That is, the flow area in a subsonic nozzle decreases in the flow direction.
  • 54. 54 A diffuser is a device that decreases fluid velocity while causing its pressure to rise; thus, d < 0, dP > 0.  V Diffuser Results dA A dP V M   2 2 1( ) Subsonic Sonic Supersonic : ( ) : ( ) : ( ) M dP M dA M dP M dA M dP M dA             1 1 0 0 1 1 0 0 1 1 0 0 2 2 2 To diffuse supersonic flow, the diffuser flow area must first decrease in the flow direction. The flow area reaches a minimum at the point where the Mach number is unity. To continue to diffuse the flow to subsonic conditions, the flow area must increase. We are most familiar with the shape of a subsonic diffuser. That is the flow area in a subsonic diffuser increases in the flow direction.
  • 55. 55
  • 56. 56 Equation of Mass Flow Rate through a Nozzle Let's obtain an expression for the flow rate through a converging nozzle at any location as a function of the pressure at that location. The mass flow rate is given by m AV   The velocity of the flow is related to the static and stagnation enthalpies.  V h h C T T C T T T o P P o      2 2 2 10 0( ) ( ) ( ) and
  • 57. 57 Write the mass flow rate as m AV o o      We note from the ideal-gas relations that o o o P RT 
  • 58. 58 What pressure ratios make the mass flow rate zero? Do these values make sense? Now let's make a plot of mass flow rate versus the static-to- stagnation pressure ratio. 0.00 0.20 0.40 0.60 0.80 1.00 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.160.16 P*/Po m[kg/s] P/Po Dia.=1 cm To=1200 K Po=1500 kPa
  • 59. 59 This plot shows there is a value of P/Po that makes the mass flow rate a maximum. To find that mass flow rate, we note The result is So the pressure ratio that makes the mass flow rate a maximum is the same pressure ratio at which the Mach number is unity at the flow cross- sectional area. This value of the pressure ratio is called the critical pressure ratio for nozzle flow. For pressure ratios less than the critical value, the nozzle is said to be choked. When the nozzle is choked, the mass flow rate is the maximum possible for the flow area, stagnation pressure, and stagnation temperature. Reducing the pressure ratio below the critical value will not increase the mass flow rate.
  • 60. 60 Using The mass flow rate becomes When the Mach number is unity, M = 1, A = A* Taking the ratio of the last two results gives the ratio of the area of the flow A at a given Mach number to the area where the Mach number is unity, A*. What is the expression for mass flow rate when the nozzle is choked?
  • 61. 61 Then 0.0 0.5 1.0 1.5 2.0 2.5 3.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 M A/A* From the above plot we note that for each A/A* there are two values of M: one for subsonic flow at that area ratio and one for supersonic flow at that area ratio. The area ratio is unity when the Mach number is equal to one.
  • 62. 62 Consider the converging nozzle shown below. The flow is supplied by a reservoir at pressure Pr and temperature Tr. The reservoir is large enough that the velocity in the reservoir is zero. Let's plot the ratio P/Po along the length of the nozzle, the mass flow rate through the nozzle, and the exit plane pressure Pe as the back pressure Pb is varied. Let's consider isentropic flow so that Po is constant throughout the nozzle. Effect of Back Pressure on Flow through a Converging Nozzle
  • 63.  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 the Fig. above.  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. 63 Effect of Back Pressure on Flow through a Converging Nozzle
  • 64.  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 steady-flow conditions, the mass flow rate through the nozzle is constant and can be expressed as  Solving for T from and for P from and substituting 64 Effect of Back Pressure on Flow through a Converging Nozzle
  • 65.  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 number.  The maximum mass flow rate can be determined by differentiating the above equation 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 65 Effect of Back Pressure on Flow through a Converging Nozzle
  • 66.  A relation for the variation of flow area A through the nozzle relative to throat area A* can be obtained by combining equations f or and for the same mass flow rate and stagnation properties of a particular fluid. This yields  Another parameter sometimes used in the analysis of one- dimensional isentropic flow of ideal gases is Ma*, which is the ratio of the local velocity to the speed of sound at the throat: 66 Effect of Back Pressure on Flow through a Converging Nozzle
  • 67.  where Ma is the local Mach number, T is the local temperature, and T* is the critical temperature.  Solving for T and for T* and substituting, we get  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. 67 Effect of Back Pressure on Flow through a Converging Nozzle
  • 69.  A plot of versus Pb /P0 for a converging nozzle is shown in Fig. below.  Notice that the mass flow rate increases with decreasing Pb /P0, reaches a maximum at Pb = P*, and remains constant for Pb /P0 values less than this critical ratio. Also illustrated on this figure is the effect of back pressure on the nozzle exit pressure Pe. We observe that 69 Effect of Back Pressure on Flow through a Converging Nozzle
  • 70.  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. 70 Effect of Back Pressure on Flow through a Converging Nozzle
  • 71. 71 1. Pb = Po, Pb /Po = 1. No flow occurs. Pe = Pb, Me=0. 2. Pb > P* or P*/Po < Pb /Po < 1. Flow begins to increase as the back pressure is lowered. Pe = Pb, Me < 1. 3. Pb = P* or P*/Po = Pb /Po < 1. Flow increases to the choked flow limit as the back pressure is lowered to the critical pressure. Pe = Pb, Me=1. 4. Pb < P* or Pb /Po < P*/Po < 1. Flow is still choked and does not increase as the back pressure is lowered below the critical pressure, pressure drop from Pe to Pb occurs outside the nozzle. Pe = P*, Me=1. 5. Pb = 0. Results are the same as for item 4. Consider the converging-diverging nozzle shown below. Effect of Back Pressure on Flow through a Converging Nozzle
  • 72. Example. Effect of Back Pressure on Mass Flow Rate  Air at 1 MPa and 600°C enters a converging nozzle, shown in Fig., 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. 72
  • 73. 73
  • 74. 74 The critical-pressure ratio is determined from Table 1 (or Eq. below) to be P*/P0 = 0.5283. That is, Pt = Pb = 0.7 MPa, and Pt /P0 = 0.670. Therefore, the flow is not choked. From Table 1 at Pt /P0 = 0.670, we read Mat = 0.778 and Tt /T0 = 0.892.
  • 75. 75
  • 76. Converging–Diverging Nozzles  When we think of nozzles, we ordinarily think of flow passages whose cross-sectional 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. 76
  • 77.  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. 77 Converging–Diverging Nozzles
  • 78.  Consider the converging– diverging nozzle shown in Fig. A fluid enters the nozzle with a low velocity at stagnation pressure P0. When Pb =P0 (case A), there is 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.78 Converging–Diverging Nozzles
  • 80. 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. 80 Converging–Diverging Nozzles
  • 81.  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 crosses the normal shock. 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 approximated as isentropic.  When Pb = PF, no shocks occur within or outside the nozzle. When Pb < PF, irreversible mixing and expansion waves occur downstream of the exit plane of the nozzle. When Pb > PF, however, the pressure of the fluid increases from PF to Pb irreversibly in the wake of the nozzle exit, creating what are called oblique shocks.81 Converging–Diverging Nozzles
  • 82. 82 Example -7 Air leaves the turbine of a turbojet engine and enters a convergent nozzle at 400 K, 871 kPa, with a velocity of 180 m/s. The nozzle has an exit area of 730 cm2. Determine the mass flow rate through the nozzle for back pressures of 700 kPa, 528 kPa, and 100 kPa, assuming isentropic flow. The stagnation temperature and stagnation pressure are T T V C o P   2 2
  • 83. 83 For air k = 1.4 and from Table or using the equation below the critical pressure ratio is P*/Po = 0.528. The critical pressure for this nozzle is P P kPa kPa o * . . ( )    0528 0528 1000 528 Therefore, for a back pressure of 528 kPa, M = 1 at the nozzle exit and the flow is choked. For a back pressure of 700 kPa, the nozzle is not choked. The flow rate will not increase for back pressures below 528 kPa.
  • 84. 84 For the back pressure of 700 kPa, P P kPa kPa P P B o o    700 1000 0700. * Thus, PE = PB = 700 kPa. For this pressure ratio Table 1 gives ME  0 7324. T T T T K K E o E o     09031 09031 09031 4161 3758 . . . ( . ) . C kRT kJ kg K K m s kJ kg m s V M C m s m s E E E E E        14 0287 3758 1000 3886 07324 3886 284 6 2 2 . ( . )( . ) . ( . )( . ) . 
  • 85. 85 E E E P RT kPa kJ kg K K kJ m kPa kg m     ( ) ( . )( . ) . 700 0 287 3758 6 4902 3 3 Then  . ( )( . ) ( ) . m A V kg m cm m s m cm kg s E E E     6 4902 730 284 6 100 134 8 3 2 2 2 For the back pressure of 528 kPa, P P kPa kPa P P E o o    528 1000 0528. *
  • 86. 86 This is the critical pressure ratio and ME = 1 and PE = PB = P* = 528 kPa. T T T T T T K K E o o E o      * . . . ( . ) . 08333 08333 08333 4161 3467 And since ME = 1,  V C kRT kJ kg K K m s kJ kg m s E E E     14 0 287 346 7 1000 3732 2 2 . ( . )( . ) .  E P RT kPa kJ kg K K kJ m kPa kg m      * * * (528 ) ( . )( . ) . 0 287 346 7 53064 3 3
  • 87. 87  . ( )( . ) ( ) . m A V kg m cm m s m cm kg s E E E     53064 730 3732 100 144 6 3 2 2 2 For a back pressure less than the critical pressure, 528 kPa in this case, the nozzle is choked and the mass flow rate will be the same as that for the critical pressure. Therefore, at a back pressure of 100 kPa the mass flow rate will be 144.6 kg/s. Example -8 Air enters a converging–diverging nozzle, shown in Fig., at 1.0 MPa and 800 K with a negligible velocity. The flow is steady, one-dimensional, 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.
  • 88. 88
  • 89. 89
  • 90. 90
  • 91. 91
  • 92. 92 Normal Shocks 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. The normal shock 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
  • 93. 93 We assume steady-flow with no heat and work interactions and no potential energy changes. We have the following Conservation of mass 1 1 2 2 2 2 2 2 AV AV V V      
  • 94. 94 Conservation of energy 2 2 1 2 1 2 1 2 1 2 2 2 : o o o o V V h h h h for ideal gases T T      Conservation of momentum Rearranging and integrating yield 1 2 2 1( ) ( )A P P m V V   Increase of entropy 2 1 0s s  Thus, we see that from the conservation of energy, the stagnation temperature is constant across the shock. However, the stagnation pressure decreases across the shock because of irreversibilities. The ordinary (static) temperature rises drastically because of the conversion of kinetic energy into enthalpy due to a large drop in fluid velocity.
  • 95. 95 We can show that the following relations apply across the shock. 2 2 1 2 1 2 2 1 12 2 1 2 2 2 2 1 2 2 1 1 ( 1)/ 2 1 ( 1)/ 2 1 ( 1)/ 2 1 ( 1)/ 2 2/( 1) 2 /( 1) 1 T M k T M k M M kP P M M k M k M M k k                The entropy change across the shock is obtained by applying the entropy-change equation for an ideal gas, constant properties, across the shock: 2 2 2 1 1 1 ln lnp T P s s C R T P   
  • 96. 96 We can combine the conservation of mass and energy relations into a single equation and plot it on an h-s 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 h-s diagram yield a curve called the Rayleigh line.
  • 97. 97 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 wave. Notice from Fig. that entropy increases, s2 >s1. This is expected since the flow through the shock is adiabatic but irreversible.
  • 99. 99
  • 100. 100 Example -9 Air flowing with a velocity of 600 m/s, a pressure of 60 kPa, and a temperature of 260 K undergoes a normal shock. Determine the velocity and static and stagnation conditions after the shock and the entropy change across the shock. The Mach number before the shock is 1 1 1 1 1 2 2 600 1000 1.4(0.287 )(260 ) 1.856 V V M C kRT m s m kJ sK kJkg K kg     
  • 101. 101 For M1 = 1.856, Table 1 gives 1 1 1 1 0.1597, 0.5921 o o P T P T   For Mx = 1.856, Table 2 gives the following results. 2 2 2 1 1 2 22 1 1 1 0.6045, 3.852, 2.4473 1.574, 0.7875, 4.931o o o P M P P PT T P P        
  • 102. 102 From the conservation of mass with A2 = A1. 2 2 1 1 1 2 2 1 600 245.2 2.4473 V V m V msV s         2 2 1 1 2 2 1 1 60 (3.852) 231.1 260 (1.574) 409.2 P P P kPa kPa P T T T K K T       1 1 2 1 1 260 439.1 0.5921 o o o T K T K T T T           1 1 1 1 60 375.6 0.1597 o o P kPa P kPa P P         
  • 103. 103 The entropy change across the shock is 2 2 2 1 1 1 ln lnP T P s s C R T P                  2 1 1.005 ln 1.574 0.287 ln 3.852 0.0688 kJ kJ s s kg K kg K kJ kg K        You are encouraged to read about the following topics in the text: •Oblique shocks 2 2 1 1 375.6 (0.7875) 295.8o o o o P P P kPa kPa P   
  • 104. Example 10. Shock Wave in a Converging–Diverging Nozzle  If the air flowing through the converging–diverging nozzle of Example 8 experiences a normal shock wave at the nozzle exit plane (see fig below), 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, one-dimensional, and isentropic flow with k = 1.4 from the nozzle inlet to the shock location. 104
  • 105. 105 Analysis (a) The fluid properties at the exit of the nozzle just before the shock (denoted by subscript 1) are those evaluated in Example 8 at the nozzle exit to be The fluid properties after the shock (denoted by subscript 2) are related to those before the shock through the functions listed in Table 2. For Ma1= 2.0, we read
  • 106. 106
  • 107. 107
  • 108. Oblique Shocks  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. Reading Assignment  Read about oblique shock waves: characteristic features, governing equations, calculation o properties;  Textbook to Read  [YUNUS A. CENGEL] Fluid Mechanics. Fundamentals and Applications 108