solutionS MANUAL FOR
Introduction to
Compressible Fluid Flow
Second Edition
by
Patrick H. Oosthuizen
William E. Carscallen
solutionS MANUAL FOR
Introduction to
Compressible Fluid Flow
Second Edition
by
Patrick H. Oosthuizen
William E. Carscallen
Boca Raton London New York
CRC Press is an imprint of the
Taylor & Francis Group, an informa business
CRC Press
Taylor & Francis Group
6000 Broken Sound Parkway NW, Suite 300
Boca Raton, FL 33487-2742
© 2014 by Taylor & Francis Group, LLC
CRC Press is an imprint of Taylor & Francis Group, an Informa business
No claim to original U.S. Government works
Printed on acid-free paper
Version Date: 20130603
International Standard Book Number-13: 978-1-4398-7795-1 (Ancillary)
This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and
information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and
publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission
to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any
future reprint.
Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic,
mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or
retrieval system, without written permission from the publishers.
For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact
the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides
licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment
has been arranged.
Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation
without intent to infringe.
Visit the Taylor & Francis Web site at
http://www.taylorandfrancis.com
and the CRC Press Web site at
http://www.crcpress.com
PREFACE
This solution manual contains complete solutions to all of the problems in the
textbook “Introduction to Compressible Fluid Flow”, second edition. The problems have
been solved using the same basic methodology as that adopted in the worked examples
contained in the textbook. Each solution in the manual begins with the statement of the
problem and each solution begins on a separate page. This arrangement will, we hope,
prove to be convenient to those using this Solution Manual. The solutions are grouped by
chapter and the solutions for each chapter are preceded by summaries of the major
equations developed in the corresponding chapter in the textbook. Instructors may wish
to provide copies of these equation summaries to their students.
The problems can all be solved using the equations and tables given in the textbook.
However, the majority of the solutions in this manual have been obtained with the aid of
the software COMPROP. The use of this software is described in an appendix in the
textbook. The software is available free of charge through the publisher to adopters of
the textbook.
The authors would like to express their sincere appreciation to Jane Paul for all of her
help in preparing the Solution Manual.
Patrick H. Oosthuizen
William E. Carscallen
Table of Contents
Preface
1. Introduction....................................................................................................................1
2. Equations for Steady One-Dimensional Compressible Fluid Flow.........................25
3. Some Fundamental Aspects of Compressible Flow ..................................................42
4. One-Dimensional Isentropic Flow ..............................................................................67
5. Normal Shock Waves.................................................................................................120
6. Oblique Shock Waves ................................................................................................200
7. Expansion Waves: Prandtl–Meyer Flow .................................................................248
8. Variable Area Flow....................................................................................................305
9. Adiabatic Flow in a Duct with Friction....................................................................437
10. Flow with Heat Transfer .........................................................................................536
11. Hypersonic Flow.......................................................................................................639
12. High-Temperature Flows ........................................................................................655
13. Low-Density Flows...................................................................................................674
14. An Introduction to Two-Dimensional Compressible Flow ..................................683
Chapter One
INTRODUCTION
SUMMARY OF MAJOR EQUATIONS
Perfect Gas Relations
R 
 RT    T

m
p
 
cp
cv
R  c p  cv
Conservation Equations for Steady Flow
Conservation of Mass:
(Rate mass enters control volume) = (Rate mass leaves control volume)
Conservation of Momentum:
(Net Force on Gas In Control Volume In Direction Considered) =
(Rate Momentum leaves Control Volume in Direction Considered)
 (Rate Momentum enters Control Volume in Direction Considered)
Conservation of Energy:
(Rate sum of Enthalpy and Kinetic Energy leave control volume)
- (Rate sum of Enthalpy and Kinetic Energy enter control volume) =
(Rate Heat is transferred into control volume) (RateWork is done by fluid in control volume)
1
(1.1)
(1.2)
(1.3)
PROBLEM 1.1
An air stream enters a variable area channel at a velocity of 30 m/s with a pressure of
120 kPa and a temperature of 10o C. At a certain point in the channel, the velocity is
found to be 250 m/s. Using Bernoulli's equation (i.e. p + ρ V 2 = constant), which assumes
incompressible flow, find the pressure at this paint. In this calculation use the density
evaluated at the inlet conditions. If the temperature of the air is assumed to remain
constant, evaluate the air density at the point in the flow where the velocity is 250 m/s.
Compare this density with the density at the inlet to the channel. On the basis of this
comparison, do you think that the use of Bernoulli's equation is justified?
SOLUTION
Bernoulli's equation gives:
p1  
V12
V2
= p2   2
2
2
This can be rearranged to give:
V12
V22 
p2 =  

  p1
2 
 2
(a)
The density, ρ , is evaluated using the initial conditions i.e. by using p, p1 /   RT1 ,
which, since air flow is being considered, gives:
=
p1
120  103
=
= 1.478 kg/m3
287  283
RT1
Substituting this back into eq. (a) then gives:
 302
2502 
p2 = 1.478  

 120  103 = 7.448  104 Pa = 74.48 kPa

2 
 2
Therefore the pressure at the point considered is 74.48 kPa.
2
Assuming that the changes in temperature can be neglected, the equation of state
gives at the exit:
2 =
p2
74.48  103
=
= 0.917 kg/m3
287  283
RT2
Since this indicates that the density changes by about 38%, the incompressible flow
assumption is not justified.
3
PROBLEM 1.2
The gravitational acceleration on a large planet is 90 ft/sec2. What is the gravitational
force acting on a spacecraft with a mass of 8000 lbm on this planet?
SOLUTION
Because:
Force  Mass  Acceleration
it follows that:
Gravitational Force  8000  90 lbm-ft/sec 2
But 32.2 lbm-ft /sec2 = 1 lbf, so this equation gives:
Gravitational Force =
800  90
 22,360 lbf
32.2
Therefore the gravitational force acting on the craft is 22,360 lbf
4
PROBLEM 1.3
The pressure and temperature at a certain point in an air flow are 130 kPa and 30° C
respectively. Find the air density at this point in kg/m3 and lbm/ft.
SOLUTION
The density, ρ , is evaluated using the perfect gas law, i.e. by using p /   RT .
Using SI units this gives since air-flow is being considered:
 
130  103
= 1 .495 kg / m3
287  303
But 1 lbm = 0.4536 kg and 1 ft = 0.3048 m so:
1.495 kg / m3 
1.495 / 0.4536
 0.0943 lbm / ft 3
3
( 1/ 0.3058)
Therefore the density is 1.459 kg/m3 or 0.0943 lbm/ft3
5
PROBLEM 1.4
Two kilograms of air at an initial temperature and pressure of 30° C and 100 kPa
undergoes an isentropic process, the final temperature attained being 850° C. Find the
final pressure, the initial and final densities and the initial and final volumes.
SOLUTION
The isentropic relations give:
 /( 1)
T 
p2   2 
 T1 
1.4/(1.4 1)
 1123 
 p1  

 303 
 100  9801 kPa 1
The initial density is given by the perfect gas law as:
1 =
100  103
= 1.15 kg/m3
287  303
The final density is given in the same way by:
2 =
9801  103
= 30.41 kg/m3
287  1123
Also, using:
Volume =
Mass
Density
It follows that the initial and final volumes are given by:
V1 =
2
= 1.739 m3
1.150
V2 
2
 0.0658 m3
30.41
and:
Therefore the initial and final volumes are 1.739 m3 and 0.0658 m3.
6
PROBLEM 1.5
Two jets of air, each having the same mass flow rate, are thoroughly mixed and then
discharged into a large chamber. One jet has a temperature of 120° C and a velocity of
100 m/s while the other has a temperature of 50° C and a velocity of 300 m/s. Assuming
that the process is steady and adiabatic, find the temperature of the air in the large
chamber.
SOLUTION
Using:
Rate Mass Enters Control Volume = Rate Mass Leaves Control Volume
and because the flow is adiabatic:
Rate Enthalpy plus
Kinetic Energy Leave
Control Volume
-
Rate Enthalpy plus
Kinetic Energy enter
Control Volume
=0
If the subscripts 1 and 2 are used to refer to conditions in the two jets and if the
subscript 3 is used to refer to conditions in the chamber then, if m refers to the mass flow
rate in each jet, the first of the above equations gives:
m3 = m1  m2 = 2mi
where mi is the mass flow rate in each of the jets, i.e., mi = m1 = m2.
The second of the above equations then gives if the velocity in the chamber is
assumed to be small because the chamber is large:

m3 h3 = m1  h1 


V12 
V22 

m
h



2
 2
2 
2 

Because air flow is involved it can be assumed that h = cpT. Therefore the above
equations together give f it is assumed that cp is constant:
7


1002 
3002 
2mi c pT3 = mi  c p  393 
  mi  c p  223 

2 
2 


Dividing through by mi cp then gives:

2T3 =  393 


1002  
3002 
   223 

2c p  
2c p 
i.e. since air flow is involved:
T3 =
1
1002  
3002  
  393 
   223 
  = 332.8K
2  
2 x 1007  
2 x 1007  
Therefore the air temperature in the chamber is 332.8 K ( = 59.8° C).
8
PROBLEM 1.6
Two air streams are mixed in a chamber. One stream enters the chamber through a 5
cm diameter pipe at a velocity of 100 m/s with a pressure of 150 kPa and a temperature of
30o C. The other stream enters the chamber through a 1.5 cm diameter pipe at a velocity
of 150 m/s with a pressure of 75 kPa and a temperature of 30o C. The air leaves the
chamber through a 9 cm diameter pipe at a pressure of 90 kPa and a temperature of 30o C.
Assuming that the flow is steady find the velocity in the exit pipe.
SOLUTION
The flow situation being considered is shown in Fig. P1.6.
Figure P1.6
The densities in the three pipes are evaluated using the perfect gas law i.e. using
p /   RT . This gives:
1 =
150  103
= 1 .725 kg / m3
287  303
9
2 =
75  103
= 0 .863 kg / m3
287  303
3 =
90  103
= 1 .035 kg / m3
287  303
Conservation of mass gives since the flow is steady:
m1  m2 = m3
i.e.:
1V1 A1  2V2 A2 = 3V3 A3
Hence:
1.725  100 

4
 0.052  0.863  150 

4
 0.0152 = 1.035  V3 x
which gives:
V3 = 54.9m/s
Therefore the velocity in the exit pipe is 54.9 m/s.
10

4
 0.092
PROBLEM 1.7
The jet engine fitted to a small aircraft uses 35 kg/s of air when the aircraft is flying at
a speed of 800 km/h. The jet efflux velocity is 590 m/s. If the pressure on the engine
discharge plane is assumed to be equal to the ambient pressure and if effects of the mass
of the fuel used are ignored, find the thrust developed by the engine.
SOLUTION
The flow relative to the aircraft is considered. The momentum equation applied to a
control volume surrounding the aircraft then gives in the direction of flight:
Net Force on Gas in
=
Control Volume
Rate Momentum Leaves
+
Control Volume
Rate Momentum Enters
Control Volume
But the pressure is equal to ambient everywhere on the surface of the control volume
and only the air that passes through the engine undergoes a velocity change, i.e., a
momentum change. Hence if F is the force exerted on the fluid by the system (this will be
equal in magnitude but opposite in direction to the force on the aircraft, i.e., to the thrust),
it follows that:
T = m Vexit  Vinlet 
This gives because Vinlet = 800 km/hr = 222.2 m/s:
T = 35   590  222.2
 = 12873 N
Therefore, the thrust developed by the engine is 12.87 kN.
11
PROBLEM 1.8
The engine of a small jet aircraft develops a thrust of 18 kN when the aircraft is flying
at a speed of 900 km/h at an altitude where the ambient pressure is 50 kPa. The air flow
rate through the engine is 75 kg/s and the engine uses fuel at a rate of 3 kg/s. The pressure
on the engine discharge plane is 55 kPa and the area of the engine exit is 0.2 m2 . Find the
jet efflux velocity.
SOLUTION
The flow relative to the aircraft is considered and the momentum equation is applied a
control volume surrounding the aircraft. This gives in the direction of flight:
Net Force on Gas in
Control Volume=
Rate Momentum Leaves
Control Volume+
Rate Momentum Enters
Control Volume
But the pressure is equal to ambient everywhere on the surface of the control volume
except on the nozzle exit plane. Hence if T is the force exerted on the flow by the system
(this will be equal in magnitude but opposite in direction to the force on the aircraft, i.e.,
to the thrust). it follows that:
T
 pexit -
 exit  mV
 inlet
pambient  Aexit = mV
This gives because Vinlet = 900 km/hr = 250 m/s:
18000   55000  50000   0.2 = (75  3)Vexit  75 x 250
Hence:
Vexit  458.3 m/s
Therefore the jet efflux velocity is 458.3 m/s.
12
PROBLEM 1.9
A small turbo-jet engine uses 50 kg/s of air and the air/fuel ratio is 90:1. The jet
efflux velocity is 600 m/s. When the afterburner is used, the overall air/fuel ratio
decreases to 50:1 and the jet efflux velocity increases to 730 m/s. Find the static thrust
with and without the afterburner. The pressure on the engine discharge plane can be
assumed to be equal to the ambient pressure in both cases.
SOLUTION
The momentum equation is applied to the control volume surrounding the engine and
gives:
Net Force on Gas in
Control Volume =
Rate Momentum Leaves
Control Volume +
Rate Momentum Enters
Control Volume
But the pressure is equal to ambient everywhere on the surface of the control volume
and the static thrust, i.e., the thrust developed when the engine is at rest, is being
considered. Hence if T is the force exerted on the flow by the system (this will be equal in
magnitude but in the opposite direction to the force on the engine, i.e., to the thrust), it
follows that:
 exit
T = mV
First consider the thrust without afterburning. In this case:
T =  m air  m fuel Vexit = (50  50 / 90)  600 = 30333 N = 30.333 kN
Next consider the thrust with afterburning. In this case:
T =  m air  m fuel Vexit = (50  50 / 50)  730 = 37230 N = 37.23 kN
Therefore the thrust without afterburning is 30.33 kN and the thrust with afterburning
is 37.23 kN.
13
PROBLEM 1.10
A rocket used to study the atmosphere has a fuel consumption rate of 120 kg/s and a
nozzle discharge velocity of 2300 m/s. The pressure on the nozzle discharge plane is 90
kPa. Find the thrust developed when the rocket is launched at sea-level. The nozzle exit
plane diameter is 0.3 m.
SOLUTION
Applying the momentum equation to a control volume surrounding the engine gives:
 exit
T   pexit  pambient  Aexit = mV
Therefore:
T   90000  101300  

4
 0.32 = 120  2300
From this it follows that:
T = 275201 N = 275.201 kN
Therefore the thrust developed is 275.2 kN.
14
PROBLEM 1.11
A solid fuelled rocket is fitted with a convergent-divergent nozzle with an exit plane
diameter of 30 cm. The pressure and velocity on this nozzle exit plane are 75 kPa and 750
m/s respectively and the mass flow rate through the nozzle is 350 kg/s. Find the thrust
developed by this engine when the ambient pressure is (a) 100 kPa and (b) 20 kPa.
SOLUTION
Applying the momentum equation to a control volume surrounding the engine gives:
 exit
T   pexit  pambient  Aexit = mV
This gives:
T   75000  pambient  

4
x 0.32  350  750
From which it follows that:
T   283706  0.0707 pambient  N
Hence, if pambient is 100 kPa:
T   283706  7070   276636 N  276.6 kN
Similarly, if pambient is 20 kPa:
T   283706  1414   282292 N  282.29 kN
Therefore the thrusts developed when the ambient pressure is 100 kPa and when it is
20 kPa are 276.6 kN and 282.29 kN respectively.
15
PROBLEM 1.12
In a hydrogen powered rocket, hydrogen enters a nozzle at a very low velocity with a
temperature and pressure of 2000° C and 6.8 MPa respectively. The pressure on the exit
plane of the nozzle is equal to the ambient pressure which is 10 kPa. If the required thrust
is 10 MN, what hydrogen mass flow rate is required? The flow through the nozzle can be
assumed to be isentropic and the specific heat ratio of the hydrogen can be assumed to be
1.4.
SOLUTION
If the subscripts 1 and 2 are used to denote conditions on the inlet and exit planes of
the nozzle respectively, then the isentropic relations give:
p 
T2   2 
 p1 
 1 /
 T2
 10000 
 

 6800000 
0.2857
 2273  352.7 K
Because the flow is adiabatic there is no heat transfer to the hydrogen in the nozzle so
the energy equation gives:
Rate Enthalpy plus
Kinetic Energy Leave
Control Volume
Rate Enthalpy plus
Kinetic Energy enter = 0
Control Volume
-
i.e.:

V2  
V2 
 cp T1  1    cp T2  2   0
2  
2 

i.e. since the kinetic energy at the inlet is negligible:
V22  2 cp T1  T2 
But:
cp  cv  R
i.e.:
16
cp 
R
1.4 (8314 / 2)

 14550 J/kg K
0.4
 1
Hence:
V22  2  14550  (2273  352.7)
i.e.:
V2  7475 m/s
Because the pressure on the exit plane is ambient, the thrust is given by:
 2
T  mV
i.e.:
10000000  m  7475 , i.e. , m  1338 kg / s
Therefore the required mass flow rate is 1338 kg/s.
17
PROBLEM 1.13
In a proposed jet propulsion system for an automobile, air is drawn in vertically
through a large intake in the roof at a rate of 3 kg/s, the velocity through this intake being
small. Ambient pressure and temperature are 100 kPa and 30° C respectively. This air is
compressed and heated and then discharged horizontally out of a nozzle at the rear of the
automobile at a velocity of 500 m/s and a pressure of 140 kPa. If the rate of heat addition
to the air stream is 600 kW, find the nozzle discharge area and the thrust developed by
the system.
SOLUTION
The energy equation applied to the system gives:
Rate Enthalpy plus
Kinetic Energy Leave Control Volume
Rate Enthalpy plus
Kinetic Energy enter=
Control Volume
Rate Heat is
Transferred into
Control Volume
But, because of the low inlet velocity, the kinetic energy at the inlet is negligible. The
above equation therefore gives since air flow is being considered:
2 

Vexit
 cp Texit 
  cp Tinlet  0  q

2 



This equation gives assuming cp = 1007 J/kg K:

5002 
1007  Texit 
  1007  303  600000
2 

which gives:
Texit  774.7 K
The exit density is evaluated using the perfect gas law, i.e. by using p /   RT . This
gives since air flow is being considered:
18
exit 
140  103
= 0.63 kg / m3
287  774.7
Then using:
m  exit Vexit Aexit
gives:
Aexit 
3
 0.00952 m2
0.63  500
Because no air enters the system in the direction of motion, the momentum equation
gives:
 exit
T   pexit  pambient  Aexit  mV
which gives:
T  140000  pambient   0.00952  3  500
This equation then gives if the ambient pressure is assumed to be 101.3 kPa = 101300
Pa:
T  1868.4 N
Therefore the nozzle discharge area is 0.00952 m2 and the thrust is 1868 N.
19
PROBLEM 1.14
Carbon dioxide flows through a constant area duct. At the inlet to the duct the
velocity is 120 m/s and the temperature and pressure are 200° C and 700 kPa
respectively. Heat is added to the flow in the duct and at the exit of the duct the velocity t.
240 m/s and the temperature is 450° C. Find the amount of heat being added to the
carbon dioxide per unit mass of gas and the mass flow rate through the duct per unit
cross-sectional area of the duct at inlet. Assume that for carbon dioxide  = 1.3.
SOLUTION
The energy equation applied to the system gives:
Rate Enthalpy plus
Kinetic Energy Leave
Control Volume
-
Rate Enthalpy plus
Kinetic Energy enter
Control Volume
=
Rate Heat is
Transferred into
Control Volume
i.e., considering the changes per unit mass flow through the system:


V2 
V2 
 cp Texit  exit    cp Tinlet  inlet   q


2 
2 


where q is here the rate of heat transfer to the gas per unit mass flow rate. Taking:
cp 
1.3  8314 / 44 
R

 818.8 J/kg K
0.3
 1
then gives:
 2402
1202 
q  818.8 x  450  200   

  226300 J/kg = 226.3 kJ/kg
2 
 2
Using:
m
 inlet Vinlet
Ainlet
and
it follows that:
20
pinlet
inlet
= RTinlet


 p

700000
m
  inlet Vinlet = 
  120  939.9 kg/s m 2


Ainlet
 8314 / 44   473 
 RTinlet 
Therefore the amount of heat added per unit mass of gas is 226.3 kJ/kg and the mass
flow rate per unit inlet cross-sectional area is 939.9 kg/s m2.
21
PROBLEM 1.15
Air enters a heat exchanger with a velocity of 120 m/s and a temperature and pressure
of 225o C and 2.5 MPa respectively. Heat is removed from the air in the heat exchanger
and the air leaves with a velocity 30 m/s at a temperature and pressure of 80o C and 2.45
MPa. Find the heat removed per kg of air flowing through the heat exchanger and the
density of the air at the inlet and the exit to the heat exchanger.
SOLUTION
The energy equation applied to the system gives:
Rate Enthalpy plus
Kinetic Energy Leave
Control Volume
-
Rate Enthalpy plus
Kinetic Energy enter
Control Volume
=
Rate Heat is
Transferred into
Control Volume
i.e. considering the changes per unit mass flow through the system:


V2 
V2 
 cp Texit  exit    cp Tinlet  inlet   q


2 
2 


where q is here the rate of heat transfer to the gas per unit mass flow rate.
The specific heat, cp , will be assumed constant and, because air flow is being
considered, it value will be taken as cp = 1007 J/kg K. Therefore:
 302
1202 
q  1007  80  220   

   134230J/kg  134.2 kJ/kg
2 
 2
using p /   RT it follows that:
inlet 
2500  103
= 17.49 kg / m3
287  498
outlet 
2450  103
= 24.18 kg / m3
287  353
22
Therefore the rate that heat is removed per kg of air is 134.2 kJ/kg and the density of
air at the inlet and the exit is 17.49 kg/m3 and 24.18 kg/m3 respectively.
23
PROBLEM 1.16
The mass flow rate through the nozzle of a rocket engine is 200 kg/s. The areas of the nozzle
inlet and exit planes are 0.7 m2 and 2.4 m2 , respectively. On the nozzle inlet plane, the pressure
and velocity are 1600 kPa and 150 m/s respectively whereas on the nozzle exit plane the pressure
and velocity are 80 kPa and 2300 m/s respectively. Find the thrust force acting on the nozzle.
SOLUTION
The momentum equation is applied to the control volume surrounding the nozzle and
gives:
Net Force on Gas in
Control Volume
=
Rate Momentum Leaves
Rate Momentum Enters
+
Control Volume
Control Volume
It is assumed that the pressure is equal to ambient everywhere on the surface of the
control volume except on the inlet and exit planes. Hence if T is the force exerted on the
flow which will be equal in magnitude but in the opposite direction to the force on the
nozzle, i.e., to the thrust, it follows that:
T   pexit Aexit  pinlet Ainlet   m Vexit  Vinlet 
Hence:
T  m Vexit  Vinlet    pexit Aexit  pinlet Ainlet 
 200  (2300  150)  (80000  2.4  1600000  0.77)   610000 N   610 kN
Therefore the thrust force on the nozzle is 610 kN this thrust force arising mainly as a
result of the pressure difference across the nozzle.
24
Chapter Two
EQUATIONS FOR STEADY
ONE-DIMENSIONAL COMPRESSIBLE
FLUID FLOW
SUMMARY OF MAJOR EQUATIONS
Relation Between Fractional Changes in Flow Variables
Mass Conservation:
d


dV
dA

 0
V
A
(2.3)
 V dV
(2.8)
Momentum:

dp

Energy:
cp d T  V d V  0
(2.17)
dp
d
dT


 0
p
T

(2.19)
   1  dp
ds
dT

 

cp
T
   p
(2.24)
Equation of State:
Entropy:
25
PROBLEM 2.1
Air enters a tank at a velocity of 100 m/s and leaves the tank at a velocity of 200 m/s.
If the flow is adiabatic, find the difference between the temperature of the air at the exit
and the temperature of the air at the inlet.
SOLUTION
Because the flow is adiabatic, the energy equation gives:
2 
2



Vexit
Vinlet
 cp Texit 
   cp Tinlet 





2
2




Hence:
1
 cp

Texit  Tinlet  
2
 Vinlet
V2 

 exit 
 2
2 

Since air flow is being considered the specific heat, cp , will be assumed to be 1007 J/
kg oC. The above equation then gives:
2002 
 1   1002
o

Texit  Tinlet  

   14.9 C
 2
1007
2



Therefore the temperature decreases by 14.9oC.
26
PROBLEM 2.2
Air at a temperature of 25° C is flowing at a velocity of 500 m/s. A shock wave (see
later chapters) occurs in the flow reducing the velocity to 300 m/s. Assuming the flow
through the shock wave to be adiabatic, find the temperature of the air behind the shock
wave.
SOLUTION
Because the flow is adiabatic, the energy equation gives if subscripts 1 and 2 refer to
conditions before and after the shock wave respectively:
cp T2 
V22
V2
 cp T1  1
2
2
Hence:
 1  V 2
V2 
3002 
 1   5002
o
  1  2   298  


  377K  104.4 C


 cp   2
2
1007
2
2





 
T2  T1  
Since air flow is being considered the specific heat cp has been assumed to be 1007 J/
kg oC.
Therefore the temperature “behind” the shock wave is 104.4o C.
27
PROBLEM 2.3
Air being released from a tire through the valve is found to have a temperature of
15°C. Assuming that the air in the tire is at the ambient temperature of 30°C find the
velocity of the air at the exit of the valve. The process can be assumed to be adiabatic.
SOLUTION
Because the flow is adiabatic, the energy equation gives if subscripts 1 and 2 refer to
conditions in the tire and at the discharge from the valve respectively:
cp T2 
V22
V2
 cp T1  1
2
2
But the velocity in the tire can be assumed to be zero so this gives:
cp T2 
V22
V2
 cp T1 , i.e., 2  cp T1  T2 
2
2
Hence, assuming that for air the specific heat cp is 1007 J/ kg o C:
V2 
2cp T1  T2   2  1007   303  288  173.8 m/s
Therefore the air leaves the valve with a velocity of 173.8 m/s.
28
PROBLEM 2.4
A gas with a molecular weight of 4 and a specific heat ratio of 1.67 flows through a
variable area duct. At some point in the flow the velocity is 180 m/s and the temperature
is 10° C. At some other point in the flow, the temperature is -10° C. Find the velocity at
this point in the flow assuming that the flow is adiabatic.
SOLUTION
Because the flow is adiabatic, the energy equation gives if subscripts 1 and 2 refer to
conditions at the first and second points considered respectively:
cp T2 
V22
V2
 cp T1  1
2
2
Hence:
V22  V12  2 cp T1  T2 
Assuming that the gas can be treated as a perfect gas:

R  cp  cv  cp 1 


cv 
 i.e.,
cp 
cp 
R
 1
Hence for the gas being considered:
cp 
1.67  (8314/ 4)
 5181 J/kg oC
1.67  1
The energy equation therefore gives:
V22  V12  2 cp T1  T2   1802  2  6181   283  263
Which gives V2 = 489.5 m/s. Therefore the velocity at the second point is 489.5 m/s.
29
PROBLEM 2.5
At a section of a circular duct through which air is flowing the pressure is 150 kPa ,
the temperature is 35 °C , the velocity is 250 m/s , and the diameter is 0.2 m. If, at this
section, the duct diameter is increasing at a rate of 0.1 m / m find dp/dx , dV/dx , and
d/dx .
SOLUTION
Because:
A 
 D2
4
it follows that:
1 dA
2 dD

A dx
D dx
Hence. at the section considered:
1 dA
2

 0.1  1m 1
0.2
A dx
But the continuity equation gives:
1 dA
1 dV
1 d


 0
 dx
A dx
V dx
But using the information supplied:
 
150000
p

 1.697 kg/m3
287  308
RT
Therefore the continuity equation gives:
1
1 dV
1 d

 0
250 dx 1.697 dx
It is next noted that the conservation of momentum equation gives:
30
(1)

1 dp
dV
 V
dx
 dx
, i.e. ,
dp
dV
  V
dx
dx
hence:
dp
dV
dV
 1.697  250 
  424.3
dx
dx
dx
(2)
The conservation of energy equation gives:
cp
dT
dV
 V
 0
dx
dx
hence again assuming that cp is equal to 1007 J / kg °C it follows that:
1007
dT
dV
 250
0
dx
dx
i.e.:
dT
dV
  0.243
dx
dx
(3)
Lastly, it is noted that from the perfect gas law, p =  R T , it follows that:
1 dp
1 d
1 dT


p dx
T dx
 dx
i.e.:
dp
1
1 d
1 dT


150000 dx
1.697 dx
308 dx
(4)
Using eq. (3), eq. (4) becomes:
1
dp
1 d
0.2483 dV


150000 dx
1.697 dx
308 dx
i.e.:
dp
d
dV
 88915
 120.9
dx
dx
dx
Substituting from eqs. (1) and (2) into eq. (5) then gives:
31
(5)
 424.3

1 dV 
dV
dV
  88915  1.697  1 
  120.9
250
dx
dx
dx


i.e.:
dV 
88915  1.697 
  424.3  120.9 
   88915  1.697
dx 
250

Hence:
dV
 502.6 s 1
dx
(6)
Using this result in eq. (1) then gives:
1 
502.6
1 d

 0
250
1.697 dx
i.e.:
d
1.697  502.6
  1.687 
 1.715 kg/m 4
dx
250
Similarly, using eq. (6) in eq. (2) gives:
dp
 424.3  502.6  213250 Pa/m
dx
Therefore the values of dp/dx , dV/dx and d/dx are 213250 Pa/m, -502.6 m/s per m,
and 1.715 kg/m4 respectively.
32
PROBLEM 2.6
Consider an isothermal air flow through a duct. At a certain section of the duct, the
velocity, temperature and pressure are 200 m/s, 25°, and 120 kPa respectively. If the
velocity is decreasing at this section at a rate of 30 per cent per m find dp/dx, ds/dx and
dp/dx.
SOLUTION
The flow considered in this problem is not adiabatic.
The conservation of momentum equation gives:

dV
1 dp
 V
, i.e. ,
 dx
dx
dp
dV
  V
dx
dx
But:
 
120000
p

 1.403 kg/m 3
287  298
RT
and:
1 dV
  0.3
V dx
so:
dp
  1.403  200  200  0.3  16836 Pa/m
dx
Because the flow is isothermal, the perfect gas law, p = ρ R T , gives:
1 dp
1 d

p dx
 dx
Hence:
33
 dp
1.403
d


 16836  0.1968 kg/m3 /m
dx
p dx 120000
Lastly since:
1 ds
1 dT   1  1 dp



cp dx
T dx    p dx
it follows that for the isothermal situation being considered:
  1  1 dp
1 ds
 

cp dx
   p dx
which gives:
1 ds
1 dp
 0.4 
 

1007 dx
 1.4  120000 dx
i.e.:
ds
1007
 0.4 
 
 16836   40.36 J/kg-K per m
 
120000
dx
 1.4 
The entropy is changing because of the heat transfer at the wall.
Therefore, the values of dp/dx, dp/dx, and ds/dx are 16.84 kPa/m, 0.1968 kg/m3 / m,
and - 40.36 J / kg-K per m respectively.
34
PROBLEM 2.7
Consider adiabatic air flow through a variable area duct. At a certain section of the
duct, the flow area is 0.1 m2, the pressure is 120 kPa, and the temperature is 15°C and the
duct area is changing at a rate of 0.1 m2/m. Plot the variations of dp/dx, dV/dx and dρ/dx
with the velocity at the section for velocities between 50 m/s and 300 m/s.
SOLUTION
The continuity equation gives:
1 dA
1 dV
1 d


 0
A dx
V dx
 dx
But using the information supplied:

p
120000

 1.452 kg/m 3
RT
287  288
Therefore the continuity equation gives:
1
1 dV
1 d
 0.1 

 0
0.1
V dx
1.452 dx
(1)
It is next noted that the conservation of momentum equation gives:

1 dp
dV
 V
 dx
dx
, i.e. ,
dp
dV
  V
dx
dx
hence:
dp
dV
  1.452 V
dx
dx
The conservation of energy equation gives:
cp
dT
dV
 V
 0
dx
dx
hence again assuming that cp is equal to 1007 J / kg °C it follows that:
35
(2)
1007
dT
dV
 V
 0
dx
dx
(3)
Lastly, it is noted that from the perfect gas law, p =  R T , it follows that:
1 dp
1 d
1 dT


p dx
 dx
T dx
i.e.:
1
1 d
1 dT
dp


120000 dx
1.452 dx
288 dx
(4)
Using eq. (3), eq. (4) becomes:
1
dp
1 d
1
dV


V
120000 dx
1.452 dx
1007  288 dx
i.e.:
dp
d
dV
 82645
 0.4138 V
dx
dx
dx
(5)
Substituting from eqs. (1) and (2) into eq. (5) then gives:
1.452 V
dV
1 dV 
dV

  82645  1.452  1 
  0.4138 V
dx
V dx 
dx

i.e.:
dV
dx
82645  1.452 

 0.4138V  1.452V 
   82645  1.452
V


i.e.:
dV
120000
 
dx
120000 / V  1.038V
Using this result in eq. (1) then gives:
36
(6)
1 d
120000
 1 
1.452 dx
120000  1.038V 2
i.e.:
d
174240
  1.452 
dx
120000  1.038V 2
(7)
Similarly, using eq. (6) in eq. (2) gives:
dp
124240V

dx
120000 / V  1.038V
For any value of V ( m/s ), eqs. (6), (7) and (8) allow dV/dx ( 1 / m), dρ/dx ( Kg / m3 / m),
and dp/dx ( Pa / m) to be found. Some results are given in the following table, these
results also being shown in Figs. 2.7a, 2.7b and 2.7c that follow the table.
V – m/s
50
75
100
125
150
175
200
225
250
275
300
dp/dx – Pa/m
3710
8585
15894
26230
40559
60480
88780
130716
197417
317159
588781
dρ/dx - Kg / m3 / m
0.0321
0.0742
0.1374
0.2268
0.3506
0.5229
0.7675
1.1305
1.7077
2.7418
5.0900
37
dV/dx - 1 / m
-51.1
-78.8
-109.5
-144.5
-186.2
-238.0
-305.7
-400.1
-543.9
-794.3
-1315.7
Figure P2.7a
Figure P2.7b
38
Figure P2.7c
39
PROBLEM 2.8
Methane flows through a circular pipe which has a diameter of 4cm. The temperature,
pressure, and velocity at the inlet to the pipe are 200 K, 250 kPa, and 30 m/s respectively.
Assuming that the flow is steady and isothermal calculate the pressure on the exit plane
and the heat added to the methane in the pipe if the velocity on the pipe exit plane is
35m/s. Assume that the methane can be treated as a perfect gas with a specific heat ratio
of 1.32 and a molar mass of 16.
SOLUTION
As shown in the textbook momentum conservation considerations give:

dp

 V dV
Using the perfect gas law this can be written as:
 RT
dp
 V dV
p
For an isothermal flow this can be integrated between any two points, 1 and 2, in the
flow to give:
p 
p 
V22 V12
1  V22 V12 

 RT  ln p2  ln p1   RT ln  2  , i.e., ln  2  
 

2
2
2 
 p1 
 p1  RT  2
Using the information provided:
R 
8314
 519.6 J/kg K
16
Hence
p 
 352 302 
1
ln  2  


  0.001564 , i.e.,
519.6  200  2
2 
 p1 
p2  p1 e0.001564  250e0.001564  250.4 kPa
40
Now since using the perfect gas law gives:
 
p
p
250000
 2.406 kg/m3
, i.e. 1  1 
RT
RT1
519.6  200
Therefore the mass flow rate through the pipe is given by:
m  1V1 A  2.406  30 

4
 0.042  0.091 kg/s
The heat added per unit mass of air flow is given by the energy equation as:

q   c p T2 

V22  
V12 
c
T


 

2   p 1
2 
i.e., since isothermal flow is being considered:
q 
V22 V12
352
302



 162.5 J/kg
2
2
2
2
Hence the rate of heat addition to the flow is given by:
Q  m q  0.091  162.5  14.79 J/s
Therefore the exit plane pressure is 250.4 kPa and the rate heat is added to the flow is
14.79 J/s.
41
Chapter Three
SOME FUNDAMENTAL ASPECTS OF
COMPRESSIBLE FLOW
SUMMARY OF MAJOR EQUATIONS
Perfect Gas Relations
Mach Number:
M 
gas velocity
V

speed of sound
a
(3.1)
Speed of Sound:
a 
p



a 

m
 RT
T
(3.2)
(3.20)
Mach Wave:
sin  
a
1

V
M
M 
1
sin 
42
(3.21)
PROBLEM 3.1
The velocity of an air flow changes by 1 per cent. Assuming that the flow is
isentropic, plot the percentage changes in pressure, temperature and density induced by
this change in velocity with flow Mach number for Mach numbers between 0.2 and 2.
SOLUTION
It was shown that in isentropic flow:
dp
dV
   M2
p
V
dT
T
  (  1) M 2
d

 M2
dV
V
dV
V
These relations give the following results for:
dV
 1 per cent
V
it being recalled that  = 1.4 for air:
Percentage Change Induced by 1 per cent Change in V
M
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
dp
p
dT
T
-0.056
-0.224
-0.504
-0.896
-1.400
-2.016
-2.744
-3.584
-4.536
-5.600
-0.016
-0.064
-0.144
-0.256
-0.400
-0.576
-0.784
-1.024
-1.296
-1.600
43
d

-0.040
-0.160
-0.360
-0.640
-1.000
-1.440
-1.960
-2.500
-3.240
-4.000
These results are shown plotted in Fig. P3.1.
Figure P3.1
44
PROBLEM 3.2
Calculate the speed of sound at 288 K in hydrogen, helium and nitrogen. Under what
conditions will the speed of sound in hydrogen be equal to that in helium?
SOLUTION
At a temperature of 288 K the speed of sound in a perfect gas is given by:
a 
 RT 
 
8314
 288
m
where m is the molar mass. It has a value of 2.016, 4.003 and 28.013 for hydrogen,
helium and nitrogen respectively. The values of  for these three gases are 1.407, 1.667
and 1.401 respectively. Hence, using the above equation:
For Hydrogen: a = 1292.7 m/s
For Helium: a = 998.6 m/s
For Nitrogen: a = 346.1 m/s
Now, because as discussed above:
a2   
8314
 T
m
Hence the speeds of sound in hydrogen and helium will be equal when the temperatures
in the two gases are such that:
 hyd Thyd
mhyd

 hel Thel
mhel
i.e., using the values of  and m for hydrogen and helium given above, the speeds of
sound in the two gases will be equal when:
Thyd
 m
1.667  2.016
 hel hyd 
 0.5967
 hyd mhel
Thel
1.407  4.003
45
Therefore the speeds of sound at a temperature of 288 K in hydrogen, helium and
nitrogen are 1292.7, 998.6 and 346.1 m/s respectively and the speed of sound in hydrogen
will be equal to that in helium when the temperature of the hydrogen is 0.5967 times the
temperature of the helium.
46
PROBLEM 3.3
Find the speed of sound in carbon dioxide at temperatures of 20 °C and 600° C.
SOLUTION
The speed of sound is given by:
a 
 RT 
 
8314
 T
m
where m is the molar mass which has a value of 44.01 for carbon dioxide. The value
of  for carbon dioxide is 1.3. Hence:
a 
1.3 
8314
 T  15.67
44.01
T
Using this equation gives:
For T = 293 K: a = 268.3 m/s
For T = 873 K: a = 463.0 m/s
Therefore the speeds of sound in carbon dioxide at temperatures of 20 °C and 600° C
are 268.3 and 463.0 m/s respectively.
47
PROBLEM 3.4
A very weak pressure wave, i.e., a sound wave, across which the pressure rise is 30
Pa moves through air which has a temperature of 30°C and a pressure of 101 kPa. Find
the density change, the temperature change and the velocity change across this wave.
SOLUTION
Across a sound wave:
dp
 a2
d
where dp and d are the pressure and density changes across the wave respectively. But:
a 
 RT 
1.4  287  303  348.7 m/s
Hence:
d 
30
dp

 8.22  106 kg/m3
2
348.7 2
a
It was also shown that across a sound wave:
d p   a dV , i.e. , d V 
dp
a
But:
 
p
101000

 1.161 kg/m3
RT
287  303
dV 
dp
30

 0.074 m/s
1.161  348.7
a
Therefore:
Lastly, it is noted that the equation of state gives:
 p 
p
( p  dp)
(1  dp / p)

 

T
(   d  )(T  d T )
  T  (1  d  /  )(1  d T / T )
i.e.:
48


dp 
d  
dT 
1 
  1 
 1 
p
T 
 


which, since the fractional changes in the variables are small, i.e., dp/ p , d / , and dT /
T are small, can be rearranged to give:
dT
dp d 



T
p
Hence:
dT
30
0.00000822


303
101000
1.161
From this it follows that:
d T  0.0879K
Therefore the changes in density, temperature and velocity across the wave are 8.22
 10- 6 kg/m3 , 0.0879 K, and 0.074 m/s respectively.
49
PROBLEM 3.5
An airplane is travelling at 1500 km/h at an altitude where the temperature is -60° C.
What is the Mach at which the airplane is flying?
SOLUTION
The speed of sound at this altitude is given by:
a 
 RT 
1.4  287  213  292.6 m/s
Hence, the Mach number is given by:
M 
1500  1000 / 3600
V

 1.42
292.6
a
Therefore the aircraft is flying at a Mach number of 1.42
50
PROBLEM 3.6
An airplane is flying at 2000 km/h at an altitude where the temperature -50° C. Find
the Mach number at which the airplane is flying.
SOLUTION
The speed of sound at this altitude is given by:
a 
 RT 
1.4  287  223  299.3 m/s
Hence, the Mach number is given by:
M 
2000  1000 / 3600
V

 1.86
299.3
a
Therefore the aircraft is flying at a Mach number of 1.86
51
PROBLEM 3.7
An airplane can fly at a speed of 800 km/h at sea - level where the temperature is
15° C. If the airplane flies at the same Mach number at an altitude where the temperature
is -44° C, find the speed at which the airplane is flying at this altitude.
SOLUTION
Because the speed of sound is proportional to the square-root of the absolute
temperature it follows that if as is the speed of sound at sea-level and if a is the speed of
sound at the altitude where the temperature is -44 °C, then:
a

as
229
 0.8917
288
Hence since:
V
800

a
as
it follows that:
V  800 
a
 800  0.8917  713.4 km/hr
as
Therefore the aircraft is flying at a speed of 713.4 km/hr.
52
PROBLEM 3.8
The test section of a supersonic wind tunnel is square in cross-section with a side
length of 1.22 meters. The Mach number in the test section is 3.5, the temperature is
-100° C, and the pressure is 20 kPa. Find the mass flow rate of air through the test
section.
SOLUTION
The speed of sound in the wind-tunnel is given by:
a 
 RT 
1.4  287  173  263.7 m/s
Hence the velocity in the wind-tunnel is given by:
V  M a  3.5  263.7  923 m/s
The density in the wind-tunnel is given by:
 
20000
p

 0.4028 kg/m3
287  173
RT
The mass flow rate through the wind-tunnel is therefore given by:
m   VA  0.4028  923  1 .22  1.22   553.4 kg/s
Hence the mass flow rate through the wind-tunnel is 553.4 kg/s.
53
PROBLEM 3.9
A certain aircraft flies at the same Mach number at all altitudes. If it flies at a speed
that is 120 km/hr slower at an altitude of 12000 m than it does at sea level, find the Mach
number at which it flies. Assume standard atmospheric conditions.
SOLUTION
Because the speed of sound is given by:
a 
i.e., because it is proportional to T
0.5
 RT
, it follows that if as is the speed of sound at sea-
level and if a is the speed of sound at an altitude of 12000 m, then since in the standard
atmosphere T = 288.16 K at sea-level and T = 216.66 K at an altitude of 12000 m:
a

as
226.66
288.16
 0.8869
Hence if Vs is the speed of the aircraft at sea-level then for the aircraft being
considered:
M 
V
V
 s
a
as
It follows that:
V
a

 0.8869
Vs
as
But:
V  Vs 
120000
 Vs  33.33 m/s
3600
From which it follows that:
V
33.33
33.33
 1
, i.e. 0.8869  1 
, i.e. Vs  294.72 m/s
Vs
Vs
Vs
54
Hence, the Mach number is given by:
M 
Vs

as
294.72
 0.8661
1.4  287  288.16
Therefore the aircraft flies at a Mach number of 0.8661.
55
PROBLEM 3.10
Air at a temperature of 45° C flows in a supersonic wind-tunnel over a very narrow
wedge. A shadowgraph photograph of the flow reveals weak waves emanating from the
front of the wedge at an angle of 35° to the undisturbed flow. Find the Mach number and
velocity in the flow approaching the wedge.
SOLUTION
Assuming that the observed waves are Mach waves, it follows that:
M 
1
1

 1.414
sin 
sin 45o
Using this then gives:
V  Ma  M
 RT  1.414 
1.4  287  318  505.4 m/s
Therefore the velocity in the wind-tunnel is 505.4 m/s.
56
PROBLEM 3.11
Air at a temperature of -10o C flows through a supersonic wind tunnel. A Schlieren
photograph of the flow reveals weak waves originating at imperfections on the walls.
These weak waves are at an angle of 40o to the flow. Find the air velocity in the wind
tunnel.
SOLUTION
Assuming that the observed waves are Mach waves, it follows that:
M 
1
1

 1.555
sin 
sin 40o
Using this then gives:
V  M a  M  RT  1.555  1.4  287  263  505.7 m/s
Therefore the velocity in the wind-tunnel is 505.7 m/s.
57
PROBLEM 3.12
A gas with a molar mass of 44 and a specific heat ratio 1.67 flows through a channel
at supersonic speed. The temperature of the gas in the channel is 10° C. A photograph of
the flow reveals weak waves originating at imperfections in the wall running across the
flow at an angle of 45 degrees to the flow direction. Find the Mach number and the
velocity in the flow.
SOLUTION
Assuming that the observed waves are Mach waves, it follows that:
M 
1
1

 1.414
sin 
sin 45o
Using this then gives:
V  M a  M  RT  1.414 
1.67 
8314
 283  422.6 m/s
44
Therefore the Mach number and velocity in the wind-tunnel are 1.414 and 422.6 m/s
respectively.
58
PROBLEM 3.13
Air at 80° F is flowing at a Mach number of 1.9. Find the air velocity and the Mach
angle.
SOLUTION
The velocity is given by:
V Ma  M
 RT  1.9 
1.4  53.3  32.2  540  2164 ft/sec
and the Mach angle is given by:
 1 
 1 
  sin 1    sin 1    31.76o
M 
 1.8 
Therefore the velocity and the Mach angle are 2164 ft/sec and 31.76° respectively.
59
PROBLEM 3.14
Air at a temperature of 25° C is flowing with a velocity of 180 m/s. A projectile is
fired into the air stream with a velocity of 800 m/s in the opposite direction to that of the
air flow. Calculate the angle that the Mach waves from the projectile make to the
direction of motion.
SOLUTION
The velocity of the air relative to the projectile is given by:
V  180  800  980 m/s
Hence:
M 
V

a
V

 RT
980
 2.832
1.4  287  298
Therefore, the Mach angle is given by:
1 
 1 
1 
o
  sin 
  20.68
M 
 2.832 
  sin 1 
Therefore the Mach waves make an angle of 20.68° to the direction of motion.
60
PROBLEM 3.15
An observer at sea-level does not hear an aircraft that is flying at an altitude of 7000m
until it is a distance of 13 km from the observer. Estimate the Mach number at which the
aircraft is flying. In arriving at the answer, assume that the average temperature of the air
between sea level and 7000 m is -10°C.
SOLUTION
It is assumed that the net disturbance produced by the aircraft is weak, i.e., that
basically what is being investigated is how far the aircraft will have travelled from the
overhead position when the sound waves emitted by the aircraft are first heard by the
observer. If the discussion of Mach waves is considered, it will be seen that the aircraft
will first be heard by the observer when the Mach wave emanating from the nose of the
aircraft reaches the observer. The assumed situation is therefore as shown in Fig. P3.15.
Figure P3.15
Now, since the temperature varies through the atmosphere, the speed of sound varies
as the sound waves pass down through the atmosphere which means that the Mach waves
from the aircraft are actually curved. This effect will be neglected here, the sound speed
61
at the average temperature between the ground and the aircraft being used to describe the
Mach wave. Since the mean air temperature between the observer and the aircraft is
given as -10°C the mean speed of sound is given by:
a 
 RT 
1.4  287  263  325.1 m/s
If α is the Mach angle based on the mean speed of sound then, since the aircraft is at
an altitude of 7000m and has travelled 13 km from the overhead position before the
sound is heard, it follows that:
tan  
7000
 0.5385
13000
This gives:
  28.3
But since sin α = 1 /M it follows that:
M 
1
 2.109
sin (28.3o )
Therefore the aircraft is flying at a Mach number of 2.109.
62
PROBLEM 3.16
An aircraft is flying at an altitude of 6 km at a Mach number of 3. Find the distance
behind the aircraft at which the disturbances created by the aircraft reach sea-level.
SOLUTION
It is assumed that the net disturbance produced by the aircraft is weak, i.e., that what
is being investigated is how far the aircraft will have travelled when the sound waves
emitted by the aircraft from the overhead position first reach the ground. Therefore, the
distance behind the nose of the aircraft at which the Mach wave emanating from the nose
of the aircraft reaches the ground is required. This is shown in Fig. P3.16.
Figure P3.16
Because the temperature varies through the atmosphere, the Mach waves from the
aircraft are actually curved. This effect will be ignored here. Since sin a = 1 /M, it follows
that:
1
  sin 1    19.47o
3
63
If d is the distance behind the aircraft at which the disturbances created by the aircraft
reaches sea-level, it follows that:
tan  
6000
d
from which it follows that:
d 
6000
tan 19.47o
 16971 m
Therefore the distance behind the aircraft at which the disturbances created by the
aircraft reach sea-level is 16.97 km.
64
PROBLEM 3.17
An observer on the ground finds that an airplane flying horizontally at an altitude of
2500 m has travelled 6 km from the overhead position before the sound of the airplane is
first heard. Assuming that, overall, the aircraft creates a small disturbance, estimate the
speed at which the airplane is flying. The average air temperature between the ground
and the altitude at which the airplane is flying is 10° C. Explain the assumptions you
have made in arriving at the answer.
SOLUTION
It is assumed that the net disturbance produced by the aircraft is weak i.e. that
basically what is being investigated is how far the aircraft will have travelled from the
overhead position when the sound waves emitted by the aircraft are first heard by the
observer. Therefore, the aircraft will first be heard by the observer when the Mach wave
emanating from the nose of the aircraft reaches the observer. The situation is therefore as
shown in Fig. P3.17.
Figure P3.17
Because the temperature varies through the atmosphere, the speed of sound varies as
the sound waves pass down through the atmosphere which means that the Mach waves
65
from the aircraft are actually curved. This effect will be neglected here and the sound
speed at the average temperature between the ground and the aircraft will be used to
describe the Mach wave.
Now the mean air temperature between the observer and the aircraft is given as 10 °C
so the mean speed of sound is given by:
a 
 RT 
1.4  287  283  337.2 m/s
If α is the Mach angle based on the mean speed of sound then since the aircraft is at
an altitude of 2500 m and has travelled 6 km from the overhead position before the sound
is heard, it follows that:
tan  
Hence:
2500
 0.4167
6000
  22.62
But since sin α = 1/M it follows that:
M 
1
 2.60
sin (22.62o )
Hence, using the mean speed of sound, it follows that the mean speed of the aircraft is
given by:
V  M a  2.6  337.2  876.7 m/s
Therefore the aircraft is flying at a Mach number of 2.60 and at a speed of 876.7 m/s.
66
Chapter Four
ONE-DIMENSIONAL ISENTROPIC FLOW
SUMMARY OF MAJOR EQUATIONS
Isentropic Flow Relations
 
p2
  2
p1
 1 
 1
 
T2
  2
T1
 1 
1
(4.2)
p 
  2
 p1 
 T 2   
a2
  2   2
a1
 T1 
 1 
1 (

 1
2
 1

(4.4)
 1
 p  2
  2
 p1 
(4.5)
 1
) M 12
T2
2

 1 2
T1
1  (
) M2
2
(4.6)

1 
p2
 
p1
1 
1
2
1
2
(  1) M 12   1

(  1) M 22 
1 
2
 
1
1 
1
2
1
2
(  1) M 12   1

(  1) M 22 
(4.7)
1
  2   V2 
A1
   
A2
 1   V1 
67
(4.8)
(4.8)
Stagnation Conditions

p0
  1 2   1

M 
 1 
p
2


(4.15)

0
  1 2   1

 1 
M 

2


(4.16)
T0
 1 2 

 1 
M 
2
T


(4.17)
 2
T*
 1 2 
M 
 

T
 1
 1

(4.25)
Critical Conditions
 2
a*
 1 2 
M 
 

a
 1
 1

1
2
(4.26)

 2
  1 2   1
p*
 

M 
 1
p
 1


 2
*
  1 2   1
M 
 


 1
  1

(4.27)
(4.28)
Relationship Between Critical and Stagnation Conditions
T*
2

T0
 1
(4.29)
a*

a0
(4.30)
2
 1
68

 2   
p*
 

p0
  1
(4.31)

 2   
*
 

0
  1 
(4.32)
For the case of air flow, these equations give:
T*
 0.833 ,
T0
p*
 0.528 ,
p0
*
 0.634
0
Maximum Escape Velocity
Vˆ 
(V 2  2c pT ) 
2c pT0 
69
 2 2a 2 
V 
 
 1 

2a02
 1
(4.34)
PROBLEM 4.1
A gas with a molar mass of 4 and a specific heat ratio of 1.67 flows through a variable
area duct. At some point in the flow the velocity is 200 m/s and the temperature is 10° C.
Find the Mach number at this point in the flow. At some other point in the flow, the
temperature is -10° C. Find the velocity and Mach number at this point in the flow
assuming that the flow is isentropic.
SOLUTION
For the gas being considered:
R 

8314

 2078.5
m
4
Hence, the Mach number at the first section is given by:
M1 
V1

a1
V1

 RT1
200
200

 0.2018
991.1
1.67  2078.5  283
Now:
 1
1 (
) M 12
T2
2

 1
T1
1 (
) M 22
2
which can be arranged to give:

  1 2   T1  
) M 1    1 
 1  (
2
  T2  

M2  



1


(
)
2




i.e.:
70
0.5
0.5

  1

2   283 
 1  ( 2 )  0.218   263  1 


M2  
  0.480
1




(
)
2


But the speed of sound at the second section will be given by:
1/2
1/ 2
T 
 263 
a2  a1  2   991.1  
  955.4 m/s
 283 
 T1 
So:
V2  M 2 a2  0.480  955.4  458.6 m/s
Therefore the Mach number at the first point is 0.2018 and the Mach number and
velocity at the second point are 0.480 and 458.6 m/s respectively.
71
PROBLEM 4.2
Air flows through a convergent-divergent duct with an inlet area of 5 cm2 and an exit
area of 3.8 cm2. At the inlet section the air velocity is 100 m/s, the pressure is 680 kPa
and the temperature is 60° C. Find the mass flow rate through the nozzle and, assuming
isentropic flow, the pressure and velocity at the exit section.
SOLUTION
Consider the inlet section. The mass flow rate is given by:
 p 
 680  103 
4
m  1V1 A1   1  V1 A1  
  100 x  5  10   0.3558 kg/s
 287  333 
 RT1 
Therefore the mass flow rate through the nozzle is 0.3558 kg/s.
Now assuming that the flow is steady the mass flow rate will be the same at all
sections of the nozzle, i.e., considering the inlet and exit sections of the nozzle:
1 V1 A1   2 V2 A2
From which it follows that:
 2 V2 A2
1
1 V1 A1
i.e., since V = M a:
 2 M 2 a2
A
 1
A2
1 M 1 a1
 2 M 2  T2 
 
1 M 1  T1 
i.e.,
0.5

A1
A2
But:
 1 2
) M1
T2
2

 1 2
T1
1 (
)M2
2
1(
and
 2 1 

1 1 
Combining the above equations then gives:
72

1
2
1
2
(   1) M  1

(   1) M 
2
1
2
2
M2
M1
1 

1 
1
1
2
1
2
(   1) M 12   1


(   1) M 22 
A1
A2
But from the given information:
M1 
V1

a1
V1

 RT1
100
100

 0.2733
347.8
1.4  287  333
Therefore since γ = 1.4 it follows that:
3
M 2 1  0.2  0.27332 
5
i.e.,

 
2
0.2733  1  0.2  M 2 
3.8
2.9073  M 2
1  0.2  M 
2 3
2
1
The variation of the left hand side of this equation with M, is shown in Fig. P4.2.
Figure P4.2
73
It will be seen from this figure that the left hand side of the equation is 1, i.e., the
equation is satisfied, at two values of M2 these two values being equal to 0.3736 and
1.9964, i.e., one solution corresponds to subsonic flow and the other to supersonic flow.
Now as discussed above the adiabatic energy equation gives:
 1 2
1 (
) M1
T2
2

 1 2
T1
1 (
)M2
2
From which it follows that:
T2

333
   1 
2
1 
  0.2733
2


   1 
2
1 
  M2
 2 
Solving this equation for T2 then gives values of 328.7K and 188.1 K corresponding
to values of M2, of 0.3736 and 1.9964 respectively.
Since the flow is by assumption isentropic it follows that:

 T   1
p2
  2  , i.e.,
p1
 T1 
 T 
p2  680   2 
 333 
3.5
Solving this equation for p2 then gives values of 649.8kPa and 92.1kPa corresponding
to values of M2 of 0.3736 and 1.9964 respectively.
Also since:
V2  M 2 a2  M 2
 RT2  M 2 
1.4  287  T2
Solving this equation for V2 then gives values of 135.8 m/s and 548.8 m/s
corresponding to values of M2 of 0.3736 and 1.9964 respectively.
74
PROBLEM 4.3
The exhaust gases from a rocket engine can be assumed to behave as a perfect gas
with a specific heat ratio of 1.3 and a molecular weight of 32. The gas is expanded from
the combustion chamber through the nozzle. At a point in the nozzle where the crosssectional area is 0.2 m2, the pressure, temperature, and Mach number are 1500 kPa,
800°C, and 0.2, respectively. At some other point in the nozzle, the pressure is found to
be 80 kPa. Find the Mach number, temperature, and cross-sectional area at this point.
Assume a one-dimensional, isentropic flow.
SOLUTION
If subscripts 1 and 2 refer to the conditions at the two sections of the nozzle
considered then the isentropic relations give:

1  12 (   1) M 12  1
p2
 
2
1
p1
1  2 (   1) M 2 
hence:

1  0.15 x 0.22  0.3
80
 
2 
1500
 1  0.15M 2 
From which it follows that:
M 2  2.5543
Also because the flow is isentropic:
T2  p2 
 
T1  p1 
 1

0.3
,
T2
 80  
o
i.e.,

 , i.e., T2  545.5 K  272.5 C
1073  1500 
The continuity equation gives:
1 V1 A1   2 V2 A2
75
From which it follows that:
 2 V2
A
 1 , i.e.,
1 V1
A2
 2 M 2 a2
A
 1 , i.e.,
1 M 1 a1
A2
 2 M 2  T2 
 
1 M 1  T1 
0.5

A1
A2
But since the flow is isentropic:
1
 p 
p 
2
T
  2  , and , 2   2 
1
T1
 p1 
 p1 
 1

Using the above results gives:
M 2  p2 
 
M 1  p1 
(  1)  2
2
A
 1 ,
A2
i.e.,
2.5543  80 


0.2  1500 
0.912

0.2
A2
which gives:
A2  0.2094 m 2
Therefore the Mach number, temperature, and area at the second section of the nozzle
are 2.554 kPa, 272.5oC, and 0.2094 m2 respectively.
76
PROBLEM 4.4
The exhaust gases from a rocket engine have a molar mass of 14. They can be
assumed to behave as a perfect gas with a specific heat ratio of 1.25. These gases are
accelerated through a nozzle. At some point in the nozzle where the cross-sectional area
of the nozzle is 0.7 m2, the pressure is 1000 kPa, the temperature is 500° C and the
velocity is 100 m/s. Find the mass flow rate through the nozzle and the stagnation
pressure and temperature. Also find the highest velocity that could be generated by
expanding this flow. If the pressure at some other point in the nozzle is 100 kPa, find the
temperature and velocity at this point in the flow assuming the flow to be onedimensional and isentropic.
SOLUTION
The mass flow rate is given by:
 1000  103 
 p 
m  1V1 A1   1  V1 A1  
  100  0.7  152.5 kg/s
RT

8314
/14
773


 1


Hence the mass flow rate through the nozzle is 152.5 kg/s. Also, the Mach number at
the first section is given by:
M1 
V1

a1
V1

 RT1
100
1.25   8314 /14   773
 0.1320
Therefore the Mach number at the first section is 0.1320
Now for γ = 1.25 and M = 0.1320 the software gives:
p0
 1.011 ,
p1
T0
 1.002
T1
hence:
p0
 1.011 , i.e., p0  1011kPa and
1000
77
T0
 1.002 , i.e., T0  774.6 K
773
Therefore the stagnation pressure and temperature are 1011 kPa and 774.6 K
respectively.
Now:
Vˆ 
 2 R 

 T0 
  1 
2 c p T0 
 2  1.25  (8314 /14) 

  774.6  2144.8 m/s
0.25
Therefore the highest velocity that can be generated by expanding the flow is 2144.8
m/s.
Now:
p0
1011

 10.11
p2
100
For this pressure ratio, the software gives for γ = 1.25 gives:
T0
 1.937
T1
M 2  2.164,
hence:
774.6
 1.937 , i.e., T2  399.9 K
T2
and since:
M2 
V2

a2
V2

 RT2
V2
1.25   8314 /14   399.9
It follows that:
2.164 
V2
1.25   8314 /14   399.9
, i.e., V2  1179.0 m/s
Therefore the temperature and velocity at the second section are 399.9 K and 1179.0
m/s respectively.
78
PROBLEM 4.5
A gas has a molar mass of 44 and a specific heat ratio of 1.3. At a certain point in the
flow where the velocity is 100 m/s, the static pressure and temperature are 80 kPa and
15° C, respectively. The gas is then isentropically expanded until its velocity is 300 m/s.
Find the pressure, temperature and Mach number that exist in the resulting flow.
SOLUTION
Applying the energy equation between the two points in the flow gives:
V12  2 c p T1  V22  2 c p T2 , i.e., V12 
2 R
2 R
T1  V22 
T
 1
 1 2
Hence, using the values given:
1002 
2 x 1.3  (8314 / 44)
2  1.3  (8314 / 44)
 288  3002 
T2
0.3
0.3
From which it follows that:
T2  239.2 K  - 33.8 C
The isentropic relations give:
1
 T   1
p2
  2
,
p1
 T1 
i.e.,
 239.2 
p2  80  

 288 
3.5
 43.08 kPa
Lastly:
M2 
V2

a2
V2

 RT2
300
300

 1.238
242.4
1.3  (8314 / 44)  239.2
Therefore the temperature, pressure and Mach number at the second section are 239.2
K, 43.08 kPa and 1.238 respectively.
79
PROBLEM 4.6
Carbon dioxide flows through a variable area duct. At a certain point in the duct the
velocity is 200 m/s and the temperature is 60° C. At some other point in the duct the
temperature is 15° C. Find the Mach numbers and stagnation temperatures at the two
points. Assume that the flow is adiabatic.
SOLUTION
For carbon dioxide (see Table 3.1 in the textbook):
  1.3 , R 

8314

 188.9 J/kg K
m
44.01
Applying the energy equation between the two points in the flow gives:
V12  2 c p T1  V22  2 c p T2 ,
i.e., V12 
2 R
2 R
T1  V22 
T
 1
 1 2
Hence, using the values given:
2002 
2  1.3  188.9
2  1.3  188.9
 333  V22 
 288
0.3
0.3
From which it follows that:
V2  337.2 m/s
The Mach numbers at the two points in the flow are then given by:
M1 
V1

a1
V1

 RT1
200
200

 0.6994
286.0
1.3  188.9  333
and:
M2 
V2

a2
V2

 RT2
337.2
337.2

 1.268
265.9
1.3  188.9  288
80
Because the flow is adiabatic, the stagnation temperature is the same at the two points
considered and is given by:
 1 2 
 1 2 


T0  T1 1 
M 1   T2 1 
M2 
2
2




i.e.:
T0  333  1  0.15  0.69942   288  1  0.15  1.2682   357.4 K
Therefore the Mach numbers at the two points in the flow are 0.6994 and 1.268
respectively and the stagnation temperature is 357.4 K ( = 84.4° C) at both points.
81
PROBLEM 4.7
At a certain point in a gas flow, the velocity is 900 m/sec, the pressure is 150 kPa and
the temperature is 60o C. Find the stagnation pressure and temperature if the gas is air
and if it is carbon dioxide.
SOLUTION
Using:
M 
V

a
V
 RT
it follows that if the gas is air:
M 
900
900

 2.461
365.8
1.4  287  333
while if the gas is carbon dioxide for which γ = 1.3 and R = 188.9 J/kg K:
M 
900
900

 3.147
286.0
1.3  188.9  333
For a gas with γ = 1.4 (air) for a Mach number of 2.461 the software gives:
p0
T
 16.08 and 0  2.211
p
T
Hence:
p0  16.08  150  2412 kPa and T0  2.211  333  736.3 K
Similarly, for a gas with γ = 1.3 (carbon dioxide) for a Mach number of 3.147 the
software gives:
p0
T
 51.70 and 0  2.486
p
T
Hence:
p0  51.70  150  7755 kPa and T0  2.486  333  827.8 K
82
Therefore the stagnation pressure and temperature are 2412 kPa and 736.3 K ( =
463.3° C) respectively if the gas is air and 7755 kPa and 827.8 K ( = 554.8° C)
respectively if the gas is carbon dioxide.
83
PROBLEM 4.8
Helium at a pressure of 120 kPa and a temperature of 20° C flows at a velocity of 800
m/s. Find the Mach number, the stagnation temperature, and the stagnation pressure.
SOLUTION
For helium (see Table 3.1 in textbook):
  1.667 ,
R 

8314

 2076.9 J/kg K
m
4.003
Therefore:
M 
V

a
V

 RT
800
800

 0.7943
1007.2
1.667  2076.9  293
Now for γ = 1.667 and M = 0.7943 the software gives:
p0
 1.612 ,
p
T0
 1.210
T
hence:
p0
 1.612 ,
120
i.e., p0  193.4 kPa and
T0
 1.210 , i.e., T0  354.5 K
293
Therefore the stagnation pressure and temperature are 193.4 kPa and 354.5 K (=
81.5o C) respectively.
84
PROBLEM 4.9
In an argon flow the temperature is 40oC and the pressure is half the stagnation
pressure. Find the Mach number and the velocity in the flow.
SOLUTION
Here:
p0
1

 2
p
0.5
But for helium (see Table 3.1 in textbook):
  1.667 ,
R 

8314

 208.2 J/kg K
m
39.94
But for γ = 1.667 and p0/p = 2 the software gives M = 0.9790. Hence:
V  Ma  M
 RT  0.9790 
1.667  208.2  313  0.9790  329.6  322.6 m/s
Therefore the Mach number and velocity in the flow are 0.9790 and 322.6 m/s
respectively.
85
PROBLEM 4.10
If an aircraft is flying at a Mach number of 2.2 at an altitude of 10,000 m in the
standard atmosphere, find the stagnation pressure and temperature for the flow over the
aircraft.
SOLUTION
At an altitude of 10000 m in the standard atmosphere the pressure and temperature
are 26.48 kPa and 223.2 K respectively. But the software or the tables give for γ = 1.4 and
M = 2.2:
p0
 10.69 ,
p
T0
 1.968
T
hence:
p0
 10.69 , i.e., p0  283.1 kPa and
26.48
T0
 1.968 , i.e., T0  439.3 K
223.2
Therefore the stagnation pressure and temperature are 283.1 kPa and 439.3 K (=
166.3o C) respectively.
86
PROBLEM 4.11
If a gas is flowing at 300 m/s and has a pressure and temperature of 90 kPa and 20o C
find the maximum possible velocity that could be generated by expansion of this gas if
the gas is air and if it is helium.
SOLUTION
The maximum possible velocity that can be generated is given by:
Vˆ 
 2
2a 2 
V


 
 1

 2
2 RT 
V 

 1 

which for the situation here being considered becomes:
Vˆ 

2  R  293 
2
 300 

 1


For air. the above equation gives:
Vˆ 
2  1.4  287  293 

2
 300 
  823.8 m/s
  1


While, since for helium (see Table 3.1 in textbook):
  1.667 , R 

8314

 2076.9 J/kg K
m
4.003
this equation gives:
Vˆ 
2  1.667  2076.9  293 

2
 300 
  1769.7 m/s
  1


Therefore the maximum possible velocities that can be generated with air and helium
are 823.8 m/s and 1769.7 m/s respectively.
87
PROBLEM 4.12
A pitot-static tube is placed in a subsonic air flow. The static temperature and
pressure in the air flow are 30° C and 101 kPa, respectively. The difference between the
pitot and static pressures is measured using a manometer and is found to be 250 mm of
mercury. Find the air velocity assuming (i) the flow to be incompressible and (ii) taking
compressibility effects into account.
SOLUTION
In subsonic flow, the pitot pressure is equal to the stagnation pressure in the flow. The
pressure difference is found from the manometer reading. This gives:
p0  p   m  g  H
where ρm is the density of the liquid in the manometer which for mercury is
13580kg/m3 . Therefore:
p0  p  13580  9.81  0.25  33.31 kPa
But:
p0
p p
33.31
hence
1  0

p
p
101
p0
 1.330
p
If compressibility effects are ignored Bernoulli's equation gives:
V 
2 ( p0  p )

Therefore since:
 
p
101000

 1.161 kg/m3
287  303
RT
the velocity if compressibility effects are ignored is given by:
88
V 
2 ( p0  p)


2  33310
 239.5 m/s
1.161
Therefore if compressibility effects are ignored, the velocity is found to be 239.5 m/s.
Turning next to the case where compressibility effects are accounted for. For air for
this value of p0 / p = 1.330, the software gives M = 0.6515. Therefore, since the
temperature of the air is 303 K, the velocity is given by:
V  M a  M  RT  0.6515 
1.4  287  303  227.3 m/s
Therefore if compressibility effects are accounted for, the velocity is found to be
227.3 m/s.
89
PROBLEM 4.13
A pitot-static tube is placed in a subsonic air flow. The static pressure and
temperature are 101 kPa and 30o C respectively. The difference between the pitot and
static pressures is measured and found to be 37 kPa. Find the air velocity.
SOLUTION
In subsonic flow, the pitot pressure is equal to the stagnation pressure in the flow.
Now:
p0
p p
37
1  0

hence
p
p
101
p0
 1.366
p
For air for this value of p0/p , the software gives M = 0.6826. Therefore, since the
temperature is equal to 303 K, the velocity can be found using:
V  Ma  M
 RT  0.6826 
Therefore the velocity is 238.2 m/s.
90
1.4  287  303  238.2 m/s
PROBLEM 4.14
A pitot tube placed in an air stream indicates a pressure of 186 kPa. If the local Mach
number is 0.8 determine the static pressure in the flow.
SOLUTION
In subsonic flow the pitot pressure is equal to the stagnation pressure in the flow.
Now for air flow at M = 0.8 the software gives p0 / p = 1.524. Hence:
p 
p0
186

 122.1 kPa
p0 / p
1.524
Therefore the pressure in the flow is 122.1 kPa.
91
PROBLEM 4.15
A pitot tube indicates a pressure of 155 kPa when placed in an air-stream in which the
temperature is 15oC and the Mach number is 0.7. Find the static pressure in the flow.
Also find the stagnation temperature in the flow.
SOLUTION
In subsonic flow the pitot pressure is equal to the stagnation pressure in the flow.
Now for air flowing at M = 0.7 the software gives p0/p = 1.387 and T0 / T = 1.098. Hence:
p
p0
155

 111.8 kPa
p0 / p 1.387
and:
T0 
T0
T  1.098  288  316.2 K
T
Therefore in the flow the pressure is 111.8 kPa and the stagnation temperature is
313.2 K ( = 43.2o C ).
92
PROBLEM 4.16
A pitot tube is placed in a stream of carbon dioxide in which the pressure is 60 kPa
and the Mach number is 0.9. What will the pitot pressure be?
SOLUTION
In subsonic flow, the pitot pressure is equal to the stagnation pressure in the flow.
Now for carbon dioxide (see Table 3.1 in textbook) γ = 1.3. For γ = 1.3 and M = 0.9, the
software gives p0 / p = 1.644. Hence:
p0  p  1.644  60  1.644  98.64 kPa
Therefore the pitot pressure is 98.64 kPa.
93
PROBLEM 4.17
Consider one-dimensional isentropic air flow through a duct. At a certain section of
this duct, the velocity is 360 m/s, the temperature is 45° C and the pressure is 120 kPa.
Find the Mach number and the stagnation temperature and pressure at this point in the
flow. If the temperature at some other point in the flow is 90° C, find the Mach number
and pressure at this point in the flow.
SOLUTION
Consider the first point. Using:
M 
V

a
V
 RT
it follows that:
M1 
360
360

 1.007
357.5
1.4  287  318
For air for M = 1.007, the software gives:
p0
 1.908 ,
p
T0
 1.283
T
Hence:
p0
 1.908 , i.e., p0  229.0 kPa and
120
T0
 1.283 , i.e., T0  408.0 K
318
Assuming that the flow is isentropic, the stagnation temperature and pressure are the
same everywhere. Therefore, at the second point considered:
T0
408.0
408.0


 1.124
T
T2
363
For air for T0 / T = 1.124, the software gives:
94
p0
 1.506 ,
p
M  0.7874
Hence:
p2 
p0
229.0

 152.1 kPa
p0 / p2
1.506
and:
M 2  0.7874
Therefore the Mach number at the first point is 1.007 while the stagnation
temperature and pressure are 408.0 K ( = 1350o C ) and 229 kPa respectively. At the
second point in the flow, the Mach number and pressure are 0.7874 and 152.1 kPa
respectively.
95
PROBLEM 4.18
A liquid fuelled rocket is fired on a test stand. The rocket nozzle has an exit diameter
of 30 cm and the combustion gases leave the nozzle at a velocity of 3800 m/s and a
pressure of 100 kPa which is the same as the ambient pressure. The temperature of the
gases in the combustion area is 2400° C. Find the temperature of the gases on the nozzle
exit plane, the pressure in the combustion area and the thrust developed. Assume that the
gases have a specific heat ratio of 1.3 and a molar mass of 9. Assume that the flow in the
nozzle is isentropic.
SOLUTION
Because the velocity in the combustion area is effectively zero, applying the energy
equation between the combustion area and the nozzle exit plane gives:
2
Vexit
 c p Texit
c p T0 
2
2
Vexit
R
R
, i.e.,

T0 
T
2
 1
 1 exit
But:
R 

8314

 923.8 J/kg K
m
9
Hence, using the values given, the energy equation gives:
1.3 x 923.8
38002
1.3 x 923.8
Texit
 (2400  273) 

1.3  1
2
1.3 1
From which it follows that Texit = 869.4 K
Because the flow is isentropic:
1
 T   1
pexit
  exit 
p0
 T0 
1
,
100
 869.4  1.31
 
i.e.,

p0
 2673 
From which it follows that p0 = 4226 kPa.
96
Because the pressure on the exit plane is equal to the ambient pressure, the thrust is
given by:
Thrust  m Vexit  ( exit Vexit Aexit ) Vexit
But the density on the nozzle exit plane is given by:
exit 
pexit
100000

 0.1245 kg/m3
923.8  869.4
R Texit
and the nozzle exit area is given by:
Aexit 

4
2
Dexit


4
 0.32  0.07068 m 2
Hence:
Thrust  ( exit Vexit Aexit ) Vexit  (0.1245  3800  0.07068)  3800  127067 N
Therefore the temperature of the gases on the exit plane is 869.4 K ( = 596.4° C ), the
pressure in the combustion area is 4226 kPa, and the thrust is 127.07kN.
97
PROBLEM 4.19
The pressure, temperature and Mach number at the entrance to a duct through which
air is flowing are 250 kPa, 26°C and 1.4 respectively. At some other point in the duct the
Mach number is found to be 2.5. Assuming isentropic flow, find the temperature,
velocity, and pressure at the second section. Also find the mass flow rate per m2 of area at
the second section.
SOLUTION
Consider the conditions at the entrance to the duct. For air for M = 1.4 the software or
isentropic tables give:
p0
 3.182 ,
p
T0
 1.392
T
Hence:
p0
 3.182 , i.e., p0  795.5 kPa and
250
T0
 1.392 , i.e., T0  416.2 K
392
Next consider the second point. For air for M = 2.5 the software or isentropic tables
give:
p0
 17.09 ,
p
T0
 2.250
T
Also, because the flow can be assumed to be isentropic the stagnation temperature
and pressure are the same everywhere. Therefore, at the second point:
p2 
p0
795.5

 46.52 kPa
17.09
17.09
and T2 
T0
416.2

 185.0 K
2.250
2.250
The velocity at the second point is given by:
V2  M a  M 2
 RT2  2.5 
1.4  287  185  681.6 m/s
The mass flow rate, which is the same at all sections, is given by:
98
 p 
m   2 V2 A2   2  V2 A2
 RT2 
Hence:
 46520 
m
2
 
  681.6  597.2 kg/s per m

A2
287
185


Therefore the temperature, velocity, and pressure on the second section are 185.0 K
( = - 88° C ), 681.6 m/s and 46.52 kPa respectively and the mass flow rate per unit area at
this section is 597.2 kg / s per m2.
99
PROBLEM 4.20
An aircraft is flying at a Mach number of 0.95 at an altitude where the pressure is 30
kPa and the temperature is -50°C. The diffuser at the intake to the engine decreases the
Mach number to 0.3 at the inlet to the engine. Find the pressure and temperature at the
inlet to the engine.
SOLUTION
Now for air flow at M = 0.95, the software or isentropic tables give:
p0
 1.787 ,
p
T0
 1.181
T
while for air flow at M = 0.3, the software or isentropic tables give:
p0
T
 1.064 , 0  1.018
p
T
If the deceleration is assumed to be isentropic, the stagnation pressure and
temperature will be the same at the two sections considered and therefore:
p2 
p0 / p1
1.787
p1 
 30  50.39 kPa
p0 / p2
1.064
T2 
T0 / T1
1.181
T1 
 223  258.7 K
T0 / T2
1.018
and in the same way:
Therefore the pressure and temperature at the inlet to the engine are 50.39 kPa and
258.7 K (= - 14.30° C ) respectively.
100
PROBLEM 4.21
A conical diffuser has an inlet diameter of 15 cm. The pressure, temperature and
velocity at the inlet to the diffuser are 70 kPa, 60oC and 180 m/s respectively. If the
pressure at the diffuser exit is 78 kPa, find the exit diameter of the diffuser.
SOLUTION
Consider the inlet. Using:
M 
V

a
V
 RT
gives:
M1 
180
180

 0.4921
365.8
1.4  287  333
For air for M = 0.4921, the software or isentropic tables give:
p0
 1.180 ,
p
T0
 1.048 ,
T
0
T
 1.126
Hence, assuming that isentropic flow can be assumed, on the exit plane of the diffuser:
p0
p p
 0 1
p2
p1 p2
from which it follows that:
p0
70
 1.180 
 1.059
p2
78
Now for air for p0 / p = 1.059, the software or isentropic tables give:
M  0.2873 ,
T0
 1.017 ,
T
The continuity equation then gives:
1 V1 A1   2 V2 A2
101
0
 1.042

i.e.:
1
a

a
M 1 1 A1  2 M 2 2 A2
0
0
a0
a0
i.e.:
0.5
0.5
T 
 T  a2
1

M 1  1  A1  2 M 2  2 
A2
0
0
 T0 
 T0  a0
i.e.:
 1 
1.126  0.4921  

 1.048 
0.5
 1 
A1  1.042  0.2873  

 1.017 
From which it follows that:
d 
A2
 1.8789 , i.e.,  2 
A1
 d1 
2
 1.8789
Hence since d1 = 15 cm:
d2
 1.87890.5 , i.e., d 2  20.56 cm
15
Therefore the diameter of the exit of the diffuser is 20.56 cm.
102
0.5
A2
PROBLEM 4.22
The control system for some smaller space vehicles uses nitrogen from a highpressure bottle. When a vehicle has to be maneuvered, a valve is opened allowing
nitrogen to flow out through a nozzle thus generating a thrust in the direction required to
maneuver the vehicle. In a typical system the pressure and temperature in the system
ahead of the nozzle are about 1.6 MPa and 30o C respectively while the pressure in the jet
on the nozzle exit plane is about 6 kPa. Assuming that the flow through the nozzle is
isentropic and the gas velocity ahead of the nozzle is negligible, find the temperature and
the velocity of the nitrogen on the nozzle exit plane. If the thrust required to maneuver the
vehicle is 1 kN, find the area of the nozzle exit plane and the required mass flow rate of
nitrogen. It can be assumed that the vehicle is effectively operating in a vacuum.
SOLUTION
Because the velocity ahead of the nozzle is effectively zero and because the flow can
be assumed to be isentropic, the stagnation temperature and pressure are 1.6 MPa and 303
K respectively. Hence, on the nozzle exit plane:
p0
1600

 266.67
p1
6
But for nitrogen:
  1.4 ,
R

8314

 296.9 J/kg K
m
28
and for γ = 1.4 and po / p = 266.67, the software gives M = 4.435 and To / T = 4.933.
Hence on the nozzle exit plane:
T 
T0
303

 61.42 K
4.933
4.933
The velocity on the nozzle exit plane is then given by:
103
V Ma  M
 RT  4.435 
1.4  296.9  61.42  4.435  159.8  708.6 m/s
The density on the nozzle exit plane is given by:
 
p
6000

 0.329 kg/m3
RT
296.9  61.42
Because the ambient pressure is zero, the thrust is given by:
  pA 
Thrust  mV
  V AV
 pA
where A is the area of the nozzle exit plane. Hence:
1000  (0.329  708.6  A)  708.6  6000  A  165196  6000   A
which gives:
A  0.00584 m 2
The mass flow rate through the nozzle is then given by:
m   V A  0.329  708.6  0.00584  1.362 kg / s
Therefore the Mach number and velocity on the nozzle exit plane are 4.435 and 708.6
m/s respectively, the nozzle exit plane area is 0.00584 m2 and the mass flow rate through
the nozzle is 0.00584 kg/s.
104
PROBLEM 4.23
Hydrogen enters a nozzle with a very low velocity and at a temperature and pressure
of 3800° R and 1000 psia respectively. The pressure on the exit plane of the nozzle is 2
psia. Calculate the hydrogen flow rate per unit nozzle exit area. The flow through the
nozzle can be assumed to be isentropic.
SOLUTION
Because the velocity at the nozzle entrance is low, the temperature and pressure on
the inlet plane can be taken as the stagnation temperature and pressure. Therefore:
p0  1000 psia , T0  3800 o R
From this it follows that, since the flow is isentropic, on the nozzle exit plane:
p0
1000

 500
p
2
Now, for hydrogen (see Table 3.1 in textbook):
  1.407 , R 

1545.3

 766.52 ft-lbf/lbm-o R
m
2.016
For γ = 1.4 and po / p = 500, the software gives M = 4.974 and To / T = 6.036. Hence
on the nozzle exit plane:
V  Ma  M
 RT  4.974 
1.407  766.52  32.2  629.6
 4.974  4675.9  23258 ft/sec
The density on the nozzle exit plane is given by:
 
p
144  2

 0.000597 lbm/ft 3
RT
766.52  629.6
The mass flow rate through the nozzle is then given by:
105
m   V A
Therefore, considering the exit plane:
m / A   V  0.000597  23258  13.89 lbm / sec-ft 2
Hence the mass flow rate through the nozzle per unit area of the exit plane is 13.89
lbm / sec-ft 2 .
106
PROBLEM 4.24
An aircraft flies at sea-level at speed of 220 m/s. What is the highest pressure that can
be acting on the surface of the aircraft?
SOLUTION
The temperature at sea-level in the standard atmosphere is 288.16 K. Hence, the
aircraft is flying at a Mach number that is given by:
M 
V

a
V

 RT
220
220

 0.6466
340.3
1.4  287  288.16
The aircraft is, therefore, flying at a subsonic velocity. As a result, the highest
pressure acting on the aircraft will be equal to the stagnation pressure in the freestream
flow ahead of the aircraft. Now for air for M = 0.6466, the software or isentropic tables
give p0 / p = 1.325. Hence, since the pressure at sea-level in the standard atmosphere is
101.33 kPa:
p0  p1  1.325  101.33  1.325  134.3 kPa
Therefore the highest pressure that can be acting on the aircraft is 134.3 kPa.
107
PROBLEM 4.25
Consider an air flow with a speed of 650 m/s, a pressure of 100 kPa, and a
temperature of 20o C. What is the stagnation pressure and the stagnation temperature in
the flow?
SOLUTION
Since air flow is involved the Mach number in the flow is obtained using:
M 
V

a
V

 RT
650
650

 1.894
343.1
1.4  287  293
For air for M = 1.894, the software gives Now for air for M = 0.6466, the software or
isentropic tables give p0 / p = 6.639 and T0 / T = 1.717. Hence:
p0  p1  6.639  100  6.639  663.9 kPa
and:
T0  T1  1.717  293  1.717  503.1 K
Therefore the stagnation pressure and temperature in the flow are 663.9 kPa and
503.1 K ( = 230.1o C ) respectively.
108
PROBLEM 4.26
When an aircraft is flying at subsonic velocity, the pressure at its nose, i.e. at the
stagnation point, is found to be 160 kPa. If the ambient pressure and temperature are 100
kPa and 25o C respectively, find the speed and the Mach number at which the aircraft is
flying.
SOLUTION
Here:
p0
160

 1.6
p
100
For air for p0 / p = 1.6 the software or isentropic tables give M = 0.8477. Hence:
V  Ma  M  RT  0.8477 
1.4  287  293  0.8477  346.0  293.3 m/s
Therefore the aircraft is flying at a Mach number of 0.8477 and a speed of 293.3 m/s.
109
PROBLEM 4.27
A body moves through air at a velocity of 200 m/s. The pressure and temperature in
the air upstream of the body are 100 kPa and 30o C respectively. Find the pressure at a
point on the body where the velocity of the air relative to the body is zero (a) accounting
for compressibility and (b) assuming incompressible flow. Assume that the flow is
isentropic.
SOLUTION
Consider the flow ahead of the body. Using the information provided:
M 
V

a
V

 RT
200
200

 0.5732
348.9
1.4  287  303
Now for air for M = 0.5732, the software or isentropic tables give p0 / p = 1.250.
Hence:
p0  p1  1.250  100  1.250  125 kPa
Next consider the result that would be obtained by assuming incompressible flow.
The density in the flow is given by:
 
p
100000

 1.150 kg/m3
RT
287  303
Bernoulli's equation gives:
p0  p 
V 2
2
1.15  2002
 100000 
 123000 Pa  123 kPa
2
Therefore the pressure at the stagnation point is 125 kPa but it would be estimated to
be 123 kPa if incompressible flow was assumed.
110
PROBLEM 4.28
Air enters a duct at a pressure of 30 psia, a temperature of 100o F and a velocity of
580 ft/sec. At some other point in the duct the pressure is found to be 12 psia. Assuming
that the flow is isentropic, find the temperature and Mach number at this point in the
flow.
SOLUTION
Consider the first point. Since for air R = 53.3 ft-lbf/lbm-oR and T1 = 560o R and
recalling that 1 lbf = 32.2lbm-ft/sec2:
M1 
V1

a1
V1

 RT1
580
580

 0.5
1160
1.4  53.3  32.2  560
For air for M = 0.5, the software or isentropic tables give p0 / p = 1.186 and T0 / T =
1.050. Hence:
p0  p1  1.186  30  1.186  35.6 psia
and:
T0  T1  1.050  560  1.050  588 o R
Since the flow can be assumed to be isentropic, the stagnation temperature and
pressure are the same everywhere. Therefore at the second point:
p0
35.6

 2.965
p2
12
For air for p0 / p = 2.965 the software or isentropic tables give M = 1.349 and To/T =
1.364. Hence:
T2 
T0
588

 431.1 o R
1.364
1.364
111
Therefore the Mach number at the second point is 1.349 and the temperature at this
point is 431.1° R (= -28.9 °F).
112
PROBLEM 4.29
Consider a rocket engine that burns hydrogen and oxygen. The combustion chamber
temperature and pressure are 3800 K and 1.5 MPa respectively, the velocity in the
combustion chamber being very low. The pressure on the nozzle exit plane is 1.5 kPa.
Assuming that the flow is isentropic, find the Mach number and the velocity on the exit
plane. Assume that the products of combustion behave as a perfect gas with γ = 1.22 and
R = 519.6 J/kg K.
SOLUTION
The pressure and temperature in the combustion chamber are assumed to be equal to
the stagnation pressure and temperature because the velocity in the combustion chamber
is low. Therefore, considering conditions on the exit plane:
p0
1500

 1000
p
1.5
For γ = 1.22 and p0 / p = 1000.0, the software gives M = 4.744 and T0 / T = 3.475.
Hence:
V  M a  M  RT  4.744 
1.22  519.6  1093.5  4.744  832.6  3949.8 m/s
Therefore the Mach number and velocity on the nozzle exit plane are 4.744 and
3949.8 m/s respectively.
113
PROBLEM 4.30
At a point in a supersonic air flow, the pressure and temperature are 5 kPa and -80° C.
If the stagnation pressure at this point is 100 kPa, find the Mach number and the
stagnation temperature.
SOLUTION
Here:
p0
100

 20.0
p
5
Now for air for p0 / p = 20.0, the software or isentropic tables give M = 2.601 and T0 /
T = 2.354. Hence:
T0  T1  2.354  (273  80)  2.354  454.3 K
Therefore the Mach number and temperature are 2.601 and 454.3 K ( = 181.3° C )
respectively.
114
PROBLEM 4.31
Air flows through a circular pipe which has a diameter of 45cm at a Mach number of 0.3. The
stagnation temperature and stagnation pressure are 500K and 250kPa respectively. Calculate the
air mass flow rate through the pipe.
SOLUTION
The mass flow rate is given by:
m   V A
But:
p0
350000

 2.439 kg/m3
RT0
287  500
0 
and:
1
o
 1 2   1

,
M 
 1 
2



i.e.,  
o
1
 1 2   1

M 
1 
2


i.e.,:
 
o
 1 2 

M 
1 
2


1
 1

2.439
1  0.2  0.3 
2
1
0.4
 2.333 kg/m3
Also:
V  Ma  M
 RT  0.3  1.4  287  500  134.5 m/s
and:
A

4
D2 

4
 0.452  0.159 m 2
115
Hence:
m   V A  2.333  134.5  0.159  49.9 kg/s
Therefore the mass flow rate through the pipe is 49.9 kg/s.
116
PROBLEM 4.32
An aircraft is flying at an altitude of 12,000 m, the atmospheric air pressure at this
altitude being 19.39 kPa. The internal volume of the aircraft is 860 m3. If the 12-cm
diameter window in an exit door failed and blew out how long would it take for the
pressure inside the aircraft take to drop from its initial value of 101 kPa to a value that is
equal to 40% of this initial value?. Assume that the hole in the door acts as a converging
nozzle and that the temperature inside the aircraft remains constant at 20°C.
SOLUTION
The time taken for the pressure in the aircraft to drop from 101 kPa to 0.4 x 101 =40.4
kPa is required. At the beginning of the process the ratio of the external pressure to the
internal pressure is 19.39 / 101 = 0.192 while at the end of the process the ratio is 19.39 /
40.4 = 0.48. Now for choking to occur, i.e., for the discharge Mach number to be 1 it is
necessary in the case of air for the ratio of the exit plane pressure to the pressure of the air
in the aircraft to be less than or equal to 0.528. Therefore at all times during the process
being considered the flow is choked.
Energy conservation requires that for the process being considered the rate of
decrease of the energy of the air in the fuselage is equal to the rate of enthalpy flux from
the fuselage, i.e., since the process undergone by the air in the fuselage is assumed to be
isothermal, energy conservation requires that:
d
 M cv T    m c p T
dt
where M is the mass of the air in the fuselage. Since the temperature of the air in the
fuselage is assumed to remain constant and since the flow is choked at the discharge
plane with the result that the temperature of the air on the discharge plane will be equal to
0.833 times the temperature of the air in the fuselage the temperature on the discharge
plane also remains constant. Therefore the above equation, in the situation being
considered, can be written as:
117
cv T
dM
  m c p T
dt
First consider the left hand side of this equation. It can be written as:
cv T
dM
d  p 
 cv T
V

dt
d t  RT 
where V is the volume of the fuselage which is a constant. Since T remains constant this
can be written as:
cv
V dp
860 d p
dp
 860 
 2151.5
R dt
dt
0.287 d t
Next consider the right hand side of the energy equation.
 p
m c p T   V Ac p T   M a Ac p T  
 RT

 M  RT Ac p T

Since the flow is choked on the exit plane, on this plane M =1 so the above equation
gives:
 p
m c p T  
 RT

 M  RT Ac p T



0.528 p

 1
 287  0.833 x 293 
1.4  287  0.833  293 

4
 0.122  1007  293
 7.867 p
p being the pressure in the fuselage.
Equating the expressions derived above for the left and right hand sides of the energy
conservation equation gives:
2151.5
dp
  7.867 p , i.e.,
dt
118
273.5
dp
 p
dt
Integrating this equation from a pressure p1 at time 0 to a pressure p2 at a later time t
gives:
273.5 ln
p2
 t
p1
Therefore the time taken for the pressure to drop from 101kPa to 40.4 kPa is given by:
t   273.5 ln
p2
40.4
  273.5 ln
 250.6 s  4.18 m
p1
101
Hence it takes 4.18 minutes for the pressure in the fuselage to drop from 101 kPa to
40.4 kPa.
119
Chapter Five
NORMAL SHOCK WAVES
SUMMARY OF MAJOR EQUATIONS
Rankine-Hugoniot Relations
   1   2

 1


p2
   1  1

 
p1
   1   2 

 
   1  1 
   1  p2

 1


2
   1  p1

 
1
   1 
p 
 2


p1 
   1 
    1  p2

 1


V1
   1  p1

 
V2
   1 

p
 2


p1 
   1 
 

T2

 
T1
 

 
p 
1
 2

1 
p1 
p 
1
 1

1 
p2 
120
(5.13)
(5.14)
(5.16)
(5.19)
Entropy Change Across Normal Shock




p
 1 
2
1 




(

1)
(

1)
 
 p   1 
s2  s1
p1
 
 ln  2  
R
 p1   (  1)  (  1) p2  

p1  



 p02 

 p01 
s02  s01   R ln 
(5.22)
(5.26)
Normal Shock Relations In Terms of Mach Number
M 22
 2  2 
M1  

 (  1) M 12  2 
   1 


 

2 M 12  (  1) 
 2  2 


 M 1  1
   1 

(5.38)
2
 a2 
[2 M 12  (  1)][2  (  1) M 12 ] 
T2




 
T1
(  1) 2 M 12
 a1 


(5.39)
p2
2 M 12  (  1)

p1
 
(5.42)
2
(  1) M 12

1
2  (  1) M 12
(5.43)
121
 / ( 1)
p02
p01


2
 (  1)

M1
 

 2 1  (  1) M 12  
2

 

 2  2    1  

 M1  





   1 


1


 1  (  1) M 2   1 

s2  s1
  2
2
1
( M 1  1)  1 
 ln  

2
R
  2  (  1) M 1  
   1


/ (  1)
(5.45)
(5.47)
Rayleigh Supersonic Pitot Tube Equation

p02

p1
[(  1) M /2]
2
1
(  1)
 2 M     1  




1


    1 

2
1
1
(  1)
(5.57)
Moving Shock Wave Relations
V1  U s , V2  U s  V
M1 
M2 
(5.58)
(5.59)
Us
 Ms
a1
Us V
U a V
a

 s 1
 M s 1  M 2
a2 a2
a1 a2 a2
a2
Ms 
Us
a1
and M 2 
122
V
a2
(5.60)
(5.61)
2 M s2  (  1)
p2

p1
(  1)
(5.62)
(  1) M s2
2

1
2  (  1) M s2
(5.63)
 a2 
[2  (  1) M s2 ][2 M s2  (  1)]
T2
   
T1
(  1) 2 M s2
 a1 
(5.64)
2( M s2  1)
[2 M s2  (  1)] 0.5 [2  (  1) M s2 ] 0.5
(5.65)
2
M 2 
V  M 2 a2 
123
2( M s2  1)
a1
(  1) M s
(5.67)
PROBLEM 5.1
A normal shock wave occurs in an air flow at a point where the velocity is 680 m/s
the static pressure is 80 kPa and the static temperature is 60 °C. Find the velocity, static
pressure, and static temperature downstream of the shock. Also find the stagnation
temperature and stagnation pressure upstream and downstream of the shock.
SOLUTION
Subscripts 1 and 2 will be used to denote conditions upstream and downstream of the
wave. Because:
a1   RT1 
1.4  287  333  365.8 m/s
it follows that:
M1 
V1
680

 1.859
a1
365.8
Now for M1 = 1.859, since air flow is involved, the software for a normal shock wave
or the normal shock tables gives for M2 = 0.6038, p2 / p1 = 3.865, T2 / T1 = 1.576, and p02
/ p01 = 0.7861. Hence:
p2  p1
p2
 80  3.865  309.2 kPa
p1
T2  T1
T2
 333  1.579  524.8 K
T1
and:
Using this value of T2 then gives:
a2 
 RT2 
1.4  287  524.8  459.2 m/s
from which it follows that:
V2  M 2 a2  0.6038  459.2  277.3 m/s
124
Now again considering the flow ahead of the shock wave for M1 = 1.859, the software
or isentropic flow tables give for isentropic flow p0 / p = 6.290, and T0 / T = 1.691.
Hence:
p10  p1
p10
 80  6.290  503.2 kPa
p1
T10  T1
T10
 333  1.691  563.1 K
T1
and:
Therefore:
p20  p10
p20
 503.2  0.7861  395.6 kPa
p10
The flow through the shock wave is adiabatic so the stagnation temperature does not
change across the shock wave so:
T20  T10 = 563.1 K
Therefore the velocity, static pressure and static temperature downstream of the shock
wave are 277.3 m/s, 309.2 kPa, and 524.8 K (= 251 .8 °C) respectively. The stagnation
temperature both before and after the shock wave is 563.1 K (= 290.1 oC) while the
stagnation pressures upstream and downstream of the shock wave are 503.2 kPa and
395.6 kPa respectively.
125
PROBLEM 5.2
The pressure ratio across a normal shock wave that occurs in air is 1.25. Ahead of the
shock wave, the pressure is 100 kPa and the temperature is 15°C. Find the velocity,
pressure, and temperature of the air behind the shock wave.
SOLUTION
Now for a normal shock in air with, p2 / p1 = 1.25, the software for normal shock
waves or the normal shock wave tables give M2 = 0.911, and T2 / T1 = 1.066. Hence:
p2  p1
p2
 100  1.25  125 kPa
p1
and:
T2  T1
T2
 288  1.066  306.9 K
T1
Using this value of T2 then gives:
a2 
 RT2 
1.4  287  524.8  459.2 m/s
from which it follows that:
V2  M 2 a2  0.911  351.2  319.9 m/s
Therefore the velocity, pressure, and temperature downstream of the shock wave are
319.9 m/s, 125 kPa and, 306.9 K (= 33.9°C) respectively.
126
PROBLEM 5.3
A perfect gas flows through a stationary normal shock. The gas velocity decreases
from 480 m/s to 160 m/s through the shock. If the pressure and the density upstream of
the shock are 62 kPa and 1.5 kg/m3, find the pressure and density downstream of the
shock and the specific heat ratio of the gas.
SOLUTION
Across the shock wave, conservation of mass gives:
 2 V2  1 V1
which gives:
 2  1
V1
V2
 1.5 
480
 4.5 kg/m3
160
Also, conservation of momentum applied across the shock wave gives:
p1  p2  1 V1 V2  V1 
i.e.:
62000  p2  1.5  480  160  480  , i.e.,
p2  292400 Pa  293.4 kPa
In order to find the specific heat ratio, γ , it is recalled that:
p2
2  M 12  (   1)

p1
(   1)
and:
2
(  1) M 12

1
2  (  1) M 12
Using the density and pressure values derived above, these equations give:
2  M 12  (  1)
292.4

 4.7161
(  1)
62
and:
127
(  1) M 12
4.5

 3
2
2  (  1) M 1
1.5
These two equations give:
M 12 
4.7161(  1)  ( 1)
2
and M 12 
3
2 
These equations together then give:
4.7161(  1)  ( 1)
3

2
2 
, i.e.,  5.7161 2  1.7161  7.4322  0
Solving this equation gives γ = 1.3.
Therefore the pressure and density downstream of the shock wave are 292.4 kPa and
4.5 kg/m3 respectively and the specific heat ratio of the gas is 1.3.
128
PROBLEM 5.4
A normal shock wave occurs in air at a point where the velocity is 600 m/s and the
stagnation temperature and pressure are 200°C and 600 kPa respectively. Find the Mach
numbers, pressures, and temperatures upstream and downstream of the shock wave.
SOLUTION
The energy equation gives for the flow upstream of the shock wave:
c p T0  c p T1 
V22
2
Therefore, since for air cp = γ R / (γ - 1) = 1007 kJ/kg K, it follows that:
1007  (200  273)  1007  T1 
6002
2
which gives T1 = 294.2 K. Hence:
M1 
V1

a1
V1

 RT1
600
600

 1.746
343.8
1.4  287  294.2
Now for M1 = 1.746 the software or the isentropic tables give since γ = 1.4 for
isentropic flow ρ0 / ρ = 5.292. Therefore:
p1  p0
p1
600

 113.4 kPa
p0
5.292
Next consider the flow through the normal shock wave. Now for M1 = 1.746, the
software or tables for a normal shock wave gives for air M2 = 0.6291, p2 / p1 = 3.390,
and T2 / T1 = 1.492. Hence:
p2  p1
p2
 113.4  3.390  38434 kPa
p1
and:
129
T2  T1
T2
 293.9  1.492  438.5 K
T1
Therefore the Mach number, velocity, pressure and temperature upstream of the
shock wave are 1.746, 113.4 kPa, and 293.9 K (20.93°C) respectively while downstream
of the shock wave the values of these quantities are 0.6291, 384.4 kPa, and 438.5 K
(165.5°C) respectively.
130
PROBLEM 5.5
Show that the downstream Mach number of a normal shock approaches a minimum
value as the upstream Mach number increases towards infinity. What is this minimum
Mach number for a gas with a specific heat ratio of 1.67?
SOLUTION
For a normal shock wave:
M 22 
(  1) M 12  2
2 M 12  ( 1)
If the numerator and denominator on the right hand side are both divided by M1 2 the
following is obtained:
2
M 12

( 1)
2 
M 12
(  1) 
M 22
As M1 approaches infinity this equation shows that M2 approaches a finite value that
is given by:
M 22 
(  1)
2
, i.e., M 2 
( 1)
2
For γ = 1.67 this gives the limiting downstream Mach number as:
M2 
(1.67  1)
 0.4479
2  1.67
Therefore as the Mach number upstream of a normal shock wave approaches infinity,
the downstream Mach number approaches the value [(γ - 1) / 2 γ ] 0.5. For γ = 1.67 this
limiting Mach number has a value of 0.4479. In reality, if the upstream Mach number is
very high the temperature attained behind the shock may be so high that the assumptions
on which the above equations are based may not be valid.
131
PROBLEM 5.6
Air is expanded isentropically from a reservoir in which the pressure is 1000 kPa to a
pressure of 150 kPa. A normal shock occurs at this point in the flow. Find the pressure
downstream of the shock wave.
SOLUTION
The flow up to the shock wave can be assumed to be isentropic. It follows that p01 =
1000 kPa. Therefore:
p01
1000

 6.6667
p1
150
Now for po / p = 6.6667 the software or the isentropic flow tables give for air (γ = 1.4)
give M = 1.897. Hence, the Mach number ahead of the shock wave is 1.897. Now for M1
= 1.897 the software or the isentropic flow tables give for air (γ = 1.4) p2 / p1 = 4.032.
Hence:
p2  p1
p2
 150  4.032  604.8 kPa
p1
Therefore the pressure downstream of the shock wave is 604.8 kPa.
132
PROBLEM 5.7
Air is expanded isentropically from a reservoir in which the pressure and temperature
are 150 psia and 60°F to a static pressure of 20 psia. A normal shock occurs at this point
in the flow. Find the static pressure, static temperature and the air velocity behind the
shock.
SOLUTION
The flow up to the shock wave can be assumed to be isentropic. It follows that p01 =
150 psia. Therefore:
p01
150

 7.5
p1
20
Now the software or the normal shock tables give for p0 / p = 7.5 for air (γ = 1.4) M =
1.973 and for To / T = 1.778. Hence:
T1  T0
T1
(60  460)

 292.5 o R
T0
1.778
Since the Mach number ahead of the shock wave is 1.973 the software or the normal
shock tables give for a normal shock wave in air (γ = 1.4) M2 = 0.582, p2/p1 =4.375, and
T2 / T1 = 1.666. Hence:
p2  p1
p2
 20  4.375  87.5 psia
p1
and:
T2  T1
T2
 292.5  1.666  487.2 o R
T1
Using this value of T2 then gives:
a2 
 RT2 
1.4  53.34  487.2  1082.4 ft/sec
from which it follows that:
133
V2  M 2 a2  0.582  1082.4  629.9 ft/sec
Therefore the velocity, pressure and temperature behind ( i.e. downstream of ) the
shock wave are 629.9 ft/sec, 87.5 psia and 487.2 °R (27.2°F) respectively.
134
PROBLEM 5.8
The exhaust gases from a rocket engine have a molar mass of 14. They can be
assumed to behave as a perfect gas with a specific heat ratio of 1.25. These gases are
accelerated through a convergent-divergent nozzle. A normal shock wave occurs in the
nozzle at a point in the flow where the Mach number is 2. Find the pressure, temperature,
density and stagnation pressure ratios across this shock wave.
SOLUTION
For M1 = 2 , the software gives the following for a normal shock in a gas with γ =
1.25: p2 / p1 = 4.333, T2 / T1 = 1.444, ρ2 / ρ1 = 3.000, and p02 / p01 = 0.6892 . Therefore the
pressure, temperature, density and stagnation pressure ratios across this shock wave are
4.333, 1.444, 3.000 and 0.6892 respectively.
135
PROBLEM 5.9
Air is expanded isentropically from a reservoir in which the pressure is 1000 kPa and
the temperature is 30°C until the pressure has dropped to 25 kPa. A normal shock wave
occurs at this point. Find the static pressure, the static temperature, the air velocity, and
the stagnation pressure after the shock wave.
SOLUTION
Because the flow ahead of the shock wave is isentropic it follows that p01 = 1000 kPa.
Therefore:
p01
1000

 40
p1
25
Now the software or the isentropic flow tables give for isentropic flow for air ( γ =
1.4 ) for p0 / p = 40:
M 1  3.057 and T0 / T1  2.869
Hence, the Mach number ahead of the shock wave is 3.057 and the temperature ahead
of the shock wave is:
T1  T0
T1
(30  273)

 105.61 K
T0
2.869
Now for a Mach number of 3.057 upstream of a normal shock wave the software or
the normal shock tables give for a normal shock wave in air (γ = 1.4) M2 = 0.4719, p2/p1
=10.74, T2 / T1 = 2.747, and p02/p1 = 12.5. Hence:
p2  p1
p2
 25  10.74  268.5 kPa
p1
and:
T2  T1
T2
 105.61  2.747  290.1 K
T1
Using this value of T2 then gives:
136
a2 
 RT2 
1.4  287  290.1  341.4 m/s
from which it follows that:
V2  M 2 a2  0.4719  341.4  161.1 m/s
Lastly:
p02 
p02
p1  12.5  25  312.5 kPa
p1
Therefore the velocity, pressure, temperature, and stagnation pressure after ( i.e.,
downstream of ) the shock wave are 161.1 m/s, 268.5 kPa, 290.1 K (17.1°C), and 312.5
kPa respectively.
137
PROBLEM 5.10
A gas with a molar mass of 4 and a specific heat ratio of 1.67 is expanded from a
large reservoir in which the pressure and temperature are 600 kPa and 35oC respectively
through a nozzle system until the Mach number is 1.5. A normal shock wave then occurs
in the flow. Find the pressure and velocity behind the shock wave.
SOLUTION
Because the flow ahead of the shock wave is isentropic it follows that p01 = 600 kPa
and that T01 = 308 K. Now the software for isentropic flow gives for γ = 1.67 and M =
1.5:
p0
 4.056 ,
p
T0
 1.754
T
Hence:
p1  p0
p1
600

 147.9 kPa
p0
4.056
and:
T1  T0
T1
308

 175.6 K
T0
1.754
Now for a Mach number of 1.5 upstream of a normal shock wave the software gives,
for γ = 1.67, M2 = 0.7158, p2/p1 =2.564, and T2 / T1 = 1.497. Hence:
p2  p1
p2
 147.9  2.564  379.2 kPa
p1
and:
T2  T1
T2
 175.6  1.497  262.9 K
T1
Also using this value of T2 and noting that the gas has a molar mass of 4, it follows
that:
138
a2 
 RT2 
1.67  (8314 / 4)  262.9  955.2 m/s
from which it follows that:
V2  M 2 a2  0.7158  955.2  683.8 m/s
Therefore the pressure and velocity behind (i.e. downstream of) the shock wave are
379.2 kPa and 683.8 m/s respectively.
139
PROBLEM 5.11
Air is expanded from a large chamber through a variable area duct. The pressure and
temperature in the large chamber are 115 psia and 100 o F respectively. At some point in
the flow where the Mach number is 2.5 a normal shock wave occurs. Find the pressure,
temperature, stagnation pressure, and velocity behind the shock wave.
SOLUTION
Assuming that the expansion flow is isentropic it follows that in the flow ahead of the
shock wave p01 = 115 psia and T01 = 100 o F. Now the software or the isentropic flow
tables give for isentropic flow of air ( γ = 1.4 ) for M = 2.5:
M 2  0.513 ,
p2 / p1  7.125 , T2 / T1  2.138 ,
p02 / p01  0.499
Hence:
p2  p01
p1 p2
1
 115 
 7.125  47.9 kPa
p01 p1
17.09
and:
T2  T01
T1 T2
1
 (460  100) 
 2.138  530 o R
T01 T1
2.259
Using this value of T2 then gives:
a2 
 RT2 
1.4  53.4  32.2  530  1128.9 ft/sec
from which it follows that:
V2  M 2 a2  0.513  1128.9  579.1 ft/sec
Lastly:
p02 
p02
p01  0.499  115  57.39 psia
p01
140
Therefore the velocity, pressure, temperature, and stagnation pressure after ( i.e.,
downstream of ) the shock wave are 579.1 ft/sec, 47.9 psia, 530 oR ( 70oF), and 57.39 psia
respectively.
141
PROBLEM 5.12
A normal shock wave occurs in an air flow at a point where the velocity is 750 m/s.
the pressure is 50 kPa and the temperature is 10oC. Find the velocity, pressure, and static
temperature downstream of the shock wave.
SOLUTION
Because:
a1 
 RT1 
1.4  287  283  337.2 m/s
it follows that:
M1 
750
 2.224
337.2
Now for a Mach number of 2.224 upstream of a normal shock wave the software or
the normal shock tables give for a normal shock wave in air (γ = 1.4) M2 = 0.5439, p2/p1
=5.604, and T2 / T1 = 1.878. Hence:
p2  p1
p2
 50  5.604  280.2 kPa
p1
and:
T2  T1
T2
 283  1.878  531.5 K
T1
Using this value of T2 then gives:
a2 
 RT2 
1.4  287  531.5  462.1 m/s
from which it follows that:
V2  M 2 a2  0.5439  462.1  251.3 m/s
Therefore the velocity, pressure and temperature downstream of the shock wave are
251.3 m/s , 280.2 kPa and 531 .5 K ( 258.5°C ) respectively.
142
PROBLEM 5.13
Air is isentropically expanded from a large chamber in which the pressure is 10,000
kPa and the temperature is 50oC until the Mach number reaches a value of 2. A normal
shock wave then occurs in the flow. Following the shock wave, the air is isentropically
decelerated until the velocity is again essentially zero. Find the pressure and temperature
that then exist.
SOLUTION
Because the flow ahead of the shock wave is isentropic it follows that p01 = 10000
kPa and T01 = 323 K. Therefore, since the stagnation temperature is unchanged by the
shock wave, when the flow is brought to rest downstream of the shock, a temperature of
323 K will again exist. Now for M1 = 2 , the software for a normal shock wave or the
normal shock tables give for air ( γ = 1.4 ), p02 / p01 = 0.7209. Hence:
p02  p01
p02
 10000  0.7209  7209 kPa
p01
Because the flow downstream of the shock wave is isentropic this is the pressure
attained after the deceleration, i.e., the pressure and temperature after the flow is
decelerated downstream of the shock wave are 7209 kPa and 323 K ( 50 oC) respectively.
143
PROBLEM 5.14
Air is expanded from a large reservoir in which the pressure and temperature are 500
kPa and 35°C through a variable area duct. A normal shock occurs at a point in the duct
where the Mach number is 2.5. Find the pressure and temperature in the flow just
downstream of the shock wave. Downstream of the shock wave the flow is brought to
rest in another large reservoir. Find the pressure and temperature in this reservoir.
Assume that the flow is one-dimensional and isentropic everywhere except through the
shock wave.
SOLUTION
The flows before and after the shock wave are assumed to be isentropic. In the flow
ahead of ( i.e., upstream of ) the shock wave therefore p01 = 500 kPa and T01 = 308 K.
Now for M = 2.5 , the software for isentropic flow or the isentropic flow tables give for
air ( γ = 1.4 ):
p0
 17.09 ,
p
T0
 2.259
T
and for M1 = 2.5 , the software for a normal shock wave or the normal shock tables
give for air ( γ = 1.4 ):
p2
 7.125 ,
p1
T2
 2.138 ,
T1
p02
 0.499
p01
Hence:
p2  p01
p1 p2
1
 500 
 7.125  208.5 kPa
p01 p1
17.09
and:
T2  T01
T1 T2
1
 308 
 2.138  292 K
T01 T1
2.259
144
and:
p02  p01
p02
 500  0.499  249.5 kPa
p01
Hence, because the flow downstream of the shock wave is assumed to be isentropic,
the pressure attained in the second reservoir will be p02, i.e., will be 249.5 kPa. The
stagnation temperature is unchanged by the shock wave. This means that the stagnation
temperature is the same everywhere. As a result the temperature in the second large
chamber will be the same as that in the first large chamber, i.e., will be equal to 308 K.
Therefore the pressure and temperature just downstream of the shock wave are 208.5
kPa and 292 K ( 19 °C ) while the pressure and temperature in the second large chamber
will be 249.5 kPa and 308 K ( 35°C ).
145
PROBLEM 5.15
Air is expanded from a reservoir in which the pressure and temperature are
maintained at 1000 kPa and 30o C. At a point in the flow at which the static pressure is
150 kPa a normal shock wave occurs. Find the static pressure, the static temperature and
the air velocity behind the shock wave. Assume the flow to be isentropic everywhere
except through the shock wave.
SOLUTION
Because the flow ahead of the shock wave is isentropic it follows that p01 =1000 kPa.
Therefore:
p01
1000

 6.6667
p1
150
Now the software or the isentropic flow tables for isentropic flow of air (γ = 1.4) give
for p0/p = 6.6667:
M 1  1.897 ,
T0
 1.720
T
Hence, the Mach number ahead of the shock wave is 1.897 and the temperature ahead
of the shock wave is given by:
T1  T0
T1
(30  273)

 176.2 K
T0
1.720
Now for a Mach number of 1.897 upstream of a normal shock wave the software or
the normal shock tables give for a normal shock wave in air ( γ = 1.4) M2 = 0.5962, p2/p1
= 4.032, and T2 / T1 = 1.606. Hence:
p2  p1
p2
 150  4.032  604.8 kPa
p1
and:
146
T2  T1
T2
 176.2  1.606  282.9 K
T1
Using this value of T2 then gives:
a2 
 RT2 
1.4  287  282.9  337.2 m/s
from which it follows that:
V2  M 2 a2  0.5962  337.2  201.0 m/s
Therefore the velocity, pressure and temperature behind ( i.e., downstream of) the
shock wave are 201 .0 m/s, 604.8 kPa and 282.9 K ( 9.9oC).
147
PROBLEM 5.16
Air is expanded through a convergent-divergent nozzle from a large chamber in
which the pressure and temperature are 200 kPa and 310 K respectively. A normal shock
wave occurs at a point in the nozzle where the Mach number is 2.5. The air is then
brought to rest in a second large chamber. Find the pressure and temperature in this
second chamber. Clearly state the assumptions you have made in arriving at the solution.
SOLUTION
It will be assumed that the flow is isentropic before and after the shock wave. The
stagnation temperature is unchanged by the shock wave. This means that the stagnation
temperature is the same everywhere. As a result the temperature in the second large
chamber will be the same as that in the first large chamber, i.e., will be equal to 310 K.
The stagnation pressure upstream of the shock wave, i.e., p0 , is 200 kPa. The shock
wave occurs at a point where the Mach number is 2.5 and for M1 = 2.5 the software for a
normal shock wave or the normal shock tables give for air, i.e., for γ = 1.4, p02/ p01 =
0.4990. Hence:
p02  p01
p02
 200  0.499  99.8 kPa
p01
Because the flow downstream of the shock wave is isentropic the stagnation pressure
is constant in the flow downstream of the shock wave. i.e. the pressure attained in the
second large chamber will be p02 , i.e., will be 99.8 kPa.
Therefore the pressure and temperature in the second large chamber will be 99.8 kPa
and 310 K ( 37 °C ).
148
PROBLEM 5.17
Air at a temperature of 10° C and a pressure of 50 kPa flows over a blunt nosed body
at a velocity of 500 m/s. Estimate the pressure acting on the front of the body.
SOLUTION
Using the value of the temperature in the flow gives the speed of sound in the flow as:
a1 
 RT 
1.4  287  283  337.2 m/s
from which it follows that the Mach number in the flow is:
M1 
500
 1.483
337.2
Therefore the flow is supersonic and a normal shock wave will form ahead of the
nose of the body.
Now for M1 = 1.483, the software for a normal shock wave or the normal shock tables
for air ( i.e., for γ = 1.4 ) give p02/p1 =3.349. Hence:
p02  p1
p02
 50  3.349  167.5 kPa
p1
Therefore if the deceleration of the flow downstream of the shock wave is assumed to
be isentropic, the pressure acting on the nose of the body will be p02, i.e., will be 167.5
kPa.
149
PROBLEM 5.18
A pitot-static tube is placed in a supersonic air flow at a Mach number of 2.0. The
static pressure and static temperature in the flow are 101 kPa and 30° C respectively.
Estimate the difference between the pitot and static pressure.
SOLUTION
In this flow p1 = 101 kPa. T1 = 303 K (this information is not used in arriving at the
answer) and M1 = 2. Now for M1 = 2 the software for a normal shock wave or the normal
shock tables for air ( i.e., for γ = 1.4 ) give p02/p1 =5.640. Hence:
p02  p1  p1
p02
 p1  101  5.640  101  468.6 kPa
p1
Therefore the difference between the pitot and the static pressures is 468.6 kPa.
150
PROBLEM 5.19
A pitot static tube is placed in a supersonic flow in which the static pressure and
temperature are 60 kPa and -20° C respectively. The difference between the pitot and
static pressures is measured and found to be 449 kPa. Find the Mach number and velocity
in the flow. Discuss the assumptions used in deriving the answers.
SOLUTION
In this flow p1 = 60 kPa, T1 = 263 K, and p02 - p1 = 449 kPa. Hence:
p02
p  p1
449
 02
 1 
 1  8.4833
p1
p1
60
This allows the Mach number ahead of the normal shock wave to be determined. An
iterative approach will be adopted here. Using the software for a normal shock wave or
the normal shock tables for air ( i.e., for γ = 1.4 ) allows different values of M1 to be
chosen and the corresponding values of p02 / p1 to be found. The value of M1 that gives
p02 / p1 = 8.4833 can then be determined. Guessed values of M1 and the corresponding
values of p02 / p1 are shown in the following tables:
M1 (Guessed)
2.0
3.0
2.5
2.4
2.49
2.495
2.493
2.494
2.4932
p02 / p1
5640
12.06
8.526
7.897
8.462
8.494
8.481
8.488
8.483
From these results it can be deduced that M1 = 2.4932. Then since:
V1  M 1 a1
and since:
151
a1 
 RT 
1.4  287  263  325.1 m/s
it follows that:
V1  2.4932  325.1  810.5 m/s
Therefore the Mach number and the velocity in the flow are 2.4932 and 810.5 m/s
respectively.
152
PROBLEM 5.20
A pitot-static tube is placed in an air flow in which the Mach number is 1.7. The static
pressure in the flow is 55 kPa and the static temperature is -5° C. What will be the
measured difference between the pitot and the static pressures?
SOLUTION
In this flow p1 = 55 kPa, T1 = 268 K (this information is not used in arriving at the
answer), and M1 = 1.7. Now for M1 = 1.7 the software for a normal shock wave or the
normal shock tables for air ( i.e., for γ = 1.4 ) give p02/p1 = 4.224. Hence:
p02  p1  p1
p02
 p1  55  4.224  55  177.3 kPa
p1
Therefore the difference between the pitot and the static pressures is 177.3 kPa.
153
PROBLEM 5.21
A pitot-static tube is placed in a supersonic flow in which the static temperature is
o
0 C. Measurements indicate that the static pressure is 80k Pa and that the ratio of the
pitot to the static pressure is 4.1. Find the Mach number and the velocity in the flow.
SOLUTION
In this situation:
p02
 4.1
p1
Assuming that air flow is involved, this allows the Mach number ahead of the normal
shock wave to be determined. An iterative approach will be adopted here. Using the
software for a normal shock wave or the normal shock tables for air ( i.e., for γ = 1.4 )
allows different values of M1 to be chosen and the corresponding values of p02 / p1 to be
found. The value of M1 that gives p02 / p1 = 4.1 can then be determined. Guessed values of
M1 and the corresponding values of p02 / p1 are shown in the following table:
M1 (Guessed)
1.2
1.3
1.5
1.7
1.6
1.65
1.67
1.673
1.671
p02 / p1
2.408
2.714
3.413
4.224
3.805
4.011
4.095
4.108
4.100
From these results it can be deduced that M1 = 1.671. Then since:
V1  M 1 a1
and since:
a1 
 RT 
1.4  287  273  331.2 m/s
154
it follows that:
V1  1.671  331.2  553.4 m/s
Therefore the Mach number and the velocity in the flow are 1.671 and 553.4 m/s
respectively.
155
PROBLEM 5.22
A pitot tube is placed in a stream of carbon dioxide in which the pressure is 60 kPa
and the Mach number is 3.0. What will the pitot pressure be?
SOLUTION
The pitot pressure is equal to the stagnation pressure downstream of the shock wave
that forms ahead of the pitot tube. Table 3.1 in the textbook gives γ = 1.3 for Carbon
Dioxide and for M1 = 3.0, the software for a normal shock wave gives for γ = 1.3:
p02
 11.44
p1
But p1 = 60 kPa. Hence:
p02 
p02
p1  11.44  60  686.4 kPa
p1
Therefore the pitot pressure is 686.4 kPa.
156
PROBLEM 5.23
A thermocouple placed in the mouth of a pitot tube can be used to measure the
stagnation temperature of a flow. Such an arrangement placed in an air flow gives the
stagnation pressure as 180 kPa, the static pressure as 55 kPa and the stagnation
temperature as 95oC. Estimate the velocity of the stream assuming that the flow is
supersonic.
SOLUTION
In this situation:
p02
180

 3.273
p1
55
This result allows the Mach number ahead of the normal shock wave to be
determined. An iterative approach will be adopted here. Using the software for a normal
shock wave or the normal shock tables for air ( i.e., for γ = 1.4 ) allows different values of
M1 to be chosen and the corresponding values of p02 / p1 to be found. The value of M1 that
gives p02 / p1 = 3.273 can then be determined. Guessed values of M1 and the
corresponding values of p02 / p1 are shown in the following table:
M1 (Guessed)
1.2
1.3
1.5
1.4
1.45
1.46
1.47
1.463
1.4625
p02 / p1
2.408
2.714
3.413
3.049
3.228
3.264
3.301
3.275
3.273
From these results it can be deduced that M1 = 1.4625. For this value of Mach
number, the software for isentropic flow or the isentropic flow for air ( i.e., for γ = 1.4 )
157
gives T0 / T = 1.428. Therefore, because the stagnation temperature remains constant
through the shock wave with the result that in the flow ahead of the tube T0 = 368K, it
follows that:
T1  T0
T1
368

 257.7 K
T0
1.428
Then since:
V1  M 1 a1
and since:
a1 
 RT 
1.4  287  257.7  321.8 m/s
it follows that:
V1  1.4625  321.8  470.6 m/s
Therefore the Mach number and the velocity in the flow are 1.4625 and 470.6 m/s
respectively.
158
PROBLEM 5.24
A shock wave propagates down a constant area duct into stagnant air at a pressure of
101.3 kPa and a temperature of 25oC. If the pressure ratio across the shock wave is 3, find
the shock speed and the velocity of the air downstream of the shock.
SOLUTION
The flow situation being considered is shown in Fig. P5.24.
Figure P5.24
Here:
p1  101.3kPa ,
T1  25o C ,
p2
 3.0
p1
For the given pressure ratio p2 / p1 of 3, M1, M2 T2 / T1 are obtained from the software for
a normal shock wave or the normal shock tables for air ( i.e., for γ = 1.4 ), their values
being:
M 1 = 1.648 ,
M 2  0.645 ,
T2
 1.421
T1
Considering the flow relative to the wave as shown in the Fig. P5.24 gives:
159
M1 
Us
a1
,
M2 
U s V
a2
Hence:
U s  M 1 a1
But:
a1 
 RT1 
1.4  287  298  346.0 m/s
so:
U s  1.648  346.0  570.3 m/s
From the equation given above, the velocity downstream of the wave is given by:
V  U s  M 2 a2  M 1 a1  M 2 a2  ( M 1  M 2
a2
) a1
a1
Hence:
V 
1.648  0.645

1.421

 346.0  304.2 m/s
Therefore the velocity of the shock wave is 570.3 m/s and the velocity of the air
behind the shock wave is 304.2 m/s.
160
PROBLEM 5.25
A normal shock wave propagates down a constant-area tube containing stagnant air at
a temperature of 300 K. Find the velocity of the shock wave if the air behind the wave is
accelerated to Mach number of 1.2.
SOLUTION
The flow situation being considered is shown in Fig. P5.25.
Figure P5.25
Here:
T1  300 K ,
V
 1.2
a2
Considering the flow relative to the wave as shown in the Fig. P5.25 gives:
M1 
Us
a1
and:
M2 
U s V
U
U a
V
V
a
 s 
 s 1 
 M 1 1  1.2
a2
a2 a2
a1 a2 a2
a2
161
The solution has been obtained by trial-and-error. The value of M1 was guessed and
the software for a normal shock wave or the normal shock tables for air ( i.e., for γ = 1.4
was used to obtain M2 and a2 / a1 [ = (T2 / T1) 0.5 ]. Using the value of a2 / a1 so obtained,
the right hand side of the above equation, i.e., M1 (a1 / a2) - 1.2 was calculated and
compared with the value of M2 given by the software or tables. This was repeated until
the value of M1 that gave M2 equal to M1 (a1 / a2) - 1.2 was obtained. Some values
obtained in this process are shown in the following table:
M1
1.5
2.0
2.5
2.51
M2
0.7011
0.5744
0.5130
0.5120
M1 (a1 / a2) - 1.2
0.1056
0.3393
0.5100
0.5130
Continuing this process gives M1 = 2.507. Hence:
U s  M s a1  M 1 a1  M 1
 RT1  2.507  1.4  287  300  870.4 m/s
Therefore the velocity of the shock wave is 870.4 m/s.
162
PROBLEM 5.26
A shock wave is moving down a constant area duct containing air. The air ahead of
the shock wave is at rest and at a pressure and temperature of 100kPa and 20°C
respectively. If the pressure ratio across the shock wave is 2.5, find the velocity, pressure
and the temperature in the air behind the shock wave.
SOLUTION
The flow situation being considered is shown in Fig. P.5.26.
Figure P5.26
Here:
p1  100 kPa ,
T1  20o C ,
p2
 2.5
p1
For the given pressure ratio p2 / p1 of 2.5, M1, M2 T2 / T1 are obtained from the
software for a normal shock wave or the normal shock tables for air ( i.e., for γ = 1.4 ),
their values being:
M 1 = 1.512 ,
M 2  0.6969 ,
163
T2
 1.328
T1
Considering the flow relative to the wave as shown in Fig. P5.26 gives:
M1 
Us
,
a1
M2 
Us V
a2
Hence:
V  U s  M 2 a2  M 1 a1  M 2 a2  ( M 1  M 2
a2
) a1
a1
But:
a1 
 RT1 
1.4  287  293  343.1 m/s
so:
U s  1.648  346.0  570.3 m/s
From the equation given above, the velocity downstream of the wave is given by:
V  U s  M 2 a2  M 1 a1  M 2 a2  ( M 1  M 2
a2
) a1
a1
so from the equation given above, the velocity downstream of the wave is given by:
V 
1.512  0.6969

1.328

343.1  243.2 m/s
The pressure downstream of the shock is given by:
p2 
p2
p1  2.5  100  250 kPa
p1
while the temperature downstream of the shock is given by:
T2 
T2
T1  1.328  293  389.1 K
T1
Therefore the velocity, pressure and temperature behind the shock wave are 243.2
m/s, 250 kPa and 389.1 K ( = 116.1°C ) respectively.
164
PROBLEM 5.27
A normal shock wave propagates at a speed of 2600 m/s down a pipe that is filled
with hydrogen. The hydrogen is at rest and at a pressure and temperature of 101.3 kPa
and 25°C respectively upstream of the wave. Assuming hydrogen to behave as a perfect
gas with constant specific heats find the temperature, pressure, and velocity downstream
of the wave.
SOLUTION
Table 3.1 in the textbook gives γ = 1.407 and m = 2.016 for hydrogen. Hence, in the
hydrogen ahead of the shock wave:
a1 
 RT1 
1.407  (8315 / 2.016)  298  1315 m/s
The shock Mach number is therefore given by:
MS 
Us
2600

 1.977
1315
a1
Therefore, using:
2( M s2  1)
V 
a1
(   1) M s
the velocity in the hydrogen behind the shock wave is given by:
V 
2  (1.977 2  1)
 1315  1607 m/s
2.407  1.977
while using:
2 M s2  (  1)
p2

p1
(  1)
the pressure in the hydrogen behind the shock wave is given by:
165
p2
2  1.407  1.977 2  (  1)

 4.4
101.3
(1.407  1)
Hence:
p2  4.4  101.3  445.7 kPa
Finally, since:
[2  (  1) M s2 ][2 M s2  (  1)]
T2

T1
(  1) 2 M s2
the temperature in the hydrogen behind the shock wave is given by:
T2
[2  (1.407  1)  1.977 2 ][2    1.977 2  (1.407  1)]

 1.68
298
(1.407  1) 2  1.977 2
Hence:
T2  1.68  298  500.5 K
Therefore the velocity, pressure, and temperature behind the shock wave are 1607
m/s, 445.7 kPa, and 500.5 K ( = 227.5°C ) respectively. These results could also have
been obtained by using the software for γ = 1.407.
166
PROBLEM 5.28
A normal shock wave, across which the pressure ratio is 1.2, moves down a duct into
still air at a pressure of 100 kPa and a temperature of 20° C.
Find the pressure,
temperature, and velocity of the air behind the shock wave. This shock wave passes over
a small circular cylinder. Assuming that the shock is unaffected by the small cylinder,
find the pressure acting at the stagnation point on the cylinder after the shock has passed
over it.
SOLUTION
The flow relative to the shock wave is first considered. Since the pressure ratio across
the shock wave is 1.2, the software for a normal shock wave or the normal shock tables
for air ( i.e., for γ = 1.4) gives:
M 1 = 1.09 ,
M 2  0.920 ,
T2
 1.059
T1
Using the given pressure ratio and above temperature ratio then gives:
p2 
p2
p1  1.2  100  120 kPa , and ,
p1
T2 
T2
T1  1.059  293  310.3 K
T1
Also:
a1 
 RT1 
1.4  287  293  343.1 m/s
a2 
 RT2 
1.4  287  310.3  353.1 m/s
and:
Now
M2 
U s V
a2
,
i.e. , V  U s  M 2 a2
167
But Ms = M1, so the above equation gives:
V  M 1 a1  M 2 a2  1.09  343.1  0.920  353.1  49.1 m/s
The cylinder is, therefore, exposed to a flow with a velocity of 49.1 m/s at a
temperature of 310.3 K and a pressure of 120 kPa. The Mach number in this flow is equal
to 49.1 / 353.1 = 0.139. Because this is subsonic, the pressure at the stagnation point on
the cylinder will be equal to the stagnation pressure in the flow. Now, isentropic flow
software or the isentropic flow tables for air ( i.e., for γ = 1.4) give for a Mach number of
0.139, p0 / p = 1.0137. Therefore, the pressure at the stagnation point on the cylinder is
1.0137 x 120 = 121.6 kPa.
Therefore the velocity, pressure and temperature behind the shock wave are 49.1 m/,
120 kPa, and 310.3 K ( = 37.3°C) respectively and the pressure at the stagnation point on
the cylinder is 121 .6 kPa.
168
PROBLEM 5.29
As a result of a rapid chemical reaction a normal shock wave is generated which
propagates down a duct in which there is air at a pressure of 100 kPa and a temperature of
30°C. The pressure behind this shock wave is 130 kPa. Half a second after the generation
of this shock wave a second normal shock wave is generated by another chemical
reaction. This second shock wave follows the first one down the duct, the pressure behind
this second wave being 190 kPa. Find the velocity of the air and the temperature behind
the second shock wave. Also find the distance between the two waves at a time of 0.7
seconds after the generation of the first shock wave.
SOLUTION
The flow situation considered is shown in Fig. P5.29.
Figure P5.29
Consider the flow relative to the first normal shock wave which is moving into still
air. For this wave the following are given:
p2
130

 1.3 ,
p1
100
p1  100 kPa ,
T1  30o C  303 K
But for a pressure ratio p2 / p1 of 1.3, M1, M2 T2 / T1 are obtained from the software
for a normal shock wave or the normal shock tables for air ( i.e., for γ = 1.4 ), their values
being:
169
M 2  0.897 ,
M 1 = 1.12 ,
T2
 1.078
T1
Therefore, using the known initial conditions:
T2 
T2
T1  1.078  303  325.5 K
T1
Also since:
M1 
U s1
a1
,
M2 
U s1  V2
a2
it follows that since:
a1 
 RT1 
1.4  287  303  348.9 m/s
and:
a2 
 RT2 
1.4  287  326.5  362.2 m/s
that:
U s1  M 1a1  1.12  348.9  390.8 m/s
and that:
V2  U s1  M 2 a2  M 1a1  M 2 a2  1.12  348.9  0.897  362.2  65.9 m/s
Next consider the second shock wave. Across this shock wave the pressure ratio is
p3 / p2 = 190 / 130 = 1.462 and for this value the software for a normal shock wave or the
normal shock tables for air ( i.e., for γ = 1.4 ) give:
M 2 = 1.182 ,
M 3  0.850 ,
T3
 1.117
T2
Hence:
T3 
T3
T2  1.117  326.5  364.8 K
T2
170
Now if Us2 is the velocity of the second shock wave then for the second shock wave:
M2 
U s 2  V2
a2
,
M3 
U s 2  V3
a3
The first of these equations gives:
U s 2  M 2 a2  V2  1.182  362.2  65.9  494.0 m/s
Also since:
a3 
 RT3 
1.4  287  364.8  382.9 m/s
the second of the above equations gives:
V3  U s 2  M 3 a3  494.0  0.850 x 382.9  168.5 m/s
Lastly, consider the distance between the shock waves at a time of 0.7 s after the
generation of the first wave, i.e., 0.2 s after the generation of the second wave. The
distance between the waves at this time will be:
s  U s1 x 0.7  U s 2  0.2  390.8  0.7  494.0  0.2  174.8 m
Therefore the velocity and temperature behind the second wave are 168.5 m/s and
364.8 K ( = 91.8°C ) respectively while the distance between the two waves at the time
considered is 174.8 m.
171
PROBLEM 5.30
A normal shock wave across which the pressure ratio is 1.25 is propagating down a
duct containing still air at a pressure of 120 kPa and a temperature of 35°C. This shock
wave is reflected off the closed end of the duct. Find the pressure and temperature behind
the reflected shock wave.
SOLUTION
The flow situation is shown in Fig. P5.30
Figure P5.30
Consider the initial shock wave before it is reflected. For this shock wave, the
following are given:
p2
 1.25 ,
p1
p1  120 kPa , T1  35o C  308 K
But for a pressure ratio p2 / p1 of 1.25, M1, M2, and T2 / T1 are obtained from the
software for a normal shock wave or from the normal shock tables for air ( i.e., for γ =
1.4 ), their values being:
M 1 = 1.10 ,
M 2  0.912 ,
Therefore, using the known initial conditions:
172
T2
 1.065
T1
p2 
p2
T
p1  1.25  120  150 kPa and T2  2 T1  1.065  308  328 K
p1
T1
Also since:
M1 
Us
,
a1
M2 
U s V
a2
it follows that since:
a1 
 RT1 
1.4  287  308  351.8 m/s
a2 
 RT2 
1.4  287  328  363.0 m/s
and since:
that:
U s  M 1a1  1.10  351.8  387.0 m/s
and that:
V  U s  M 2 a2  M 1a1  M 2 a2  1.10  351.8  0.912  363.0  55.9 m/s
Next consider the wave that is “reflected” off the closed end. The strength of this
wave must be such that it brings the flow to rest. Hence, the Mach numbers of the air
flow upstream and downstream of the reflected wave relative to this wave are:
M up 
U sR  V
a2
,
M down 
U sR
a3
where UsR is the velocity of the reflected wave and a3 is the speed of sound in the
flow downstream of the reflected wave. Hence, since a2 = 363 m/s and V = 55.9 m/s, it
follows that:
M up 
U sR  V
U
U
55.9
 sR 
 sR  0.154
363
a2
a2
a2
and:
M down 
U sR
U a
U
 s R 2  sR
a3
a2 a3
a2
Combining these equations gives:
173
T2
T3
M down   M up  0.154 
T2
T3
While there are more elegant methods of obtaining the solution, a trial-and-error
approach will be adopted here. A series of values of Mup is selected and then for each of
these values the software for normal shock waves or the normal shock tables for air ( i.e.,
for γ = 1.4 ), is used to find Mdown and T3 / T2 . These values are then used to derive the
value of Mdown = (Mup – 0.154)( T3 / T2)0.5. The correct value of Mup is that which has the
value of Mdown as given directly by the software or tables for normal shock waves equal to
that given by the above equation, i.e., by Mdown = (Mup – 0.154)( T3 / T2)0.5. This correct
value can be deduced from the results obtained for various guessed values of Mup. Results
for various values of Mup are shown in the following table.
Mup
(guessed)
1.000
1.100
1.200
1.050
1.080
1.090
T3 / T2
(software or tables)
1.000
1.065
1.128
1.033
1.052
1.059
Mdown
(software or tables)
1.000
0.912
0.842
0.953
0.928
0.920
(Mup – 0.154)( T3 / T2)0.5
0.846
0.917
0.985
0.882
0.903
0.910
From these results, it can be deduced that the correct values of Mup is 1.097. Now for
an upstream Mach number of 1.097, the software or the tables for normal shock waves
gives p3 / p2 = 1.238 and T3 / T2 = 1.063. Hence, the pressure and temperature behind the
reflected shock are given by:
p3 
p3
T
p2  1.238  150  186 kPa and T3  3 T2  1.063  328  349 K
p2
T2
Therefore the pressure and temperature behind the “reflected” shock wave are 186
kPa and 349 K ( = 76°C ) respectively.
174
PROBLEM 5.31
A normal shock wave is propagating down a duct in which p = 110 kPa and T =
30°C. The pressure ratio across the shock is 1.8. Find the velocity of the shock wave and
the air velocity behind the shock. This moving shock strikes a closed end to the duct.
Find the pressure on the closed end after the shock reflection.
SOLUTION
The flow situation being considered is shown in Fig. P5.31.
Figure P5.31
Consider the initial shock wave before it is reflected. For this shock wave the
following are given:
p2
 1.8 ,
p1
p1  110 kPa ,
T1  30o C  303 K
But for a pressure ratio p2 / p1 of 1.8, assuming that the gas involved is air, M1, M2
T2 / T1 are obtained from the software for a normal shock wave or the normal shock
tables for air ( i.e., for γ = 1.4 ), their values being:
M 1 = 1.30 ,
M 2  0.786 ,
175
T2
 1.191
T1
Therefore, using the known initial conditions:
p2 
p2
T
p1  1.8  110  198 kPa and T2  2 T1  1.191  303  360.8 K
p1
T1
Also since:
M1 
Us
,
a1
M2 
Us V
a2
It follows that since:
a1 
 RT1 
1.4  287  303  348.9 m/s
and since:
 RT2 
a2 
1.4  287  360.8  380.8 m/s
that:
U s  M 1a1  1.30  348.9  453.6 m/s
and that:
V  U s  M 2 a2  M 1a1  M 2 a2  1.30  348.9  0.786  380.8  154.3 m/s
Next consider the wave that is “reflected” off the closed end. The strength of this
wave must be such that it brings the flow to rest. Hence, the Mach numbers of the air
flow upstream and downstream of the reflected wave relative to this wave are:
M up 
U sR  V
,
a2
M down 
U sR
a3
where UsR is the velocity of the reflected wave and a3 is the speed of sound in the
flow downstream of the reflected wave. Hence, since a2 = 380.8 m/s and V = 154.3 m/s,
it follows that:
M up 
U sR  V
U
U
154.3
 sR 
 sR  0.405
380.8
a2
a2
a2
176
and:
M down 
U sR
U a
U
 sR 2  sR
a3
a2 a3
a2
T2
T3
Combining these equations gives:
M down 
M
up
 0.405 
T2
T3
While there are more elegant methods of obtaining the solution, a trial-and-error
approach will be adopted here. A series of values of Mup are selected and then for each of
these values the software for normal shock waves or the normal shock tables for air ( i.e.,
for γ = 1.4 ), is used to find Mdown and T3 / T2 . These values are then used to derive the
value of Mdown = (Mup – 0.405)( T3 / T2)0.5. The correct value of Mup is that which has the
value of Mdown as given directly by the software or tables for normal shock waves equal to
that given by the above equation, i.e., by Mdown = (Mup – 0.405)( T3 / T2)0.5. This correct
value can be deduced from the results obtained for various guessed values of Mup. Results
for various values of Mup are shown in the following table.
Mup
(guessed)
1.000
1.100
1.200
1.300
1.250
1.270
1.280
T3 / T2
(software or tables)
1.000
1.065
1.128
1.191
1.159
1.172
1.178
Mdown
(software or tables)
1.000
0.912
0.842
0.786
0.813
0.802
0.796
(Mup – 0.405)( T3 / T2)0.5
0.595
0.674
0.749
0.820
0.785
0.799
0.806
From these results, it can be deduced that the equation is satisfied when Mup = 1.272.
Now for an upstream Mach number of 1.272, the software or the tables for normal shock
waves gives p3 / p2 = 1.72. Hence, the pressure behind the reflected shock is given by:
177
p3 
p3
p2  1.72  198  341 kPa
p2
Therefore the velocity of the initial shock wave and the air velocity behind this initial
shock wave are 453.6 m/s and 154.3 m/s respectively and the pressure behind the
“reflected” shock wave is 341 kPa.
178
PROBLEM 5.32
A normal shock wave is moving down a duct into still air in which the pressure is 100
kPa and the temperature is 20°C. The pressure ratio across the shock is 1.8. Find the
velocity of the shock and the velocity of the air behind the shock. If this shock strikes the
closed end of the duct find the pressure on this closed end after the shock reflection.
SOLUTION
The flow situation being considered is shown in Fig. P5.32 .
Figure P5.32
Consider the initial shock wave before it is reflected. For this shock wave the
following are given:
p2
 1.8 ,
p1
p1  100 kPa , T1  20o C  293K
But for a pressure ratio p2 / p1 of 1.8, assuming that the gas involved is air, M1, M2
T2 / T1 are obtained from the software for a normal shock wave or the normal shock
tables for air ( i.e., for γ = 1.4 ), their values being:
M 1 = 1.30 , M 2  0.786 ,
179
T2
 1.191
T1
Therefore, using the known initial conditions:
p2 
p2
T
p1  1.8  100  180 kPa and T2  2 T1  1.191  293  349.0 K
p1
T1
Also since:
M1 
Us
,
a1
M2 
Us V
a2
It follows that since:
a1 
 RT1 
1.4  287  293  343.1 m/s
a2 
 RT2 
1.4  287  349.0  374.5 m/s
and since:
that:
U s  M 1a1  1.30  343.1  446.0 m/s
and that:
V  U s  M 2 a2  M 1a1  M 2 a2  1.30  343.1  0.786  374.5  151.7 m/s
Next consider the wave that is “reflected” off the closed end. The strength of this
wave must be such that it brings the flow to rest. Hence, the Mach numbers of the air
flow upstream and downstream of the reflected wave relative to this wave are:
M up 
U sR  V
a2
,
M down 
U sR
a3
where UsR is the velocity of the reflected wave and a3 is the speed of sound in the
flow downstream of the reflected wave. Hence, since a2 = 374.5 m/s and V = 151.7 m/s, it
follows that:
M up 
U sR  V
U
U
151.7
 sR 
 sR  0.405
374.5
a2
a2
a2
180
and:
M down 
U sR
U a
U
 sR 2  sR
a3
a2 a3
a2
T2
T3
Combining these equations gives:
M down 
M
up
 0.405 
T2
T3
While there are more elegant methods of obtaining the solution, a trial-and-error
approach will be adopted here. A series of values of Mup are selected and then for each of
these values the software for normal shock waves or the normal shock tables for air ( i.e.,
for γ = 1.4 ), is used to find Mdown and T3 / T2 . These values are then used to derive the
value of Mdown = (Mup – 0.405)( T3 / T2)0.5. The correct value of Mup is that which has the
value of Mdown as given directly by the software or tables for normal shock waves equal to
that given by the above equation, i.e., by Mdown = (Mup – 0.405)( T3 / T2)0.5. This correct
value can be deduced from the results obtained for various guessed values of Mup. Results
for various values of Mup are shown in the following table.
Mup
(guessed)
1.000
1.100
1.200
1.300
1.250
1.270
1.280
T3 / T2
(software or tables)
1.000
1.065
1.128
1.191
1.159
1.172
1.178
Mdown
(software or tables)
1.000
0.912
0.842
0.786
0.813
0.802
0.796
(Mup – 0.405)( T3 / T2)0.5
0.595
0.674
0.749
0.820
0.785
0.799
0.806
From these results, it can be deduced that the equation is satisfied when Mup = 1.272.
Now for an upstream Mach number of 1.272, the software or the tables for normal shock
waves gives p3 / p2 = 1.72. Hence, the pressure behind the reflected shock is given by:
p3 
p3
p2  1.72  180  309.6 kPa
p2
181
Therefore the velocity of the initial shock wave and the air velocity behind this initial
shock wave are 446 m/s and 151.7 m/s respectively and the pressure behind the
“reflected” shock wave is 309.6 kPa.
182
PROBLEM 5.33
A shock across which the pressure ratio is 1.18 moves down a duct into still air at a
pressure of 100 kPa and a temperature of 30°C. Find the temperature and velocity of the
air behind the shock wave. If instead of being at rest, the air ahead of the shock wave is
moving towards the wave at a velocity of 75 m/s, what would be the velocity of the air
behind the shock wave?
SOLUTION
First consider the shock wave moving into still air. For this shock wave, the following
are given:
p2
 1.18 ,
p1
p1  100 kPa ,
T1  320o C  303 K
But for a pressure ratio p2 / p1 of 1.18, M1, M2, and T2 / T1 are obtained from the
software for a normal shock wave or the normal shock tables for air ( i.e., for γ = 1.4 ),
their values being:
M 1 = 1.075 ,
M 2  0.932 ,
T2
 1.049
T1
Therefore, using the known initial conditions:
T2 
T2
T1  1.049  303  317.8 K
T1
Also since:
M1 
Us
a1
,
M2 
Us V
a2
and:
a1 
 RT1 
1.4  287  303  348.9 m/s
183
and:
a2 
 RT2 
1.4  287  317.8  357.3 m/s
it follows that:
U s  M 1a1  1.075  348.9  375.1 m/s
and that:
V  U s  M 2 a2  M 1a1  M 2 a2  1.075  348.9  0.932  357.3  42.1 m/s
Next consider the shock wave moving into air flowing at a velocity of 75 m/s, the
situation being as shown in Fig. P5.33.
Figure P5.33
Because the pressure ratio across the shock is the same as with no flow, the pressures
and temperatures on the two sides of the wave are the same as when there is no flow. The
Mach numbers relative to the wave on the upstream and downstream sides of the wave in
this case are given by:
M1 
U s  75
,
a1
M2 
Us V
a2
The first of these equations gives:
U s = M 1 a1 - 75 = 1.075  348.9 - 75 = 300.1 m/s
184
and the second of the above equations then gives:
V = U s - M 2 a2 = 300.1 - 0.932  357.3 = - 32.9 m/s
The minus sign indicates that the flow behind the shock is moving in the opposite
direction to the shock, i.e., the flow behind the shock is moving in the same direction as
the air flow ahead of the shock.
Therefore the velocity and temperature behind the shock wave moving into still air
are 42.1 m/s and 317.8 K ( = 44.8°C ) respectively and the velocity behind the shock
wave moving into the air stream is - 32.9 m/s.
185
PROBLEM 5.34
Air at a pressure of 105 kPa and a temperature of 25°C is flowing out of a duct at a
velocity of 250 m/s. A valve at the end of the duct is suddenly closed. Find the pressure
acting on the valve.
SOLUTION
Consider the flow relative to the shock wave. Because the wave must bring the air to
rest, it will be seen that if Us is the velocity of the shock wave:
M1 
Us V
a1
,
M2 
Us
U a
 s 1
a2
a1 a2
Combining these equations gives:
M1  M 2
a2
V

a1
a1
But the speed of sound ahead of the wave is:
a1 
 RT1 
1.4  287  298  346.0 m/s
and V = 250 m/s so the above equation gives:
M1  M 2
a2
a
250

 M 2 2  0.722
a1
346
a1
Because the normal shock relations determine M2 and a2 / a1 as functions of M1 the
above equation together with the normal shock relations can be used to determine M1 .
While there are more elegant methods of obtaining the solution, the brute force approach
is to choose a series of values of M1 and then use the software for normal shock waves or
the normal shock tables for air ( i.e., for γ = 1.4 ) to find the right hand side, i.e., to find
the value of:
186
M2
a2
 0.722
a1
for each of these values of M1. The value of M1 that makes the right hand equal to M1 can
then be deduced from these results. A set of results is shown in the following table, the
values of M2 and a2 / a1 being obtained by using the software for normal shock waves or
the normal shock tables for air.
M1
(guessed)
1.000
1.400
1.500
1.600
1.550
1.540
1.520
M2
(software or tables)
1.000
0.740
0.701
0.668
0.684
0.687
0.694
a2 / a1
(software or tables)
1.000
1.120
1.149
1.178
1.164
1.161
1.155
M2 (a2 / a1) + 0.722
(calculated)
1.722
1.551
1.528
1.509
1.518
1.520
1.524
From these results, it can be deduced that the equation is satisfied when M1 = 1.522.
Now for an upstream Mach number of 1.522, the software or the tables for normal shock
waves gives p2 / p1 = 2.536. Hence, the pressure behind the reflected shock is given by:
p2 
p2
p1  2.536  105  266.3 kPa
p1
Therefore the pressure acting on the valve is 266.3 kPa.
187
PROBLEM 5.35
Air is flowing out of a duct at a velocity of 250 m/s. The static temperature and
pressure in the flow are 0°C and 70 kPa. A valve at the end of the duct is suddenly
closed. Estimate the pressure acting on this valve immediately after it is closed.
SOLUTION
Consider the flow relative to the shock wave. Because the wave must bring the air to
rest, it will be seen that if Us is the velocity of the shock wave:
M1 
Us V
a1
,
M2 
Us
U a
 s 1
a2
a1 a2
Combining these equations gives:
M1  M 2
a2
V

a1
a1
But the speed of sound ahead of the wave is:
a1 
 RT1 
1.4  287  273  331.2 m/s
and V = 250 m/s so the above equation gives:
M1  M 2
a2
a
250

 M 2 2  0.755
a1
331.2
a1
Because the normal shock relations determine M2 and a2 / a1 as functions of M1 the
above equation together with the normal shock relations can be used to determine M1 .
While there are more elegant methods of obtaining the solution, the brute force approach
is to choose a series of values of M1 and then to use the software for normal shock waves
or the normal shock tables for air ( i.e., for γ = 1.4 ) to find the right hand side of the
above equation, i.e., to find the value of:
188
M2
a2
 0.755
a1
for each of these values of M1. The value of M1 that makes the right hand equal to M1 can
then be deduced from these results. A set of results is shown in the following table, the
values of M2 and a2 / a1 being obtained by using the software for normal shock waves or
the normal shock tables for air.
M1
(guessed)
1.000
1.400
1.500
1.600
1.550
1.560
M2
(software or tables)
1.000
0.740
0.701
0.668
0.684
0.681
a2 / a1
(software or tables)
1.000
1.120
1.149
1.178
1.164
1.167
M2 (a2 / a1) + 0.755
(calculated)
1.755
1.584
1.561
1.542
1.551
1.550
From these results it can be deduced that the equation is satisfied when M1 = 1.551.
Now for an upstream Mach number of 1.551, the software or the tables for normal shock
waves gives p2 / p1 = 2.639. Hence, the pressure behind the reflected shock is given by:
p2 
p2
p1  2.639  70  184.7 kPa
p1
Therefore the pressure acting on the valve is 184.7 kPa.
189
PROBLEM 5.36
Air is flowing down a duct at a velocity of 200 m/s. The pressure and the temperature
in the flow are 85 kPa and 10°C respectively. If a valve at the end of the duct is suddenly
closed, find pressure acting on the valve immediately after closure.
SOLUTION
Consider the flow relative to the shock wave. Because the wave must bring the air to
rest, it will be seen that if Us is the velocity of the shock wave:
M1 
Us V
a1
,
M2 
Us
U a
 s 1
a2
a1 a2
Combining these equations gives:
M1  M 2
a2
V

a1
a1
But the speed of sound ahead of the wave is:
a1 
 RT1 
1.4  287  283  337.2 m/s
and V = 200 m/s so the above equation gives:
M1  M 2
a2
a
200

 M 2 2  0.593
a1
337.2
a1
Because the normal shock relations determine M2 and a2 / a1 as functions of M1 the
above equation together with the normal shock relations can be used to determine M1 .
While there are more elegant methods of obtaining the solution, the brute force approach
is to choose a series of values of M1 and then to use the software for normal shock waves
or the normal shock tables for air ( i.e., for γ = 1.4 ) to find the right hand side of the
above equation, i.e., to find the value of:
190
M2
a2
 0.593
a1
for each of these values of M1. The value of M1 that makes the right hand equal to M1 can
then be deduced from these results. A set of results is shown in the following table, the
values of M2 and a2 / a1 being obtained by using the software for normal shock waves or
the normal shock tables for air.
M1
(guessed)
1.000
1.400
1.500
1.450
1.420
M2
(software or tables)
1.000
0.740
0.701
0.720
0.731
a2 / a1
(software or tables)
1.000
1.120
1.149
1.135
1.126
M2 (a2 / a1) + 0.593
(calculated)
1.593
1.422
1.399
1.410
1.416
From these results, it can be deduced that the equation is satisfied when M1 = 1.418.
Now for an upstream Mach number of 1.418, the software or the tables for normal shock
waves gives p2 / p1 = 2.719. Hence, the pressure behind the reflected shock is given by:
p2 
p2
p1  2.719  85  185.2 kPa
p1
Therefore the pressure acting on the valve is 185.2 kPa.
191
PROBLEM 5.37
A piston in a pipe containing stagnant air at a pressure of 101 kPa and a temperature
of 25°C is suddenly given a velocity of 100 m/s into the pipe causing a normal shock
wave to propagate through the air down the pipe. Find the velocity of the shock wave and
the pressure acting on the piston.
SOLUTION
The flow situation considered is shown in Fig. P5.37.
Figure P5.37
Considering the flow relative to the shock wave it will be seen that if Us is the
velocity of the shock wave:
M1 
Us
a1
and
M2 
Us  V
U a
V a1
 s 1 
a2
a1 a2
a1 a2
Combining these equations gives:

V  a 
M 2   M1    1 
a1   a2 

But the speed of sound ahead of the wave is:
a1 
 RT1 
1.4  287  298  346.0 m/s
192
and V = 100 m/s so the above equation gives:
 a1 
100   a1 

M 2   M1 
     M 1  0.289   
346   a2 

 a2 
Because the normal shock relations determine M2 and a2 / a1 as functions of M1 the
above equation together with the normal shock relations can be used to determine M1 .
While there are more elegant methods of obtaining the solution, the brute force approach
is to choose a series of values of M1 and then to use the software for normal shock waves
or the normal shock tables for air ( i.e., for γ = 1.4 ) to find the right hand side of the
above equation, i.e., to find the value of:
 a1 

 a2 
 M 1  0.289  
for each of these values of M1. The value of M1 that makes the right hand equal to M2 can
then be deduced from these results. A set of results is shown in the following table, the
values of M2 and a2 / a1 being obtained by using the software for normal shock waves or
the normal shock tables for air.
M1
(guessed)
1.000
1.400
1.200
1.100
1.160
1.180
1.190
M2
(software or tables)
1.000
0.740
0.842
0.912
0.868
0.855
0.849
a2 / a1
(software or tables)
1.000
1.120
1.062
1.032
1.050
1.056
1.059
(M1 – 0.289)(a1 / a2)
(calculated)
0.711
0.992
0.858
0.786
0.830
0.844
0.851
From these results, it can be deduced that the equation is satisfied when M1 = 1.189.
Hence:
U s  M 1 a1  1.189  346  411 m/s
193
Also since for an upstream Mach number of 1.189, the software or the tables for
normal shock waves gives p2 / p1 = 1.49, the pressure behind the reflected shock is given
by:
p2 
p2
p1  1.49  101  150 kPa
p1
Therefore the velocity of the shock wave is 411 m/s and the pressure acting on the
piston is 150 kPa.
194
PROBLEM 5.38
Air flows at a velocity of 90 m/s down a 20 cm diameter pipe. The air is at a pressure
of 120 kPa and a temperature of 30°C. A valve at the end of the pipe is suddenly closed.
This valve is held in place by eight mild steel bolts each with a diameter of 12 mm. Will
the bolts hold the valve without yielding? Assume that the pipe is discharging the air to
the atmosphere and that the ambient pressure is 101 kPa. It will be necessary to look up
the yield strength of the steel in order to answer this question.
SOLUTION
A normal shock wave propagates down the pipe following the closure of the valve.
Consider the flow relative to the shock wave. Because the wave must bring the air to rest
it will be seen that if Us is the velocity of the shock wave then:
M1 
Us  V
a1
M2 
and
Us
U a
 s 1
a2
a1 a2
Combining these equations gives:
M1  M 2
a2
V

a1
a1
But the speed of sound ahead of the wave is:
a1 
 RT1 
1.4  287  303  348.9 m/s
and V = 90 m/s so the above equation gives:
M1  M 2
a2
a
90

 M 2 2  0.258
a1
348.9
a1
Because the normal shock relations determine M2 and a2 / a1 as functions of M1 the
above equation together with the normal shock relations can be used to determine M1 .
While there are more elegant methods of obtaining the solution, the brute force approach
is to choose a series of values of M1 and then to use the software for normal shock waves
195
or the normal shock tables for air ( i.e., for γ = 1.4 ) to find the right hand side of the
above equation, i.e., to find the value of:
M2
a2
 0.258
a1
for each of these values of M1. The value of M1 that makes the right hand equal to M1 can
then be deduced from these results. A set of results is shown in the following table, the
values of M2 and a2 / a1 being obtained by using the software for normal shock waves or
the normal shock tables for air.
M1
(guessed)
1.000
1.400
1.200
1.100
1.180
1.160
M2
(software or tables)
1.000
0.740
0.842
0.912
0.855
0.868
a2 / a1
(software or tables)
1.000
1.120
1.062
1.032
1.056
1.050
M2 (a2 / a1) + 0.258
(calculated)
1.258
1.087
1.152
1.200
1.160
1.170
From these results, it can be deduced that the equation is satisfied when M1 = 1.17.
Since for an upstream Mach number of 1.17, the software or the tables for normal shock
waves gives p2 / p1 = 1.43, the pressure behind the shock is given by:
p2 
p2
p1  1.43  120  171.6 kPa
p1
Therefore the pressure difference across the valve is ( 171.6 - 101) = 70.6 kPa. The
total force on the valve is therefore, because the pipe diameter is 0.2 m, given by:
F  70.6 

4
 0.22  2.22 kN
Because there are eight 12 mm diameter bolts, the stress in the bolts is therefore:
196
 
8 

2.22
4
 0.012
 2454 kPa
2
But the yield stress of mild steel is over 200 MPa so yield failure of the bolts will not
occur.
197
PROBLEM 5.39
A cannon fires a shell that causes a projectile to move down the barrel at a velocity of
740m/s. What is the speed of the normal shock proceeding down the barrel in front of the
projectile if the undisturbed air in the barrel is at 101kPa and 20°C? How fast would the
projectile have to be moving down the barrel if the velocity of the shock ahead of the
projectile was 2.5 times the velocity of the projectile?
SOLUTION
Part 1
The velocity behind the shock wave must equal the velocity of the projectile, i.e.,
must equal 740m/s. Now equation (5.67) for a moving shock wave gives:
2( M s2  1)
V
a1
(   1) M s
Where V is the velocity behind the shock wave, Ms is the shock Mach number, and a1
is the speed of sound ahead of the shock wave. Now:
a1 
 RT1  1.4  287  293  343.1 m/s
Substituting this result into the above equation gives:
740 
2( M s2  1)
 343.1 , i.e., since  = 1.4,
(   1) M s
2M s2  5.176 M s  2  0
Solving this quadratic equation gives:
M s  2.929
Hence if Us is the velocity of the shock wave:
U s  M s a1  2.929  343.1  1005.1 m/s
198
Therefore the velocity of the shock wave is 1005.1 m/s. If the software or tables had
been used instead of the equation for Ms the iterative approach discussed before would
have been adopted.
Part 2
Here:
U s  2.5V
Therefore since:
Ms 
Us
2.5V

a1
a1
Substituting this result into the equation for V used in Part 1 gives:
 2.5V 2 
2 
  1
 a1 

2( M s2  1)
V 
a1 
(   1) M s
 2.5V 
2.4  

 a1 
Since, from Part 1, a1 = 343.1 m/s, this equation gives on rearrangement:
V  190 m/s
Therefore the speed of the projectile is 190 m/s.
199
Chapter Six
OBLIQUE SHOCK WAVES
SUMMARY OF MAJOR EQUATIONS
Changes Across Oblique Shock Wave in Terms of Mach Number
p2
2 12 sin 2   (  1)

p1
 1
(6.10)
2
(  1) M 12 sin 2 

1 2  (  1) M 12 sin 2 
(6.11)
T2
[2  (  1) M 12 sin 2  ] [2 M 12 sin 2   (  1)]

T1
(  1) 2 M 12 sin 2 
(6.12)
M 22 sin 2 (    ) 
M12 sin 2   2 / (  1)
2 M12 sin 2  / (  1)  1
(6.13)
Oblique Shock Wave Angle Equation
tan  
2 cot  ( M 12 sin 2   1)
2  M 12 (  cos 2  )
200
(6.19)
PROBLEM 6.1
Air is flowing over a flat wall. The Mach number, pressure and temperature in the airstream are 3, 50 kPa, and -20o C respectively. If the wall turns through an angle of 4o
leading to the formation of an oblique shock wave, find the Mach number, the pressure
and the temperature in the flow behind the shock wave.
SOLUTION
Here, the Mach number upstream of the shock wave is M1 = 3.0 and the flow is
turned through an angle δ of 4°. Using the software for an oblique shock wave or the
oblique shock chart and the normal shock tables for air ( i.e., for γ = 1.4 ) gives for these
values:
M 2 = 2.799 ,
p2
 1.352 ,
p1
T2
 1.091
T1
Hence:
p2 
p2
T
p1  1.352  50  67.6 kPa and T2  2 T1  1.091  253  276 K
p1
T1
Therefore the Mach number, pressure and temperature “behind”, i.e., downstream of,
the oblique shock wave are 2.799, 67.6 kPa and, 276 K ( = 3° C ) respectively.
201
PROBLEM 6.2
An air flow in which the Mach number is 2.5 passes over a wedge with a half - angle
of 10°, the wedge being symmetrically placed in the flow. Find the ratio of the stagnation
pressures before and after the oblique shock wave generated at the leading edge of the
wedge.
SOLUTION
Here, the Mach number upstream of the shock wave is Ml = 2.5 and the flow is turned
through an angle δ = 10°. Using the software for an oblique shock wave or the oblique
shock chart and the normal shock tables for air ( i.e., for γ = 1.4 ) gives for these values:
p02
 0.9759
p01
Hence:
p01
1
1


 1.025
p02
p02 / p01
0.9759
Therefore the ratio of the stagnation pressures before and after the oblique shock
wave is 1.025.
202
PROBLEM 6.3
A symmetrical wedge with a 12° included angle is placed in an air flow in which the
Mach number is 2.3 and the pressure is 60 kPa. If the center-line of the wedge is at an
angle of 4° to the direction of flow, find the pressure difference between the two surfaces
of the wedge.
SOLUTION
As shown in Fig. P6.3, the flow over the upper surface of the wedge, which has a half
angle of 6°, is turned through an angle of 2° while the flow over the lower surface is
turned through an angle of 10°.
Figure P6.3
First consider the flow over the upper surface. For this surface, the Mach number
upstream of the shock wave is M1= 2.3 and the flow is turned through an angle δ = 2°.
Using the software for an oblique shock wave or the oblique shock chart and the normal
shock tables for air ( i.e., for γ = 1.4 ) gives for these values:
p2
 1.132
p1
203
Therefore, the pressure on the upper surface is given by:
pupper 
p2
p1  1.132  60  67.9 kPa
p1
Next consider the flow over the lower surface. For this surface, the Mach number
upstream of the shock wave is M1 = 2.3 and the flow is turned through an angle δ = 10°
Using the software for an oblique shock wave or the oblique shock chart and the normal
shock tables for air ( i.e., for γ = 1.4 ) gives for these values:
p2
 1.796
p1
Therefore, the pressure on the lower surface is given by:
plower 
p2
p1  1.796  60  107.8 kPa
p1
The pressure difference between the lower and upper surface is given by:
plower  pupper  107.8  67.9  39.9 kPa
Therefore the difference between the pressures on the lower and upper surfaces is
39.9 kPa.
204
PROBLEM 6.4
Air flows over a wall at a supersonic velocity. The wall turns towards the flow
generating an oblique shock wave. This wave is found to be at angle of 50° to the initial
flow direction. A scratch on the wall upstream of the shock wave is found to generate a
very weak wave that is at angle of 30° to the flow. Find the angle through which the wall
turned.
SOLUTION
The scratch is assumed to generate a Mach wave. Therefore, the Mach angle α in the
flow ahead of the oblique shock wave is 30° . The Mach number in the flow ahead of the
oblique shock wave is therefore given by:
M1 
1
1

 2
sin 
sin 30o
Hence, the Mach number upstream of the shock wave is Ml = 2.0 and the shock angle
β = 50°. Using the software for an oblique shock wave or the oblique shock wave gives
for these values:
  18.1o
Therefore the wall turns through an angle of 18.1 °.
205
PROBLEM 6.5
Air flows over a plane wall at a Mach number of 3.5, the pressure in the flow being
100 kPa. The wall turns through an angle leading to the generation of an oblique shock
wave whose strength is such that the pressure downstream of the corner is 548 kPa. Find
the turning angle of the corner.
SOLUTION
The pressure ratio across the oblique shock wave is given by:
p2
548

 5.48
p1
100
Hence, the Mach number upstream of the shock wave is M1 = 3.5 and the pressure
ratio across the oblique shock wave is 5.48. Now for a normal shock wave with a
pressure ratio of 5.48 the software or the normal shock tables for air ( i.e., for γ = 1.4 )
gives for these values the Mach number ahead of the shock wave as 2.20. Hence:
M N 1  2.20
i.e.:
M 1 sin   2.2 , i.e., 3.5sin   2.2
from which it follows that:
  38.95o
Hence, the Mach number upstream of the shock wave is M1 = 3.5 and the shock angle
β = 38.95°. Using the software for an oblique shock wave or the oblique shock wave
graph for air ( i.e., for γ = 1.4 ) gives for these values:
  23.6
Therefore the turning angle of the corner is 23.6°.
206
PROBLEM 6.6
Air, flowing down a plane walled duct at a Mach number of 3, passes over a wedge.
What is the largest included angle that this wedge can have if the oblique shock wave that
is generated is attached to the wedge. Sketch the flow pattern that will exist if the wedge
angle is greater than this maximum value.
SOLUTION
For an upstream Mach number M1 = 3 by entering an arbitrary turning angle δ of say
5°, the software gives the maximum possible turning angle that will give an attached
shock wave as 34.07°. This value can also be obtained using the oblique shock wave
chart for air ( γ = 1.4 ). The maximum included angle that the wedge can have and still
give an attached shock wave is therefore 2  34.07 = 68.14°. If the wedge angle is
greater than this value, a curved detached shock wave such as that shown in Fig. P6.6 will
be formed.
Figure P6.6
207
PROBLEM 6.7
A uniform air flow at a Mach number of 2.5 passes around a sharp concave corner in
the wall that turns the flow through an angle of 10° and leads to the generation of an
oblique shock wave. The pressure and temperature in the flow upstream of the corner are
70 kPa and 10° C respectively. Find the Mach number, the pressure, the temperature and
the stagnation pressure downstream of the oblique shock wave. How large would the
corner angle have to be before the shock became detached from the corner?
SOLUTION
The Mach number upstream of the shock wave is M1 = 2.5 and the flow is turned
through an angle δ = 10°. Using the software for an oblique shock wave or the oblique
shock chart and the normal shock tables for air ( i.e., for γ = 1.4 ) gives for these values:
M 2  2.086 ,
 max  29.8°
and:
p2
 1.863 ,
p1
T2
 1.203 ,
T1
p02
 0.9759
p01
Also, considering the flow upstream of the shock wave where the Mach number is
2.5, the software for isentropic flow or the table for isentropic flow of air ( i.e., for γ =
1.4 ) gives for this Mach number value:
p01
 17.09
p1
Using the above results gives:
p2 
p2
p1  1.8633  70  130.4 kPa ,
p1
and:
208
T2 
T2
T1  1.203  283  340.5 K
T1
p02 
p02 p01
p1  0.9759  17.09  70  1168 kPa
p01 p1
Therefore the Mach number, pressure, temperature and stagnation pressure behind the
shock wave are 2.086, 130.5 kPa, 340.5 K ( = 67.5° C ) and 1168 kPa respectively. If the
corner angle is greater than 29.8° the shock wave will be detached.
209
PROBLEM 6.8
Find the minimum values of the Mach number for which the oblique shock wave
generated at the leading edge of a wedge placed in a supersonic air flow remains attached
to the wedge for deflection angles of 15°, 25° and, 40°.
SOLUTION
Figure P6.8
As indicated schematically in Fig. P6.8, the Mach number corresponding to any
specified maximum deflection angle, δmax , can be obtained directly from an oblique
shock chart. For the three specified values of δmax the values of M1 shown in the
following table are obtained in this way.
δmax
15o
25o
40o
M1
1.61
2.15
4.50
210
PROBLEM 6.9
A wedge shaped body symmetrically placed in a supersonic air stream is to be used to
determine the Mach number in the flow. This will be done by using optical methods to
measure the angle that the oblique shock wave attached to the leading edge of the wedge
makes to the upstream flow. If the total included angle of the wedge that is to be used is
45°, find the Mach number range over which this method can be used.
SOLUTION
The flow is turned through an angle of 22.5° by the wedge because it is placed
symmetrically with respect to the approaching flow. Using the same procedure as in
Problem 6.8, the oblique shock chart can be used to find the value of M1 that has a
maximum turning angle of 22.5°. This procedure gives M1 = 1.98. If the wedge is placed
in a flow with a Mach number that is less than this value, the shock wave will be
detached and it will be difficult to determine the Mach number. Hence, this method can
be used provided the flow Mach number is greater than 1.98.
211
PROBLEM 6.10
Air flowing at a Mach number of 2.5 passes over a wedge that turns the flow through
an angle of 5° . Find the pressure ratio across the oblique shock wave that is generated. If
this oblique shock wave is reflected off a plane surface, find the overall pressure ratio.
SOLUTION
The situation being considered is shown in Fig. P6.10.
Figure P6.10
The conditions downstream of the initial wave, i.e., in region 2, are first obtained. The
Mach number upstream of this incident shock wave is M1 = 2.5 and the flow is turned
through an angle δ = 5°. Using the software for an oblique shock wave or the oblique
shock chart and the normal shock tables for air ( i.e., for γ = 1.4 ) gives for these values:
M 2 = 2.292 ,
212
p2
 1.379
p1
Conditions behind the reflected wave, i.e., in region 3 will next be derived by
considering the changes from region 2 to region 3. The reflected wave also turns the flow
through 5° and the Mach number ahead of this wave is 2.292, i.e., for the reflected wave:
M 2 = 2.292 ,
  5o
Using the software for an oblique shock wave or the oblique shock chart and the
normal shock tables for air ( i.e., for γ = 1.4 ) gives for these values:
p3
 1.352
p2
The overall pressure ratio across the reflected shock wave is then given by:
p3
p
p
 2  3  1.379  1.352  1.864
p1
p1
p2
Therefore the pressure ratio across the incident wave is 1.379 and the overall pressure
ratio across the reflected wave system is 1 .864.
213
PROBLEM 6.11
An oblique shock wave with a wave angle of 26° in an air stream in which the Mach
number is 2.7, the pressure is 100 kPa and the temperature is 30° C impinges on a straight
wall. Find the Mach number, pressure, temperature and stagnation pressure downstream
of the reflected wave.
SOLUTION
The Mach number in the initial flow is 2.7. For this value of Mach number, the
software for isentropic flow or the isentropic flow tables for air ( γ = 1.4 ) give:
p0
 23.28
p1
Therefore the stagnation pressure in the initial flow is given by:
p1 
p0
p1  23.28  100  2328 kPa
p1
The conditions downstream of the initial wave are next obtained. The Mach number
upstream of this incident shock wave is Ml = 2.7 and the wave angle is β = 26°. Using the
software for an oblique shock wave or the oblique shock chart and the normal shock
tables for air ( i.e., for γ = 1.4 ) gives for these values:
M 2 = 2.449 ,   5.624o ,
p2
 1.469 ,
p1
T2
 1.469 ,
T1
p02
 0.9942
p01
Conditions behind the reflected wave will next be derived. This reflected wave also
turns the flow through 5.624° and the Mach number ahead of this wave is 2.449, i.e., for
the reflected wave:
M 2 = 2.449 ,   5.624o
214
Using the software for an oblique shock wave or the oblique shock chart and the
normal shock tables for air ( i.e., for γ = 1.4 ) gives for these values:
M 3 = 2.219 ,
p3
 1.425 ,
p2
T3
 1.108 ,
T2
p03
 0.9955
p02
Therefore:
p3 
p
p2
 3  p1  1.469  1.425  100  209.3 kPa
p1
p2
and:
T3 
T
T2
 3  T1  1.118  1.108  303  375.3 K
T1
T2
and:
p03 
p02
p
 03  p01  0.9942  0.9955  2328  2304 kPa
p01
p02
Therefore the Mach number, pressure, temperature and stagnation pressure behind the
reflected shock wave are 2.219, 209.3 kPa, 375.3 K ( = 102.3° C ) and 2304 kPa
respectively.
215
PROBLEM 6.12
Air flows down a duct at a Mach number of 1.5. The top wall·of the duct turns
towards the flow leading to the generation of an oblique shock wave which strikes the flat
lower wall of the duct and is reflected from it. What is the smallest turning angle that
would give a Mach reflection off the lower wall?
SOLUTION
A Mach reflection occurs when the maximum turning angle corresponding to the
Mach number downstream of the incident wave, i.e., M2 ,is not enough to bring the flow
in region 3 parallel to the wall, i.e., is less than the turning angle produced by the upper
wall.
A simple iterative type approach will be adopted here. A turning angle will be
assumed. The maximum turning angle that can be produced by the reflected wave will be
calculated and the difference between this maximum value and the required turning angle
will be calculated. This is termed Δδ in the following table. When Δδ is negative, a Mach
reflection will exist. The turning angle that makes Δδ = 0 can then be deduced. To
illustrate how the results in the table are derived, consider the case where δ is assumed to
be 5°. Now for:
M 1  1.5 and   5o
the software for an oblique shock wave or the oblique shock chart and the normal shock
tables for air ( i.e., for γ = 1.4 ) give:
M 2  1.325
Then for an initial Mach number of 1.325 the software or the chart for an oblique
shock wave for air ( i.e., for γ = 1.4 ) give δmax = 7.355°. Hence:
    max    7.355  5  2.355o
Using this approach, the values listed in the following table have been derived.
216
δ – Degrees
(Guessed)
10
3
5
7
6
6.5
6.2
6.1
M2
(Software or
Chart/Table)
1.114
1.397
1.325
1.249
1.288
1.269
1.280
1.284
δmax – Degrees
(Software or
Chart)
1.823
9.345
7.355
5.259
6.330
5.806
6.109
6.220
Δδ – Degrees
(Calculated)
-8.886
+6.345
+2.355
-1.741
+0.330
-0.194
-0.091
+0.120
Interpolating between the results given in the above table then indicates that Δδ = 0
approximately when δ =6.15°. Therefore if the turning angle produced by the wedge is
more than about 6.15°, a Mach reflection will occur.
217
PROBLEM 6.13
A two-dimensional wedge with an included angle of 10° is placed in a wind tunnel
which has parallel walls. If the Mach number in the freestream ahead of the wedge is 2,
and if the center-line of the wedge is inclined at an angle of 2° to the direction of the air
flow, find the Mach numbers upstream and downstream of the oblique shock waves after
they are reflected off the upper and lower walls of the tunnel. Also sketch the flow pattern
marking the angles the waves make to the tunnel walls.
SOLUTION
As shown in Fig. P6.13a, the flow over the upper surface of the wedge, which has a
half angle of 5°, is turned through an angle of 3° while the flow over the lower surface is
turned through an angle of 7°.
Figure P6.13a
First consider the flow over the upper surface of the wedge. For this surface, the
Mach number upstream of the shock wave is Ml = 2 and the flow is turned through an
angle δ = 3°. Using the software for an oblique shock wave or the oblique shock chart
and the normal shock tables for air ( i.e., for γ = 1.4 ) gives for these values:
218
M 2U  1.892 and 1U  32.5o
where the subscript U denotes the upper surface.
Now consider the reflected wave from the upper surface. This wave also turns the
flow 3°. Thus, for this wave, the Mach number upstream of the shock wave is 1.892 and
the flow is turned through an angle δ = 3°. Using the software for an oblique shock wave
or the oblique shock chart and the normal shock tables for air ( i.e., for γ = 1.4 ) gives for
these values:
M 3U  1.787 and  2U  34.5o
Next consider the flow over the lower surface. For this surface, the Mach number
upstream of the shock wave is M1 = 2 and the flow is turned through an angle δ = 7°.
Using the software for an oblique shock wave or the oblique shock chart and the normal
shock tables for air ( i.e., for γ = 1.4 ) gives for these values:
M 2L  1.750 and 1L  34.5o
where the subscript L denotes the upper surface.
Now consider the reflected wave from the upper surface. This wave also turns the
flow through 7o. Thus, for this wave, the Mach number upstream of the shock wave is
1.750 and the flow is turned through an angle δ = 7°. Using the software for an oblique
shock wave or the oblique shock chart and the normal shock tables for air ( i.e., for γ =
1.4 ) gives for these values:
M 3L  1.509 and  2L  41.9o
Therefore the Mach numbers behind the reflected shock waves from the upper and
lower surfaces of the wedge are 1.787 and 1.509 respectively. Noting that the shock angle
β is measured relative to the direction of the flow upstream of the wave, the wave angles
relative to the walls will be as indicated in Fig. P6.13b.
219
Figure P6.13b
220
PROBLEM 6.14
Air in which the pressure is 60 kPa is flowing down a plane walled duct at a Mach
number of 2.5. The air-stream passes over a wedge with an included angle of 10° . The
oblique shock wave that is generated by the wedge is reflected off the flat wall of the
duct. Find the pressure and Mach number after the reflection.
SOLUTION
It is assumed that the wedge is symmetrically placed in the flow so the shock waves
produced by the wedge both turn the flow through 5°. The conditions downstream of the
initial wave are first obtained. The Mach number upstream of this incident shock wave is
M1 = 2.5 and the turning angle is δ = 5°. Using the software for an oblique shock wave or
the oblique shock chart and the normal shock tables for air ( i.e., for γ = 1.4 ) gives for
these values:
M 2 = 2.292 ,
p2
 1.379
p1
Conditions behind the reflected wave will next be derived. This reflected wave also
turns the flow through 5o and the Mach number ahead of this wave is 2.292, i.e., for the
reflected wave:
M 2 = 2.292 ,
  5o
Using the software for an oblique shock wave or the oblique shock chart and the
normal shock tables for air ( i.e., for γ = 1.4 ) gives for these values:
M 3 = 2.098 ,
p3
 1.352
p2
Hence:
p3 
p3
p
 2  p1  1.379  1.352  60  111.9 kPa
p2
p1
221
Therefore the pressure and Mach number behind the reflected shock wave are 111.9
kPa and 2.098 respectively.
222
PROBLEM 6.15
Air at a pressure and temperature of 40 kPa and -30° C flows at a Mach number of 3
down a wide duct. The upper wall of the duct turns sharply through an angle of 5° leading
to the formation of an oblique shock wave as shown in Fig. P6.15a. Find the Mach
number, temperature and pressure behind this shock wave. As shown in the figure, this
shock wave strikes the lower wall of the duct exactly at a point where the lower wall
turns away from the flow through an angle of 2°. Find the Mach number, pressure and
temperature behind the “reflected” wave.
Figure P6.15a
SOLUTION
Here, the Mach number upstream of the initial shock wave is M1 = 3 and the flow is
turned through an angle δ = 5°. Using the ·software for an oblique shock wave or the
oblique shock chart and the normal shock tables for air ( i.e., for γ = 1.4 ) gives for these
values:
M 2 = 2.750 ,
p2
 1.455 ,
p1
Using these results gives:
223
T2
 1.115
T1
p2 
p2
 p1  1.455  40  58.2 kPa
p1
T2 
T2
 T1  1.115  243  271.0 K
T1
and:
Therefore, the Mach number, temperature and pressure behind the initial shock wave
are 2.750, 271.0 K ( = -2° C) and 58.2 kPa respectively.
Next consider the “reflected” shock wave. The flow downstream of this wave must be
parallel to the wall. Hence, as shown in Fig. P6.15b, this wave produces a turning angle
of 5° - 2° = 3°.
Figure P6.15b
Hence, the Mach number upstream of the reflected shock wave is M2 = 2.75 and the
reflected wave turns the flow through an angle δ = 3°. Using .the software for an oblique
shock wave or the oblique shock chart and the normal shock tables for air ( i.e., for γ =
1.4 ) gives for these values:
M 3 = 2.612 ,
p3
 1.235 ,
p2
224
T3
 1.062
T2
Using these results gives:
p3 
p3
 p2  1.235  58.2  71.88 kPa
p2
and:
T3 
T3
 T2  1.062  271  287.8 K
T2
Therefore the Mach number, temperature and pressure behind the “reflected” shock
wave are 2.612, 287.8 K ( = 14.8°C ) and 71.88 kPa respectively.
225
PROBLEM 6.16
Find the pressure ratio p3 / p1 for the flow situation shown in Fig.P6.16.
Figure P6.16
SOLUTION
The Mach number upstream of the initial shock wave is M1 = 3 and the flow is turned
through an angle δ = 4°. Using the software for an oblique shock wave or the oblique
shock chart and the normal shock tables for air ( i.e., for γ = 1.4 ) gives for these values:
M 2 = 2.799 ,
p2
= 1.352
p1
Next consider the “reflected” shock wave. The flow downstream of this wave must be
parallel to the wall so this wave produces a turning angle of 4° + 1° = 5°. Hence, since
the Mach number upstream of the reflected shock wave is M = 2.799 and the flow is
turned through an angle δ = 5° . Using the software for an oblique shock wave or the
oblique shock chart and the normal shock tables for air ( i.e., for γ = 1.4 ) gives for these
values:
p3
= 1.422
p2
226
Hence:
p3
p
p
= 3  2 = 1.352  1.422 = 1.923
p1
p2
p1
Therefore the pressure ratio p3 / p1 across the shock wave system is 1.923.
227
PROBLEM 6.17
Find the pressure ratio p4 / p1 for the flow situation shown in Fig. P6.17.
Figure P6.17
SOLUTION
Air flow will be assumed. The conditions behind the two initial shock waves will first
be considered, i.e., the conditions in regions 2 and 3 shown in Fig. P6.17 will first be
derived. First consider region 2. Using the software for oblique shock waves or the
oblique shock chart and the normal shock tables for air ( i.e., for γ = 1.4 ) for M =2.8 and
δ = 7° gives:
M 2 = 2.477 ,
p2
= 1.627
p1
Next consider region 3. Using the software for oblique shock waves or the oblique
shock chart and the normal shock tables for air ( i.e., for γ = 1.4 ) for M =2.8 and δ = 5 °
gives:
M 3 = 2.613 ,
p3
= 1.328
p1
Next consider region 4. The strengths of the shock waves after the intersection must
be such that the pressure and the flow direction is the same throughout this region. It is
228
convenient to consider two parts of region 4: (a) Region 42 which is downstream of
Region 2 and (b) Region 43 which is downstream of Region 3. The solution must be such
that the flow directions and pressures in Regions 42 and 43 are the same. While there are
elegant ways of obtaining the solution, a very simple approach will be adopted here. A
flow direction that is the same in both Regions 42 and 43 will be assumed. This direction
will be specified by the angle Δ which is the flow direction relative to the flow upstream
of the initial shock waves. The turning angle produced by the oblique shock wave
between Regions 2 and 42 is given by:
 = 7o - 
Similarly, the turning angle produced by the oblique shock wave between Regions 3
and 43 is given by:
 = 4o + 
For this chosen flow direction, the pressure ratios in the two regions, i.e., p42 / p1 and
p43 / p1 can then be calculated. These two pressure ratios will not in general be the same
because the value of Δ has been guessed. The procedure is repeated for several different
values of Δ and the value of Δ that makes p42 / p1 = p43 / p1 can then be deduced.
To illustrate the procedure, consider the case where the guessed value of Δ is 1o. In
this case the turning angle between region 2 and region 42 is δ = 6° and, it will be
recalled, M2 = 2.477. Using the software for an oblique shock wave or the oblique shock
chart and the normal shock tables for air ( i.e., for γ = 1.4 ) for this value of M and δ
gives:
p42
= 1.463
p2
Next consider the change between region 3 and region 43. For Δ = 1°, the turning
angle between region 3 and region 43 is δ = 5° and, it will be recalled, M2 = 2.613. Using
the software for an oblique shock wave or the oblique shock chart and the normal shock
tables for air ( i.e., for γ = 1.4 ) for this value of M and δ gives:
229
p43
= 1.395
p3
Using these results then gives:
p42
p
p
= 42  2 = 1.463  1.627 = 2.380
p1
p2
p1
and:
p43
=
p1
p43
p
 3 = 1.395  1.328 = 1.853
p3
p1
This procedure is repeated for several other values of Δ giving the following results:
Δ- degrees
(Chosen)
0
1
2
3
p42
p2
1.556
1.463
1.376
1.294
p43
p3
1.307
1.395
1.488
1.564
p42
p1
2.532
2.380
2.239
2.105
p43
p1
1.736
1.853
1.976
2.104
It will be seen from the results given in the above table that p42 / p1 = p43 / p1
approximately when Δ = +3° and that:
p4
= 2.105
p1
Therefore the overall pressure ratio is 2.105.
230
PROBLEM 6.18
If the Mach number and pressure ahead of the oblique shock wave system shown in
Fig. P6.18 are 3 and 50 kPa respectively, find the pressure in the region 4 downstream of
the wave intersection.
Figure P6.18
SOLUTION
It will be assumed that air flow is involved. The conditions behind the two initial
shock waves will first be considered, i.e., the conditions in the regions 2 and 3 will first
be derived.
First consider region 2. Using the software for oblique shock waves or the oblique
shock chart and the normal shock tables for air ( i.e., for γ = 1.4 ) for M =3 and δ = 9 °
gives:
M 2 = 2.554 ,
p2
= 1.922
p1
Next consider region 3. Using the software for oblique shock waves or the oblique
shock chart and the normal shock tables for air ( i.e., for γ = 1.4 ) for M =3 and δ = 5 °
gives:
231
M 3 = 2.750 ,
p3
= 1.455
p1
Next consider region 4. The strengths of the shock waves after the intersection must
be such that the pressure and the flow direction is the same throughout this region. It is
convenient to consider two parts of region 4: (a) Region 42 which is downstream of
Region 2 and (b) Region 43 which is downstream of Region 3. The solution must be such
that the flow directions and pressures in Regions 42 and 43 are the same. While there are
elegant ways of obtaining the solution, a very simple approach will be adopted here. A
flow direction that is the same in both Regions 42 and 43 will be assumed. This direction
will be specified by the angle Δ which is the flow direction relative to the flow upstream
of the initial shock waves. The turning angle produced by the oblique shock wave
between Regions 2 and 42 is given by:
 = 9o - 
Similarly, the turning angle produced by the oblique shock wave between Regions 3
and 43 is given by:
 = 5o + 
For this chosen flow direction, the pressure ratios in the two regions, i.e., p42 / p1 and
p43 / p1 can then be calculated and the values of p42 and p43 can then be found. These two
pressures will not in general be the same because the value of Δ has been guessed. The
procedure is repeated for several different values of Δ and the value of Δ that makes p42 =
p43 = p4 can then be deduced.
To illustrate the procedure, consider the case where the guessed value of Δ is 3°. In
this case the turning angle between region 2 and region 42 is δ = 6° and, it will be
recalled, M2 = 2.554. Using the software for an oblique shock wave or the oblique shock
chart and the normal shock tables for air ( i.e., for γ = 1.4 ) for these values of M and δ
gives:
p42
= 1.477
p2
232
Next consider the change between region 3 and region 43. For Δ = 1°, the turning
angle between region 3 and region 43 is δ = 8° and, it will be recalled, M2 = 2.750. Using
the software for an oblique shock wave or the oblique shock chart and the normal shock
tables for air ( i.e., for γ = 1.4 ) for this value of M and δ gives:
p43
= 1.455
p3
Using these results then gives:
p42 =
p42
p
 2  p1 = 1.477  1.922  50 = 141.9 kPa
p2
p1
p43 =
p43
p
 3  p1 = 1.724  1.455  50 = 125.4 kPa
p3
p1
and:
This procedure is repeated for several other values of Δ giving the following results:
Δ- degrees
(Chosen)
+2
+3
+4
+5
p42
p2
1.573
1.477
1.387
1.302
p43
p3
1.404
1.724
1.839
1.956
p42
p43
kPa
151.2
141.9
133.3
125.1
kPa
102.1
125.4
133.8
142.2
It will be seen from the results given in the above table that p42 = p43 approximately
when Δ = +3.9° and that p4 = 133.5 kPa.
Therefore the pressure in region 4 is 133.5 kPa.
233
PROBLEM 6.19
Air is flowing at a Mach number of 2 and a pressure of 70 kPa in a two-dimensional
channel. The upper wall turns towards the flow through an angle of 5° and the lower wall
turns towards the flow through an angle of 3°, two oblique shock waves thus being
generated. These two shock waves intersect each other. Find the pressure in the region
just downstream of the shock intersection.
SOLUTION
The flow situation being considered is shown in Fig. P6.19.
Figure P6.19
The conditions behind the two initial shock waves will first be considered, i.e., the
conditions in the regions 2 and 3 will first be derived.
First consider region 2. Using the software for oblique shock waves or the oblique
shock chart and the normal shock tables for air ( i.e., for γ = 1.4 ) for M = 2 and δ = 5°
gives:
M 2 = 1.821 ,
234
p2
= 1.315
p1
Next consider region 3. Using the software for oblique shock waves or the oblique
shock chart and the normal shock tables for air ( i.e., for γ = 1.4 ) for M =2 and δ = 3°
gives:
M 3 = 1.892 ,
p3
= 1.182
p1
Next consider region 4. The strengths of the shock waves after the intersection must
be such that the pressure and the flow direction is the same throughout this region. It is
convenient to consider two parts of region 4: (a) Region 42 which is downstream of
Region 2 and (b) Region 43 which is downstream of Region 3. The solution must be such
that the flow directions and pressures in Regions 42 and 43 are the same. While there are
elegant ways of obtaining the solution, a very simple approach will be adopted here. A
flow direction that is the same in both Regions 42 and 43 will be assumed. This direction
will be specified by the angle Δ which is the flow direction relative to the flow upstream
of the initial shock waves. The turning angle produced by the oblique shock wave
between Regions 2 and 42 is given by:
 = 5o - 
Similarly, the turning angle produced by the oblique shock wave between Regions 3
and 43 is given by:
 = 3o + 
For this chosen flow direction, the pressure ratios in the two regions, i.e., p42 / p1 and
p43 / p1 can then be calculated and the values of p42 and p43 can then be found. These two
pressures will not in general be the same because the value of Δ has been guessed. The
procedure is repeated for several different values of Δ and the value of Δ that makes p42 =
p43 = p4 can then be deduced.
To illustrate the procedure, consider the case where the guessed value of Δ is 1°. In
this case the turning angle between region 2 and region 42 is δ = 4° and, it will be
recalled, M2 = 1.821. Using the software for an oblique shock wave or the oblique shock
235
chart and the normal shock tables for air ( i.e., for γ = 1.4 ) for these values of M and δ
gives:
p42
= 1.232
p2
Next consider the change between region 3 and region 43. For Δ = 1°, the turning
angle between region 3 and region 43 is δ = 4° and, it will be recalled, M2 = 1.892. Using
the software for an oblique shock wave or the oblique shock chart and the normal shock
tables for air ( i.e., for γ = 1.4 ) for this value of M and δ gives:
p43
= 1.237
p3
Using these results then gives:
p42 =
p42
p
 2  p1 = 1.232  1.315  70 = 113.4 kPa
p2
p1
p43 =
p43
p
 3  p1 = 1.237  1.182  70 = 102.3 kPa
p3
p1
and:
This procedure is repeated for several other values of Δ giving the following results:
Δ- degrees
(Chosen)
+1.0
+1.5
+2.0
+2.5
p42
p2
1.232
1.202
1.172
1.139
p43
p3
1.237
1.271
1.305
1.339
p42
p43
kPa
113.4
110.6
107.9
104.9
kPa
102.3
105.2
108.0
110.8
It will be seen from the results given in the above table that p42 = p43 approximately
when Δ = +2° and that p4 = 108 kPa.
Therefore the pressure in region 4 is 108 kPa.
236
PROBLEM 6.20
An air stream in which the Mach number is 3 and the pressure is 80 kPa flows
between two parallel walls. The upper wall turns sharply through an angle of 18° and the
lower wall turns sharply through an angle of 12° leading to the generation of two oblique
shock waves which intersect each other. Sketch the flow pattern and find the flow
direction, the Mach number and the pressure immediately downstream of the shock
intersection.
SOLUTION
The flow situation being considered is shown in Fig. P6.20a.
Figure P6.20a
The conditions behind the two initial shock waves will first be considered, i.e., the
conditions in the regions 2 and 3 will first be derived.
First consider region 2. Using the software for oblique shock waves or the oblique
shock chart and the normal shock tables for air ( i.e., for γ = 1.4 ) for M = 3 and δ = 18o
gives:
M 2 = 2.100 ,
237
p2
= 3.270
p1
Next consider region 3. Using the software for oblique shock waves or the oblique
shock chart and the normal shock tables for air ( i.e., for γ = 1.4 ) for M =3 and δ = 12o
gives:
M 3 = 2.406 ,
p3
= 2.341
p1
Next consider region 4. The strengths of the shock waves after the intersection must
be such that the pressure and the flow direction is the same throughout this region. It is
convenient to consider two parts of region 4: (a) Region 42 which is downstream of
Region 2 and (b) Region 43 which is downstream of Region 3. The solution must be such
that the flow directions and pressures in Regions 42 and 43 are the same. While there are
elegant ways of obtaining the solution, a very simple approach will be adopted here. A
flow direction that is the same in both Regions 42 and 43 will be assumed. This direction
will be specified by the angle Δ which is the flow direction relative to the flow upstream
of the initial shock waves. The turning angle produced by the oblique shock wave
between Regions 2 and 42 is given by:
 = 18o - 
Similarly, the turning angle produced by the oblique shock wave between Regions 3
and 43 is given by:
 = 12o + 
For this chosen flow direction, the pressure ratios in the two regions, i.e., p42 / p1 and
p43 / p1 can then be calculated and the values of p42 and p43 can then be found. These two
pressures will not in general be the same because the value of Δ has been guessed. The
procedure is repeated for several different values of Δ and the value of Δ that makes p42 =
p43 = p4 can then be deduced.
To illustrate the procedure, consider the case where the guessed value of Δ is 4o In
this case the turning angle between region 2 and region 42 is δ = 14o and, it will be
recalled, M2 = 2.100. Using the software for an oblique shock wave or the oblique shock
238
chart and the normal shock tables for air ( i.e., for γ = 1.4 ) for these values of M and δ
gives:
M 42 = 1.578 and
p42
= 2.130
p2
Next consider the change between region 3 and region 43. For Δ = 4o the turning
angle between region 3 and region 43 is δ = 16o and, it will be recalled, M2 = 2.406.
Using the software for an oblique shock wave or the oblique shock chart and the normal
shock tables for air ( i.e., for γ = 1.4 ) for this value of M and δ gives:
M 43 = 1.755 and
p43
= 2.539
p3
Using these results then gives:
p42 =
p42
p
 2  p1 = 2.130  3.370  80 = 574.3 kPa
p2
p1
p43 =
p43
p
 3  p1 = 2.539  2.341  80 = 416.1 kPa
p3
p1
and:
This procedure is repeated for several other values of Δ giving the following results:
Δ- degrees
(Chosen)
+4.0
+5.0
+5.8
+6.0
p42
p43
kPa
574.3
545.4
524.1
518.2
kPa
416.1
501.2
521.8
527.4
M 42
M 43
1.578
1.617
1.649
1.656
1.756
1.711
1.674
1.666
Interpolating between the results given in the above table then indicates that p42 =
p43 approximately when Δ = +5.85o and that p4 = 523 kPa. Under these circumstances,
M42 = 1.65 and M43 = 1.67. The flow direction is shown in Fig. P6.20b.
239
Figure P6.20b
240
PROBLEM 6.21
Consider an air stream flowing at a Mach number of 3.2 with a pressure of 60 kPa.
Consider two cases. In the first case, the stream passes through a single normal shock
wave and is then isentropically decelerated to a very low velocity. In the second case, the
flow first passes through an oblique shock that turns the flow through 25° and then passes
through a normal shock wave before being isentropically decelerated to a very low
velocity. Compare the pressures attained in the two cases. Do the results indicate which
arrangement should be used in decelerating a flow from supersonic to subsonic velocities
at the inlet to a turbojet engine in a supersonic aircraft?
SOLUTION
The Mach number is 3.2 in the flow upstream of the wave system. The software for
isentropic flow or the isentropic flow tables for air ( i.e., for γ = 1.4 ) gives for these
values:
p01
= 49.22
p1
Consider the first case where there is a single normal shock wave. For a normal shock
wave with an upstream Mach number of 3.2 the software or the normal shock tables give:
p02
= 0.2762
p01
Because the flow is isentropically decelerated downstream of the shock wave, p02 will
be the pressure attained as a result of the deceleration. It is given by:
p02 =
p02
p
 01  p1 = 0.2762  49.22  60 = 815.6 kPa
p01
p1
Next consider the second case where there is an oblique shock wave followed by a
normal shock wave. The Mach number upstream of the oblique shock wave is 3.2 and the
flow is turned through an angle δ = 25° Using the software for an oblique shock wave or
241
the oblique shock chart and the normal shock tables for air ( i.e., for γ = 1.4 ) gives for
these values:
M 2 = 1.830 ,
p02
 0.6446
p01
Now consider the normal shock wave that occurs downstream of the oblique shock
wave. For a normal shock wave with an upstream Mach number of 1.830 the software or
the oblique shock chart and the normal shock tables for air ( i.e., for γ = 1.4 ) give:
p03
= 0.7993
p02
Because the flow is isentropically decelerated downstream of the shock wave system
p03 will be the pressure attained as a result of the deceleration. It is given by:
p03 =
p03
p
p
 02  01  p1 = 0.7993  0.6446  49.22  60 = 1521.6 kPa
p02
p01
p1
Therefore a pressure of 815.6 kPa is attained with the first arrangement while a
pressure of 1521.6 kPa is attained with the second arrangement. The maximum attainable
pressure would be p01 which is equal to 49.22  60 = 2953.2 kPa. The first arrangement
has a “pressure recovery factor” of 815.6 / 2953.2 = 0.276 ( 27.6 % ) while the second
arrangement has a “pressure recovery factor” of 1521.6/2953.2 = 0.515 ( 51.5 % ) .
Because as high a pressure as possible is required at the entrance to the engine, these
results indicate that the use of a diffuser that involves oblique shock waves is preferable
to the use of a diffuser that involves a single normal shock wave.
242
PROBLEM 6.22
A ram-jet engine is fitted to a small aircraft that cruises at a Mach number of 4 at an
altitude where the pressure is 30 kPa and the temperature is -45°C. The air entering the
engine is slowed to subsonic velocities by passing it through two oblique shock waves
each of which turn the flow through 15° and by then passing it through a normal shock
wave. Following the normal shock, the flow is isentropically decelerated to a Mach
number of 0.1 before it enters the combustion zone. Find the values of the pressure and
the temperature at the inlet to the combustion zone. What values would have been
attained if initial deceleration had been through a single normal shock wave instead of
through the combination of oblique shocks and a normal shock?
SOLUTION
The Mach number upstream of the first oblique shock wave is 4.0 and the flow is
turned through an angle δ = 15°. Using the software for an oblique shock wave or the
oblique shock chart and the normal shock tables for air ( i.e., for γ = 1.4 ) gives for these
values:
M 2 = 2.929 ,
p2
= 3.698 ,
p1
T2
= 1.547
T1
The Mach number upstream of the second oblique shock wave is 2.929 and the flow
is turned through an angle δ = 15° . Using the software for an oblique shock wave or the
oblique shock chart and the normal shock tables for air ( i.e., for γ = 1.4 ) gives for these
values:
M 3 = 2.203 ,
p3
= 2.768 ,
p2
T3
= 1.378
T2
Now consider the normal shock wave. For a normal shock wave with an upstream
Mach number of 2.203 the software for a normal shock wave or the normal shock tables
for air ( i.e., for γ = 1.4 ) give:
243
M 4 = 0.5467 ,
p4
= 5.495 ,
p3
T4
= 1.860
T3
The flow is isentropically decelerated downstream of the normal shock wave until the
Mach number is 0.1. For a Mach number M4 of 0.5467, the software or the isentropic
flow tables for air ( i.e., for γ = 1.4 ) give:
p04
= 1.225 ,
p4
T04
= 1.060
T4
and for a Mach number M5 of 0.1, the software or the isentropic flow tables for air ( i.e.,
for γ = 1.4 ) give:
p05
= 1.007 ,
p5
T05
= 1.002
T5
Because the flow downstream of the shock wave system is isentropic p05 = p04 and
T05 = T04, the stagnation temperature, in fact, being the same everywhere in the flow.
Hence:
p5 
p
p
p
p2
p
 3  4  04  5  p1
p1
p2
p3
p4
p05
i.e.:
p5 = 3.698  2.768  5.495  1.225 
1
 30 = 2053 kPa
1.007
Similarly:
T5 =
T
T
T
T2
T
 3  4  04  5  T1
T1
T2
T3
T4
T05
i.e.:
T5 = 1.547  1.378  1.860  1.060 
1
 228 = 956 K
1.002
This result could have been obtained somewhat more simply by recalling that, as
noted above, the stagnation temperature does not change across a shock wave so T05 = T01
244
The above results show that the pressure and temperature at the entrance to the
combustion system are 2053 kPa and 956 K ( = 683° C) respectively.
Next consider the situation where the deceleration takes place through a single normal
shock wave. For a normal shock wave with an upstream Mach number of 4 the software
or the isentropic flow tables for air ( i.e., for γ = 1.4 ) give:
M 2 = 0.435 ,
p2
= 18.506 ,
p1
T2
= 4.0476
T1
The flow is again isentropically decelerated downstream of the normal shock wave
until the Mach number is 0.1. For a Mach number M2 of 0.435, the software for isentropic
flow or the isentropic flow tables for air ( i.e., for γ = 1.4 ) give:
p02
= 1.139 ,
p2
T02
= 1.038
T2
and for a Mach number of 0.1, the software or the isentropic flow tables for air ( i.e., for γ
= 1.4 ) give as noted before for isentropic flow:
p03
= 1.007 ,
p3
T03
= 1.002
T3
Because the flow downstream of the shock wave system is isentropic, p03 = p02 and
T03 = T02 ( = T01 ). Hence:
p3 =
p
p
p2
1
 02  3  p1 = 18.50  1.139 
 30 = 627.8 kPa
p1
p2
p03
1.007
and similarly:
T3 =
T
T
T2
1
 02  3  T1 = 4.407  1.038 
 228 = 956 K
T1
T2
T03
1.002
The above results show that when there is only a single normal shock wave the
pressure and temperature at the entrance to the combustion system are 627.8 kPa and 956
K ( = 683° C) respectively. The temperature is, of course, the same as with the first
arrangement because there are no stagnation temperature changes across a shock wave.
245
Therefore the pressures at the inlet to the combustion zone with the two arrangements
are 2053 kPa and 627.8 kPa. With both arrangements, the temperature at the inlet to the
combustion zone is 956 K ( = 683° C ).
246
PROBLEM 6.23
The Mach number in a supersonic airstream is determined by measuring the angle of
the attached shock wave generated by a wedge placed in the airstream, the wedge being
symmetrically aligned with the flow. If the included angle of the wedge probe is 34°
determine the Mach number range over which the probe can be effectively used.
SOLUTION
At low Mach numbers the oblique shock wave will be detached from the wedge and
the device cannot be used. Therefore the minimum Mach number at which the device can
be used is that at which the shock wave is first attached to the leading edge of the wedge,
i.e., at which the Mach number is just equal to that at which δmax is equal to the turning
angle produced by the wedge, i.e., is equal to 34/2 = 17°.
The software or the oblique shock chart for air (γ = 1.4) gives δmax = 17° when the
Mach number is 1.7. Therefore the lowest Mach number at which the device can be used
is 1.7. There is, of course, no upper Mach number limit on the use of the device as long as
the oblique shock relations can be assumed to apply. Hence the wedge probe can be used
if M is equal to or greater than 1.7.
247
Chapter Seven
EXPANSION WAVES PRANDTL - MEYER FLOW
SUMMARY OF MAJOR EQUATIONS
Steady Expansion Wave
M
M 2  1 dM
  1 1   1 2

tan
( M  1)  tan 1 M 2  1 (7.14)
 1 2 M
 1
 1
1 1
M
2
 
Unsteady Expansion Wave
a2
   1  V2
 1 

a1
 2  a1
1


 2a1    pa  2  
V2  
 1    


1

   p1  


248
(7.26)
(7.27)
PROBLEM 7.1
Air is flowing over a flat wall. The Mach number, pressure and temperature in the airstream are 3, 50 kPa and -20 °C respectively. If the wall turns “away” from the flow
through an angle of 10° leading to the formation of an expansion wave, what will be the
Mach number, pressure and temperature in the flow behind the wave?
SOLUTION
For M1 = 3 the software or tables for isentropic flow of air give:
1 = 49.76o ,
p01
T
= 36.73 , 01 = 2.80
p1
T1
Therefore, after the expansion wave:
 2  1  10  49.76  10  59.76o
For this value of θ , the software or the tables for isentropic flow can be used to give:
M 2 = 3.58 ,
p02
= 85.40 ,
p01
T02
= 2.80
T2
Hence, because the flow through the expansion wave is isentropic so that p02 = p01
and T02 = T01 it follows that:
p2 =
p01
p2
1

 p1 = 36.73 
 50 = 21.5 kPa
p1
p02
85.40
and:
T2 =
T01
T
1
 2  T1 = 2.80 
 ( 273  20 ) = 199.0 K =  74o C
T1
T02
3.56
Therefore the Mach number, pressure and temperature downstream of the expansion
wave are 3.58, 21.5 kPa and 199 K (- 74 °C) respectively.
249
PROBLEM 7.2
Air flows along a flat wall at a Mach number of 3.5 and a pressure of 100 kPa. The
wall turns toward the flow through an angle of 25° leading to the formation of an oblique
shock wave. A short distance downstream of this, the wall turns away from the flow
through an angle of 25° leading to the generation of an expansion wave causing the flow
to be parallel to its original direction. Find the Mach number and pressure downstream of
the expansion wave.
SOLUTION
The flow situation being considered is shown in Fig. P7.2.
Figure P7.2
First consider the changes through the shock wave. For M1 = 3.5 and δ = 25° the
software for an oblique shock wave or the oblique shock chart together with the normal
shock tables for air give:
M 2 = 1.98 ,
250
p2
= 5.91
p1
Next consider the expansion wave. For M2 = 1.98 the software for isentropic flow
gives:
 2 = 25.83o ,
p02
= 7.59
p2
Therefore, after the expansion wave:
3 =  2  25 = 25.83  25 = 50.83o
For this value of θ the software for isentropic flow or the isentropic flow tables for air
gives:
M 3 = 3.06 ,
p03
= 40.0
p3
Hence, because the flow through the expansion wave is isentropic so that p03 = p02 , it
follows that:
p3 =
p02
p
p
1
 3  2  p1 = 7.59 
 5.91  100 = 112.1 kPa
p2
p03
p1
40
Therefore the Mach number and pressure downstream of the expansion wave are 3.06
and 112.1 kPa respectively.
251
PROBLEM 7.3
Air is flowing at a Mach number of 2 at a temperature and pressure of 100 kPa and
0°C down a duct. One wall of this duct turns through an angle of 5° away from the flow
leading to the formation of an expansion wave. This expansion wave is reflected off the
flat opposite wall of the duct. Find the pressure and temperature behind the reflected
wave.
SOLUTION
The flow situation being considered is shown in Fig. P7.3.
Figure P7.3
The initial wave and the reflected wave both turn the flow through 5°. Now, in the
initial flow at Ml = 2, the software or table for isentropic flow of air give:
1 = 26.38o ,
p01
= 7.82 ,
p1
T01
= 1.80
T1
Hence:
3 = 1  5  5 = 26.38  10 = 36.38o
For this value of θ3 the software for isentropic flow or the table for isentropic flow of
air gives:
M 3 = 2.384 ,
p03
= 14.26 ,
p3
252
T03
= 2.136
T3
Hence, because the flow through the expansion wave is isentropic so that p03 = p01
and T03 = T01 it follows that:
p3 =
p01
p
1
 3  p1 = 7.82 
 100 = 54.84 kPa
p1
p03
14.26
and:
T3 =
T01
T
1
 3  T1 = 1.80 
 273 = 230 K =  43o C
T1
T03
2.136
Therefore the pressure, temperature and Mach number behind the reflected wave are
54.84 kPa, 230 K ( = -43°C ) and 2.38 respectively.
253
PROBLEM 7.4
Air is flowing through a wide channel at a Mach number of 1.5, the pressure being
120 kPa. The upper wall of the channel turns through an angle of 4° “away” from the
flow leading to the generation of an expansion wave. This expansion wave “reflects” off
the flat lower surface of the channel. Find the Mach number and pressure after this
reflection.
SOLUTION
The initial wave and the reflected wave both turn the flow through 4°. Now in the
initial flow at Ml = 1.5 the software for isentropic flow or the table for isentropic flow of
air gives:
1 = 11.91o ,
p01
= 3.67
p1
Hence:
3 = 1  4  4 = 11.91  8 = 19.91o
For this value of θ3 the software for isentropic flow or the table for isentropic flow of
air gives:
M 3 = 1.77 ,
p03
= 5.50
p3
Hence, because the flow through the expansion wave is isentropic so that p03 = p01 it
follows that:
p3 =
p01
p
 3  p1 = 3.67  5.50  120 = 80.1 kPa
p1
p03
Therefore the pressure and Mach number behind the reflected wave are 80.1 kPa, and
1.77 respectively.
254
PROBLEM 7.5
Air flowing at a Mach number of 3 is turned through an angle that leads to the
generation of an expansion wave across which the pressure decreases by 60%. Find the
angle that the upstream and downstream ends of the expansion wave make to the initial
flow direction.
SOLUTION
Subscripts 1 and 2 will be used to denote conditions before and after the expansion
wave respectively. The upstream and downstream ends of the expansion wave are at the
Mach angle to the local flow direction. Therefore, the upstream end of the wave is at an
angle of:
 1 
1  1 
o
 = sin   = 19.5
3
 M1 
1 = sin 1 
Now for M1 = 3 the software for isentropic flow or the table for the isentropic flow of
air gives:
1 = 49.76o ,
p01
= 36.73
p1
Because the pressure decreases by 60%, it follows that:
p2 = 0.4 p1
Hence, because the flow through the expansion wave is isentropic so that p02 = p01 it
follows that:
p02
p
p01
36.73
 1   p01 
= 01 =
= 
= 91.83
 =

p2
p2
0.4 p1
0.4
 0.4   p1 
For p02 / p2 = 91.83 the software for isentropic flow or the table for isentropic flow of
air gives:
 2 = 60.7 o ,
255
M 2 = 3.64
The corner therefore turns the flow through an angle of:
 2  1 = 60.7  49.8 = 10.9o
The downstream end of the wave is at an angle of:
 1 
1  1 
o
 = sin 
 = 15.95
 3.64 
 M2 
 2 = sin 1 
to the local flow direction. Therefore, the downstream end of the wave is at an angle of
15.95 - 10.9 = 5.05° to the initial flow direction.
Therefore the upstream and downstream ends of the wave are at angles of 19.5° and
5.05° to the initial flow direction. These angles are shown in Fig. P7.5.
Figure P7.5
256
PROBLEM 7.6
An airstream flowing at a Mach number of 4 is expanded around a concave corner
with an angle of 15° leading to the generation of an expansion wave. Some distance
downstream of this the air flows around a concave corner leading to the generation of an
oblique shock wave and returning the flow to its original direction. If the pressure in the
initial flow is 80 kPa, find the pressure downstream of the oblique shock wave.
SOLUTION
First consider the expansion wave. For M1 = 4 the software for isentropic flow or the
table for isentropic flow of air gives:
1 = 65.79o ,
p01
= 151.84
p1
Therefore, after the expansion wave:
 2 = 1  15 = 65.79  15 = 80.79o
For this value of θ the software for isentropic flow or the table for isentropic flow of
air give:
M 2 = 5.45 ,
p02
= 875
p2
Next consider the changes through the shock wave. Because the flow is returned to its
original flow direction by the oblique shock wave the turning angle produced by the
shock wave, i.e., δ, is also 15°. For M2 = 5.45 and δ = 15°, the software for oblique shock
waves or the oblique shock chart and the normal shock tables for air gives:
M 3 = 3.73 ,
p3
= 5.337
p2
Hence, because the flow through the expansion wave is isentropic so that p02 = p01 it
follows that:
257
p3 =
p01
p
p
1
 2  3  p1 = 151.84 
 5.337  80 = 74.9 kPa
p1
p02
p2
875
Therefore the Mach number and pressure downstream of the shock wave are 3.73 and
74.9 kPa respectively.
258
PROBLEM 7.7
An oblique shock wave occurs in an air flow in which the Mach number is 2.5, this
shock wave turning the flow through 10°. The shock wave impinges on a free boundary
along which the pressure is constant and equal to that existing upstream of the shock
wave. The shock is “reflected” from this boundary as an expansion wave. Find the Mach
number and flow direction downstream of this expansion wave.
SOLUTION
The flow situation being considered is shown in Fig. P7.7.
Figure P7.7
First consider the oblique shock wave. For M = 2.5 and δ = 10°, the software for
oblique shock waves or the oblique shock chart and the normal shock tables for air gives:
M 2 = 2.086 ,
p2
= 1.863
p1
Next consider the “reflected” expansion wave. The software for isentropic flow or the
table for isentropic flow of air gives for M2 = 2.086:
 2 = 28.7o ,
259
p02
= 8.94
p2
The strength of the “reflected” expansion waves must be such that p3 = p1, i.e., such
that:
p03
p
p
p
= 03  02  2
p3
p02
p2
p1
Hence, because the flow through the expansion wave is isentropic so that p03 = p02 it
follows that:
p03
= 1  8.94  1.863 = 16.66
p3
For this value of p03 / p3 the software for isentropic flow or the table for isentropic
flow of air gives:
3 = 38.76o ,
M 3 = 2.485
Therefore the Mach number downstream of the “reflected” expansion wave is 2.485.
The expansion wave turns the flow through an angle of θ3 - θ2 = 38.76 - 28.7 = 10° and
the total turning angle relative to the direction of the initial flow is therefore 20°.
260
PROBLEM 7.8
Air is flowing through a wide channel at a Mach number of 2 with a pressure of 140
kPa. The upper wall of the channel turns through an angle of 8° “away” from the flow
leading to the generation of an expansion wave while the lower wall of the channel turns
through an angle of 6° “away” from the flow also leading to the generation of an
expansion wave. The two expansion waves interact and “pass through” each other. Find
the Mach number, flow direction and pressure just downstream of this interaction.
SOLUTION
The entire flow is isentropic and the flow downstream of the waves all has the same
flow direction and is at the same pressure. All the flow that passes through the wave
system will have been turned isentropically through a total angle of 8 + 6 = 14° Now, for
M1 = 2 the software for isentropic flow or the table for isentropic flow of air gives:
1 = 26.38o ,
p01
= 7.82
p1
Therefore, after the expansion wave:
3 = 1  14o = 26.38  14o = 40.38o
For this value of θ it can be deduced using the software for isentropic flow or the
table for isentropic flow of air:
M 3 = 2.385 ,
p03
= 14.23
p3
Because the flow is isentropic so that p03 = p01 it follows that:
p3 =
p01
p
1
 3  p1 = 7.82 
 140 = 76.9 kPa
p1
p03
14.23
Therefore the Mach number and pressure downstream of the wave system are 2.39
and 76.9 kPa respectively. All portions of the flow are turned upward through 8 ° and
261
downward through 6°. Hence, the flow downstream of the wave system will be flowing at
an angle of 2° upwards relative to the initial flow direction.
262
PROBLEM 7.9
A symmetrical double-wedge shaped body with an included angle of 15° is aligned
with an air flow in which the Mach number is 3 and the pressure is 20 kPa. The flow
situation is, therefore, as shown in Fig. P7.9. Find the pressures acting on the surfaces of
the body.
Figure P7.9
SOLUTION
Because the body is symmetrically placed relative to the oncoming flow the pressures
acting on the upper surfaces will be the same as the pressures on the corresponding lower
surfaces.
Oblique shock waves will form at the leading edge of the body which turn the flow
parallel to the forward surfaces of the body, i.e., these shock waves turn the flow through
an angle of 15/2 = 7.5°. Expansion waves then form at the top and bottom corners of the
body, these waves turning the flow parallel to the rear surfaces of the body i.e. these
expansion waves turn the flow through an angle of 15°.
First consider the shock waves. If subscript 1 denotes conditions ahead of the body
and if subscript 2 denotes conditions on the forward surfaces of the body then for the
shock waves M1 is 3 and δ is 7.5°. For these values, the software for oblique shock waves
or the oblique shock chart and the normal shock tables for air give:
M 2 = 2.628 ,
263
p2
= 1.736
p1
From this it follows that:
p2 =
p2
 p1 = 1.736  20 = 34.72 kPa
p1
Next consider the expansion waves. If subscript 3 denotes conditions on the rear
surfaces of the body then since for surfaces 2 where the Mach number, M2 , is 2.628, the
software for isentropic flow or the table for isentropic flow of air give:
 2 = 42.04o ,
p02
= 20.84
p2
and since the flow is turned through 15o by the expansion waves, it follows that:
3 =  2  15 = 42.04  15 = 57.04o
Using this value of θ3 , the software for isentropic flow or the table for isentropic flow
of air give:
M 3 = 3.408 ,
p03
= 66.91
p3
It therefore follows that, since the flow through the expansion wave is isentropic
which means that p03 = p02:
p3 =
p3
p
1
 02  p2 =
 20.84  34.72 = 10.81 kPa
p03
p2
66.91
Therefore the pressure acting on the forward surfaces of the body is 34.72 kPa while
the pressure acting on the rear surfaces of the body is 10.81 kPa.
264
PROBLEM 7.10
A simple wing may be modelled as a 0.25m wide flat plate set at an angle of 3° to an
airflow at a Mach number of 2.5, the pressure in this flow being 60 kPa. Assuming that
the flow over the wing is two-dimensional, estimate the lift and drag force per m span due
to the wave formation on the wing. What other factors cause drag on the wing?
SOLUTION
An expansion wave forms on the upper surface at the leading edge. This wave turns
the flow parallel to the upper surface of the plate. Similarly, an oblique shock wave forms
on the lower surface at the leading edge. This wave turns the flow parallel to the lower
surface of the plate.
First consider the expansion wave which turns the flow parallel to the upper surface,
the region adjacent to the upper surface being designated as 2. Now, in the freestream,
i.e., in region 1, where the Mach number, M1 is 2.5, the software for isentropic flow or
the table for isentropic flow of air gives:
1 = 39.12o ,
p01
= 17.09
p1
Hence, since the flow is turned through 3° by the expansion wave, it follows that:
3 =  2  3 = 39.12  3 = 42.12o
Using this value of θ2 , the software for isentropic flow or the table for isentropic flow
of air gives:
M 2 = 2.63 ,
p02
= 20.96
p2
It therefore follows that, since the flow through the expansion wave is isentropic which
means that p02 = p01:
265
p2 
p
p2
1
 01  p1 
 17.09  60  48.49 kPa
p02
p1
20.96
Therefore the pressure acting on the upper surface of the plate is 48.49 kPa .
Next, consider the oblique shock wave which turns the flow parallel to the lower
surface, the region adjacent to the lower surface being designated as 3. Now, since M1 is
2.5 and the turning angle, δ, produced by the oblique shock wave is 3° , the software for
oblique shock waves or the oblique shock chart and the normal shock tables for air give:
M 3 = 2.374 ,
p3
= 1.216
p1
From this it follows that:
p3 =
p3
 p1 = 1.216  60 = 72.98 kPa
p1
Therefore, the pressure acting on the lower surface of the plate is 72.98 kPa.
The lift is the net force acting on the plate normal to the direction of initial flow while
the drag is the net force on the plate parallel to the direction of initial flow. Therefore,
since the plate area per m span is 0.3 m2, it follows that:
Lift per m span = ( p3  p2 ) A cos 3o =  72.98  48.49   0.25  0.999 = 6.12 kN / m span
Drag per m span = ( p3  p2 ) A sin 3o =  72.98  48.49   0.25  0.0523 = 0.32 kN / m span
Therefore the lift and drag due to the waves per m span are 6.12 and 0.32 kN
respectively. The viscous shear stress acting on the plate can also be a major contributor
to the drag on the plate.
266
PROBLEM 7.11
Consider two-dimensional flow over the double-wedge airfoil shown in Fig. P7.11a.
Figure P7.11a
Find the lift and drag per m span acting on the airfoil and sketch the flow pattern.
How does the pressure vary over the surface of the airfoil?
SOLUTION
Consider the angles and surfaces shown in Fig. P7 .11b.
Figure P7.11b
267
The flow pattern is as shown in Fig. P7.11b. Expansion waves thus occur at the
leading edge and at the corners on the upper surface while an oblique shock wave will
occur at the leading edge of the lower surface and expansion waves will occur at the
corners on the lower surface. The angles of turning produced by the waves are as follows:
Expansion Wave A ( Between Regions 1 and 2 ) - Angle of Turn = 2.5°
Expansion Wave B ( Between Regions 2 and 3 ) - Angle of Turn = 7.5°
Expansion Wave C ( Between Regions 3 and 4 ) - Angle of Turn = 7.5°
Shock Wave 0 ( Between Regions 1 and 5 ) - Angle of Turn = 17.5°
Expansion Wave E ( Between Regions 5 and 6 ) - Angle of Turn = 7.5°
Expansion Wave F ( Between Regions 6 and 7 ) - Angle of Turn = 7.5°
First consider the flow over the upper surface. Consider Expansion Wave A which
separates regions 1 and 2. Ahead of the wave, i.e., in region 1, where the Mach number,
M1, is 3, the software for isentropic flow or the table for isentropic flow of air gives:
 2 = 49.76o ,
p01
= 36.73
p1
Hence, since the flow is turned through 2.5° by the expansion wave, it follows that:
 2 = 1  2.5 = 49.76  2.5 = 52.26o
Using this value of θ2 , the software for isentropic flow or the table for isentropic flow
of air give:
M 2 = 3.133 ,
p02
= 44.79
p2
It therefore follows that, since the flow through the expansion wave is isentropic
which means that p02 = p01:
p2 =
p
p2
1
 01  p1 =
 36.73  20 = 16.40 kPa
p02
p1
44.79
268
Next, consider Expansion Wave B which separates regions 2 and 3. Because the flow
is turned through an angle of 7.5° by this wave and because θ2 is 52.26° it follows that:
3 =  2  7.5 = 52.26  7.5 = 59.76o
Using this value of θ3 , the software for isentropic flow or the table for isentropic flow
of air give:
p03
= 85.39
p3
M 3 = 3.58 ,
It therefore follows that, since the flow through the expansion wave is isentropic
which means that p03 = p02:
p3 =
p3
p
1
 02  p2 =
 44.79  16.40 = 8.60 kPa
p03
p2
85.39
Next consider Expansion Wave C which separates regions 3 and 4. Because the flow
is turned through an angle of 7.5° by this wave and because θ3 is 59.76° it follows that:
 4 = 3  7.5 = 59.76  7.5 = 67.26o
Using this value of θ4 , the software for isentropic flow or the table for isentropic flow
of air give:
M 4 = 4.11 ,
p04
= 176.5
p4
It therefore follows that, since the flow through the expansion wave is isentropic
which means that p04 = p03:
p4 =
p
p4
1
 03  p3 =
 85.39  8.60 = 4.16 kPa
p04
p3
176.5
Next consider the flow over the lower surface. Consider Shock Wave D which
separates regions 1 and 5. Ahead of the wave, i.e., in region 1, where the Mach number,
M1, is 3, the software for an oblique shock wave or the oblique shock chart with the
normal shock tables for the flow of air give:
269
M 5 = 2.126 ,
p5
= 3.273
p1
Also for M5 = 2.126, the software for isentropic flow or the table for isentropic flow
of air give:
5 = 29.79o ,
p05
= 9.524
p5
Next consider Expansion Wave E which separates regions 5 and 6. Because the flow
is turned through an angle of 7.5o by this wave and because θ5 is 29.79° it follows that:
 6 = 5  7.5 = 29.79  7.5 = 37.29o
Using this value of θ6 , the software for isentropic flow or the table for isentropic flow
of air give:
M 6 = 2.42 ,
p06
= 15.38
p6
It therefore follows that, since the flow through the expansion wave is isentropic which
means that p06 = p05:
p6 =
p6
p
1
 05  p5 =
 9.54  65.46 = 40.54 kPa
p06
p5
15.38
Next consider Expansion Wave F which separates regions 6 and 7. Because the flow
is turned through an angle of 7.5° by this wave and because θ6 is 37.29° it follows that:
 7 = 6  7.5 = 37.29  7.5 = 44.79o
Using this value of θ7 , the software for isentropic flow or the table for isentropic flow
of air give:
M 7 = 2.76 ,
270
p07
= 25.32
p7
It therefore follows that, since the flow through the expansion wave is isentropic which
means that p07 = p06:
p7 =
p7
p
1
 06  p6 =
 15.38  40.54 = 24.63 kPa
p07
p6
25.32
Hence:
p2 = 16.40 kPa, p3 = 8.60 kPa, p4 = 4.16 kPa
p5 = 65.46 kPa, p6 = 40.54 kPa, p7 = 24.63 kPa
The lengths of the inclined surfaces of the airfoil are equal to 0.2/ cos ( 7.5°) = 0.2017
m. Therefore, the net upward force, F, acting on the airfoil at right angles to the center
line per m span is:
F = (65.46 - 16.40) x 0.2017  cos 7.5 + (40.54 - 8.60)  0.2
+ (24.63 - 4.16) x 0.2017  cos 7.5 = 9.81 + 6.39 + 4.09 = 20.29 kN
Hence:
Lift = F cos 10° = 20.29 cos 10° = 19.98 kN
Drag = F sin 10° = 20.29 sin 10° = 3.52 kN
Therefore the lift and drag acting on the airfoil per m span as a result of the waves are
19.98 kN and 3.52 kN. The wave pattern is shown in Fig. P7.11b. The pressure is
constant on each segment of the surface of the airfoil but varies discontinuously from
segment to segment. On the three segments that make up the upper surface, starting with
the one adjacent to the leading edge, the pressures are 16.40, 8.60 and 4.16 kPa while on
the three segments that make up the lower surface, starting with the one adjacent to the
leading edge, the pressures are 65.46, 40.54 and 24.63 kPa.
271
PROBLEM 7.12
For the double wedge airfoil shown in Fig. P7.12a, find the lift per meter span if the
Mach number and pressure in the uniform air flow ahead of the air foil are 3 and 40 kPa
respectively.
Figure P7.12a
SOLUTION
Consider the angles and surfaces shown in Fig. P7.12b.
Figure P7.12b
272
Consider the flow pattern shown in Fig. P7.12b. Expansion waves occur at the
leading edge and at the corner on the upper surface while an oblique shock wave occurs
at the leading edge of the lower surface and an expansion wave occurs at the corner on
the lower surface. The angles of turning produced by the waves are as follows:
Expansion Wave A ( Between Regions 1 and 2 ) - Angle of Turn = 3o
Expansion Wave B ( Between Regions 2 and 3 ) - Angle of Turn = 12.5o
Shock Wave C ( Between Regions 1 and 4 ) - Angle of Turn = 13o
Expansion Wave D ( Between Regions 4 and 5 ) - Angle of Turn = 12.5o
First consider the flow over the upper surface. Consider Expansion Wave A which
separates regions 1 and 2. Ahead of the wave, i.e., in region 1, where the Mach number,
M1, is 3, the software for isentropic flow or the table for isentropic flow of air give:
 2 = 49.76o ,
p01
 36.73
p1
Hence, since the flow is turned through 3o by the expansion wave, it follows that:
 2 = 1  3 = 49.76  3 = 52.76o
Using this value of θ2 , the software for isentropic flow or the table for isentropic flow
of air give:
M 2 = 3.161 ,
p02
 46.61
p2
It therefore follows that, since the flow through the expansion wave is isentropic which
means that p02 = p01:
p2 
p
p2
1
 01  p1 
 36.73  40  31.52 kPa
p02
p1
46.61
Next, consider Expansion Wave B which separates regions 2 and 3. Because the flow
is turned through an angle of 12.5o by this wave and because θ2 is 52.76o it follows that:
 3 =  2  12.5 = 52.76  12.5 = 65.26o
273
Using this value of θ3 , the software for isentropic flow or the table for isentropic flow
of air give:
M 3 = 3.96 ,
p03
 143.93
p3
It therefore follows that, since the flow through the expansion wave is isentropic which
means that p03 = p02:
p3 
p3
p
1
 02  p2 
 46.61  31.52  10.21 kPa
p03
p2
143.93
Next consider the flow over the lower surface. Consider Shock Wave C which
separates regions 1 and 4. Ahead of the wave, i.e., in region 1, where the Mach number,
M1, is 3 and since the shock wave turns the flow through an angle of 13o, the software for
an oblique shock wave or the oblique shock chart with the normal shock tables for the
flow of air give:
M 4 = 2.356 ,
p4
 2.493
p1
From this it follows that:
p4 
p4
 p1  2.493  40  99.72 kPa
p1
Also for M5 = 2.356, the software for isentropic flow or the table for isentropic flow
of air give:
 4 = 35.67o ,
p04
= 13.65
p4
Next consider Expansion Wave D which separates regions 4 and 5. Because the flow
is turned through an angle of 12.5o by this wave and because θ4 is 35.67o it follows that:
 5 =  4  12.5 = 35.67  12.5 = 48.17 o
274
For this value of θ5 , the software for isentropic flow or the table for isentropic flow of
air give:
p05
= 32.51
p5
M 5 = 2.92 ,
It therefore follows that, since the flow through the expansion wave is isentropic which
means that p05 = p04:
p5 
p5
p
1
 04  p4 
 13.65  99.72  41.87 kPa
32.51
p05
p4
hence:
p2 = 31.52 kPa ,
p3 = 10.21 kPa ,
p4 = 99.72 kPa ,
p5 = 41.87 kPa
Consideration of the figures given above shows that the distance, L, of the corner
from the leading edge measured along the center-line of the airfoil is such that:
L sin 5o  (0.8  L) sin 7.5o ,
i.e., L 
0.8sin 7.5o
 0.481
sin 5o  sin 7.5o
Therefore, the force, F, acting on the airfoil per m span at right angles center-line is:
F  (99.72  31.52)  0.4805  (41.87 10.21)   0.8  0.4805  42.89 kN
Hence
Lift = F cos (8o) = 42.47 kN
, and, Drag = F sin (8°) = 5.97 kN
Therefore the lift acting on the airfoil per m span as a result of the waves is 42.47 kN.
275
PROBLEM 7.13
A uniform air flow over a plane wall at a Mach number of 2.2 and with a temperature
of 325K encounters a symmetrical triangular “bump” on the wall. The upstream and
downstream faces of this bump are at an angle of 15o to the plane wall over which the air
is flowing. The situation being considered is as shown in Fig. P7.13. Find the Mach
numbers and temperatures in the flows over the upstream and downstream faces of the
“bump” and in the flow over the wall downstream of the bump, i.e., find the Mach
numbers and temperatures in the flow regions 2, 3, and 4 shown in the figure.
Bump
Flow
1
2
o
o
15
15
3
4
Figure P7.13
SOLUTION
First consider conditions in region 1. For M1 = 2.2, the software or the isentropic flow
tables for air give:
T0
 1.986
T1
Next consider conditions in region 2. Using the software or the oblique shock chart
and the normal shock tables for air gives for δ = 15o:
M 2  1.631 ,
T2
 1.287 .
T1
For this Mach number the software or the isentropic flow tables for air give:
 2  15.783o
Next consider conditions in region 3. Since the expansion wave turns the flow
through 30o it follows that:
276
 3   2  30  15.783o  30o  45.783o
For this value of θ3 the software or the isentropic flow tables for air give:
T0
M 3  2.80 ,
 2.568
T3
Lastly consider conditions in region 4. Using the software or the oblique shock chart
and the normal shock tables for air gives for δ = 15o and : M3 = 2.8:
T4
 1.360
T3
Since the stagnation temperature is the same everywhere in the flow, the flow through
M 4  2.115 ,
the oblique shock waves being adiabatic and the flow through the expansion wave being
isentropic it follows that:
T0
T1  1.986 T1  1.986  325  645.5 K
T1
T
T2  2 T1  1.287 T1  1.287  325  418.3 K
T1
T0
645.5
T3 

 251.4 K
T0 / T3
2.568
T
T4  4 T3  1.360 T3  1.360  251.4  341.9 K
T3
Therefore the Mach numbers and temperatures in regions 1, 2, 3, and 4 are 2.2 and
T0 
325K, 1.632 and 418.3K, 2.80 and 251.4K, and 2.115 and 341.9K respectively.
277
PROBLEM 7.14
A uniform air flow over a plane wall at a Mach number of 2.2 and with a temperature
of 270K and a pressure of 90kPa is deflected downwards by the changes in the wall
direction shown in Fig. P7.14a. Determine the Mach numbers, the pressures, and the
stagnation pressures in regions 2 and 3 shown in Fig. P7.14a. Sketch the oblique shock
waves generated by the changes in wall direction and on this sketch indicate the angles
that the shock waves make to the initial flow direction. Determine the maximum turning
angle that the second wall segment can have (15o in the above calculation) if the second
shock wave is to remain attached to this wall segment.
10o
15o
1
2
Initial Flow
3
Figure P7.14a
SOLUTION
Consider conditions in region 2. Using the software or the oblique shock chart and the
normal shock tables for air gives for M1 = 2.2 and δ = 10o:
M 2  1.8228 ,
p2
 1.7641 ,
p1
12  35.785o
For this Mach number the software or the isentropic flow tables for air give:
p2
 0.1681
p02
Next consider conditions in region 3. Using the software or the oblique shock chart
and the normal shock tables for air gives for M2 = 1.8228 and δ = 15o:
278
M 3  1.2694 ,
p3
 2.1409 ,  23  50.493o
p2
For this Mach number the software or the isentropic flow tables for air give:
p3
 0.3762
p03
Therefore:
p2 
p2
p1  1.7641  90  158.8 kPa ,
p1
p02 
p02
158.8
p2 
 944.5 kPa
p2
0.1681
and:
p3 
p3
p2  2.1409  158.8  340.0 kPa ,
p2
p03 
p03
340.0
p3 
 903.7 kPa
p3
0.3762
The angles that the oblique shock waves make to the initial flow direction are shown
in Fig. P7.14b.
35.8o
Initial Flow Direction
FIGURE P7.14b
279
60.5o
Since:
M 2  1.8228
The maximum turning angle is that corresponding to this Mach number. Using the
software or the oblique Shock chart for air give:
 23Max  19.31o
Therefore in region 2 the pressure and stagnation pressure are 158.8.1kPa and 944.5
kPa respectively while in region 3 the pressure and stagnation pressure are 340 kPa and
903.7 kPa respectively. The maximum turning angle for the second oblique shock wave is
19.31o.
280
PROBLEM 7.15
A thin flat plate is placed in a uniform air flow in which the Mach number is 2.3 and
the pressure is 10 kPa. The plate is set at an angle to the flow and as a result an oblique
shock wave originating at the leading edge of the lower surface of the plate is generated
and an expansion wave originating at the leading edge of the upper surface of the plate is
generated. The oblique shock wave makes an angle of 40o to the direction of the
undisturbed flow upstream of the plate. Find the angle at which the plate is set relative to
the undisturbed flow direction and the pressures acting on the upper and lower surfaces of
the plate.
SOLUTION
Considering the oblique shock, the upstream Mach number, M1 , is 2.3 and the wave
angle relative to the flow, β , is 40o. For these values the software or the oblique shock
wave chart for air gives:
Turning Angle,   15.3o
This is the angle at which the plate is set to the freestream flow.
Now for a Mach number of 2.3, a wave angle of 40o, and a turning angle of 15.3o the
software or the normal shock tables for air give:
p2
 2.383
p1
Hence:
p2  2.383 p1  2.383  10  23.83 psia
Next consider the expansion wave on the upper surface. For M1 = 2.3 the software or
the isentropic flow tables for air give:
p1
 0.0800
p0
Therefore since the flow is turned through an angle of 15.3o by the expansion wave it
1  34.28o ,
follows that:
281
 2  1  15.3  34.28  15.3  49.58o
For this value of θ the software or the isentropic flow tables for air give:
p3
M 3  2.991 ,
 0.0276
p0
Since the flow through the expansion wave is isentropic the stagnation pressure
remains constant through the wave. Therefore:
p3 
p3 / p0
0.0276
p1 
 10  3.449 psia
p1 / p0
0.0800
Therefore the plate is set at an angle of 15.3o to the upstream flow and the pressures
acting on the lower and upper surfaces are 28.83 psia and 3.449 psia respectively.
282
PROBLEM 7.16
A safety diaphragm at the end of 3 m long pipe containing air at a pressure of 200 kPa
and a temperature of 10°C suddenly ruptures causing an expansion wave to propagate
down the pipe. Find the velocity at which the air is discharged from the pipe, the velocity
of the front and the back of the wave and the time taken for the front of the wave to reach
the end of the pipe. Assume that the ambient pressure of the air surrounding the pipe is
100 kPa.
SOLUTION
The velocity at which the air will leave the pipe is given by:
1


 2a1    pa  2  
V2  
 1    
   1    p1  


In the situation here being considered T1 = 10oC = 283K. Hence:
a1 
 R T1  1.4  287  283  337.2 m/s
Therefore:
(1.4 1)/(2 1.4)

 2  337.2    100 
V2  
  159.0 m/s
 1  

 1.4  1    200 

Therefore, the velocity at which the air is discharged from the pipe is 159 m/s. The
front of the wave propagates at the local speed of sound in the undisturbed air, i.e., at a1
i.e. at 337.2 m/s. The tail of the wave propagates at the local speed of sound behind the
wave, i.e., a2 , relative to gas behind the wave, i.e., at a2 - V2 relative to the pipe. But:
a2
   1   V2 
 1.4  1   159 
 1 
   1  

  0.9057
a1
 2   a1 
 2   337.2 
Hence:
283
a2 = 0.9057 a1 = 0.9057  337.2 = 305.4 m/s
Therefore, the velocity of the tail of the wave relative to the walls of the pipe is 305.4 159.0 = 146.4 m/s.
Because the front of the wave is moving at a velocity of 337.2 m/s, the time taken for
the front of the wave to reach the end of the pipe is given by
t 
3
 0.0089 s
337.2
Therefore the time for the head of the wave to reach the end of the pipe is 0.0089 s.
284
PROBLEM 7.17
An unsteady expansion wave propagates down a duct containing air at rest at a
pressure of 800 kPa and a temperature of 2000°C. The pressure behind the wave is 300
kPa. Find the velocity and the Mach number in the flow that is induced behind the wave
in the duct.
SOLUTION
The velocity induced by the expansion wave is given by:
1




pa 2  
 2a1  
V2  
 1    
   1    p1  


In the situation here being considered T1 = 2000°C = 2273K. Hence:
a1 =
 R T1 =
1.4  287  2273 = 995.7 m/s
Therefore:
(1.4 1)/(2 1.4)

 2  955.7    300 
V2 = 
 = 624.8 m/s
 1  

 1.4  1    800 

The flow through the expansion wave is isentropic. Therefore, across the wave:
 a2 
 p2 
  =  
 a1 
 p1 
 1
2
Therefore:
 1
 1.4 1 


 p2  2
 300  21.4 
a2 = a1  
= 955.7  
= 830.7 m/s

 800 
 p1 
The Mach number behind the wave is therefore given by:
285
M2 =
V2
624.8
=
= 0.752
a2
830.7
Therefore the velocity and Mach number in the air behind the wave are 624.8 m/s and
0.752 respectively.
286
PROBLEM 7.18
Air is flowing through a long pipe at a velocity of 50 m/s at a pressure and
temperature of 150 kPa and 40°C respectively. Valves at the inlet and exit to this pipe are
suddenly and simultaneously closed. Discuss the waves that are generated in the pipe
following valve closure and find the pressures acting on each valve immediately
following the valve closure.
SOLUTION
The air in contact with both valves must be brought to rest when they are closed. A
normal shock wave must propagate into to the moving air from the downstream (exit)
valve, this shock wave increasing the air pressure and bringing the air to rest. Similarly,
an expansion wave must propagate into the moving air from the upstream (inlet) valve,
this wave decreasing the air pressure and bringing the air to rest. The flow pattern is,
therefore, as shown in Fig. P7.18.
Figure P7.18
First consider the expansion wave. Across this wave:
1


 2a1    pa  2  
V2  
 1    
   1    p1  


In the present case since T1 = 40°C = 313 K, it follows that:
a1 =
 R T1 =
1.4  287  313 = 354.6 m/s
287
Hence:
(1.4 1)/(2 1.4)

 2  354.6    p2 

50 = 
1
 , i.e.,
 


 1.4  1    150 
p2 = 122.8 kPa
Therefore the pressure acting on the upstream valve is 122.8 kPa.
Next consider the shock wave. Consider the flow relative to the shock wave. Because
the wave must bring the air to rest, it will be seen that if Us is the velocity of the shock
wave:
M1 =
Us  V
a1
and:
M2 =
Us
U a
= s 1
a2
a1 a2
Substituting this into the equation for M1 then gives:
M1 = M 2
a2
V

a1
a1
But the speed of sound ahead of the wave was previously shown to be a1 = 354.6 m/s and
V = 50 m/s. Therefore the above equation gives:
M1 = M 2
a2
a
50

= M 2 2  0.141
a1
354.6
a1
Because the normal shock relations determine M2 and a2 / a1 as functions of M1 , the
above equation together with the normal shock relations can be used to determine M1 .
The solution has here been obtained by trial-and-error. The value of M1 was guessed and
the software for a normal shock wave or the normal shock tables for air ( i.e., for γ = 1.4 )
is used to obtain M2 and a2 / a1 [ = (T2 / T1)0.5 ]. Using the values so obtained, the right
hand side of the above equation, i.e., M2 (a2 / a1) + 0.141 is calculated and compared
with the value of M1 given by the software or tables. This procedure was repeated until
the value of M1 that gave M1 equal to M2 (a2 / a1) + 0.141 was obtained. Some values
obtained in carrying out this process are shown in the following table:
288
M1
1.00
1.40
1.20
1.10
1.08
M2
1.000
0.740
0.842
0.912
0.928
a2 / a1
1.000
1.120
1.062
1.032
1.026
M2 (a2 / a1) + 0.141
1.141
0.970
1.035
1.082
1.093
From these results it will be seen that M1 = 1.09. The software for a normal shock
wave or the normal shock tables for air ( i.e., for γ = 1.4) then gives for this Mach number
p2 / p1 = 1.22. Therefore:
p2 =
p2
p1 = 1.22  150 = 183 kPa
p1
Therefore the pressure acting on the upstream valve is 183 kPa.
289
PROBLEM 7.19
A long pipe is conveying air at a pressure and temperature of 150 kPa and 100°C
respectively at a velocity of 160 m/s. Valves are fitted at both the inlet and the exit to the
pipe. Discuss what waves will be developed and what pressure will act on the valve if (i)
the inlet valve is suddenly closed (ii) the exit valve is suddenly closed.
SOLUTION
(i) If the inlet valve is suddenly closed an expansion wave must propagate into the
moving air from the valve, this wave decreasing the air pressure and bringing the air to
rest. Across this expansion wave:
1


 2a1    pa  2  
V2  
 1    
   1    p1  


In the present case since T1 = 100°C = 373 K, it follows that:
a1 
 R T1  1.4  287  373  387.1 m/s
Hence:
 2  387.1    p2 
160  
 1  

 1.4  1    150 
(1.4 1)/(2 1.4)

 , i.e.,

p2  82.0 kPa
Therefore the pressure acting on the inlet valve is 82.0 kPa.
(ii) If the exit valve is suddenly closed a normal shock wave must propagate into the
moving air from this valve, this shock wave increasing the air pressure and bringing the
air to rest. Consider the flow relative to the shock wave. Because the wave must bring the
air to rest, it will be seen that if Us is the velocity of the shock wave:
290
M1 
Us V
a1
and:
M2 
Us
U a
 s 1
a2
a1 a2
Substituting this into the equation for M1 then gives:
M1  M 2
a2
V

a1
a1
But the speed of sound ahead of the wave was previously shown to be a1 = 387.1 m/s and
V = 160 m/s. Therefore the above equation gives:
M1  M 2
a2
a
160

 M 2 2  0.413
a1 387.1
a1
Because the normal shock relations determine M2 and a2 / a1 as functions of M1 , the
above equation together with the normal shock relations can be used to determine M1 .
The solution has here been obtained by trial-and-error. The value of M1 was guessed and
the software for a normal shock wave or the normal shock tables for air ( i.e., for γ = 1.4 )
was used to obtain M2 and a2 / a1 [ = (T2 / T1)0.5 ]. Using these values the right hand side
of the above equation, i.e., M2 (a2 / a1) + 0.413 was calculated and compared with the
value of M1. This was repeated until the value of M1 that gave M1 equal to M2 (a2 / a1) +
0.413 was obtained. Some values obtained in carrying out this process are shown in the
following table:
M1
1.00
1.40
1.30
1.20
1.28
1.27
M2
1.000
0.740
0.786
0.842
0.796
0.802
a2 / a1
1.000
1.120
1.091
1.062
1.085
1.083
291
M2 (a2 / a1) + 0.413
1.413
1.240
1.271
1.307
1.277
1.281
From these results it will be seen that M1 = 1.279. The software for a normal shock
wave or the normal shock tables for air ( i.e., for γ = 1.4) then gives for this Mach number
p2 / p1 = 1.74. Therefore:
p2 
p2
p1  1.74  150  261 kPa
p1
Therefore the pressure acting on the exit valve is 261 kPa.
292
PROBLEM 7.20
A closed tube contains air at a pressure and temperature of 200 kPa and 30°C
respectively. One end of the tube is suddenly opened to the surrounding atmosphere. At
what velocity does the air leave the open end of the tube if the ambient pressure is 100
kPa?
SOLUTION
An expansion wave will propagate into the tube decreasing the pressure from 200 kPa
to 100 kPa and accelerating the air to a velocity V2. The value of this velocity induced by
the expansion wave is given by:
 1


 2a1    pa  2 
V2  
 1  
   1    p1  


Now in the present case since T1 = 30°C = 303 K it follows that:
a1 
 R T1 
1.4  287  303  348.9 m/s
Hence:
(1.4 1)/(2 1.4)

 2  348.9    100 
V2  
 164.5 m/s

 1  

 1.4  1    200 

Therefore the velocity at which the air leaves the tube is 164.5 m/s.
293
PROBLEM 7.21
A long pipe containing air is separated into two sections by a diaphragm. The
pressure on one side of the diaphragm is 500 kPa and the pressure on the other side of the
diaphragm is 100 kPa, the air temperature being 20°C in both sections. If the diaphragm
suddenly ruptures causing a shock wave to move into the low pressure section and an
expansion wave to move into the high pressure section, find the pressure and air velocity
in the region between the two waves.
SOLUTION
The conditions in the low and high pressure sides will be designated by subscripts 1
and 2 respectively and the conditions behind the expansion wave and behind the shock
waves will be designated by subscripts 3 and 4 respectively. The speed of sound in the
undisturbed air, i.e., in sections 1 and 2 is given by:
a1  a2 
 RT 
1.4  287  293  343.1 m/s
The strengths of the shock wave and expansion wave must be such that the pressure
and velocity in the region between the shock wave and the expansion wave is everywhere
the same, i.e., the strengths of the waves must be such that p3 = p4 and V3 = V4. A very
simple trial-and-error method of obtaining the solution will be adopted here. In this
approach the pressure between the two waves will be guessed i.e., the value of p3 = p4
will be guessed. The air velocity behind the shock and behind the expansion wave will
then be separately calculated. Because the value of the pressure is guessed, these two
values will not in general be equal. Calculations will then be undertaken with different
values of the guessed pressure and then the pressure that makes the air velocity behind
the shock wave and behind the expansion wave equal will be deduced. To illustrate the
procedure, assume that:
p3  p4  150 kPa
First consider the expansion wave. Across this wave:
294
 1


 2a1    pa  2 
V2  
 1  
   1    p1  


For the specified conditions, i.e., for:
p1  500 kPa , a1  343.1 m/s
Hence for p3 = 150 kPa, this equation gives:
(1.4 1)/(2 1.4)

 2  343.1    150 
V3  

1
  271.1 m/s
 


 1.4  1    500 
Next consider the flow across the shock wave. For p4 = 150 kPa:
p4
150

 1.5
p2
100
For this pressure ratio, the software for normal shock waves or the normal shock tables
for air give:
M 2  1.197 , M 4  0.844 ,
T4
 1.126
T2
But M2 is the shock Mach number, i.e.:
M2 
Us
Us

a2
343.1
Hence for the guessed pressure:
U s  M 2 a2  343.1  1.197  410.7 m/s
Also:
T 
a4
  4
a2
 T2 
0.5
 1.1260.5  1.061
Hence:
295
a4  1.061  343.1  364 m/s
But:
M4 
U s  V4
a4
from which it follows that:
V4  U s  M 4 a4  410.7  0.844  364  103.5 m/s
Hence, when it is guessed that p3 = p4 = 150 kPa, it is found that V3 = 271.1 m/s and
V4 = 103.5 m/s.
Calculations of this type have been carried out for a number of other values of the
guessed pressures, the results of some of these calculations being given in the following
table.
p3 = p4
kPa
120
200
300
230
215
p4 / p2
M1
M4
a4 / a2
1.20
2.00
3.00
2.30
2.15
1.083
1.363
1.648
1.455
1.409
0.926
0.755
0.654
0.718
0.736
1.027
1.274
1.192
1.136
1.123
a4
m/s
352.2
437.1
408.9
389.8
385.3
Us
m/s
371.6
467.6
565.4
499.2
483.4
V4
m/s
45.4
137.6
298.0
219.0
199.8
V3
m/s
316.4
210.5
120.7
180.0
194.9
V4 - V3
m/s
-271
-73
+177
+39
+5
By interpolation between these results it can be deduced that V3 = V4 when p3 = p4 =
214 kPa and that at this pressure V3 = V4 = 193 m/s, i.e., the pressure and velocity of air
between the moving shock wave and the moving expansion wave are 214 kPa and 193
m/s respectively.
296
PROBLEM 7.22
A shock tube containing air has initial pressures on the two sides of the diaphragm of
400 kPa and 10 kPa, the temperature of the air being 25°C in both sections. If the
diaphragm separating the two sections is suddenly ruptured, find the velocity, pressure
and temperature of the air between the moving shock wave and the moving expansion
wave that are generated.
SOLUTION
The conditions in the low and high pressure sides will be designated by subscripts 1
and 2 respectively. The conditions behind the expansion wave and behind the shock
waves will be designated by subscripts 3 and 4 respectively. The speed of sound in the
undisturbed air, i.e., in sections 1 and 2 is given by:
a1  a2   R T 
1.4  287  298  346.0 m/s
The strengths of the shock wave and expansion wave must be such that the pressure
and velocity in the region between the shock wave and the expansion wave is everywhere
the same, i.e., the strengths of the waves must be such that p3 = p4 and V3 = V4. A very
simple trial-and-error method of obtaining the solution will be adopted here. In this
approach the pressure between the two waves will be guessed, i.e., the value of p3 = p4
will be guessed. The air velocity behind the shock and behind the expansion wave will
then be separately calculated. Because the value of the pressure is guessed, these two
values will not in general be equal. Calculations will then be undertaken with different
values of the guessed pressure and then the pressure that makes the air velocity behind
the shock wave and behind the expansion wave equal will be deduced. To illustrate the
procedure, assume that:
p3  p4  100 kPa
First consider the expansion wave. Across this wave:
297
 1


 2a1    pa  2 
V2  
 1  
   1    p1  


For the specified conditions, i.e., for:
p1  400 kPa ,
a1  346.0 m/s
Hence for p3 = 100 kPa, this equation gives:
(1.4 1)/(2 1.4)

 2  346.0    100 
V3  

 310.8 m/s
1





 1.4 1    400 

Next consider the flow across the shock wave. For p4 = 100 kPa:
p4
100

 10
p2
10
For this pressure ratio, the software for normal shock waves or the normal shock tables
for air give:
M 2  2.951 , M 4  0.478 ,
T4
 2.623
T2
But M2 is the shock Mach number, i.e.:
M2 
Us
Us

a2
346.0
Hence for the guessed pressure:
U s  M 2 a2  346.0  2.951  1021 m/s
Also:
T 
a4
  4
a2
 T2 
0.5
 2.6230.5  1.620
Hence:
a4  1.620  346.1  560 m/s
298
But:
M4 
U s  V4
a4
from which it follows that:
V4  U s  M 4 a4  1021  0.478  560  753 m/s
Hence, when it is guessed that p3 = p4 = 100 kPa, it is found that V3 = 311 m/s and V4
= 753 m/s.
Calculations of this type have been carried out for a number of other values of the
guessed pressures, the results of some of these calculations being given in the following
table.
p3 = p4
kPa
80
60
40
45
p4 / p2
M1
M4
a4 / a2
8.0
6.0
4.0
4.5
2.645
2.299
1.890
1.455
0.501
0.534
0.598
0.577
1.512
1.395
1.265
1.299
a4
m/s
523.2
482.7
437.7
449.5
Us
m/s
915.2
795.5
654.0
692.0
V4
m/s
653.1
537.7
392.6
432.6
V3
m/s
355.4
410.7
484.9
463.8
V4 - V3
m/s
+298
+127
-92
-31
By interpolation between these results it can be deduced that V3 = V4 when p3 = p4 =
47 kPa and that at this pressure V3 = V4 = 452 m/s, i.e., the pressure and velocity of air
between the moving shock wave and the moving expansion wave are 47 kPa and 452 m/s
respectively.
The pressure ratio across the shock wave is therefore 47/10 = 4.7. With this pressure
ratio, the software for normal shock waves or the normal shock tables for air give the
temperature ratio across the shock wave as 1.72. Hence the temperature behind the shock
is 1.72  298 = 512 K (= 240°C). The pressure ratio across the expansion wave is 47/400
= 0.118 and, because the flow through the expansion wave is isentropic, the temperature
ratio across the expansion wave is given by:
299
p 
T3
  3
T1
 p1 
 1

1.4 1
1.4
 0.118
 0.543
Hence, the temperature behind the expansion wave is 0.543  298 = 162 K ( =-111° C).
Therefore the pressure and velocity between the shock wave and the expansion wave
are 47 kPa and 452 m/s. The portion of the air between the waves that has been traversed
by the shock wave is at a temperature of 512 K ( = 240°C) while the portion of the air
between the waves that has been traversed by the expansion wave is at a temperature of
162 K ( = -111° C).
300
PROBLEM 7.23
The air pressure in the high and low pressure sections of a constant diameter shock
tube are 600 kPa and 20 kPa respectively. The temperatures in both sections are 30° C.
After the diaphragm that separates the two sections is ruptured, a shock wave propagates
into the low pressure section and an expansion wave propagates into the high pressure
section. Find the air velocity and temperatures between the two waves and the velocity of
the shock wave.
SOLUTION
The conditions in the low and high pressure sides will be designated by subscripts 1
and 2 respectively. The conditions behind the expansion wave and behind the shock
waves will be designated by subscripts 3 and 4 respectively. The speed of sound in the
undisturbed air, i.e., in sections 1 and 2 is given by:
a1  a2 
 R T  1.4  287  303  348.9 m/s
The strengths of the shock wave and expansion wave must be such that the pressure
and velocity in the region between the shock wave and the expansion wave is everywhere
the same, i.e., the strengths of the waves must be such that p3 = p4 and V3 = V4. A very
simple trial-and-error method of obtaining the solution will be adopted here. In this
approach the pressure between the two waves will be guessed, i.e., the value of p3 = p4
will be guessed. The air velocity behind the shock and behind the expansion wave will
then be separately calculated. Because the value of the pressure is guessed, these two
values will not in general be equal. Calculations will then be undertaken with different
values of the guessed pressure and then the pressure that makes the air velocity behind
the shock wave and behind the expansion wave equal will be deduced. To illustrate the
procedure, assume that:
p3  p4  300 kPa
First consider the expansion wave. Across this wave:
301
 1


 2a1    pa  2 
V2  
 1  
   1    p1  


For the specified conditions, i.e., for:
p1  600 kPa ,
a1  348.9 m/s
Hence for p3 = 300 kPa, this equation gives:
(1.4 1)/(2 1.4)

 2  348.9    300 
V3  

1
  164.5 m/s
 


 1.4  1    600 
Next consider the flow across the shock wave. For p4 = 300 kPa:
p4
300

 15
p2
20
For this pressure ratio, the software for normal shock waves or the normal shock tables
for air give:
M 2  3.606 , M 4  0.447 ,
T4
 3.461
T2
But M2 is the shock Mach number, i.e.:
M2 
Us
Us

a2
348.9
Hence for the guessed pressure:
U s  M 2 a2  348.9  3.606  1258 m/s
Also:
T 
a4
  4
a2
 T2 
0.5
 3.4610.5  1.860
Hence:
302
a4  1.860  348.9  649 m/s
But:
M4 
U s  V4
a4
from which it follows that:
V4  U s  M 4 a4  1258  0.447  649  968 m/s
Hence, when it is guessed that p3 = p4 = 300 kPa, it is found that V3 = 164.5 m/s and
V4 = 968 m/s.
Calculations of this type have been carried out for a number of other values of the
guessed pressures, the results of some of these calculations being given in the following
table.
p3 = p4
kPa
150
100
80
85
p4 / p2
M1
M4
a4 / a2
7.5
5.0
4.0
4.25
2.565
2.105
1.890
1.946
0.507
0.560
0.598
0.587
1.484
1.332
1.265
1.282
a4
m/s
517.9
464.8
441.3
447.4
Us
m/s
894.9
734.4
659.4
669.0
V4
m/s
632.4
474.2
395.5
416.3
V3
m/s
313.4
394.0
436.3
425.0
V4 - V3
m/s
+319
+80
-41
-9
By interpolation between these results it can be deduced that V3 = V4 when p3 = p4 =
86 kPa and that at this pressure V3 = V4 = 421 m/s, i.e., the pressure and velocity of air
between the moving shock wave and the moving expansion wave are 86 kPa and 421 m/s
respectively. The pressure ratio across the shock wave is therefore 86/20 = 4.3. With this
pressure ratio, the software for normal shock waves or the normal shock tables for air
give the temperature ratio across the shock wave as 1.65. Hence the temperature behind
the shock is 1.65  303 = 500 K (= 227° C). With this pressure ratio, the software for
normal shock waves or the normal shock tables for air also give the shock Mach number
as 1.956. Hence, the velocity of the shock wave is 1.956  348.9 = 683 m/s.
303
The pressure ratio across the expansion wave is 86/600 = 0.143 and, because the flow
through the expansion wave is isentropic, the temperature ratio across the expansion
wave is given by:
p 
T3
  3
T1
 p1 
 1

 0.143
1.4 1
1.4
 0.574
Hence, the temperature behind the expansion wave is 0.574  303 = 174 K ( =-99° C).
Therefore the pressure and velocity between the shock wave and the expansion wave
are 86 kPa and 421 m/s and the velocity of the shock wave is 683 m/s. The portion of the
air between the waves that has been traversed by the shock wave is at a temperature of
500 K ( = 227° C) while the portion of the air between the waves that has been traversed
by the expansion wave is at a temperature of 174 K ( = -99° C).
304
Chapter Eight
VARIABLE AREA FLUID FLOW
SUMMARY OF MAJOR EQUATIONS
Variable Area Flow Equations
1
1


 2
 2    p0    p1    
V1  
   1     
1



  0 
 p0   




A2
A1

1  

 p1   
   
 p2   
1  
 
1

 
p2
  2
p1
 1 
p1
p2

, i.e.,
1T1
2T2
 
T2
  2
T1
 1 
1
305



1 
p2   
 
p0  
p1 

p0 
1

(8.19)
1
2
(8.22)

(4.2)
T2
p 
 2 1
T1
p1  2
p 
  2
 p1 
1

(4.3)
(4.4)
1
 T 2   
a2
  2   2
a1
 T1 
 1 
1
2
1
 p  2
  2
 p1 
 1 2
1 (
)M1
T2
2

T1 1  (   1) M 2
2
2
(4.5)
(4.6)

1  12 (   1) M 12  1
p2
 
2
1
p1
1  2 (   1) M 2 
1  (   1) M 
2
 

1
1  (   1) M 
2
1
2
2
1
2
1
2
1
1
(4.7)
(4.8)
:
  2  V2 
A1
   
A2
 1  V1 
(4.9)
Variable Area Flow Equations in Terms of Critical Conditions
V *2 
2 2
a0
 1
(8.23)
2
 a* 
T*
2


 
T0
 1
 a0 
306
(8.24)

 2  
p*
 

p0
  1 
 2 
m


p0  0    1 
A 
*
1
(8.25)

1
2( 1)
(8.26)
1
 2  2( 1)    1  2


  1 
A
 2 



1
A*
1 2
2


 p  γ  p   
     
 p0   p0  


(8.27)
1
1


2
 2    p0    pe    
Ve  
   1     
   1    0    p0   



(8.28)
1
1


2
 pe   2    p0    pe    
m   0 Ae   

  1  
 p0     1    0    p0   



1

1
1
 2  2( 1)    1  2


  1 
Ae
 2 



1
A*
1 2
2


 pe    pe   
     
 p0   p0  


307
(8.29)
(8.30)
Variable Area Flow Equations in Terms of Mach Number

V  Ma0 1 

2
 a

   1 

 a0 

1
 M 1  
A2
 

A1
 M 2  1 

  1  2 

M 
 2 

  1  2 

M 
 2 


1
2
(8.31)
1
1
   1  2  2( 1)

 M2 

 2 

  1  2 
M

 1
 2 

(8.35)
1

   1  2  2( 1)
1
1


M 
2( 1)




A
1
2
1





2
 1  




2
  



 M 
 1  


1
A*
M
M


1
2


  1 


 


 

2

 
(8.36)

p0
  1 2  1

M 
 1 
p
2


(4.15)

0
  1 2  1

 1 
M 

2


(4.16)
T0
 1 2 

 1 
M 
T
2


(4.17)
308
PROBLEM 8.1
Air is discharged from a large reservoir in which the pressure and temperature are 0.8
MPa and 25°C respectively through a convergent nozzle with an exit diameter of 5 cm.
The nozzle discharges to the atmosphere. Find the mass flow rate through the nozzle and
the pressure and temperature on the nozzle exit plane.
SOLUTION
The flow is assumed to be isentropic. The reservoir is large so the velocity in it will
be small and the pressure and temperature in it will, as a result, effectively be the
stagnation pressure and temperature, i.e.:
p0 = 800 kPa
and
T0 = 298 K
Because a convergent nozzle is involved the highest Mach number that can exist on the
nozzle exit plane is 1. When this situation exists, the software for isentropic flow or the
isentropic flow tables for air gives:
p0
p
 0  1.893
pe
p*
where the subscript e denotes conditions on the exit plane. Using the above results gives,
if the nozzle is choked:
pe 
p0
800

 422.6 kPa
p0 / pe
1.893
Because this is greater than the atmospheric pressure (101.3 kPa) it shows that the nozzle
is choked, i.e., the Mach number on the exit plane is 1. In this case the software for
isentropic flow or the isentropic flow tables for air gives:
T0
T
 0  1.200
Te
T*
309
Hence:
Te 
T0
298

 248.3 K
T0 / Te
1.200
The mass flow rate through the nozzle is then given by:
m  e Ve Ae
But because the nozzle is choked so that Me = 1 it follows that Ve= ae. Hence:
m  e ae Ae
Now:
e 
pe
422600

 5.930 kg/m3
287  248.3
RTe
and:
ae 
 R Te 
1.4  287  248.3  315.9 m/s
and:
Ae 

4
De2 

4
 0.052  0.001964 m 2
The mass flow rate is therefore given by:
m  e ae Ae  5.930  315.9  0.001964  3.678 kg/s
Therefore the exit plane pressure and temperature are 422.6 kPa and 248.3 K ( = 24.7°C ) respectively while the mass flow rate through the nozzle is 3.678 kg/s.
310
PROBLEM 8.2
A supersonic nozzle possessing an area ratio of 3.0 is supplied from a large reservoir
and is allowed to exhaust to atmospheric pressure. Determine the range of reservoir
pressures over which a normal shock will appear in the nozzle. For what value of
reservoir pressure will the nozzle be perfectly expanded with supersonic flow at the exit
plane? Find the minimum reservoir pressure to produce sonic flow at the nozzle throat.
Assume isentropic flow except for shocks with γ = 1.4.
SOLUTION
It will be assumed that the back pressure, i.e., atmospheric pressure, is 101.3 kPa. It
will also be assumed that the supply reservoir is large enough to be able to assume that
the reservoir pressure is the stagnation pressure. The nozzle has an area ratio, i.e., Ae /
A* , Ae being the exit area, of 3. For this area ratio the software for isentropic flow or the
isentropic flow tables for air give either:
M e = 0.1974 ,
p0
= 1.028
pe
M e = 2.637 ,
p0
= 21.14
pe
or:
When there is a normal shock wave at the exit, the flow in the nozzle is shock-free
with a supersonic velocity at the exit. Hence, if there is a shock wave at the exit, the
Mach number ahead of the shock wave will be 2.637. For this Mach number, the software
for normal shock waves or the normal shock wave tables for air give the pressure ratio
across the wave as:
pb
= 7.946
pe
311
Here it has been noted that with a normal shock wave at the exit the flow downstream of
the shock wave will be subsonic and the pressure downstream of the shock wave must
therefore be equal to the back pressure, pb. But pb = 101.3 kPa. Hence, when there is a
normal shock wave at the exit:
pe 
pb
101.3

 12.75 kPa
7.946
pb / pe
As noted above, when there is a normal shock wave at the exit, the flow in the nozzle
is shock free with a supersonic velocity at the exit. Hence, with a normal shock wave on
the exit plane:
p0
= 21.14
pe
Therefore, with a normal shock at the exit:
p0 =
p0
pe = 21.14  12.75 = 269.5 kPa
pe
When the nozzle is perfectly expanded there are no shock waves in the flow and pe =
pb. Therefore, because the exit flow is supersonic, under these conditions:
pe = 101.3 kPa and
p0
= 21.14
pe
Hence:
p0 =
p0
pe = 21.14  101.3 = 2141.5 kPa
pe
When sonic flow is first reached at the throat, the flow in the divergent section of the
nozzle will be subsonic and pe = pb . In this case then the subsonic isentropic flow values
given above apply, i.e.:
M e = 0.1974 ,
312
p0
= 1.028
pe
Hence in this situation:
p0 =
p0
pe = 1.028  101.3 = 104.1 kPa
pe
Therefore a normal shock wave exists on the exit plane when the reservoir pressure is
269.5 kPa, the nozzle is perfectly expanded with a supersonic velocity at exit when the
reservoir pressure is 214.5kPa and a sonic velocity is first reached when the reservoir
pressure is 104.1 kPa. From these results, it follows that:
1. If the reservoir pressure is less than 104.1 kPa, the flow is subsonic throughout.
2. If the reservoir pressure is between 104.1 kPa and 269.5 kPa, a normal shock wave
will exist in the divergent portion of the nozzle.
3. If the reservoir pressure is greater than 269.5 kPa, supersonic flow exists in the
nozzle but there are no shock waves in the nozzle.
313
PROBLEM 8.3
A converging-diverging nozzle is designed to operate with an exit Mach number of
1.75. The nozzle is supplied from an air reservoir at 1000 psia. Assuming onedimensional flow calculate:
•
maximum back pressure to choke the nozzle
•
range of back pressures over which a normal shock will appear in the nozzle
•
back pressure for the nozzle to be perfectly expanded to the design Mach number
•
range of back pressures for supersonic flow at the nozzle exit plane
SOLUTION
It will be assumed that the supply reservoir is large enough to be able to assume that
the reservoir pressure is the stagnation pressure, i.e.:
p0  1000 psia
The nozzle is designed to generate an exit Mach number of 1.75. For this Mach
number the software for isentropic flow or the isentropic tables for air give:
Ae
= 1.386 ,
A*
p0
= 5.324
pe
Ae being the nozzle exit area and pe being the pressure on the nozzle exit plane.
When the nozzle is operating at this design condition, i.e., when the nozzle is
perfectly expanded, there are no shock waves in the flow and pe = pb Therefore, when the
nozzle is operating at the design condition:
pe 
p0
1000

 187.8 psia
5.324
p0 / pe
The highest back pressure, pb , that will give choking at the throat is that which gives
a Mach number of one at the throat and which involves subsonic flow in the divergent
section of the nozzle with the result that pe = pb , pe being the pressure on the nozzle exit
314
plane. For this case then the subsonic isentropic flow software or the isentropic flow
tables for air give for A / A * = 1.386:
p0
= 1.169
pe
Hence in this case:
pe 
p0
1000

 855.4 psia
1.169
p0 / pe
When there is a normal shock wave at the exit, the flow in the nozzle is shock free
with a supersonic velocity at the exit. Hence, if there is a shock wave at the exit, the
Mach number ahead of the shock wave will be 1.75 and pe = 187.8 psia. For a Mach
number of 1.75, the software for normal shock waves or the normal shock wave tables for
air give the pressure ratio across the wave as:
p0
= 3.406
pe
Now it will be noted that with a normal shock wave at the exit the flow downstream of
the shock wave will be subsonic and the pressure downstream of the shock wave must
therefore be equal to the back pressure, pb . Hence, when there is a normal shock wave at
the exit:
pb 
pb
pe  3.406  187.8  639.7 psia
pe
Normal shocks will appear in the nozzle for back pressure between that required to
first choke the nozzle and that which gives a normal shock wave at the exit, i.e., for back
pressures between 855.4 psia and 639.7 psia.
The flow will be supersonic on the exit plane at back pressures below that required to
give a normal shock wave on the exit plane, i.e., for back pressures below 639.7 psia.
Therefore the maximum back pressure to choke the nozzle is 855.4 psia, a normal
shock wave will occur in the divergent section of the nozzle for back pressure between
639.7 psia and 855.4 psia, the nozzle will be perfectly expanded when the back pressure
315
is 187.8 psia and the flow will be supersonic on the exit plane at back pressures below
639.7 psia.
316
PROBLEM 8.4
A convergent-divergent nozzle is designed to expand air from a chamber in which the
pressure is 700kPa and the temperature is 35oC to give a Mach number of 1.6. The mass
flow rate through the nozzle under design conditions is 0.012 kg/so Find (1) the. throat
and exit areas of the nozzle (2) the design back pressure and the temperature of the air
leaving the nozzle with this back pressure (3) the lowest back pressure for which there
will be no supersonic flow in the nozzle (4) the back pressure below which there are no
shock waves in the nozzle.
SOLUTION
It will be assumed that the supply chamber is large enough to be able to assume that
the chamber pressure is the stagnation pressure and that the chamber temperature is the
stagnation temperature, i.e.:
p0  700 kPa ,
T0  308 K
The nozzle is designed to generate an exit Mach number of 1.6. For this Mach
number the software for isentropic flow or the isentropic tables for air give:
Ae
= 1.250 ,
A*
p0
= 4.250 ,
pe
T0
= 1.512
Te
Ae , pe , and Te being the nozzle exit area, the pressure on the nozzle exit plane, and the
temperature on the nozzle exit plane respectively.
Part (1)
When the nozzle is operating at the design conditions the Mach number will be one at
the throat. The mass flow rate through the nozzle will therefore be given by:
m   *V * A *
317
But because when operating under design conditions the nozzle is choked so that M * = 1
it follows that V *= a *. Hence:
m   * a * A *
i.e.:
A* 
m
 *a*
For a Mach number of 1 the software for isentropic flow or the isentropic flow tables for
air give:
T0
 1.200 ,
T*
p0
 1.893
p*
Hence:
T* 
T0
308

 256.7 K
T0 / T * 1.200
p* 
p0
700

 369.8 kPa
p0 / p * 1.893
and:
Therefore:
* 
p*
369800

 5.020 kg/m3
RT * 287  256.7
and:
a* 
 R T* 
1.4  287  256.7  321.2 m/s
Therefore, since:
m  0.012 kg/s
It follows that:
A* 
m
0.012

 0.000007422 m 2
5.020  321.2
 *a*
318
This is the throat area of the nozzle. The exit area is given by:
Ae 
Ae
A *  1.250  0.000007422  0.000009303 m 2
A*
Therefore the throat and exit areas of the nozzle are 7.422 x 10-6 m2 and 9.303 x 10-6
m2 respectively.
Part (2)
When the nozzle is operating at the design conditions, i.e. when the nozzle is
perfectly expanded, there are no shock waves in the flow and pe = pb and as noted above:
p0
= 4.250 ,
pe
T0
= 1.512
Te
Therefore, when the nozzle is operating at the design conditions:
Te 
T0
308

 203.7 K
1.512
T0 / Te
and:
pe 
p0
700

 164.7 kPa
4.250
p0 / pe
Therefore, the design back pressure is 164.7 kPa and the temperature of the air
leaving the nozzle at the design conditions is 203.7 K ( = -69.3°C ).
Part (3)
The highest back pressure, pb , that will give choking at the throat is that which gives
a Mach number of one at the throat and which involves subsonic flow in the divergent
section of the nozzle with the result that pe = pb , pe being the pressure on the nozzle exit
plane. For this case then the subsonic isentropic flow software or the isentropic flow
tables for air give for
Ae
= 1.250
A*
319
the following:
p0
= 1.231
pe
Hence in this case:
pe 
p0
700

 568.6 kPa
1.231
p0 / pe
If the pressure is below this value, there will be a region of supersonic flow in the
divergent section of the nozzle. Therefore, the lowest back pressure at which there is no
supersonic flow in the nozzle is 568.6 kPa.
Part (4)
When there is a normal shock wave at the exit, the flow in the nozzle is shock free with a
supersonic velocity at the exit. Hence, if there is a shock wave at the exit, the Mach
number ahead of the shock wave will be 1.6 and pe = 164.7 kPa. For a Mach number of
1.6, the software for normal shock waves or the tables for a normal shock wave give the
pressure ratio across the wave as:
pb
= 2.820
pe
it having been noted that, with a normal shock wave at the exit, the flow downstream of
the shock wave will be subsonic and the pressure downstream of the shock wave must
therefore be equal to the back pressure, pb. Hence, when there is a normal shock wave at
the exit:
pb 
pb
pe  2.820  164.7  464.5 kPa
pe
If the back pressure is lower than this all waves that occur in the flow will be outside
the nozzle.
Therefore there will be supersonic flow on the nozzle exit plane when the back
pressure is less than 464.5 kPa.
320
PROBLEM 8.5
A converging-diverging nozzle is designed to generate an exit Mach number of 2.
The nozzle is supplied with air from a large reservoir in which the pressure is kept at 6.5
MPa. Assuming one-dimensional isentropic flow find: (1) the maximum back pressure at
which the nozzle will be choked, (2) the range of back pressures over which there will be
a shock in the nozzle, (3) the design back pressure, (4) the range of back pressures over
which there is supersonic flow on the nozzle exit plane.
SOLUTION
It will be assumed that the supply chamber is large enough to be able to assume that
the chamber pressure is the stagnation pressure, i.e.:
p0  6500 kPa
The nozzle is designed to generate an exit Mach number of 2. For this Mach number
the software for isentropic flow or the isentropic flow tables for air give:
Ae
= 1.688 ,
A*
Ae
p0
= 7.829
pe
and pe being the nozzle exit area and the pressure on the nozzle exit plane
respectively. Therefore when the nozzle is operating at design conditions:
pe 
p0
6500

 830.3 kPa
7.829
p0 / pe
Part (1)
The highest back pressure, pb , that will give choking at the throat is that which gives
a Mach number of one at the throat and which involves subsonic flow in the divergent
section of the nozzle with the result that pe = pb , pe being the pressure on the nozzle exit
plane. For this case then the subsonic isentropic flow software or the isentropic flow
tables for air give for Ae / A* = 1.688:
321
p0
= 1.100
pe
Hence in this case:
pe 
p0
6500

 5909 kPa  5.909 MPa
1.100
p0 / pe
Therefore the maximum back pressure at which the nozzle is choked is 5.909 MPa.
Part (2)
When there is a normal shock wave at the exit, the flow in the nozzle is shock free
with a supersonic velocity at the exit. Hence, if there is a shock wave at the exit, the
Mach number ahead of the shock wave will be 2 and pe = 830.3 kPa. For a Mach number
of 1.6, the software for normal shock waves or the tables for a normal shock wave for
give the pressure ratio across the wave as:
pb
= 4.500
pe
it having been noted that, with a normal shock wave at the exit, the flow downstream of
the shock wave will be subsonic and the pressure downstream of the shock wave must
therefore be equal to the back pressure, pb. Hence, when there is a normal shock wave at
the exit:
pb 
pb
pe  4.500  830.3  3736 kPa  3.736 MPa
pe
If the back pressure. is higher than this there will be a shock wave in the divergent
portion of the nozzle provided that the back pressure is less than that which gives
subsonic flow throughout the nozzle, this value having been found to be 5.905 MPa in
Part (1).
Therefore there will be a normal shock in the nozzle for back pressures that are less
than 5.905 MPa and greater than 3.736 MPa.
322
Part (3)
When the nozzle is operating at this design condition, i.e., when the nozzle is
perfectly expanded, there are no shock waves in the flow and pe = pb and, as noted above:
p0
= 7.829
pe
Under these conditions it was found above that pe = 830.3 kPa. Therefore the design back
pressure is 830.3 kPa.
Part (4)
If the back pressure is less than that required to give a normal shock wave on the exit
plane all waves that exist will be outside the nozzle. Therefore the flow on the exit plane
will be supersonic for back pressures that are less than 3.736 MPa.
323
PROBLEM 8.6
A convergent-divergent nozzle is designed to expand air from a chamber in which the
pressure is 800 kPa and the temperature is 40°C to give a Mach number of 2.5. The throat
area of the nozzle is 0.0025 m2. Find (1) the flow rate through the nozzle under design
conditions, (2) the exit area of the nozzle, (3) the design back pressure and the
temperature of the air leaving the nozzle with this back pressure, (4) the lowest back
pressure for which there is only subsonic flow in the nozzle, (5) the back pressure at
which there is a normal shock wave on the exit plane of the nozzle, (6) the back pressure
below which there are no shock waves in the nozzle, (7) the range of back pressures over
which there are oblique shock waves in the exhaust from the nozzle, (8) the range of back
pressures over which there are expansion waves in the exhaust from the nozzle, (9) the
back pressure at which a normal shock wave occurs in the divergent section of the nozzle
at a point where the nozzle area is half way between the throat and the exit plane areas.
SOLUTION
It will be assumed that the supply chamber is large enough to be able to assume that
the chamber pressure is the stagnation pressure and that the chamber temperature is the
stagnation temperature i.e.:
p0  800 kPa , T0  313 K
Part (1)
When the nozzle is operating at the design conditions the Mach number will be one at
the throat. The mass flow rate through the nozzle will therefore be given by:
m   *V * A *   * a * A *
For a Mach number of 1 the software for isentropic flow or the isentropic tables for
air give:
324
p0
= 1.893 ,
pe
T0
= 1.200
Te
Hence:
T* 
T0
313

 260.8 K
T0 / T * 1.200
and:
p* 
p0
800

 422.6 kPa
p0 / p * 1.893
Therefore:
* 
p*
422600

 5.646 kg/m3
RT * 287  260.8
and:
a* 
 R T* 
1.4  287  260.8  323.7 m/s
Therefore, since:
A *  0.0025 m 2
It follows that:
m   * a * A *  5.646  323.7  0.0025  4.659 kg/s
Therefore the mass flow rate through the nozzle is 4.659 kg/s
Part (2)
The nozzle is designed to generate an exit Mach number of 2.5. For this Mach
number the software for isentropic flow or the isentropic flow tables for air give:
Ae
= 2.637 ,
A*
p0
= 17.09 ,
pe
T0
= 2.250
Te
Ae , pe , and Te being the nozzle exit area, the pressure on the nozzle exit plane, and the
temperature on the nozzle exit plane respectively.
325
Hence:
Ae 
Ae
A *  2.637  0.0025  0.0066 m 2
A*
Therefore the exit plane area of the nozzle is 0.0066 m2.
Part (3)
When the nozzle is operating at this design condition, i.e., when the nozzle is
perfectly expanded, there are no shock waves in the flow and pe = pb and as noted above:
p0
= 17.09 ,
pe
T0
= 2.250
Te
Therefore, when the nozzle is operating at the design conditions:
Te 
T0
313

 139.1 K
2.250
T0 / Te
and:
pe 
p0
800

 46.81 kPa
17.09
p0 / pe
Therefore, the design back pressure is 46.81 kPa and the temperature of the air
leaving the nozzle at the design conditions is 139.1 K ( = -133.9°C ).
Part (4)
The highest back pressure, pb , that will give choking at the throat is that which gives
a Mach number of one at the throat and which involves subsonic flow in the divergent
section of the nozzle with the result that pe = pb , pe being the pressure on the nozzle exit
plane. For this case then the subsonic isentropic flow software or the isentropic flow
tables for air give for
Ae
= 2.637
A*
326
the following:
p0
= 1.036
pe
Hence in this case:
pe 
p0
800

 772.2 kPa
1.036
p0 / pe
Therefore the lowest back pressure at which there is only subsonic flow in the nozzle
is 772.2 kPa.
Part (5)
When there is a normal shock wave at the exit, the flow in the nozzle is shock free
with a supersonic velocity at the exit. Hence, if there is a shock wave at the exit, the
Mach number ahead of the shock wave will be 2.5 and pe = 48.61 kPa. For a Mach
number of 2.5, the software for normal shock waves or the tables for a normal shock
wave for give the pressure ratio across the wave as:
pb
= 7.125
pe
it having been noted that, with a normal shock wave at the exit, the flow downstream of
the shock wave will be subsonic and the pressure downstream of the shock wave must
therefore be equal to the back pressure, pb. Hence, when there is a normal shock wave at
the exit:
pb 
pb
pe  7.125  46.81  333.5 kPa
pe
If the back pressure is lower than this, all waves that occur in the flow will be outside
the nozzle.
Therefore there will be supersonic flow on the nozzle exit plane when the back
pressure is less than 464.5 kPa.
327
Part (6)
There are no shock waves in the nozzle for back pressures below that required to give
a normal shock on the exit plane. Hence, there are no shock waves in the nozzle for back
pressures below 333.5 kPa.
Part (7)
Oblique shock waves will occur in the exhaust from the nozzle for back pressures that
are below that which will give a normal shock wave on the exit plane and above the
design back pressure, i.e., for back pressures that are between 333.5 kPa and 46.81 kPa.
Part (8)
Expansion waves will occur in the exhaust from the nozzle for back pressures that are
below the design back pressure, i.e., for back pressures that are less than 46.81 kPa.
Part (9)
Because the nozzle has an area ratio of:
Ae
= 2.637
A*
the normal shock wave occurs at a point in the divergent section of the nozzle at which:
0.5  Ae  A *
A
A
=
= 0.5 e  0.5 = 0.5  2.637  0.5 = 1.819
A*
A*
A*
Using this area ratio, the isentropic flow software or the isentropic flow tables for air
give the Mach number and pressure ratio just upstream of the shock wave as:
M s = 2.089 ,
p0
= 8.992
ps
the subscript s denoting conditions just upstream of the shock wave. Hence, the pressure
just upstream of the shock wave is given by:
328
ps 
p0
800

 88.97 kPa
8.992
p0 / ps
Next consider the flow across the shock wave. For a Mach number of 2.089 the
software for normal shock waves or the tables for a normal shock wave give:
M d = 0.5629 ,
pd
= 4.925
ps
the subscript d denoting conditions just downstream of the shock wave.
Hence, the pressure just downstream of the shock wave is given by:
pd 
pd
ps  4.925  88.97  438.2 kPa
ps
For the flow just downstream of the shock wave where the Mach number is 0.5629
the software for isentropic flow or the isentropic flow tables for air give:
Ad
= 1.236 ,
Ad *
p0d
= 1.240
pd
Using this area ratio value gives:
Ae
A Ad
A / A * Ad
2.637
= e
= e
=
 1.236 = 1.792
Ad *
Ad Ad *
Ad / A * Ad *
1.819
For this area ratio the software for isentropic flow or the isentropic flow tables for air
give for subsonic flow:
p0d
= 1.087
pe
Hence:
pe 
p0 d / pd
1.240
pd 
 438.2  499.9 kPa
p0 d / pe
1.087
Because the flow downstream of the shock wave is subsonic this will be equal to the
back pressure.
329
Therefore the normal shock wave will occur at the specified position in the nozzle
when the back pressure is 499.9 kPa.
330
PROBLEM 8.7
A variable area diffuser is fitted to an aircraft designed to operate at a Mach number
of 3.5. If the shock wave is “swallowed” at this Mach number, find the ratio of the throat
area for shockless operation to the throat area at which the shock wave is swallowed.
SOLUTION
Consider the situation where there is a normal shock wave on the inlet plane, i.e., the
situation just before the shock is swallowed. The Mach number ahead of the shock wave
is 3.5. The software for normal shock waves or the tables for normal shock wave flow for
air then give the Mach number downstream of the shock wave as 0.4512. The flow
downstream of the shock wave is subsonic and software for isentropic flow at subsonic
speeds gives for M = 0.4512:
Ai
A
= i = 1.446
A*
At
the subscripts i and t referring to conditions on the inlet plane and at the throat
respectively.
Next consider the situation when the shock has been swallowed and the throat area
reduced until there are no shock waves in the diffuser. The software for isentropic flow or
the isentropic flow tables for air give for an inlet Mach number Mi = 3.5:
Ai
A
= i = 6.790
A*
At
Therefore with a shock at the inlet the throat area is given by:
Ats =
Ai
1.446
while with no shock waves the throat area is given by:
At =
Ai
6.790
331
The ratio of the throat area of the diffuser with no shocks to that during starting
therefore is:
At
A / 6.790
1.446
= i
=
= 0.2130
Ats
Ai /1.446
6.790
Therefore the ratio of the throat area with shockless operation to that required in order
to swallow the shock during starting is 0.2130.
332
PROBLEM 8.8
A nozzle is designed to expand air from a chamber in which the pressure and
temperature are 800 kPa and 40°C respectively to a Mach number of 2.5. The throat area
of this nozzle is to be 0.1 m2. Find (1) the exit area of the nozzle, (2) the mass flow rate
through the nozzle when operating at design conditions, (3) the back pressure at which
there will be a normal shock wave on the exit plane of the nozzle, (4) the range of back
pressures over which expansion waves will occur outside the nozzle.
SOLUTION
It will be assumed that the supply chamber is large enough to be able to assume that
the chamber pressure is the stagnation pressure and that the chamber temperature is the
stagnation temperature, i.e.:
p0  800 kPa , T0  313 K
Part (1)
The nozzle is designed to generate an exit Mach number of 2.5. For this Mach
number the software for isentropic flow or the isentropic tables for air give:
Ae
= 2.637 ,
A*
p0
= 17.09 ,
pe
T0
= 2.250
Te
Ae , pe , and Te being the nozzle exit area, the pressure on the nozzle exit plane, and the
temperature on the nozzle exit plane respectively.
Hence:
Ae 
Ae
A *  2.637  0.1  0.2637 m 2
A*
Therefore the exit plane area of the nozzle is 0.2637 m2 .
333
Part (2)
When the nozzle is operating at the design conditions the Mach number will be one at
the throat. The mass flow rate through the nozzle will therefore be given by:
m   *V * A *
But because when operating under design conditions the nozzle is choked so that M * = 1
it follows that V *= a *. Hence:
m   * a * A *
For a Mach number of 1 the software for isentropic flow or the isentropic flow tables
for air give:
T0
 1.200 ,
T*
p0
 1.893
p*
Hence:
T* 
T0
313

 260.8 K
T0 / T * 1.200
and:
p* 
p0
800

 422.6 kPa
p0 / p * 1.893
Therefore:
* 
p*
422600

 5.646 kg/m3
RT * 287  260.8
and:
a* 
 R T* 
1.4  287  260.8  323.7 m/s
Therefore, since:
A *  0.1 m 2
it follows that:
334
m   * a * A *  5.646  323.7  0.1  182.8 kg/s
Therefore the mass flow rate through the nozzle is 182.8 kg/s.
Part (3)
When the nozzle is operating at this design condition, i.e., when the nozzle is
“perfectly” expanded, there are no shock waves in the flow and pe = pb and, as noted
above,:
p0
= 17.09
pe
Hence, when the nozzle is operating at the design condition:
pe 
p0
800

 46.81 kPa
17.09
p0 / pe
Therefore the design back pressure is 46.81 kPa.
When there is a normal shock wave at the exit, the flow in the nozzle is shock free
with a supersonic velocity at the exit. Now, if there is a shock wave at the exit, the Mach
number ahead of the shock wave will be 2.5 and pe = 46.81 kPa. For a Mach number of
2.5, the software for normal shock waves or the tables for a normal shock wave for air
give the pressure ratio across the wave as:
pb
= 7.125
pe
Here it has been noted that with a normal shock wave at the exit the flow downstream
of the shock wave will be subsonic and the pressure downstream of the shock wave must
therefore be equal to the back pressure, pb . Hence, when there is a normal shock wave at
the exit:
pb 
pb
pe  7.125  46.81  333.5 kPa
pe
Therefore there is a normal shock wave on the nozzle exit plane when the back
pressure is 333.5 kPa.
335
Part (4)
Expansion waves will occur in the exhaust from the nozzle for back pressures that are
below the design back pressure, i.e., for back pressures that are less than 46.81 kPa.
336
PROBLEM 8.9
A rocket nozzle is designed to operate supersonically with a chamber pressure of 500
psia and an ambient pressure of 14.7 psia. Find the ratio between the thrust at sea level to
the thrust in space (0 psia). Assume a constant chamber pressure with a chamber
temperature of 2500oR. Assume the rocket exhaust gases behave as a perfect gas with γ =
1.4 and R = 20 ft-lbf/lbm oR.
SOLUTION
It will be assumed that the pressure and temperature in the combustion chamber are
effectively the stagnation pressure and the stagnation temperature i.e. that:
p0  500 psia , T0  2960 o K
Because the engine is designed to operate at an ambient pressure of 14.7 psia, on the
nozzle exit plane:
p0
500
=
= 34.01
14.7
pe
For this pressure ratio, the software for supersonic isentropic flow gives:
Ae
= 4.033 ,
A*
T0
= 2.739 ,
Te
M e = 2.949
Hence:
Te 
T0
2960

 1081 o R
T0 / Te
2.739
Therefore on the exhaust plane:
e 
pe
14.7  144

 0.09791 lbm/ft 3
RTe
20  1081
and:
337
ae 
 R Te 
1.4  20  32.2  1081  987.2 ft/sec
Hence:
Ve  M e ae  2.949  987.2  2911 ft/sec
If a control volume drawn around the engine is considered then, at design ambient
conditions, the pressure will be atmospheric everywhere on the surface of this control
volume and therefore if Fd is the engine thrust:
 e   e Ve2 Ae  0.9791  29112  Ae / 32.2  25770 Ae
Fd  mV
the conversion factor 1 lbf = 32.2 lbm ft / sec 2 having been used.
When the engine is operating in space, the pressure on the exhaust plane will still be
equal 14.7 psia. If a control volume drawn around the engine is considered then, under
these conditions, the pressure will be zero everywhere on the surface of this control
volume except on the exhaust plane and therefore if Fs is the engine thrust under these
conditions:
 e  ( pe  0) Ae   e Ve2 Ae  pe Ae
Fd  mV
 0.9791  29112  Ae / 32.2  144  14.7  Ae  27880 Ae
The ratio of the thrust under the design sea-level conditions to the thrust in space is
therefore given by:
Fd
25770

 0.9243
27880
Fs
Therefore the ratio of the thrust under the design sea-level conditions to the thrust in
space is 0.9243.
338
PROBLEM 8.10
Air flows through a convergent-divergent nozzle that has an inlet area of 0.0025m2.
The inlet temperature and pressure are 50°C and 550 kPa respectively and the velocity at
the inlet is 80 m/s. If the flow is assumed to be isentropic, and if the exit pressure is 120
kPa, find the throat and exit areas and the exit velocity.
SOLUTION
Because the flow is assumed to be isentropic there can be no shock waves in the flow.
Subscripts i and e will be used to denote conditions at the inlet and exit respectively.
Here:
Ai = 0.0025 m 2 , Ti  323 K , pi  550 kPa , Vi  80 m/s and pe  120 kPa
At the inlet:
Mi 
Vi

ai
Vi

 R Ti
80
 0.2221
1.4  287  323
For this Mach number the software for isentropic flow or the isentropic tables for air
give:
Ai
= 2.683 ,
A*
p0
= 1.035 ,
pi
T0
= 1.010
Ti
Hence:
A* 
Ai
0.0025

 0.0009318 m 2
Ai / A *
2.683
p0 
p0
pi  1.035  550  569.3 kPa
pi
and:
and:
339
T0 
T0
Ti  1.010  323  362.2 K
Ti
Therefore:
p0
569.3

 4.744
pe
120
For this value of p0 / p the software for isentropic flow or the isentropic tables for air
give:
Ae
= 1.313 ,
A*
T0
= 1.560 ,
Te
M e = 1.674
Using these results gives:
Ae 
Ae / A *
1.313
Ai 
 0.0025  0.001223 m 2
Ai / A *
2.683
and:
Te 
T0
326.2

 209.1 K
T0 / Te
1.560
Hence:
Ve  M e ae  Ve
 R Te  1.674  1.4  287  209.1  1.674  289.9  485.3 m/s
Therefore the throat and exit areas are 0.0009318 m2 and 0.001223 m2 respectively
while the exit velocity is 485.3 m/s.
340
PROBLEM 8.11
Air flows through a convergent divergent passage. The passage inlet area is 5 cm2, the
minimum area is 3 cm2 and the exit area is 4 cm2. The air velocity at the inlet to the
passage is 120 m/sec, the pressure is 700 kPa and the temperature is 40°C. Assuming that
the flow is isentropic, find the mass flow rate through the passage, the Mach number at
the minimum area section, and the velocity and pressure at the exit section.
SOLUTION
Subscripts i, t and e will be used to denote conditions at the inlet, at the throat, and at
the exit respectively. Here:
Ai = 0.0005 m 2 , Ti  313 K , pi  700 kPa , Vi  120 m/s , At  0.0003 m 2 , Ae  0.0004 m 2
The mass flow rate through the nozzle is given by:
m  i Vi Ai
But:
i 
pi
700000

 7.792 kg/m3
287  313
RTi
Hence:
m  i Vi Ai  7.792  120  0.0005  0.4676 kg/s
At the inlet:
Mi 
Vi

ai
Vi

 R Ti
120
120

 0.3384
354.6
1.4  287  313
For this Mach number the software for isentropic flow or the isentropic tables for air
give:
Ai
= 1.832 ,
A*
p0
= 1.082 ,
pi
341
T0
= 1.023
Ti
Hence:
At
A A
 0.0003 
 t i 
  1.832  1.099
A*
Ai A *  0.0005 
For this value of A / A* the software for isentropic flow or the isentropic tables for air
give:
M t = 0.6937
Therefore the Mach number does not reach a value of one at the throat and the flow
remains subsonic throughout the nozzle.
Next consider the flow at the exit. Here:
Ae
A A
 0.0004 
 e i 
  1.832  1.466
A*
Ai A *
 0.0005 
For this value of A / A* the software for isentropic flow or the isentropic tables for air
give:
M e = 0.4431 ,
p0
= 1.144 ,
pe
T0
= 1.039
Te
Therefore:
pe 
p0 / pi
1.082
pi 
 700  662.1 kPa
p0 / pe
1.144
and:
Te 
T0 / Ti
1.023
Ti 
 313  308.2 K
T0 / Te
1.039
Finally:
Ve  M e ae  Ve
 R Te  0.4431  1.4  287  308.2  0.4431  351.9  155.9 m/s
Therefore the mass flow rate through the nozzle is 0.4676 kg/s, the Mach number at
the nozzle throat is 0.6937 and the velocity and pressure on the exit are 155.9 m/s and
662.1 kPa respectively.
342
PROBLEM 8.12
Air flows through a convergent-divergent nozzle from a large reservoir in which the
pressure and temperature are maintained at 700 kPa and 60°C respectively. The rate of air
flow through the nozzle is 1 kg/s. On the exit plane of the nozzle the stagnation pressure
is 550 kPa and the static pressure is 500 kPa. A shock wave occurs in the nozzle and the
flow can be assumed to be isentropic everywhere except through the shock wave. Find
the nozzle throat area, the Mach numbers before and after the shock, the nozzle areas at
the point where the shock occurs and on the exit plane and the air density on the exit
plane of the nozzle.
SOLUTION
It will be assumed that the reservoir is large enough for the reservoir pressure to be
taken as the stagnation pressure upstream of the shock wave and for the reservoir
temperature to be taken as the stagnation temperature i.e.:
p0  700 kPa , T0  333 K
Subscripts s, d, and e will be used to denote conditions just upstream of the shock
wave, just downstream of the shock wave and at the exit respectively. Because a shock
wave occurs in the flow, the velocity must go from subsonic in the convergent section to
supersonic in the divergent section. Hence, the Mach number at the throat must be one.
The mass flow rate through the nozzle will therefore be given by:
m   *V * A *   * a * A *
For a Mach number of 1 the software for isentropic flow or the isentropic tables for
air give:
p0
= 1.893 ,
pe
Hence:
343
T0
= 1.200
Te
T* 
T0
333

 277.5 K
T0 / T * 1.200
p* 
p0
700

 369.8 kPa
p0 / p * 1.893
and:
Therefore:
* 
p*
369800

 4.643 kg/m3
RT * 287  277.5
and:
a* 
 R T* 
1.4  287  277.5  333.9 m/s
Therefore, since:
m  1 kg/s
it follows that:
A* 
m
1

 0.000645 m 2
4.643  333.9
 *a*
The difference between the stagnation pressures at the inlet and the exit all occurs
across the shock wave, i.e.:
p0d
550

 0.7857
p0s
700
The software for normal shock waves or the normal shock wave tables for air give for
this stagnation pressure ratio:
M s = 1.860 ,
M d = 0.6036 ,
344
Td
= 1.577
Ts
Therefore, the shock wave occurs at a point in the divergent section of the nozzle at
which the Mach number is 1.860. The software for isentropic flow or the isentropic tables
for air give for this Mach number:
As
 1.507
A*
Hence the nozzle area at the point where the shock wave occurs is given by:
As  A *
As
 0.000645  1.507  0.000974 m 2
A*
On the exit plane:
p0d
550

 1.1
500
pe
The software for isentropic flow or the isentropic tables for air give for this stagnation
pressure ratio:
Ae
= 1.690 ,
Ad *
T0
= 1.028
Te
The Mach number downstream of the shock wave is 0.6036 and the software for
isentropic flow or the isentropic tables for air give for this Mach number:
Ad
 1.184
Ad *
Also, because:
Ad  As  0.000974 m 2
it follows that:
Ae 
Ae / Ad *
1.690
Ad 
 0.000974  0.001390 m 2
Ad / Ad *
1.184
Also:
345
Te 
T0
333

 323.9 K
1.028
T0 / Te
Therefore:
e 
pe *
500000

 5.379 kg/m3
RTe
287  323.9
Therefore the nozzle throat area is 0.000645 m2, the Mach numbers upstream and
downstream of the shock are 1.860 and 0.6036, the nozzle areas at the point where the
shock occurs and at the exit are 0.000974 m2 and 0.001390 m2 and the density on the exit
plane is 5.379 kg / m3.
346
PROBLEM 8.13
Air flows through a converging-diverging nozzle. The air has a Mach number of 0.50
and a pressure and a temperature of 280 kPa and 100oC respectively at the inlet to the
nozzle. The nozzle throat area is 6.5  10- 4 m2 and ratio of the exit area to the throat area
is 4. If the pressure on the exit plane of the nozzle is 170 kPa, find the Mach number and
the temperature on the nozzle exit plane and the nozzle area at the point in the nozzle at
which the normal shock wave occurs.
SOLUTION
Conditions at the inlet, just upstream of the shock wave, just downstream of the shock
wave, and at the exit will be designated by the subscripts i, s, d and e respectively. For the
given inlet Mach number of 0.5, the software for isentropic flow or the isentropic tables
for air give:
p0
= 1.186 ,
pi
T0
= 1.050
Ti
Therefore:
p0 
p0
pi  1.186  280  332.1 kPa
pi
and:
T0 
T0
Ti  1.050  283  297.2 K
Ti
The information provided also gives:
A *  0.00065 m 2 ,
Ae 
Ae
A *  4  0.00065  0.002600 m 2
A*
A simple trial-and-error solution procedure will be used to find the nozzle area at
which the shock wave occurs. This involves the following steps:
347
1. Guess the area, As , at which the shock wave occurs. The value must be between
the throat and the exit plane areas, i.e., between 0.00065 m2 and 0.0026 m2.
2. Calculate the area ratio for the flow just upstream of the shock wave, i.e. find:
As
As
=
0.00065
A*
and then use the software for isentropic flow or the isentropic tables for air to give
the values of:
Ms
and
p0
ps
corresponding to this area ratio. Then calculate:
ps 
p0
332.1

kPa
p0 / ps
p0 / ps
3. For the value of Ms obtained in (2), use the software for normal shock waves or
the normal shock tables for air to obtain the values of:
Md
and
pd
ps
Then find the pressure just downstream of the shock wave using:
pd  ps
pd
ps
4. Next consider the flow downstream of the shock wave. For the value of Md
obtained in (3), use the software for isentropic flow or the isentropic tables for air
to give the values of:
p0d
pd
and
5. Then calculate:
348
Ad
Ad *
Ae
A A
0.0026 Ad
 e d 
Ad *
As Ad *
As
Ad *
it having been noted that:
Ad  As
6. Use the value of the area ratio Ae / Ad*obtained in (5) with isentropic flow
software or the isentropic tables for air to give the value of:
p0d
pe
then find the value of pe corresponding to the chosen value of As using:
pe 
p0d / pd
pd
p0d / pe
The value of As that gives pe = 170 kPa can then be deduced from the results so
obtained. Typical results obtained using this procedure are shown in the following table:
As – m2
0.00100
0.00200
0.00175
0.00155
Ms
1.886
2.664
2.522
2.392
pe - kPa
250.5
132.7
152.2
171.1
Hence the shock wave effectively occurs at a point in the nozzle where the area is
0.00156 m 2, the Mach number ahead of the shock then being 2.400 and:
Ae
 2.163
Ad *
For this value of the area ratio, the software for isentropic flow or the isentropic flow
tables for air give:
M e  0.279 and
349
T0
 1.016
Te
Because the stagnation temperature remains constant in the flow:
Te 
T0
297.2

 292.5 K
1.016s
T0 / Te
Therefore the Mach number and temperature on the exit plane are 0.279 and 292.5 K
( = 19.5oC ) respectively and the area of the nozzle at the point where the shock wave
occurs is 0.00156 m2.
350
PROBLEM 8.14
Air is supplied to a convergent-divergent nozzle from a large tank in which the
pressure and temperature are kept at 700 kPa and 40°C respectively. If the nozzle has an
exit area that is 1.6 times the throat area and if a normal shock occurs in the nozzle at a
section where the area is 1.2 times the throat area, find the pressure, temperature and
Mach number at the nozzle exit. Assume one-dimensional, isentropic flow.
SOLUTION
The flow is assumed to one-dimensional and isentropic everywhere except through
the shock wave. It will be assumed that the supply tank is large enough to be able to
assume that the tank pressure is the stagnation pressure and that the tank temperature is
the stagnation temperature, i.e.:
p0  700 kPa , T0  313 K
Conditions just upstream of the shock, just downstream of the shock and on the exit
plane will be denoted by subscripts s, d and e respectively. The normal shock wave
occurs at a point in the divergent section of the nozzle at which:
As
 1.2
As *
The software for isentropic flow or the isentropic tables for air give for this area ratio
the Mach number just upstream of the shock wave as:
M s = 1.534
Next consider the flow across the shock wave. The software for normal shock waves
or the tables for a normal shock in air give for a Mach number of 1.534:
M d = 0.6894 ,
351
p0d
= 0.9187
p0s
The software for isentropic flow or the isentropic tables for air give for a Mach number of
0.6894:
Ad
 1.102
Ad *
Hence since As = Ad:
Ae
A A
A A
A / A * Ad
1.6
 e d  e d  e e

 1.102  1.469
Ad *
Ad Ad *
As Ad *
As / As * Ad * 1.2
The software for isentropic flow or the isentropic tables for air give for this value of
the area ratio:
M e = 0.4418 ,
p0d
= 1.143 ,
pe
T0d
= 1.039
Te
Therefore:
pe 
p0d / p0s
0.9187
p0s 
 700  562.6 kPa
p0d / pe
1.143
There is no change in the stagnation temperature across the shock wave, i.e.:
T0d  T0s  T0  313 K
Hence:
Te 
T0d
313

 301.3 K
T0d / Te
1.039
Therefore the pressure, temperature and Mach number on the nozzle exit plane are
562.6 kPa, 301.3 K ( = 28.3°C ) and 0.4418 respectively.
352
PROBLEM 8.15
Air enters a convergent-divergent nozzle at a Mach number of 0.2. The stagnation
pressure is 700 kPa and the stagnation temperature is 5oC. The throat area of the nozzle is
46  10-4 m2 and the exit area is 230  10-4 m2. If the pressure at the exit to the nozzle is
500 kPa, determine if there is a shock in the divergent portion of the nozzle. If there is a
shock wave, determine the nozzle area at which the shock occurs and the Mach number
and pressure just before and just after the shock wave.
SOLUTION
If there is no shock wave in the nozzle the flow will be isentropic throughout. In this
case:
Ae
0.0230

5
A * 0.0046
For an area ratio of 5, the software for supersonic isentropic flow or the isentropic tables
for air give:
p0
= 47.64
pe
Therefore if there are no shock waves in the nozzle and the flow is supersonic in the
divergent portion of the nozzle:
pe 
p0
700

 14.69 kPa
47.64
p0 / pe
Similarly for an area ratio of 5, the software for subsonic isentropic flow or the isentropic
tables for air give:
p0
= 1.010
pe
Therefore if there are no shock waves in the nozzle and the flow is subsonic in the
divergent portion of the nozzle:
353
pe 
p0
700

 693.1 kPa
1.010
p0 / pe
The actual back pressure, i.e., 500 kPa, lies between the above two values of pe which
indicates that there is a normal shock wave in the divergent section of the nozzle.
A simple trial-and-error solution procedure will be used to find the nozzle area at
which the shock wave occurs. Conditions at the inlet, just upstream of the shock wave,
just downstream of the shock wave and at the exit will be designated by subscripts i, s, d,
and e respectively. The solution procedure then involves the following steps:
1. Guess the area, As at which the shock wave occurs. The 'value must be between
the throat and the exit plane areas, i.e., between 0.0046 and 0.0230m2.
2. Calculate the area ratio for the flow just upstream of the shock wave, i.e., find:
As
As

A * 0.0046
and then use the software for isentropic flow or the isentropic tables for air to give
the values of:
Ms
and
p0
ps
corresponding to this area ratio. Then calculate:
ps 
p0
700

kPa
p0 / ps
p0 / ps
3. For the value of Ms obtained in (2), use the software for normal shock waves or
the normal shock tables for air to obtain the values of:
Md
and
pd
ps
Then find the pressure just downstream of the shock wave using:
pd  ps
354
pd
ps
4. Next consider the flow downstream of the shock wave. For the value of Md
obtained in (3), use the isentropic flow software or the isentropic tables for air to
give the values of:
p0d
ps
and
Ad
Ad *
5. Then calculate:
Ae
A A
0.0230 Ad
 e d 
Ad *
As Ad *
As
Ad *
it having been noted that:
Ad  As
6. Use the value of the area ratio Ae / Ad* obtained in (5) with isentropic flow
software or the isentropic tables for air to obtain the value of:
p0d
pe
then find the value of pe corresponding to the chosen value of As using:
pe 
p0d / pd
pd
p0d / pe
The value of As that gives pe = 500 kPa can then be deduced from the results so obtained.
Typical results obtained using this procedure are shown in the following table:
As – m2
0.0150
0.0100
0.0075
Ms
2.725
2.290
1.958
pe - kPa
273.4
399.8
509.2
From results such as these it can be deduced that the shock wave occurs at a point in
the nozzle where the area is 0.00766 m2, the Mach number ahead of the shock then being
1.986.
355
For Ms equal to 1.986, the software for normal shock waves or the normal shock
tables for air give:
M d  0.5798 and
pd
 4.435
ps
Also, for this value of Ms, the software for isentropic flow or the isentropic tables for air
give:
p0
 7.656
ps
Therefore:
ps 
p0
700

 91.4 kPa
7.656
p0 / ps
hence:
pd 
pd
ps  4.435  91.4  405.5 kPa
ps
Therefore the shock wave occurs at a point in the nozzle at which the area of the
nozzle is 0.00766 m2 and the Mach number is 1.986. The pressures just upstream and just
downstream of the shock wave are 91.4 kPa and 405.5 kPa respectively.
356
PROBLEM 8.16
Air at a temperature 20°C and a pressure 101 kPa flows through a convergentdivergent nozzle at the rate of 0.5 kg/s. The exit area of the nozzle is 1.355 times the
inlet area. If the air leaves the nozzle at a static temperature of 20°C and a stagnation
temperature of 30°C, calculate the inlet and exit Mach numbers, the increase in entropy
(if any) between the inlet and the outlet, the area at which the shock (if any) occurs and
the stagnation pressure at the exit.
SOLUTION
It will be assumed that the flow is adiabatic. In this case, even if a shock wave occurs
in the flow, the stagnation temperature remains the same everywhere in the nozzle, i.e.:
T0 = 303 K
Conditions on the inlet plane will be denoted by the subscript i. On the inlet plane:
T0
303
=
= 1.034
Ti
293
The software for isentropic flow or the isentropic flow tables for air give for this
value of the stagnation temperature ratio:
M i = 0.4123 ,
Ai
= 1.552 ,
A*
p0
= 1.124
pi
Therefore:
p0 
p0
pi  1.124  101  113.5 kPa
pi
It is next recalled that:
m  i Vi Ai  i M i ai Ai
Hence:
357
Ai 
m
i M i ai
But:
i 
pi
101000

 1.201 kg/m3
RTi
287  293
and:
ai 
 R Ti 
1.4  287  293  343.1 m/s
and:
M i = 0.4123 ,
m = 0.5 kg/s
Hence:
Ai 
m
0.5

 0.002943 m 2
1.201  0.4123  343.1
i M i ai
The throat area of the nozzle is then given by:
At  A * 
Ai
0.002943

 0.001896 m 2
1.552
Ai / A *
Next consider conditions at the exit, these conditions being designated by the
subscript e. Here:
Te  Ti
Therefore, since the stagnation temperature is the same at all points in the nozzle:
T0
T
= 0
Ti
Te
Hence:
M i = M e = 0.4123
But:
358
Ae = 1.355 Ai
i.e., the Mach numbers at inlet and exit are the same but the flow areas are different. This
can only be the case if there is a shock wave in the flow.
The above results indicate that if As is the area of the nozzle at which the shock wave
occurs and if subscripts s and d refer to conditions just upstream and just downstream of
the shock wave then since:
As = Ad
it follows that:
As A *
A Ad *
Ai  d
Ae
A * Ai
Ad * Ae
where A* is the actual throat area .and Ad* is the critical area for the flow downstream of
the shock wave.
Now because Me = Mi:
A *
A*
 d
Ai
Ae
the above equation gives:
As
A
 1.355 d
A*
Ad *
i.e., the shock wave must occur at a point in the divergent section of the nozzle at which
the Mach number is such that the above relation is satisfied. This value of the Mach
number Ms will be found using a trial-and-error procedure that involves the following
steps:
1. Guess the value of Ms.
2. Use software for isentropic flow or the isentropic flow tables for air to find the
value of As / A * corresponding to this value of Ms.
3. Use the software for normal shock waves or the normal shock wave tables for air
to find the value of Md corresponding to this value of Ms.
359
4. Use isentropic software or the isentropic flow tables for air to find the value of
Ad/Ad * corresponding to this value of Md .
5. Evaluate 1.355 Ad / Ad * and compare the result with the value of As / A *. The
value of Ms that gives:
As
A
 1.355 d
A*
Ad *
can then be deduced. Some typical results are shown in the following table.
Ms
1.500
2.000
1.800
1.900
1.950
1.965
Md
0.7011
0.5774
0.6165
0.5956
0.5862
0.5835
As / A *
1.176
1.688
1.439
1.555
1.619
1.639
1.355 Ad / Ad *
1.482
1.648
1.584
1.617
1.633
1.637
Hence the shock wave effectively occurs at a point in the nozzle where the Mach
number is 1.963, the value of As / A * corresponding to this value of Ms being 1.637. At
this Mach number, the software for normal shock waves or the normal shock wave tables
for air give:
pd
= 4.329 ,
ps
Td
= 1.658 ,
Ts
p0d
= 0.7381
p0s
The nozzle area at the point where the shock occurs is given by:
As  A *
As
 0.001896  1.637  0.003104 m 2
A*
Further, because p0s = p0 = 113.5 kPa, it follows that:
p0e  p0d 
p0d
p0s  0.7381  113.5  83.78 kPa
p0s
The changes in entropy across the shock wave, Δs , is given by:
360
 1
1.4 1





1.4




T
p
1
s


d
s



 ln    
 ln 1.658  
   0.08695
 Ts   pd  

4.329
cp






Hence:
s  0.08695 cp  0.08695  1007  87.55 J / kg K
Therefore the Mach number at both the inlet and the exit of the nozzle is 0.4123, the
area of the nozzle at the section where the shock wave occurs is 0.003104 m2, the
increase in entropy in the nozzle is 87.55 J/kg K and the stagnation pressure at the exit is
83.78 kPa.
361
PROBLEM 8.17
Air flows from a large reservoir in which the temperature and pressure are 80°C and
780 kPa through a convergent-divergent nozzle which has a throat diameter of 2.5 cm.
When the back pressure is 560 kPa, a shock wave is found to occur at a location in the
nozzle where the static pressure is 210 kPa. Find the exit area, exit temperature, the exit
Mach number, the area at .which the shock wave occurs and the pressure ratio across the
shock.
SOLUTION
It will be assumed that the reservoir is large enough to be able to assume that the
reservoir pressure is the stagnation pressure and that the reservoir temperature is the
stagnation temperature ahead of the shock wave, i.e.:
p0  780 kPa , T0  353 K
The throat area of the nozzle is given by:
A* 

4
D *2 

4
 0.0252  0.0004909 m 2
It has been noted that because there is a shock wave in the in the divergent portion of the
nozzle, the nozzle must be choked, i.e., the Mach number must be one at the nozzle
throat.
Conditions just upstream of the shock, just downstream of the shock and on the exit
plane will be denoted by subscripts s, d and e respectively. The normal shock wave
occurs at a point in the divergent section of the nozzle at which:
ps  210 kPa
i.e., at which:
p0
780

 3.714
210
ps
362
The software for isentropic flow or the isentropic tables for air give for this pressure
ratio:
M s = 1.508 ,
As
= 1.182
A*
Hence:
As  A *
As
 0.0004909  1.182  0.0005802 m 2
*
A
Next consider the flow across the shock wave. The software for normal shock waves
or the normal shock wave tables for air give for an upstream Mach number of 1.508:
M d = 0.6036 ,
p0d
= 0.9272 ,
p0s
pd
= 2.486
ps
Now, for a Mach number of 0.6893, the software for isentropic flow or the isentropic
tables for air give:
Ad
 1.096
Ad *
Using the above results, the stagnation pressure behind the shock wave is given by:
p0d =
p0d
p0s = 0.9272  780 = 723.2 kPa
p0s
Hence on the exit plane:
p0d
723.2

 1.2915
560
pe
The software for isentropic flow or the isentropic tables for air give for this stagnation
pressure ratio:
M e = 0.6157 ,
Ae
T
= 1.170 , 0d  1.076
Ad *
Te
Therefore, because there is no change in the stagnation temperature across the shock
wave:
363
Te 
T0
353

 328.1 K
1.076
T0 d / Te
Also, since As = At:
Ae 
Ae / Ad *
A /A *
1.170
 e d As 
 0.0005802  0.0006194 m 2
Ad / Ad *
Ad / Ad *
1.096
Therefore the area, temperature and Mach number on the nozzle exit plane are
0.0006194 m2 , 328.1 K ( = 55.1oC ) and 0.6157 respectively, the shock wave occurs at a
point in the divergent section where the area is 0.0005802 m2 and the pressure ratio
across the shock wave is 2.486.
364
PROBLEM 8.18
Air flows from a reservoir in which the pressure is kept at 124 kPa through a
convergent-divergent nozzle and exhausts to the atmosphere where the pressure is 101.3
kPa. Under these conditions the nozzle is choked and the flow is subsonic on both sides
of the throat. To what value must the pressure in the reservoir be raised so that there is a
normal shock on the nozzle exit plane?
SOLUTION
It will be assumed that the supply reservoir is large enough to be able to assume that
the chamber pressure is the stagnation pressure, i.e., that:
p0  124 kPa
Because the flow is subsonic in the nozzle, the pressure on the exit plane of nozzle is
equal to the back pressure, i.e.:
pe  101.3 kPa
It therefore follows that:
p0
124

 1.224
pe
101.3
For a pressure ratio of 1.224 the software for isentropic flow or the isentropic tables
for air give:
Ae
= 1.262
A*
When there is a normal shock wave at the exit, the flow in the divergent portion of the
nozzle is supersonic and shock free. There is therefore a supersonic velocity at the exit.
Because the area ratio of the nozzle is 1.262, the software for isentropic flow or the
isentropic flow tables for air give for this area ratio:
365
M e = 1.614 ,
p0
= 4.342
pe
Hence, the Mach number ahead of the shock wave will be 1.614. For a Mach number
of 1.614, the software for normal shock waves or the normal shock wave tables for air
give the pressure ratio across the wave as:
pb
= 2.872
pe
But, with a normal shock wave at the exit, the flow downstream of the shock wave will
be subsonic and the pressure downstream of the shock wave must therefore be equal to
the back pressure, pb , i.e., will be equal to 101.3 kPa. Hence, when there is a normal
shock wave at the exit:
pe 
pb
101.3

 35.27 kPa
2.872
pb / pe
The pressure in the supply reservoir, i.e., p0 , with a normal shock on the nozzle exit
plane is therefore given by:
p0 
p0
pe  4.342  35.27  153.1 kPa
pe
Therefore when there is a normal shock wave on the exit plane of the nozzle the
pressure in the supply reservoir must be increased to 153.1 kPa.
366
PROBLEM 8.19
Air flows through a convergent-divergent nozzle. The nozzle exit and throat areas are
0.5 and 0.25 m2 respectively. If the inlet stagnation pressure is 200 kPa and the back
pressure is 120 kPa, determine the nozzle area at which the normal shock wave is located.
What is the increase in entropy across the shock? At what back pressure will the shock
wave be located on the nozzle exit plane?
SOLUTION
Conditions just upstream and just downstream of the shock wave will be designated
by subscripts s and d respectively. The information provided gives:
p0s  200 kPa ,
pe  120 kPa
it having been noted that because the flow downstream of the shock wave is subsonic, the
exit plane pressure will be equal to the back pressure.
A simple trial-and-error solution procedure will be adopted. This involves the
following steps:
1. Guess the area, As at which the shock wave occurs. The value must be between
the throat and the exit plane areas, i.e., between 0.5 and 0.25 m2.
2. Calculate the area ratio for the flow just upstream of the shock wave, i.e., find:
As
A
 s A*  4 A*
A * 0.25
and then use the software for isentropic flow or the isentropic flow tables for air to
give the values of:
Ms
and
p0
ps
corresponding to this area ratio. Then calculate:
367
ps 
p0
200

kPa
p0 / ps
p0 / ps
3. For the value of Ms obtained in (2), use the software for normal shock waves or
the normal shock wave tables for air to obtain the values of:
Md
pd
ps
and
Then find the pressure just downstream of the shock wave using:
pd  ps
pd
ps
4. Next consider the flow downstream of the shock wave. For the value of Md
obtained in (3), use the isentropic flow software or the isentropic flow tables for
air to give the values of:
p0d
ps
and
Ad
Ad *
5. Then calculate:
Ae
A A
0.5 Ad
 e d 
Ad *
As Ad *
As Ad *
it having been noted that:
Ad  As
6. Use the value of the area ratio Ae / Ad* obtained in (5) with isentropic flow
software or the isentropic flow tables for air to give the value of:
p0d
pe
then find the value of pe corresponding to the chosen value of As using:
pe 
p0d / pd
pd
p0d / pe
368
The value of As that gives pe = 120 kPa can then be deduced from the results so
obtained.
Typical results obtained using this procedure are shown in the following table:
As – m2
0.300
0.400
0.440
0.420
0.430
0.439
Ms
1.534
1.935
2.050
1.995
2.023
2.047
pe - kPa
162.2
129.2
119.8
124.7
122.8
119.9
Hence the shock wave effectively occurs at a point in the nozzle where the area is
0.439 m2, the Mach number ahead of the shock being 2.047. At this Mach number, the
software for normal shock waves or the normal shock wave tables for air gives:
pd
= 4.722 ,
ps
Td
= 1.726
Ts
The changes in entropy across the shock wave, Δs , is given by:
 1
1.4 1





1.4




T
p
1
s


d
s



 ln    
 ln 1.726  
   0.1023
 Ts   pd  

cp
 4.722  




Hence:
s  0.1023 cp  0.1023  1007  103.0 J / kg K
When there is a normal shock wave at the exit, the flow in the divergent portion of the
nozzle is supersonic and shock free. There is therefore a supersonic velocity at the exit.
Because the area ratio of the nozzle is 2, the software for supersonic isentropic flow or
the isentropic flow tables for air give:
M e  2.197 and
369
p0
 10.65
pe
Hence:
pe 
p0
200

 18.78 kPa
10.65
p0 / pe
The Mach number ahead of the shock wave is 2.197. With a normal shock wave at
the exit the flow downstream of the shock wave will be subsonic and the pressure
downstream of the shock wave must therefore be equal to the back pressure, pb. Hence,
for this Mach number of 2.197, the software for normal shock waves or the normal shock
wave tables for air give the pressure ratio across the wave as:
pb
 5.465
pe
Hence:
pb 
pb
pe  5.465  18.78  102.6 kPa
pe
Therefore when there is a normal shock wave in the nozzle it occurs at a position at
which the area in the nozzle is 0.439 m2 . The change in entropy across this shock wave
is 103.0 J / kg K. When there is a normal shock wave on the exit plane of the nozzle the
back pressure is 102.6 kPa.
370
PROBLEM 8.20
Air at a pressure of 350 kPa, a temperature of 80°C and a velocity of 180 m/s enters a
convergent-divergent nozzle. A normal shock occurs in the nozzle at a location where the
Mach number is 2. If the air mass flow rate through the nozzle is 0.7 kg/s and if the
pressure on the nozzle exit plane is 260 kPa, find the nozzle throat area, the nozzle exit
area, the temperatures upstream and downstream of the shock wave, and the change in
entropy through the nozzle.
SOLUTION
First consider the conditions on the inlet plane, the conditions on this plane being
designated by the subscript i. On this plane:
Mi 
Vi
ai
But:
ai 
 R Ti 
1.4  287  353.4  376.6 m/s
Therefore:
Mi 
180i
 0.4780
376.6
Also, because:
m  i Vi Aii
and:
i 
pi
350000

 3.455 kg/m3
287  353
RTi
it follows that:
371
Ai 
m
0.7

 0.001126 m 2
3.455  180
i Vi
For a Mach number of 0.4780, the software for isentropic flow or the isentropic flow
tables for air give:
Ai
= 1.384 ,
A*
T0
= 1.046 ,
Ti
p0
= 1.169
pi
Therefore:
p0 
p0
pi  1.169  350  409.2 kPa
pi
and:
T0 
T0
Ti  1.046  353  369.2 K
Ti
Next consider the shock wave. Conditions just upstream and just downstream of this
wave will be designated by subscripts sand d respectively. Now the software for
isentropic flow or the isentropic flow tables for air give for a Mach number of 2:
As
= 1.688 ,
A*
T0
= 1.800 ,
Ts
p0
= 7.824
ps
Therefore:
ps 
p0
409.2

 52.3 kPa
7.824
p0 / ps
and:
Ts 
T0
369.2

 205.1 K
1.800
T0 / Ts
and:
As 
As
A *  1.688  0.000814  0.001374 m 2
A*
372
This is of course also equal to Ad.
Now consider the changes across the shock wave. The software for normal shock
waves or the normal shock wave tables for air give for a Mach number of 2:
M d = 0.5774 ,
Td
= 1.688 ,
Ts
pd
= 4.500
ps
Hence, the pressure and temperature just downstream of the shock wave are given by:
pd
ps  4.500  52.3  235.4 kPa
ps
pd 
and:
Td 
Td
Ts  1.688  205.1  346.2 K
Ts
Next consider the flow downstream of the shock wave. Isentropic flow software or
the isentropic flow tables for air give for the flow just downstream of the 'shock wave
where the Mach number is 0.5774:
Ad
= 1.216 ,
Ad *
p0d
= 1.253
pd
Therefore:
p0d
p p
235.4
 0d d  1.253 
 1.134
pe
pd pe
260
Software for isentropic flow or the isentropic flow tables for air give for this value of p0 /
p:
Ae
= 1.507
Ad *
Hence:
Ae =
Ae / Ad *
1.507
Ad =
 0.001374 = 0.001703 m 2
Ad / Ad *
1.216
373
The changes in entropy in the nozzle all occur across the shock wave. Hence, if Δs is
the change in entropy in the nozzle:
 1
1.4 1





1.4




T
p
s
346.2
52.30




d
s



  0.09372
 ln    
 ln 

 Ts   pd  
 205.1   235.4  
cp




Hence:
s  0.09372 cp  0.09372  1007  94.38 J / kg K
Therefore the nozzle throat and exit areas are 0.000814 and 0.001703 m2 respectively,
the temperatures upstream and downstream of the shock wave are 205.1 K ( - 67.9°C )
and 346.2 K ( 73.2°C ) respectively and the change in entropy across the shock wave is
94.38 J / kg K.
374
PROBLEM 8.21
Air with a stagnation pressure and temperature of 100 kPa and 1500oC is expanded
through a convergent-divergent nozzle that is designed to give an exit Mach number of 2.
The nozzle exit plane area is 30 cm2. Find the mass rate of flow through the nozzle when
operating at design conditions and the exit plane pressure under these design conditions.
Also find the exit plane pressure if a normal shock wave occurs in the divergent portion
of the nozzle at a section where the area is half way between the throat and the exit plane
areas.
SOLUTION
Here:
p0 = 100 kPa
and
T0 = 423 K
When the nozzle is operating at the design conditions the Mach number is 2 on the
exit plane. For a Mach number of 2, the software for isentropic flow or the isentropic
flow tables for air give:
Ae
= 1.773 ,
A*
T0
= 1.600 ,
Te
p0
= 7.665
pe
Ae being the nozzle exit area and Te and pe being the temperature and pressure on the
nozzle exit plane respectively. Therefore:
pe 
p0
100

 13.05 kPa
p0 / pe
7.665
and:
Te 
T0
423

 264.4 K
1.600
T0 / Te
Under design conditions the pressure on the exit plane is equal to the back pressure,
i.e., the back pressure under these conditions is 13.05 kPa.
The mass flow rate through the nozzle under design conditions is given by:
375
m   e Ve Ae   e M e ae Ae
But:
e 
pe
13050

 0.1867 kg/m3
287  264.4
RTe
and:
ae 
 R Te 
1.4  287  264.4  325.9 m/s
and:
Me = 2
and
Ae = 0.003 m 2
Hence:
m   e M e ae Ae = 0.1867  2  325.9  0.003 = 0.3651 kg/s
Next consider the flow when there is the shock wave in the divergent portion of the
nozzle. Because the nozzle has an area ratio of:
Ae
= 1.773
A*
the normal shock wave occurs at a point in the divergent section of the nozzle at which:
0.5  Ae  A *
A
 A

=
= 0.5  e  1 = 0.5  1.773  1 = 1.387
A*
A*
 A*

The software for isentropic flow or the isentropic flow tables for air give, using this area
ratio, the Mach number and pressure ratio just upstream of the shock wave as:
M s = 1.717 ,
p0
= 4.886
ps
the subscript s denoting conditions just upstream of the shock wave. Hence, the pressure
just upstream of the shock wave is given by:
ps 
p0
100

 20.47 kPa
4.886
p0 / ps
Next consider the flow across the shock wave. Software for normal shock waves or
the normal shock tables for a Mach number of 1.717 give:
376
M d = 0.6258 ,
pd
= 3.202
ps
subscript d denoting conditions just downstream of the shock wave. Hence, the pressure
just downstream of the shock wave is given by:
pd 
pd
ps  3.202  20.47  65.53 kPa
ps
Isentropic flow software or the isentropic flow tables for air give for the flow just
downstream of the shock wave where the Mach number is 0.6258:
Ad
= 1.164 ,
Ad *
p0d
= 1.281
pd
Using this value of the area ratio gives:
Ae
A A
A A
A / A * Ad
1.773
 1.164 = 1.488
= e d = e d = e
=
Ad *
Ad Ad *
As Ad *
As / A * Ad *
1.387
For this area ratio the software for subsonic isentropic flow or the isentropic flow
tables for air give:
p0d
= 1.131
pe
Therefore:
pe 
p0d / pd
1.281
pd 
 65.53  74.22 kPa
p0d / pe
1.131
Because the flow downstream of the shock wave is subsonic this will be equal to the back
pressure.
Therefore the mass flow rate through the nozzle under design conditions is 0.3651
kg/s and the back pressure under these conditions is 13.05 kPa. The normal shock wave
will occur at the specified point when the back pressure is 74.22 kPa.
377
PROBLEM 8.22
Air flows through a convergent-divergent nozzle with an exit-to-throat area ratio of
4.0. If a normal shock wave occurs in the nozzle at a location where the area ratio is 2.5
times the throat area, find the Mach number on the exit plane of the nozzle.
SOLUTION
Subscripts s, d and e will be used to refer to conditions just upstream of the shock,
just downstream of the shock and at the nozzle exit respectively. The flow is supersonic
ahead of the shock wave which occurs at a point where A/A* = 2.5. For this value of A/A*
the software for isentropic flow or the isentropic flow tables for air give:
M s = 2.443
This is the Mach number ahead of the shock wave. For this Mach number, the
software for normal shock waves or the normal shock tables for air give:
M d = 0.5186
The Mach number then decreases in the remainder of the divergent section of the
nozzle. Now the software for isentropic flow or the isentropic flow tables for air give for
a Mach number of 0.5186:
Ad
= 1.306
Ad *
Ad* being the throat area that would give a Mach number of one in the flow downstream
of the shock wave.
At the exit of the nozzle:
Ae
A / A * Ad
4
 1.306 = 2.090
= e
=
Ad *
Ad / A * Ad *
2.5
The software for subsonic isentropic flow gives for this area ratio:
378
M e = 0.2912
Therefore the Mach number on the nozzle exit plane is 0.2912.
379
PROBLEM 8.23
Air flows through a convergent-divergent nozzle. The stagnation temperature of the
supply air is 200°C and the nozzle has an exit area of 2 m2 and a throat area of 1 m2. If
the air from the nozzle is discharged to an ambient pressure of 70 kPa, find the minimum
supply stagnation pressure required to produce choking in the nozzle and the mass flow
rate through the nozzle when it is choked. Also find the supply stagnation pressure that
exists if a normal shock wave occurs in the divergent portion of the nozzle at a section
where the area is 1.5 m2.
SOLUTION
The nozzle has an area ratio that is given by:
Ae
2
=
= 2
At
1
subscripts e and t referring to conditions at the exit and at the throat respectively. Also the
supplied information gives T0 = 573K
The lowest supply stagnation pressure, p0 , that will give choking at the throat is that
which gives a Mach number of one at the throat and which involves subsonic flow in the
divergent section of the nozzle with the result that pe = pb. For this case then the subsonic
isentropic flow software or the isentropic flow tables for air give for A/A* = 2:
M e = 0.3059 ,
T0
= 1.019 ,
Te
p0
= 1.067
pe
Hence in this case because pe = 70 kPa:
p0 
p0
pe  1.067  70  74.69 kPa
pe
The mass flow rate through the nozzle under these conditions is therefore be given
by:
380
m   e Ve Ae  e M e ae Ae
But using the results given above:
Te 
T0
573

 562.3 K
1.019
T0 / Te
Therefore:
e 
pe
70000

 0.4338 kg/m3
RTe
287  562.3
and:
ae 
 R Te 
1.4  287  562.3  475.3 m/s
and since:
Ae = 2 m 2 ,
M e = 0.3059
it follows that:
m   e M e ae Ae = 0.4338  0.3059  475.3  2 = 126.1 kg/s
Next consider the flow when there is a normal shock wave in the divergent section of
the nozzle. Subscripts s and d will be used to refer to conditions just upstream and just
downstream of the shock wave. The flow is supersonic ahead of the shock wave and
occurs at a point where:
As
A
1.5
= s =
= 1.5
1
A*
At
For this value of As / A * the software for supersonic isentropic or the isentropic flow
tables for air flow give:
M e = 1.854 ,
p0
= 6.243
ps
Thus the Mach number ahead of the shock wave is 1.854. For this Mach number, the
software for normal shock waves or the normal shock tables for air give:
381
M d = 0.6049 ,
pd
= 3.844
ps
The Mach number then decreases in the remainder of the divergent section of the
nozzle.
Now the software for isentropic flow or the isentropic flow tables for air flow give for
a Mach number of 0.6049:
p0d
= 1.280 ,
pd
Ad
= 1.182
Ad *
Ad* being the throat area that would give a Mach number of one in the flow downstream
of the shock wave and p0d being the stagnation pressure downstream of the shock wave.
At the exit of the nozzle:
Ae
A A
2
 1.182 = 1.576
= e d =
Ad *
Ad Ad * 1.5
The software for subsonic isentropic flow or the isentropic flow tables for air flow give
for this area ratio:
M e = 0.4044 ,
p0d
= 1.119
ps
Now pe = 70 kPa. Therefore:
p0 
p0 ps p0d / pe
1
1.119
pe  6.243 

 70  99.39 kPa
ps pd p0d / pd
3.844
1.280
Therefore the minimum supply pressure to produce choking is 74.69 kPa and the
mass flow rate through the nozzle under these circumstances is 126.1 kg/s while the
supply pressure that exists when there is a normal shock at the specified position in the
divergent section of the nozzle is 99.39 kPa.
382
PROBLEM 8.24
Air flows from a large reservoir in which the pressure is 450 kPa through a
convergent-divergent nozzle. A normal shock wave occurs in the divergent portion of the
nozzle at a point where the nozzle area is twice the throat area. Find the Mach numbers
on each side of this shock wave. If the Mach number on the exit plane of the nozzle is
0.2, find the back pressure required to maintain the shock at this location.
SOLUTION
It will be assumed that the reservoir is large enough to be able to assume that the
reservoir pressure is the stagnation pressure before the shock wave occurs, i.e.:
p0  450 kPa
Subscripts s, d, and e will be used to denote conditions just upstream of the shock
wave, just downstream of the shock wave, and at the exit respectively. The normal shock
wave occurs at a point in the divergent section of the nozzle at which:
As
= 2
A*
Isentropic software or the isentropic flow tables for air give, using this area ratio, the
Mach number and pressure ratio just upstream of the shock wave as:
M s = 2.197 ,
p0
= 10.65
ps
Hence, the pressure just upstream of the shock wave is given by:
ps 
p0
450

 42.25 kPa
10.65
p0 / ps
Next consider the flow across the shock wave. The software for normal shock waves
or the normal shock tables for air give for a Mach number of 2.197:
383
M d = 0.5475 ,
pd
= 5.465
ps
Hence, the pressure just downstream of the shock wave is given by:
pd 
pd
ps  5.465  42.25  230.9 kPa
ps
For the flow just downstream of the shock wave where the Mach number is 0.5475,
isentropic flow software or the isentropic flow tables for air give:
p0d
= 1.226
pd
Similarly on the exit plane where the Mach number, Me , is 0.2, isentropic flow
software or the isentropic flow tables for air give:
p0d
= 1.028
pe
Hence:
pe 
p0d / pd
1.226
pd 
 230.9  275.4 kPa
p0d / pe
1.028
Because the flow downstream of the shock wave is subsonic this will be equal to the
back pressure.
Therefore the Mach numbers upstream and downstream of the normal shock wave are
2.197 and 0.5475 respectively and the back pressure with the shock in the prescribed
position is 275.4 kPa.
384
PROBLEM 8.25
The stagnation pressure and temperature at the inlet to a supersonic wind-tunnel are
100 kPa and 30°C respectively. The Mach number in the test section of the tunnel is 2. If
the cross-sectional area of the test section is 1.2 m2, find the throat areas of the nozzle and
diffuser.
SOLUTION
Here:
p0  100 kPa and T0  303 K
The nozzle is designed to generate a Mach number of 2. For this Mach number the
software for isentropic flow or the isentropic flow tables for air give:
At
= 1.688
An *
At being the test section area and An* being the throat area of the nozzle. Hence:
An * 
At
1.2

 0.7109 m 2
At / Ae * 1.688
During starting there is a normal shock wave in the working section which reduces
the Mach number from a value of 2 ahead of the shock to a subsonic value. The Mach
number then increases in value in the convergent section of the diffuser to a value of one
at the throat before decreasing again in the divergent section of the diffuser. The software
for normal shock waves or the normal shock tables for air give for an upstream Mach
number of 2, a Mach number downstream of the shock wave of 0.5774. The flow
downstream of the shock wave is subsonic and the software for isentropic flow at
subsonic speeds or the isentropic flow tables for air give for M = 0.5774:
At
= 1.216
Ad *
Ad * being the throat area of the diffuser.
385
Hence:
Ad * 
At
1.2

 0.9868 m 2
At / Ad * 1.216
Therefore the nozzle and diffuser throat areas are 0.7109 and 0.9868 m2 respectively.
386
PROBLEM 8.26
Air flows through a convergent nozzle. At a section within this nozzle at which this
cross-sectional area is 0.01 m2, the pressure is 300 kPa and the temperature is 30oC. If the
velocity at this section of the nozzle is 150 m/s, find the Mach number at this section, the
stagnation temperature and pressure, and the mass flow rate through the nozzle. If the
nozzle is choked, find the area, pressure and temperature at the exit of the nozzle.
SOLUTION
At the first section considered where conditions will be denoted by the subscript 1:
M1 =
V1
=
a1
V1
 RT1
Hence:
M1 =
150
= 0.4299
1.4  287  303
For a Mach number of 0.4299, the software for isentropic flow or the isentropic flow
tables for air flow give:
T0
= 1.037 ,
T1
p0
= 1.136
p1
Hence:
T0 
T0
T1  1.037  303  314.2 K
T1
and:
p0 
p0
p1  1.136  300  340.8 kPa
p1
The mass flow rate through the nozzle is then given by:
m  1 V1 A1  1 M 1 a1 A1
387
Now:
1 
p1
300000

 3.450 kg/m3
287  303
RT1
and:
a1 
 R T1 
1.4  287  303  348.9 m/s
and since:
A1 = 0.01 m 2
it follows that the mass flow rate is given by:
m  1 M 1 a1 A1 = 3.450  0.4299  348.9  0.01 = 5.175 kg/s
Next consider the nozzle exit when the nozzle is choked and the Mach number is
equal to one. For this situation, the software for isentropic flow or the isentropic flow
tables for air flow give:
p0
p
T
T
= 0 = 1.893 , 0 = 0 = 1.200
pe
p*
Te
T*
Hence:
Te 
T0
314.2

 261.8 K
1.200
T0 / Te
and:
pe 
p0
340.8

 180.0 kPa
p0 / pe
1.893
Now the mass flow rate is given by:
m   e Ve Ae
i.e., because the nozzle is choked so that Me = 1:
Ve  ae
388
hence:
m  e ae Ae
i.e.:
Ae 
m
e ae
Now:
e 
pe
180000

 2.396 kg/m3
RTe
287  261.8
and:
ae 
 R Te 
1.4  287  261.8  324.3 m/s
Hence:
Ae 
m
5.175

 0.00666 m 2
2.396  324.3
e ae
Therefore the Mach number at the initial point considered is 0.4299, the stagnation
pressure and temperature are 340.8 kPa and 314.2 K ( = 41.2°C ) respectively, and the
mass flow rate through the nozzle is 5.175 kg/s. On the exit plane the area, pressure and
temperature are 0.00666 m2 , 180.0 kPa, and 261.8 K ( = -11.2°C ) respectively.
389
PROBLEM 8.27
Carbon dioxide flows through an 8 cm inside diameter pipe. In order to determine the
mass flow rate, a venturi meter with a throat diameter of 5 cm is installed in the pipe. The
pressure and temperature just upstream of the venturi meter are 600 kPa and 40°C,
respectively. The difference between the pressure just upstream of the venturi meter and
the pressure at the throat of the venturi meter is 15 kPa. Find the mass flow rate of carbon
dioxide.
SOLUTION
Conditions in the pipe upstream of the venturi meter will be designated by the
subscript 1 while conditions at the throat of the venturi meter will be designated by the
subscript 2. The information provided gives:
T1 = 313 K ,
p1 = 600 kPa ,
p2 = p1 15 = 585 kPa
and:
D1  0.08 m , D2  0.05 m
The flow will be assumed to be isentropic. Equation (8.22) therefore gives:


1 1 

 p  
A2
  1  
A1
 p2   
1  
 
1
2
p1  

p0  
1 
p2   
 
p0  
1

i.e.,
1
1
2 

2
 A2 
 p1    p0   p1  
      1
1 
 A1 
 p2   

 p0  p2 
For carbon dioxide:
390
R 
8314.3
 189 ,   1.3
44
Hence, because:
2
4
4
 A2 
 D2 
 0.05 
  
 
  0.1526
 0.08 
 A1 
 D1 
The above equation therefore gives:
1
1
2 

 600    p0   600  
0.1526  

1
1 
 585   


p
585
 0

from which it follows that:
p0


 4.380 , i.e.,
p0  602.4 kPa
Therefore in the pipe:
p0
602.4

 1.0039
p1
600
For this value of p0 / p1 the software for isentropic flow gives for γ = 1.3:
M 1  0.0774
The mass flow rate is then given by:
m  1 V1 A1  1 M 1 a1 A1
Now:
1 
p1
600000

 10.14 kg/m3
RT1 189  313
and:
a1 
 R T1  1.3  189  313  277.3 m/s
and since:
391
A1 =

4
D12 =

4
 0.082 = 0.005027 m 2
it follows that the mass flow rate is given by:
m  1 M 1 a1 A1 = 10.14  0.00774  277.3  0.005027 = 1.094 kg/s
Therefore the mass flow rate of carbon dioxide is 1.094 kg/s.
392
PROBLEM 8.28
A large rocket engine designed to propel a satellite launcher has a thrust of one
million lbf when operating at sea-level, the exit plane pressure being equal to the ambient
pressure under these conditions. The combustion chamber pressure and temperature are
500 psia and 4500oF respectively. If the products of combustion can be assumed to have
the properties of air and if the flow through the nozzle can be assumed to be isentropic,
find the throat and exit diameters of the nozzle.
SOLUTION
It will be assumed that the pressure and temperature in the combustion chamber are
effectively the stagnation pressure and the stagnation temperature. i.e., it will be assumed
that:
p0 = 500 psia ,
T0 = 4960 o R
It will also be assumed that the ambient pressure is 14.7 psia.
On the nozzle exit plane therefore:
p0
500
=
= 34.01
pe
14.7
For this pressure ratio, the software for isentropic flow or the isentropic flow tables
for supersonic air flow give:
Ae
T
= 4.033 , 0 = 2.739 , M e  2.949
A*
Te
Hence:
Te 
T0
4960

 1811 o R
T0 / Te
2.739
If a control volume drawn around the engine is considered, the pressure will be
atmospheric everywhere on the surface of this control volume and therefore if F is the
engine thrust:
393
 e   e Ve2 Ae =  e M e2 ae2 Ae
F  mV
i.e.:
Ae =
F
e M e2 ae2
Now:
e 
pe
14.7  144

 0.0219 lbm/ft 3
RTe
53.3  1811
and:
ae 
 R Te  1.4  53.3  32.2  1811  2088 ft/sec
and:
F = 1000000 lbf
Hence:
Ae 
F
32200000

 38.85 ft 2
2 2
e M e ae
0.0219  2.9492  20862
The diameter of the nozzle at the exit is then given by:
De 
4 Ae


4  38.85

 7.034 ft
The throat area of the nozzle is then obtained by noting that:
A* 
Ae
38.84

 9.631ft 2
Ae / A * 4.033
The diameter of the throat of the nozzle is then given by:
D* 
4A*


4  9.631

 3.502 ft
Therefore the throat and exit diameters of the nozzle are 3.502 ft and 7.034 ft
respectively.
394
PROBLEM 8.29
Air, at a pressure of 700 kPa and a temperature of 80°C flows through a convergingdiverging nozzle. The inlet area is 0.005 m2 and the pressure on the exit plane is 40 kPa.
If the mass flow rate through the nozzle is 1 kg/sec, find, assuming one-dimensional
isentropic flow, the Mach number, temperature and velocity of the air on the discharge
plane.
SOLUTION
On the inlet section, where conditions are designated by the subscript i, the density is
given by:
i 
pi
700000

 6.909 kg/m3
RTi
287  353
Hence, since:
m  i Vi Ai
it follows that:
Vi 
m
1

 28.95 m/s
i Ai
6.909  0.005
from which it follows that:
Mi =
Vi
=
ai
Vi
 RTi
28.95
28.95
=
= 0.07687
376.6
1.4  287  353
For a Mach number of 0.07687 the software for isentropic flow or the isentropic flow
tables for air flow give:
T0
= 1.001 ,
Ti
Hence:
395
p0
= 1.004
pi
T0 
T0
Ti  1.001  353  353.4 K
Ti
and:
p0 
p0
pi  1.004  700  702.8 kPa
pi
Because the flow is isentropic, the stagnation pressure and temperature are the same
throughout the flow, i.e., the above values apply everywhere in the flow.
Next consider the exit plane where conditions are designated by subscript e. Here:
p0
702.8
=
= 17.57
pe
40
For this value of p0 / pe , the software for isentropic flow or the isentropic flow tables
for air flow give:
M e = 2.518 ,
T0
= 2.268
Te
Hence:
Te 
T0
353.4

 155.8 K
T0 / Te
2.268
Using these results then gives:
Ve = M e ae = M e
 RTe = 2.518  1.4  287  155.8 = 2.518  250.2 = 630.0 m/s
Therefore Mach number, temperature and velocity on the discharge plane are 2.518,
155.8 K ( = - 117.2°C ) and 630.0 m/s.
396
PROBLEM 8.30
A small jet aircraft designed to cruise at a Mach number of 1.5 has an intake diffuser
with a fixed area ratio. Find the ideal area ratio for this diffuser and the Mach number to
which the aircraft must be taken in order to swallow the normal shock wave if the
diffuser has this ideal area ratio.
SOLUTION
First, consider the diffuser operating under design conditions with no shock waves.
The software for isentropic flow or the isentropic flow tables for supersonic air flow give
for a Mach number M = 1.5:
Ai
= 1.176
A*
the subscript i referring to conditions on the inlet plane. Therefore:
Ainlet
= 1.176
Athroat
Next consider the situation where there is a normal shock wave on the inlet plane, i.e.,
just before the shock is swallowed. The flow downstream of the shock wave is subsonic
and the software for isentropic flow or the isentropic flow tables for air flow give for
A/A* = 1 .176:
M i = 0.6106
This is the Mach number downstream of the shock wave. But the software for normal
shock waves or the normal shock wave tables for air give for a downstream Mach
number, M2 = 0.6106:
M 1 = 1.827
This means that if the Mach number of the aircraft exceeds this value, the shock wave
will be swallowed.
397
Therefore the throat to inlet area ratio of the diffuser is 1 / 1.176 = 0.8502 and the
aircraft must be taken to a Mach number of 1.827 in order to swallow the shock.
398
PROBLEM 8.31
A fixed supersonic converging-diverging diffuser is designed to operate at a Mach
number of 1.7. To what Mach number would the inlet have to be accelerated in order to
swallow the shock during start-up?
SOLUTION
First consider the diffuser operating under design conditions with no shock waves.
The software for isentropic flow or the isentropic flow tables for supersonic air flow give
for a Mach number M = 1.7:
Ai
= 1.338
A*
the subscript i referring to conditions on the inlet plane. Therefore the diffuser has an
inlet-to-throat area ratio of 1.338.
Next consider the situation where there is a normal shock wave on the inlet plane, i.e.,
just before the shock is swallowed. The flow downstream of the shock wave is subsonic
and the software for isentropic flow or the isentropic flow tables for air flow give for
A/A* = 1 .338:
M i = 0.5010
This is the Mach number downstream of the shock wave. But the software for normal
shock waves or the normal shock wave tables for air give for a downstream Mach
number, M2 = 0.5010:
M 1 = 2.634
This means that if the Mach number of the aircraft exceeds this value, the shock wave
will be swallowed.
Therefore the aircraft must be taken to a Mach number of 2.634 in order to swallow
the shock.
399
PROBLEM 8.32
A jet aircraft is designed to fly at a Mach number of 1.9. It is fitted with a variablearea diffuser. If the diffuser just “swallows” the shock wave at the design Mach number
and the throat area is then reduced to give a “shockless” flow, find the percentage
reduction in diffuser throat area that is required.
SOLUTION
Consider the situation where there is a normal shock wave on the inlet plane, i.e., just
before the shock is swallowed. The Mach number ahead of the shock wave is 1.9. The
software for normal shock waves or the normal shock tables for air give for an upstream
Mach number of 1.9, a Mach number downstream of the shock wave of 0.5956. The flow
downstream of the shock wave is subsonic and the software for isentropic flow at
subsonic speeds or the isentropic flow tables for air give for M = 0.5956:
Ai
A
= i = 1.193
A*
At
the subscripts i and t referring to conditions on the inlet plane and at the throat
respectively.
Next consider the situation when the shock has been swallowed and the throat area
reduced until there are no shock waves in the diffuser. The software for isentropic flow or
the isentropic flow tables for air give for an inlet Mach number Mi = 1.9:
Ai
A
= i = 1.555
A*
At
Therefore with a shock at the inlet the throat area is given by:
Ats =
Ai
1.193
and with no shock waves the throat area is given by:
400
At =
Ai
1.555
The percentage reduction in throat area is then given by:
Ats  At
A / A  At / Ai
1/1.193 - 1/1.555
 100 = ts i
 100 =
 100 = 23.28 %
Ats
Ats / Ai
1/1.193
Therefore the required percentage reduction in diffuser throat area is 23.28 per cent.
401
PROBLEM 8.33
A small jet aircraft designed to cruise at a Mach number of 3.0 has an intake diffuser
with a variable area ratio. Find the ratio of the throat area under these cruise conditions to
the throat area required when the aircraft is flying at a Mach number of 3. Assume the
diffuser intake area does not change.
SOLUTION
The software for isentropic flow or the isentropic flow tables for air give A / A* =
4.235 for a Mach number of 3 and A / A* = 1.176 for a Mach number of 1.5. Hence:
For M = 3:
Ai
= 4.235
At
For M = 1.5:
Ai
= 1.176
At
the subscripts i and t referring to conditions on the inlet plane and at the throat
respectively.
Since the intake area Ai is the same at the two Mach numbers, it follows that:
At at M  3
1/ 4.235 1.176


 0.2777
At at M  1.5 1/1.176
4.235
Therefore the throat area required under cruise conditions is 0.2777 of that required at
a Mach number of 1.5.
402
PROBLEM 8.34
A converging-diverging supersonic diffuser is to be used at Mach 3.0. The diffuser is
to use a variable throat area so as to swallow the starting shock. What per cent increase in
throat area will be necessary?
SOLUTION
Consider the situation where there is a normal shock wave on the inlet plane of the
diffuser, i.e., just before the shock is swallowed. The Mach number ahead of the shock
wave is 3. The software for normal shock waves or the normal shock tables for air give
for an upstream Mach number of 3, a Mach number downstream of the shock wave of
0.4752. The flow downstream of the shock wave is subsonic and the software for
isentropic flow at subsonic speeds or the isentropic flow tables for air give for M =
0.4752:
Ai
A
= i = 1.390
A*
At
the subscripts i and t referring to conditions on the inlet plane and at the throat
respectively.
Next consider the situation when the shock has been swallowed and the throat area
reduced until there are no shock waves in the diffuser. The software for isentropic flow or
the isentropic flow tables for air give for an inlet Mach number Mi = 3:
Ai
A
= i = 4.235
A*
At
Therefore with a shock at the inlet the throat area is given by:
Ats =
Ai
1.390
and with no shock waves the throat area is given by:
403
At =
Ai
4.235
The percentage increase in throat area required in order to allow the shock wave to be
swallowed is then given by:
Ats  At
A / A  At / Ai
1/1.390 - 1/4.235
 100 = ts i
 100 =
 100 = 204.7 %
Ats
Ats / Ai
1/4.235
Therefore the percentage increase in diffuser throat area required in order to allow the
shock wave to be swallowed is 204.7 per cent.
404
PROBLEM 8.35
A wind-tunnel designed for a test section Mach number of 4, is fitted with a variable
area diffuser. Find the ratio of the diffuser throat area when operating under ideal running
conditions to the diffuser throat area during starting when there is a shock wave in the
working section, the Mach number ahead of this shock being 4.
SOLUTION
During starting there is a normal shock wave in the working section which reduces
the Mach number from a value of 4 ahead of the shock to a subsonic value. The Mach
number then increases in value in the convergent section of the diffuser to a value of one
at the throat before decreasing again in the divergent section of the diffuser. The software
for normal shock waves or the normal shock tables for air give for an upstream Mach
number of 4, a Mach number downstream of the shock wave of 0.4350. The flow
downstream of the shock wave is subsonic and the software for isentropic flow at
subsonic speeds or the isentropic flow tables for air give for M = 0.4350:
Aw
A
= w = 1.487
A*
At
the subscripts w and t referring to conditions on the inlet plane and at the throat
respectively.
Next consider the situation when the shock has been swallowed and the throat area
reduced until there are no shock waves in the system. The software for isentropic flow or
the isentropic flow tables for air give for an inlet Mach number Mw = 4:
Aw
A
= w = 10.72
A*
At
Therefore with a shock wave in the working section the throat area is given by:
Ats =
Aw
1.487
405
and with no shock waves the throat area is given by:
At =
Aw
10.72
Therefore the ratio of the diffuser throat area with no shock waves to that required during
the starting process is given by:
At
A /A
1/10.72
1.487
= t w  100 =
=
= 0.1387
Ats
Ats / Aw
1/1.487
10.72
Therefore the ratio of the diffuser throat area with no shock waves to that required
during the starting process is 0.1387.
406
PROBLEM 8.36
Air flows from a tank in which the pressure is kept at 750 kPa and the temperature is
kept at 30oC through a converging nozzle which discharges the air to the atmosphere. If
the throat area of this nozzle is 0.6 cm2, find the rate at which the air is discharged from
the tank in kg/s.
SOLUTION
The flow is assumed to be isentropic. The tank is assumed to be large so the velocity
in it will be small and the pressure and temperature in it will, as a result, effectively be
the stagnation pressure and temperature i.e.:
p0 = 750 kPa
and
T0 = 303 K
Because a convergent nozzle is involved the highest Mach number that can exist on
the nozzle exit plane is 1. When this situation exists, the software for isentropic flow or
the isentropic flow tables for air give:
p0
p
 0  1.893
pe
p*
where the subscript e denotes conditions on the exit plane.
Using the above results gives, if the nozzle is choked:
pe 
p0
750

 396.2 kPa
p0 / pe
1.893
Because this is greater than the atmospheric pressure (101.3 kPa) it shows that the nozzle
is choked, i.e., the Mach number on the exit plane is 1. In this case the software for
isentropic flow or the isentropic flow tables for air give:
T0
T
 0  1.200
Te
T*
407
Hence:
Te 
T0
303

 252.5 K
T0 / Te
1.200
The mass flow rate through the nozzle is then given by:
m   e Ve Ae
But because the nozzle is choked so that Me = 1 it follows that Ve= ae. Hence:
m   e ae Ae
Now:
e 
pe
396200

 5.467 kg/m3
RTe
287  252.5
and:
ae 
 R Te  1.4  287  252.5  318.5 m/s
and:
Ae  0.00006 m 2
The mass flow rate is therefore given by:
m   e ae Ae  5.467  318.5  0.00006  0.1045 kg/s
Therefore the mass flow rate through the nozzle is 0.1045 kg/s.
408
PROBLEM 8.37
In transonic wind-tunnel testing, the small area decrease caused by placing the model
in the test section, i.e., by the model blockage, can cause relatively large changes in the
flow in the test section. To illustrate this effect, consider a tunnel that has an empty test
section Mach number of 1.08. The test section has an area of 1 m2 and the stagnation
temperature of the air flowing through the test section is 25oC. If a model with a crosssectional area of 0.005m2 is placed in this test section, find the percentage change in test
section velocity. Assume one-dimensional isentropic flow.
SOLUTION
First consider the empty wind-tunnel. For a Mach number of 1.08, the software for
isentropic flow or the isentropic flow tables for air flow give:
T0
= 1.2333 ,
T
A
= 1.005
A*
From which it follows that in the test section:
T 
T0
298

 241.6 K
T0 / T
1.2333
Hence:
V = M a = M  RT = 1.08 
1.4  287  241.6 = 1.08  311.6 = 336.5 m/s
When the model is in the tunnel, the minimum flow area is:
A  1  0.005  0.995 m 2
In this situation then:
A
0.995
 1.005 = 1.000
= 1.005 =
A*
1
For this area ratio, of course:
409
M =1 ,
T0
= 1.2
T
In this case then:
T 
T0
298

 248.4 K
T0 / T
1.2
Hence in this case:
V =M a =M
 RT = 1.000  1.4  287  248.4 = 1.000  315.9 = 315.9 m/s
The percentage change in the velocity induced by the model is then given by:
336.5  315.9
 100  6.12 %
336.5
Therefore the percentage change in the velocity is 6.12 per cent. This is induced by a
0.5 per cent change in area.
410
PROBLEM 8.38
Consider one-dimensional isentropic flow through a convergent-divergent nozzle that
has a throat area of 10 cm2 . The pressure at the throat is 310 kPa and the flow goes from
subsonic to supersonic velocities in the nozzle. Find the pressures and Mach numbers at
points in the nozzle upstream and downstream of the throat where the nozzle crosssectional area is 29 cm2.
SOLUTION
The conditions at the sections that are upstream and downstream of the throat will be
designated by subscripts u and d respectively. Using the given values of area:
Au
A
29
= d =
= 2.9
A*
A * 10
Because the flow is isentropic and goes from subsonic to supersonic velocities in the
nozzle, the software for isentropic flow or the isentropic flow tables for air give for the
above values of A / A * :
M u = 0.2046 ,
p0
= 1.030
pu
and:
M d = 2.601 ,
p0
= 20.0
pu
At the throat the Mach number will be one and the software for isentropic flow or the
isentropic flow tables for air give for a Mach number of one:
p0
p
= 0 = 1.893
pt
p*
the subscript t being used to indicate conditions at the throat.
411
Hence, because the flow is isentropic and the stagnation pressure is the same at all
points in the nozzle, it follows that:
pu 
p0 / pt
1.893
 310  569.7 kPa
pt 
p0 / pu
1.030
and:
pd 
p0 / pt
1.893
 310  29.34 kPa
pt 
p0 / pd
20.0
Therefore the pressure and Mach number at the upstream point are 569.7 kPa and
0.2046 respectively while those at the downstream point are 29.34 kPa and 2.601
respectively.
412
PROBLEM 8.39
A moving piston forces air from a well-insulated 15 cm diameter pipe though a
convergent nozzle fitted to the end of the pipe. The nozzle has an exit diameter of 4 mm
and the air is discharged to the atmosphere. If the force on the piston is 3700 N and the
air temperature is 30oC, estimate the velocity on the nozzle exit plane, the piston velocity
and the mass flow rate at which the air is discharged from the nozzle.
SOLUTION
It is assumed that the flow is steady and isentropic. Subscript p will be used to denote
conditions in the pipe. Ambient pressure will be assumed to be 101.3 kPa. The pressure
acting on the piston is:
pp 
Force
3700

 209400 Pa  209.4 kPa

Piston Area
2
 0.15
4
Since the flow is assumed to be isentropic, this will be the pressure existing
everywhere in the pipe. Therefore, the pressure ratio across the convergent nozzle is:
pp
pa
=
209.4
= 2.067
101.3
If the nozzle is choked, i.e., if the Mach number on the exit plane is one and the
pressure on the exit plane is higher than the ambient pressure, the pressure ratio across
the nozzle is given by software for isentropic flow or the isentropic flow tables for air
flow for a Mach number of 1 as:
p0
= 1.893
p*
Because the·pressure in the pipe, pp ,will be close to and slightly below the stagnation
pressure, the above results together indicate that the pressure ratio across the nozzle is
greater than that required to choke the nozzle, i.e., the nozzle will be choked.
413
The continuity equation applied across the nozzle gives, because the flow on the exit
plane is choked:
 p Vp Ap =  *V * Ae
Ae being the area of the nozzle exit.
Because V* = a*, this equation can be written as:
 p Vp ap Ap
 * a*
=
0 ap a0 Ae
0 a0
Therefore, since software for isentropic flow or the isentropic flow tables for air flow
give:
T0
0
= 1.577 ,
= 1.2
T*
*
and because:
2
2
D 
 0.15 
=  p =
 = 1406
Ae
 0.004 
 De 
Ap
it follows that:
p
M
0 p
Tp Ap
T0 Ae
=
1

1.577
1
1

= 0.0004117
1.2 1406
Because ρ0/ ρp and T0/ Tp are both functions of Mp , this is an equation for Mp. A
consideration of the magnitude of the right hand side of this equation indicates that Mp
will be very small and that as a consequence both ρ0/ ρp and T0/ Tp will effectively be
equal to one. Therefore the above equation effectively gives:
M p = 0.0004117
Therefore, because Tp = 303 K, it follows that:
Vp = M p ap = M p
 RTp = 0.0004117  1.4  287  303
= 0.0004117  348.9 = 0.1437 m/s
414
Also because:
p 
and because:
Ap 
pp
RTp

4

209400
 2.408 kg/m3
287  303
Dp2 

4
 0.152  0.01767 m 2
the mass flow rate out of the nozzle is given by:
m   p Vp Ap = 2.408  0.1437  0.01767 = 0.006115 kg/s
Therefore piston velocity, which will be equal to the air velocity in the pipe, is 0.1437
m/s and the mass flow rate of air out of the nozzle is 0.006115 kg/s
415
PROBLEM 8.40
A convergent-divergent nozzle with an exit area to throat area ratio of 3 is supplied
with air from a reservoir in which the pressure is 350 kPa. The air from the nozzle is
discharged into another large reservoir. It is found that the flow leaving the nozzle exit is
directed inwards at an angle of 4o to the nozzle centre-line. The velocity on the nozzle
exit plane is supersonic. What is the pressure in the second reservoir?
SOLUTION
It will be assumed that the supply reservoir is large enough to ensure that the reservoir
pressure is effectively the stagnation pressure. i.e.:
p0  350 kPa
The area ratio of the nozzle is:
Ae
= 3
A*
Ae being the nozzle exit area.
Because the flow on the nozzle exit plane is supersonic, it will be assumed that the
flow in the nozzle is isentropic. For an area ratio of 3, software for isentropic flow or the
isentropic flow tables for air give:
M e = 2.637 ,
p0
= 21.14
pe
the subscript e denoting conditions on the exit plane.
Hence:
pe 
p0
350

 16.56 kPa
p0 / pe
21.14
The flow is turned inwards through an angle of 4° as indicated in Fig. P8.40.
416
Figure P8.40
This turning must be produced by an oblique shock wave as indicated in Fig. P8.40.
In the region immediately downstream of this shock wave the pressure will be equal to
the pressure in the downstream reservoir, i.e., equal to the back pressure pb ,which must
therefore be greater than the exit plane pressure pe . It will be assumed that the nozzle
diameter is big enough to ensure that near the edge of the nozzle on the discharge plane
this shock wave can be treated as a plane oblique shock wave. This wave occurs in a flow
in which the Mach number is 2.637 and it turns the flow through 4o. The software for
oblique shock waves or the oblique shock chart and normal shock tables for air give for
these values:
pb
= 1.311
pe
Hence:
pb 
pb
pe  1.311  16.56  21.71 kPa
pe
Therefore the pressure in the second reservoir is 21.71 kPa.
417
PROBLEM 8.41
A converging-diverging nozzle has an exit to throat area ratio of 4. It is supplied with
air from a large reservoir in which the pressure is kept at 500 kPa and it discharges into
another large reservoir in which the pressure is kept at 10 kPa. Expansion waves form at
the exit edges of the nozzle causing the discharge flow to be directed outwards. Find the
angle that the edge of the discharge flow makes to the axis of the nozzle.
SOLUTION
It will be assumed that the supply reservoir is large enough to ensure that the reservoir
pressure is effectively the stagnation pressure i.e.:
p0  500 kPa
The area ratio of the nozzle is:
Ae
= 4
A*
Ae being the nozzle exit area.
Because there are expansion waves in the discharge, the flow at the nozzle exit is
supersonic and it will consequently be assumed that the flow in the nozzle is isentropic.
For an area ratio of 4, the software for isentropic flow or the isentropic flow tables for air
give:
 e  48.59o
the subscript e denoting conditions on the exit plane and θ being the Prandtl-Meyer angle.
It will be assumed that the nozzle diameter is big enough to ensure that near the edge
of the nozzle on the discharge plane the expansion wave that is formed can be treated as a
plane oblique expansion wave. Also because the flow though the expansion wave is
isentropic, the stagnation pressure downstream of the wave is also 500 kPa. Hence,
downstream of the wave:
418
pb
500

 50
pb
10
the subscript b denoting conditions downstream of the expansion wave.
For p0 / p = 50, the software for isentropic flow or the isentropic flow tables for air
give:
 d  53.61o
Therefore the flow is turned through the following angle by the expansion wave:
 =  d -  e = 53.61 - 48.59 = 5.02o
Therefore the flow is turned outwards through an angle of 5.02° by the expansion
wave as indicated in Fig. P8.41.
Figure P8.41
419
PROBLEM 8.42
A small meteorite punches a 3 cm diameter hole in the skin of an orbiting space
laboratory. The pressure and temperature in the laboratory are 80 kPa and 20°C
respectively. Estimate the initial rate at which air flows out of the laboratory. State the
assumptions you make in arriving at your estimate.
SOLUTION
It is assumed that:
1. The flow out of the hole is isentropic
2. The conditions in the laboratory can be taken as the stagnation conditions
3. The pressure outside of the laboratory is extremely low and the flow out of the hole
is therefore choked
4. The size of the laboratory is large enough to ensure that there is only a slow drop in
pressure and that the flow can therefore be treated as steady
5. The flow can be assumed to be one-dimensional
Because the conditions in the laboratory are taken as the stagnation conditions, it
follows that:
p0 = 80 kPa
and
T0 = 293 K
Because the flow is assumed to be choked with the Mach number on the exit plane then
being equal to one, the software for isentropic flow or the isentropic flow tables for air
flow give:
p0
p
T
T
= 0 = 1.893 and 0 = 0 = 1.200
pe
p*
Te
T*
where the subscript e denotes conditions on the exit plane.
Using the above results gives:
pe =
p0
T0
80
293
=
= 42.26 kPa and Te =
=
= 244.2 K
1.893
1.200
p0 / pe
T0 / Te
420
The mass flow rate through the nozzle is then given by:
m  e Ve Ae
i.e., because the nozzle is choked so that Me = 1, so that:
Ve  ae
hence:
m  e ae Ae
Now:
e 
pe
42260

 0.6029 kg/m3
RTe
287  244.2
and:
ae =
 RTe =
and:
Ae 

4
De2 
1.4  287  244.2 = 313.2 m/s

4
 0.032  0.0007069 m 2
The mass flow rate is then given by:
m   e ae Ae = 0.6029  313.2  0.0007069 = 0.1335 kg/s
Therefore the initial mass flow rate through the hole is 0.1335kg/s.
421
PROBLEM 8.43
A jet engine is running on a test bed. The stagnation pressure and stagnation
temperature just upstream of the convergent nozzle fitted to the rear of this engine are
found to be 700 kPa and 700°C respectively. The exit diameter of the nozzle is 0.5 m. If
the test is being run at an ambient pressure of 101 kPa, find the mass flow rate through
the engine, the jet exit velocity and the thrust that is developed by the engine. Assume the
gases have the properties of air.
SOLUTION
The flow is assumed to be isentropic with:
p0 = 700 kPa
and
T0 = 973 K
Because a convergent nozzle is involved the highest Mach number that can exist on the
nozzle exit plane is 1. When this situation exists, the software for isentropic flow or the
isentropic flow tables for air flow give:
p0
p
T
T
= 0 = 1.893 and 0 = 0 = 1.200
pe
p*
Te
T*
where the subscript e denotes conditions on the exit plane.
Using the above results gives if the nozzle is choked:
pe =
p0
700
=
= 369.8 kPa
p0 / pe
1.893
Because this is greater than the atmospheric pressure (101 kPa) it shows that the nozzle is
choked, i.e., the Mach number on the exit plane is 1.
Similarly, the above result for temperature ratio gives:
Te =
T0
973
=
= 810.8 K
1.200
T0 / Te
Because the nozzle is choked so that Me = 1:
422
Ve  ae
Hence:
 RTe =
Ve = ae =
1.4  287  810.8 = 570.8 m/s
The mass flow rate through the nozzle is given by:
m  e ae Ae
Now:
e 
pe
369800

 1.589 kg/m3
RTe
287  810.8
and:
Ae 

4
De2 

 0.52  0.1964 m 2
4
The mass flow rate is therefore given by:
m  e ae Ae = 1.586  570.8  1964 = 178.1 kg/s
If a control volume around the engine is considered, the pressure on the surface of
this control volume will be equal to 101 kPa everywhere except on the nozzle exit plane.
The thrust, F, will therefore be given by:
F = m Ve + (pe - 101) Ae = 178.1  570.8 + (369.8 - 101)  0.1964
= 101680 + 52.79 = 101733 N
Therefore the mass flow rate through the engine is 178.1 kg/s, the jet efflux velocity
is 570.8 m/s and the thrust is 101.7 kN.
423
PROBLEM 8.44
Air flows through a converging-diverging nozzle from a large reservoir in which the
pressure is 300 kPa and the temperature is 100oC. The nozzle has a throat area of 1 cm2
and an exit area of 4 cm2. The nozzle discharges into another large reservoir in which the
pressure can be varied. For what range of back pressures will the mass flow rate through
the nozzle be constant and what will the mass flow rate be under these circumstances.
SOLUTION
It will be assumed that the supply chamber is large enough to be able to assume that
the chamber pressure is the stagnation pressure and that the chamber temperature is the
stagnation temperature, i.e.:
p0 = 300 kPa
and
T0 = 373 K
The area ratio of the nozzle is:
Ae
4
=
= 4
A*
1
Ae being the nozzle exit area.
The mass flow rate will be constant for all back pressures for which the nozzle is choked,
i.e., for which the throat Mach number is one. The highest back pressure, pb , that will
give choking at the throat is that which gives a Mach number of one at the throat and
which involves subsonic flow in the divergent section of the nozzle with the· result that pe
= pb, pe being the pressure on the nozzle exit plane. For this case then the subsonic
isentropic flow software or the isentropic flow tables for air flow give for A / A * = 4:
p0
= 1.015
pe
Hence in this case:
424
pe =
p0
300
=
= 295.6 kPa
p0 / pe
1.015
If the back pressure is less than or equal to this value the nozzle will be choked. When
the nozzle is choked the Mach number is one at the throat. The mass flow rate through
the nozzle will therefore be given by:
m   *V * A *   * a * A *
For a Mach number of 1 the software for isentropic flow or the isentropic flow tables
for air flow give:
T* =
T0
373
=
= 310.8 K
T0 / T * 1.200
and:
p* =
p0
300
=
= 158.5 kPa
p0 / p * 1.893
Therefore since:
* 
p*
158500

 1.777 kg/m3
RT * 287  310.8
and:
a* =
 RT * =
1.4  287  310.8 = 353.4 m/s
and:
A *  0.0001 m 2
The mass flow rate is given by:
m   * a * A * = 1.777  353.4  0.0001 = 0.06280 kg/s
Therefore the mass flow rate through the nozzle will be constant if the back pressure
is less than or equal to 295.6 kPa and under these circumstances the mass flow rate
through the nozzle is 0.06280 kg/s.
425
PROBLEM 8.45
Air flows through a convergent-divergent nozzle, the flow becoming supersonic in
the divergent section of the nozzle. A normal shock wave occurs in the divergent section.
If the static pressure behind this shock wave is equal to the static pressure at the throat of
the nozzle, find the ratio of the nozzle area at which the shock wave occurs to the nozzle
throat area.
SOLUTION
Conditions just upstream and just downstream of the shock wave will be designated
by subscripts s and d respectively. The flow ahead of the shock wave must be supersonic
so the pressure at the nozzle throat must be equal to the critical pressure in the flow ahead
of the shock wave p * . The pressure downstream of the shock wave is given by:
pd = ps
pd
ps
But pd is equal to the pressure at the throat of the nozzle, i.e.:
pd = p *
The above equation can therefore be written as:
p * = ps
pd
ps
which can be rearranged to give:
pd / ps
p*
=
p0 / ps
p0
where p0 is the stagnation pressure upstream of the shock wave. Software for
isentropic flow or the isentropic flow tables for air flow give:
p0
= 1.893
p*
426
Therefore the above equation gives:
p0 / ps
= 1.893
pd / ps
The quantities p0/ ps and pd/ ps are both functions of the Mach number upstream of the
shock wave Ms , i.e., the above equation determines Ms. Here the solution will be
obtained in a very simple way. A series of values of Ms will be selected. For each of these
values, the software for isentropic flow or the isentropic flow tables for air flow will be
used to get the value of p0 / ps and the software for normal shock waves or the normal
shock wave tables for air flow will be used to get the value of pd / ps. For each value of
Ms the value of the left hand side (LHS), i.e., of:
LHS =
p0 / ps
pd / ps
will be evaluated. The value of Ms that gives LHS = 1.893 will then be deduced. Some
typical results are given in the following table.
Ms
2.000
3.000
2.200
2.100
2.150
2.151
p0 / ps
7.824
38.73
10.69
9.145
9.888
9.904
pd / ps
4.500
10.33
5.480
4.978
5.226
5.231
LHS
1.739
3.556
1.951
1.837
1.892
1.893
From the above results it will be seen that the pressure just downstream of the shock
is equal to the throat pressure when the Mach number just upstream of the shock wave is
given by:
M s  2.151
For this value of Mach number, the software for isentropic flow or the isentropic flow
tables for air flow give:
427
As
= 1.920
A*
Therefore the shock wave occurs at a point where the nozzle area is 1.92 times the
nozzle throat area.
428
PROBLEM 8.46
A nozzle is designed to expand air from a chamber in which the pressure and
temperature are 800 kPa and 40°C respectively to a Mach number of 2.5. The throat area
of this nozzle is to be 0.05 m2. Find:
1. The exit area of the nozzle.
2. The mass flow rate through the nozzle when operating at design conditions.
3. The back pressure at which there will be a normal shock wave on the exit plane of
the nozzle.
4. The range of back pressures over which there will be a normal shock wave in the
nozzle.
5. The range of back pressures over which oblique shock waves will occur outside the
nozzle.
6. The range of back pressures over which expansion waves will occur outside the
nozzle.
SOLUTION
It will be assumed that the supply chamber is large enough to be able to assume that
the chamber pressure is the stagnation pressure and that the chamber temperature is the
stagnation temperature, i.e.:
p0  800 kPa , T0  313 K
Part (1)
The nozzle is designed to generate an exit Mach number of 2.5. For this Mach
number the software for isentropic flow or the isentropic tables for air give:
Ae
= 2.637 ,
Ad *
T0
= 2.250 ,
Te
p0
= 17.09
pe
Ae being the nozzle exit area and Te and pe being the temperature and pressure on the
nozzle exit plane respectively.
Hence:
429
Ae 
Ae
A *  2.637  0.05  0.1319 m 2
A*
Therefore the exit plane area of the nozzle is 0.1319 m2.
Part (2)
When the nozzle is operating at the design conditions the Mach number will be one at
the throat. The mass flow rate through the nozzle will therefore be given by:
m   *V * A *   * a * A *
For a Mach number of 1 the software for isentropic flow or the isentropic tables for
air give:
T0
= 1.200 ,
T*
p0
= 1.893
p*
Hence:
T* 
T0
313

 260.8 K
T0 / T * 1.200
p* 
p0
800

 422.6 kPa
p0 / p * 1.893
and:
Therefore:
* 
p*
422600

 5.646 kg/m3
RT * 287  260.8
and:
a* 
 R T* 
1.4  287  260.8  323.7 m/s
and:
A *  0.05 m 2
Hence:
m   * a * A * = 5.646  323.7  0.05 = 91.38 kg/s
430
Therefore the mass flow rate through the nozzle is 91.38 kg/s.
Part (3)
When there is a normal shock wave at the exit, the flow in the nozzle is shock free
with a supersonic velocity at the exit. Hence, if there is a shock wave at the exit, the
Mach number ahead of the shock wave will be 2.5 and:
pe =
p0
800
=
= 46.81 kPa
p0 / pe
17.09
For a Mach number of 2.5, the software for normal shock waves or the normal shock
wave tables for air give the pressure ratio across the wave as:
pb
= 7.125
pe
Here it has been noted that with a normal shock wave at the exit the flow downstream
of the shock wave will be subsonic and the pressure downstream of the shock wave must
therefore be equal to the back pressure, pb . Hence, when there is a normal shock wave at
the exit:
pb =
pb
pe = 7.125  46.81 = 333.5 kPa
pe
Therefore there is a normal shock wave on the nozzle exit plane when the back
pressure is 333.5 kPa.
Part (4)
The highest back pressure, pb, that will give choking at the throat is that which gives a
Mach number of one at the throat and which involves subsonic flow in the divergent
section of the nozzle with the result that pe = pb , pe being the pressure on the nozzle exit
plane. For this case the subsonic isentropic flow software or the isentropic tables for air
give for Ae / A * = 2.637:
pb
= 1.036
pe
431
Hence in this case:
pe =
p0
800
=
= 772.2 kPa
p0 / pe
1.036
If the back pressure is below this value a region of supersonic flow terminated by a
normal shock wave will develop in the divergent section of the nozzle. As the back
pressure is decreased this normal shock wave will move towards the nozzle exit plane
and when the back pressure has dropped to 333.5 kPa the shock wave will be on the
nozzle exit plane. Any further reduction in back pressure will cause the shock wave to
move outside the nozzle. Therefore a normal shock wave will occur in the nozzle when
the back pressure is between 333.5 kPa and 772.2 kPa.
Part (5)
When the nozzle is operating at this design condition, i.e., when the nozzle is
“perfectly” expanded, there are no shock waves in the flow and pe = pb and, as noted
above:
p0
= 17.09
pe
Therefore, when the nozzle is operating at the design condition:
pe =
p0
800
=
= 46.81 kPa
p0 / pe
17.09
i.e., the design back pressure is 46.81 kPa.
As shown above, a normal shock wave occurs on the exit plane of the nozzle when
the back pressure is 333.5 kPa. Therefore oblique shock waves will occur in the discharge
from the nozzle when the back pressure is between 46.81 kPa and 333.5 kPa.
Part (6)
Expansion waves will occur in the exhaust from the nozzle for back pressures that are
below the design back pressure, i.e., for back pressures that are less than 46.81 kPa.
432
PROBLEM 8.47
A convergent-divergent nozzle through which air flows is designed to generate a
Mach number of 2.5. The supply reservoir pressure is 320 kPa. What is the design back
pressure? If the nozzle operates with a back pressure of 100 kPa the nozzle will be overexpanded and oblique shock waves will exist at the nozzle exit. Find the Mach number
and flow direction just downstream of the exit plane oblique shock waves under these
conditions.
SOLUTION
When operating at the design conditions there are no shock waves in the nozzle and
the nozzle flow can be assumed to be isentropic throughout. Also assuming that the
supply reservoir is large p0 for the nozzle flow is 320 kPa.
If the subscript e is used to denote conditions on the exit plane then since Me = 2.5 the
software or the isentropic flow tables for air give:
p0
 17.086
pe
Hence:
pe 
p0
320

 18.73 kPa
17.086 17.086
Therefore the design back pressure is 18.73kPa.
Turning next to the flow when the back pressure is 100kPa which is less than the
design back pressure of 18.73 kPa the nozzle flow is over-expanded and oblique shock
waves will form at the exit. The exit plane Mach number and pressure under these
conditions are the design values since the oblique shock waves lie outside of the nozzle,
i.e., Me = 2.5 and pe=18.73 kPa. The oblique shock wave must therefore increase the
pressure form 18.73 kPa to 100 kPa, i.e., if the subscript d is used to denote conditions
downstream of the oblique shock wave then:
pb
100

 5.339
18.73
pe
433
Using the software or the oblique shock chart and the normal shock tables for air give
M1 = 2.5 and p2/p1 =5.339:
M 2   M d   1.068 and   29.26 o
Therefore downstream of the oblique shock wave the Mach number is 1.068 and the
flow is directed towards the nozzle center-line at an angle of 29.26o.
434
PROBLEM 8.48
A convergent-divergent nozzle through which air flows is designed to generate a
Mach number of 2.2. The supply reservoir pressure is 2.2 MPa and the flow is discharged
into ambient air at a pressure of 101 kPa. Under these conditions the nozzle will be
under-expanded and as a result expansion waves will exist at the nozzle exit. Find the
flow direction just downstream of the exit plane expansion waves under these conditions.
What effect does the presence of the expansion waves have on the net axial thrust
produced by the nozzle?
SOLUTION
Because the expansion wave lies outside of the nozzle the flow within the nozzle will
be that existing under design conditions, i.e., no waves exist in the nozzle and the flow
through the nozzle can be assumed to be isentropic. Also, assuming that the supply
reservoir is large, then p0 for the flow is 2.2 MPa.
Since the nozzle exit plane Mach number is 2.2, the software or the isentropic flow
tables for air give if the subscript e denotes conditions on the exit plane:
p0
 10.693 and e  32.249o
pe
Now downstream of the expansion wave the pressure is 101 kPa. Hence since the
flow through the expansion wave is isentropic and there is therefore no change in the
stagnation pressure across the expansion wave it follows that downstream of the
expansion wave where conditions are denoted by the subscript d:
p0
2200

 21.782
pd
101
For this value of p0/p the software or the isentropic flow tables for air give:
M d  2.657 and  d  42.667 o
435
Therefore the change in flow direction across the expansion wave is given by:
   d   e  42.667  32.249  10.428 o
Hence downstream of the expansion wave the flow is outwards away form the nozzle
center-line at an angle of 10.428o.
Because the expansion waves lie downstream of the nozzle exit plane, the flow in the
nozzle and on the nozzle exit plane is exactly the same as exist under design conditions.
Therefore since the thrust depends only on the mass flow rate through the nozzle, the exit
plane velocity, and the difference between the pressure on the nozzle exit plane and the
ambient pressure, the thrust with the expansion waves present will be the same as that
exiting under design conditions.
436
Chapter Nine
ADIABATIC FLOW IN A DUCT
WITH FRICTION
SUMMARY OF MAJOR EQUATIONS
Hydraulic Diameter
DH 
4 (Area)
4A

Perimeter
P
(9.2)
Friction Factor
Laminar Flow:
f 
16
Re
Turbulent Flow:
  
5.74  
f  0.0625  log 


Re0.9  
  3.7 DH
2
Mach Number Change
4f
1 1
1 
  1 M 12 (1  12 (  1) M 22 )
l   2  2 
ln 2
DH
  M1
M2 
2
M 2 (1  12 (  1) M 12 )
Values Relative to Those at M2 = 1
437
(9.26)
 1 M 2 
4f *
 1
(  1) M 2

l  
ln
2 
DH
2
2(1  12 (  1) M 2 )
 M 

p
1 
(  1)/2


*
2
p
M 1  (  1) M /2 
(9.27)
1/ 2
(9.28)
T
(  1)/2

*
T
1  (  1) M 2 /2
(9.29)
1
p0
1 1  (  1) M 2 /2)  2( 1)


p0*
M 
(  1)/2

Fanno Line Equation
1
1


  T  T  2


s  s1
T
0


 ln   
 T1   T0  T1  
cp


438
(9.30)
(9.32)
PROBLEM 9.1
Air flows through a duct with a constant cross-sectional area. The pressure,
temperature, and Mach number at the inlet to the duct are 180 kPa, 30°C and 0.25
respectively. If the Mach number at the exit of the duct has risen to 0.75 as a result of
friction, determine the pressure, temperature, and velocity at the exit. Assume that the
flow is adiabatic
SOLUTION
Conditions at the inlet and outlet of the duct will be denoted by subscripts 1 and 2
respectively. Using the software or the friction flow tables, the following are obtained.
At the inlet where M = 0.25:
p1
 4.355 ,
p*
T1
 1.185
T*
p2
 1.385 ,
p*
T2
 1.079
T*
At the outlet where M = 0.75:
Hence, because p1 = 180 kPa and T1 = 303 K:
p2 
p2 /p*
1.385
p1 
 180  57.2 kPa
*
p1 /p
4.355
T2 
T2 /T *
1.079
T 
 303  275.9 K
* 1
T1 /T
1.185
and:
The velocity at the exit is given by:
V2  M 2 a2  M 2  RT2  0.75  1.4  287  275.9  249.7 m/s
439
Therefore the velocity, pressure, and temperature at the outlet are 249.7 m/s, 57.2 kPa
and 275.9 K ( = 2.9° C) respectively.
440
PROBLEM 9.2
Air flows through a well insulated 4 in. diameter pipe at the rate of 500 lbm/min. The
pressure drops from 50 psia at the inlet to the pipe to a value of 40 psia at the exit. If the
temperature at the inlet is 200oF, find the Mach number at the exit of the pipe.
SOLUTION
Conditions at the inlet and outlet of the duct will be denoted by subscripts 1 and 2
respectively.
At the inlet, the density is given by:
1 
p1
144  50

 0.2047 lbm/ft 3
RT1
53.3  660
Hence, the velocity at the inlet is given by:'
V1 
m
m
500 / 60


 466.5 ft/sec
2
0.2047  ( / 4)  (4 /12) 2
1 A 1 ( / 4) D
The Mach number at the inlet is then given by:
M1 
V1

a1
V1

 RT1
466.5
 0.3705
1.4  53.3  32.2  660
For this value of M1 the following is obtained using the software or the tables for air
flow:
p1
 2.917
p*
Therefore at the exit:
p2
p p
40
 2 1* 
 2.917  2.344
*
p
p1 p
50
441
For this value of p2 / p*, the following is obtained using the software or the friction
flow tables for air:
M 2  0.460
Therefore the Mach number at the outlet of the pipe is 0.460.
442
PROBLEM 9.3
Consider compressible flow through a long, well-insulated duct. At the inlet to the
duct the Mach number, pressure, and temperature are 0.3, 100 kPa and 30°C respectively.
Assuming that the flow is adiabatic and that the pipe is sufficiently long to ensure that the
flow is choked at the exit, find the velocity and temperature at the pipe exit.
SOLUTION
Air flow will be assumed. For the inlet Mach number of 0.3, the following are
obtained using the software or from the friction flow tables for air:
p1
 3.619 ,
p*
T1
 1.179
T*
Because the flow is choked at the exit of the pipe, at the exit:
p2  p * ,
T2  T *
Therefore:
p2 
p1
100

 27.63 kPa
*
3.619
p1 / p
and:
T2 
T1
303

 257.0 K
*
T1 / p
1.179
Because the flow at the exit of the pipe is choked, M2 = 1. Hence:
V2  M 2 a2 
 RT2 
1.4  287  257.0  321.3 m/s
Therefore the velocity, pressure, and temperature at the outlet are 321.3 m/s, 27.6 kPa
and 257.0 K ( = -16°C) respectively.
443
PROBLEM 9.4
Air flows through a 5 cm diameter pipe. Measurements indicate that at the inlet to the
pipe the velocity is 70 m/s, the temperature is 80°C and the pressure 1 MPa. Find the
temperature, the pressure, and the Mach number at the exit to the pipe if the pipe is 25 m
long. Assume that the flow is adiabatic and that the mean friction factor is 0.005.
SOLUTION
Conditions at the inlet and outlet of the duct will be denoted by subscripts 1 and 2
respectively. At the inlet:
M1 
V1

a1
V1

 RT1
70
 0.1859
1.4  287  353
For this value of M1, the following are obtained using the software or the friction flow
tables for air:
p1
 5.872 ,
p*
T1
 1.192 ,
T*
4 fl1*
 17.22
D
But:
4 fl2*
4 fl1*
4 fl1 2


D
D
D
Hence:
4 fl2*
4  0.005  25
 17.22 
 7.22
D
0.05
For this value of 4 f l2* / D the following are obtained using the software or friction
flow tables:
M 2  0.2665 ,
p2 /p*  4.082 , T2 /T *  1.183
444
Using these values gives, because the pressure and temperature at the inlet are 1000
kPa and 353 K respectively:
p2 /p*
4.082
p2 
 p1 
 1000  695.2 kPa
*
p1 /p
5.872
and:
T2 
T2 /T *
1.183
T2 
 353  350.3 K
T2 /T *
1.192
Therefore the Mach number, pressure, and temperature at the exit are 0.2665,
695.2 kPa and 350.3 K ( = 77.3°C ) respectively
445
PROBLEM 9.5
Air flows down a pipe with a diameter of 0.15m. At the inlet to the pipe the Mach
number is equal to 0.1, the pressure is equal to 70 kPa, and the temperature is equal to
35°C. If the flow can be assumed to be adiabatic and if the mean friction factor is 0.005
determine the length of the pipe if the Mach number at the exit is 0.6. Also find the
pressure and temperature at the exit to the pipe.
SOLUTION
Conditions at the inlet and outlet of the duct will be denoted by subscripts 1 and 2
respectively. Using the software or the friction flow tables for air the following are
obtained:
At the inlet where M = 0.1:
p1
 10.94 ,
p*
T1
 1.198 ,
T*
4 fl1*
 66.92
D
T2
 1.119 ,
T*
4 fl2*
 0.49
D
At the outlet where M = 0.6:
p2
 1.763 ,
p*
Now:
4 fl1 2
4 fl1*
4 fl2*


 66.92  0.49  66.43
D
D
D
from which it follows that:
l1 2 
66.43D
66.43  0.15

 498.2 m
4f
4  0.005
Hence the length of the pipe is 498.2 m.
Also because p1 = 70 kPa and T1 = 308 K:
446
p2 
p2 /p*
1.763
 p1 
 70  11.28 kPa
*
p1 /p
10.94
and:
T2 
T2 /T *
1.119
T 
 308  287.7 K
* 2
T2 /T
1.198
Therefore the length of the pipe is 498.2 m and the pressure and temperature at the
exit are 11.28 kPa and 287.7 K ( = 14.7o C ) respectively.
447
PROBLEM 9.6
Air flows from a large tank through a well-insulated 12 mm diameter pipe. If the air
enters the pipe at a Mach number of 0.2 and leaves at a Mach number of 0.6, find the
length of the pipe. Assume a mean friction factor of 0.005. How much longer must the
pipe be if the exit Mach number is 1? If the pipe is 75 cm longer than this latter value and
if the same conditions exist in the supply chamber, what reduction in the flow rate will
occur?
SOLUTION
The situation being considered is shown in Fig. P9.6.
Figure P9.6
It is assumed that the flow in the pipe is adiabatic and the flow from the tank into the
pipe is isentropic. As shown in Fig. P9.6, conditions at the inlet and outlet of the pipe are
denoted by subscripts 1 and 2 respectively. Using the software for Fanno flow or the
Fanno flow tables for air the following are obtained.
At the inlet where M = 0.2:
4 fl1*
 14.53
D
At the outlet where M = 0.6:
448
4 fl2*
 0.4908
D
Now:
4 fl1 2
4 fl1* 4 fl2*


 14.53  0.4908  14.04
D
D
D
From which it follows that:
l1 2 
14.04 D
14.04  0.012

 8.42 m
4f
4  0.005
Hence the length of the pipe is 8.42 m.
If the flow is choked at the exit:
4 fl12
4 fl1*

 14.53
D
D
From which it follows that:
l1 2 
14.53D
14.53  0.012

 8.72 m
4f
4  0.005
Hence, in this case, the pipe is 8.72 - 8.42 = 0.3 m longer than in the first case considered.
If the pipe is 75 cm longer that in the second case considered, the flow will be choked
at the exit of the pipe and therefore in this case:
l1 2  l1*  8.72  0.75  9.47 m
In this case therefore:
4 fl1*
4  0.005  9.47

 15.78
D
0.012
For this value of 4 f l1* / D the following are obtained using the software for Fanno
flow or the Fanno flow tables for air:
449
M 1  0.193
The mass flow rate through the pipe is given by:
m  1 V1 A 
1
a
M 1 1 A  0 a0
a0
0
subscript 0 referring to conditions in supply tank.
Because the conditions in the tank are the same with both lengths of pipe, it follows
that:
m L

m S
 1 / 0 L
 1 / 0 S
M 1L  a1 / a0 L
M 1S  a1 / a0 S
the subscripts Sand L referring to the conditions with the original pipe length and with the
increased pipe length. The software for isentropic flow or the isentropic flow tables for
air give for the two cases:
When M1 = 0.2:
p0 /p  1.020 , T0 /T  1.008
When M1 = 0.193:
p0 /p  1.019 , T0 /T  1.007
Hence:
m L
1.020
0.193



1.019
0.2
m S
1.008
 0.966
1.007
Therefore the mass flow rate with the longer pipe is 0.966 of the mass flow rate with
the shorter pipe, i.e., there is a 3.4% reduction in the mass flow rate.
450
PROBLEM 9.7
Air flows from a large tank, in which the pressure and temperature are 100 kPa and
30°C respectively, through a 1.6m long pipe with a diameter of 2.5 cm. The pipe is
connected to a short convergent nozzle with an exit diameter of 2.1 cm. The air from this
nozzle is discharged into a large tank in which the pressure is maintained at 35 kPa.
Assuming that the friction factor is equal to 0.002, find the mass flow rate through the
system. The flow in the nozzle can be assumed to be isentropic and the pipe can be
assumed to be heavily insulated.
SOLUTION
The situation being considered is shown in Fig. P9.7.
Figure P9.7
It is assumed that the flow in the pipe is adiabatic and that the flow from the supply
tank to the pipe is isentropic.
As shown in Fig. P9.7, conditions at the inlet and exit of the pipe are denoted by
subscripts 1 and 2 respectively while those at the exit of the discharge nozzle are denoted
by subscript 3.
To start the calculation, it will be assumed that the flow at the exit of the discharge
nozzle is choked, i.e., that M3 = 1. The validity of this assumption will be ascertained by
checking that the value of p3 obtained using this assumption is greater than or equal to the
pressure in the discharge tank.
Now, if the flow is choked at the discharge nozzle exit, then:
451
A *  A3
and for M3 = 1, software for isentropic flow or the isentropic flow tables for air give:
p0 /p3  1.893
Hence, because the flow across the discharge nozzle is assumed to isentropic:
A2
A
D2
2.52
 2  22 
 1.4172
A*
A3
D3
2.12
For this value of the area ratio, the software for isentropic flow or the isentropic flow
tables for air give:
M 2  0.4408 ,
p0 /p2  1.143
Hence:
p3
p /p
1.143
 0 2 
 0.6038
p2
p0 / p1
1.893
Next consider the flow in the pipe. At the exit where M2 = 0.4408 the software for
Fanno flow or the Fanno flow tables for air give:
p2
 2.438 ,
p*
4 fl2*
 1.681
D
But:
4 fl1*
4 fl2*
4 fl1 2
4  0.002  1.6


 1.681 
 1.681  0.512  2.193
D
D
D
0.025
For this value of 4 f l1* / D the following are obtained using the software for Fanno
flow or the Fanno flow tables for air:
M 1  0.193 ,
hence:
452
p1
 2.651
p*
p3
p p / p*
2.438
 3 2
 0.6038 
 0.5552
p1
p2 p1 / p *
2.651
Now consider the flow through the nozzle at the inlet to the pipe. Because the Mach
number at the exit of this nozzle is M1, i.e., 0.4065, the software for isentropic flow or the
isentropic flow tables for air give:
p0 /p1  1.121 , T0 /T1  1.033 , 0 /1  1.085
Then:
p3 
p3 p1
0.5552
p0 
 100  49.5 kPa
p1 p0
1.121
This is greater than the pressure in the discharge tank, i.e., greater than 35 kPa. This
shows that the assumption that the flow is choked at the outlet of the discharge nozzle is
correct.
The mass flow rate through the system is given by:
m  1 V1 A1 
1
a
M 1 1 A1 0 a0
0
a0
i.e., using the values at point 1 given above:
m 
1
 0.4065 
1.085
1
A1 0 a0  0.3686 A1  0 a0
1.033
But:
a0 
 RT0 
1.4  287  303.0  348.9 m/s
and:
0 
p0
100000

 1.1499 kg/m3
RT0
287  303.0
and:
A1 

4
D12 

4
 0.0252  0.0004908 m 2
453
Hence:
m  0.3686 A1 0 a0  0.3686  0.0004908  1.1499  348.9  0.0726 kg/s
Therefore the mass flow of air through the system is 0.0726 kg/s.
454
PROBLEM 9.8
Air at an inlet temperature of 60°C flows with a subsonic velocity through an
insulated pipe having an inside diameter of 5 cm and a length of 5 m. The pressure at the
exit to the pipe is 101 kPa and the flow is choked at the end of the pipe. If the average
friction factor is 0.005, determine the inlet and exit Mach numbers, the mass flow rate
and the change in temperature and pressure through the pipe.
SOLUTION
Conditions at the inlet and outlet of the duct will be denoted by subscripts 1 and 2
respectively. Because the flow is choked at the exit, the length of the pipe is equal to l1*.
Hence:
4 f l1*
4  0.005  5

 2.00
D
0.05
For this value of 4 f l1* / D the following are obtained using the software for Fanno
flow or the Fanno flow tables for air:
M 1  0.4183 ,
p1 /p*  2.574 , T1 /T *  1.159
Using these values gives because:
p2  p *  101kPa , T1  303 K
the following:
p1 
p1
 p2  2.574  101  260.0 kPa
p*
and:
T2 
T*
333
 T1 
 287.3 K
T1
1.159
The change in pressure and temperature down the pipe are therefore:
455
 p  p1  p2  260  101  159 kPa and  T  T1  T2  333  287.3  45.7 K
The mass flow rate through the pipe is given by:
m  1 V1 A1  1M1 a1 A1
But:
a1 
 RT1 
1.4  287  333.0  365.8 m/s
and:
1 
p1
260000

 2.72 kg/m3
RT1
287  333.0
and:
A1 

4
D12 

4
 0.052  0.00196 m 2
Hence:
m  1M1 a1 A1  2.72  0.4183  365.8  0.00196  0.818 kg/s
Therefore the inlet and outlet Mach numbers are 0.4183 and 1 respectively, the mass
flow rate is 0.818 kg/s and the pressure and temperature changes down the pipe are 159
kPa and 45.7 K respectively.
456
PROBLEM 9.9
Hydrogen flows through a 50 mm diameter pipe. The inlet pressure is 400 kPa, the
inlet velocity is 300 m/s, and the inlet temperature is 30°C. How long is the pipe if the
flow is choked at the exit end? Assume a mean friction factor of 0.0058 and that the flow
is adiabatic.
SOLUTION
For hydrogen it will be assumed that the molar mass is 2.016 and the specific heat
ratio, γ , is 1.407. Conditions at the inlet and outlet of the duct will be denoted by
subscripts 1 and 2 respectively.
At the inlet:
M1 
V1

a1
V1

 RT1
300
300

 0.2263
1326
1.407  (8314 / 2.016)  303.0
For this value of M1 and for γ = 1.407, the software for Fanno flow gives:
4 fl1*
 10.78
D
But the flow is choked at the exit so l1* is the length of the pipe, i.e., equals l1-2 . Hence:
l1 2 
10.78 D
10.78  0.05

 23.2 m
4f
4  0.0058
Therefore the pipe has a length of 23.2 m.
457
PROBLEM 9.10
Air flows through a 0.15 m  0.25 m rectangular duct. The Mach number, pressure
and temperature at a certain section of the duct are found to be 2, 75 kPa, and 5°C
respectively. Assuming the mean friction factor of 0.006, find the maximum length of
duct that can be installed downstream of this section if no shock wave is to occur in the
duct. Also find the exit pressure and temperature that will exist with this maximum length
of duct.
SOLUTION
Conditions at the inlet and outlet of the duct will be denoted by subscripts 1 and 2
respectively. The maximum duct length exists when the flow at the exit is choked.
Because the flow is choked at the exit:
p2  p * ,
T2  T * ,
l1 2  l1*
For the inlet Mach number of 2, the software for Fanno flow or the Fanno flow tables
for air give:
p1
 0.4082 ,
p*
T1
 0.6667 ,
T*
4 fl1*
 0.305
D
Hence:
l1 *  l1 2 
0.305 D
4f
But D is taken as the hydraulic diameter, i.e.:
D  4
Area
 0.15  0.25 
 4  
  0.1875 m
Perimeter
 2   0.15  0.25  
and so:
458
l1 2 
0.305 D
0.305  0.1875

 2.383 m
4f
4  0.006
The pressure and temperature at the exit are given by:
p2  p* 
p1
75

 183.7 kPa
*
p1 / p
0.4082
and:
T2  T * 
T1
278

 417.0 K
*
T1 / T
0.6667
Therefore the maximum duct length is 2.383 m and the exit pressure and temperature
with this duct length are 183.7 kPa and 417 K ( = 144°C ) respectively.
459
PROBLEM 9.11
Air is stored in a tank at a pressure and temperature of 1.6 MPa and 20°C
respectively. What is the maximum possible mass rate of flow from the tank through a
pipe with a diameter of 1.2 cm and a length of 30 cm? The pipe discharges to the
atmosphere and the atmospheric pressure is 101 kPa. The average friction factor can be
assumed to be 0.006 and the flow in the pipe can be assumed to be subsonic and
adiabatic.
SOLUTION
Conditions at the inlet and outlet of the duct will be denoted by subscripts 1 and 2
respectively. The maximum flow rate exists when the flow at the exit is choked. Because
the flow is choked at the exit:
p2  p * ,
T2  T * ,
l1 2  l1 *
But, using the given information:
4 f l1 2
4 f l1*
4  0.006  0.3


 0.6
D
D
0.012
For this value of 4 f l1* / D the following are obtained using the software for Fanno
flow or the Fanno flow tables for air:
M 1  0.5748 ,
p1 /p*  1.846
Using these values gives because:
p2  p *  101 kPa
the following:
p1 
p1
 p2  1.846  101  186.5 kPa
p*
460
But the flow in the supply nozzle is assumed to be isentropic. The software for
isentropic flow or the isentropic flow tables for air give for a Mach number of 0.5748:
p0 /p1  1.251 , T0 /T1  1.066
Hence:
p0
1600

 1279.0 kPa
p0 / p1
1.251
p1 
and:
T1 
T0
293

 274.9 K
1.066
T0 / T1
Therefore:
p2  p * 
p1
1279.0

 692.8 kPa
p1 / p *
1.846
This is greater than the back pressure of 101 kPa so there will be expansion waves in
the discharge from the pipe.
The mass flow rate through the system is given by:
m  1 V1 A1  1M 1 a1 A1
But:
a1 
 RT1 
1.4  287  274.9  332.3 m/s
and:
1 
and:
A1 
p1
692800

 8.7811 kg/m 3
R T1
287  274.9

4
D12 

4
 0.0122  0.000113 m 2
Hence:
m  1M 1 a1 A1  8.7811  0.5748  332.3  0.000113  0.1895 kg/s
Therefore the mass flow of air through the system is 0.1895 kg/s.
461
PROBLEM 9.12
Air flows through a 12 m long pipe which has a diameter of 25 mm. At the inlet to the
pipe the air velocity is 80 m/sec, the pressure is 350 kPa and the temperature is 50oC. If
the mean friction factor is 0.005, find the velocity, pressure, and temperature at the end of
the pipe. Assume the flow to be adiabatic.
SOLUTION
Conditions at the inlet and outlet of the duct will be denoted by subscripts 1 and 2
respectively. At the inlet:
M1 
V1

a1
V1

 RT1
80
80

 0.2221
360.3
1.4  287  323
For this value of M1 the following are obtained using the software for Fanno flow or
the Fanno flow tables for air:
p1
 4.908 ,
p*
T1
 1.188 ,
T*
4 fl1*
 11.33
D
But:
4 f l2*
4 f l1* 4 f l1 2
4  0.005  12


 11.33 
 1.73
D
D
D
0.025
For this value of 4 f l2*/D the following are obtained using the software for Fanno
flow or the Fanno flow tables for air:
M 2  0.4371 ,
p2
 2.460 ,
p*
T2
 1.156
T*
Using these values gives because the pressure and temperature at the entrance are 350
kPa and 323 K respectively:
462
p2 
p2 / p*
2.460
p1 
 350  175.4 kPa
*
p1 / p
4.908
and:
T2 
T2 / T *
1.156
T 
 323  314.3 K
* 1
T1 / T
1.188
Hence:
V1  M 1 a1  M 1  RT1  0.4371  1.4  287  314.3  151.8 m/s
Therefore the velocity, pressure and temperature at the exit are 151.8 m/s, 175.4 kPa,
and 314.3 K ( = 41.3°C ) respectively.
463
PROBLEM 9.13
Air is expanded from a large reservoir in which the pressure and temperature are 200
kPa and 30°C respectively through a convergent nozzle which gives an exit Mach
number of 0.2. The air then flows down a pipe with a diameter of 25 mm, the Mach
number at the end of this pipe being 0.8. Assuming that the flow in the nozzle is
isentropic and the flow in the pipe adiabatic, find the length of the pipe and the pressure,
at the exit of the pipe. The friction factor in the pipe can be assumed to be 0.005.
SOLUTION
Conditions at the inlet and outlet of the pipe: will be denoted, by subscripts 1 and 2
respectively. Consider the isentropic flow through the nozzle. Because the exit Mach
number, M1, is equal to 0.2, the software for isentropic flow or the isentropic flow tables
for air give:
p0
 1.028
p1
Hence:
p1 
p0
200

 194.6 kPa
p0 / p1
1.028
Now consider the flow in the pipe. At the inlet and exit of the pipe, the software for
Fanno flow or the tables for Fanno flow of air give:
At the inlet where M = 0.2:
p1
 5.455 ,
p*
4 fl1*
 14.53
D
At the outlet, where M = 0.8:
4 fl2*
 0.07229
D
p2
 1.289 ,
p*
Now:
464
4 fl1 2
4 fl1*
4 fl2*


 14.53  0.07229  14.46
D
D
D
From which it follows that:
l1 2  14.46
D
14.46  0.025

 18.07 m
4f
4  0.005
Also because p1 = 194.6 kPa it follows that:
p2 
p2 / p *
1.289
p1 
 194.6  46.0 kPa
p1 / p *
5.455
Therefore the length of the pipe is 498.2 m and the pressure at the exit of the pipe is
46.0 kPa.
465
PROBLEM 9.14
Air flows through a 4 cm diameter pipe. At the inlet to the pipe the stagnation
pressure is 150 kPa, the stagnation temperature is 80°C and the velocity is 120 m/s. If the
mean friction factor is 0.006, and if the flow can be assumed to be adiabatic, find the
maximum duct length before choking occurs.
SOLUTION
Consider the flow at the inlet to the pipe. The energy equation gives:
cp T0  cp T1 
V12
2
Hence noting that:
cp  cv  R ,i.e., cp 
R
 1
the energy equation can be written:
  1  2
T0  T1  
 V1
 2 R 
Hence, at the inlet:


  1  2
1.4 1
2
T1  T0  
  120  345.8 K
 V1  353  


2

R
2
1.4
287




Using this result then gives:
M1 
V1

a1
V1

 RT1
120
 0.3219
1.4  287  345.8
For this value of M1 the following are obtained using the software for Fanno flow or
the Fanno flow tables for air:
466
4 f l1*
 4.375
D
Because the flow is choked at the outlet of the pipe l* = l1-2 so:
l1 2 
4.375 D
4.375  0.04

 7.29 m
4f
4  0.006
Therefore the length of the pipe is 7.29 m.
467
PROBLEM 9.15
Air f1ows through a 0.5 inch diameter pipe at subsonic velocities. The pipe is 20 feet
long and the pressure and temperature at the inlet to the pipe are 60 psia and 130oF. The
pipe is discharged into a large vessel in which the pressure is kept at 20 psia. If the mean
friction factor is assumed to be 0.0055 and if the flow is assumed to be adiabatic, find the
mass flow rate through the pipe.
SOLUTION
Conditions at the, inlet and outlet of the pipe will be denoted by subscripts l and 2
respectively.
The mass flow rate through the pipe must be such that the pressure on the exit plane is
20 psia. Now:
4 f l2*
4 f l1* 4 f l1 2
4 f l1* 4  0.0055  20
4 f l1*





 10.56
D
D
D
D
0.5 /12
D
One simple way of obtaining the solution involves the following steps:
(1) Guess the value of M1.
(2) Using this value of M1 use the software for Fanno flow or the tables for Fanno
flow of air to determine p1/p* and 4 f l1*/ D.
(3) Use the above equation to find 4 f l2*/ D.
(4) Use this value of 4 f l2*/ D with the software for Fanno flow or the tables for
Fanno flow of air to determine p2/p*.
(5) With this value of p2/p* find p2 using:
p2 
p2 / p *
p1
p1 / p *
(6) Compare this value with the required value of p2 , i.e., 20 psia.
(7) Repeat the above procedure with different values of M1 and deduce from the
results the value that makes p2 = 20 psia.
468
Some results obtained using this procedure are shown in the following table:
M1
0.200
0.180
0.220
0.225
0.226
4 f l1*/ D
14.53
18.54
11.60
10.99
10.87
p1/p*
5.46
6.07
4.96
4.84
4.82
4 f l2*/ D
3.97
7.98
1.04
0.43
0.31
p2/p*
3.25
4.25
4.12
1.71
1.60
p2 - psia
35.75
42.02
25.69
21.23
19.95
From these results it will be seen that M1 is approximately equal to 0.226.
The mass flow rate through the system is given by:
m  1 V1 A1  1M 1 a1 A1
But:
a1 
 RT1 
1.4  (53.3  32.2)  590  1191 ft/sec
and:
1 
p1
144  60

 0.275 lbm/ft 3
RT1
53.3  590
and:

2

 0.5 
2
A1  D   
  0.0014 ft
4
4  12 
2
1
Hence:
m  1M 1 a1 A1  0.275  0.226  1191  0.0014  0.1008 lbm/sec
Therefore the mass flow of air through the system is 0.1008 lbm / sec.
469
PROBLEM 9.16
Air flows through a circular pipe at a rate of 8.3 kg/s. The Mach number at the inlet to
the pipe is 0.15 and at the exit to the pipe is 0.5. The pressure and temperature at the inlet
are 350 kPa and 38°C respectively. Assuming the flow to be adiabatic, and the mean
friction factor to be 0.005, find the, length and the diameter of the duct and the pressure
and temperature at the exit of the duct.
SOLUTION
Conditions at the inlet and outlet of the pipe will be denoted by subscripts 1 and 2
respectively. The mass flow rate is given by:
m  1 V1 A1  1M1 a1 A1
But:
a1 
RT1 
1.4  287  311  353.5 m/s
and:
1 
p1
350000

 3.921 kg/m3
RT1
287  311
and:
A1 

4
D2
Because the mass flow rate is 8.3 kg/s:
m  8.3  1M 1 a1 A1  1M 1 a1

4
D12  3.921  0.15  353.5 
Hence:
D1 
4  8.3
 0.2255 m
3.921  0.15  353.5
470

4
D12
At the inlet and exit of the pipe, the software for Fanno flow or the tables for Fanno
flow of air give:
At the inlet where M = 0.15:
p1
 7.287 ,
p*
T1
 1.195 ,
T*
4 fl1*
 27.93
D
T2
 1.143 ,
T*
4 fl2*
 1.069
D
At the outlet, where M = 0.5:
p2
 2.138 ,
p*
Now:
4 fl1 2
4 fl1* 4 fl2*


 27.93  1.069  26.86
D
D
D
From which it follows that:
l1 2  26.86
D
26.86  0.2255

 302.9 m
4f
4  0.005
Also because p1 = 350 kPa and T1 = 311K it follows that:
p2 
p2 / p *
2.138
p1 
 350  102.7 kPa
p1 / p *
7.287
and:
T2 
T2 / T *
1.143
T1 
 311  297.5 K
T1 / T *
1.195
Therefore the length and diameter of the pipe are 302.9 m and 22.55 cm respectively
and the pressure and temperature at the exit of the pipe are 102.7 kPa and 297.5 K ( =
24.5°C ) respectively.
471
PROBLEM 9.17
Air is expanded from a large reservoir in which the pressure and temperature are 250
kPa and 30°C respectively through a convergent nozzle which gives an exit Mach
number of 0.3. The air from the nozzle flows down a pipe having a diameter of 5 cm. The
Mach number at the end of this pipe is 0.95. Find the length of the pipe and the pressure
at the end of the pipe. If the actual pipe length was only 0.75 of this length, find the Mach
number and the pressure that would exist at the end of the pipe. The flow in the nozzle
can be assumed to be isentropic and the friction factor in the pipe can be assumed to .be
0.005.
SOLUTION
Conditions at the inlet and outlet of the pipe will be denoted by subscripts 1 and 2
respectively.
Consider the isentropic flow through the nozzle. Because the exit Mach number, M1 ,
is equal to·0.3, software for isentropic flow or the isentropic flow tables for air give:
p0
 1.064
p1
Hence:
p1 
p0
250

 235.0 kPa
p0 / p1
1.064
Now consider the flow in the pipe. At the inlet and exit of the pipe, the software for
Fanno flow or the tables for Fanno flow of air give:
At the inlet where M = 0.3:
4 fl1*
 5.299
D
p1
 3.619 ,
p*
At the outlet, where M = 0.95:
472
p2
 1.061 ,
p*
4 fl2*
 0.0033
D
Now:
4 fl1 2
4 fl1* 4 fl2*


 5.299  0.0033  5.296
D
D
D
From which it follows that:
l1 2  5.296
D
5.296  0.05

 13.24 m
4f
4  0.005
Also because p1 = 235 kPa it follows that:
p2 
p2 / p *
1.061
p1 
 235.0  68.9 kPa
3.619
p1 / p *
Therefore the length of the pipe is 13.24 m and the pressure at the exit of the pipe is
68.9 kPa.
Consideration is next given to the conditions existing at the end of the pipe when its
length is:
l1 2 = 0.75  13.24 = 9.93 m
It will be assumed that the conditions at the inlet to the pipe are unchanged. With the
new pipe length:
4 fl12
4  0.005  9.93

 3.972
D
0.05
Therefore:
4 fl2*
4 fl1* 4 fl1 2


 5.299  3.972  1.327
D
D
D
For this value of 4 f l2* / D the following are obtained using the software for Fanno
flow or the tables for Fanno flow of air:
473
p2
 2.272 , M 2  0.4717
p*
Using these values gives because the pressure at the entrance is assumed to still be
235 kPa:
p2 
p2 / p *
2.272
p1 
 235  147.5 kPa
p1 / p *
3.619
In this situation therefore the Mach number and the pressure at the exit of the pipe are
0.4717 and 147.5 kPa respectively.
474
PROBLEM 9.18
Air flows at a steady rate at subsonic velocity through a pipe with an internal diameter
of 26 mm and a length of 15 m. The pressure and temperature in the air at the inlet to pipe
are 140kPa and 120°C respectively. Assuming that the flow is adiabatic and using an
average friction factor for the flow of 0.005 find the maximum possible mass flow rate
through the pipe. Also find the temperature and pressure at the exit of the pipe when the
Mach number at the exit is equal to 1.
SOLUTION
Conditions at the inlet and outlet of the duct will be denoted by subscripts 1 and 2
respectively. The maximum flow rate exists when the flow at the exit is choked.
Therefore because the flow is choked at the exit:
p2  p * ,
T2  T * ,
l1 2  l1 *
But, using the given information:
4 f l1*
4  0.005  15

 11.54
D
0.026
For this value of 4 f l1* / D the following are obtained using the software for Fanno
flow or the Fanno flow tables for air:
M 1  0.2204 ,
p1 /p*  4.945 ,
T1 /T *  1.188
The mass flow rate is then given by:
m  1 V1 A1  1M1 a1 A1
But:
a1 
 RT1 
1.4  287  393  397.4 m/s
and:
475
1 
p1
140000

 1.241 kg/m 3
RT1
287  393
and:
A

4
D2 

4
 0.0262  0.0005309 m2
Hence:
m  1M 1 a1 A1  1.241  0.2204  397.4  0.0005309  0.5771 kg/s
The pressure and temperature at the exit are given by:
p2  p * 
p1
140

 28.3 kPa
p1 / p * 4.945
T2  T * 
T1
393

 330.8 K
T1 / T * 1.188
and:
Therefore the maximum flow rate is 0.5771 kg/s and the exit pressure and
temperature are 28.3 kPa and 330.8 K ( = 57.8°C ) respectively.
476
PROBLEM 9.19
Air flows down a 20 mm diameter pipe which has a length of 0.8 m. If the velocity at
the inlet to the pipe is 200 m/s and the temperature is 30°C, find the average friction
factor if the flow is choked at the exit to the pipe. Assume the flow to be adiabatic.
SOLUTION
Conditions at the inlet to the pipe will be denoted by subscript 1.
Using the given inlet conditions gives:
M1 
V1

a1
V1

 RT1
200
200

 0.573
348.9
1.4  287  303
For this value of M1 the following are obtained using the software for Fanno flow or the
Fanno flow tables for air:
4 fl1*
 0.6084
D
But because the flow is choked at the exit:
l1 2  l *  0.8 m
Hence from the above two results:
4 x f  0.8
 0.6084
D
i.e.:
f 
0.6084 D
0.6084  0.02

 0.0038
4  0.8
4  0.8
Therefore the average friction factor is 0.0038.
477
PROBLEM 9.20
Air enters an insulated pipe with a diameter of 7.5 cm at a Mach number of 3.0. As a
result of friction, the Mach number decreases to a value of 1.5 at the exit of the pipe. If
the mean friction factor is equal to 0.002, find the length of the pipe.
SOLUTION
Conditions at the inlet and outlet of the pipe will be denoted by subscripts 1 and 2
respectively.
At the inlet and exit of the pipe, the software for Fanno flow or the tables for Fanno
flow of air give:
At the inlet where M = 3:
4 fl1*
 0.5222
D
At the outlet where M = 1.5:
4 fl2*
 0.1361
D
Now:
4 fl1 2
4 fl1* 4 fl2*


 0.5222  0.1361  0.3861
D
D
D
From which it follows that:
l1 2  0.3861
D
0.3861  0.075

 3.62 m
4f
4  0.002
Therefore the pipe has a length of 3.62 m.
478
PROBLEM 9.21
A converging-diverging nozzle supplies air to a well-insulated constant area duct. At
the inlet to the duct the Mach number is 2, the pressure is 140 kPa, and the temperature is
-100°C. If the Mach number is 1 at the exit to the duct, determine the pressure and
temperature at the duct exit.
SOLUTION
Conditions at the inlet and outlet of the duct will be· denoted by subscripts t and 2
respectively. Because the flow is choked at the exit:
p2  p * ,
T2  T *
Hence, the pressure and temperature at the exit are given by:
p2  p * 
p1
140

 343.0 kPa
p1 / p * 0.4082
and:
T2  T * 
T1
173

 259.5 K
T1 / T * 0.6667
Therefore the exit pressure and temperature are 343.0 kPa and 259.5 K ( = -13.5°C )
respectively.
479
PROBLEM 9.22
Air flows down a constant area pipe which has a diameter of 5 cm. The Mach number
at the inlet to the pipe is 2 and the inlet pressure and temperature are 80 kPa and 200° C
respectively. The flow in the pipe can be assumed to be adiabatic. If the pipe is 0.6 m
long and if the average friction factor is 0.005, find the Mach number, pressure and
temperature at the exit of the pipe. If, on leaving the pipe, the air flows through a
convergent-divergent nozzle which has an exit area that is three times the throat area and
if the air stream leaves the nozzle at a subsonic velocity, find the pressure and the Mach
number at the exit of the nozzle if the flow in the nozzle can be assumed to be isentropic.
SOLUTION
Conditions at the inlet and outlet of the pipe will be denoted by subscripts 1 and 2
respectively while conditions at the exit of the nozzle will be denoted by subscript 3.
First consider the flow in the pipe. For the inlet Mach number Ml of 2, the software
for Fanno flow or the Fanno flow tables for air flow give:
p1 / p *  0.4082 ,
T1 / T *  0.6667 ,
4 f l1 *
 0.305
D
Now:
4 fl2*
4 fl1* 4 fl1 2
4  0.005  0.6


 0.0.305 
 0.065
0.05
D
D
D
For this value of 4 f l2* / D the following are obtained using the software for Fanno
flow or the Fanno flow tables for air:
M 2  1.301 ,
p2 /p *  0.7281 ,
T2 /T *  0.8967
Hence the pressure and temperature at the exit are given by:
p2 
p2 / p *
0.7281
p1 
 80  142.7 kPa
p1 / p *
0.4082
480
and:
T2 
T2 / T *
0.8967
T1 
 293  394 K
T1 / T *
0.6667
Therefore the Mach number, pressure and temperature at the exit of the pipe are
1.301, 142.7 kPa and 394.1 K ( = 121.1°C ) respectively.
Next consider the flow in the nozzle. The flow through the nozzle is assumed to be
isentropic so no shock waves exist in the nozzle. The Mach ·number at the inlet to the
nozzle is 1.301. For this value the software for isentropic flow or the isentropic flow
tables for air give:
p0
 2.775
p2
And the area ratio of the nozzle is given as:
A3
3
A*
Because the flow leaves the nozzle at a subsonic velocity, the software for isentropic
flow o the isentropic flow tables for air give for this value of A3 / A*:
M 3  0.1974 ,
p0 /p3  1.028
Hence:
p3 
p0 / p2
2.775
p2 
 142.7  385.2 kPa
p0 / p3
1.0282
Therefore the Mach number and pressure at the exit of the nozzle are 0.1974 and
385.2 kPa respectively.
481
PROBLEM 9.23
Air with a stagnation pressure of 600 kPa and a stagnation temperature of 150°C
flows through a convergent-divergent nozzle, the Mach number being greater than 1 at
the nozzle exit. The throat area of the nozzle is 1 cm2. The flow from the nozzle enters a
duct which has a constant area of 3cm2, the flow being choked at the duct exit. If the flow
in the nozzle can be assumed to be isentropic and if the flow in the duct can be assumed
to be adiabatic, and if the mean friction factor is 0.004, find the pressure and temperature
on the exit plane of the duct.
SOLUTION
Conditions at the inlet and outlet of the pipe will be denoted by subscripts 1 and 2
respectively. Consider the flow in the nozzle. The area ratio of the nozzle is:
A1
3
 3
A* 1
Because the flow leaves the nozzle at a supersonic velocity, the software for
isentropic flow or the isentropic tables for the flow of air give for this value of A1 / A*:
M 1  2.637 ,
p0
 21.14 ,
p1
T0
 2.391
T1
Hence:
p1 
p0
600

 28.38 kPa
p0 / p1
21.14
and:
T1 
T0
423

 176.9 K
T0 / T1
2.391
The values of the Mach number, pressure and, temperature at the inlet to the duct
have thus been established.
482
Now for a Mach number Ml of 2.637, the software for Fanno flow or the tables for
Fanno flow of air give:
p1
 0.2687 ,
p*
T1
 0.5019
T*
But because the flow is choked at the exit:
p2  p * ,
T2  T *
Hence, the pressure and temperature at the exit are given by:
p2  p * 
p1
28.38

 105.6 kPa
p1 / p * 0.2687
and:
T2  T * 
T1
176.9

 352.5 K
T1 / T * 0.5019
Therefore the exit pressure and temperature are 105.6 kPa and 352.5 K ( = 79.5°C )
respectively.
483
PROBLEM 9.24
Air enters a pipe having a diameter of 0.1 m and a length of 1 m with a Mach number
of 2 and a pressure of 90 kPa. Assuming the flow to be adiabatic and the mean friction
factor to be 0.005, plot a graph of the pressure variation along the length of the duct.
SOLUTION
Conditions at the inlet to the pipe will be denoted by the subscript 1. For the inlet
Mach number Ml of 2, the software for Fanno flow or the Fanno flow tables for air flow
give:
p1 / p *  0.4082 ,
4 f l1 *
 0.3050
D
Now at any other point at a distance x m down the pipe:
4 f l1* 4 f x
4f l*
4  0.005
x  0.305  0.2 x


 0.305 
D
D
D
0.1
For any selected value of x, this equation gives the value of 4 f l* / D. The software for
Fanno flow for supersonic flow or the Fanno flow tables for air then give the value of p/
p* corresponding to this value of 4 f l* / D. The pressure at distance x down the pipe is
then given by using:
p 
p / p*
p / p*
p
p1 
 90  220.5
kPa
p1 / p *
0.4082
p*
Using this, procedure, the pressures for various values of x up to 1 m have been
determined and are shown in the following table:
484
x-m
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
4 f l* / D
0.305
0.285
0.265
0.245
0.225
0.205
0.185
0.165
0.145
0.125
0.105
p / p*
0.4082
0.4284
0.4491
0.4706
0.4930
0.5163
0.5407
0.5664
0.5937
0.6228
0.6544
p - kPa
90.0
94.5
99.0
103.8
108.7
113.8
119.2
124.9
130.9
137.3
144.3
The variation of pressure with distance down the pipe given by these results is shown
in Fig. P9.24.
Figure P9.24
485
PROBLEM 9.25
An air stream enters a 2.5 cm diameter pipe with a Mach number of 2.5 and a
pressure and temperature of.30 kPa and -15°C respectively. The average friction factor
can be assumed to be 0.005. Determine the maximum possible length of tube if are to be
no shock waves in the flow. Also find the values of the pressure and the temperature at
the tube exit for this maximum length. Assume the flow to be adiabatic.
SOLUTION
Conditions at the inlet and outlet of the duct will be denoted by subscripts 1 and 2
respectively. Because there are no shock waves in the flow, i.e., the flow remains
supersonic, the maximum duct length will be that which gives an exit Mach number of 1,
i.e., will cause the flow to be choked at the exit. Because the Mach number is 1 at the
exit, i.e., because flow is choked at the exit:
l1 2  l1 * ,
p2  p * ,
T2  T *
For the specified value of the inlet Mach number, M1 , i.e., 2.5, the following are
obtained using the software for Fanno flow or the Fanno flow tables for air flow:
p1 / p *  0.2921 ,
T1 / T *  0.5333 ,
4 f l1 *
 0.4320
D
Hence:
4 f l1 2
4 f l1 *

 0.4320
D
D
from which it follows that
l1 2  0.4320
D
0.4320  0.025

 0.54 m
4f
4  0.005
The pressure and temperature at the exit are given by:
486
p2  p * 
p1
30

 32.57 kPa
p1 / p * 0.2921
and:
T2  T * 
T1
258

 483.8 K
T1 / T * 0.5333
Therefore the maximum length of the pipe is 0.54 m and, with this length, the exit
pressure and temperature are 32.57 kPa and 483.8 K ( = 210.8°C ) respectively.
487
PROBLEM 9.26
Air flows from a large reservoir in which the pressure and temperature are 1 MPa and
30°C respectively through a convergent-divergent nozzle and into a constant area duct.
The ratio of the nozzle exit area to its throat area is 3.0 and the length-to-diameter ratio of
the duct is 15. Assuming that the flow in the nozzle is isentropic, that the flow in the duct
is adiabatic, and that the average friction factor is 0.005, find the back pressure for a
shock to appear at the exit to the duct.
SOLUTION
The situation being considered is shown in Fig. P9.26.
Figure P9.26
Because there is a shock wave on the exit plane of the duct, the flow in the duct must
be supersonic. Conditions at the inlet to the duct will be denoted by subscript 1 while
conditions at the exit of the duct just before the shock wave will be denoted by subscript
2. Conditions just after the shock wave at the exit of the duct will be denoted by subscript
3.
Consider the flow in the nozzle. The area ratio of the nozzle is:
A1
 3
A*
488
Because the flow leaves the nozzle at a supersonic velocity, the software for
isentropic flow or the isentropic tables for the flow of air give for this value of A1 / A*:
M 1  2.637 ,
p0
T
 21.14 , 0  2.391
p1
T1
Hence:
p1 
p0
1000

 47.62 kPa
p0 / p1
21.14
and:
T1 
T0
303

 126.7 K
T0 / T1
2.391
The values of the Mach number, pressure and, temperature at the inlet to the duct
have thus been established.
Now for a Mach number Ml of 2.637, the software for Fanno flow or the tables for
Fanno flow of air give:
p1 / p *  0.2687 ,
4 f l1 *
 0.4599
D
Hence since l1-2 / D = 15:
4 fl2*
4 fl1* 4 fl1 2


 0.4599  4  0.005  15  0.1599
D
D
D
For this value of 4 f l2* / D the following are obtained using the software for Fanno
flow or the Fanno flow tables for air:
M 2  1.566 ,
p2 /p*  0.5732
Hence the pressure at the exit is given by:
p2 
p2 / p *
0.5732
p1 
 47.62  101.6 kPa
p1 / p *
0.2687
489
Now consider the changes across the normal shock wave at the exit of the duct. The
Mach number ahead of the shock is M2 , i.e., 1.566 and for this Mach number the software
for normal shock waves or the tables for normal shock waves in air give:
p3 /p2  2.694
Hence:
p3 
p3
p2  2.694  101.6  273.7 kPa
p2
Because the flow downstream of the shock wave is subsonic, this will be equal to the
back pressure, i.e., the back pressure is 273.7 kPa.
490
PROBLEM 9.27
Air enters a pipe at a Mach number of 2.5, a temperature of 40°C and a pressure of 70
kPa. The pipe has a diameter of 2.0 cm and the flow can be assumed to be adiabatic. A
shock occurs in the pipe at a location where the Mach number is 2. If the Mach number at
the exit from the pipe is 0.8 and if the average friction factor is 0.005, find the distance of
the shock from the entrance to the pipe and the total length of the pipe. Also find the
pressure at the exit of the pipe.
SOLUTION
The situation being considered is shown in Fig. P9.27.
Figure P9.27
As shown in Fig. P9.27, the conditions at the inlet, upstream of the shock,
downstream of the shock and at the outlet of the pipe are denoted by subscripts 1, 2, 3
and 4 respectively. Using the software for Fanno flow or the tables for Fanno flow of air,
the following are obtained.
At the inlet where M = M1 = 2.5:
p1 / p *  0.2921 ,
Upstream of the shock where M = M2 = 2:
491
4 f l1 *
 0.4320
D
p2 / p *  0.3080 ,
4 f l2 *
 0.4082
D
Hence:
p2 / p *
0.4082
p1 
 70  97.82 kPa
p1 / p *
0.2921
p2 
Also:
4 f l1 2
4 f l1* 4 f l2*


 0.4320  0.3050  0.127
D
D
D
From which it follows that:
l1 2 
0.127 D
0.127  0.02

 0.127 m
4f
4  0.005
Therefore the normal shock occurs at a distance of 0.127 m downstream of the inlet to the
pipe.
Next consider the changes across the normal shock wave. The Mach number ahead
of the shock is M2, i.e., is 2 and for this Mach number the software for normal shock
wave or the tables for normal shock waves in air give:
M 3  0.5774 ,
p3 / p2  4.500
Hence:
p3 
p3
p2  4.500  97.82  440.2 kPa
p2
For the flow downstream of the shock wave where M = M3 = 0.8, the software for
Fanno flow or the tables for Fanno flow of air give:
p3 / p *  1.837 ,
4 f l3 *
 0.5876
D
At the exit of the pipe where M = M4 = 0.8, the software for Fanno flow or the tables
for Fanno flow of air give:
492
p4 / p *  1.837 ,
4 f l4 *
 0.5876
D
Now:
4 f l3 4
4 f l3*
4 f l4*


 0.5876  0.07229  0.5153
D
D
D
From which it follows that
l3 4 
0.5153D
0.5153  0.02

 0.5153 m
4f
4  0.005
Hence the length of the pipe downstream of the shock wave is 0.5153 m. Therefore
the total length of the pipe is 0.127 + 0.5153 = 0.6423 m.
Also:
p4 
p4 / p *
1.289
p3 
 440.2  308.4 kPa
p3 / p *
1.837
Therefore the normal shock occurs at a distance of 0.127m from the inlet to the pipe,
the total length of the pipe is 0.6423 m, and the pressure at the exit of the pipe is 308.4
kPa .
493
PROBLEM 9.28
Air with a stagnation pressure of 700 kPa flows through a convergent-divergent
nozzle with an exit area to throat area ratio of 3. The flow in this nozzle can be assumed
to be isentropic. The air from the nozzle enters a well insulated duct with a length-todiameter ratio of 20. The mean friction factor is 0.002. The air from the duct is
discharged into a large reservoir in which the pressure is 100 kPa. Find the Mach
numbers at the inlet and exit of the duct.
SOLUTION
It will be assumed that the flow is at a supersonic velocity at the exit of the nozzle.
Conditions at the inlet and outlet of the duct will be denoted by subscripts 1 and 2
respectively.
Consider the flow in the nozzle. The area ratio of the nozzle is:
A1
3
A*
Because the flow leaves the nozzle at a supersonic velocity, the software for
isentropic flow or the isentropic tables for the flow of air give for this value of A1 / A*:
p0
 21.14
p1
M 1  2.637 ,
Hence:
p1 
p0
700

 33.11 kPa
p0 / p1
21.14
Now for a Mach number Ml of 2.637, the software for Fanno flow or the tables for
Fanno flow of air give:
p1 / p *  0.2687 ,
4 f l1 *
 0.4599
D
494
Hence, if there are no shock waves in the duct:
4 f l2*
4 f l1* 4 f l1 2


 0.4599  4  0.002  20  0.2999
D
D
D
For this value of 4 f l2* / D the following are obtained using the software for Fanno
flow or the Fanno flow tables for air:
M 2  1.983 ,
p2 /p*  0.4133
Hence the pressure at the exit is given by:
p2 
p2 / p *
0.4133
p1 
 33.11  50.93 kPa
0.2687
p1 / p *
This is less than the pressure in the discharge reservoir, i.e., less than 100 kPa.
Therefore a shock wave occurs in the system. Assume for the moment that a normal
shock wave lies on the exit plane of the duct and let the conditions downstream of this
shock be denoted by subscript 3. The Mach number ahead of the shock is M2 , i.e., 1.983
and for this Mach number the software for normal shock waves or the normal shock wave
tables for air give:
p3 /p2  4.421
Hence:
p3 
p3
p2  4.421  50.93  225.16 kPa
p2
This is greater than the pressure in the discharge reservoir. This means that the back
pressure is less than that required to give a normal shock wave on the exit plane but
greater than required for a shock free flow. Therefore the flow will involve oblique shock
waves outside the duct in the discharge into the reservoir. This means that the Mach on
the discharge plane of the duct is equal to M2 , i.e., the Mach numbers at the inlet and exit
of the duct are 2.637 and 1.983 respectively.
495
PROBLEM 9.29
Air with a stagnation pressure of 300 kPa and a stagnation temperature of 30oC enters
a constant-area duct at Mach number of 3. The duct has a length-to-diameter ratio of 60.
The flow can be assumed to be adiabatic and the average friction factor is 0.0025. If the
pressure at the exit to the duct is 50 kPa, determine the Mach number at the exit of the
duct and the location of the shock down the duct in diameters. (Hint: Apply an iterative
solution using guessed values of the shock wave position.)
SOLUTION
Conditions at the inlet, upstream of the shock, downstream of the shock and at the
outlet of the duct are denoted by subscripts 1, 2, 3 and 4 respectively. The Mach number
at the inlet of the duct is 2 and for this value of M the software for isentropic flow or the
isentropic tables for the flow of air give:
p0
 36.73
p1
Hence:
p1 
p0
300

 8.168 kPa
p0 / p1
36.73
Thus, at the inlet to the duct, the Mach number and pressure are 3 and 8.168 kPa
respectively. Now for this inlet Mach number Ml, the software for Fanno flow or the
tables for Fanno flow of air give:
4 f l1 *
 0.5220
D
p1 / p *  0.2182 ,
The procedure used to obtain the solution then involves the following steps:
(1) Guess the position of the shock wave in diameters down the duct, i.e., guess the
value of l1-2 / D .
(2) Use:
496
4 f l2*
4 f l1* 4 f l1 2
4 f l1 2


 0.5222 
D
D
D
D
to find the value of 4 f l2* / D.
(3) Use the software for Fanno flow or the Fanno flow tables for air to find the values
of M2 and p2 / p* corresponding to this value of 4 f l2* / D.
(4) Use:
p2 
p2 / p *
p / p*
p1  2
 8.168  37.43  p2 / p * kPa
p1 / p *
0.2182
to find the value of p2.
(5) Use the derived value of M2 with the software for normal shock waves or the
normal shock wave tables for air to get the values of p3 / p2 and M3 and find the
value of p3 using:
p3
p2
p2
p3 
(6) Use the software for Fanno flow or the Fanno flow tables for air to find the values
of p3 / p* and 4 f l3* / D corresponding to the derived value of M3.
(7) Use:
4 f l3* 4 f l3 4
4 f l3*
4 f l4*
l 




 4 f  60  1 2 
D
D
D
D
D 

to find the value of: 4 f l4* / D.
(8) Use the software for Fanno flow or the Fanno flow tables for air to find the values
of M4 and p4 / p* corresponding to this value of 4 f l4* / D.
(9) Use:
p4 
p4 / p *
p3
p3 / p *
to find the value of p4.
(10) Compare the value of p4 so derived with the required value of 50 kPa.
(11) Repeat the process with a new guessed value of the shock wave position and use
the results to deduce the shock wave position that gives p4 = 50 kPa.
497
Some results obtained using this procedure are given in the following table:
l1-2 / D
30
31
29
M4
0.7883
0.8007
0.7753
p4 - kPa
49.07
48.01
49.98
From these results, it will be seen that the shock wave occurs at approximately 29
diameters from the inlet to the duct and that the Mach number at the exit of the duct, M4 ,
is approximately equal to 0.775.
498
PROBLEM 9.30
Air flows at a steady rate through a 0.08m diameter 1.5 m long pipe. A convergentdivergent nozzle expands the air to a Mach number of 2.25 and a pressure of 40 kPa at
the inlet to the pipe. The air from the pipe is discharged into a large chamber in which the
pressure can be varied. Assuming a friction factor of 0.003, find the pressure in this
chamber if (1) there are no shock or expansion waves in the flow, and if (2) there is a
normal shock wave on the exit plane of the pipe.
SOLUTION
For the specified value of the pipe inlet Mach number, M1 , i.e. 2.25, the following are
obtained using the software for Fanno flow or the tables for Fanno flow of air:
p1 / p *  0.3432 ,
4 f l1 */ D  0.3738
Case 1 - No Waves
Conditions at the inlet and outlet of the duct will be denoted by subscripts 1 and 2
respectively. Hence:
4 f l2*
4 f l1* 4 f l1 2
4  0.003  1.5


 0.3738 
 0.1448
D
D
D
0.08
For this value of 4 f l2* / D the software for supersonic Fanno flow or the tables for
Fanno flow of air give:
M 2  1.535 ,
p2 / p *  0.5884
Hence:
p2 
p2 / p *
0.5884
p1 
 40  68.58 kPa
p1 / p *
0.3432
499
Because there are no waves in the flow, this will be the pressure in the discharge
chamber, i.e., in this case the back pressure is 68.58 kPa.
Case 2 - Normal Shock Wave at Exit
Conditions at the inlet of the duct, at the outlet of the duct just upstream of the normal
shock wave, and just downstream of the normal shock wave will be denoted by subscripts
1, 2 and, 3 respectively.
The conditions at the exit of the duct for Case 1 apply just upstream of the shock in
the present case, i.e.:
M 2  1.535 ,
p2  68.58 kPa
Consider the changes across the normal shock wave at the exit of the duct. The Mach
number ahead of the shock is M2 , i.e., 1.535 and for this Mach number the software for
normal shock waves or the tables for normal shock waves in air give:
p3
 2.582
p2
Hence:
p3 
p3
p2  2.582  68.58  177.1 kPa
p2
Because the flow downstream of the shockwave is subsonic, this will be equal to the
pressure in the discharge chamber, i.e., in this case the back pressure is 177.1 kPa.
500
PROBLEM 9.31
Air is expanded from a large reservoir in which the pressure and temperature are 300
kPa and 20°C respectively through a convergent-divergent nozzle which gives an exit
Mach number of 1.5. The air from the nozzle flows down a pipe having a diameter of 5
cm to a reservoir in which the pressure is 140 kPa. A normal shock wave occurs at the
end of the pipe. Find the length of the pipe. Discuss what will occur if the pressure in the
reservoir at the discharge end of the pipe is decreased. The flow in the nozzle can be
assumed to be isentropic and that in the pipe can be assumed to be adiabatic. The friction
factor in the pipe can be assumed to be 0.002.
SOLUTION
Conditions at the inlet to the pipe, upstream of the shock at the pipe exit, and
downstream of the shock at the pipe exit will be denoted by subscripts 1, 2 and 3
respectively.
Consider the flow in the nozzle. The Mach number at the exit of the nozzle is 1.5 and
for this value of M the software for isentropic flow or the isentropic tables for the flow of
air give:
p0
 3.671
p1
Hence:
p1 
p0
300

 81.72 kPa
p0 / p1
3.671
Thus, at the inlet to the duct, the Mach number and pressure are 3 and 81.72 kPa
respectively. For this inlet Mach number Ml, the software for Fanno flow or the tables for
Fanno flow of air give:
p1 / p *  0.6065 ,
4 f l1 *
 0.1361
D
501
A trial-and-error solution will be used. A series of pipe lengths will be guessed and
the outlet pressure downstream of the normal shock wave will be calculated. These
results will then be used to deduce the pipe length that gives an exit pressure that is equal
to the specified downstream reservoir pressure of 140 kPa. To calculate the exit pressure
for any assumed pipe length l1-2 it is recalled that:
4 f l2*
4 f l1* 4 f l1 2
4  0.002  l1 2


 0.1361 
 0.1361  0.16 l1 2
D
D
D
0.05
Using the value of 4 f l2* / D obtained by using the above equation, the software for
Fanno flow or the tables for Fanno flow of air can be used to obtain the values of M2 and
p2 / p*. The value of p2 can then be obtained using:
p2 
p2 / p *
p / p*
p1  2
 81.72  134.7   p2 / p * kPa
p1 / p *
0.6065
Lastly, consider the changes across the normal shock wave at the exit of the pipe. The
Mach number ahead of the shock is M2 and for this Mach number the software for normal
shock waves or the normal shock wave tables for air give the value of p3 / p2. The value
of p3 is then obtained using:
p3 
p3
p2
p2
Results obtained with various values of the assumed pipe length, l1-2 , are shown in
the following table:
l1-2 - m
0.30
0.40
0.70
0.85
0.83
0.84
4 f l2* / D
0.0881
0.0721
0.0241
0.0001
0.0033
0.0017
M2
1.367
1.321
1.164
1.009
1.055
1.039
p2 / p*
0.6835
0.7137
0.8344
0.9893
0.9396
0.9559
p2 - kPa
92.07
96.14
112.4
133.3
126.4
128.8
p3 / p2
2.013
1.869
1.414
1.021
1.132
1.093
p3 - kPa
185.3
179.7
158.9
136.1
143.1
140.7
Because the flow downstream of the shock wave is subsonic, the pressure
downstream of the shock, p3, must be equal to the back-pressure, i.e., equal to 140 kPa.
502
Interpolation between the above results indicates that this is the case when l1-2 is
approximately 0.841 m. Therefore the length of the pipe is approximately 0.841 m.
If the back-pressure is decreased below 140 kPa, the shock wave will move outside
the pipe and become an oblique shock wave attached to the exit edge of the pipe. As the
back-pressure is further reduced, the strength of this shock wave system is reduced. When
the back-pressure is reduced to the value of p2 there are no shock waves in the system and
the pressure on the exit plane of the pipe is equal to the back-pressure. For lower values
of the back-pressure there will be expansion waves in the discharge from the pipe.
503
PROBLEM 9.32
Air at an initial temperature of 45°C is to be transported through a 50 m long, wellinsulated pipe. If the mean friction factor in the pipe can be assumed to be 0.025, find the
minimum pipe diameter that can be used to carry the flow without choking occurring if
the inlet air velocity is 50 m/s, 100 m/s and 400 m/s.
SOLUTION
For each specified inlet velocity, the inlet Mach number, M1, can be calculated using:
M1 
V1

a1
V1

 RT1
V1
V1

357.5
1.4  287  318
For each value of M1 so obtained, the software for Fanno flow or the Fanno flow
tables for air can be used to get the value of:
4 f l1*
D
Because the flow is, by assumption, choked at the outlet, l1* is the length of the pipe,
i.e., 50 m. Hence:
4 fl1*
4  0.025  50
5


,
D
D
D
i.e. , D 
5
(4 fl1* / D)
m
This allows the value of the pipe diameter D corresponding to the specified inlet
velocity V1 to be found. Results for the three specified values of V1 are given in the
following table.
V1 – m/s
50
100
400
M1
0.1399
0.2797
1.1189
4f l1* / D
32.56
6.375
0.0136
504
D-m
0.1536
0.7843
367.7
Therefore the diameters required for inlet velocities of 50, 100, and 400 m/s if there is
choking at the pipe exit are 0.1536, 0.7843 and 367.7 m respectively.
505
PROBLEM 9.33
Air flows from a supersonic nozzle with a throat diameter of 6.5 cm into a 13 cm
diameter pipe. The stagnation pressure at the inlet to the pipe is 700 kPa. At distances of
5 diameters and 33 diameters from the inlet to the pipe the static pressures are measured
and found to be 24.5 kPa and 50 kPa respectively. Determine the Mach numbers at these
two sections and the mean friction factor between the two sections. The flow in the
nozzle can be assumed to be isentropic and the flow in the pipe can be assumed to be
adiabatic.
SOLUTION
Conditions at the inlet, at the first measurement point, and at the second measurement
point will be denoted by subscripts 1, 2 and 3 respectively.
Consider the flow in the nozzle. The area ratio of the nozzle is:
2
2
A1
D 
 13 
  1  
 4
A*
 D*
 6.5 
Now the software for isentropic flow or the isentropic tables for the flow of air give for
this value of A1 / A*:
M 1  2.94 ,
p0
 33.57
p1
Hence:
p1 
p0
700

 20.85 kPa
p0 / p1
33.57
Thus the values of the Mach number and pressure at the inlet to the duct have been
established.
Now for a Mach number Ml of 2.94, the software for Fanno flow or the tables for
Fanno flow of air give:
p1 / p *  0.2256
506
At the first measurement point, therefore, because p2 = 24.5 kPa:
p2
p p
24.5
 2 1 
 0.2256  0.2651
p*
p1 p *
20.85
For this value of p / p* the software for Fanno flow or the tables for Fanno flow of air
give:
4 f l2 *
 0.4682
D
Similarly at the second measurement point, because p3 = 50 kPa:
p3
p p
50
 3 1 
 0.2256  0.5410
p*
p1 p *
20.85
For this value of p / p* the software for Fanno flow or the tables for Fanno flow of air
give:
4 f l3 *
 0.1848
D
hence:
4 f l2  3
D

4 f l2* 4 f l3*

 0.4682  0.1848  0.2834
D
D
But:
l2  3
D

l3
l
 2  33  5  28
D D
Hence since:
4 f l2  3
D
 0.2834
It follows that:
4  f  28  0.2834 ,
i.e.,
f 
0.2834
 0.00253
4  28
Therefore the mean friction factor between the two measurement points is 0.00253.
507
PROBLEM 9.34
An apparatus that is used for determining friction factors with air flow through a pipe
consists of a reservoir connected to a convergent-divergent nozzle which in turn is
connected to the pipe. The nozzle has a throat diameter of 0.6 cm and the diameter of the
pipe, which is well insulated, is 0.9 cm. In one experiment, the pressure and temperature
in the reservoir are 1.7 MPa and 40°C respectively and the pressure at a distance of 0.15
m from the inlet to the pipe is 340 kPa. Calculate the average friction factor in the pipe.
Assume that the flow in the nozzle is isentropic.
SOLUTION
It will be assumed that the flow in the pipe is supersonic and adiabatic. Conditions at
the inlet and at the measurement point will be denoted by subscripts 1 and 2 respectively.
Consider the flow in the nozzle. The area ratio of the nozzle is:
2
2
A1
D 
 0.9 
 1  
  2.25
A*  D * 
 0.6 
The software for isentropic flow or the isentropic tables for the flow of air give for this
value of A1 / A*:
M 1  2.328 ,
p0
 13.07
p1
Hence:
p1 
p0
1700

 130.1 kPa
p0 / p1
13.07
Thus the values of the Mach number and pressure at the inlet to the pipe have been
established.
Now for a Mach number Ml of 2.328 the software for Fanno flow or the tables for
Fanno flow of air give:
508
p1
 0.3260 ,
p*
4 f l1 *
 0.3930
D
At the measurement point, because p2 = 340 kPa:
p2
p p
340
 2 1 
 0.3260  0.8522
p*
p1 p *
130.1
For this value of p / p* , the software for Fanno flow or the tables for Fanno flow of air
give:
4 f l2 *
 0.01918
D
But:
4 f l1 2
D

4 f l1* 4 f l2*

 0.3930  0.01918  0.3738
D
D
Hence since l1-2 = 0.15 m and D = 0.009 m the above equation gives:
4  f  0.15
 0.3738 , i.e.,
0.009
f 
Therefore the mean friction factor is 0.00561.
509
0.3738  0.009
 0.00561
4  0.15
PROBLEM 9.35
Air is expanded from a large reservoir in which the pressure and temperature are 600
kPa and 30oC respectively through a convergent-divergent nozzle which gives an exit
Mach number of 2.0. The air from the nozzle flows down a pipe having a diameter of 3.5
cm. (1) If the Mach number at the end of this pipe is 1.2, find the length of the pipe. (2) If
the back pressure is changed until a normal shock wave occurs half way between the
nozzle exit plane and the exit plane of the pipe, find the back pressure. The flow in the
nozzle can be assumed to be isentropic and the friction factor in the pipe can be assumed
to be 0.005.
SOLUTION
Adiabatic flow will be assumed in the pipe.
First consider the flow in the nozzle. The Mach number at the exit of the nozzle is 2
and for this value of M the software for isentropic flow or the isentropic tables for the
flow of air give:
p0
 7.824
p1
Hence:
p1 
p0
600

 76.69 kPa
p0 / p1
7.824
Thus, at the inlet to the duct, the Mach number and pressure are 2 and 76.69 kPa
respectively.
Flow Situation 1
For this inlet Mach number Ml of 2 the software for Fanno flow or the tables for
Fanno flow of air give:
p1 / p *  0.4082 ,
4 f l1 *
 0.3050
D
510
while for the exit Mach number of 1.2, the software for Fanno flow or the tables for
Fanno flow of air give:
4 f l2 *
 0.03364
D
From these results it follows that:
4 f l1 2
4 f l1* 4 f l2*


 0.3050  0.03364  0.2714
D
D
D
Hence:
l1 2 
0.2714 D
0.2714  0.035

 0.4750 m
4f
4  0.005
Therefore the length of the pipe is 0.4750 m.
Flow Situation 2
The conditions at the inlet, upstream of the shock, downstream of the shock and at the
outlet of the pipe will be denoted by subscripts 1, 2, 3 and 4 respectively.
Using:
4 f l2*
4 f l1* 4 f l1 2


D
D
D
it follows because:
l1 2 
0.4750
 0.2375 m
2
that:
4 f l2*
4 f l1* 4 f l1 2
4  0.005  0.2375


 0.3050 
 0.1693
D
D
D
0.035
For this value of 4 f l2* / D the software for supersonic Fanno flow or the tables for Fanno
flow of air can be used to obtain the following values of M2 and p2 / p*:
M 2  1.592 ,
p2 / p *  0.5608
511
Hence:
p2 
p2 / p *
0.5608
p1 
 76.69  105.4 kPa
p1 / p *
0.4082
Now consider the changes across the normal shock wave. The Mach number ahead of
the shock is M2 , i.e., 1.592 and for this Mach number the software for normal shock
waves or the normal shock wave tables for air give:
M 3  0.6709 ,
p3 / p2  2.790
Hence:
p3 
p3
p2  2.790  105.4  285.4 kPa
p2
Next consider the flow downstream of the normal shock wave. For Mach number M3
of 0.6709, the software for Fanno flow or the tables for Fanno flow of air give:
p3 / p *  1.564 ,
4 f l3 */ D  0.2708
But:
4 f l3*
4 f l3 4
4 f l4*


D
D
D
and:
l1 2 
0.4750
 0.2375 m
2
Therefore:
4 f l4*
4  0.005  0.2375
 0.2708 
 0.1351
D
0.035
For this value of 4 f l* / D, the software for Fanno flow gives or the tables for Fanno
flow of air give because the flow is subsonic downstream of the shock wave:
p4 / p *  1.397
512
Hence:
p4 
p2 / p *
1.3978
p3 
 285.4  254.9 kPa
p3 / p *
1.5642
Therefore in this situation, the pressure at the exit of the pipe is 254.9 kPa.
513
PROBLEM 9.36
Air enters a linearly converging duct with a circular cross-section at a Mach number
of 0.6. The inlet diameter of the duct is 10 cm and the wall makes an angle of 10° to the
axis of the duct. Using numerical integration, produce a plot of Mach number along the
duct up to the point where the Mach number reaches a value of 1. The mean friction
factor can be assumed to be equal to 0.005 and the flow can be assumed to be adiabatic.
SOLUTION
The situation being considered is shown in Fig. P9.36a. As shown in this figure, x is
the distance measured along the axis from the inlet and D is the diameter at any value of
x.
Figure P9.36a
If ΔD is the decrease in diameter between the inlet and any distance x down the duct,
i.e., if the diameter at distance x from the inlet is Din – ΔD, then:
D / 2
 tan (10o )
x
i.e.:
514
D  2 x tan (10o )  0.3527 x
Therefore, the variation of diameter with x is given by:
D  Din  0.3527 x  0.1  0.3527 x m
Now because the cross-sectional area A is equal to π D 2 / 4 , it follows that:
1 dA
1 d D2
2 dD
 2

A dx
D dx
D dx
But using the expression for the variation of D with x derived above gives:
dD
  0.3527
dx
Therefore:
1 dA
0.7053
 
A dx
D
Now since the variation of Mach number along the duct is described:
1  (  1) M 2 /2  d A 1  (  1) M 2 /2   M 2 4 f d x
dM
 
 A 
 2
M
1 M 2
1 M 2
DH




then because air flow is involved so γ = 1.4 the following applies:
1  0.2M 2  d A 1  0.2M 2 
dM
4f dx
 

0.7 M 2


2
2
M
D
 1 M  A
 1 M 
Substituting the derived expression for (1 / A ) ( d A / d x) and the given value of f into
this equation then gives:
1  0.2M 2  0.7053
1  0.2M 2 
4  0.005
d M  
M
d
x

0.7 M 3
dx



2
2
D
D
 1 M 
 1 M 
i.e.:
2
 M   1  0.2 M 
d M   0.7053  0.014 M 2    
 dx
2
 D  1 M

515
This equation can be solved using available software or by using a simple explicit finitedifference procedure, i.e., by writing the equation as:
M
 M   0.7053  0.014 M 2  
D
2
  1  0.2 M 
 x

2
 1 M

where ΔM is the finite change in M over a finite distance Δx along the duct axis. The
solution starts with the specified value of M at the inlet, i.e., 0.6. This procedure can be
easily implemented in EXCEL and this has been done here. The variation of M with x
given using this procedure is shown in Fig. P9.36b.
Figure P9.36b
516
PROBLEM 9.37
A high speed wind-tunnel is supplied with air from a collection of interconnected
compressed air tanks situated outside of the laboratory. The air is delivered from the
tanks to the tunnel through a long 100 cm diameter pipe. The pressure at the inlet to this
supply tank system is 10 MPa and the air is to be delivered to the tunnel at a pressure of 1
MPa. How long can this supply pipe be if choking is not to occur? Assume adiabatic
subsonic one-dimensional flow and a mean friction factor of 0.005.
SOLUTION
In a situation such as this, the flow in the pipe will be subsonic. If the maximum pipe
length is used, the flow will be choked at the exit of the pipe. In this case, in view of the
fact that the pressure at the exit of the pipe is 1 MPa, this means that for the flow in the
pipe:
p* = 1 MPa
Therefore, because the pressure at the inlet is 10 MPa, at the inlet:
p1
10

 10
p*
1
subscript 1 being used to denote the conditions on the inlet plane.
For this value of p / p* , the software for Fanno flow or the tables for Fanno flow of
air give:
4 f l1 *
 55.31
D
Hence:
l1 * 
55.31 D
55.31  0.1

 276.5 m
4f
4  0.005
Because the flow at the outlet of the pipe is choked, l1*, will be the length of the pipe.
Therefore, the maximum possible length that the pipe can have is 276.5 m.
517
PROBLEM 9.38
Air flows down an adiabatic constant area circular pipe which has a diameter of
1.5cm. At the inlet to the pipe the Mach number is 0.16. Determine the pipe lengths that
would give exit Mach numbers of 0.56 and 1. Assume that the friction factor, f , is 0.004.
If the stagnation conditions at the inlet to the pipe are maintained at the same values find
the percentage change in the flow rate through the pipe for the two cases considered if the
length of the pipe is increased by 1m.
SOLUTION
Conditions at the inlet and outlet of the duct will be denoted by subscripts 1 and 2
respectively. Using the software or the friction flow tables for air the following are
obtained:
At the inlet where M = 0.16:
1
 5.720 ,
*
V1
 0.175 ,
V*
4 fl1*
 24.198
D
At the outlet when M = 0.56:
2
 1.681 ,
*
V2
 0.595 ,
V*
4 fl2*
 0.674
D
At the outlet when M = 1
2
 1,
*
V2
 1,
V*
4 fl2*
 0
D
First consider the case where M2 = 0.56. Using the values given above:
4 fl1 2
4 fl1* 4 fl2*


 24.198  0.674  23.524
D
D
D
from which it follows that:
518
l1 2 
23.524 D
23.524  0.015

 22.1 m
4f
4  0.004
Hence the length of the pipe in this case is 22.1 m.
Next consider the case where M2 = 1. Using the values given above:
4 fl1 2
4 fl1* 4 fl2*


 24.198  0  24.198
D
D
D
from which it follows that:
l1 2 
24.198D
24.198  0.015

 22.69 m
4f
4  0.004
Hence the length of the pipe in this case is 22.69 m.
Next consider the effect of increasing the duct length by 1m on the mass flow rate, the
exit Mach numbers being the same. First consider the case where M2 = 0.56. In this case
using the values given above:
l1 2  22.1  1  23.1m
Therefore:
4 fl1*
4 fl1 2
4 fl2*
4  0.004  23.1



 0.674  25.314
D
D
D
0.015
For this value the software or the friction flow tables for air give:
1
 5.735 ,
*
V1
 0.171
V*
Therefore the since the mass flow rate ratio is given by:
m   VA 
 V
 *V * A
* V *
Because the same inlet Mach number, i.e., 0.16, is being considered for the shorter and
longer pipes ρ* and V* are the same in the two situations. Therefore in this case the mass
flow rate ratio is given by:
519
 1 /  *longer V1 / V *longer 5.735 0.171
m longer



 0.980
m shorter
 1 /  *shorter V1 / V *lshorter 5.720 0.175
In this case therefore the mass flow rate is reduced by 2%.
Next consider the case where M2 = 1. In this case using the values given above:
l1 2  22.69  1  23.69 m
Therefore:
4 fl1*
4 fl1 2
4 fl2*
4  0.004  23.269



 0  24.820
D
D
D
0.015
For this value the software or the friction flow tables for air give:
1
 5.820 ,
*
V1
 0.166
V*
Therefore the mass flow rate ratio is given by:
m longer

m shorter
 1 /  *longer V1 / V *longer
 1 /  *lshorter V1 / V *lshorter

5.820
0.171

 0.994
5.720
0.175
In this case therefore the mass flow rate is reduced by 0.6%.
520
PROBLEM 9.39
Air flows through the circular duct system shown in Fig. P9.39. Find the mass flow
rate through this system. Assume that the flow is adiabatic and therefore that Fanno flow
exists in the 20m long constant diameter section, the friction factor, f , being 0.008. Also
assume that the flows through the convergent sections at the inlet and exit of the duct are
isentropic. Use an iterative solution procedure in which the duct exit pressures with
various values of the discharge Mach number, M3 , are calculated and the value of M3 that
gives the specified exit plane pressure is deduced.
p = 25kPa
p = 125kPa
T = 25oC
Diameter = 2.5cm
Diameter = 2cm
20m
Figure P9.39
SOLUTION
It will be assumed that a circular duct system is involved. It will be noted that since
the flow from region 0 to section 1 and from section 2 to section 3 is assumed to be
isentropic and because the flow from section1 to section 2 is assumed to be adiabatic the
stagnation temperature is the same everywhere in the flow system, i.e., that T0 = 298K
throughout the system.
It will be assumed as an initial guess that the flow is choked at the exit from the
system, i.e., it will be assumed that M3 = 1.
Since the flow at section 3 is assumed to be choked it follows that A* = A3. Therefore:
 D22 / 4
D22
A2
0.0252



 1.563
0.022
 D32 / 4
A*
D32
521
For this value of A/A* the software or the isentropic tables for air give:
M 2  0.426 ,
p2
 2.52 ,
p*
p02
 1.122
p2
It is then noted that the software or the friction flow tables for air give for M2 = 0.426:
4 fl2*
 1.88 ,
D
4 fl1 2
4 fl1* 4 fl2*


,
D
D
D
i.e.,
p2
 2.49
p*
4 fl1*
4 fl1 2
4 fl2*
4  0.01  20



 1.88  37.08
D
D
D
0.025
For this value the software or the friction flow tables for air give:
M 1  0.1367 ,
p1
 7.997
p*
For this Mach number value the software or the isentropic tables for air give:
p0
 1.013
p1
Using the results obtained above then gives:
p1 
p0
125

 123.4 kPa
p0 / p1
1.013
and:
p2 
p2 p *
1
p1  2.49 
 123.4  36.06 kPa
p * p1
8.52
Therefore since the flow between sections 2 and 3 is isentropic so that p03 = p02 and
because M3= 1, the software or the isentropic tables for air give p3/p03 = 0.5283:
p3 
p3 p03 p02
p2  0.5283  1  1.126  36.06  21.45 kPa
p03 p02 p2
But for an exit Mach number of 1or less p3 must equal the back pressure, i.e., equal
25kPa. Therefore the guess that the Mach number is 1 on plane 3 is not correct.
522
As a second guess it will be assumed that M3 =0.9. For this Mach number the
software or the isentropic tables for air give:
A3
 1.009 ,
A*
p03
 1.691
p3
Hence:
 D22 / 4
D22
A2 A2 A3
0.0252
1.009
1.009






 1.009  1.577
0.022
 D32 / 4
A * A3 A *
D32
For this value the software or the isentropic tables for air give:
M 2  0.4042 ,
p02
 1.119
p2
It is then noted that the software or the friction flow tables for air give for M2 = 0.4042:
4 fl2*
 2.234 ,
D
4 fl1 2 4 fl1* 4 fl2*


,
D
D
D
i.e.,
p2
 2.667 ,
p*
p02
 1.577
p*
4 fl1*
4 fl1 2
4 fl2*
4  0.01  20



 2.234  34.234
D
D
D
0.025
For this value the software or the friction flow tables for air give:
M 1  0.1367 and
p1
 7.997
p*
For this Mach number value the software or the isentropic tables for air give:
p0
 1.013
p1
Using the results obtained above then gives:
p1 
p0
125

 123.4 kPa
p0 / p1
1.013
and:
p2 
p2 p *
1
p1  2.667 
 123.4  41.15 kPa
p * p1
7.997
523
and:
p3 
p3 p03 p02
1
p2 
 1  1.119  41.15  27.23 kPa
p03 p02 p2
1.691
But, as discussed above, for an exit Mach number of 1or less p3 must equal the back
pressure, i.e., equal 25kPa. Therefore the guess that the Mach number is 0.9 on plane 3 is
not correct. However comparing the exit plane pressures obtained by assuming exit plane
Mach numbers of 1 and 0.9 indicates that the correct exit plane Mach number must lie
between these two values. Therefore the above procedure was used to obtain the exit
plane pressure when an exit plane Mach number of 0.965 is assumed. The exit plane
pressure so obtained is 25.17kPa which is very close to the required value of 25kPa and it
will therefore be assumed that the correct exit plane Mach number is 0.965.
Now in the process of finding the exit plane pressure for an exit plane Mach number
is 0.965 it is found that M1 = 0.1369. For this Mach number the software or the isentropic
flow tables for air give:
p0
T
 1.013 and 0  1.004
p1
T1
Therefore:
p1 
p0
T0
125
298

 123.4 kPa and T1 

 296.8 K
p0 / p1 1.013
T0 / T1 1.004
Hence:
 p 
m  1 V1 A1   1  M 1
 RT1 


 RT1  D12 
4

 123.4


  0.1369 
 287  296.8 


1.4  287  296.8   0.0252 
4

 0.0336 kg/s
Therefore the mass flow rate through the system is 0.0336kg/s.
524
PROBLEM 9.40
Air flows through a variable area circular duct. The flow can be assumed to be
adiabatic. The Mach number, temperature and pressure at the inlet to the duct are 0.4,
450 K, and 550 kPa respectively. The duct has an increasing cross-sectional area
designed to ensure that despite the effects of friction the air temperature remains constant
along the duct. If the distance between the inlet and the outlet of the duct is equal to 125
times the inlet duct diameter and if the friction factor is assumed to be 0.005 find the
Mach number and pressure at the exit of the duct and the exit diameter to inlet diameter
ratio.
SOLUTION
If the subscripts 1 and 2 apply to conditions at the inlet and the exit of the duct then,
because the temperature remains constant in the flow, T2 = T1 and because the flow is
adiabatic T02 = T01.
Applying the energy equation to the flow through the duct and recalling that the flow
is adiabatic, i.e., that there is no heat transfer to or from the flow, gives:
cpT 
V2
 constant
2
Therefore since T remains constant this equation shows that V remains constant in the
flow. But:
V  Ma  M
 RT , i.e., M 
V
 RT
so since V and T remain constant this equation shows that M must also remain constant,
i.e., M2 = M1 = 0.4.
Now it was shown in the text that:
dA
2 f  M 2A

dx
D
525
i.e., since A = πD2/4 this equation can be written as:
dD
 fM2
dx
Integrating this equation between the inlet and the exit then gives:
D2  D1  f  M 2 L
where L is the length of the duct. The above equation can be written as:
D2
L
 1  fM2
,
D1
D1
D2
L
 1  fM2
D1
D1
i.e.,
Therefore since L/D1 =125:
D2
L
 1 fM 2
 1  0.02  1.4  0.42  125  1.56
D1
D1
To determine p2 it is noted that the continuity equation gives:
 p1 
 p2 
 V1 A1  
 V2 A2 ,
 RT1 
 RT2 
1 V1 A1   2 V2 A2 , i.e., 
p 
p 
i.e.,  1  V1 D12   2  V2 D22
 T1 
 T2 
Hence since the temperature and the velocity remain constant in the flow the above
equation gives:
2
p2 D  p1 D
2
2
2
1
,
i.e.,
2
D 
 1 
p2  p1  1   550  
  226.0 kPa
 1.56 
 D2 
Therefore Mach number and pressure at the exit of the duct are 0.4 and 226kPa
respectively and the exit diameter to inlet diameter ratio is 1.56.
526
PROBLEM 9.41
Air flows from a large chamber in which the pressure and temperature are maintained
at 725 kPa and 725 K respectively through a convergent-divergent nozzle and then into a
constant diameter insulated pipe which in turn discharges into another large chamber in
which the pressure is kept constant. The ratio of the nozzle exit area to the throat area of
the nozzle is 2.4. The pipe has a diameter of 1.25 cm and a length of 30 cm. A friction
factor of 0.005 can be assumed for the pipe flow. Find:
1. The mass flow rate through the system if the pressure in the reservoir into which
the flow is discharged is kept at 0kPa.
2. The pressure in the reservoir into which the flow is discharged if there is a normal
shock wave on the nozzle exit plane.
3. The maximum pressure in the reservoir into which the flow is discharged at which
choked flow will exist in the nozzle-pipe system.
SOLUTION
It will be assumed that the flow in the nozzle is isentropic and that the flow in the
pipe is adiabatic. Therefore the stagnation temperature, T0, has the same value throughout
the flow.
Using the software or the isentropic flow tables for air gives the discharge Mach
number from a nozzle with an exit area to the throat area (i.e., A/A*) of 2.4 as 2.3986.
This then is the Mach number at the inlet to the pipe. For this Mach number the software
or the friction flow tables for air give:
4f l*
 0.4096
D
This equation gives:
l *  0.4096
D
4f
 0.4096 
527
0.0125
 0.256 m
4  0.005
This is less than the length of the pipe which is 0.3m. It follows that there must be a
shock wave in the pipe.
Part 1.
The mass flow rate will therefore determined by the conditions at the nozzle throat where
the Mach number is 1. Hence:
 p
m   T VT AT   T
 R TT

 AT
 M T aT 

 AE

 pT
 AE  

 R TT

 AT   2
 M T aT 
 DT

 AE  4
where the subscripts T and E refer to conditions at the throat and at the exit of the nozzle.
Now the Mach number at the throat is 1 so the software or the isentropic flow tables for
air give:
pT
T
 0.5287 and T  0.8333
p0
T0
Therefore the above equation gives:
p / p p 
 A 
m   T 0 0  M T aT  T  DE2
 AE  4
 R TT / T0  T0 
 0.5286  725000 

1
 
 1  1.4  287  0.833  725 

 0.01252

2.4 4
 287  0.8333  725 
 0.05567 kg/s
Hence the mass flow rate is 0.05567kg/s.
Part 2.
The Mach number ahead of the normal shock wave at the exit of the nozzle will be
the design Mach number which was shown above to be equal to 2.3986, i.e., M2 =
2.3986, the subscript 2 denoting conditions just upstream of the normal shock wave. At
this Mach number the software or the isentropic flow tables for air give:
528
p2
 0.06855
p0
where, since the nozzle flow is assumed to isentropic, p0 = 725kPa.
Next consider the changes across the shock wave. Using the value of the upstream
Mach number, i.e., M2 = 2.3986, the software or the normal shock tables for air give:
p3
 6.5455
p2
M 3  0.5233 ,
subscript 3 denoting conditions downstream of the shock wave.
The flow downstream of the shock wave in the pipe will next be considered. Using
the known inlet Mach number of 0.5233, the software of the friction flow tables for air
give:
4 fl3*
 0.8945
D
p3
 1.7888 ,
p0
Now, if the subscript 4 is used to denote conditions on the exit plane of the pipe, then:
4 fl3 4
4 fl3* 4 fl4*


,
D
D
D
i.e.,
4 fl3* 4 fl3 4
4 fl4*
4  0.005  0.3


 0.8945 
 0.4145
D
D
D
0.0125
The software or the friction flow tables for air give for this value of 4f l* / D:
M 4  0.6208 ,
p4
 1.7003
p*
Therefore:
p4 
p4 p3 p2
p p * p3 p2
p0  4
p0
p3 p2 p0
p * p3 p2 p0
 1.7003 
1
 6.5455  0.06855  725  309.2 kPa
1.7888
Because the Mach number at the end of the pipe is subsonic the pressure on the nozzle
discharge plane will be equal to the back pressure, pb. Therefore:
529
pb  309.2 kPa
Part 3.
Choking will occur at the throat of the nozzle. Therefore choking will first occur
when the Mach number reaches a value of 1 at the throat and the Mach number then
decreases to subsonic values in the divergent section of the nozzle. Using the software or
the isentropic flow tables for air give the discharge Mach number from a nozzle with an
exit area to the throat area (i.e., A/A*) of 2.4 when there is subsonic flow in the divergent
section of the nozzle as 0.2503. For this Mach number the software or the isentropic flow
tables for air give on the exit plane:
p2
 0.9574
p0
The Mach number at the inlet to the pipe is therefore 0.2503. For this Mach number the
software or the friction flow tables for air give:
4 f l2 *
 8.4581 ,
D
p2
 4.3494
p*
Therefore, as before:
4 fl23
4 fl2* 4 fl3*
,


D
D
D
i.e.,
4 fl3*
4 fl2* 4 fl3 4
4  0.005  0.3


 8.4506 
 7.9781
D
D
D
0.0125
The software or the friction flow tables for air give for this value of 4f l* / D:
M 3  0.5382 ,
p3
 1.9790
p*
Therefore:
p3 
p3 p2 p2
p p * p2
p0  3
p0
p2 p3 p0
p * p2 p0
 1.9790 
1
 0.9574  725  315.8 kPa
4.3494
530
Since the flow is subsonic this will be equal to the back pressure. From this it follows that
if the back pressure is lower than 315.8 kPa then a region of supersonic flow terminated
by a normal shock wave will develop in the flow system.
Therefore for the situation considered in Part 1 the mass flow rate through the system
is 0.05567 kg/s, for the situation considered in Part the back pressure is 309.2kPa, while
for the situation considered in Part 3 if the back pressure is lower than 315.8 kPa then a
region of supersonic flow terminated by a normal shock wave will develop in the flow
system
531
PROBLEM 9.42
Air flows from a large chamber in which the pressure and temperature are maintained
at 100 psia and 500 °R respectively through a convergent-divergent nozzle and then into
a constant diameter insulated pipe which in turn discharges into another large chamber in
which the pressure is kept constant. The ratio of the nozzle exit area to the throat area of
the nozzle is 3. The pipe has a diameter of 1ft and a length of that is equal to 15 times its
diameter.
A friction factor of 0.005 can be assumed for the pipe flow. The pipe
discharges into another large chamber in which the pressure, the back-pressure, is kept
constant. A normal shock wave occurs in the pipe at a distance equal to 3 times the pipe
diameter downstream of the nozzle exit. Assuming isentropic flow in the nozzle and
Fanno flow in the pipe determine the back-pressure. Sketch the process that occurs in
this flow system on a Fanno line. Also sketch the pressure variation along the length of
the flow system.
SOLUTION
The subscripts 1 and 2 will be used to denote conditions at the nozzle throat and at the
nozzle exit respectively while the subscripts 3 and 4 will be used to denote conditions
upstream and downstream of the shock wave. The subscript e will be used to denote
conditions at the exit of the pipe.
The flow in the nozzle is assumed to be isentropic so the stagnation pressure is
everywhere the same in the nozzle. Since M1 = 1, software or isentropic tables for air
flow give:
p1
T
 0.5283 and 1  0.8333
p0
T0
Therefore since p0 = 100 psia and T0 = 500 oR it follows that:
p1 
p1
T
p0  0.5283  100  52.83psia , T1  1 T0  0.8333  500  416.6 o R
p0
T0
532
Next consider conditions on the nozzle exit plane. Because A2/A1 = A2/A* = 3 the
software or isentropic tables for air flow give:
p2
T
 0.04730 and 2  0.4182
p0
T0
Therefore:
p2 
p2
T
p0  0.04730  100  4.73psia , T2  2 T0  0.4182  500  209.1 o R
p0
T0
Next consider the flow between the nozzle exit and the normal shock wave. The
subscript 3 is used to denote conditions at the upstream face of the shock wave. Since the
Mach number at the nozzle exit, M2, is equal to 2.6374, the software of the friction flow
tables for air give:
4 f l2 *
 0.4560 ,
D
p2
 0.2686 ,
p*
T2
 0.5019
T*
Now:
4 fl23
4 fl2* 4 fl3*


, i.e.,
D
D
D
4 fl3*
4 fl2* 4 fl23
4  0.005  3D


 0.4599 
 0.3999
D
D
D
D
the fact that the shock wave occurs at a distance of 3 diameters downstream of the nozzle
exit plane having been used.
For the above calculated value of 4 f l3*/D the software or the friction flow table for
air give:
M 3  2.3569 ,
p3
 0.3199 ,
p*
T3
 0.5684
T*
Therefore:
p3 
T3 
p3 p *
1
 4.73  5.633psia
p2  0.3199 
p * p2
0.2686
T3 T *
1
 209.1  236.9 o R
T2  0.5684 
T * T2
0.5019
533
Using the software or the normal shock wave tables for air gives, for an upstream
Mach number of 2.3569:
M 4  0.5278 ,
p4
T
 6.3144 , 4  0.4182
p3
T3
Therefore:
p4 
p4
T
p3  6.3144  5.6331  35.570 psia , T4  4 T3  1.9996  236.85  473.61 o R
p3
T3
The software or the friction flow tables for air give for M4 = 0.5278:
4 f l4 *
 0.8641 ,
D
p4
T4
 2.020 ,
 1.1367
p*
T*
At the exit of the pipe:
4 fl4 e 4 fl4* 4 fle*
,


D
D
D
i.e.,
4 fle* 4 fl4* 4 fl4 e
4  0.005  (12  3) D


 0.8641 
 0.6841
D
D
D
D
For the above calculated value of 4 f l4*/D the software or the friction flow table for air
give:
M e  0.5580 ,
pe
Te
 1.9047 ,
 1.1297
p*
T*
Therefore:
pe 
Te 
pe p *
1
 35.570  33.54 psia
p4  1.9047 
p * p4
2.020
Te T *
1
 473.61  480.4 o R
T4  1.1297 
T * T4
1.1367
Because the Mach number at the exit of the pipe is less than 1 the back pressure will
be the same as the exit plane pressure. Hence the back pressure is 33.54psia.
The process that occurs in the flow system is shown on the Fanno line in Fig. P9.42a
while the form of the pressure variation along the flow system is shown in Fig. P9.42b.
534
4
2
T
e
M=1
3
2
2
s2
Figure P9.42a
60
Pressure - psia
1
0
4
3
30
0
2
0
5
3
3
0
Distance along duct –nozzle system
Figure P9.42b
535
Chapter Ten
FLOW WITH HEAT TRANSFER
SUMMARY OF MAJOR EQUATIONS
Heat Transfer in External Flows
Twad
 1 2
 1  r
M
2
T
For Laminar Flow:
r  Pr1/2
For Turbulent Flow:
r  Pr1/3
Q  h A ( Tw  Twad )
Tprop  T f  0.5 (TW  T f )  0.22 (TWad  T f )
(10.7)
(10.9)
(10.16)
One-Dimensional Flow in a Constant Area Duct Neglecting Viscous Effects
p2
1   M 12

1   M 22
p1
(10.32)
p02
[1   M 12 ] [1  (  1) M 22 /2]  /1

p01
[1   M 22 ] [1  (  1) M 12 /2]  /1
(10.34)
T2
M 22 (1   M 12 ) 2

T1
M 12 (1   M 22 ) 2
(10.39)
536
V2
M 22 (1   M 12 )

V1
M 12 (1   M 22 )
(10.40)
 2 M 12 (1   M 22 )

1 M 22 (1   M 12 )
(10.41)
T02
M 22 (1   M 12 ) 2 [1  (  1) M 22 /2]

T01
M 12 (1   M 22 ) 2 [1  (  1) M 12 /2]
(10.42)
T0max
(1   M 12 ) 2
1

T01
2(1   ) M 12 [1  (  1) M 12 /2]
(10.43)
( 1)
 2

2



s2  s1
M
1

M

2
1


 ln
 M 12  1   M 22 

cp


(10.44)
One-Dimensional Flow in a Constant Area Duct with Variables Expressed Relative
to those at M = 1
T0
2(  1) M 2 [1  (  1) M 2 /2]

T0*
(1   M 2 ) 2
(10.47)
T
(1   ) 2 M 2

T*
(1   M 2 ) 2
(10.48)
p
(1   )

*
p
(1   M 2 )
(10.49)

p0
 1     2  
  1) 2   1
1
M 




2  
p0*
2

 1   M     1  
537
(10.50)
V
(1   ) M 2

V*
(1   M 2 )
(10.51)

(1   M 2 )

*
(1   M 2
(10.52)
1





s  s*
1


2


 ln M 
2 


cp
1

M

 


(10.53)
Rayleigh Line
2
(1   ) 2  4 (T /T * ) 
s  s*
 T     1   (1   )

ln
 ln  *   

 
cp
2
 T      2

(10.56)
Temperature Maximum
1
M 

Tmax
(1   ) 2

T*
4
(10.57)
(10.58)
Variable Area Flow
dA
( M 2  1)
dq
( M 2  1)
dM



0
2
2
A
[1  (  1) M /2] c pT [1  (  1) M /2] M
(10.67)
M = 1 corresponds to the point at which:

dA
(  1)
dq

0
[1  (  1)/2] c pT
A
538
(10.68)
Constant Area Flow with Heat Transfer and Friction
DH 
4 (Area)
4A

P
Perimeter
 1  M 2 
dq
 M 2 f dx



2
2
DH
 2  c pT [1  (  1) M /2]

 dM
(1   M 2 )
(  1) M 2
 (1   M 2 ) 

2
2
[1  (  1) M /2]  M

(10.74)
(10.81)
Isothermal Constant Area Flow
1   M 2 
4 fl *
2


ln[

M
]
 M2 
DH



p
V*
M*




*
p*
V
M
1
M
(10.105)
(10.106)
T0
2 
 1 2 

M 
1 
*
T0
3  1 
2

Combustion Waves
m 2 
V1 
1
1
p2  p1
(1/1 )  (1/ 2 )
p2  p1
(1/1 )  (1/ 2 )
,
539
V2 
(10.109)
1
2
p2  p1
(1/1 )  (1/ 2 )
(10.110)
1
1
V22  V12  ( p1  p2 ) 

1 
 2
 p2   1 
 2    p2  1   1q
1     1  
 1      
   1    p1   2   p1
 p1    2 
(10.112)
(10.114)
 p 1/    
s2  s1
 ln  2   1  
cp
 p1    2  
(10.115)
 1 
 ( p2 /p1 )
  
(  1)( p2 /p1 )  1
 2 
(10.119)
p  
 1  p /p  1
1
M 12    2 1
   (  1)  2   1
   1  1 / 2
  
 p1  
Strong Detonation Waves
1


 1
2
(10.121)
p2
 (  1)( 1q /p1 )
p1
(10.122)
V12 
p2 /1
1  1 / 2
540
PROBLEM 10.1
A rocket ascends vertically through the atmosphere with a velocity that can be
assumed to increase linearly with altitude from zero at sea-level to 1800m/s at an altitude
of 30,000m. If the surface of this rocket is assumed to be adiabatic, estimate the variation
of the skin temperature with altitude at a point on the surface of the rocket a distance of
3m from the nose of the rocket. Use the flat plate equations given in this chapter and
assume that at the distance from the nose considered, the Mach number and temperature
outside the boundary layer are the same as those in the freestream ahead of the rocket.
SOLUTION
It will be assumed that the boundary layer over the surface of the rocket is turbulent.
Therefore assuming that the Prandtl number, Pr, is equal to 0.7 at all altitudes considered,
the recovery factor is:
r  Pr1/ 3  0.71/ 3  0.89
The surface temperature is then given by:
Twad
  1  2
 1.4  1  2
2
 1 r
 M  1  0.89  
 M  1  0.178 M
Tamb
 2 
 2 
where Tamb is the ambient temperature.
If V is the velocity of the rocket then the Mach number is given by:
M 
V

a
V

 RTamb
V

1.4  287  Tamb
V
0.0499 V

1.4  287  Tamb
Tamb
Since the velocity varies linearly with altitude, h, it follows that:
V 
h
 1800  0.06 h m/s
30000
541
Therefore for any selected altitude the properties of the standard atmosphere give the
value of Tamb and the above equation gives V. The Mach number can then be calculated
and used to determine Twad. Some typical results are shown in the following table:
Table P10.1
h-m
0
5000
10,000
15,000
20,000
25,000
30,000
Tamb - K
288.16
255.69
223.26
216.66
216.66
216.66
231.24
V – m/s
0
300
600
900
1200
1500
1800
M
0
0.936192
2.003763
3.051081
4.068108
5.085134
5.906655
Twad - K
288.16
295.58
382.8198
575.6696
854.8994
1213.909
1667.279
Because of the increasing velocity, and therefore the increasing Mach number, with
altitude the surface temperature will be seen to increase from ambient, i.e., 288.16K, at h
=0 to 1667.28K at h = 30,000m.
542
PROBLEM 10.2
A flat plate, which can be assumed to be adiabatic, is placed in a wind tunnel, the
plate being aligned with the flow. The test section Mach number is 3 and the static
temperature is 15oC. During the start-up of the tunnel, a normal shock wave occurs in the
test section half-way along the plate. The boundary layer on the plate is turbulent both
before and after the shock wave. What are the plate temperatures before and after the
shock wave under these circumstances? Discuss the result obtained, considering the
stagnation temperatures before and after the shock wave.
SOLUTION
Since the boundary layer flow is turbulent and assuming that for air the Prandtl
number, Pr, is equal to 0.7 it follows that the recovery factor is given by:
r  Pr1/ 3  0.71/ 3  0.89
In the situation being considered it therefore follows that, if γ for air is assumed to be 1.4,
the surface temperature is given by:
Twad
  1  2
 1.4  1  2
2
 1  r
 M  1  0.89  
 M  1  0.178 M
2
2
T




Upstream of the shock wave T∞ = 15oC = 288K so the above equation gives for this
region:
Twad  T 1  0.178 M 2   288 1  0.178  32   749 K
For a normal shock with an upstream Mach number of 3 the software or the normal
shock wave tables for air give:
T2
 2.679 and M 2  0.475
T1
543
Hence downstream of the shock wave the freestream Mach number is 0.475 and the
freestream temperature is 2.679 x 288 = 772K. Therefore using the equation derived
above for the wall temperature gives:
Twad  T 1  0.178 M 2   772 1  0.178  0.4752   803 K
Therefore the wall temperatures upstream and downstream of the shock wave are
749K and 803K respectively. To understand this result it is necessary to recall that the
flow through the shock wave is adiabatic and that the stagnation temperature, T0, is
therefore not changed by the shock wave. Upstream of the shock wave the temperature
rise from 288K to Twad occurs entirely across the boundary layer and the process is not
adiabatic. However downstream of the shock wave part of the temperature rise from
288K to Twad occurs across across the shock wave (an adiabatic process) and part of the
temperature rise occurs across the boundary layer ( a non-adiabatic process) . Hence the
wall temperature downstream of the shock wave will be closer to T0 than the wall
temperature upstream of the shock wave. The stagnation temperature throughout the
freestream flow is, it should be noted, given by:
T0
  1  2
 1.4  1  2
2
 1 
M  1  
 M  1  0.2 M
T
 2 
 2 
i.e.:
T0  T 1  0.2 M 2   288 1  0.2  32   806 K
544
PROBLEM 10.3
Air, at a Mach number of 3 and a temperature of -30oC flows over a flat plate that is
aligned with the flow. The plate is kept at a temperature of 25oC. Find the mean heat
transfer rate from the plate surface per unit area.
SOLUTION
The adiabatic wall temperature will first be determined. It will be assumed that the
boundary layer flow is turbulent. Therefore assuming that the Prandtl number, Pr, is equal
to 0.7 it follows that recovery factor is:
r  Pr1/3  0.71/3  0.89
Therefore assuming that for air γ = 1.4 it follows that:
Twad
  1  2
 1.4  1 
2
 1  r
 M  1  0.89  
  3  2.602
T
2
2




Hence:
Twad  2.602T  2.602  243  632 K
Since the surface of the plate is being maintained at a temperature of 298K, the mean
heat transfer rate from the plate to the air per unit surface area is given by:
q  h Tw  Twad   1000   298 - 632    334290 W/m 2
The negative sign means that heat is being transferred fro the air to the plate. Hence
the heat transfer rate to the plate – 334.29kW/m2.
545
PROBLEM 10.4
Air is expanded through a convergent-divergent nozzle from a Mach number of 0.2 to
a Mach number of 2. The overall nozzle length is 0.7m and the throat of the nozzle is at a
distance of 0.25m from the inlet. The convergent and divergent sections are of such a
shape that the cross-sectional area of the nozzle varies linearly with distance in both
sections. The initial temperature and pressure are 1000kPa and 800K respectively. It can
be assumed that the boundary layer on the walls of the nozzle is turbulent and that it
effectively originates at the entrance section of the nozzle and it can be assumed that the
properties of this boundary layer are described by the flat plate equations. If the walls of
the nozzle are effectively adiabatic, estimate the variation of wall temperature with
distance along the nozzle. Assume that the variation of the properties of the freestream
flow outside the boundary layer can be obtained by assuming one-dimensional isentropic
flow and the effect of the boundary layer on the effective flow area can be ignored.
(HINT: Calculate the Mach number and temperature variation in the freestream with
distance along the nozzle. Then use the definition of the recovery factor applied at each
section to find the adiabatic wall temperature.)
SOLUTION
The Mach number at the throat of the nozzle will of course be 1 since the flow goes
from subsonic flow to supersonic flow in the nozzle. The area of the throat of the nozzle
is therefore A*. Now because the Mach numbers at the inlet and exit of the nozzle are 0.2
and 2 respectively the software or the table for the isentropic flow of air give:
Ai
 2.963 and
A*
Ae
 1.688
A*
Where the subscripts i and e refer to conditions on the inlet and exit sections of the
nozzle. At the throat:
At
 1
A*
546
Now the cross-sectional area of the nozzle varies linearly with distance in both the
convergent and divergent sections of the nozzle. Hence since the length of the convergent
section is 0.25m and the length of the divergent section is 0.45m it follows that if x is the
distance measured along the nozzle from the inlet section the cross-sectional area
variations in the convergent and divergent sections are given by:
Convergent Section:
 x 
A  Ai  
  Ai  At 
 0.25 
that is:
A 
A
A
A
 x   Ai
 x   Ai

 x 
 i 
 t   i 
 1  2.963  


1.963
A*
A *  0.25   A * A * 
A *  0.25   A *

 0.25 
Divergent Section:
 x  0.25 
A  A*  
  Ae  A*
 0.7  0.25 
that is:
A


 x  0.25   Ae
 x  0.25   Ae
 x  0.25 
1 
 1  1  
 1  1  


 0.688
A*
0.45
A
*
0.45
A
*




 0.45 


These above two equations give the variations of A / A* with distance along the
nozzle. Using these values of A / A* the software or the tables for the isentropic flow of
air can be used to find the variations of M and T0/T with distance along the nozzle. The
using:
T 
T / T0
T / T0
T
Ti 
Ti 
 800
Ti
Ti / T0
Ti / T0
Using this equation the variation of the temperature of the air outside the boundary layer
with distance along the nozzle can be found. It is then noted that because air flow is
involved and because the boundary layer flow is assumed to be turbulent:
Twad  T 1  0.178 M 2 
547
This equation then allows the variation of the wall temperature with distance along the
nozzle to be found. Some typical results are shown in the following table.
Table P10.4
x-m
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7.
M
0.2
0.23
0.28
0.35
0.47
1
1.32
1.46
1.58
1.67
1.75
1.82
1.89
1.94
2
T-K
800
797.7
794.3
787.1
772.3
672
598
564.9
539
518.8
501.2
485.3
470.4
459.7
448
548
Twad - K
805.7
805.4
805.1
804.3
802.6
791.6
783.5
779.8
777
774.8
772.8
771.1
769.4
768.3
767
PROBLEM 10.5
If instead of being adiabatic, the wall of the nozzle described in Problem 10.4 is kept
at a uniform temperature of 200oC, estimate the variation of the local wall heat transfer
rate , q , per unit wall area with distance along the nozzle assuming that the value of the
heat transfer coefficient is 1000 W/m2K.
SOLUTION
Since the surface of the plate is being maintained at a temperature of 473K, the mean
heat transfer rate from the plate to the air per unit surface area is given by:
q  h Tw  Twad   1000   473 - Twad  W/m 2
The variation of the adiabatic wall temperature with distance along the nozzle was
derived in the solution to Problem 10.4. Using these values the local heat transfer rate per
unit wall area can be found using the above equation. Some typical values are shown in
the following table.
Table P10.5
x-m
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7.
M
0.2
0.23
0.28
0.35
0.47
1
1.32
1.46
1.58
1.67
1.75
1.82
1.89
1.94
2
Twad - K
805.7
805.4
805.1
804.3
802.6
791.6
783.5
779.8
777
774.8
772.8
771.1
769.4
768.3
767
549
q – W/m2
-332700
-332440
-332070
-331280
-329650
-318620
-310480
-306840
-303990
-301760
-299800
-298080
-296440
-295260
-294000
The minus sign on the heat transfer rate values again indicates that heat is being
transferred to the wall from the air.
550
PROBLEM 10.6
A flat plate with a length of 0.8m and a width of 1.2m is placed in the working
section of a wind-tunnel in which the Mach number is 4, the temperature is - 70oC and
the pressure is 3kPa. If the surface temperature of the plate is kept at 30oC by an internal
cooling system, find the rate at which heat must be added to or removed from the plate.
Consider both the top and the bottom of the plate and assume a mean heat transfer
coefficient of 1000 W/m2K.
SOLUTION
The adiabatic wall temperature will first be determined. It will be assumed that the
boundary layer flow is turbulent. Therefore assuming that the Prandtl number, Pr, is equal
to 0.7 it follows that recovery factor is:
r  Pr1/3  0.71/3  0.89
Therefore assuming that for air γ = 1.4 it follows that:
Twad
  1  2
 1.4  1 
2
 1  r
 M  1  0.89  
  4  3.848
T
2
2




Hence:
Twad  3.848T  3.848  203  781.1 K
Since the surface of the plate is being maintained at a temperature of 303K, the mean
heat transfer rate from the plate to the air from both sides of the plate is given by the
following equation, A being the surface area of one side of the plat:
Q  h 2 A Tw  Twad   1000  2   0.8  1.2  303 - 781.1   917950 W
The negative sign means that heat is being transferred from the air to the plate. Hence
the heat transfer rate to the plate – 917.95 kW.
551
PROBLEM 10.7
Air flows though a constant-area duct. The air has a temperature of 20°C and a Mach
number of 0.5 at the entrance to the duct. It is desired to transfer heat to the duct such that
at the exit of the duct the stagnation temperature is 1180°C. Is this possible? If not, what
adjustment must be made to the Mach number at the entrance in order to give a discharge
stagnation temperature of 1180°C? Ignore the effects of friction.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2. Now
for a Mach number of 0.5, the software for isentropic flow or the tables for isentropic
flow of air give:
T01
 1.050
T1
hence:
T01  1.050  (20  273)  307.7 K
The software for Rayleigh flow or the Tables for Rayleigh flow of air give for M1 =
0.5:
T01
 0.6914
T0 *
Using this value gives:
T02
T T
1453
 02 01 
 0.6914  3.265
T0 *
T01 T0 *
307.7
The highest value that T02 / T0* can have is 1 so this flow is not possible. The highest
value that M1 can have is that which gives M2 equal to 1, i.e., which makes T02 = T0*. If
this is the case, T0* = 1453 K and so:
T01
293

 0.2017
T0 *
1453
552
The software for Rayleigh flow or the Tables for Rayleigh flow of air give for this
value of T01 / T0*:
M 1 = 0.2175
Therefore for the given inlet and outlet temperatures, the inlet Mach number must not
exceed 0.2175.
553
PROBLEM 10.8
Air with a stagnation temperature of 430oC and a stagnation pressure of 1.6 MPa
enters a constant-area duct in which heat is transferred to the air. The Mach number at the
inlet to the duct is 3 and the Mach number at the exit to the duct is 1. Determine the
stagnation temperature and the stagnation pressure. Ignore the effects of friction.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2. The
software for Rayleigh flow or the Tables for Rayleigh flow of air give for M1 = 3:
T01
 0.6540 ,
T0 *
p01
 3.424
p0 *
At the exit because M2. = 1:
T02  T0 * ,
p02  p0 *
hence:
T02  T0* 
T01
703

 1074.9 K
*
T01 /T0
0.6540
and:
p02  p0* 
p01
1600

 467.3 kPa
*
p01 /p0
3.424
Therefore the exit stagnation temperature and pressure are 1074.9 K ( = 801.9°C ) and
467.3 kPa respectively.
554
PROBLEM 10.9
Air at a temperature of 100°C with a pressure of 101 kPa enters a constant-area
combustion chamber at a velocity of 130 m/s. Determine the maximum amount of heat
that can be transferred to the air flow per unit mass of air. Assume friction effects are
negligible.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2
respectively.
At the inlet:
M1 
V1

a1
V1

 RT1
130
130

 0.3358
387.1
1.4  287  373
Now for a Mach number of 0.5 the software for isentropic flow or the tables for
isentropic flow of air give:
T01
 1.023
T1
hence:
T01  1.023  373  381.6 K
Now the software for Rayleigh flow or the Tables for Rayleigh flow of air give for M1
= 0.3358:
T01
 0.4128
T0 *
If the maximum amount of heat is transferred the flow is choked at the exit, i.e., M2 =
1 and hence:
T02  T0 *
and so:
555
T02  T0* 
T01
381.6

 924.4 K
*
T01 /T0
0.4128
The amount of heat transferred per unit mass of air is therefore given by:
q  cp (T02 - T01 ) = 1.007  (924.4 - 381.6) = 546.6 kJ / kg
Therefore the maximum amount of heat that can be transferred is 546.6 kJ/kg.
556
PROBLEM 10.10
Heat is supplied to air flowing through a short tube causing the Mach number to
increase from an initial value of 0.3 to a final value of 0.6. If the initial air temperature is
40° C and if the effects of friction are neglected, find the heat supplied per unit mass.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2.
At the inlet where the Mach number, M1 , is 0.3, the software for isentropic flow or
the tables for isentropic flow of air give:
T01
 1.018
T1
hence:
T01  1.018  313  318.6 K
Now the software for Rayleigh flow or the Tables for Rayleigh flow of air give
for M1 = 0.3:
T01
 0.3469
T0 *
At the exit where the Mach number, M2, is 0.6, the software for Rayleigh flow or the
Tables for Rayleigh flow of air give for this value of Mach number:
T02
 0.9167
T0 *
Hence:
T02 
T02 /T0*
0.9167
T 
 318.6  841.9 K
* 01
T01 /T0
0.3469
The amount of heat transferred per unit mass of air is therefore given by:
557
q  cp (T02 - T01 ) = 1.007  (841.9 - 318.6) = 527 kJ / kg
Therefore the amount of heat transferred per unit mass of air is 527 kJ/kg.
558
PROBLEM 10.11
Air flows through a 10 cm diameter pipe at a rate of 0.18 kg/s. If the air enters the
pipe at a temperature of 0oC and a pressure of 60 kPa, how much heat can be added to the
air per kg without choking the flow? The effects of friction are negligible.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2.
Because:
m  1 V1 A , i.e., V1 
m
1 A
and because:
1 
p1
60000

 0.7658 kg/m3
RT1
287  273
and:
A =

4
D2 =

4
0.12 = 0.007854 m 2
it follows that:
V1 
m
0.18

 29.93 m/s
1 A
0.7658  0.007854
At the inlet therefore:
M1 
V1

a1
V1

 RT1
29.93
29.93

 0.0904
331.2
1.4  287  273
Now for a Mach number of 0.0904, the software for isentropic flow or the tables for
isentropic flow of air give:
T01
 1.002
T1
559
hence:
T01  1.002  273  273.6 K
The software for Rayleigh flow or the Tables for Rayleigh flow of air give for M1 =
0.0904:
T01
 0.03841
T0 *
If the flow is choked at the exit, i.e., if M2 = 1 and then:
T02  T0 *
and so:
T02  T0* 
T01
273.6

 7121.7 K
T01 /T0*
0.03841
The amount of heat transferred per unit mass of air is therefore given by:
q  cp (T02 - T01 ) = 1.007  (7121.7 - 273.6) = 6896 kJ / kg
Therefore the maximum amount of heat that can be transferred is 6896 kJ/kg.
560
PROBLEM 10.12
Air enters a constant-area combustion chamber at a pressure of 101 kPa and a
temperature of 70° C with a velocity of 130 m/s. By ignoring the effects of friction,
determine the maximum amount of heat that can be transferred to the flow per unit mass
of air.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2.
At the inlet:
M1 
V1

a1
V1

 RT1
130
130

 0.3502
371.2
1.4  287  343
Now for a Mach number of 0.0904, the software for isentropic flow or the tables for
isentropic flow of air give:
T01
 1.025
T1
hence:
T01  1.025  343  351.6 K
The software for Rayleigh flow or the Tables for Rayleigh flow of air give for M1 =
0.3502:
T01
 0.4393
T0 *
If the maximum amount of heat is transferred the flow is choked at the exit, i.e., M2 =
1, and therefore:
T02  T0 *
and so:
561
T02  T0* 
T01
351.6

 800.3 K
*
0.4393
T01 /T0
The amount of heat transferred per unit mass of air is therefore given by:
q  cp (T02 - T01 ) = 1.007  (800.3 - 351.6) = 451.9 kJ / kg
Therefore the maximum amount of heat that can be transferred is 451.9 kJ/kg.
562
PROBLEM 10.13
Air flows through a 0.25m diameter duct. At the inlet the velocity is 300 m/s and the
stagnation temperature is 90oC. If the Mach number at the exit is 0.3, determine the
direction and the rate of heat transfer. For the same conditions at the inlet determine the
amount of heat that must be transferred to the system per unit mass of air if the flow is to
be sonic at the exit of the duct.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2.
Using:
V12
T01  T1 
2 cp
gives:
T1  T01 
V12
3002
 363 
 273.6 K
2 cp
2  1007
Therefore:
M1 
V1

a1
V1

 RT1
300
300

 0.9048
331.6
1.4  287  273.6
Because the Mach number at exit, M2 , is 0.3, the Mach number is decreasing along
the duct. This indicates that heat is being removed from the flow.
The software for Rayleigh flow or the Tables for Rayleigh flow of air give for M1 =
0.9048:
T01
 0.9929
T0 *
and for M2 = 0.3 they give:
T02
 0.3469
T0 *
563
Using these values gives:
T02  T0* 
T02 / T0*
0.3469
T01 
 363  126.8 K
*
T01 / T0
0.9929
The amount of heat transferred per unit mass of air is therefore given by:
q  cp (T02 - T01 ) = 1.007  (126.8 - 363) = - 237.8 kJ / kg
The negative sign indicates that heat is being removed from the flow.
If the flow is sonic at the exit, i.e., if M2 = 1, then:
T02  T0*
and so:
T02  T0* 
T01
363

 365.6 K
*
T01 / T0
0.9929
Thus in this situation the amount of heat transferred per unit mass of air is given by:
q  cp (T02 - T01 ) = 1.007  (365.6 - 363) = 2.62 kJ / kg
Therefore heat is removed from the system at a rate of 237.8 kJ / kg of air. However,
if, on the other hand, the flow is sonic at the exit heat would have to be added to the
system at a rate of 2.62 kJ / kg of air.
564
PROBLEM 10.14
Air flows through a constant-area combustion chamber which has a diameter of 0.15
m. The inlet stagnation temperature is 335K, the inlet stagnation pressure is 1.4 MPa, and
the inlet Mach number is 0.55. Find the maximum rate at which heat can be added to the
flow. Neglect the effects of friction.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2.
The software for Rayleigh flow or the Tables for Rayleigh flow of air give for M1 =
0.55:
T01
 0.7599
T0 *
If the maximum amount of heat is transferred the flow is choked at the exit, i.e., M2 =
1 and hence:
T02  T0*
and so:
T02  T0* 
T01
335

 440.9 K
*
T01 / T0
0.7599
The maximum amount of heat transferred per unit mass of air is therefore given by:
q  cp (T02 - T01 ) = 1.007  (440.9 - 335) = 106.6 kJ / kg
The mass flow rate of air is given by:
m  1 V1 A  1 M 1 a1 A
Now for a Mach number of 0.55, the software or the tables for air flow give for
isentropic flow:
565
T01
 1.061 ,
T1
p01
 1.228
p1
hence:
T1 
T01
335

 315.7 K
T01 / T1
1.061
and:
p01
1.4

 1.140 MPa
p01 / p1
1.228
p1 
Therefore:
1 
p1
1400000

 15.45 kg/m3
RT1
287  315.7
and:
a1 
 RT1 
1.4  287  315.7  337.0 m/s
and:
A

4
D2 

4
 0.152  0.01767 m 2
The mass flow rate of air is then given by:
m  1 M 1 a1 A  15.45 x 0.55 x 337.0 x 0.01767  50.6 kg/s
The rate of heat transfer is then given by:
Q  m q = 50.6  106.6 = 5.394 MW
Therefore rate of heat transfer to the system is 5.394 MW.
566
PROBLEM 10.15
Air enters a constant area duct at a Mach number of 0.15, a pressure of 200 kPa, and a
temperature of 20o C. Heat is added to the air as it flows through the pipe at a rate of 60
kJ/kg of air. Assuming that the flow is steady and that the effects of wall friction can be
ignored, find the temperature, pressure, and Mach number at which the air leaves the
duct. Assume that the air behaves as a perfect gas.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2.
For the inlet Mach number of 0.15, the software or the isentropic flow tables for air
give for isentropic flow:
T01
 1.005
T1
hence:
T01  1.005  293  294.5 K
The software for Rayleigh flow or the Tables for Rayleigh flow of air give for M1 =
0.15:
T01
 0.1020 ,
T0 *
T1
 0.1218 ,
T*
p1
 2.327
p*
Also since:
q  cp (T02 - T01 )
it follows that:
60  1.007  ( T02  294.5) ,
i.e., T02  294.5 
60
 354.1 K
1.007
Therefore:
T02
T T
354.1
 02 01 
 0.1020  0.1226
T0 *
T01 T0 *
294.5
567
The software for Rayleigh flow or the Tables for Rayleigh flow of air give for T02 /
T0* = 0.1226:
M 2  0.1655 ,
T2
 0.1463 ,
T*
p2
 2.311
p*
Hence:
T1 
T2 / T *
T1 
T1 / T *
p1 
p2 / p *
2.311
p1 
 200  198.6 kPa
p1 / p *
2.327
0.1463
 293  351.9 K
0.1218
and:
Therefore the temperature, pressure, and Mach number at which the air leaves the
duct are 351.9 K ( = 78.9°C), 198.6 kPa, and 0.1655 respectively.
568
PROBLEM 10.16
At the inlet to a constant-area combustion chamber the Mach number is 0.2 and the
stagnation temperature is 120°C. What is the amount of heat transfer to the gas per unit
mass if the Mach number is 0.7 at the exit of the chamber? What is the maximum
possible amount of heat transfer? The gas can be assumed to have the properties of air.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2.
For the inlet Mach number of 0.2, the software for Rayleigh flow or the Tables for
Rayleigh flow of air give:
T01
 0.1736
T0 *
while for the outlet Mach number of 0.7, the software for Rayleigh flow or the Tables for
Rayleigh flow of air give:
T02
 0.9085
T0 *
Hence:
T02
T02 / T0*

T01 
T01 / T0*
0.9085
 393  2057 K
0.1736
From this it follows that:
q  cp (T02 - T01 ) = 1.007  (2057 - 393) = 1675 kJ / kg
If the maximum amount of heat is transferred the flow is choked at the exit, i.e., M2 =
1 and hence:
T02  T0*
and so:
569
T02  T0* 
T01
393

 2264 K
*
T01 / T0
0.1736
The maximum amount of heat transferred per unit mass of air is therefore given by:
q  cp (T02 - T01 ) = 1.007  (2264 - 393) = 1884 kJ / kg
Therefore the actual heat transfer is 1675kJ / kg of air while the maximum possible
heat transfer is 1884 kJ / kg of air.
570
PROBLEM 10.17
Air flows through a constant-area duct. At the inlet to the duct the stagnation pressure
is 600 kPa and the stagnation temperature 200°C. If the Mach number at the inlet of the
duct is 0.5 and if the flow is choked at the exit of the duct, determine the heat transfer per
unit mass and the exit temperature. Assume that friction effects can be neglected.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2. For
the specified inlet Mach number, M1, of 0.5, the software for Rayleigh flow or the Tables
for Rayleigh flow of air give:
T01
 0.6914
T0 *
Because the flow is choked at the exit, i.e., because M2 = 1, it follows that:
T02  T0*
and so:
T02  T0* 
T01
473

 684.1 K
*
T01 / T0
0.6914
The amount of heat transferred per unit mass of air is therefore given by:
q  cp (T02 - T01 ) = 1.007  (684.1 - 473) = 212.6 kJ / kg
Also, because the flow is choked at the exit, i.e., because M2 = 1, isentropic flow
software or the tables for the isentropic flow of air give for this value of Mach number:
T02
 1.2
T2
and so:
T2  T0* 
T02
684.1

 570.1 K
T02 / T2
1.2
571
Therefore the heat transfer per unit mass of air is 212.6 kJ / kg and the exit
temperature is 570.1 K ( = 297.1°C).
572
PROBLEM 10.18
Air enters a constant diameter pipe at a pressure of 200 kPa. At the exit of the pipe the
pressure is 120 kPa, the Mach number is 0.75, and the stagnation temperature is 330°C.
Determine the inlet Mach number and the heat transfer per unit mass of air.
SOLUTION
Conditions at the inlet and exit of the pipe will be denoted by subscripts 1 and 2.
For the outlet Mach number of 0.75, the software for Rayleigh flow or the Tables for
Rayleigh flow of air give:
T02
 0.9401 ,
T0 *
p2
 1.343
p*
Hence:
p1
p p
200
 1 2 
 1.343  2.238
p*
p2 p *
120
200 x 1.343 = 2.238
By trial-and-error the software for Rayleigh flow or the Tables for Rayleigh flow of
air give for this value of p1 / p*:
T01
 0.2182 ,
T0 *
M 1  0.2275
Hence:
T01 
T01 / T0 *
0.2182
T02 
 603  140.0 K
T02 / T0 *
0.9401
From this it follows that:
q  cp (T02  T01 )  1.007   603  140   466.2 kJ / kg
573
Therefore the heat transfer per unit mass of air is 466.2 kJ / kg and the inlet Mach
number 0.2275.
574
PROBLEM 10.19
Air flows through a 4 inch diameter pipe. The pressure and temperature at the inlet to
the pipe are 15 psia and 70o F. The velocity at the inlet is 200 ft/sec. If the temperature at
the exit to the pipe is 1300oF, how much heat must be added to the air and what will be
the exit pressure, velocity and Mach number? Ignore the effects of friction.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2.
At the inlet:
M1 
V1

a1
V1

 RT1
200
200

 0.1772
1128.5
1.4  53.3  32.2  (70  460)
Now for a Mach number of 0.1772, the software for isentropic flow or the tables for
isentropic flow of air give:
T01
 1.006
T1
hence:
T01  1.006  530  533.2o R
The mass flow rate is given by:
m  1 V1 A
So, because:
1 
p1
144  15

 0.07646 lbm/ft 3
53.3  530
RT1
and:

2

 4
A
D 
    0.08727 ft 2
4
4
 12 
2
575
The mass flow rate of air is then given by:
m  1 V1 A  0.07646  200  0.08727  1.335 lbm/sec
The software for Rayleigh flow or the tables for Rayleigh flow of air give for the inlet
Mach number, M1, of 0.1772:
p1
 2.299 ,
p*
T01
 0.1392 ,
T0 *
T1
 0.1660
T*
At the exit where T2 = 1300oF = 1760oR it follows that:
T2
T T
1760
 2 1* 
 0.1660  0.5512
*
T
T1 T
530
From the software for Rayleigh flow g or the tables for Rayleigh flow of air it can
then be deduced that for this value of T2 / T*:
M 2  0.3680 ,
T02
 0.4718
T0 *
Hence:
T02 
T02 / T0*
0.4718
T 
 533.2  1807.2o R
* 01
T01 / T0
0.1392
Alternatively, the isentropic flow software or the tables for isentropic flow of air give for
M2 = 0.3680:
T02
 1.027
T2
T02 / T2 = 1 .027
which in turn gives:
576
T02  1.002T2  1.027  1760  1807 o R
The same result, of course, as before.)
Using the above results then gives:
Q  m cp (T02  T01 )  1.335  0.24  1807.2  533.2   408.2 btu/sec
it having been noted that for air cp = 0.24 btu /lbm o R.
It also follows that:
p2 
p2 / p *
2.0178
p1 
 15  13.162 psia
p1 / p *
2.2992
and that:
V2  M 2 a2  M 2
 RT2  0.3680  1.4  53.3  32.2  1760  756.8 ft/sec
Therefore the rate of heat addition is 408.2 btu/sec and the exit pressure, velocity, and
Mach number are 13.16 psia, 756.8 ft/sec and 0.3680 respectively.
577
PROBLEM 10.20
Air flows though a constant area duct. The air enters the duct at a pressure of 1 MPa,
a Mach number of 0.5, and stagnation temperature of 45o C. The Mach number at the
duct exit is 0.90 and the stagnation temperature at this point is 160o C. Find the amount
of heat transferred per unit mass to or from the air in the duct. Also find the pressure at
the duct exit. Assume that the effects of friction are negligible.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2.
Because the stagnation temperature increases, heat is being transferred to the flow.
The amount of heat transferred per unit mass of air is given by:
q  cp (T02  T01 )  1.007   433  318   115.8 kJ / kg
Next consider the pressure change. The software for Rayleigh flow or the Tables for
Rayleigh flow of air give for the inlet Mach number M1 = 0.15:
p1
 1.778
p*
Similarly, the software for Rayleigh flow or the Tables for Rayleigh flow of air give
for the outlet Mach number M2 = 0.9:
p2
 1.778
p*
Hence:
p2 
p2 / p *
1.125
p1 
 1000  632.7 kPa
p1 / p *
1.778
Therefore the rate of heat addition is 115.8 kJ/kg of air and the pressure at the outlet
is 632.7 kPa.
578
PROBLEM 10.21
Air enters a constant area duct at a pressure of 620 kPa and a temperature of 300oC,
the velocity at the inlet being 100 m/s. If the velocity at the exit to the duct is 210 m/s,
determine the pressure, temperature, stagnation pressure, and stagnation temperature at
the exit to the duct. Also find the heat transfer per unit mass in the duct. Assume that the
effects of friction are negligible.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2.
At the inlet:
M1 
V1

a1
V1

 RT1
100
100

 0.2084
479.8
1.4  287  573
Now for a Mach number of 0.2084, the software for isentropic flow or the tables for
isentropic flow of air give:
T01
 1.009 ,
T1
p01
 1.031
p1
hence:
T01  1.009  573  578.2o R
and:
p01 = 1.031  620 = 639.2 kPa
Now the software for Rayleigh flow or the tables for Rayleigh flow of air give M1 =
0.2084:
T01
 0.11869 ,
T0 *
p01
 1.232 ,
p0 *
The Mach number at exit is given by:
579
T1
 0.2223 ,
T*
p1
 2.262
p*
M2 
V2

a2
V2
V V a
V
 2 1 1  2 M1
V1 a1 a2
V1
 RT2
T1
T2
Hence:
M2 
V2
M1
V1
T1 / T *
210
0.2223
0.2063

 0.2084 

T2 / T *
100
T2 / T *
T2 / T *
Because T2/T* is a function of M2, the above equation determines M2. A very simple
approach to solving this equation is to use a trial-and-error method in which M2 is
guessed and the software for Rayleigh flow or the tables for Rayleigh flow of air is then
used to find the corresponding value of T2/T*. The right hand side (RHS) of the above
equation, i.e.:
0.2063
T2 / T *
can then be evaluated. This process can be repeated with various values of M2 and the
value that makes M2 = RHS can then be deduced. Some typical results are given in the
following table.
M2
0.3000
0.4000
0.3200
0.3100
0.3150
0.3120
0.3130
0.3126
T2/T*
0.4089
0.6151
0.4512
0.4300
0.4406
0.4343
0.4364
0.4355
RHS
0.3227
0.2631
0.3072
0.3147
0.3109
0.3131
0.3124
0.3126
Therefore the exit Mach number, M2 is equal to 0.3126. For this Mach number,
software for Rayleigh flow or the tables for Rayleigh flow of air give:
580
T02
 0.3700 ,
T0 *
p02
 1.193 ,
p0 *
T2
 0.4355 ,
T*
p2
 2.111
p*
Hence:
p2 
T2 
p02 
p2 / p *
2.111
p1 
 620  578.6 kPa
p1 / p *
2.262
T2 / T *
0.4355
T1 
 573  1122.5 K
T1 / T *
0.2223
p02 / p0 *
1.193
p01 
 639.2  619.0 kPa
p01 / p0 *
1.232
It then follows that the amount of heat transferred per unit mass of air is given by:
q  cp (T02  T01 )  1.007  1144.6  578.2   570.4 kJ/kg
Therefore the pressure, temperature, stagnation pressure, and stagnation temperature
at the outlet are 578.6 kPa, 1122.5 K ( = 849.5°C ), 619.0 kPa and, 1144.6 K ( =
871.6°C ) respectively while the rate of heat addition is 570.4 kJ/kg of air.
581
PROBLEM 10.22
Air enters a pipe at a pressure of 200 kPa. At the exit of the pipe, the pressure is 120
kPa, the Mach number is 0.75, and the stagnation temperature is 300°C. Determine the
inlet Mach number and the heat transfer rate to the air assuming that the effects of friction
can be ignored.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2.
The software for Rayleigh flow or the tables for Rayleigh flow of air give for the
specified exit Mach number, M2 = 0.75:
T02
 0.9401 ,
T0 *
p2
 1.343
p*
Hence:
p1
p p
200
 2 1  1.343 
 2.238
p*
p * p2
120
The software for Rayleigh flow or the tables for Rayleigh flow of air can be used to
deduce that this value of p1 /p * corresponds to an inlet Mach number of:
M 1 = 0.2275
and that at this Mach number:
T01
 0.2182
T0 *
Hence:
T01 
T01 / T0 *
0.2182
T02 
 573  133 K
T02 / T0 *
0.9401
It then follows that the amount of heat transferred per unit mass of air is given by:
582
q = cp (T02 - T01 ) = 1.007  (573 - 133) = 402.8 kJ/kg
Therefore the inlet Mach number is 0.2275 and the rate of heat addition is 402.8 kJ/kg
of air.
583
PROBLEM 10.23
Air with a stagnation pressure of 600 kPa and a stagnation temperature of 200°C
enters a constant-area pipe with a diameter of 2cm. Heat is transferred to the air as it
flows through the duct at a rate of 100 kJ/kg. The air is then isentropically brought to rest
in a large chamber. Plot the mass flow rate of air as a function of the back pressure ( i.e.,
the chamber pressure ) for the range 0 to 400 kPa. Assume frictionless flow.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2.
Because the flow from the end of the pipe is brought to rest isentropically, the back
pressure is equal to p02 .
Considering the heat addition gives:
q  cp (T02  T01 )
Hence:
100  1.007  T02  473 , i.e.,
T02  473 
100
 572.3 K
1.007
The mass flow rate is given by:
m  1 V1 A
So, because:
1 
p1
144  15

 0.07646 lbm/ft 3
53.3  530
RT1
and:

2

 4
A  D      0.08727 ft 2
4
4  12 
2
The mass flow rate of air is then given by:
584
m   2 V2 A   2 M 2 a2 A
so, because:
2 
p2
RT2
and:
A

4
D2 

4
 0.022  0.0003142 m 2
and:
a2 
 RT2
it follows that:
 p 
m   2  M 2
 RT2 
 RT2  0.0003142
 p2 M 2
 p2 
5
287
0.0003142
2.194
10

M



T





 2
2
 T
 287 T2 
2


 kg/s


It will be assumed that the flow is subsonic.
The procedure used here involves:
(1) Select a M2 value.
(2) Using the software for Rayleigh flow or the tables for Rayleigh flow of air find,
for this value of M2, the values of T02/T0* and p02 / p0*.
(3) Calculate T01 /T0* using:
T01
T T
T
473 T02
 01 02* 
 0.8265 02*
*
*
T0
T02 T0
572.3 T0
T0
(4) Using the software for Rayleigh flow or the tables for Rayleigh flow of air find,
for this value of T01 /T0*, the values of M1 and p01 / p0*.
(5) Calculate p02 using:
p02 
p02 / p0 *
p /p *
p01  600  02 0 kPa
p01 / p0 *
p01 / p0 *
This is the back pressure as discussed before.
585
(6) Using the software for isentropic flow or the tables for isentropic flow of air find
the values of T0 /T2 and p0 / p2. Hence, find T2 and p2.
(7) Calculate the mass flow rate corresponding to the chosen. M2 using:
pM
m  2.194  105   2 2
 T
2


 kg/s


Some typical results obtained using this procedure are given in the following table:
M2
p02
kPa
0.4
0.6
0.8
0.9
491.8
452.4
370.6
329.3
m dot
kg/s
2.31
2.58
2.89
2.93




10-4
10-4
10-4
10-4
The variation of mass flow rate with back pressure p02 obtained using this procedure
is shown in Fig. P10.23.
Figure P10.23
586
PROBLEM 10.24
Air flows through a rectangular duct with a 10 cm by 16 cm cross-sectional area. The
air velocity, pressure, and temperature at the inlet to the duct are 90 m/sec, 105 kPa, and
25°C, respectively. Heat is added to the air as it flows through the duct and it leaves the
duct with a velocity of 200 m/s. Find the pressure and temperature at the exit and the total
rate at which heat is being added to the air. Ignore the effects of friction.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2.
At the inlet:
M1 
V1

a1
V1

 RT1
90
90

 0.260
1.4  287  298 346
Now for a Mach number of 0.260 the software gives for isentropic flow or the tables
for isentropic flow of air give:
T01
 1.014
T1
hence:
T01  1.014  298  302.2 K
The software for Rayleigh flow or the tables for Rayleigh flow of air give for the inlet
Mach number, M1, of 0.260:
p1
 2.193 ,
p*
T01
 0.2745 ,
T0 *
T1
 0.3250
T*
The Mach number at exit is then given by:
M2 
V2
V V a
V
 2 1 1  2 M1
a2
V1 a1 a2
V1
587
 RT1
V
 2 M1
V1
 RT2
T1
T2
hence:
M2 
V2
M1
V1
T1 / T *
200

 0.260 
90
T2 / T *
0.325

T2 / T *
0.3924
T2 / T *2
Because T2 / T* is a function of M2, the above equation can be used to determine M2.
A very simple approach to solving the equation is to use a trial-and-error approach in
which M2 is guessed and the software for Rayleigh flow or the tables for Rayleigh flow of
air is then used to find the corresponding value of T2 / T*. The right hand side (RHS) of
the above equation, i.e.:
0.3924
T2 / T *2
can then be evaluated. This process can be repeated with various values of M2 and the
value that makes M2 = RHS can then be deduced. Some typical results are given in the
following table.
M2
0.4000
0.5000
0.4200
0.4100
0.4150
0.4130
0.4120
0.4121
0.4122
T2 / T*
0.6151
0.7901
0.6535
0.6345
0.6440
0.6402
0.6383
0.6385
0.6387
RHS
0.4200
0.3706
0.4075
0.4135
0.4105
0.4117
0.4123
0.4122
0.4122
Therefore the exit Mach number, M2 is equal to 0.4122. For this Mach number,
software for Rayleigh flow or the tables for Rayleigh flow of air give:
p2
 1.939 ,
p*
T02
 0.5503 ,
T0 *
Hence:
588
T2
 0.6387
T*
p2 
p2 / p *
1.939
p1 
 105  92.84 kPa
p1 / p *
2.193
T2 
T2 / T *
0.6387
T1 
 298  585.6 K
T1 / T *
0.3520
p02 
p02 / p0 *
1.193
p01 
 639.2  619.0 kPa
p01 / p0 *
1.232
T02 
T02 / T0 *
0.5503
T01 
 302.2  605.8 K
T01 / T0 *
0.2745
It then follows that rate at which heat is transferred to the air is given by:
Q  m cp (T02  T01 )  1  V1  A  cp  (T02  T01 )
So, because:
1 
p1
105000

 1.228 kg/m3
287  298
RT1
and:
A  0.1  0.16  0.016 m 2
it follows that:
Q  1.228  590  0.016  1.007   605.8  302.2   540.6 kW
Therefore the pressure and temperature at the exit are 92.84 kPa and 585.6 K ( =
312.6° C ) respectively while the rate of heat addition is 540.6 kW.
589
PROBLEM 10.25
Air is heated as it flows through a constant-area duct by an electric heating coil
wrapped uniformly around the duct. The air enters the duct at a velocity of 100 m/s, a
temperature of 20° C, and a pressure of 101.3 kPa, and the heat transfer rate is 40 kJ/kg
per unit length of duct. Plot the variations of exit Mach number, exit temperature and exit
pressure with the length of the duct. Neglect the effects of fluid friction.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2. The
length of the duct ( m ) will be denoted by L.
At the inlet:
M1 
V1

a1
V1

 RT1
100
100

 0.2915
343.1
1.4  287  293
Now for a Mach number of 0.2915 the software gives for isentropic flow or the tables
for isentropic flow of air give:
T01
 1.017
T1
hence:
T01  1.017  293  298.0 K
The heat addition per unit mass of air is given by:
q = 40 L
Hence since:
q = cp (T02 - T01 )
it follows that:
590
40 L = 1.007  (T02 - 298) , i.e.,
T02 = 298 + 39.72 L ( K )
The software for Rayleigh flow or the tables for Rayleigh flow of air give for the inlet
Mach number, M1, of 0.2915:
p1
 2.145 ,
p*
T01
 0.3313 ,
T0 *
T1
 0.3909
T*
It then follows that:
T02
T T
T
 02 01*  02  0.3313  0.001112 T02
*
T0
T01 T0
298
and that:
p2 
p2 / p *
p / p*
p
p1  2
 101.3  47.23 2
2.145
p1 / p *
p*
and that:
T2 
T2 / T *
T /T *
T
T1  2
 293  749.6 2
T1 / T *
0.3909
T*
The procedure used to obtain the results involves the following steps:
(1) Select a duct length L.
(2) Calculate T02 using:
T02 = 298 + 39.72 L ( K )
(3) Calculate T02 / T0* using:
T02
 0.001112 T02
T0*
(4) For this value of T02 / T0* use Rayleigh flow software or the tables for Rayleigh flow
of air to get the corresponding values of p2 / p * and T2 / T* .
(5) Calculate p2 and T2 using:
591
p2  47.23
p2
p*
T2  749.6
T2
T*
and:
Some typical results obtained using this procedure are given in the following table.
The maximum tube length is that which makes M2 = 1, i.e., which makes:
T02
1
T0*
which means that in this case:
T02 
1
 899.3 K
0.001112
and so in this case:
899.3 = 298 + 39.72 L ,
i.e.,
L =
899.3 - 298
= 15.14 m
39.72
In this limiting case p2 / p * = T2 / T* = 1 so p2 = 47.23 kPa and T2 = 749.6 K.
L –m
0.00
0.20
1.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
14.50
15.14
M2
0.2915
0.2964
0.3156
0.3395
0.3882
0.4393
0.4955
0.5607
0.6424
0.7685
0.8208
1.0000
p2 - K
101.3
100.9
99.47
97.62
93.61
89.22
84.35
78.71
71.84
62.06
58.33
47.23
T2 - K
293.0
300.7
331.2
369.0
443.7
516.6
587.2
654.6
715.8
763.8
770.6
749.6
The variations of M2, T2, and p2 with L are shown in Figs. P10.25a, P10.25b and P10.25c
respectively. The fact that T2 reaches a maximum when M 2  1/  should be noted.
592
Figure P 10. 25a
Figure P10.25b
593
Figure P10.25c
594
PROBLEM 10.26
Air enters a combustion chamber at a velocity of 100 m/s, a pressure of 90 kPa, and a
temperature of 40°C. As a result of the combustion, heat is added to the air at a rate of
500 kJ/kg. Find the exit velocity and Mach number. Also find the heat addition that
would be required to choke the flow. Neglect the effects of friction and assume that the
properties of the gas in the combustion chamber are the same as those of air.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2.
At the inlet:
M1 
V1

a1
V1

 RT1
100
100

 0.2820
354.6
1.4  287  313
Now for a Mach number of 0.2820, the software for isentropic flow or the tables for
isentropic flow of air give:
T01
 1.016
T1
hence:
T01  1.016  313  318 K
The software for Rayleigh flow or the tables for Rayleigh flow of air give for the inlet
Mach number, M1, of 0.2820:
T01
 0.3140 ,
T0 *
T1
 0.3709
T*
The amount of heat transferred per unit mass of air is therefore given by:
q  cp (T02  T01 )
hence:
595
T02 
q
500
 T01 
 318  814.5 K
cp
1.007
If the maximum amount of heat is transferred the flow is choked at the exit, i.e., M2 =
1 and hence:
T02 = T0 * =
T01
351.6
=
= 800.3 K
T01 / T0 *
0.4593
From this it follows that:
T02
T T
814.5
= 02 01 =
 0.3140 = 0.8043
318
T0 *
T01 T0 *
The software for Rayleigh flow or the tables for Rayleigh flow of air give for this
value of T02 / T0*:
M 2  0.5868 ,
T2
 0.9030
T*
From this it follows that:
T2 
T2 / T *
0.9030
T1 
 313  762.0 K
T1 / T *
0.3709
and that:
V2  M 2 a2  M 2
 RT2  0.5868  1.4  287  762  0.5868  553.3  324.7 m/s
If the flow is choked at the exit:
T02 = T0 *
and so:
T02  T0 * 
T01
318

 1012.7 K
0.3140
T01 / T0 *
596
The amount of heat transferred per unit mass of air in this case is therefore given by:
q = cp (T02 - T01 ) = 1.007  ( 1012.7 - 318) = 699.6 kJ/kg
Therefore the velocity and Mach number at the exit are 324.7 m/s and 0.5868
respectively while the heat addition required to choke the flow is 699.6 kJ/kg.
597
PROBLEM 10.27
A fuel-air mixture, which can be assumed to have the properties of air, enters a
constant area combustion chamber at a velocity of 10 m/s and a temperature of 100°C.
What amount of heat must be added per unit mass to cause the flow at the exit to be
choked? Also find the exit Mach number and temperature if the actual heat addition due
to combustion is 1000 kJ/kg.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2.
At the inlet:
M1 
V1

a1
V1

 RT1
10
10

 0.0258
387.1
1.4  287  373
Now for a Mach number of 0.0258 the software for isentropic flow or the tables for
isentropic flow of air give:
T01
 1.0
T1
hence:
T01  373 K
The software for Rayleigh flow or the tables for Rayleigh flow of air give for the inlet
Mach number, M1, of 0.0258:
T01
 0.003190 ,
T0 *
T1
 0.003827
T*
If flow is choked at the exit, i.e., if M2 = 1 then:
T02 = T0 *
Therefore:
T02 = T0 * =
T01
373
=
= 116928 K
T01 / T0 *
0.00319
598
From this it follows that the amount of heat transferred per unit mass of air in this
case is:
q = cp (T02 - T01 ) = 1.007  ( 116928 - 373) = 117371 kJ/kg
If the actual heat addition per unit mass of air is 1000 kJ/kg then using:
q = cp (T02 - T01 )
it follows that:
1000 = 1.007  ( T02 - 373) , i.e., T02 = 373 +
1000
= 1366.1 K
1.007
Therefore:
T02
T T
1366.1
= 02 01 =
 0.00319 = 0.01168
373
T0 *
T01 T0 *
The software for Rayleigh flow or the tables for Rayleigh flow of air give for this
value of T02 / T0*:
M 2  0.04948 ,
T2
 0.01401
T*
From this it follows that:
T2 
T2 / T *
0.01401
T1 
 373  1365.5 K
T1 / T *
0.003827
Therefore the heat addition required to choke the flow per unit mass of air is 117371
kJ/kg while for the actual heat transfer the temperature and Mach number at which the air
leaves the duct are 1365.5 K ( = 1092.5o C ) and 0.01401 respectively. For the situation
considered, because of the low inlet Mach number, there is little possibility of choking
occurring and, with the actual heat transfer, compressibility effects are essentially
negligible.
599
PROBLEM 10.28
Fuel and air are thoroughly mixed in the proportion of 1:40 by mass before entering a
constant-area combustion chamber. The pressure, temperature and velocity at the inlet to
the chamber are 50 kPa, 30°C and 80 m/s respectively. The heating value of· the fuel is
40 MJ/kg of fuel. Assuming steady flow and that the properties of the gas mixture are the
same as those of air, determine the pressure, the stagnation temperature and the Mach
number at the exit of· the combustion chamber. Neglect the effects of friction.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2.
At the inlet:
M1 
V1

a1
V1

 RT1
80
80

 0.2293
348.9
1.4  287  303
Now for a Mach number of 0.2293 the software for isentropic flow or the tables for
isentropic flow of air give:
T01
 1.011
T1
hence:
T01  1.011  303  306.3 K
The software for Rayleigh flow or the tables for Rayleigh flow of air give for the inlet
Mach number, M1, of 0.2213:
T01
 0.2213 ,
T0 *
600
p1
 2.235
p*
Next consider the combustion. Now 1 kg of fuel gives 40 MJ so, because the fuel-air
ratio is 1:40, it follows that 1 kg of fuel + 40 kg of air give 40 MJ. Hence the heat release
per unit mass of the mixture is:
q 
40000
 975.6 kJ/kg
40  1
Hence, using:
q = cp (T02 - T01 )
it follows that:
975.6 = 1.007  ( T02 - 306.3) ,
i.e., T02 = 306.3 +
975.6
= 1275.2 K
1.007
Therefore:
T02
T T
1275.2
= 02 01 =
 0.2213 = 0.9213
306.3
T0 *
T01 T0 *
The software for Rayleigh flow or the tables for Rayleigh flow of air give for this
value of T02 / T0*:
M 2  0.7187 ,
p2
 1.393
p*
From this it follows that:
p2 
p2 / p *
1.393
p1 
 50  31.16 kPa
p1 / p *
2.235
Therefore the pressure, the stagnation temperature and the Mach number at the exit
are 31.16 kPa, 1275.2 K (=1000.2°C ), and 0.7187 respectively.
601
PROBLEM 10.29
An air-fuel mixture enters a constant-area combustion chamber at a velocity of 100
m/s, a pressure of 70 kPa, and a temperature of 150o C. Assuming that the fuel-air ratio is
0.04, that the heating value of the fuel is 30 MJ/kg and that the mixture has the properties
of air, calculate the Mach number of the gases after combustion is completed and. the
change of stagnation temperature and stagnation pressure across the combustion chamber.
Neglect the effects of friction and assume that the properties of the gas in the combustion
chamber are the same as those of air.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2.
At the inlet:
M1 
V1

a1
V1

 RT1
100
100

 0.2426
412.3
1.4  287  423
Now for a Mach number of 0.2426 the software for isentropic flow or the tables for
isentropic flow of air give:
T01
 1.012 ,
T1
p01
 1.042
p1
hence:
p01  1.042  70  72.94 kPa
T01  1.012  423  428.1 K
The software for Rayleigh flow or the tables for Rayleigh flow of air give for the inlet
Mach number, M1, of 0.2426:
T01
 0.2440 ,
T0 *
602
p1
 1.220
p*
Next consider the combustion. Now 1 kg of fuel gives 30 MJ so, because the fuel-air
ratio is 0.04, it follows that 1 kg of fuel + (1 / 0.04) kg of air give 30 MJ. Hence the heat
release per unit mass of the mixture is:
q 
30
30

 1.154 MJ/kg
1  (1/ 0.04)
26
Hence, using:
q = cp (T02 - T01 )
it follows that:
1154 = 1.007  ( T02 - 428.1) ,
i.e., T02 = 428.1 +
1154
= 1574.1 K
1.007
Therefore:
T02
T T
1574.1
= 02 01 =
 0.2440 = 0.8972
428.1
T0 *
T01 T0 *
The software for Rayleigh flow or the tables for Rayleigh flow of air give for this
value of T02 / T0*:
M 2  0.6848 ,
p02
 1.047
p*
From this it follows that
p02 =
p02 p0 *
1.047
 72.94 = 62.60 kPa
p01 =
p0 * p01
1.220
The increases in stagnation pressure and temperature are therefore given by:
p0 = p02 - p01 = 62.60 - 72.94 = - 10.34 kPa
the negative sign indicating a decrease, and:
T0 = T02 - T01 = 1574.1 - 428.1 = 1146 K
603
Therefore the Mach number at the exit of the chamber is 0.6848, the decrease in
stagnation pressure across the chamber is 10.34 kPa, and the increase in stagnation
temperature across the chamber is 1146 K.
604
PROBLEM 10.30
An air-fuel mixture flows through a constant-area combustion chamber. The velocity,
pressure and temperature at the entrance to the chamber are 130 m/s, 170 kPa, and 120°C
respectively. If the enthalpy of reaction is 600 kJ/kg of mixture, find the Mach number
and pressure at the exit of the chamber. Neglect the effects of friction and assume that the
properties of the gas in the combustion chamber are the same as those of air.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2.
At the inlet:
M1 
V1

a1
V1

 RT1
130
130

 0.3272
397.4
1.4  287  393
Now for a Mach number of 0.3272 the software for isentropic flow or the tables for
isentropic flow of air give:
T01
 1.021
T1
hence:
T01  1.021  393  401.3 K
The software for Rayleigh flow or the tables for Rayleigh flow of air give for the inlet
Mach number, M1, of 0.3272:
T01
 0.3970 ,
T0 *
p1
 2.087
p*
Next, consider the combustion. Because:
q = cp (T02 - T01 )
605
and because q =600 kJ / kg, it follows that:
600 = 1.007  ( T02 - 401.3) , i.e., T02 = 401.3 +
600
= 997.1 K
1.007
Therefore:
T02
T T
997.1
= 02 01 =
 0.3970 = 0.9865
401.3
T0 *
T01 T0 *
The software for Rayleigh flow or the tables for Rayleigh flow of air give for T02 / T0*
=0.9865:
M 2  0.8719 ,
p2
 1.163
p*
From this it follows that
p2 =
p2 / p *
1.163
p1 =
 170 = 94.73 kPa
p1 / p *
2.087
Therefore the Mach number and pressure at the exit are 0.8719 and 94.73 kPa
respectively.
606
PROBLEM 10.31
An air-fuel mixture enters a constant-area combustion chamber at a Mach number of
0.2, a pressure of 70 kPa, and a temperature of 35°C. If the heat transfer to the gases in
the combustion chamber is 1.2 MJ/kg of mixture, determine the Mach number at the exit
of the chamber and the change in stagnation temperature through the chamber. Neglect
the effects of friction and assume that the properties of the gas in the combustion chamber
are the same as those of air.
SOLUTION
Conditions at the inlet and exit of the chamber will be denoted by subscripts 1 and 2.
For the specified inlet Mach number, M1, of 0.2, the software for isentropic flow or
the tables for isentropic flow of air give:
T01
 1.008
T1
hence:
T01  1.008  308  310.5 K
The software for Rayleigh flow or the tables for Rayleigh flow of air give for the inlet
Mach number, M1, of 0. 2:
T01
 0.1736
T0 *
Next, consider the combustion. Because:
q = cp (T02 - T01 )
and because q =1200 kJ / kg, it follows that:
607
1200 = 1.007  ( T02 - 401.3) , i.e., T02 = 310.5 +
1200
= 1502.2 K
1.007
Therefore:
T02
T T
1502.2
= 02 01 =
 0.1736 = 0.8399
310.5
T0 *
T01 T0 *
The software for Rayleigh flow or the tables for Rayleigh flow of air give for T02 / T0*
=0.8399:
M 2  0.6201
The increase in stagnation temperature is given by:
T0 = T02 - T01 = 1502.2 - 310.5 = 1191.7 K
Therefore the exit Mach number is 0.6201 and the increase in stagnation temperature
is 1191.7 K.
608
PROBLEM 10.32
In a gas turbine plant, air from the compressor enters the combustion chamber at a
pressure of 420 kPa, a temperature of 110°C, and a velocity of 80 m/s. Fuel having an
effective heating value of 35,000 kJ/kg is sprayed into the air stream and burnt. Two
types of injection systems are available. One gives a fuel-to-air mass ratio of 0.015, the
other a ratio of 0.021. The temperature entering the turbine should not be less than 750°C
but should not exceed the temperature determined by the metallurgical limit of the blade
material. Which of the two injection systems should be used?
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2.
At the inlet:
M1 
V1

a1
V1

 RT1
80
80

 0.2413
331.5
1.4  287  383
Now for a Mach number of 0.2413 the software for isentropic flow or the tables for
isentropic flow of air give:
T01
 1.012
T1
hence:
T01  1.012  383  387.6 K
The software for Rayleigh flow or the tables for Rayleigh flow of air give for the inlet
Mach number, M1, of 0.2413:
T01
 0.2417
T0 *
609
Next, consider the combustion. If FAR is the fuel/air ratio then since 1 kg of fuel
gives 35 MJ it follows that 1 kg of fuel + ( 1 / FAR) kg of air give 35 MJ . Therefore the
heat release per unit mass of the mixture is:
q 
35
MJ / kg
1  1/ FAR 
hence using:
q = cp (T02 - T01 )
it follows that:
35000
 1.007  ( T02 - 387.6)
1  1/ FAR 
i.e.:
T02  387.6 
35000
K
1.007  1  1/ FAR  
Applying this for the two situations being considered gives:
For Case 1:
In this case FAR = 0.015 so the above equation gives:
T02  387.6 
35000
 901.3 K
1.007  1  1/ 0.015  
Therefore:
T02
T T
901.3
= 02 01 =
 0.2417 = 0.5620
387.6
T0 *
T01 T0 *
The software for Rayleigh flow or the tables for Rayleigh flow of air give for T02 / T0*
=0.5620:
610
T2
 0.6515
T*
From this it follows that
T2 =
T2 / T *
0.6515
T1 =
 383 = 1032 K
T1 / T *
0.2417
For Case 2:
In this case FAR = 0.021 so the above equation gives:
T02  387.6 
35000
 1102.5 K
1.007  1  1/ 0.021 
Therefore:
T02
T T
1102.5
= 02 01 =
 0.2417 = 0.6875
387.6
T0 *
T01 T0 *
The software for Rayleigh flow or the tables for Rayleigh flow of air give for T02 / T0*
=0.6875:
T2
 0.7861
T*
From this it follows that
T2 =
T2 / T *
0.7861
T1 =
 383 = 1246 K
T1 / T *
0.2417
Therefore, the first system gives a turbine inlet temperature of 1032 K ( = 759.4° C )
and the second system gives a turbine inlet temperature of 1246 K ( = 972.7° C). Hence
both systems give a turbine inlet temperature that is above the minimum required value
and below the material limit which is in excess of 1000° C.
The second system should, therefore, be used because it gives the higher but still
acceptable turbine inlet temperature and so will give the higher engine efficiency.
611
PROBLEM 10.33
Air enters a constant-area duct with a Mach number of 2.5, a stagnation temperature
of 300°C, and a stagnation pressure of 1.2 MPa. If the flow is choked, determine the
stagnation pressure and stagnation temperature at the exit to the duct and the heat transfer
per unit mass if there is a normal shock at the inlet of the duct.
SOLUTION
Conditions upstream and downstream of the normal shock wave at the inlet will be
denoted by subscripts 1 and 2 while those at the outlet of the duct will be denoted by
subscript 3.
The software for normal shock waves or the normal shock wave tables for air give:
p02
 0.4990
p01
M 2  0.5130 ,
Because there is no change in stagnation temperature across the shock wave, it
follows that:
T02  T01
Using the above results gives:
p02  0.499 x1200  598.9 kPa , T02  573K
The software for Rayleigh flow or the tables for Rayleigh flow of air give M2 = 0.513:
T02
 0.7101 ,
T0 *
p02
 1.109
p0 *
Since the flow is choked at the exit M3 = 1. Therefore:
T03  T0 * ,
p03  p0 *
Hence:
612
T03 = T0 * =
T02
573
=
= 806.9 K
T02 / T0 *
0.7101
p03 = p0 * =
p02
598.9
=
= 540.0 kPa
p02 / p0 *
1.109
and:
The amount of heat transferred per unit mass of air is given by:
q = cp (T03 - T02 ) = 1.007  (806.9  573) = 235.5 kJ/kg
Therefore the stagnation pressure and stagnation temperature at the outlet are 540 kPa
and 806.9 K ( = 533.9°C) respectively and the heat added per unit mass of air is 235.5
kJ/kg.
613
PROBLEM 10.34
Air at a temperature of 300 K and a Mach number of 1.5 enters a constant-area duct
which feeds a convergent nozzle. At the exit of the nozzle the Mach number is 1.0 and
the ratio of the nozzle exit area to the duct area is 0.98. If a normal shock wave occurs in
the duct just upstream of the nozzle inlet, calculate the amount and direction of the heat
exchange with the air flow through the duct. Ignore the effect of friction on the flow in
the duct and assume that the flow downstream of the shock to be isentropic.
SOLUTION
Conditions at the inlet and exit of the duct will be denoted by subscripts 1 and 2.
Conditions downstream of the normal shock will be denoted by subscript 3 and those at
the outlet of the nozzle be denoted by subscript 4.
Now for the inlet Mach number of 1.5, the software for isentropic flow or the
isentropic flow tables for air give:
T01
 1.450
T1
hence:
T01  1.450  300  435 K
Consider the flow through the nozzle. Because M4 = 1 and A3 / A4 = 1 / 0.98 = 1.0204,
it follows using isentropic flow software or the isentropic flow tables for air that since A4
= A* and therefore that A3 / A* = 1.0204:
M 3  0.8509
Next consider the normal shock wave. Because the Mach number downstream of the
shock wave is 0.8509, normal shock wave software or the normal shock wave tables for
air give:
614
M 2  1.186
The Mach numbers at the inlet and exit of the duct are now known. Using these
values, the software for Rayleigh flow or the tables for Rayleigh flow of air give:
T01
 0.9093 ,
T0 *
T02
 0.9812
T0 *
Hence:
T02 =
T02 / T0 *
0.9812
 435 = 469.4 K
T01 =
T01 / T0 *
0.9093
The amount of heat transferred per unit mass of air is given by:
q = cp (T02 - T01 ) = 1.007  (469.4  435) = 34.64 kJ/kg
Therefore heat added per unit mass of air is 34.64 kJ / kg.
615
PROBLEM 10.35
A jet engine is operating at an altitude of 7000 m. The mass of air passing through the
engine is 46 kg/sec and the heat addition in the combustion chamber is 500 kJ/kg. The
cross-sectional area of the combustion chamber is 0.5 m2 and the air enters the chamber
at a pressure of 80 kPa and a temperature of 80°C. After the combustion chamber, the
products of combustion, which can be assumed to have the properties of air, are expanded
through a convergent nozzle to match the atmospheric pressure at the nozzle exit.
Estimate the nozzle exit diameter and the nozzle exit velocity assuming the flow in the
nozzle is isentropic. State the assumptions that have been made.
SOLUTION
Conditions at the inlet and exit of the combustion chamber will be denoted by
subscripts 1 and 2 and conditions at the exit of the nozzle will denoted by subscript 3.
Because:
m  1 V1 A
and because:
1 
p1
80000

 0.992 kg/m 3
RT1
287  281
and:
A  0.5 m 2
it follows that:
V1 
m
45

 90.73 m/s
1 A 0.992  0.5
Using this result then gives:
M1 
V1

a1
V1

 RT1
90.73
90.73

 0.270
336
1.4  287  281
616
Now for a Mach number of 0.270 the software for isentropic flow or the tables for
isentropic flow of air give:
T01
 1.014
T1
Hence:
T01  1.014  281  284.9 K
The software for Rayleigh flow or the tables for Rayleigh flow of air give for the inlet
Mach number, M1, of 0.270:
T01
 0.292 ,
T0 *
T1
 0.345 ,
T*
p1
 2.177
p*
Next, consider the combustion. Because:
q = cp (T02 - T01 )
and because q = 500 kJ / kg, it follows that:
500  1.007  T02  284.9  , i.e., T02  284.9 
500
= 781.4 K
1.007
Therefore:
T02
T T
781.4
= 02 01 =
 0.292 = 0.801
T0 *
T01 T0 *
284.9
The software for Rayleigh flow or the tables for Rayleigh flow of air give for T02 / T0*
=0.801:
M 2  0.585 ,
T2
 0.901 ,
T*
617
p2
 1.622
p*
From this it follows that
p2 =
p2 / p *
1.622
p1 =
 80 = 59.61 kPa
p1 / p *
2.177
T2 =
T2 / T *
0.901
T1 =
 281 = 733.9 K
T1 / T *
0.345
and:
Also for a Mach number of 0.585, the software for isentropic flow or the tables for
isentropic flow of air give:
T02
 1.068 ,
T2
p02
 1.261
p2
At the nozzle exit, the pressure is ambient which at 7000 m is 35.7 kPa, i.e.:
p3  35.7 kPa
Therefore, because the nozzle flow is assumed to be isentropic:
T03  T02
and
p03  p02
it follows that:
p03
p p
59.61
= 02 2 = 1.261 
= 2.106
p3
p2 p3
35.7
For this value of p03 / p3 , the isentropic flow software or the tables for isentropic
flow of air give:
M 3  1.09 ,
The nozzle exit temperature is then given by:
618
T03
 1.237
T3
T3 =
T02 / T2
1.068
T2 =
 733.9 = 633.6 K
T03 / T3
1.237
It therefore follows that:
V3  M 3 a3  M 3  RT3  1.09 
1.4  287  633.6  1.09  504.6  550 m/s
Then because:
3 
p3
35700

 0.196 kg/m3
RT3
287  633.6
A3 
m
45

 0.417 m 2
3 V 3 0.196  550
it follows that:
therefore:

4
D32  A3  0.417 ,
hence, D3 
4  0.417

 0.729 m
Therefore the nozzle exit velocity and diameter are 550 m/s and 0.729 m respectively.
It has been assumed that the flow is steady and one-dimensional, that the gases have the
properties of air, that the nozzle flow is isentropic, and that there are no heat losses from
the combustion chamber.
619
PROBLEM 10.36
Air enters a 7.5 cm diameter pipe at a pressure of 1.3 kPa, a temperature of 200°C,
and a Mach number of 1.8. Heat is added to the flow as a result of a chemical reaction
taking place in the duct. Find the heat transfer rate necessary to choke the flow in the
pipe. Assume that the air behaves as a perfect gas with constant specific heats and neglect
changes in the composition of the gas stream due to the chemical reaction.
SOLUTION
Conditions at the inlet and exit of the pipe will be denoted by subscripts 1 and 2.
The software for isentropic flow or the tables for isentropic flow of air give for M1 =
1.8:
T01
 1.648
T1
Hence:
T01  1.648  473  779.5 K
The software for Rayleigh flow or the tables for Rayleigh flow of air give for the inlet
Mach number, M1, of 1.8:
T01
 0.8363
T0 *
At the exit, because the flow is choked, M2 = 1:
T02  T0 *
hence:
T02 = T0 * =
T01
779.5
=
= 932.1 K
T01 / T0 *
0.8363
The amount of heat transferred per unit mass of air is given by:
620
 cp (T02 - T01 ) = 1 V1 Acp (T02 - T01 )
Q = m
So, because:
p1
1300

 0.009576 kg/m3
RT1
287  473
1 
and because:
V1  M 1 a1  M 1  RT1  1.8  1.4  287  473  784.7 m/s
and since:
A 

4
D2 

4
 0.0752  0.004418 m 2
Using the above results then gives:
Q = 1 V1 Acp (T02 - T01 )
= 0.009576  784.7  0.004418  1.007 
 932.1
Therefore the required heat transfer rate is 5.102 kW.
621
 779.5

 5.102 kW
PROBLEM 10.37
Air enters a 15 cm diameter pipe at a pressure of 1.3 MPa and a temperature of 20°C
and with a velocity of 60 m/s. Assuming that the friction factor is 0.004 and that the flow
is effectively isothermal, find the Mach number at a point in the pipe where the pressure
is 300 kPa and the length of the pipe to this point.
SOLUTION
Conditions at the inlet and downstream section of the pipe will be denoted by
subscripts 1 and 2. The distance between the inlet and the second point is designated by
l 1-2.
At the inlet:
M1 
V1

a1
V1

 RT1
60
60

 0.1749
343.1
1.4  287  293
Now for a Mach number of 0.1749 the software for isothermal flow in a constant area
duct or the tables for the isothermal flow of air in a constant area duct give:
T01
 0.8880 ,
T0 *
p1
 4.78 ,
p*
4 f l1 *
 19.20
D
At the downstream point where the pressure is 300 kPa:
p2
p p
300
= 1 2 = 4.78 
= 1.103
1300
p*
p* p1
For this value of p2 / p1 the software for isothermal flow the software for isentropic
flow in a constant area duct or the tables for the isothermal flow of air in a constant area
duct give:
M 2  0.767 ,
4 f l2 *
 0.020
D
622
To find the length of the pipe, it is recalled that:
l2 *  l1 *  l12
from which it follows that:
4 f l2 *
4 f l1 * 4 f l12


D
D
D
i.e.,
0.020  19.20 
4 f l12
,
D
i.e.,
4 f l12
 19.18
D
i.e.,
4  0.004  l12
19.18  0.15
 19.18 , i.e., l12 
 179.2 m
0.15
4  0.004
Therefore the Mach number at the second point is 0.767 and the length of the pipe to
this point is 179.2 m.
623
PROBLEM 10.38
Oxygen is to be pumped through a 125 m long, 25 mm diameter pipe by a compressor
which delivers the oxygen to the pipe at a pressure of 1.3 MPa. The mean friction factor
can be assumed to be 0.0045 and the flow in the pipe can be assumed to be isothermal,
the oxygen temperature being 30oC. Find the mass flow rate.
SOLUTION
It will be assumed for oxygen that:
  1.4 and R  8314 / 32  259.8 J/kg K
It will also be assumed that at the exit of the pipe:
M 
1

Therefore, if subscript 1 denotes the conditions at the inlet pipe:
4 f l1*
4  0.0045  12.5

9
D
0.025
The tables of the software for isothermal flow give for this value:
M 1  0.242
The mass flow rate is then given by:
m  1 V1 A  1 M 1 a1 A 
p1
M1
R T1
 RT1

4
D2
Hence since the temperature is 303 K throughout the flow:
m 
p1

1300000

 0.242  1.4  259.8  303 
 0.0252  0.651 kg/s
M 1  RT1 D 2 
R T1
4
259.8  303
4
Therefore the mass flow rate is 0.651 kg/s.
624
PROBLEM 10.39
Consider subsonic air flow through a pipe with a diameter of 2.5 cm which is 3 m
long. At the inlet to the pipe the pressure is 200 kPa and the Mach number is 0.35. If the
mean value of the friction factor is assumed to be equal to 0.006, determine the exit Mach
number assuming that the flow is isothermal.
SOLUTION
Conditions at the inlet and exit of the pipe will be denoted by subscripts 1 and 2 and
the length of the pipe is designated by l 12.
At the inlet Mach number, M1 , of 0.35 the software for isothermal flow in a constant
area duct or the tables for the isothermal flow of air in a constant area duct give:
4 f l1 *
 3.068
D
It is next recalled that:
l2 *  l1 *  l12
from which it follows that:
4 f l2 *
4 f l1 *
4 f l12


D
D
D
i.e.,
4 f l2 *
4  0.006  3
 3.068 
 0.1880
D
0.15
For this value of 4 f l 2* / D, the software for isothermal flow in a constant area duct
or the tables for the isothermal flow of air in a constant area duct give:
M 2  0.6399
Therefore the, Mach number at exit is 0.6399.
625
PROBLEM 10.40
Air at a pressure of 550 kPa and a temperature of 30°C flows through a pipe with a
diameter of 0.3 m and a length of 140 m. If the mass flow rate is at its maximum value,
find, assuming that the flow is isothermal and that the average friction factor is 0.0025,
the Mach number at the inlet to the pipe and the mass flow rate through the pipe.
SOLUTION
Conditions at the inlet and outlet of the pipe will be denoted by subscripts 1 and 2.
The length of the pipe is designated by l 1-2 . The flow will be assumed to be subsonic.
The maximum flow rate occurs when the flow at the exit is choked, i.e., when:
l12  l1 *
Hence:
4 f l1 *
4 f l12
4  0.0025  140


 4.667
D
D
0.3
For this value of 4 f l *2 / D, the software for isothermal flow in a constant area duct
or the tables for the isothermal flow of air in a constant area duct give:
M 1  0.3044 ,
p1
 2.89
p*
Therefore:
p2 = p* =
p1
550
=
= 190.3 kPa
p1 / p* 2.89
The mass flow rate is next calculated using:
m = 1 V1 A = 1 M 1 a1 A
So, because:
1 
p1
550000

 6.325 kg/m3
RT1
287  303
626
and:
A 

4
D2 

4
 0.32  0.07069 m 2
and:
a1 
 RT1 
1.4  287  303  348.9 m/s
it follows that:
m = 1 M 1 a1 A = 6.325  0.3044  348.9  0.07069  47.45 kg / s
Therefore the Mach number at the inlet is 0.3044 and the mass flow rate through the
pipe is 47.45 kg/s.
627
PROBLEM 10.41
In long pipelines, such as those used to convey natural gas, the temperature of the gas
can usually be considered constant. In one such case, the gas leaves a pumping station at
pressure of 320 kPa and a temperature of 25°C with a Mach number of 0.10. At some
other point in the flow, the pressure is measured and found to be 130 kPa. Calculate the
Mach number of the flow at this section and determine how much heat has been added to
or removed from the gas per unit mass between the pumping station and the point where
the measurements are made. The gas can be assumed to have the properties of methane.
SOLUTION
Conditions at the inlet and downstream section of the pipe will be denoted by
subscripts 1 and 2. It will be assumed that the gas has the following properties:
R  519.6 ,   1.3 ,
cp 
R
 2.25 kJ / kg K
 1
Using the isentropic flow software for γ = 1.3 gives for the inlet where M1 = 0.1
gives:
T01
 1.0015
T1
Hence:
T01 
T01
T1  1.0015  298  298.5 K
T1
Also for this inlet Mach number, M1 , of 0.1:
p1

p*
1

 M1
1
 8.771
1.3 x 0.1
and:
T01
2 
 1 2 
2  1.3 
  1

M1  

 0.12   0.8980
1 
1 
*
T0
3  1 
2
3  1.3  1 
2


628
At the downstream point where the pressure is 130 kPa:
p2
p p
130
 1 2  8.771 
 3.563
320
p*
p * p1
But:
p2

p*
1
, i.e., M 2 
 M2
1

  p2 / p* 
1
 0.2462
1.3  3.563
For this value of M2 :
T0
2 
 1 2 
2  1.3
   1

M  

 1 
 0.24622   0.9047
1 
*
T0
3  1 
2
3  1.3  1 
2


The above results then give:
T02 =
T02 / T0 *
0.9047
T01 =
 298.5 = 300.7 K
T01 / T0 *
0.8980
The amount of heat transferred per unit mass of air is given by:
q = cp (T02 - T01 ) = 2.25  (300.7  298.5) = 4.95 kJ/kg
Because this is positive, heat has been added to the gas.
Therefore the Mach number at the downstream point is 0.2462 and the heat added to
the gas per unit mass of gas is 4.95 kJ/kg.
629
PROBLEM 10.42
Natural gas is to be pumped through a 36 inch diameter pipe connecting two
compressor stations 40 miles apart. At the upstream station the pressure is 100 psig and
the Mach number is 0.025. Find the pressure at the downstream station. Assume that
there is sufficient heat transfer through the pipe to maintain the gas at a temperature of
70 °F. The gas can be assumed to have a specific heat ratio of 1.3 and the friction factor
can be assumed to be 0.004.
SOLUTION
Conditions at the inlet and downstream stations will be denoted by subscripts 1 and 2.
For the inlet Mach number, M1 , of 0.025:
p1

p*
1

 M1
1
 35.08
1.3  0.025
and:
1   M 2 
1  1.3  0.0252 
4 fl1*
2
M
 


 ln[1.3  0.0252 ]  1222.7
ln[
]



2
2 
D
M

1.3
0.025





from which it follows that:
4 f l2 *
4 f l1 * 4 f l12
4  0.004  40  5280


 1222.7 
 96.3
D
D
D
3
From this it follows that at the downstream station:
1   M 2 2 
1  1.3  M 2 2 
4 fl2*
2
M
 96.3  


 ln[1.3  M 2 2 ]
ln[

]
2


2
2 
D
  M2 
 1.3  M 2 
i.e.:
1  1.3  M 2 2 
 ln[1.3  M 2 2 ]  96.3

2 
 1.3  M 2 
630
Solving this equation gives M2 = 0.08371. Therefore:
p2

p*
1

 M2
1
 10.48
1.3  0.08371
Hence:
p2
p2 / p *
10.48
p1 
 100  29.88 psia
p1 / p *
35.08
Therefore the pressure at the downstream station is 29.88 psia.
631
PROBLEM 10.43
A 3 km long pipeline with a diameter of 0.10 m is used to transport methane at a rate
of 1.0 kg/s. If the gas remains essentially at a constant temperature of 10°C and if the
pressure at the exit to the pipe is 150 kPa, find the inlet velocity. Assume an average
friction factor of 0.004. Methane has a molal mass of 16 and a specific heat ratio of 1.3.
SOLUTION
Conditions at the inlet and downstream section of the pipe will be denoted by
subscripts 1 and 2. The distance between the inlet and the second point is denoted by l 12
Using:
m =  2 V2 A , i.e.,
V2 =
m
2 A
and noting that:
R 
8314.3
 519.6 ,   1.3
16
it follows that:
2 
p2
150000

 1.020 kg/m3
RT2 519.6  283
and:
A 

4
D2 

4
 0.12  0.007854 m 2
Hence:
M2 
V2

a2
V2

 RT2
124.8
124.8

 0.285
437.2
1.3  519.6  283
632
For this outlet Mach number, M2 , of 0.285, using the software for isothermal flow in
a constant area duct with friction gives for γ = 1.3:
4 f l2 *
 6.222
D
Therefore using:
4 f l1 *
4 f l2 * 4 f l12
4  0.004  3000


 6.222 
 486.2
D
D
D
0.1
For this value of 4 f l *1 / D, using the software for isothermal flow in a constant area
duct with friction gives for γ = 1.3 gives:
M 1  0.03948
From this it follows that, because a1 = a2 :
V1  M 1a1  0.03948  437.2  17.26 m / s
Therefore the inlet velocity is 17.26 m/s.
633
PROBLEM 10.44
Air enters a convergent duct at a Mach number of 0.75, a pressure of 500 kPa and a
temperature of 35° C. The exit area of the duct is half the inlet area. If, as a result of heat
transfer, the Mach number remains constant at 0.75, find the heat transfer rate per kg of
air. Neglect the effects of friction.
SOLUTION
Conditions at the inlet and outlet of the duct will be denoted by subscripts 1 and 2.
Using the prescribed inlet conditions:
 1 2 

T01  T1 1 
M   308  1  0.2  0.752 
2


Now:
dA
( M 2  1)
dq
( M 2  1)
dM


2
2
A
[1  (  1) M /2] c pT
[1  (  1) M /2] M
Now since the Mach number remains constant, i.e., since d M = 0 , this equation gives
in the situation being considered:
dA
( M 2  1)
dq

2
A
[1  (  1) M /2] c pT
But:
dq  c p dT0
Hence the above equation gives:
dT0
dA
( M 2  1)

2
A
[1  (  1) M /2] T
But:
634
   1 M 2 , i.e., T  T 1    1 M 2 
T0
 1


0
2
2
T


Therefore combining the above equations give:
dT
dA
 ( M 2  1) 0
A
T0
Integrating this equation between the inlet and the outlet gives:
T 
A 
ln  2   ( M 2  1) ln  02 
 A1 
 T01 
Using the prescribed conditions then gives:
 T 
ln  0.5   (1.4  0.752  1) ln  02 
 364.3 
i.e.:
ln  0.5 
 T 
ln  02  
  0.3878
(1.4  0.752  1)
 364.3 
which gives:
T02  247.2 K
Using this value then gives:
q = cp (T02 - T01 ) = 1.007  (247.2  342.7) = -96.17 kJ/kg
the negative sign indicating that heat is removed from the flow.
Therefore the amount of heat removed per unit mass of air is 96.17 kJ/kg.
635
PROBLEM 10.45
A tube containing a combustible gas mixture is contained in a long insulated pipe at a
pressure of 150 kPa and a temperature of 30°C. The gas is ignited at one end of the tube
leading to the propagation of a detonation wave down the pipe. If the combustion causes
a heat “release” of 1 MJ/kg of gas, find the pressure and temperature behind the
detonation wave and the velocity at which the wave is moving down the pipe. Assume
that the gas mixture has the properties of air.
SOLUTION
The initial conditions and those behind the wave will be denoted by subscripts 1 and 2
respectively.
The equation of state gives:
1 
p1
150000

 1.725 kg/m3
RT1
287  303
Now for a detonation wave:
 p2   1 
 2    p2  1   1q
1     1  
 1       
   1    p1   2   p1
 p1    2 
But here:
1 q
p1

1.725  1000000
 11.50
15000
The above equation therefore gives because γ = 1.4:
  p2   1  
 p2   1 
1     1  7 1        11.50
 p1    2 
  p1    2  
But for a detonation wave:
636
 1 
 ( p2 /p1 )
  
(  1)( p2 /p1 )  1
 2 
These two equations together determine the values of ρ1 / ρ2 and p2 / p1 . Here, a
series of values of p2 / p1 have been chosen and the second of the above equations has
used to determine the corresponding values of ρ1 / ρ2 . Using these values, the left and
right hand sides of the first of the above equations were computed and these values
compared and the value of p2 / p1 that made the two sides of the first equation equal was
deduced. The procedure gave the following results:
p2
 6.124 ,
p1
1
 0.6259
2
Using these values gives:
p2  6.125  150  918.6 kPa
The pressure behind the wave is therefore 918.6 kPa.
Now, the equation of state gives:
T2
p 
p 
 2 1 , i.e., T2  T1 2 1  303  6.124  0.6259  1161.4 K
T1
p1  2
p1  2
The temperature behind the wave is, therefore, 1161.4 K.
Finally, since:
 1   p /p  1 
 1   6.124  1 
M 12     2 1   

 , i.e., M 1  3.128
    1  0.6259 
    1  1 / 2 
and so:
V1  M 1 a1  M 1
 RT1  3.128  1.4  287  303  1091.4 m/s
This is the velocity relative to the wave upstream of the wave and is, therefore, equal
to the velocity at which the wave is propagating, i.e., the wave is moving at a speed of
1091.4 m/s.
637
Therefore the pressure and temperature behind the wave are 918.6 kPa and 1161.4 K (
= 888.4 °C ) respectively and the velocity of the wave is 1091.4 m/s.
638
Chapter Eleven
HYPERSONIC FLOW
SUMMARY OF MAJOR EQUATIONS
Newtonian Theory
C p  2 sin 2 
(11.4)
p
 1   M 2 sin 2 
p
(11.5)
Modified Newtonian Theory
Cp
C pS
C pS 
N
 sin 2 
(11.9)
N
[(  1)/2] / ( 1)
[2 /(  1)1/ ( 1) [ /2]]
(11.13)
For γ = 1.4 this gives:
C p  1.839 sin 2 
(11.14)
D  ( pAB  p ) W
(11.16)
CD  C pS sin 2 
(11.18)
Forces on a Body
639
where β is the half-angle of the wedge.
The Newtonian model gives:
CD  2sin 2 
640
(11.19)
PROBLEM 11.1
A flat plate is set at an angle of 3° to a flow at a Mach number of 8 in which the
pressure is 1 kPa. Estimate the pressure acting on the lower surface of this plate.
SOLUTION
Using the Newtonian theory, the pressure on the lower surface will be given by:
p  p 1   M 2 sin 2  
In the situation here being considered θ = 3°, M∞ = 8, and p∞ = 1 kPa.
Hence:
p  1  1   82  sin 2 3o   1.245 kPa
Therefore, according to Newtonian theory, the pressure on the lower surface of the
plate is 1.245 kPa.
641
PROBLEM 11.2
Air at a pressure of 10 Pa and flowing at a Mach number of 8 passes over a body
which has a semi-circular leading edge with a radius of 0.15 m. Assuming the flow to be
two-dimensional, find the pressure acting on this nose portion of the body at a distance of
0.1m around the surface measured from the leading edge of the body.
Figure P11.2
SOLUTION
The angle, φ , that a radial line through the point being considered makes to the
center-line of the body is given by:
 
0.1
 0.6667 radians
0.15
The angle that the surface makes at the point being considered to the direction of the
undisturbed flow is therefore given by:
 

2
 

2
 0.6667  0.9041 radians
If Newtonian theory is used, the pressure on the surface at the point considered is
given by:
642
p  p 1   M 2 sin 2  
i.e.:
p  10  1    82  sin 2 0.9041  563.4 Pa
Alternatively, if the Modified Newtonian theory is used, the pressure on the surface at
the point considered is given by:
1.839  2 2 
1.839 x 1.4



p  p 1 
M  sin    10  1 
 82  sin 2 0.9041  518.8 Pa
2
2




Therefore, Newtonian theory indicates that the pressure at the point considered is
563.4 Pa while the Modified Newtonian theory indicates that the pressure is 518.8 Pa at
this point. The actual pressure will be between these two values.
643
PROBLEM 11.3
Using the Newtonian model, estimate the pressures acting on surfaces 1, 2 and 3 of
the body shown in the following figure.
Figure P11.3
SOLUTION
The angles that the surfaces make to the direction of the undisturbed flow are:
θ1 = 30 + 10 = 40°
θ2 = 90 – 10 = 80°
θ3 = 30 – 10 = 20°
Hence, if Newtonian theory is used, the pressure on surface 1 is given by:
p  p 1   M 2 sin 2  
i.e.:
p1  4  1    92  sin 2 40o   191.4 Pa
Similarly:
644
p2  4  1    92  sin 2 80o   443.9 Pa
and:
p3  4  1    92  sin 2 20o   57.1 Pa
Hence, the pressures on surfaces 1, 2 and 3 are 191.4 Pa, 443.9 Pa and 57.1 Pa
respectively.
645
PROBLEM 11.4
Consider hypersonic flow over the body shape indicated in the following figure.
Figure P11.4a
Using Newtonian theory, derive an expression for the pressure distribution around the
surface of this body in terms of the distance from the stagnation point, S.
SOLUTION
At distance S around the surface stagnation point, the angle φ shown in Fig. P11.4b is
given by:
 
S
2S

R
D
Figure P11.4b
646
Therefore, at this distance around the surface the angle that the surface makes to the
direction of the undisturbed flow is:
 

2
 

2

2S
D
But:
2S 

 2S 
sin  
  cos 

D
2
 D
Hence, if Newtonian theory is used, the pressure on surface is given by:
p  p 1   M 2 sin 2  
i.e.:

 2 S 
p  p 1   M 2 cos 2 

 D 

This allows the pressure at a point distance S around the surface to be found.
647
PROBLEM 11.5
Consider two-dimensional air flow at a Mach number of 7 over the body shown in the
following figure. The pressure in the flow ahead of the body is 12 Pa. Using the
Newtonian method, find the pressures acting on the surfaces 1, 2 and 3 indicated in the
figure.
Figure P11.5
SOLUTION
The angles that surfaces 1 and 2 make to the direction of the undisturbed flow are:
θ1 = 30 + 30 = 60°
θ2 = 30 - 30 = 0°
Surface 3 is in the “shadow” of surface 1.
Hence, if Newtonian theory is used, the pressure on surface 1 is given by:
p1  p 1   M 2 sin 2 1 
i.e.:
p1  12  1    7 2  sin 2 60o   629.4 Pa
648
Because θ2 = 0° it follows that:
p1  p  12 Pa
and because surface 3 is “shadowed”:
p3  p  12 Pa
Hence, the pressures on surfaces 1, 2 and 3 are 629.4 Pa, 12 Pa, and 12 Pa
respectively.
649
PROBLEM 11.6
A wedge shaped body has an included angle of 40° and a base width of 1.5 m. Using
the Newtonian model and assuming two-dimensional flow, find the drag on the wedge
per m width when it is moving through air in which the ambient pressure is 10 Pa at a
Mach number of 7.
SOLUTION
The flow situation being considered is shown in the following figure:
Figure P11.6
For the top and bottom surfaces:
θ1 = θ2 = 20°
Hence:
p1  p2  p 1   M 2 sin 2 1 
i.e.:
p1  p2  10  1    7 2  sin 2 20o   90.25 Pa
Surface 3 is in the “shadow” of the other two surfaces so:
p1  p  10 Pa
650
The drag, D, per m width of the body is then given by:
D  p1 S sin1  p2 S sin 2  p3 B  2 p1 S sin1  p3 B
where, as shown in Fig. P11.6, S is the length of the top and bottom surfaces of the body
and B is the height of the base. But:
sin 1 
B/2
S
Hence:
D  2 p1 S
B
 p3 B  p1 B  p3 B
2S
i.e.:
D  ( p1  p3 ) B   90.25  10   1.5  120.4 N
Therefore, the drag on the body per m width is 120.4 N.
651
PROBLEM 11.7
The axi-symmetric body shown in the following figure is an approximate model of
some earlier spacecraft. Using the Newtonian model, derive an expression for the drag
coefficient for this body in hypersonic flow.
Figure P11.7a
SOLUTION
The downstream surface is entirely “shadowed” by the upstream surface so the
pressure on the downstream surface is everywhere equal to p∞.
Consider any point P on the surface such as that shown in Fig. P11.7b.
Figure P11.7b
652
The angle of the surface to the undisturbed flow at this point is:
 

2

Therefore:


sin   sin      cos 
2

i.e.:
R2  r 2
sin   cos  
R2
2
2
Consider the strip on the surfaces shown in Fig. P11.7b. The drag force due to the
pressure forces on this strip will be equal to (p - p∞ ) x projected area in flow direction,
i.e., equal to (p - p∞) 2 π r dr. Therefore, the total drag on the body is given by:
D /2
  p  p  2 r dr
D 

0
Now, Newtonian theory gives:
p  p 1   M 2 sin 2  
hence:
 R2  r 2 
p  p   p M 2 sin 2    p M 2 cos 2    p M 2 

2
 R 
Therefore the drag is given by:
D   p M
2


D /2
0
2
 R2  r 2 
D2 
1 D 
2
2 r 
1    
 d r   p M  
2
8 
2  R  
 R 
Since the frontal area of the body is π D 2 / 4 , the above result can be written in terms
of the drag coefficient by noting that:
CD 
D
  p M / 2  D 2 / 4 
2

Hence:
653
CD 
1
2
 1  D 2 
1    
 2  R  
This expression gives the drag coefficient for the body in hypersonic flow.
654
Chapter Twelve
HIGH TEMPERATURE FOWS
SUMMARY OF MAJOR EQUATIONS
Specific Heats for a Monatomic Gas
cv  32 R
(12.9)
c p  52 R
(12.10)

(12.11)
5
3
Specific Heats for a Diatomic Gas
2
evib /T
 vib 
cv  R  R 
2
  /T
 T  [e vib  1]
5
2
(12.19)
2
evib /T
 vib 
cp  R  R 
2
  /T
 T  [e vib  1]
(12.20)
 2  vib  2  /T  /T

2
vib
vib

e
e
[
1]
1  

 7  T 

  1.4 

2
1  2  vib  evib /T [e vib /T  1]2 
 5  T 

(12.21)
7
2
655
cv  32 , R, c p  52 R,   53
T very low:
T intermediate values: cv  52 R,
c p  72 R,   75
cv  72 R,
T high:
(12.24)
c p  92 R,   79
Stagnation Conditions in a Diatomic Gas
 2   7 T
1
1
  

 vib /T
M 2      0  1   vib   vib /T0

1 e
 1 
  T e
  2  T
p0  T0 

p  T 
7/ 2
(12.28)
  vib   evib /T0    vib   evib /T   (12.31)
 evib /T0  1 
  vib /T0
 vib /T
 exp 



1 
 1   T   evib /T  1  
 T0   e
e
Normal Shock Wave in a Diatomic Gas
  
p2  p1  1V12 1  1 
 2 
2
V12   1  
1      h2  h1
2   2  




1
1
 
 
h2  h1  R  72 (T2  T1 )   vib  vib /T2
    vib /T1
 
1  e
 1  
 e

 p2  1   T2 
     
 p1   2   T1 
656
(12.36)
(12.37)
(12.39)
(12.40)
Perfect Gas Law
a

 p  2  (v  b)  RT
v 

p
where:
RT (1   )
A
(
v
B
)


v2
v2
A  A0 (1  a /v ) ,
B  B0 (1  b /v ) ,   c /vT 3
(12.43)
(12.50)
(12.51)
Dissociation of a Gas
N
p   pi
(12.53)
pi ni

p nT
(12.54)
iA A  iB B  iC C
(12.55)
i 1
Kp 
iB
iC
n 
n 
K p   B  p iB  C  p iC
 nT 
 nT 
pBiB pCiC
p iAA
(12.56)
i
A
 nA  iA  nBiB nCiC  p iB  iC iA
 n  p   iA  iB  iC iA
 T
 nA  nT
657
(12.57)
Dissociation and Ionization of Air
p  pO2  pO  pN2  pN
pO2
 K pO 2
pO2
pN2
 K pN2
pN 2
2 pN 2  p N
2 pO2  pO

658
a
( 4)
b
(12.58)
(12.59)
(12.60)
(12.67)
PROBLEM 12.1
During the entry of a space, vehicle into the Earth's atmosphere, the Mach number at
a given point on the trajectory is 38 and the atmospheric temperature is 0° C. Calculate
the temperature at the stagnation point of the vehicle assuming that a normal shockwave
occurs ahead of the vehicle and assuming that the air behaves as a calorically perfect gas
with γ = 1.4. Do you think that the value so calculated is accurate? If not, why?
SOLUTION
Across a normal shock if the gas is calorically perfect:
[2 M 12  (  1)] [2  (  1) M 12 ] 
T2
 

T1
(  1) 2 M 12


Hence, since γ = 1.4 for air and T
1
= 273 K, the temperature downstream of the shock
will be given by:
2.8M 12  0.4  2  0.4 M 12 

T2

273
5.76 M 12
i.e.:
T2  273 
 2.8  38
2
 0.4  2  0.4 x 382 
5.76  382
 76910 K
The Mach number downstream of the shock wave is given by:
 (  1) M 12  2 
0.4 M 12  2
0.4  382  2
M 22  



2
2.8M 12  0.4
2.8  382  0.4
 2 M 1  (  1) 
This equation gives M2 = 0.143. Hence the temperature at the stagnation point (i.e., the
stagnation temperature) is given by:
659
 1 2 

T0  T 1 
M   76910  1  0.2  0.1432   77225 K
2


At a temperature as high as this, dissociation and ionization of the air molecules and
atoms will occur. As a result the value for T 2 found above will not be accurate.
660
PROBLEM 12.2
Oxygen, kept at a pressure of 10.1 kPa, is heated to a temperature of 4000 K.
Determine the relative amounts of diatomic and monatomic oxygen that are present after
the heating.
SOLUTION
The value of Kp at a temperature of 4000 K is given in the textbook as:
T = 4000 K : K p = 100.95 = 8.91
The reaction being considered is:
O2  a O2 + b O
Mass balance requires that:
2 = 2a + b
while the definition of the equilibrium constant, Kp , gives:
 b /a + b  p 2
 a /a + b  p
2
2
K p = p0 / p02 =
=
b2
p
a  a + b
But p = 10.1 / 101.1 = 0.1 atm so:
Kp =
b2
0.1b 2
p =
a  a + b
a  a + b
But the mass balance equation given above gives b = 2 (1 - a), the above equation gives:
0.4 1  a 
b2
Kp =
p =
a a + b
2a - a 2
Hence:
661
2
0.4 1  a 
 8.91
2a - a 2
2
i.e.:
9.31a 2 - 18.62 a + 0.4 = 0
Solving this equation for a gives:
a = 0.0217
Hence:
b = 2(1 - a ) = 2 (1 - 0.0217) = 1.957
Therefore at the temperature considered the oxygen consists of approximately 1.1 %
molecules of O2 and 98.9 % molecules of O.
662
PROBLEM 12.3
At a point in an air flow system at which the velocity is extremely low, the pressure is
10 MPa and the temperature is 8000 K. At some other point in the flow system the
pressure is 100 kPa. Assuming that the flow is isentropic, find the temperature and
velocity at this second point.
SOLUTION
Because of the temperature involved ( 8000 K ), there is a possibility that significant
dissociation exists. The solution will therefore be obtained using the Mollier chart in the
textbook.
Now, at point 1 in the flow:
p = 10000 / 101.3 = 98.7 atm ;
T = 8000 K ; V = 0
From the Mollier chart for air for this temperature and pressure:
h1
 240
RTr
At point 2 in the flow:
p = 100 / 101.3 = 0.99 atm
Therefore, because the process is being assumed to be isentropic the Mollier chart for
air for this gives:
T2  4700 K ,
h2
 110
RTr
Hence, since Tr = 273 K:
h1 = 240 R Tr = 240  287  273 = 18.8  106 J / kg (m 2 /s 2 )
and:
663
h2 = 110 R Tr = 110  287  273 = 8.6  106 J / kg (m 2 /s 2 )
The energy equation gives:
h1 = h2 +
V22
2
so, using the above derived values for the enthalpies gives:
V22
V22
18.8  10 = 8.6  10 +
, i.e.,
= 10.2  106 , i.e., V2 = 4520 m/s
2
2
6
6
Therefore the temperature and velocity at the second point are 4700 K and 4520 m/s
respectively.
664
PROBLEM 12.4
Nitrogen at a static temperature of 800 K and a pressure of 70 kPa is flowing at Mach
3. Determine the pressure and temperature that would exist if the gas is brought to rest
isentropically.
SOLUTION
It will be assumed that the effects of dissociation and ionization can be ignored. The
adequacy of this assumption will be checked later. At the temperatures involved in this
flow, the vibrational excitation will have to be allowed for. For nitrogen θvib = 3340 K.
Now:
 2  7
M 2   
   2
1
1
 T0    vib  

 T  1   T   evib /T0  1  evib /T  1  

 


But at a temperature of 800 K:

1 

  1.4 
1 

2

2  vib   vib /T vib /T
2
e
e
[
1]


7  T 


2
2  vib   vib /T vib /T
2
 1]
e
[e

5  T 
i.e. because:
 vib
T

3340
 4.175
800
it follows that:
2 4.175
 2

[e 4.175  1]2 
1  7  4.175 e
  1.4 
  1.360
1  2  4.1752 e 4.175 [e 4.175  1]2 
 5

Hence using the Mach number equation given above:
665
 T
1



 0.0156  
32  1.471 3.5  0  1   4.175   3340/T0
1
e

  800 
Solving this equation iteratively for T0 gives:
T0 = 1993 K
This is low enough to justify the neglect of dissociation.
The pressure is then given by using:
p0  T0 

p  T 
7/ 2
  vib   evib /T0    vib
 evib /T0  1 
exp

  vib /T0   
 vib /T

1
e
T

 1  T



 0   e
 /T
 
  e vib
  vib /T  
 1  
 e
i.e., because:
 vib
T0

3340
 1.676
1993
it follows that:
1993 
p0  70  
 800 
7/2

 e 4.175  1 
 e1.676 
 e 4.175  

exp
1.676
4.175

 e1.676  1 
 e1.676  1 
 e 4.175  1    1241 kPa






 
Therefore the temperature and pressure after the gas is brought to rest isentropically
are 1993 K and 1241 kPa respectively.
666
PROBLEM 12.5
Air at a pressure of 101 kPa and a temperature 20° C has its temperature raised to
4000 K in a constant-pressure process. Determine the composition of the air at this
elevated temperature. Assume the air to initially consist of 3.76 mols of nitrogen per mol
of oxygen.
SOLUTION
The equilibrium constants for O2 and N2 at 4000K are:
K pO2 = 100. 95 = 8.913
K pN2 = 10-12.62 = 2.399  10-13
Therefore:
pO2 = pO 2 / 8.913
pN 2 = pN 2 /
(i)
 2.399  10 
-13
Hence:
 pO2

pN 2
+ pN = 3.76 
 pO 
-13
1.2  10
 4.457

and:
pO 2
pN 2
+ pO +
+ pN = p
8.913
2.399  1013
i.e., beause p = 1 atm:
667
(ii)
 pO / p 
2
 p   pN / p 
p 
 3.76  O  
- N = 0
-13
 p  1.2  10
 p 
2
p 
 pN / p  +  pN  = 1
  O 


-13
 p  2.399  10
 p 
1.185
2
and:
 pO / p 
8.913
2
Solving between these two equations for pO / p and pN / p then gives:
pO
 0.361 ,
p
pN
 3.87  107
p
Substituting these values into equations (i) and (ii) above then gives:
pO2
p
 0.015 ,
pN 2
p
 0.624
From these results it will be seen that, at the temperature considered, the oxygen is
essentially all dissociated while hardly any of the nitrogen is dissociated, the air then
consisting essentially of 3.76 N2 + 1.92 O + 0.04 O2 .
668
PROBLEM 12.6
As a result of an explosion, a normal shock wave moves at a velocity of 6000 m/s
through still air at a pressure and temperature of 1.01 kPa and -25°C respectively. Find
the pressure, temperature and air velocity behind the wave.
SOLUTION
The following equations apply across the normal shock wave:

 
p2  p1  1V12 1  1 
2 

and:
2
V12   1  
1      h2  h1
2   2  


Here V1 = 6000 m/s, p1 = 101 kPa, and T1 = 298 K. Using the perfect gas law then
gives:
1 
101000
p1

 1.181 kg/m3
287  298
RT1
Hence, the first of the above equations gives:
  

 
p2  101000  1.181  60002 1  1   101000  42516000 1  1  Pa
 2 
 2 
i.e., expressing the pressure in atmospheres:
669
  
p2  1  420.95 1  1 
 2 
The second of the above equations gives:
2
   2 
   2 
h2  h1
V12   1  
60002
1
1     
1      230 1   1  

RTr
2 RTr    2  
2  287  273    2  
   2  




The simplest method of finding the solution, although not very elegant, is to use a
trial-and-error approach. One possible such procedure involves the following steps:
1. Use the Mollier chart given in the textbook to find h1 / R T.
2. Guess a value of ρ1 / ρ2.
3. Use the first of the above two equations to calculate the value of p2.
4. Use the second the above two equations to calculate h2 / R T.
5. These two values together define a point on the Mollier chart. Establish the xand y- coordinates of this point on the chart.
6. Find the value of ρ2 / ρr corresponding to this point on the Mollier chart by using
the second of the two charts given in the textbook.
7. Find the corresponding value of ρ1 / ρ2 using:
1
 
1
 1 r 
2
r 2
2 / r
8. Compare the value of the density ratio so obtained with the initial guessed value.
9. Repeat the procedure with different initial guessed values until the two values
agree.
Using this procedure gives p2 = 37000 kPa and T2 = 10000 K. Due to .the coarse
scales used, .it is not possible to get the result very accurately using the Mollier charts
given in the textbook. Also, using:
670
V12  V22 
h2  h1
 230
RTr
2 RTr
gives V2 = 3600 m/s.
Therefore the pressure, temperature and air velocity behind the wave are
approximately 37 MPa, 10000 K, and 3600 m/s respectively.
671
PROBLEM 12.7
A blunt-nosed body is moving through air at a velocity of 5000 m/s. The pressure and
the temperature of the air are 22 kPa and 43° C respectively. The shock wave that exists
ahead of the body can be assumed to be normal in the vicinity of the stagnation point.
Find the pressure behind this shock wave.
SOLUTION
The following equations apply across the normal shock wave:
  
p2  p1  1V12 1  1 
 2 
and:
2
V12   1  
1      h2  h1
2   2  


Here V1 = 5000 m/s, p1 = 22 kPa, and T1 = 316 K. Using the perfect gas law then
gives:
1 
22000
p1

 0.243 kg/m3
287
316
RT1

Hence, the first of the above equations gives:
  
  
p2  22000  0.243  50002 1  1   22000  6075000 1  1  Pa
 2 
 2 
i.e., expressing the pressure in atmospheres:
672
  
p2  0.218  60.15 1  1 
 2 
The second of the above equations gives:
h2  h1
V12

RTr
2 RTr
   2 
   2 
   2 
50002
1
1
1     
1      319 1   1  
2  287  273    2  
   2  
   2  


The simplest method of finding the solution, although not very elegant, is to use a
trial-and-error approach. One possible such procedure involves the following steps:
1. Use the Mollier chart given in the textbook to find h1 / R T.
2. Guess a value of ρ1 / ρ2.
3. Use the first of the above two equations to calculate the value of p2.
4. Use the second the above two equations to calculate h2 / R T.
5. These two values together define a point on the Mollier chart. Establish the xand y- coordinates of this point on the chart.
6. Find the value of ρ2 / ρr corresponding to this point on the Mollier chart by using
the second of the two charts given in the textbook.
7. Find the corresponding value of ρ1 / ρ2 using:
1
 
1
 1 r 
2
r 2
2 / r
8. Compare the value of the density ratio so obtained with the initial guessed value.
9. Repeat the procedure with different initial guessed values until the two values
agree.
Using this procedure gives p2 = 10000 kPa. Due to .the coarse scales used, it is not
possible to get the result very accurately using the Mollier charts given in the textbook.
673
Chapter Thirteen
LOW DENSITY FLOWS
SUMMARY OF MAJOR EQUATIONS
Knudsen Number
Kn 
Kn 

(13.1)
L
M
Re
(13.7)
In situations in which a distinct boundary layer exists:
Kn 

 Re0.5
M



a L
Re0.5
(13.9)
Slip Flow Criteria
Slip flow exits roughly in the following ranges:
If Re  1: 0.01 
M
M
 0.1 , If Re  1: 0.01 
 0.1
0.5
Re
Re
(13.10)
Free Molecular Flow Criteria
Free molecular flow exits roughly when
M
 3
Re
674
(13.11)
Slip Flow Result
2
 u
us    1 
d
 y
(13.14)
y0
Free Molecular Flow Result
CD 
675
4
S
(13.27)
PROBLEM 13.1
A small rocket probing the atmosphere has a length of 3 m. It is fired vertically
upward through the atmosphere, its average velocity being 1000 m/s. Consider the flow
over the rocket at this average velocity at altitudes of 30,000 m and 80,000 m. Can the air
flow over the rocket be assumed to be continuous at these two altitudes? At an altitude
of 30,000 m, the air has a temperature, pressure, and viscosity of -55°C, 120 Pa, and 1.5
 10-5 kg/m-s respectively while an altitude of 80,000 m the air has a temperature,
pressure, and viscosity of -34°C, 0.013 Pa, and 1.7  10-5 kg/m-s respectively.
SOLUTION
At 30,000 m:
T = -55° C ( = 218 K); p = 120 Pa ;  = 1.5  10-5 kg / m s
Using these values gives:
a 
 RT 
1.4  287  218  296.0 m/s
and:
 
120
p

 0.00192 kg/m3
287  218
RT
Hence, since the length of the body is 3m and assuming the velocity is 1000 m/s, at
this altitude:
Re 
 V L 0.00192  1000  3

 3.84  105
0.000015

and:
M 
V
1000

 3.378
a
296
676
Because Re > 1 in order to determine whether the flow is in the slip flow region
consider the quantity:
M
3.378

 0.0055
0.5
3840000.5
Re
This is less than 0.01 so at an altitude of 30,000 m the flow can be assumed to be
continuous.
Next consider conditions at 80,000 m where:
T = -34°C ( = 239 K);
p = 0.013 Pa ;
 = 1.7  10-5 kg / m s
Using these values gives:
a
 RT 
1.4  287  239  309.9 m/s
and:
 
p
0.013

 0.000000189 kg/m3
287  239
RT
Also, again assuming the velocity is 1000 m/s, at this altitude:
Re 
 V L 0.000000189  1000  3

 33.4
0.000017

and:
M 
V
1000

 3.227
a
309.9
Because at this altitude, as at the lower altitude, Re > 1 in order to determine whether
the flow is in the slip flow region consider the quantity:
M
3.227

 0.558
0.5
Re
33.40.5
677
Since this is greater than 0.01 at an altitude of 80,000 m the flow cannot be assumed
to be continuous. Indeed, because M/Re
0.5
, > 0.1, the flow is not even in the slip flow
region. However, because M /Re = 3.227 / 33.4 = 0.097, which is less than 3, the flow is
not in the free molecular region. Hence, at this altitude the flow is transitional.
Therefore the flow can be assumed to be continuous at an altitude of 30,000 m but
this assumption cannot be used at an altitude of 80,000 m.
678
PROBLEM 13.2
A small research vehicle with a length of 4 ft travels at a Mach number of 15 at
altitudes of 100,000 ft and 250,000 ft. Determine whether, at these two altitudes, the
missile is in the continuum, slip, transition or free molecular flow regimes. It can be
assumed that at an altitude of 100,000 ft, the pressure, temperature and viscosity are 22
psf, 340° R and 96 x 10-7 lbm/ft-sec respectively while at an altitude of 250,000 ft, they
are 0.11psf, 450° R and 100 x 10-7 lbm/ft-sec respectively.
SOLUTION
At 100,000 ft:
T = 340° R ; p = 22 psf ;  = 96 x 10-7 lbm /ft-s
Using these values and recalling that for air R = 53.3 ft-lbf / lbm-° R gives:
 
p
22

 0.00121 lbm/ft 3
RT
53.3  340
and, recalling that 1 lbf = 32.2 lbm-ft / sec2:
a 
 RT 
1.4  53.3  32.2  340  903.9 ft/sec
Hence, since the length of the body is 4 ft and the velocity is given by:
V  M a  15  903.9 = 13557 ft/sec
and therefore at this altitude:
Re 
V L
0.00121  13557  4

 6.84  106

0.0000096
Because Re > 1 in order to determine whether the flow is in the slip flow region
consider the quantity:
M
15

 0.00574
0.5
Re
68400000.5
679
This is less than 0.01 so at an altitude of 100,000 ft the flow can be assumed to be
continuous.
Next consider conditions at 250,000 ft where:
T = 450° R ;
p = 0.11 psf ;
 = 100  10-7 lbm /ft-s
Using these values gives:
 
p
0.11

 0.000004586 lbm/ft 3
RT
53.3  450
and:
a 
 RT 
1.4  53.3  32.2  450  1039.8 ft/sec
Hence, the velocity is given by:
V  M a  15  1039.8 = 15597 ft/sec
and therefore at this altitude:
Re 
V L
0.000004586  15597  4

 28611

0.00001
Because at this altitude, as at the lower altitude considered, Re > 1 in order to
determine whether the flow is in the slip flow region consider the quantity:
M
15

 0.0887
0.5
Re
286110.5
This is greater than 0.01 so at an altitude of 250,000 ft the flow cannot be assumed to
be continuous. Because M / Re 0.5 < 0.1, the flow is in the slip flow region.
Therefore the flow can be assumed to be continuous at an altitude of 100,000 ft and to
be in the slip flow region at an altitude of 250,000 ft.
680
PROBLEM 13.3
Find the drag force per unit length on a 0.5 cm diameter cylinder placed in an air-flow in
which the temperature is 800 K, the density is 8  10-9 kg/m3, and the velocity 10,000 m/s.
SOLUTION
The speed of sound in the flow is given by:
a 
 RT  1.4  287  800  567 m/s
Hence the Mach number is given by:
V 10000

 17.6
a
567
M 
At the temperature being considered the viscosity of the air will be assumed to be 420
x 10-7 N s / m2. The Reynolds based on the cylinder diameter is then given by:
Re 
VD
0.000000008  10000  0.005

 0.0095

0.000042
Therefore Figure 13.7 in the text indicates that the flow in the situation being considered
is Free Molecular. It will be assumed, therefore, that the drag coefficient is given by:
CD 
5
S
As discussed in this chapter it will be assumed that the mean molecular velocity is
given by:
cm 
2 RT 
2  287  800  678 m/s
So:
S 
V
10000

 14.8
cm
678
681
and therefore:
CD 
5
5

 0.34
S 14.8
The on the cylinder per unit length is then given by:
D  CD 
1
1
V 2 A  CD  V 2  D  1
2
2
the frontal projected area having been used in defining CD. This equation gives:
D  0.34 
1
 0.000000008  100002  0.005  1  0.00068 N
2
Therefore the drag force on the cylinder per unit length is 0.00068 N.
682
Chapter Fourteen
AN INTRODUCTION TO TWODIMENSIONAL COMPRESSIBLE FLOWS
SUMMARY OF MAJOR EQUATIONS
Governing Equations


( u) 
(  v)  0
x
y
u
u
u
u
p
 v
 
x
y
x
u
v
v
p
 v
 
x
y
y
 
(u 2  v 2 ) 
 
(u 2  v 2 ) 
c
T


v
c
T

p
p

  0
x 
2
y 
2


 u12  v12 
 u 2  v2 
c pT  
  c pT1  
  c pT0
 2 
 2 
(14.2)
(14.5)
(14.6)
(14.9)
(14.14)
Velocity Potential
u 

,
x
v 
683

y
(14.26)
  2  2 
     

 2  2   
  0
y 
y y 
 x x
 x
(14.27)
2
2
     2 
 2Ф  2
1     2 
     2

 2 
2

  0 (14.32)


a  x  x 2
x 2
y 2
x y x y  y  y 2 
2
2
   1          
a  a 
 
 
 
 2   x   y  
2
2
0
(14.35)
Linearized Velocity Potential
p
up 
(1  M 2 )
x
 2 p
x
tan  

2
vp 
,
 2 p
y
2
vp
dy

dx s
u  u p
vp
u
Cp  
 0
(14.37)
y
(14.44)
(14.45)
s
dy
dx s
(14.47)
2 p
u x
(14.54)
 y
(14.57)

s
p
Linearized Subsonic Flow
  x,
684
(14.58)
p   p
2  p

 2
Cp 
2  p
(14.59)
 0
 2
1  2  p 


  u d 
Cp 
1

(14.64)
C p0
C p0 
1  M 2
CL 0
CL 
(14.65)
(14.67)
1  M 2
Linearized Supersonic Flow
Cp 
2


Cp 
CD 

upper surface
(14.78)
M 2  1
2
(14.81)
M 1
C p d ( x /c) 
685
2
2


lower surface
C p d ( x /c)
(14.85)
PROBLEM 14.1
Air flows with a Mach number of 2.8 over a flat plate which is set at an angle of 7° to
the upstream flow. The pressure in the upstream flow is 100 kPa. Find the lift and drag
coefficients using linearized theory.
SOLUTION·
First consider the upper surface. The angle that the upper surface makes to the flow is
-7° = -0.1222 radians. Hence:
2 upper
Cp upper 
M 2  1
2    0.1222 

2.82  1
  0.09345
Next consider the lower surface. The angle that the lower surface makes to the flow is
+ 7° = + 0.1222 radians. Hence:
Cp lower 
2 lower
M 2  1

2  0.1222
2.82  1
 0.09345
The net force , F, on the plate normal to the plate is given by:
F 
p
lower
 pupper  A
where A is the surface area of the plate. This equation can be written as:
F   plower – p  –
p
upper
– p   A
The lift, L, is related to F by:
L  F cos
Because linearized theory is being used, this equation can be approximated by:
L  F
Hence:
686
L

2
1
2   u A
F
2
1
2   u A
i.e.:
CL  Cp lower  Cp upper
From which it follows that:
CL  0.09345 

 0.09345   0.1869
Similarly, the drag, D, is related to F by:
D  F sin
Because linearized theory is being used, this equation can be approximated by:
D  F
Hence:
1
2
D

 u2 A
1
2
F
 u2 A
i.e.:
CD 
C
p lower
 Cp upper 
From which it follows that:
CD  0.09345 

 0.09345    0.1222  0.02284
Therefore the lift and drag coefficients are 0.1869 and 0.02284 respectively.
687
PROBLEM 14.2
A thin symmetrical supersonic airfoil has parabolic upper and lower surfaces with a
maximum thickness occurring at midchord. Using linearized theory compute the drag
coefficient on this airfoil when it is set at an angle of attack of 0°.
SOLUTION
Figure P14.2
The shape of the upper and lower surfaces is described by an equation that has the
form:
y  A  Bx  Cx 2
For the upper surface this equation must satisfy the following conditions:
x  0 : y  0,
x  c / 2 : y  t / 2,
x  c: y  0
Applying these to the equation for the shape of the surface gives:
0  A,
t
Bc
C c2
 A

, 0  A  Bc  Cc 2
2
2
4
Solving between these equations gives:
A0 , B 
2t
2t
, C  2
c
c
Therefore, the shape of the upper surface is given by:
688
 2t 
 2t 
y    x   2  x2
c
c 
i.e.:
y
x x
    
2t  c   c 
2
Consider any point on this upper surface. The angle the surface makes to the flow is
given by:
dy
dx
tan  
i.e., because linearized theory is being used and θ is thus being assumed to be small:
dy
dx
 
Using the equation that describes the variation of y with x for the upper surface then
gives:
 2t  
 x 
    1  2   
 c 
 c 
Hence:
Cp 
2

M 1
2

4 t / c  
 x 
1  2  c  
2
 
M 1 
This equation was derived by considering the upper surface. By symmetry, the same
equation must apply to the lower surface, i.e., on both surfaces:
Cp 
2
M 1
2


4 t / c  
 x 
1  2  c  
2
 
M 1 
Hence, since:
689
CD 

C p d  x / c  
upper surface

C p d  x / c 
lower surface
i.e., due to symmetry:
1
CD  2  C p d  x / c 
0
From this it follows that:
CD  16
t / c 
M 2
2
1

 x  
 x 
1  2  c   d( x / c)
 

 1  2  c  
1 
0
i.e.:
CD  16
t / c
2
4
4
16

1   

2
3
3
M 2  1 
t / c 
This expression gives the drag coefficient for the airfoil.
690
2
M 2  1
PROBLEM 14.3
The pressure coefficient at a certain point on a two-dimensional airfoil in a very low
Mach number air flow is found to be -0.5. Using linearized theory, estimate the pressure
coefficients that would exist at the same point on this airfoil in flows at Mach numbers of
0.5 and 0.8.
SOLUTION
If Cp0 is the pressure coefficient at the point considered at very low Mach numbers,
when 1 - M∞2 is effectively equal to 1, then at any larger Mach number, M∞ , the pressure
coefficient at this point is given by:
Cp
Cp0

1
1  M 2
, i.e., Cp 
Cp0
1  M 2
In the present situation where, Cp0 = -0.5, this equation gives:
Cp 
 0.5
1  M 2
Hence, when M∞ = 0.5:
Cp 
 0.5
1  0.52
  0.577
and when M∞ = 0.8:
Cp 
 0.5
1  0.82
  0.833
Therefore the pressure coefficients at Mach numbers of 0.5 and 0.8 are -0.577 and
-0.833 respectively.
691
PROBLEM 14.4
A thin airfoil can be approximated as a flat plate. The airfoil is set at an angle of 10°
to an airflow with a Mach number of 2, a temperature of -50° C, and a pressure of 50 kPa.
Using linearized theory, find the pressures on the upper and lower surfaces of this wing.
SOLUTION
First consider the upper surface. The angle that the upper surface makes to the flow is
-10° = - 0.1745 radians. Hence:
Cp upper 
2 upper
M 1
2


2    0.1745 
22  1
  0.2015
i.e.:
pupper  p
1
2
 u2
  0.2015
But:
1
2
 u
2

p  2
 p M 2

u 
2 p
2
So:
pupper  p
1
2
 u2

pupper  p
 p M 2 / 2
  0.2015 , i.e.,
pupper  p 
0.2015  p M 2
2
Therefore:
pupper  p 
0.2015  p M 2
0.2015  1.4  50  4
 50 
 21.79 kPa
2
2
Next consider the lower surface. The angle that the lower surface makes to the flow is
+10° = +0.1745 radians. Hence:
692
Cp lower 
2 lower
2  0.1745

M 1
2

22  1
 0.2015
i.e.:
pupper  p
1
2
 u2
 0.2015
Hence using the same procedure as used in dealing with the upper surface:
plower  p
 0.2015 , i.e.,
 p M 2 / 2
plower  p 
0.2015  p M 2
2
Hence:
plower  50 
0.2015  1.4  50  4
 78.21 kPa
2
Therefore the pressures on the upper and lower surfaces are 21.79 kPa and 78.21 kPa
respectively.
693
PROBLEM 14.5
An airfoil has a triangular cross-sectional shape. The lower surface of the airfoil is flat
and the ratio of the maximum thickness to the chord is 0.1. The maximum thickness
occurs at a distance of 0.3 times the chord downstream of the leading edge. If this airfoil
is placed with its lower surface at an angle of attack of 2° to an airflow in which the
Mach number is 3, use linearized theory to determine the distribution of the pressure
coefficient over the surface of the airfoil.
SOLUTION
Figure P14.5
The situation being considered is shown in Fig. P14.5. From this figure it will be seen
that:
tan 1 
0.1c
 0.3
c/3
Hence, because linearized theory is being used, it will be assumed that:
 1  0.3 radians
Similarly:
694
0.1c
 0.15
2c / 3
tan 2 
Hence, again because linearized theory is being used, it will be assumed that:
1  0.15 radians
Therefore for the three surfaces that make up the surface of the airfoil, the angles to
the flow ahead of the airfoil are given by:
1 = 2° = 0.0349 radians
 2 = - 1 + 1 = - 0.0349 + 0.3 = 0.2651 radians
3 = - 1 + 2 = - 0.0349 + 0.15 = 0.1849 radians
For each surface of the airfoil:
Cp 
2
M 2  1
2

32 1
 0.7071
Hence:
Cp1 = 0.7071  0.0349 = 0.0247
Cp2 = 0.7071  0.2651 = 0.1875
Cp3 = 0.7071  ( - 0.0349) = - 0.1307
Therefore, according to linearized theory, the pressure coefficients on the lower, the
forward upper and the rear upper surfaces are 0.0247, 0.1875 and -0.1307 respectively.
695
PROBLEM 14.6
A symmetrical double-wedge airfoil has a maximum thickness equal to 0.05 times the
chord. This airfoil is placed at an angle of attack of 5° to an airstream with a Mach
number of 2, a pressure of 50 kPa, and a temperature of -50° C. Find the lift and drag
acting on the airfoil using linearized theory and using shock wave and expansion wave
results.
SOLUTION
Figure P14.6a
The situation being considered is shown in Fig. P14.6a. From this figure it will be
seen that:
tan  
0.025 c
 0.05
0.5 c
Because linearized theory is being used, it will be .assumed that:
  0.05 radians
For the surfaces indicated in Fig. P13.6a that make up the surface of the airfoil, the
angles to the flow ahead of the airfoil are given by:
696
θ1 = - 5° + φ = - 0.08727 + 0.05 = - 0.03727 radians
θ2 = + 5° + φ = 0.08727 + 0.05 = 0.1373 radians
θ3 = - 5° - φ = - 0.08727 - 0.05 = - 0.1373 radians
θ4 = + 5° - φ = 0.08727 - 0.05 = 0.03727 radians
For each surface of the airfoil:
2
Cp 
M 2  1
i.e.:
p  p

2
1
2   u
2
M 2  1
But:
 
1
1
 u2   p  
2
2
  p
 2
1
 p M 2
 u 
2

Hence:
 p M2 
p  p     
2 

2
M 2  1
so:
 1.4  50  22 
p  50  
  50  80.83 kPa
2


Using this result gives for the four surfaces that make up the airfoil surface:
p1 = 50 + 80.83 x ( - 0.03727) = 46.99 kPa
p2 = 50 + 80.83 x 0.1373 = 61.10 kPa
p3 = 50 + 80.83 x ( - 0.1373) = 38.90 kPa
p4 = 50 + 80.83 x 0.03727 = 53.01 kPa
The net force, F, on the airfoil at right angles to the center-line per m span is given
by:
697
F  p2 
c/2
c/2
c/2
c/2
 cos   p4 
 cos   p1 
 cos   p3 
 cos 
cos 
cos 
cos 
cos 
i.e.:
F   p2  p4  p1  p3 
c
c
  61.10  53.01  46.99  38.9   14.11 c
2
2
The lift, L, is related to F by:
L = F cos 5°
Because linearized theory is being used, this equation can be approximated by:
L = F
hence:
L  14.11c kN
Similarly, the drag, D, is related to F by:
D = F sin 5°
But 5° = 0.08727 radians so, because linearized theory is being used, this equation can be
approximated by:
D = F  0.08727 = 14.11  0.08727  c = 1.23 c kN
Hence, linearized theory gives the lift and drag on the airfoil per m span as 14.11c kN
and 1.23c kN respectively.
Next, consider the application of shock-expansion theory. The assumed wave pattern
is shown in Fig. P14.6b.
698
Figure P14.6b
As before:
tan  
0.025 c
 0.05
0.5 c
Hence:
 = tan -1 0.5 = 0.04996 radians = 2.863°
The angle of the center-line to the approaching flow is 5° = 0.08727 radians.
Expansion waves thus occur at the leading edge and at the corner on the upper surface
while an oblique shock wave will occur at the leading edge of the lower surface and an
expansion wave occurs at the corner on the lower surface. The angles of turning produced
by the waves are as follows:
Expansion Wave A: Angle of Turn = 5 - 2.863 = 2.137°
Expansion Wave B: Angle of Turn = 2 φ = 5.724°
Shock Wave C: Angle of Turn = 5 +2.863 = 7.863°
Expansion Wave D: Angle of Turn = 2 φ = 5.724°
699
First consider the flow over the upper surface. Consider Expansion Wave A. Ahead
of the wave the Mach number is 2. For this Mach number, the software for isentropic
flow or the isentropic flow tables for air give:
p0
 7.824 ,   26.38o
p
Hence, since the flow is turned through 2.137° by the expansion wave, it follows that:
1 = 26.38 + 2.137 = 28.52o
Using this value of θ 1, the software for isentropic flow or the isentropic tables give:
M 1  2.078 ,
p01
 8.84
p1
It therefore follows that, since the flow through the expansion wave is isentropic
which means that p01 = p0 :
p1 
p1 p0
7.824
p 
 50  44.25 kPa
p01 p
8.84
Next consider Expansion Wave B which separates regions 1 and 3. Because the flow
is turned through an angle of 5.724° by this wave and because θ1 is 28.52° it follows that:
3 = 28.52 + 5.724 = 34.24o
Using this value of θ3, the software for isentropic flow or the isentropic tables for air
give:
M 3  2.298,
p03
 12.45
p3
It therefore follows that, since the flow through the expansion wave is isentropic
which means that p03 = p01 :
700
p3 
p3 p01
8.84
p1 
 44.25  31.42 kPa
p03 p1
12.45
Next consider the flow over the lower surface. Consider Shock Wave C. Ahead of the
wave, the Mach number is 2 and the shock wave turns the flow through an angle of
7.863°. The software for oblique shock waves or the oblique shock chart in conjunction
with the normal shock tables for air give for these values:
M 2  1.719 ,
p2
 1.530
p
From this it follows that:
p2 
p2
p  1.530  50  76.50 kPa
p
Also for M 2. = 1.719 the software for isentropic flow or the isentropic tables for air
give:
 2  18.36o ,
p02
 5.08
p2
Next consider Expansion Wave D. Because the flow is turned through an angle of
5.724° by this wave it follows that:
3 = 18.36 + 5.724 = 24.08o
Using this value of θ4, the software for isentropic flow or the isentropic tables for air
give:
M 4  1.914 ,
p04
 6.85
p4
It therefore follows that, since the flow through the expansion wave is isentropic
which means that p04 = p02 :
701
p4 
p4 p02
5.08
p2 
 76.50  56.73 kPa
p04 p2
6.85
Hence:
p1 = 44.25 kPa, p2 = 76.50 kPa , p1 = ·31.42 kPa, p4 = 56.73 kPa
It was shown above that the net force, F, on the airfoil at right angles to the centerline per m span is given by:
F 
 p2
 p4  p1  p3 
c
c
  76.50  56.73  44.25  31.42   28.78 c kN
2
2
The lift, L, is then given by:
L = F cos 5o = 28.67 c kN
Similarly, the drag, D, is given by:
D = F sin 5° = 2.51c kN
Therefore shock-expansion theory gives the lift and drag per m span as 28.67c and
2.51c kN respectively. These values are very different from those given by linearized
theory. This is because the pressure changes involved are actually relatively large.
702
K13746
ISBN: 978-1-4398-7983-2
90000
w w w. c rc p r e s s . c o m
9 781439 879832