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PART
APPLICATIONS OF FLUID MECHANICS
PART 3
I
I
APPLICATIONS OF
FLUID MECHANICS
PART 3
First edition
5th print
C. F. MEYER
M.Eng. (Mech.), G.C .C., MSAIME, Dip. Ter. Ed.
Principal Lecturer, Mechanical Engineering,
Tshwane University of Technology
ISBN 0-620-3078 1-1
O 2003: THE AUTHOR. No part of this book may be
reproduced, in any form or by any means, electronic,
mechanical, photocopying or otherwise.
Publisher:
C F Meyer,
Ambassador 69,
14 Squirrel ave,
Monumentpark,
Pretoria 0 181,
RSA
PREFACE
This book is the fourth in a series of Fluid Mechanics books - the first in the
series being "Principles of Fluid Mechanics", followed by "Applications of
Fluid Mechanics Part 1" and "Applications of Fluid Mechanics Part 2".
The series is mainly intended for use by Mechanical Engineering students
studying at the Universities of Technology in South M i c a , but due to the
universality of the topics it would be useful for other students as well as
persons in industry requiring a refresher course or a basic reference book in
the field of Fluid Mechanics.
The advent of personal computers in the last few years has opened up a
powerful tool to engineering practitioners. Many powerful design and
analytical packages are available in industry to help the technician or
engineer to perform his or her job more effectively. I am of the opinion that
these packages are to be used only after the student has acquired a thorough
knowledge of the basic methods used in engineering calculations. Having
said this I think it is also of the utmost importance that students be exposed
to the use of computers in solving engineering problems. For this reason I
have included many problems that require computer solutions. They can all
be done using spreadsheet programs like EXCEL or QUATTRO PRO.
The topics in this book are somewhat advanced in the sense that they might
not all be frequently encountered in the everyday industrial world. However,
they do have a very important place in the mechanical engineering world,
albeit somewhat specialized. Most of the chapters can be expanded to cover
much more work. I have tried to condense the topics into a concise format
that treats all the essential principles of the topic. Further reading should not
pose a problem as the basic foundation has been well laid.
Finally, I want to apologize for errors that might occur, and would welcome
criticism and advice to improve future editions.
August 2003
C. F. Meyer
List of symbols
area
breadth, width
diameter
Fanning's friction factor
gravitational acceleration
pressure head
mass
pressure
power
volumetric flow rate
radius
gas constant
seconds
time
free stream velocity
velocity
watt
alpha, angle
beta, angle
gamma, ratio of specific heats
delta, small increment
more or less equal to
rho, density
sigma, surface tension
tau, shearing stress
theta, angle
mu, dynamic (absolute) viscosity
nu, kinematic viscosity
Contents
Chapter 1 Pipe Networks
1.1 Introduction
1.2 Flow resistance in pipes
1.3 Principles of network analysis
1.4 The correction value AQ
1.5 Pressures in a pipe network
Problems
Chapter 2 Dimensional Analysis and Model Studies
2.1 Introduction
2.2 Dimensions and Units
2.3 Fundamentals of Dimensional Analysis
2.4 The Buckingham Pi method
2.5 The Rayleigh method
2.6 The Inspection method
2.7 Model Studies and Similarity
Problems
17
17
19
20
24
25
26
31
Chapter 3 Boundary Layers
3.1 Introduction
3.2 Fundamental principles and the Navier-Stokes equations
3.3 The formation of boundary layers
3.4 Equations of motion for boundary layers
3.4.1 Laminar boundary layers
3.4.2 Turbulent boundary layer
3.4.3 Laminar and Turbulent boundary layers
Problems
36
36
39
40
40
51
55
59
Chapter 4 External Flow over Bodies
4.1
4.2
4.3
4.4
4.5
Introduction
Drag on Flat Plates
Drag on Two and Three Dimensional Bodies
Forces on Streamlined Bodies
Forces on rotating Bodies
63
63
66
75
80
4.6 Aerodynamic forces on road vehicles
Problems
Chapter 5 Compressible Flow
5.1 Introduction
5.2 Basic thermodynamic relationships
5.3 Sonic velocity, Mach number, and stagnation properties
5.4 Normal shock waves
5.5 Isentropic flow through a conduit
5.5.1 Subsonic flow
5.5.2 Supersonic flow
5.6 Compressible flow with friction
5.7 Compressible flow with heat transfer
5.8 Compressible flow with constant temperature
Problems
Appendices
130
130
Figure A1 Dynamic viscosity of common fluids vs. temperature
131
Figure A2 Moody diagram for Fanning's friction factor f
Table A1 Properties of water
132
133
Table A2 Properties of Air at standard atmospheric pressure
Table A3 The International Standard atmosphere
134
Table A4 Physical properties of common liquids at specified
temperature
135
Table A5 Physical properties of common gases at standard
atmospheric pressure (101,3 kPa) and temperature (15'C) 136
137
Table A6 Isentropic flow for gas with y = 1,4
145
Table A7 Normal Shock Tables for a Gas with y=1,4
155
Table A8 Fanno flow - friction flow for gases with y = 1.4
162
Table A9 Rayleigh flow - frictionless with heat transfer y = 1,4
References and Farther reading
171
Answers to questions
172
Chapter 1
Pipe Networks
1.1
1.2
1.3
1.4
1.5
Introduction
Flow resistance in pipes
Principles of network analysis
The correction value AQ
Pressures in a pipe network
1.1 Introduction
A pipe network is a system of pipes interconnected such that a particle
could flow through the network by any of a variety of paths. The
network could be the water distribution system to a city suburb or the
water distribution in a large factory or mine. It could also be the
lubrication system of a complicated piece of machinery of the fuel supply
to a large engine or furnace. The most common application is to water
flowing in large piping systems.
The analysis of piping networks differ from that of electrical networks
because of the fact that piping friction is not linearly proportional to the
flow rate, whereas in electrical circuits the resistance is generally directly
proportional to the current flowing through a resistor. We therefore
cannot produce a set of linear equations for solving by normal algebra or
matrix algebra. We have to obtain solutions by numerical iteration. One
such method is the one developed by Professor Hardy Cross in 1946 - the
so-called Hardy Cross Method, which we shall be using.
After studying the chapter, you should be able to:
* Identify pipe networks
* Calculate the flow rates in a simple pipe network manually by an
iterative calculation.
* Design a computer spreadsheet program to solve networks that are
more complex.
1.2 Flow Resistance in pipes
The pressure head loss due-to fiiction through a pipe of length L and
diameter d is given by
where f is Fanning's fiiction factor, V the mean velocity, Q the
volumetric flow rate, and g the gravitational constant.
The value of the fiiction factor f is of course not necessarily constant; it
would only be in the fully developed turbulent region of the Moody
diagram. Therefore, we could initially make a guess and readjust it once
we have established a stable flow rate. If a programmable calculator or a
computer program is used, we could easily incorporate a subprogram to
compute the f-values by employing for instance the Churchill-Usagi
equations.
I
'
[I
I
1
where A and B are
is the stirjke roughness as given in table 1.1 andsld is the relative
roughness.
E
Table 1.1 Surface roughnesses of pipes
(mm)
Concrete
Cast iron
Galvanized steel
Commercial steel
Drawn tubing, copper, aluminium
Glass, plastic, PVC, HDPE
0,3 - 3
0,26
0,15
0,045
0,001 5
0 (smooth)
The Reynolds number is
with p the density, p the dynanzic viscosity, and v the kinematic
viscosity.
If the expected flow in the pipes would be turbulent (which is usually the
case with water in large diameter pipes), we could take the initial guess in
the Reynolds number region of 10' to lo7. This is usually a good bet. For
strange fluids, it is better to investigate what the Reynolds number could
possibly be, and then obtain an f value from a Moody diagram, or calculate
it by the above formula.
Note:
Beware or the difference between Fanning's f and Darcy's J:
American textbooks generally use Darcy's J; which is four times
bigger than Fanning'sJ
Other formulas are available to calculate f, for example the Colebrook-White
formulas. The advantage of the Churchill-Usagi equations is that they span
all flow regimes, i.e. fiom laminar to fully turbulent and all relative
roughness values.
1.3 Principles of Network Analysis
(
Suppose we have a network as shown in Figure 1.1. All the pipes could have
different lengths, diameters, and surface roughness that are known. If we
have an inflow QA at point A and outflows at points E and F, the various
flow resistances of the different pipes would influence the flow rate in each
individual pipe. The objective is thus to find the flow rate in each individual
pipe.
Figure 1.1 A pipe network
The basic strategy for solving the flow in the network by the Hardy Cross
method is:
Assume a certain flow through each pipe in the network, such that
(a)
the algebraic sum of the flows at all the junctions is zero, i.e. at every
junction of two or more pipes the total outflow must be equal to the total
inflow.
So, suppose in Figure 1.1 that QA was given as 5 m3/s, then we can take Q A ~
as 3 m3/s and QAD= 2 m3/s. Similarly for junction D we would say because
QAD= 2 m3/s, we guess QDc = 1,5 m3/s and QDF= 0,5 m3/s, and so on. We
may choose the flow rates way off fiom what it eventually would turn out to
be, the assumed flow direction may even be in the wrong direction, it does
not matter, our method will rectify this eventually, but it would cause the
solution to take longer.
(b)
We now calculate the pressure drop in each pipe by using Eq. (1.I).
(c) Using the results of (b) above, we take the loops formed by the pipes
and calculate the algebraic sum of the pressure head drops around the loops.
Ideally it should be zero (we should end up with the same pressure we
started of with), but initially it probably will not be, because our assumed or
guessed values, designate them Qo, were not correct.
(d) We now make corrections to our initial Qo values by calculating a
correction AQ for each loop and adding that to our initial Qo values.
(e) We repeat the steps fiom (b) until we find the AQ have become
negligibly small.
1.4 The correction value A Q
We had
This can be written as h, = kQ" where k = jZ/3,03dS for the specific pipe and
n = 2. The SI unit of k is s2.
-
Note, some American textbooks use the Hazen- Williamsformula where n
1,85. One could also expressf as an exponentialfunction ofQ and combine
it with Q to obtain a d~flerentpower of Q. Usually it is better to retain Q 2
and adjust f after each iteration ifneed be. We shall do that.
If the initial Qo is now adjusted to a new value
amount AQ ,then for each pipe,
Q1
by adding a small
h, = k ~ =; k ( ~+, AQ)"
= k(Q," +nQ,""AQ+.....)
Neglecting the terms containing higher powers of AQ because they are
negligibly small compared to Q o we have
Therefore, because we should have the pressure drops around a loop adding
up to zero,
C h , =CkQ," =CkQ," + C ~ Q ; - ' A Q = O
1
Thus
~l
Assuming we use equation 1.1 then n = 2 , then
1
Which is
We inserted the absolute value sign (modulus) to ensure the expression is in
accordance with our sign convention (see next paragraph). This value of AQ
is calculated and all the initial pipe flows in the loops are adjusted.
~
11
SIGN CONVENTION
Ifwe travel clockwise around the loop and the direction of the flow in a
certain pipe is with us, then h, in that pipe is positive.
Now,
-when going around a loop, if the sum of the pressure drops gives us a
> o ) it means the flows that are in a clockwise
positive value (i.e. C
I~Q, ~ Q,
direction must be reduced (i.e. add a negative quantity to the initial flow),
and the flows anti-clockwise must be increased. This will happen when we
add AQ to our previous Q values. The reason for the absolute value sign is
to make the pressure drops dependent on the flow direction.
If a loop 1 has a pipe (or pipes) common to another loop 2, add the AQ
calculated by loop 1's pipes, and subtract the AQ calculated by loop 2's
pipes. Thus, the effect of all the loops will influence each other, and that
will eventually cause the flows to converge to a final constant value.
After we have adjusted the flow rates by adding the AQ 's, we calculate new
AQ 's and so on.
To summarize the steps:
1 . Assume a reasonably good distribution of flows that satisfies continuity
at the junctions.
2. Calculate the k values for each pipe. Assume a reasonable f value, or
better, calculate the f value by the Churchill-Usagi formula considering
the assumed flow (Use Q to obtain V, to obtain Re, to obtain f).
3. Calculate the sum of the pressure drops i.e. C
I ~ Q, / Q ,
4. Calculate 2x1 kQl wbich will always be positive.
5. Calculate AQ .
6. Add the AQ algebraically to all the pipe flows in the loop under
consideration, and subtract the AQ of the adiacent loo^ from the flow in
tbe common pipe only.
7. Repeat until AQ approaches zero.
Example 1.1
The following simple network consists of pipes for which the f values can be
assumed to be constant at f = 0,005. All the pipe diameters are 500 mm. The
lengths vary as shown. Calculate the flow rate in each pipe.
Solution:
Assume a flow rate in each pipe to satisfy continuity. We choose QAB= QDC
= 0,l m3/s, QAC= 0,08 m3/s, QAD= 0,12 m3/s and QDc= 0,02 m3/s
Next, we calculate the k values
Similarly, k ~ =c 15,8 s2 , kAc = 15,8 s2 ,km
We update our diagram to the following.
= 10,6 s2 ,kcD = 5,3
s2
Now tabulate the results. Taking two loops i.e. ABCA and CDAC, we find
Loop ABCA:
Loop CDAC:
These values are added to theJirst Q values i.e.
QAB+ AQ = 0,l + (-0,021) = 0,079
Similarly
Note now, the pipe that is common to both loops is adjusted by the AQ of
both loops as follows
QCA+ AQADCA- AQCDAC= -0,080 + (- 0,021) - (0,010) = -0,111
And similarly for loop CDAC
QAc + AQCDAC- AQmcA = 0,080 + (0,010) - (-0,021) = 0,111
We get the following values.
Loop ABCA:
Loop CDAC:
Continuing like this the Q values are gradually changed and AQ becomes
smaller until it approaches zero. At that point, we say the Q values have
converged.
Thus, we get,
Loop ABCA:
Loop CDAC:
We have obtained both AQ 's as 0,OO. Of course, if we had worked with
more decimals, they would still not be zero. In practice, one would have to
make the choice how accurate the results need to be. The specific
circumstances will dictate it. The Q values in these last two tables are the
solution to the problem.
Exercises:
1. Create the above example on a computer spreadsheet like Excel or
Quattro Pro.
2. Now modify your spreadsheet to calculate the f values by the
Churchill Usagi formulas. Assume the pipes are galvanized steel and
the fluid is water.
(Hint: Extend your tables to include columns for diameter d,
roughness value E, relative roughness dd, velocity V, jluid density p,
fluid viscosity p, Reynolds number Re, Churchill Usagifactors A and
B and then$ Zhe f values as well as the k values will now constantly
be varying because Q influences them).
1.5 Pressures in the pipe network
Once we have the flow rates, the pressure at any point can be determined
provided the pressure and height at any other point in the system is known.
For instance if the pressure at point A is PA, and point C is at the same height
as A, then the pressure drop from A to C would be given by
If the height of point C is lower than A by an amount AH,the pressure due
to this would be higher at C by an amount
Example 1.2:
For the network of example 1.1, calculate the pressure at point C if it is 5 m
is 400'k~a
lower than point A where the
PC=P~-/?!&+B&=PA-B-
flAcQ:c
3,03d:,
+
mm
+
= 400x10~- 103x9,810~005x200X0~096210~~9,8981~5
= 4441kPa -t
3,03~0,5~
Note the positive and negative signs.
Problems
1.1
Create a computer spreadsheet, to calculate Fanning's friction factor f
if the pipe dimensions, surface roughness, and the fluid properties of
density and viscosity are given.
1.2
Create a computer spreadsheet that will calculate the pressure head
drop hf if the flow rate Q is supplied along with the other necessary
fluid properties and pipe particulars.
1.3
Create a graph depicting the hydraulic gradient fib)-vs. flow rate.
Use water as the conveyed fluid and various standard pipe sizes as the
different curves. The flow rate should vary fiom a velocity of 0,2 mls
up to 5 mls.
1.4
Determine the flow rate in each pipe of the following network.
Assume the f value is 0,004 and keep it constant for all iterations. Do
the problem manually with a calculator-,and then by a computer
spreadsheet.
!
I
I
50 m 0,lOm diam
I
c
100 m 0,2 m diam
150 m 0,076 m diam
40 m 0,10 m diam
D
1.5
Determine the flow rates in the following network: The pipes are
galvanized steel and the fluid water. For clarity reasons the pipe
length is given followed by the diameter, both in metres.
1.6
Determine the flow rates in the following network. The pipes are
galvanized steel and the fluid water.
The flow coming into the network at A is 2,5 in3/s, and at E it is 9,5 m3/s.
The outflows at all the other junctions are 1,O m3/s. The lengths and
diameters of the pipes, in metres, are:
1.7
The drawing shows the fuel supply to a small gas turbine. The 4
injection nozzles each passes 5 litres of kerosene fuel per minute. The
pipes are smooth and 10 min in diameter. The lengths are:
Determine the pressure drop between F and C.
16
1.8
Derive the formula for AQ. Explain the steps where necessary.
Chapter 2
Dimensional Analysis and Model Studies
2.1
2.2
2.3
2.4
2.5
2.6
2.7
In troduction
Dimensions and Units
Fundamentals of Dimensional Analysis
The Buckingham Pi method
The Rayleigh method
The Inspection method
Model Studies and Similarity
2.1 Introduction
Dimensional analysis is a useful tool which one can use to organise the
variables that play a role in a certain physical phenomenon, in such a way
that groups of dimensionless numbers are formed. It is traditionally
studied in fluid mechanics, but finds wide applications in subjects like
heat transfer and others. Dimensional analysis helps in planning and
minimising experimental work because the effects of fewer variables
need to be investigated. Furthermore, when model studies need to be
done, it can be used to decide on a size for the model and to predict the
performance of prototypes from tests on scale models. It will not provide
us with all the formulas required but it will help to group the variables
together and with further experimental studies we can then establish the
necessary relationships.
2.2 Dimensions and Units
Any physical situation can be described in tenns of certain variables or
properlies pertaining to that phenomenon. For instance if a certain body
is moving this movement can be described in terms of the mass of the
body, its velocity, acceleration, physical size, etc. These measurable
properties are refemed to as dimensions. (Note t h ~ sis not to be confused
with the dimensioning that is done on engineering drawings, they are
dimensions too, but we are referring to a wider concept here.)
If one wants to quantify the dimension, we use certain basic quantities
called units and the amount of the dimension is then expressed in
fractions or multiples of the units. Thus we have in the System
International (SI) for the dimension of length [L] the unit metre (m), and
for the dimension of time [TI the unit seconds (s), for dimension of mass
the unit kilogram (kg), and for temperature [el the unit kelvin (K).
These dimensions are called primary dimensions and we can express any
other property in terms of them. For a person who is familiar with the SI,
the primary dimensions are very similar to the definition of the various
units. (In the American system of units the definitions of units are much
more complicated). Some dimensional analysis texts use also force as a
primary dimension, the FLT system, but as we use only SI units in this
book, we keep to the abovementioned primary dimensions and the MLT
system. There are a limited number of SI secondary units like Pascal also
in use. Refer table 2.1.
[w
Let's take for example the property, area. It can be expressed as [length]'.
We read for the square brackets "dimensions of'. Similarly, force would
be
[force] = [mass] x [length] l [time]'
or
[force] = [mass] x [length] x [time]-'
thus
[force] = [MLT-~]
Similarly
[viscosity] = [ML"T-'1
Table 2.1 Dimensions in the MLTsystem
2.3 Fundamentals of Dimensional Analysis
Dimensional analysis uses as a foundation the requirement that any
equation must be numerically correct as well as dimensionally correct. It
is nonsensical to say
2 seconds x 3 metres = 7,67 dogs
or
2Pa+6N=8kg/s
The dimensions in the LHS
be the same as in the RHS.
In mathematical formulas, the exponents don't have any dimensions, as
they are pure numbers. Integrals and derivatives have the dimensions as
normal.
Thus, if y refers to a length, then
Similarly,
length.
[dtdy] = [L-'I.
[jY2&] = [L'] if both x and y have dimensions of
To appreciate the usefulness of dimensional analysis, consider the well
known Moody chart for the determination of fiiction head loss in a pipe.
The pressure head loss hf in a pipe is found fiom experience to be
dependent on the fluid velocity V, the density p, the dynamic viscosity p,
the diameter d and the roughness e of the pipe. Note, we say from
experience, thereby implying some prior knowledge about the problem
under consideration. This knowledge won't be supplied by dimensional
analysis, but needs to be acquired f?om other sources like experiments,
theory or even common sense.
If we had to do experiments to determine the effect of each of the above
factors on the pressure head loss hr, we would have to do experiments
varying one factor at a time and keeping all the others constant. This
would need a lot of experimentation and the results would be so extensive
that it would indeed not be easy to use.
Dimensional analysis would help us in grouping these .factors in
dimensionless groups so it can be depicted by a rather simple diagram
like the Moody diagram, which is a graphical representation of only three
different dimensionless groups instead of the original six. We will use
this example to explain a method employed in dimensional analysis.
There are three main methods employed in dimensional analysis. They
are:
(a) The Buckingham Pi method
(b) The Rayleigh method
(c) The inspection method
2.4 The Buckingham Pi method
Buckingham stated that ifwe have n variables, which are playing a role
in a certain phenomenon, and these n variables contain m different
dimensions among them, then n-m independent dimerzsionless groups
can be formedfrom them.
The Buckingham ll method (ll,because in the normal mathematical use
Il implies a dimensionless ratio) involves the following steps:
Step I : Identify all the variables that affect the situation and determine the
number of dimensionless groups that will be formed.
Step 2: Choose repeating variables, together they must contain all the
different dimensions. These repeating variables will be used in all the
groups. It is a good idea to choose one variable that has to do with the
size of the problem e.g. diameter or length, another that is a property of
the fluid or material, e.g. density, and another one connected to the
kinematics of the problem, e.g. velocity. These variables should be
independent from each other - e.g. do not choose a length and an area.
Step 3: Write the different I7 groups consisting of the repeating variables
with unknown exponents assigned and each time also choosing one of the
remaining variables with an exponent of one assigned.
Step 4: Assign the correct dimensions to each variable.
Step 5: Solve the equations for the unknown exponents. Remember the II
group is per definition dimensionless therefore its dimensions will be
[ M ~ T ~ L ~We
B ~ ]thus
. compare the exponents of like dimensions and
solve. It is helpful to start solving the simplest equation first - usually
time, and the more complicated ones like length last.
Step 6: Obtain the various dimensionless groups and manipulate them if
necessary to obtain well-known dimensionless groups like Re, Ma, etc.
Let's do an example as per these steps to illustrate the method.
Example 2.1
Use the Buckingham Pi method to analyse the pressure drop in a pipeline.
Solution:
Step 1:
Variables are: Ap, L, d, p, V, p, E, i.e. 7 variables
containing [M, L, TI i.e. 3 dimensions,
thus 7 - 3 = 4 dimensionless groups
Choose 4 p and V as repeating variables and each
time one of the remaining.
HI= d"p b ~ =" [L]"[ML-~]~[LT-~]~[L]
~
= [MYLO]
Steps 2 , 3 , 4 :
Step 5:
M:
T:
L:
:.
b=0
-c= 0
a-3b+c+ 1=0
a+O+O+l=Oaa=-1
Similarly for the next groups
*2-d
-
-I -1
p ~ ~ ~ = p / ~ V(see
d note 3 below)
n2= pvd/p +
& = dapbVL=[L]" [ M L - ~[LT"]~
]~
[L]
Note:
1. It is good policy to always check whether your obtained
dimensionless group is in fact dimensionless.
2. Dimensional analysis cannot tell us what the relationships
between the groups would be, for that experimental work might
be able to designate a simple value to them, but they could also
form complicated relationships that can be depicted only by
graphs. (Recall how complicated the formulas of ChurchillUsagi mentioned in chapter 1 were).
3. We may combine the obtained dimensionless groups with each
other to get a new dimensionless group, we may also invert
them or raise or lower their power or multiply with a constant
(which is dimensionless). These operations are often performed
so as to obtain the well known dimensionless groups.
For instance:
IT3 = A ~ / =~ pvg h~f / p ~ 2
Multiply this group with the inverse of the &I group and
multiply by 214 to obtain a new IT3 group.
This group is the well-known Fanning'sfiiction factorJ;
(Of course, knowing the result beforehand helped in
choosing the correct manipulations, normally one would
perhaps need some 'mathematical intuition'!)
4. We have obtained three significant groups; ~ l dthe relative
roughness, pVd/p the Reynoldr number and thefiicfion facfor f.
5. Plotting results ftom experiments by using these three
dimensionless groups, we obtain a compact, powerful and easy
to use tool - the Moody diagram.
2.5 The Rayleigh method
This method is also referred to as the indicia1 method, because we group
together variables which have the same exponent or index. We explain it
stmightaway by an example.
Example 2.2
The thrust force F of a boat's propeller is known to depend upon the
rotational speed N, the diameter D, the forward velocity V, the fluid
density p and the viscosity p Determine a relationship between these
quantities and F.
Solution:
The relationship must be F = f(N,D,V,p,p) where f is some function or
relationship that needs to be found.
Let us assume it is F = ~ N ' D ~ Vdp""~
where k, a, b, c, 4 e are unknown numerical constants. Since the equation
has to be numerically as well as dimensionally true, we can compare the
dimensions on both sides and thus determine the exponents.
Now,
[F] = [force] = [MLT~]
[Nl = [rotational speed] = [TI]
[Dl = [diameter] = [L]
[!!I = [velocity] = [LT-'1
[p] = [density] = [ML-3]
[p] = [ML-IT']
Substituting these dimensions into the above equation,
Now equate the powers of the dimensions and solve. We won't be able to
solve them all because we have 5 unknowns and only 3 equations, but the
rest are expressed in terms of each other:
Substituting this into (1)
Regroup the variables according to identical numerical or alphabetic
exponents,
Again we have three dimensionless groups and we can manipulate them
to give more sensible groups.
V/DN is a velocity ratio
(~ND'/p) x (V/DN) = pVD/p which is a Reynolds number
and
(pN2D4)x (VIDN)' = (pv2DZ)which is a dynamic force
Thus we have the relationship
F/ p
~
2
=~flpVD/p,
Z
VIDN)
Where f means an unknown function which would incorporate the k, c,
and e unknowns, and which would have to be established by experimental
work.
2.6 The inspection method
This method is usefil only when we have a small number of simple
variables. Remember, our objective is to obtain dimensionless groups, so
if you can arrange the variables by simply 'looking at them' it is also
valid. If one were studying a fluid mechanics problem, Reynolds numbers
2.5 The Rayleigh method
This method is also referred to as the indicia1 method, because we group
together variables which have the same exponent or index. We explain it
straightaway by an example.
Example 2.2
The thrust force F of a boat's propeller is known to depend upon the
rotational speed N, the diameter D, the forward velocity V, the fluid
density p and the viscosity p. Determine a relationship between these
quantities and F.
The relationship must be F = f(N,D,V,p,p) where f is some h c t i o n or
relationship that needs to be found.
Let us assume it is F = ~ N " D ~ dp"
V"~
where k, a, b, c, 4 e are unknown numerical constants. Since the equation
has to be numerically as well as dimensionally true, we can compare the
dimensions on both sides and thus determine the exponents.
Now,
[F] = [force] = [MLT~]
N= [rotational speed] = [TI]
[Dl = [diameter] = [L]
[velocity] = [LT']
[p] = [density] = [ML"]
[p] = [ML-IT']
[v=
Substitutingthese dimensions into the above equation,
Now equate the powers of the dimensions and solve. We won't be able to
solve them all because we have 5 unknowns and only 3 equations, but the
rest are expressed in terms of each other:
Substituting this into (1)
Regroup the variables according to identical numerical or alphabetic
exponents,
Again we have three dimensionless groups and we can manipulate them
to give more sensible groups.
V/DN is a velocity ratio
( p m Ip) x (VIDN) = pVDIp which is a Reynolds number
and
(pNZD4)x (VIDN)' = (~v'D') which is a dynamic force
Thus we have the relationship
Where f means an unknown function which would incorporate the k, c,
and e unknowns, and which would have to be established by experimental
work.
2.6 The inspection method
This method is useful only when we have a small number of simple
variables. Remember, our objective is to obtain dimensionless groups, so
if you can arrange the variables by simply 'looking at them' it is also
valid. If one were studying a fluid mechanics problem, Reynolds numbers
would very often be applicable, as would be the ratios of sizes. The
number of dimensionless groups can still be found by Buckingham's
statement.
A bit of advice that is worth mentioning here is when presenting
information by graphs, it is very often advisable to create dimensionless
graphs, e.g. temperature ratios vs. pressure ratios. This usually reduces
the amount of graphs and makes the information displayed by the graphs
more powerful. An example is Figure 4.15, giving a velocity ratio.
2.7 Model Studies and Similarity
Many designs ideas in engineering need to be tested on scale models first
to check the feasibility and to establish certain performance parameters.
This is especially true in the fields of fluid mechanics and heat transfer
because of the rather complex phenomenon of turbulent flow.
The problem is then how to size the model and how to interpret the
results that are obtained fiom this model to implement it in the design of a
full-scale prototype. We therefore have to establish some rules first
before we may say we have similarity or similitude between the model
and full-scale prototype.
For a model to be representative of a full-scale prototype, it needs to be
(a) Geometrically similar, and
@) Dynamically similar.
Geometric similarity
This means that all dimensions (in the drawing sense) need to be at an
appropriate scale. Thus if a propeller is made at 1/10 of M1-scale, the
diameter, blade cord length, maximum thickness, etc. all need to be at
1110 of full-scale.
This requirement is sometimes disregarded, for instance in model studies
of rivers or harbours. Then the vertical scale could be changed to
compensate for certain influences like wave action. These so-called
distorted models require deeper analysis that is beyond the scope of this
book.
Dynamic similarity
This means that all theforces acting on the body must in the same ratio to
the forces on the full-scale prototype. This ratio would not be the same as
the geometric ratio or scale. It also implies that the directions of the
forces also need to be identical to the full-scale. It implies further that if a
model is tested in a fluid, the fluid forces should be of a similar nature,
which means the nature of theflow should be the same - i.e. laminar or
turbulent, supersonic or subsonic, etc.
Consider a hypothetical case in a flow field where the relationships
between the different variables are to be investigated. Assume the
variables are pressure drop Ap, velocity V, length between the two points
L, density p, viscosity p, bulk modulus of compressibility K, surface
tension o,and gravitational constant g.
We could employ our dimensional analysis methods and regroup these
variables into dimensionless groups, each one which is the ratio of two
similar dimensions. We would end up with the following groups.
Cp = Ap / '/zPv2
(2.1)
the pressure coefficient = pressure drop / dynamic pressure.
The pressure coefficient is important in all dynamic problems where
velocity and dynamic pressure play a role.
Re = pVL /p
(2.2)
the Reynolds number = dynamic forces / viscous forces .
This can be seen by manipulating the expression as follows:
Now p~~ is mass, or rather, it has the same dimensions as mass, and
(v2I L ) has the same dimensions as acceleration, and as mass multiplied
by acceleration gives force, the numerator of equation (2.2) constitutes
the inertia forces. The denominator has been split into a viscous stress
(recall Newton's law on viscosity F/A = pdV/dy) times an area, giving a
viscous force. Thus equation (2.2) is a ratio of inertia to viscous forces.
At high flow velocities the inertia or dynamic forces become large, the
viscous forces relatively weak, and our flow would be turbulent - the
viscous forces are not strong enough to order the flow into neat laminar
flow. The Reynolds number is important in all internal flow situations as
in pipes as well as external flow situations as over vehicles - thus where
the type of flow, laminar or turbulent, plays a role. We shall examine the
Reynolds number effects intensively in chapter 4.
Similarly in another flow analysis we could apply dimensional analysis to
obtain:
Ma = ~ l ( K / p ) ' ~= V / ~ ( ~ R T=J V/a
(2.3)
the Mach number = relative velocity / velocity of sound.
By squaring equation (2.3) and multiplying by pLZ12 the numerator is the
dynamic force and the denominator is the elastic force in a solid. It can
also be seen as a ratio of inertial forces to dynamic forces at sonic
velocity in a gas. The Mach number is important near the speed of sound
and above, i.e. where the fluid velocity is so high that the dynamic
pressure is comparable to the dynamic pressure at sonic velocity. The
speed of sound in a gas is a = \I(yi?lJ with y the ratio of the specific heat
at constant pressure and at constant volume, R the gas constant and T the
absolute temperature. We shall examine the Mach number effects
intensively in chapter 5.
the Weber number
w e = p ~ L/Zo
(2.4)
= dynamic force / surface tension force.
Multiplying equation 2.4 with WL we would get in the denominator twice
the dynamic pressure times an area - as above, the inertia or dynamic
force. The numerator would be d which is the surface tension force the force that holds a droplet of liquid in a spherical shape. The Weber
number is important where surface tension of a liquid plays a role, e.g.
when liquid droplets are moving in a gas. This could be applicable when
atomising paint by a spray gun or when fuel is injected into an oil burner.
Fr = v / ( L ~ ) " ~
(2.5)
the Froude number = wave force 1 gravitational force.
When the RHS of equation (2.5) is squared and multiplied by p ~ ' /p ~ Z ,
we obtain pv2L' in the numerator. This is an inertia force, for instance
the force of a wave. The denominator is p ~ 3 which
g
is the gravitational
force or weight. The Froude number plays a role where the dynamic force
of the fluid and the weight of a fluid are important. This would be cases
where we disturb the fiee surface of a liquid, for instance the flow of
water around the bow of a ship or the flow in and from open canals.
When doing model studies all these numbers should ideally be the same
for both model and prototype to obtain full dynamic similarity. However,
this is not always possible. To obtain for instance both Reynolds number
and Froude number similarity we would need a fluid with a specific
viscosity and density that is just not available (see example 2.4).
Thus very often, we would ignore some of the similarity requirements
and get some data from analytical studies, and use the model study to
acquire the more difficult information. For instance, in chapter 3 we shall
study boundary layer theory as applied to fluid liiction. Thus, when
testing the hull shape of a boa< one could use Froude similarity
experiments to get information on the bow wave resistance, and use skin
fiction fonnulas to obtain values on the skin drag.
For aircraft design the Mach number is important, but only at high
velocities. Therefore one would not be concerned about Ma number
similarity when designing a glider, but certainly when designing a
supersonic jet fighter plane.
A large valve to be installed in a pipe in a hydroelectric station is to be
tested to establish the pressure loss coefficient of it. The prototype will
have a diameter of 2m and the flow through it will be 20 m3/s. The model
valve would have a diameter of 250 mm. What flow rate is required to
simulate the prototype conditions?
Solution:
We use Reynolds number equality because it is internal flow. Therefore
Re,
~i
i
~
I?
= R%
We are interested in the flow rate quantities, but Q a VA a V L ~ ,
Therefore,
VCOQIL~
substitute
but water will be used in both, so the kinematic viscosities are the same
substituting we get
a=?
2
0,25
a&=2,5a3/s
Note: In this example the required velocity is very large, probably
impractical. If we calculate the Reynolds number we see it to be about 50
x 10"based on the diameter). Investigating typical loss coefficient vs. Re
number graphs we see that the loss coefficients generally drop until a
value of around 50 000 and then stays more or less constant. We thus
deduce that it would not be necessary to obtain full Re number similarity
but could probably test it at a lower flow velocity. Alternatively an even
smaller mode1 should be considered.
A 1/40 scale model of a ship needs to be tested in a towing tank.
Determine the required kinematic viscosity of the model fluid so that both
Reynolds number and the Froude number are the same for model and
prototype. Assume the prototype fluid to be seawater with a kinematic
viscosity of 1,04x1o ' ~m2/s.
Solution:
For Fr similarity
thus
a-Jz
VP
--v,
V, = 0,l 58Vp
For Re similarity
-V m L ,0 ,
V,L,
up
Substitute the velocity relationship
0,158V p L " =-V p4 0 L ,
urn,
thus
u, = 0 , 0 0 3 9 5 ~=~0,0039 x 1,04x104 = 4,06 x lo-' m2 / s
Looking at tables or graphs of kinematic viscosity, we don't find such a
fluid, and we will have to alter our requirements, i.e. as previously
discussed, by only obtaining Froude similarity.
Problems
2.1 The equation for the velocity V in a pipe with diameter d and length L
under laminar conditions is given by the equation V = ~~d~ /32@, with
Ap the pressure drop and p the viscosity of the fluid. Determine whether
the equation is dimensionally consistent by inspecting the dimensions on
both sides.
2.2 The velocity of sound c in a gas is assumed to be a function of the gas
density p, pressure p, and dynamic viscosity p. Determine a relationship
by the Rayleigh method.
2.3 Find by dimensional analysis a relationship for a lubricated journal
bearing, between the resisting torque T, the journal diameter D, the
clearance C, the length L, the speed of rotation N, the lubricating oil's
viscosity and the axial load W.
2.4 Viscosity is to be studied with a falling sphere apparatus. A sphere of
mass m is dropped in the liquid and the terminal velocity of the sphere is
measured. Perform a dimensional analysis on the situation assuming the
viscosity p is a function of sphere diameter D and mass m, local gravity g
and liquid density p.
2.5 The maximum rise of a liquid in a tube is dependent on the diameter
of the tube, the surface tension o, the density of the liquid and
gravitational constant g. Determine an expression for the rise Ah.
2.6 Observations show that the side thrust F of a rotating ball in a fluid is
dependent on the ball diameter, the free stream velocity V, the density p,
the viscosity p, the surface roughness E, and the rotational speed w.
Determine the dimensionless groups that would be important in studying
this phenomenon.
2.7 The flow of gas and solid particles in a furnace causes erosion on the
walls. Material properties that may be of significance in this problem are
modulus of elasticity E, Brine11 hardness number Br (a dimensionless
group), and ultimate strength (stress) o. The velocity V, particle diameter
d, particle mass flow rate m (kg/s) and tube diameter are also factors to
be considered. Perform a dimensional analysis for the erosion rate R that
has units kg/m2s.
2.8 Spray painting needs to be applied at just the right thichess or the
paint will run, too thin and the gloss h i s h is not obtained. An optimum
thickness s is thought to exist and is a function of the droplet diameter d,
the initial yield stress of the paint q. surface tension o, viscosity p,
density p of the paint, and volumetric flow rate Q. Use the Buckingham
pi method to develop an expression for the thickness.
2.9 The thrust of a jet engine is a function of the air inlet density pa, the
air pressure pa, pressure after the compressor p,, energy content in the
fuel E, exhaust gas pressure p., and velocity V. Use the Buckingham pi
method to develop an expression for the thrust.
2.10 The torque T required for a centrifugal pump is thought to depend on
the flow rate Q, the total head pumped H, the liquid density p, rotational
speed N and efficiency q. Determine an expression for the torque.
2.1 1 An aeroplane wing has a chord length c, maximum thickness t, and
angle of attack a.The fluid density is p, fluid viscosity p, sonic velocity a
and the lift force L. Determine a dimensionless relationship among these
variables.
2.12 The power P required to drive a fan is a hnction of density p,
viscosity p, fan rotor diameter D, mass flow rate M, and rotational speed
N. Use the Rayleigh method to determine an expression for power.
2.13 Use dimensional analysis to obtain expressions for the force F on a
body of a given shape of length L, moving with a velocity V in a fluid
with density p and dynamic viscosity p when:
(a) The fluid is incompressible and the body is totally submerged,
@)The fluid is incompressiblebut the body is partially submerged,
(c) The fluid is compressible, having a bulk modulus K, and the body is
submerged.
1.14 The characteristics of a ship 20 m in length is to be studied with a 2
m long model. The ship velocity is 25 kmh. What is the required velocity
of the model for dynamic similarity? Measurements on the model indicate
a drag force of 20 N. What is the expected drag on the prototype?
Assume water to be the fluid in both cases and neglect viscous (fiction)
effects.
2.15 A certain fuel, benzene, is used as a fuel in an engine, and is sprayed
into the intake of the engine. The spray nozzle is 3 mm in diameter and
the fuel concentration is assumed to be small enough so the density of the
mixture is close enough to that of air. The system is to be modelled with
an orifice opening that is 6 rnm and is spraying water. Determine the
average velocity of the mixtures for dynamical similarity between the
two.
2.16 A 1R-scale model of a missile has a drag coefficient of 2,5 at twice
the speed of sound. What would the model / prototype drag ratio be in an
atmosphere of the same temperature but at density one-third of that of the
model tests and at the same Mach number?
2.17 A ship model built to 1/25 scale has a wave resistance of 15 N at its
nominal speed. Calculate the wave resistance to be expected in the
prototype.
2.18 Tests on a truck-trailer combination are to be made. For actual
conditions, the truck will be travelling at 100 kmh and be pulling a trailer
of length 14m. It is desired to model this situation with a 2 m long model
in a wind tunnel. What is the required flow velocity in the wind tunnel for
dynamic similar condition? Suppose the model could-be submerged in a
water tunnel. What should the water velocity then be to achieve
similarity?
2.19 A model of an air ship is to be studied. The prototype is envisaged to
fly at 40 kmih and be 8 m long with a diameter of 750 rmn. The model is
to be 750 mm long. What should the diameter of the model be? What
should the velocity in a wind tunnel be to simulate actual conditions? If
the drag on the full model is found to be x N, what would the drag on the
prototype be?
2.20 A yacht is 7 m long and a model of it is 300 mrn long. The yacht will
sail at 17 lan/h. What velocity of the model is required for dynamic
similarity between the two if the fluid is water? What velocity is required
if glycerine is used?
2.2 1 An airplane is designed to fly at 260 m/s at 10 000 m. If a one-tenthscale model is tested in a pressurised wind tunnel at 20'C, what should
the tunnel pressure be to obtain both Reynolds and Mach number
similarity?
Chapter 3
Boundary Layers
3.1 I~~troduction
3.2 F~indamentalprinciples and the Navier-Stokes eqi~ation.~
3.3 Theformation of boundary layers
3.4 Equations o f motion,for boundac lavers
3.1 Introduction
In this chapter we look at what is happening in the thin layer of fluid that
occurs near to the surface of an object that is exposed to a fluid flowing past
it. We start with some equations that represent the mathematical model of
flow. and then look at some boundary layer theories, and the equations that
are valid in the boundary layer.
It is important to understand how boundary layers originate and how to
control them as they have a big influence on the flow pattern of the fluid that
is flowing past the surface. They play a big role in the resistance against
motion of ships and aircraft, and in the heat transfer of a solid body when it
is exposed to a fluid at a different temperature.
3.2 Fundamental principles and the Navier-Stokes
equations
Figure 3.1 shows an imaginary elemental control volume in a flow field. In
other words. we have a fluid that is flowing somewhere and we draw a cube
in this fluid, and then we investigate how the fluid is flowing through this
block and what the conditions of the flow are where it goes into each face
and again where it is coming out at a face. The sides of the block are dx, dy,
and dz long, and the arrows are the velocity vectors of the components of the
flow perpendicular to the faces.
Figure 3.1 An elenzental block in aflowjield.
At each face of the control volume the fluid would have a certain velocity,
pressure, and certain properties as density, temperature, viscosity, etc. The
stream of fluid would therefore constitute a certain amount of momentum
coming in, and where it goes out it could have a different momentum if the
velocity or density had changed inside the control volume. This change in
momentum could have been caused by a pressure differential over the
control volume or body forces like gravity. Also, the fluid flow component
streaming past the faces in a direction parallel to the faces would exert a
shearing action on the face, thereby actually affecting the flow inside the
control volume.
If one applies the normal laws of conservation of mass and momentum to the
elemental block, the following partial differential equations can be derived.
Their derivation is lengthy, involved and beyond the scope of this text as
well as most undergraduate books.
1
The continuity equation:
The momentum equations combined with Newton's viscosity law for the
three principal directions give after rather involved mathematical operations
what is called the NavierStokes equations:
X component:
Y component:
Z component:
I
!
I
1
Similarly the energy conservation equations can also be applied to
incorporate thermal effects. This type of analysis forms the basis for so
called Computational Fluid Dynamics (CFD) computer programs. These
programs are commercially available and are becoming more and more
powerful. They operate on the principle that the whole flow field is divided
into a mesh or small elemental blocks (cubes, prisms, cylinders, etc.) and
formulas like the above applied to each block. At some places the conditions
like velocities and pressures are specified by the user and the effect on the
next layers of blocks are calculated. By highly sophisticated numerical
mathematical methods, the flow field is analyzed through an iterative
process. CFD modelling has replaced a lot of development work which
previously had to be performed by model and wind tunnel work, however, it
has not replaced it altogether, because the computational model usually has
to be verified experimentally.
3.3 The formation of boundary layers
When a moving fluid is in contact with a stationary wall the flow velocity
near the wall is influenced by the wall and a so-called boundary layer is
formed in the fluid. Refer figure 3.2 which shows a stationary flat plate in an
air stream which is flowing at a uniform velocity towards the plate.
turbulent
laminar boundary layer
laminar sublayery
Figure 3.2 Flow over aflat surface
At position (a) the velocity in the air stream is U and is the same
everywhere. This is called a uniform flow or an idealflow pattern. As the air
arrives at the plate though, the air molecules which make contact with the
plate are brought to a standstill by the microscopic roughness of the wall
surface. These stationary molecules retard the ones adjacent to them by
viscous action, i.e. the resistance against shearing motion, and also by the
constant exchange of momentum between adjacent molecules or layers of
the flow stream. At position (b) we thus find a thin layer of fluid that is
moving slower than the rest of the free stream. This affected layer becomes
thicker in the direction of flow as shown by the dotted line. This is called a
laminar boundary layer. The velocity in the boundary layer increases from
zero at the wall up to the velocity of the unaffected free stream. We define
the boundary layer thickness 6 as the distancefiom the wall up to the point
where the flow velocity is 99% of thefiee stream velocity. The flow in this
laminar boundary layer is smooth and the layers slide over one another
without apparent eddies or swirls. Newton's viscosity law T = -pdv/dy is
applicable here, with z the shearing stress in the fluid at a distance y from
the wall and the velocity at this point is v. The dynamic viscosity is p.
Outside of this laminar boundary layer the free stream is still moving
unaffected at a constant velocity V. After a certain distance from the leading
edge, the laminar boundary layer becomes so thick that it becomes unstable
and the flow pattern becomes chaotic. It is now called a turbulent boundary
layer and because of the bigger interchange of momentum it grows much
faster in thickness. Near the wall there is still a thin layer of flow that is
laminar and this is called the laminar sub-layer. The higher the velocity is
at a certain position from the leading edge, the thinner the boundary layer
would be at that place. However, the boundary layer keeps on growing
thicker in the direction of the flow. As we shall see later on from
calculations, the laminar sub-layer is typically a fraction of a millimetre
thick, whilst the turbulent boundary layer over the body of say a motor car is
typically a few centimetres thick. The transition from laminar to turbulent
boundary layer typically occurs within a few decimetres.
3.4 Equations of motion for boundary layers
3.4.1 Laminar boundary layers
Blasius equations
From the differential equations of paragraph 3.1 we can obtain the following
equations for two-dimensional flow of a boundary layer of a Newtonian
fluid with constant properties l i e viscosity and density. We assume the flow
is steady, i.e. aVlat = 0 and incompressible, aplat = 0. If we also assume the
rate of change of velocity and pressure in the y direction is much less than
the rate of change in the x direction, the equations simplify a great deal and
we get:
From Continuity:
X component fiom Navier-Stokes:
Y component from Navier-Stokes:
av, + 5 3= I ap + v-a2v,
v, ax
Q
pax
ity2
*
=0
Q
Bernoulli's equation applied to the region outside the boundary layer:
(3.6)
(3.7)
P +u2+ gv = constant
P
2
Differentiating U to x
Furthermore, equation 3.7 tells us that the pressure is constant in the y
direction i.e. no pressure variation along the y direction. If we also assume
uniform flow in the fiee stream, then aU/& = 0. Then equation 3.8 gives us
a constant pressure also in the x direction, i.e. dpl& = 0.
It is important to stress at this point that one must realize the velocity V,
inside the boundary layer changes along the flow path in the x direction as
well as in the y direction. Similarly V, is also not the same at dzferent
distances j?om the wall. (V, would be zero for incompressible flow, but
would varyfor compressibleflow.)
We thus are left with a set of partial differential equations that needs to be
solved with the help of the boundary conditions which are:
Ify=O
V,=O
Ify=O
V,=O
Ify=co
V,=U
1
I
This rather complicated problem was solved in 1908 by Blasius by doing a
coordinate transformation to a coordinate q. The solution, i.e. the velocity Vx
at any given point, is given in the form of a function of q and the values
calculated by Blasius for that function, f(q), which are given in table 3.1.
The table is used in the following way. For any particular values of v, U, x
and y, calculate q and then find from the table the corresponding value of
f(q) from which V, is readily calculated. Interpolation can be performed for
bigger accuracy. (Note the table consists of two columns only, the last half is
shown to the right of the first half.)
The boundary thickness is as previously stated where the velocity is 99% of
the free stream velocity, that is where V, /U = 0,99. This occurs where is
about 5.
If we define the local Reynolds number for the flow at a point a distance x
from the leading edge as
I
I
Then at the interface of the boundary layer and the free stream, that is where
y=6
I
Thus the boundary layer thickness 6 = -- 5 0 -- 5,Ox
(UIMC)~'~
(UX/V)~'~
Table 3.1 Blasius solution for velocity proflie laminar boundary layer
U
From the above formulas we see the boundary layer gets thicker in the flow
direction, and the thickness of the boundary layer at a certain position is
thinner the greater the velocity is.
Because some of the flow over the plate is moving slower (in the boundary
layer), the rest of the flow is shifted a small distance away from the solid
surface to compensate for this slower moving flow, so as to satisfy the
conservation of mass. This amount is called the displacetnetlt tltickness d*.
Consider a point in a flow field which is initially a distance H away from the
surface of a flat plate. Refer fig 3.3. This point will be displaced by an
amount 6* as it moves along the plate. This displacement thickness 6* is less
than the boundary layer thickness 6 .
E"zk-z:p
g
vx
O+X
x
6.
&militeq
effect
Figure 3.3 Boundary layer tirickness and displacement tl~ickness
In figure 3.3 the mass flow rate at point 0 must be the same as the mass flow
rate at point x, so
The width b and the density p can be cancelled out. Evaluate the left ktegral
and add and subtract a U in the right integrand. So,
The change in the upper boundary of the integral is valid because outside of
the boundary i.e. in the region (H + 6*) - 6 the velocity is constant and V, =
u
thus
!(I
16*= -%)dy/
(3.12)
If one would now substitute the Blasius values for the ratios Vx/U at
increments along the distance from 0 to 6 and perform the integration, one
would get
The flowing fluid causes a slteari~tgstress on the plate surface and this can
be evaluated by applying Newton's law at the surface
From the Blasius solution, the slope of Vx/U at the wall is
0,06641/0,2 = 0,332. Thus,
The wall shear stress therefore becomes
At any point the shearing stress divided by the dynamic pressure of the free
stream gives the so called local drag coefficiertt or skin friction coefficient
cd
1 &I
0,664
C, = -
Substituting and simplifying gives
(3.15)
The shearing stress causes a slrearing force Df along the length of a plate
with width b. The shearing stress is not everywhere the same, of course. This
shearing force, also called the skin friction drag, is the integral of the stress
times the elemental area of the plate along the length L.
giving
where
1-
I
1
Re, = -
We can now define a skin friction drag coefficient CD as
Substituting equation 3.16 gives
so that
Example 3.1
Crude oil of dynamic viscosity 6x10" Pa.s and relative density 0,86 flows
past a plate at a free stream velocity of 3 d s . The plate is 1,2 m wide and
2m long in the direction of flow.
a) Determine the boundary layer thickness and shear stress at a point 0,65 m
from the leading edge.
b) Determine the skin friction drag exerted on one side of the plate.
c) Create a graph showing the distance from the wall versus velocity in the
X direction at a point 1,s m from the leading edge.
Solution:
a) First determine the Reynolds number
Substituting this value in equation 3.11 gives
The shearing stress at this point is
b) The skin friction drag is (use Reynolds number at L = 2 m)
D, = 0,664~@,&
= 0,664 x 3 x
6 x 10" x 1,2
Jw
= 13,30 1 N
c) We create a table as follows
Choose a few q values and the associated f(q) values fiom table 3.1. Using
the formulas calculate the velocity V, and the distance fiom the wall y.
E.g.
if q = 2, f(q) = 0,62977 then V, = U x f (7)= 3 x 0,62977 = 1,889 and
Plot the values y versus V, to obtain the graph.
Distance from wall vs Vx velocity
0.01
0.009
0.008
0.007
5
.=
0.006
0.005
0.004
0.003
0.002
0.001
0
0
0.5
1
1.5
Vx (mls)
2
2.5
3
3.5
The momentum integral equations
!
The Blasius equations are exact equations derived by simplifying the
Navier-Stokes equations and solving them. Another method is to integrate
equations 3.5 and 3.6 to obtain the so called momentum integral equation.
We do not go into the mathematical operations here but give only the results.
The momentum integral equation is
It incorporates the conservation of momentum and continuity and is valid for
laminar as well as turbulent conditions. To obtain the integrands we have to
choose a relationship between V, ,U, 6 , and y. For instance, if we choose a
linear velocity profile
we would get for the boundary layer thickness
The displacement thickness is given by
The skin friction or local drag coefficient is
The total drag coefficient is
Comparing these equations with those of Blasius we see they don't differ
much. Another comparison is given in figure 3.4.
Figure 3.4 Comparison of different velocityprojiles
I
From figure 3.4 it is clear that the parabolic profile closely fits the Blasius
profile. Table 3.2 gives a comparison between the various velocity profiles
and their boundary layer thicknesses, displacement thicknesses, and the drag
coefficients.
Table 3.2 Monzentunz itrtegral resultsfor various assurned velocityprofiles
Sine wave
v, = sin. 3Y
.U
26
4,80x
-
1,74x
-
JRe
1,310
6
3.4.2 Turbulent boundary layer
Up to a Reynolds value of about 5 x lo5 the boundary layer over aj7at
stnooth plate can be regarded as laminar. After this point turbulent flow
occurs and we have to apply different laws. Turbulent flow would also
appear sooner if the surface is rough or if the flow is induced or trbped by a
thin wire across the flow path. We again use the momentum integral
equation, equation 3.18, but instead of substituting an equation for a
parabolic velocity profile into it, the so-called one-seventh- power law is
used. This is a realistic velocity profile which has been proven
experimentally.
for 5 ~ 1 0 ~ 1 ~ e f i 1 0 ~(3.23)
52
The results are then
0,368~
for 5x1 0 ~ 5 R e ~ 1 1 0 ~
7
for 5 x l 0 ~ 5 ~ e ~ 5 1 0
7
for 5 ~ 1 0 ~ 5 ~ e ~ 1 1 0
for 5 ~ 1 0 ' 5 ~ e ~ 75 1 0
for 5x1 0'5ReL<10
7
A slightly different formula results if we modify it to agree better with
experimental results.
r
l
for 5 x l 0 ~ 5 ~ e ~ 5 1 0 '
C , =-
I
I
C , =- 0'0305 for 2,9x1075ReL55x10'
( ~ e ), li
(3.31)
All the above results are applicable to smooth surfaces. Results for rough
and smooth surfaces are shown in figure 3.4. This chart (by F A White) is
the flat-plate analogy of the Moody diagram for pipes. Note the references to
some curve fit formulas in the diagram.
Figure 3.5 Drag coefficientsfor smooth and rouglz flaiplates.
Example 3.2
A supertanker is cruising at 9 knots (1 knot is 0,515 d s ) . It has a length of
220 m, is 45 m wide and its draft is 20 m. Estimate the skin friction drag and
the power to overcome this drag in salt water (density 1025 kg/m3, viscosity
1,07x10" Pas) .
Solution:
We calculate the area of the ship subjected to a shearing stress by
approximating the hull to be more or less rectangular.
Then the effective width is
b = 20 + 45 + 20 = 85 m
Giving an effective area of
A= 85 x 220 = 18 700 m2
The velocity of the water flowing past the hull is taken as that of the ship (it
would probably be slightly different),
pUL =
The Reynolds number is Re = P
The drag force is
1025 x 4,635 x 220 = 9,768
1,07 x 10"
F=C~XAX'/Z~U~
I
But from figure 3.4 the CDvalue is about 0,0016.
I
The power required is
Thus
F = 0,0016 x 18 700 x '/z x 1025 x 4,635'
= 329.4
kN+
P = F x U = 329,4 x lo3 x 4,635.= 1.527 MW+
3.43 Laminar and Turbulent boundary layers.
In some instances we are interested in the combined effect of the initial
laminar flow as well as the subsequent turbulent flow. This would be the
case where each contributes an appreciable drag to the final total drag. It
would for instance not be applicable to a large ship because the drag would
be by far mostly due to that of the turbulent boundary layer.
Initially a laminar boundary layer would exist until Re = 500 000. Thereafter
the turbulent boundary layer occurs up to the end of the plate. The question
arises whether this turbulent boundary layer will have the effect of one with
the origin at the leading edge of the plate (where the laminar one originates),
or of one with the origin at the critical point. Experiments reveal that it is the
former. Thus we have to calculate the two resistances separately and
combine them to get the net effect. We therefore calculate the resistance of
the turbulent part of the boundary layer as if the entire boundary layer is
turbulent and then subtract from that the resistance that would have occurred
on the plate up to the transition zone.
We thus have
D, = (-x 1,328
~e,"'
0 074
+-L--
0 074
~ e , " ~ Re,
115
xcr
'
-
We have defined the average drag force coefficient as
-
Thus
But
1,328 Re,
Re,
Re,
Substituting
C,
=--+----
But
Re,
=
0,074
Re,
0,074 Re,
~ e , " ' Re,
500 000
Substituting and simplifying we get
An expression that fits the experimental data over a wide range is the
Prandtl-Schligting equation.
1-
C, =
--
(log ~ e)2.',"
for lo7 5 Rer 5 lo9
(3.34)
The variation of drag coefficients for a smooth flat parallel to the flow is
graphically displayed in figure 3.5. As can be seen, the drag coefficient is at
its lowest at high Reynolds number in the laminar region and at high
Reynolds number in the turbulent region. One way of decreasing skin drag is
thus to keep the boundary layer laminar for as long as possible, by having
the surface very smooth. As we shall see in chapter 4 though, we sometimes
would rather prefer the boundary layer to be turbulent to reduce total drag!
0 01
0 008
-
-
I
I
I
I
I
I
I
I
I
I
I
Turbulent
-
-
Turbulent
--rT1*'
1- -
0.002-
1-
1w
2
5
lor
2
I
I
I
I
5
10
2
5
I@
Rynolds numhx Re'
Figure 3.6 Variation of drag coefficientfor a smooth flat plate parallel to
tlreflow.
Example 3.3
Air with a density kinematic viscosity of 1,49 x 10" m2/s flows past a flat
plate 3m long and 0,5 m wide with a velocity of 30 m/s. Assume the
boundary layer is first laminar and then changes into a turbulent boundary
layer. Determine the average skin fiiction drag coefficient, the total shearing
force by both sides of the plate, and the percentage of the total drag
developed in the laminar region.
Solution:
The total resistance is
We first calculate the Reynolds number
Now
0 074 1700
cD=L--=
Re,"'
Re,
0,074 --= 1700
0,00326- 0,000282= 0,00298
(6,04x106)"' 6,04xlo6
The total shearing force per side is thus
Df = c D ~ p u 2 =
/ 20,00298 X 3 X 0,5 X 1,l X 302/2 = 2,21 N
For two sides
Determine x,
The laminar resistance is
Thus the laminar force as a fraction of the total force is
0,231 1 4,425 = 5,25 %-+
q versus f(q) from the values in table 3 . l .
3.2 Create by computer spreadsheet graphs depicting the various velocity
profiles similar to figure 3.4.
3.3 Water flows over a flat smooth plate. Create by computer the velocity
dstribution at a point 0,l m downstream of the edge of a plate if the fiee
stream velocity of water is 2,5 m/s (X axis to be distance fiom plate and Y
axis to be velocity). Create a graph showing the boundary layer thickness
and the displacement thickness along the plate length, and a graph showing
the shearing stress along the plate.
3.4 Create by computer spreadsheet 5 curves depicting the Blasius boundary
layer profile at various distances fiom the leading edge. The first curve
should depict the conditions more or less at a position fiom the leading edge
equal to the square of the boundary layer thickness, and the last one at about
25 times the first distance. The other curves should be at any intermediate
distances you choose. (Hint, the y axis of your graph should depict y with a
maximum value of about 3 or 4 times the boundary layer thickness 61, where
61 is the boundary layer thickness for curve 1, at position xl where 61 4x1 .
The x axis would depict velocity V, with a maximum value of the fieestream
velocity U. The intermediate graphs could be depicting conditions at
distances 4x1 ,9x1 , and 16x1).
-
3.5 Derive equation 3.16 from the preceding equation showing all the steps.
3.6 A small radio controlled yacht is 500 mm long, the wetted width is 120
lnm and it is sailing at 0,9 m/s. Plot the variation in boundary layer
thickness, the local shearing stress, and calculate the total skin friction drag
on the hull. Assume laminar conditions prevail.
-
3.7 An aircraft flies at a speed of 600 kmlh at an altitude of 10 000 m on a
standard day. Assuming the boundary layer behaves as on a flat smooth
surface, at what distance from the leading edge will transformation occurs?
3.8 Certain regions on the wings of missiles might be turbulent at low
altitudes and be changing to laminar at higher altitudes. Explain possible
reasons.
3.9 A wind tunnel has a test section which is 300 mm square by 600 mm
long. The nominal wind speed is U = 28 mls at the entrance to the test
section and the boundary layer thickness is 20 mm at the entrance and 30
mm at the exit from the test section. The boundary layer profile is of the one
seventh power law shape. Determine the freestream velocity at the exit from
the test section.
3.10 Air at standard conditions flows over a flat plate. The freestream speed
is 15 mls. Find 6 and ,T at 1 m from the leading edge for completely
laminar flow assuming a parabolic velocity profile and (b) completely
turbulent flow assuming a 117 power law velocity profile.
3.11 Calculate the drag force on a flat plate with dimensions 0,75 m by 0,75
m when it is aligned to a flow of air with a freestream velocity of 1,75 d s .
3.12 A towboat for barges is to be tested in a towing tank. The towboat
model is built at a scale ratio of 1:13,5. Dimensions of the model is overall
length 3,5 m, beam (width) 1 m, and draft (depth) 190 mm. Estimate the
average drag on the model at a speed relative to the water of 5 kmh.
3.13 A supertanker has a length of 300 m, a beam of 80 m and a draft of 25
m. Calculate the displacement in tomes. The ship steams at 25 kmlh through
seawater at a mean temperature of 10'C. Calculate the thickness of the
boundary layer at the stem (rear) of the ship, (b) the total skin friction drag
on the ship, (c) the power to overcome this drag and (d) estimate the distance
required for the ship to stop. (Hint; treat the deceleration as if the velocity is
reducing in a number of intervals or steps. For each interval calculate the
-
deceleration and assume it stavs constant for that int'erval only and calculate
the distance required to deceleiate to the next velocity region).
3.14 A thin plate 450 mm by 900 mm is immersed in a stream of glycerine at
20'C and a velocity of 6 m/s. Calculate the viscous drag if the plate side that
is parallel to the stream is (a) the short side (b) the long side.
3.15 A four bladed helicopter blade rotates at 200 rlmin. If each blade is 4 m
long and 0,45 m wide, estimate the torque needed to overcome friction on
the blades if they act as flat plates.
3.16 A thin smooth sign 6m long by 1,5 m high is attached to the side of a
truck. Estimate the friction drag on the sign when the truck is travelling at
110 lunlh..
3.17 A plate 5 m wide by1 1 m long is towed at 4 m/s through sea water at
20'C. Estimate the tow force and power required if the plate is (a) smooth
and (b) rough with E = 0,0004 in.
3.18 A boat has a wetted area of 8500 m2and is 200 m long. Estimate (a) the
power required to propel the ship at 25 km/h if the hull is smooth, (b) the
power required if the hull has a roughness value of E = 5 mm and (c)
estimate the speed that can be obtained if the same power calculated in (a) is
now available for the rough hull.
3.19 A jet airplane flies at 950 km/h at 10 000 in standard altitude. It has a
smooth wing 7,5 m long and 55 m wide to the direction of flight. Estimate
the power required to overcome the fiiction drag. If the wing is rough and
requires 13 MW to overcome friction, estimate the wing roughness in
inillimetres.
3.20 A torpedo 600 mm diameter and length 5 in moves in seawater at a
speed of 90 kmlh. Estimate the power required to overcome skin drag if
E = 0,5 rnrn. State the assumptions you made.
-
-- - .
--7-F- '
.
-
-
3.21 A ship is 150 m long and has a wetted area of 5000 m2. It is encrusted
with barnacles, the ship requiring 5000 kW to overcome fnction drag when
moving at 25 ktnh in seawater at 20 'C.What is the average roughness of
the barnacles? How fast would the ship be moving with the same power if
the hull was cleaned? Ignore wave drag.
-
3
k
1
-
-
-
-
I
CHAPTER 4
External Flow over Bodies
4.1
4.2
4.3
4.4
4.5
4.6
Introduction
Drag on Flat Plates
Drag on Two and Three Dimensional Bodies
Forces on Streamlined Bodies
Forces on rotating Bodies
Aerodynamic forces on road vehicles
4.1 Introduction
In this chapter we will be looking at the forces developed on an object
submerged in a fluid whch is moving relatively to it. These conditions find
many applications in the engineering world i.e. road vehicles, airplanes,
submarines and also stationary objects like buildings, advertisement boards
and also sports equipment like balls or racquets. A body immersed in a
flowing fluid experience friction forces on it as well as pressure forces.
Friction forces are due to the viscous fiiction stress acting on the body by the
relatively moving fluid and have been taken care of in chapter 3. What we
are concerned with in this chapter are the forces due to a difference in
pressure on the body. This force can generally be decomposed in a drag
force acting parallel to the direction of motion, and a lift force acting
perpendicular to the motion. This lift force could be upwards, as on an
airplane's wing, or downward as on a racing car's pressure wings.
We start off by looking at some standard simple objects that is set up in a
flow field, and then to some basic aerodynamic principles on streamlined
shapes and to road vehicles.
4.2 Drag on Flat Plates
Consider a long flat plate which is held perpendicular to the flow of a fluid
with density p as is shown in figure 4.1. At point A on the centre line near
,
1,
I
NIg
1
1
1
1
1
1
,
-
--
-
-
-
-
-
-----
--
- --
-
-
.
I
-
-
I
-
the surface the fluid is not moving, i.e. it is stagnant and the pressure there is
called the stagnation pressure. In absolute terms it is equal do the static
pressure p, plus the dynamic pressure of the free stream velocity V
Figure 4.1 Flow around a longflatplate
A little distance away from the centre point, say point B, the fluid will be
flowing outwards towards the edge of the plate and because it now also has a
dynamic pressure, its static pressure i.e. the pressure acting on the disc
surface at point B, is less than at A. The pressure reduces further towards the
edge and near the edge it is zero (ambient). At the back or downwind side
the pressure is less than the ambient and is more or less constant
everywhere. The velocity parallel to the plate at the back is also very small
- there are only weak eddy currents of flow. Some distance behind the disc
the streamlines would join up again and we would get a more or less
uniform flow pattern. The pressure distribution is shown in figure 4.2. The
area integral of the pressure distribution gives the drag force.
The drag force FD is proportional to the projected or frontal area A and the
dynamic pressure
prestire distnibution
on upstream side
presrrre disirubtrtion
jd
on downstream
O
side
v
+ -
+1.0
+0.5
0
-0.5
-1.0
-1.5
Figure 4.2 Pressure dktribution on a long flat plate in a stream
The ratio of the drag force and the product of dynamic pressure times area is
the drag coefficient i.e.
FD
CD = ----A ; pv2
For long rectangular flat plates the drag coefficients are fairly constant for
the range of 104<~e<106
and is equal to about 2. For plates other than
circular and long rectangular flat plates the ratio of the dimensions also plays
a role in the value of the drag coefficient. See figure 4.3 which gives CD
values for a rectangular flat plate.
Figure 4.3 Drag
coefficientsfor a
rectangular flat
plate
-
-
4.3 Drag on Two and Three Dimensional Bodies
Two dimensional bodies are bodies like long cylinders where the end effects
are so small as to be negligible. When we have to take end effects into
account we refer to it as three dimensional bodies. The total drag resistance
is now a combination of fiiction drag and pressure drag, the latter also
referred to as form drag. The fiction part of the drag can be calculated by
using the theory in chapter 3, provided of course we know what the velocity
is at all points over the body.
Re=?
,,IR~=!$
Figure 4.4 Drag coefficients for various objects
If the body has a curved surface the velocity near the surface is not constant
and is difficult to predict at any specific place. The total drag is therefore
mainly found experimentally in wind tunnels and water tunnels or from CFD
models. See figure 4.4 which gives CDvalues for a variety of two and three
dimensional bodies. Table 4.1 gives approximate CD values for various
bodies at high Reynolds numbers.
Table 4.1 Approximate CDvalues for various bodies
-:
T
l
Type of Body
d2
b
0
T
+-I+
Length Ratio
Re
CD
I/b = I
>lo4
1.18
Rectangular
plate
I/b = 5
I / b = 10
116 = 20
>lo4
> 10'
>lo4
1.20
1.30
1.50
Circular
Ild
I/d
I/d
I/d
= 0.5
= 1
=2
=
>lo'
>lo'
0
>lo4
>lo4
1.17
1.15
0.90
0.85
0.87
Square rod
x
>lo'
2.00
Square rod
m
>lo'
1.50
axis
// to Row
Triangular
cylinder
Semicircular
shell
Semicircular
shell
Hemispherical
shell
Hemispherical
shell
=
O(disk)
68
I
-
-
-
-
-
-
Table 4.1 (conL) Approximate Co valuesfor various bodiesCube
Cube
I
0.81
Cone-60'
vertex
Parachute
To understand form drag we need to focus again on the boundary layers.
Consider a curved surface over which a fluid is flowing. Refer figure 4.5.
Figure 4.5 Flow in the boundary layer with rising pressure gradient
If there is a slight curvature on the surface which has the effect that the flow
stream would be expanding or diffusing, this would cause a reduction in
velocity of the flow, which again would cause the static pressure in the
stream to increase due to Bernoulli's principle. The static pressure in the
fluid at (b) is higher than at (a). This higher pressure is of course transmitted
through the fluid into the boundary layer. The result is that the boundary
layer between (a) and (b) experiences a forward action due to the viscous
action of the adjacent stream, but also a reverse action due to the higher
pressure at (b). The result is that the boundary layer becomes thicker and
slows down especially at the surface. The flow can become stagnant and the
flow direction can even reverse as shown in (c). When this reverse flow
--
I
-
=
-occurs, we say flow separation has taken place, and the main flow is no
more following the curvature of the surface but continues in a more or less
straight line, with the result that no static pressure regain takes place. The
flow in the separation region is highly turbulent. The boundary layer is
dependent on the Reynolds number, therefore at higher velocity the
boundary layer will transform to turbulent quicker and this turbulent
boundary layer will continue fhther along the curvature before separation
occurs because of the mixing effect of the turbulent layer which is, as it
were, sweeping away the stagnant layer. A rough surface would also cause
the boundary layer to become turbulent sooner and this would also cause a
reduction in drag coefficient. That explains why the drag coefficient is not
constant but is a function of the Reynolds number and the surface roughness.
Re < I
No separation
Re= 10
Steady separation bubble
2 5 0 < ~ e< 2 x 1 6
Oscillating von Krjrmcin vortex street wake
Re =2xloJ
Laminar boundary layer, wide wake
Re 4x16
Turbulent boundary layer narrow wake
Figure 4.6 Typicalflow patterns around a cylinder at various Re numbers
-
At Reynolds numbers of less than 1, the flow around a long cylinder is
nearly similar to ideal flow (that is flow with no viscosity), and no
separation is taking place. The pressure drag is very small and nearly all
drag is due to skin friction. Refer figure 4.6.
At an increased Re of between 2 and 30, separation occurs and the pressure
difference starts becoming the mayor contributor to drag. The separated
region is rather narrow and two symmetrical eddies form at the back of the
cylinder. Further increase of Re tends to elongate the fixed eddies, which
then begin to oscillate and break away alternating from side to side. This
process is intensified by a M h e r increase in Re and this alternating
shedding of the eddies create two rows of separate vortices which is known
as a von Kcirmhn vortex street.
This shedding of the vortices causes a lateral force on the cylinder which
could then begin to oscillate in harmony with the shedding frequency. This
could be catastrophic for structures like chimneys or smoke stacks. One way
to solve this problem is to fit a helix plate strip around the chimney to
disturb the flow and prevent a resonant frequency building up. The singing
of telephone wires is also due to this phenomenon.
The frequency of such forced vibrations, is given by a formula from
Strouhal. The Strouhal number is defined by
fd and ~&=0,198(1-19,7/~e) for250<Re<2x lo5
So-=-
v
where f is the frequency, d the cylinder diameter and V the free stream
velocity. The last part of equation 4.4 is empirical and is valid in the
indicated range.
The pressure distribution around a cylinder in a uniform flow field is shown
in figure 4.7. At the upstream centre line we have the stagnation pressure of
the free stream. This positive pressure rapidly reduces and at about 20" from
this point the pressure on the surfaces changes to less than ambient reaching
a maximum 'suction' pressure at around 80'. From about 120' the pressure
stays more or less constant because separation has then occurred and the rest
of the circumference at the back of the cylinder is then in the wake region.
Figure 4.7 A polar plot of the pressure distribution around a cylinder
immersed in a uniformflow.
Figure 4.8 Drag coefficients of a smooth and rough long cylinder
Figure 4.9 Drag coefficients of a sphere
Flow past a sphere
For very low Reynolds numbers the drag coefficient is given by an exact
equation relating CDand Re. For Re < 1 the flow around a sphere is laminar
and can be investigated analytically. Flow with Re < 1 is referred to as
Stokesjlow.
The so called Stokes law, for the drag on a sphere is
Note this drag force is depended on the fust power of the velocity, which is
typical for laminar flow. Combining equations 4.3 and 4.4 it can be shown
that
for Re < 1
I
-
---- r - - --Determine the drag on a 10 mm sphere that moves at a velocity of 90 mmls
in an oil of viscosity 0,l P a s and a density of 850 kg/m3.
Solution:
Thus
Terminal velocity of falling objects
When a body is immersed in a fluid the buoyancy forces are opposing
gravity forces. If the body is free to move it will be going upwards or
downwards depending on which of the two afore mentioned forces wins.
The resulting motion will cause a drag force to develop which will always
oppose the motion. The resultant of these three forces would cause the body
to accelerate. When the body has obtained its maximum speed, its terminal
velociw, acceleration will cease. Consider the case of a solid ball of a
material with a high density p,, and mass m, that has been dropped into a
container with a fluid of viscosity ,u and density p. When the ball has reached
terminal velocity V, the net force on it is zero, then
The sphere volume v*,= lrd/6, and the projected frontal area A
Substituting and rearranging we find
Solving for the terminal velocity V, we find
= nd/4.
-
-
-(4.7)
If the buoyancy force were bigger than the gravity force the object would
rise upwards and the above equation would not be valid.
Example 4.2
A steel ball of diameter 2,5 rnm falls into a container with glycerine (p =
0,95 Pa.s and p = 1263 kg/m3). Determine its terminal velocity.
Solution:
We have the equation
However, the CDvalue in the RHS is not known. We therefore have to guess
a value for the velocity, use it to calculate the Reynolds number, use it to
find the CDvalue from figure 4.9, substitute this value back into the above
equation and find the velocity. Use this calculated value to again calculate
Re, CD,and thenV. Continue until the V values converge to a stable value.
To ease these repetitive calculations we first simplie the formulas by
substituting the known values,
and
Doing the iteration in an orderly fashion speeds it up, and lessens the chance
of errors.
Assume:
V = 1 mls, then: Re = 3,324 CD= 8
V = 0,0654 mls
for
Clearly, Re is less than 1 so we can rather substitute CD= 24/Re and
calculate the answer.
Thus
i.e.
Note:
(a) ifRe stayed >1 we would have had to continue with the
iteration until the V values converge.
(6) We could also have guessed a CDvalue or an Re number,
the point is, the others then have to be calculated and the
process repeated through iteration.
4.4 Forces on Streamlined Bodies
Various applications in the engineering world require that a shape have the
minimum of drag e.g. airplanes, ships, missiles, submarines, etc. This can be
achieved by reducing or eliminating the separated flow region behind the
body by streamlining or faring the body. By tapering the body gradually
from the widest section, we aim to prevent separation and increase the static
pressure back to the ambient pressure at the trailing edge. Remember the
pressure is the lowest where the velocity is the highest and vice versa.
However, adding a slowly tapering shape increases the wetted area of the
body and ads to skin friction. The optimum shape is thus the one that causes
the minimum of total drag.
Lead~ng
edge
An& o i attack
\
Camber lone
Figure 4.10 Terminology of aerofoils
Y
A properly designed streamlined section could have a drag of less than 10%
of that of a cylinder with diameter the same as the maximum thickness of the
streamlined shape. Thus the CD values of aerofoil sections are typically
around 0,01 to 0,02. The CD value is also dependent on the angle of attack.
Refer figure 4.1 1. Often it is required that a streamlined shape should not
only have a low CD value (as for a submarine) but should also produce a
large lift force (as for an aircraft wing).
4.
The lift force is defined as
where A is the plan area of the wing i.e. the wing length times the cord
length and CLthe lift coefficient.
Another important definition associated with wings and turbine blades is the
aspect ratio, it being the ratio of wing span b to cord length c, so
Generally the lift obtained from a high aspect ratio wing is greater than that
of a low aspect ratio wing of the same plan area.
One measure of the usefihess of a wing section of an aircraft or the blade
section of a turbine is the lift to drag ratio. The higher this ratio the better.
The CD and CL values are typically displayed versus the angle of attack as
shown in figure 4.11. High performance aerofoils develop lift that could be
100 times the drag. Another way is to present it as a lift and drag polar
diagram, see figure 4.12. The most efficient angle of attack is found by
drawing a line from the origin at a tangent to the C L - C ~
curve, as shown in
figure 4.12.
Angle of attack a,degrees
Figure 4.11 Typical li@ and drag coefflccients vs. angle of attack
At high angles of attack the aerofoil stalls, then the lift decreases and the
drag increases tremendously - possibly with catastrophic consequences in
aircraft.
Aerofoil sections are often classified according to the so called NACA (An
American organisation, the National Advisory Committee for Aeronautics)
designated method. According to this numbering system the first digit
indicates the maximum camber in percentage of the cord length, the second
digit the position of this maximum curvature as a percentage of the cord
length fiom the leading edge, and the last two digits the maximum section
thickness as a percentage of cord distance from the leading edge.
Figure 4.12 A lift and drag polar diagram with angle of attack indicated
(jhn Janna)
The development of lift on an aerofoil can be explained in various ways.
One way isto realise that due to the curvature of the upper section of say an
aeroplane's wing, the air flowing over it experiences a centrifugal action.
The pressure between the air stream and the wing section is therefore
reduced, causing a lifting action. This air stream as well as the air passing
below the wing is thus deflected downwards. The result of these two streams
being deflected is, because the direction of velocity was changed, and thus
the momentum vector, an upward force equal to the change in momentum
resulted on the wing section.
Another explanation is by the Kutta-Joukowski law, which states that the l i i
can only develop if there is circulation around the section. The circulation is
(capital Greek letter 'gamma')
The integral is that of the velocity at a point on a circumferential line drawn
around the wing section times a small distance ds which is part of this line.
Refer figure 4.13. The dotted lines would be the path along which the
integral (velocity times the distance element is performed). The approaching
wind velocity is V, and because the velocity over the wing is faster than
under the wing, and behind the wing there is a downward component in the
velocity vector, the net result of the integral is a certain value with units
m2/s. The lift force that develops per meter length is given by
The units are
Figure 4.13 Circulation around an aerofoil and the starting vortex Cfrom
Douglas)
A consequence of this theory is that whenever lift is produced on an object,
it will be associated with the formation of vortices near the object. In fact,
when a aerofoil starts moving a starting vortex is formed which stays behind
on the ground in the case of an aircraft. At the tips of the wings of aircraft
vortices are continuously formed because the air tends to flow from the
higher pressure region below the wing to the lower pressure region on top of
the wing. The vortex trails behind the moving wing and these so-called tip
vortices or trailing vortices tend to exist for some time (typically five
minutes for large aircraft) before it dies down due to viscous effects from the
surrounding air. This is a mayor concern for air traffic controllers at airports
because of the danger to following aircraft. This phenomenon also explains
why the aspect ratio of wings influence the lift of the wing.
4.5 Forces on Rotating Bodies
When a cylinder or a sphere is spun in still air, no lift effect is produced.
However, if they are spun whilst an air stream is blowing over them, a lift is
produced on the rotating objects. This is referred to as the Magnus effect.
This is commonly seen in ball sports where the ball is given a spin to
influence its direction during its flight. It has not found much engineering
applications, although experimental sailing boats have been constructed
where the sails were replaced by a rotating cylinder. Figure 4.14 shows how
the stream lines of a flow field are changed by the rotation of a cylinder. In
figure 4.14 (a) we have inviscid flow with no rotation, in (b) we have
rotation without uniform flow, and in (c) we have the result of both rotation
and uniform flow taking place.
(C)
Figure 4.14 Flow past a rotating cylinder CfromMunson)
Figure 4.15 shows theoretical and experimental values for drag and lift vs.
velocity ratios of a rotating cylinder. For smooth spheres it has been found
that the CDand CLvalues are about 0,6 and 0,35 respectively for ratios of
peripheral speed / forward speed from about 1,8 to 4.
Velocity ratio, ?
!?
!'
u,
Figure 4.15 Theoretical and exgerimerrtal valuesfor drag and lift vs.
velocity ratios of a rotating cylinder @om White)
4.6 Aerodynamic forces on road vehicles
It is becoming more and more important to reduce drag on road vehicles due
to rising fuel prises and due to the increased speeds required from passenger
as well as goods vehicles. Alternative fuel vehicles generally require a low
drag because of the limited power available. Also, due to the higher speeds
of inotor cars, the development of a downward force on the body is
necessary to aid road holding.
For road vehicles the resistance against motion is part rolling resistance and
part drag force. At low speeds the rolling resistance is predominant but after
more or less directly proportional to velocity but air resistance is mainly
proportional to the square of the velocity. The slun friction part is not nearly
as big as the form drag.
Figure 4.16 gives some drag coefficients for a variety of road vehicles
1932 Fiat Balillo
Volkswagen "Bug"
Volkswagen Van
Volkswagen Scirocco
Mercury Topaz
Toyota Celica
Chevrolet Corvette
Dodge Daytona n r b o
Citroen
Figure 4.16 Drag coefficients of varioiis road vehicles worn Roberson)
At present the practical limit to the drag coefficient appears to be around 0,3
or slightly less, and in future this will undoubtedly be reduced further,
especially with electric or alternative fuel vehicles. In this regard it is
necessary to stress that one should not concentrate to lower the CD value
only, but rather on the product CDA.Thus the ergonomic factors of the body
of the vehicle should be considered as well as the aerodynamic factors.
Figure 4.17 shows the pressure distribution around a motor car on its centre
line. Note it is mainly an upward or suction pressure.
Figure 4.1 7 Pressure distribution on a motor car (from Roberson)
For vehicles with large side areas like buses, a major consideration would
also be stability duning cross wind conditions. Such sideway stability is
sometimes achieved only at the expense of the drag coefficient.
The effect of small body details on the overall drag coefficient have been
investigated and are shown in figure 4.18. This figure should be read in
conjunction with table 4.2. The various Ni values are read off from the table
(second column), and inserted into the equation
Table 4.2 Values for C D ~
A: olan view. front end
A- 1 1
Approximately semicircular
A-2 2
Well-rounded outer quarters
A-3 3
Rounded corners without protuberances
A-4 4
A-5 5
A-6 6
Rounded corners with protuberances
Squared tapering-in corners
Square constant width front
-
B: plan view, windshield
B-1 1
Full wraparound (approx semicircular)
B-2 2
Wraparound ends
B-3 3
Bowed
Flat
B-4 4
C: Plan view, roof
C-1 1
Well- or medium-tapered to rear
C-2 2
Tapering to front and rear or approx constant
C-3 3
Tapering to front (maximum width at rear)
D: Plan view, lower rear end
D-1 1
Well- or medium-tapered rear end
D-2 2
Small taper to rear or constant width
D-3 3
Outward taper (or flared out fins)
E: Side elevation, front end
E-1 1
Low, rounded front, sloping up
E-2 1
High, tapered, rounded hood
E-3 2
Low, squared front, sloping up
High, tapered, squared hood
E-4 2
E-5 3
Medium-height, rounded front, sloping up
E-6 4
Medium-height, squared front, sloping up
E-7 4
High, rounded front, with horizontal hood
E-8 5
High, squared front, with horizontal hood
F: Side elevation, windshield peak
F-1 1
Rounded
F-2 2
Squared (including flanges or gutters)
F-3 3
Forward-projecting peak
G: Side elevation. rear roof/trunk
-
G-1
G-2
G-3
G-4
G-5
G-6
1
2
3
4
4
5
Fastback (roofline continuous to tail)
Semi fastback (with discontinuity in line to tail)
Squared roof with trunk rear edge squared
Rounded roof with rounded trunk
Squared roof with short or no trunk
Rounded roof with short or no trunk
H: Front elevation, cowl and fender cross section at windshield
H-1 1
Flush hood and fenders, well rounded body sides
H-2 2
High cowl, low fenders
Hood flush with rounded-top fenders
H-3 3
H-4 3
Hood cowl with rounded-top fenders
Hood flush with square-edged fenders
H-5 4
H-6 5
Depressed hood with high squared-edged fenders
Component
A Fmnt end (top)
m
A-1
A-2
A-3
I
Shape
A-4
A-5
m
A-6
B Windshield
C Roof
D Lower rear end
D-1
E Fmnt end (ride)
D-2
0-3
&
E- 1
F Windshield peak
G Rear roofltrunk
H Cowllfender cross
section at windshield
H-1
Figure 4.18 Body details that affect the drag coeffient
To increase road holding of racing cars some additional wings need to be
installed to provide negative lift. This downward force could be of the order
of the weight of the car.
Example 4.3
The pressure wing on a racing car is at an angle of attack of 17' and has
characteristics like the aerofoil of figure 4.11. The wing is 1,6 m wide and
has a chord length of 250 mm. Estimate the downward force generated at a
speed of 290 km/h and the contribution to the drag. How much power is
needed from the engine to achieve this additional downward force for better
road holding. Assume atmospheric pressure 90 kPa and temperature 30'C.
Solution:
From figure 4.11, at 17', CL= 1,4 and CD = 0,13.
V= 290/3,6 = 80,56 m/s
!i
At the given atmospheric conditions the density of the air is
The lift force is
The drag force is FD = C D A fV 2= 0,13xl,6xo,25xfxl,O35x80,56'= 174,6 N +
The power needed to overcome this additional drag force is
P = FDxV = 174,6x80,56 = 14,07 kW+
Horse-truck vehicles often consist of a large rectangular projected area a
poor streamlining. This can be improved with fairings, air deflectors and the
like. Figure 4.19 shows some alterations and the effect on the CD value. It
has obvious streamlining effects for the ideal case of relative wind direction
from straight ahead. These ideal conditions are seldom experienced in
practice, so proper wind tunnel testing should be done at angles also other
than from straight ahead, and the results well studied before a decision can
be taken regarding the benefits of adding various fairings. The so called gap
seal type offers some insensitivity to cross wind effects. Better still are
designs where the cab is extended back and the effect is to reduce the
crosswind that passes through between the horse and the truck.
Tractor-trailer trucks
a
Standard
Fairing
Withfairing
a
G ~ seal
D
h i Q n d
gap seal
Figure 4.19 Alterations to horse-truck combinatiorrs @om Munsorr)
Figure 4.20 gives some drag coefficients for some biking configurations.
Note the effect of the area.
A
=
0,51 n12
Upright commuter
& Racing
&& Drafting
f i Streamlined
Figure 4.20 Some CDvaluesfor biking conjigurations(from Murrson)
I- -
88
-
-
-
-
-
-
-- --
Problems
4.1 Use equation 4.3 and 4.4 to derive equation 4.5.
4.2 A circular road sign of diameter 450 mm is subjected to a wind of 120
km/h. Determine the force on the sign. Assume standard atmospheric
conditions. Also calculate the Reynolds number by using the diameter as the
nominal dimension.
4.3 The pressure distribution on a circular disk of diameter 0,5 m is given in
the table below. The pressure behind the disc is constant at -3,14 kPa.
Determine the drag force and the drag coefficient on the disc by numerically
integrating the pressure over the area. (Hint: the (gauge) stagnation pressure
at the centre would be equal to the dynamic pressure of the fiee stream
velocity). Estimate the velocity of the wind if the density of the wind is 1,2
kg /m3. Use modelling laws to estimate the drag force on a 5m diameter disc
-
4.4 Calculate the force of a wind of 120 k m h on an advertisement board
which is rectangular 10 m by 4 m. The board is erected at an altitude of 1500
m above sea level. Assume standard atmospheric conditions.
4.5 Estimate the wind force on a square tubing gate. The outside frame is
made from 38 rnm tubing has inner dimensions 6,l m by 2 m. Vertical bars
inside of the outer frame are spaced 150 mm apart and are made fiom 25
mm square tubing and are oriented flat (in other words their sides parallel to
the plane of the gate.). Determine how the force would be influenced if the
25 mm tubing were turned through 45'. Use standard atmospheric conditions
and wind velocity of 120 krnlh.
4.6 The following data has been obtained fiom wind tunnel tests on the
pressure distribution around a cylinder in a free stream of air moving at 36
m/s. The atmospheric pressure is 86 kPa and temperature 20 "C. It gives the
water manometer readings h (being the difference between the pressure head
on the surface and the free stream pressure head) vs. the angle a measured
fiom the stagnation point. The cylinder diameter is 63 mm and the length
300 mm. Draw a polar plot similar to Figure 4.7 of the pressure vs. the
angle. Determine the form drag force on the cylinder by numerical
integration of the pressure and the surface area. The force on the cylinder
was measured as 16.3 N. How does that compare to the numerical
integration value and to the value you would obtain from the graphs in this
chapter?
4.7 An experimental aircraft have landing gear of which the structural
elements are made fiom 20 mm diameter tubing. The cruising speed of the
aircraft would be about 100 kmlh. Would there be a difference in the drag
between smooth and rough pipes? Explain. If the total length (measured
perpendicular to the flight direction) of the tubing is 10 m, what would the
reduction in drag force be if the tubes would be streamlined with a thin s k i
to obtain a CDvalue of 0,05. What would the power reduction be?
4.8 Derive an equation similar to equation 4.6 for the case where the object
was moving upwards in a fluid at terminal speed.
4.9 Use the same fluid as example 4.2 but let the diameter of the steel ball be
25 mm.
4.10 A glass ball diameter 10 mm, density 2700 kg/m3 falls into oil of
viscosity 0,0331 Pa.s and density 870 kg/m3. Calculate its terminal velocity.
4.1 1 A droplet of petrol is injected by a carburettor into a downward moving
air stream flowing at 2 m/s. The droplet can be assumed to be spherical with
diameter 500 p.Take the density of the petrol as 850 kglm3 and
atmospheric density 1,2 kg/m3 and viscosity 18,2 x 1 r 6Pas. Calculate the
terminal absolute velocity of the droplet.
4.12 A lead ball 25 rnrn diameter hangs from a thii string in a wind tunnel.
The air velocity past the ball is 250 kmlh. Determine the angle that the string
makes with the vertical. Take air density as 1,2 kg/m3, viscosity 18,2 x 10"
Pa.s. Relative density of lead is 11,35.
4.13 A 9 m long pole for lighting has a square section. At its base the pole is
150 mm x 150 mm,and it tapers to the top to 50mm x 50 mm. The wind past
the pole is 100 kmth. Determine the bending moment at the base of the pole
for both the major wind directions. (Hint: Use the midpoint dimensions as
the nominal dimensions).
4.14 A water tower consists of a cylindrical pillar 15 m long and 5m in
diameter, with a spherical tank of diameter 12,s m on top. Determine the
I
blowing. For Re =lo7 the Cu value for spheres and cylinders are about 0,3
and 0,7 respectively. Specify your assumptions regarding atmosphericconditions.
-
4.1 5 Calculate the frequency of the vortex shedding that would be occurring
from a 0,1 m diameter periscope of a submarine which is travelling at 10
kmh. Take the density of the water as 1030 kglm3 and the viscosity as 1,3 x
10" Pa.s. Also calculate the force per metre length on the periscope.
4.16 A chimney is 0,4 m in diameter and 50 m in length. What will be the
minimum wind speed and the maximum wind speeds at which alternating
vortex shedding could appear. Also calculate the minimum and maximum
frequencies of this phenomenon. Assume practical atmospheric conditions.
Use 250 < Re <lo5 as the shedding region.
4.17 A certain aircraft has a mass of 800 kg, a cruising speed of 185 kmlh,
and a wing area of 18 m2. Determine the lift coefficient at this speed. If the
engine delivers 150 kW of power, and if 60 % of this is used to overcome
body drag and propeller losses, what is the drag coefficient of the wings?
4.18 A 2 m diameter cylinder which is 5 m long is rotated at 1800 rlmin in
an air stream flowing at 30 mls. Calculate what the theoretical lift and drag
forces would be as well as the actual lift and drag forces from experimental
data.
4.19 Choose a motor car shape and use table 4.2 to determine what the CD
value of the car is. Compare it with published CD values.
4.20 A truck has a width of 2,4 m and height of 5 m. The height of the tyres
protruding between the body and road is 0,5 m and the width of the tyres is
0,4 m (per side). The truck travels 209 000 km per year at an estimated
average speed of 80 kmlh. The engine has an overall efficiency of 25%. The
file1 has a specific energy of 32 MJAcg. Density of the fuel is 860 kg/m3.~he
fuel costs R4,50/litre. How much could be saved per year if the CDvalue
could be reduced from 0,96 to 0,75?
I
-
92
-
.-
rn
--
-
b
-
-
-
-- -
4.21 A streamlined pedal car with a frontal area of 0,5 m2 and a expected CD
value of 0,13 is being build by students. What is the expected maximum
speed that could be expected if the average power expended by the driver is
assumed to be 200 W? Assume a rolling resistance coefficient of 0,015 and a
mass of driver plus car of 115 kg.
Projects
P4.1 Define the requirements for a specific application of an aerofoil section
and search the net and or books for a suitable profile. Compile a report,
setting out your objectives, search methodology, findings and critical
evaluation of your results. Nominal time to be spent on the project, ten
hours.
P4.2 Design a hand held wind speed indicator based on the theory presented
in this chapter for wind speeds up to 80 kmlh. Make all drawings necessary
to manufacture it. Show all calculations. Present the results as a technical
report that will be used to determine whether to commercialise it.
P4.3 Create a spreadsheet program that will be able to predict the horizontal
distance travelled by a subsonic projectile through the atmosphere. Values
that the users would input would be firing velocity, fuing angle, mass
(constant), CD value, projected area, atmospheric conditions, etc. Set up a
laboratory experiment with a ping pong ball and verify your computer
predictions.
,
-
I
Compressible Flow
5.1 Introduction
5.2 Basic thermodynamic relationships
5.3 Sonic velocity, Mach number, and stagnation properties
5.4 Nonnal shock waves
5.5 Zsenhopic flow through a conduit
5.6 Compressibleflow with fiiction
5.7 Compressibleflow with heat transfer
5.8 Compressibleflow with constant temperature
5.1 Introduction
Compressible flow has to do with flow of air and other gasses inside
conduits or over objects in situations where we cannot ignore density
changes. At low speeds, the effect of the velocity would be minimal and
no appreciable change in density would usually occur. However, when
approaching the velocity of sound the effects become so pronounced that
compressibility effects and thus the changes in density, temperature, and
pressure need to be taken into account. Different laws and formulas are
therefore used. The topic is also referred to as gas dynamics. With long
pipelines, the effect of tliction on the pressure and density could also be
rather much and we need to look at that as well. Examples of
compressible flow would thus be natural gas piped fiom producer to
consumer, flow of steam through a turbine, flow of air through a
compressor and gas turbine of a turbo-aircraft, flow of gas fiom a missile
engine, flow over an aircraft near sonic velocity, etc.
5.2 Basic thermodynamic relationships
The ideal gas law is given by
---
94
-
-
-- -
b
-
where p is the pressure of a gas, p the density, T the absolute temperature
in kelvin and R the gas constant for the specific gas under consideration
(for air 287 J/kg.K)
This formula is valid for all gasses under all conditions and is dependent
only on the conditions. In other words, R is the constant that determines
the relationship between p, p and T.
Enthalpy H or specific enthalpy h is the property that combines the
pressure energy and internal energy into a single property.
where u is the specific internal energy (Jkg), p the pressure (Pa), v the
specific volume (m3/kg).
The specific heat at constant volume is given by
The specific heat at constantpressure is given by
The above specific heats are related to the gas constant by
In an adiabatic process the relationship between the volume and pressure
is given by
where
Other relationships that are also often used are
-
cp
=Elr - l
Entropy is a property of the gas that is related to the conditions present
and can be expressed in terms of the other properties. We are usually
more interested in the entropy change that occurs in a process, so let's
examine this aspect closer. Entropy change is associated with the change
in the probability of the state of the gas, or the amount of change in the
chaos of the molecules in a gas, but also to the amount of energy that is
available fiom the gas.
Consider an amount of gas in an insulated container at a certain
temperature and pressure. If that gas is expanded through a small opening
into another container of the same size that has been evacuated, the gas in
the both containers will eventually, after the pressures and temperatures
have equalised, be at a lower pressure but at a higher temperature. The
enthalpy of the gas would still be exactly the same though. In other
words, the energy content of the gas (pressure plus heat) is still the same.
However, the quality of the energy has deteriorated to a lower quality; it
is less useful to us to do work with. The entropy of the gas has in fact
increased. This increase in entropy is normal whenever a natural energy
conversion process takes place. Very often it can be explained by the
molecular action of the fiiction between the molecules. Therefore, a
helpful way to think about an increase of entropy is to imagine that it is
the process where, due to internal friction, some of the pressure energy of
a gas is converted into heat energy that is less useful to us.
If the same amount of gas were expanded in an insulated fiictionless
cylinder and piston arrangement, then we could get work done from the
piston. The same (or nearly the same) amount of work would be required
to compress the gas to its original condition. This process, a reversible
adiabatic process is isentropic, i.e. no change in the entropy of the gas. In
this process, we converted the pressure energy of the gas into mechanical
5.3 Sonic velocity, Mach number, and stagnation
properties
The velocity of sound is such an important phenomenon that it is
worthwhile spending some time deriving a formula for it.
It is helpid to think about gasses, or rather, to model gasses as consisting
of many masses (e.g. heavy balls) connected to each other with springs.
Refer figure 5.1.
figure 5.1 Model of a compressiblefluid
If one would push against a mass it will be displaced, causing the spring
connecting it to the next mass to compress, this compressed spring will
then displace the next mass a bit, and so on. If one visualises a long row
of stationary masses then if one would push against the fist mass with a
constant speed AV and keep on pushing at the same speed more and more
of the masses would be moving at this speed AV. This compression wave
would be moving ahead at a different speed than AV, let's say at speed a,
compressing more and more of the uncompressed springs.
Now, keeping in mind this model, lets consider a stationary fluid in which
a disturbance was made and the pressure wave is moving, orpropagafing,
at a velocity a. Refer figure 5.1.
Fluid here stationary
Figure 5.2 A sound wave travelling in a stationaryjluid
1
E K n l of-te wave the conditions are p an2 p, and behind the wave the
,
conditions are p+Ap and p +Ap and the velocity AV. If we now deduct
from all the above velocities the velocity a, then we get a stationary wave
towards which the upstream fluid is approaching with velocity a and
behind the wave (to the left of it) the fluid will be flowing away with
velocity a-AV (i.e. a little slower than to the right of the wave). Refer
figure 5.3. Thus, we now have a steady system and we draw a control
volume around the wave and apply some conservation laws to the fluid
that is going into our control volume and coming out of it. We take the
area of the stream as A.
- - . Control volume
,
I
I
PyP
+Fluid
here moving back
Wave stationary with velocity a
Figure 5.3 Stationary wave with movingfluid
Conservation of mass dictates that mass flow in = mass flow out.
-
pz.4 = ( p + Ap)(a AV)A
(5.10)
Multiplying and ignoring the higher order terms as negligibly small (i.e.
ApAv, this reduces to
The momentum equation dictates that the net force causes a change in
momentum.
I
tp+ &)A
- PA = @;[a -(a
Ap = paAV
giving
+A V ) ~
-
Substituting Equation 5.1 1 in place of AV and rearranging we get
From this we see that the speed of propagation is dependent on the
pressure and density changes.
The relationship between pressure and density over the wave is as
follows.
The wave propagates so fast that there is very little time for the heat that
develops due to the compression effect, to flow away. Thus, the process
is adiabatic, hence
-p-
- const. = C,
p7
Differentiating this to p we get
From the ideal gas law we get
P=RT
dp
thus -=@T
P
dP
Substituting into Equation 5.12, we get
Thus, the speed of sound is not constant; it is dependent on the type of
fluid and the conditions like temperature or pressure and density.
The ratio of a relative velocity (i.e. aircraft velocity V relative to
atmospheric velocity) to the velocity of sound in the atmosphere, is called
the Mach number, Ma
Subsonic flow
Transonic flow
Supersonic flow
Hypersonic flow
Stagnation properties
When an air stream is approaching an aircraft's wing, the air is deflected
around the wing. At some small narrow area on the leading edge of the
wing, there is no relative flow between the air and the wing and that
stationary point is called a stagnation point and the conditions there are
stagnation conditions. The pressure at this point is higher than the free
stream pressure because the kinetic energy was changed to pressure
energy in a reversible adiabatic process, and similarly the temperature
there is higher due to the compression effect. Refer figure 5.4. Stagnation
properties are thus the sum of the static and dynamic properties.
Sometimes, e.g. with flow inside pipes, it makes more sense to use the
word total instead of stagnant.
Stagnation point
.-
Figure 5.4 Stagnation point on a wing travelling through air
rn
-- A
-t
-
-- - - - -
--
the stagnation point the stagnation temperature To is
Where T is the static temperature, which is the temperature that would be
measured by a thermometer in the stationary fluid or moving with the
fluid such that there isn't any relative velocity with the fluid. The
subscript 0 indicates stagnation.
Now,
C , =- YR a n d a = @
Y-1
substituting,
and
thus
We have thus expressed the stagnation temperature in terms of the static
temperature, y, and the Mach number, the last two being dimensionless
numbers.
At the stagnation point, the stagnation pressure po can be derived
similarly to the above derivation by using also Equation 5.9. We would
get,
Similarly, one could also get an expression for the stagnation dens@ po
The above Equations 5.16, 5.17, and 5.18 are applicable under all ,
circumstances and simply give the stagnation properties in terms of the
B
static properties and the Mach numbers.
5.4 Normal shock waves
Normal shock waves are wave fionts that are normal (perpendicular) to
the flow direction. They occur when supersonic flow is reduced over a
very small distance (in a fiaction of a millimetre) to subsonic flow with
an associated increase in static temperature, pressure, and density. Refer
figure 5.5.
9
0
8
8
I
I
I
I
-
Ma> l
Upstream
before shock
I
PI
TI
vl
PI
;,
,:
,
I
:,
,
I
I
I
I
0
I
I
I
Ma<l
I
i----+
:
:,
0
vz
I
8
P2
T2
:
m
Downstream
Mer shock
0
I
I
I
I
Figure 5.5 Control volume around a shock wave
To analyse a normal shock wave we draw a control volume around the
shock wave and apply our conservation laws to the incoming and
outgoing streams of fluid.
Conservation of mass dictates that inass flow in = mass flow out.
The momentum equation dictates that the net force causes a change in
inoinentum
As no energy was added or flowed out, the energy equation requires the
total enthalpy to stay constant
~ O =
I
(5.21)
h,
To, = T, because h = C,T
This implies
(5.22)
Using the equation for the speed of sound, Equation 5.13, and the gas
law, Equation 5.1, the continuity equation, Equation 5.19, is rewritten as
The Mach number is introduced into Equation 5.20 as follows
( p , - p 2 ) A = pIV,A(V2 -V,)= p$'zA(V2
-V,)
PI - P2 = ~2V?2 h y 2
Rearranging, and inserting the gas law,
inserting the speed of sound and Mach numbers it becomes
or in terms of the static pressure ratio over the shock wave,
This value will always be >1.
We can write Equation 5.22 as
then
Substituting Equations 5.26 and 5.27 into 5.23, we obtain an equation for
the Mach numbers over a normal shock wave
The soIution for this equation has the trivial solution of Mal = Ma2,
which would be the case for subsonic conditions, however for supersonic
conditions, we would get
For Ma, = I, this equation gives Ma2 = I, which is the case for a sound
wave, meaning also that across a sound wave the increase in pressure and
temperature is infinitesimal. This equation can also be used to solve for
Ma1 in terms ofMa2by simply exchanging the subscripts.
Example 5.2
A normal shock wave occurs in air flowing at a Mach number of 1,6. The
static pressure and temperature is 100 kPa and 20'C. Determine the Mach
number, pressure and temperature ofthe air downstream of the shock.
Solution:
Exercise:
Calculate the stagnationpressure before and afier the shock wave using
Equation 5.1 7.
If one would calculate the stagnation pressure before and after the shock
wave, one would find they are not the same. This is because the shock is
not an isentropic process. In fact, it is associated with severe internal
friction between the molecules, pressure energy thus being converted to
internal energy, with an increase in entropy.
It can be shown that the change in entropy across a shock wave is
Exercise:
Calculate the change in entropy across the shock wave of Example 5.2
Normal shock waves would be formed in front of a blunt body moving at
supersonic speed in air. The shock wave would bend backwards as
depicted in figure 5.6, forming so-called oblique shock waves. The same
relationships exist for these oblique waves as derived above for normal
waves; the difference being that one would use the velocity components
normal to the wave.
Figure 5.6 Shock wave in front of a blunt body (from Roberson).
The oblique shock waves continue to bend in the downstream direction
until the Mach number of the component normal to the wave becomes
unity, after which a Mach wave is formed, which is an ordinary sound
wave. Oblique waves would also form if a wedge-like body moves with
the sharp end leading at supersonic speeds. The oblique wave that would
form in this case is shown in figure 5.7. Calculations on oblique shock
waves are beyond the scope of this book.
Figure 5.7 Oblique shock waveforming at a wedge-like body (from
Douglas).
5.5 Isentropic flow through ra nozzle
'
--
I
5.5.1 Subsonic flow
We know for subsonic flow, when air passes through a nozzle the air is
accelerated in the converging part whilst the pressure is reduced. In the
diverging part the velocity would be decreased and some of the pressure
recovered, the main principle being that of Bernoulli's. The efficiency of
this recovery section depends very much on factors like the divergence
angle and Reynolds numbers, etc.
When the velocity of flow is high the change in velocity and the
subsequent change in pressure and density could be appreciable. We have
to take this into account when using a venturi or plate orifice to determine
the flow rate. Refer Figure 5.8.
Fig 5.8 Flow through a venturi
We assume an adiabatic process is occurring, and that we have to take the
compressibility of the air into account. The densities at the entrance to the
venturi and the throat would therefore not be the same.
The Euler equafion for frictionless flow along a streamline is
*+vd~+~dz=o
P
The Bernoulli equation is the integration of Euler's equation and is
I $+
+ gz = con.7,
(5.3 1)
s
But or adiabatic process~e-
= comr = k thus p = (plk)I17
and density is
inserting this into Equation 5.31
P'
Integrating and putting k = p 1 p7
Or for two points at the sane level (zl= zz)
But for adiabatic flow, pl 1p: = p2 l pi
thus
By conservation of mass
Thus
Substituting equations (5.33) and (5.34) in (5.32) and rearranging we get
km
This equation has been derived for frictionless flow, i.e. with no losses.
To compensate for the friction losses that would occur, we introduce a
coefficient of discharge, Cd, and the mass flow rate becomes
The ratio EL would very often be measured with a manometer and pi
PI
would be calculated after measuring the pressure and temperature of the
gas. The same formula is applicable for orifice plates and flow nozzles the appropriate coefficient needs to be used though. These coefficients
are empirically obtained and are functions of the diameter ratios and the
pipe Reynolds numbers. For orifices with comer taps, i.e. the manometer
tubes are connected in the comers upstream and downstream of the
orifice plate, a formula to calculate the discharge coefficient is given by
where 3/ = d/D the orifice diameter to the pipe diameter, and Re the
Reynolds number of the flow in the pipe.
Example 5.3
A venturi meter with inlet diameter 75 mm and coefficient of discharge of
0,96 has a throat diameter of 25 mm is used for measuring the flow rate
of air through a pipe. A mercury U-tube is used to measure the pressure
differential between inlet and throat and it indicates 100 mrn. The
absolute pressure inside the pipe is 135 kPa, and the temperature of the
air in the pipe is 20°C. Determine the mass flow rate of the air.
Solution:
.
{( 1 (
rir = C, ~ , p 2,
The density of the air is
-
Y-1
- 1 - EL
P,
pl
]l7],/[[::
p - 135x10'
p = -= 1,605 kglm'
RT 287 x 293
-
pry
PI The pressure ratio -m
= 0,9012
x f l x--
PI
135x103,
,,
-
.
-
m = 0,0927 kgls +
1.5.2 Supersonic flow
In supersonic flow things are not as simple. Let's investigate by first
developing an equation to predict the dependence of the Mach number on
the area variation. We consider a duct of varying area as in figure 5.8,
through which a gas is flowing with a certain mass flow ratem.
Figure 5.8 A duct with varying area
The mass flow rate is
where A is the area at the position x and V the velocity at that position. If
we would write this equation in the differential form (by taking the In on
both sides and differentiating that), we get
l d p 1dA
--+--+--=o
p d x Adx
1dV
Vdx
Along. a horizontal streamline with no viscous losses, the Euler equation
Using Equation 5.12 to get
= a'
dp
We can manipulate the Mach number relationship and the above
relationships to get
This relationship gives a lot of information on what to expect in a conduit
of varying area.
Subsonic Flow
In this case (M2-1) is negative, so for converging flow ( W k< 0) the
velocity would increase (dV/dr > 0).
For diverging flow, the velocity would decrease.
Supersonic flow
Now (M2-I)is positive and for converging flow decreasing area would be
giving a decreasing velocity, and increasing area increasing velocity. This
is the principle behind the combustion chamber design for supersonic
aircraft. Refer figure 5.9.
-.
,.,
+
1
!
Combustor
!
Figure 5.9 The engine of a supersonic aircrafl.
In the converging part the flow is decelerated to be slowest in the parallel
section where combustion takes place, heat being generated, and then the
gasses are accelerated out in the diffuser part of the engine, the change in
momentum causing the required thrust.
Consider a converging - diverging nozzle as shown in Figure 5.10. If we
would connect the diverging outlet with a vacuum pump and decrease the
pressure there, air would flow through the nozzle and the bigger the
pressure difkrence between the inlet and outlet is, the b t e r the air flow.
The mass flow rate would increase up to a point where the velocity in the
throat is equal to sonic speed. Further reduction in the exit pressure will
have no effect on the massflow rate because this lower pressure can not
be propagated back to the high pressure side because in the fhroat we
already have Ma = 1. This condition is referred to as chokedflow. We
also call it the critical state and denote conditions where Ma = 1 with an
asterisk sign, *.
Figure 5.10 A converging - diverging nozzle
The ratio of the pressure for choked flow is called the critical pressure
ratio and is obtained by substituting Ma =1 in equation 5.17 obtaining
Similarly we get for the
IT"
y+ll
and for the density ratio
Substituting the value of y = 1,4 into equation (5.38) we see that the static
pressure at the throat would be 52,8 % of the atmosphericpressure.
d
---
-
<
-
9
- -
,
D
Exercise.
Show that for air the critical temperature ratio is 0,833 and the critical .
density ratio 0,634.
The only way to increase the mass flow rate through the nozzle would be
to change the conditions at the entrance, for instance increasing the
pressure because then the density increases or if the temperature is
increased the throat velocity is increased.
The ratio of the areas and the Mach number is
This equation is valid for all Mach numbers - subsonic, transonic, and
supersonic. The area ratio MA* is the ratio of the area at the position
where the Mach number is Ma to the area where the Mach number is
unity.
Example 5.4
A supersonic wind tunnel is to produce a Mach number of 3. The throat
area has a diameter of 30 mm. What should the diameter of the test
section be?
Solution:
The throat area is A =
%0.03'
= 706,9x104 m 2
Thus the diameter of the test section should be
Equation 5.41 is an implicit equation for Ma. If the value of Ma needs to
be found for a specific area ratio, we would have to find it by iteration or
else through interpolation using the tables A6 in the appendix.
The rnassflow rate through a convergent - divergent nozzle is found by
taking the conditions at the throat, then
It is more convenient to express the mass flow rate in terms of the
stagnation conditions because they are usually known. The stagnation
temperature and density are gven by equations 5.16 and 5.18. For sonic
velocity they become
and
When substituted into the previously given equation we get
The stagnation density can be eliminated by using the gas law to get
Exercise:
Show that for gases with y = 1,4the above equation becomes
A supersonic wind tunnel with a square test section of 150 mm by 150
mm is to operate at Mach number 3 using air. The static temperatures and
pressure in the test section are -20'C and 50 kPa. Calculate the mass flow
rate.
Solution:
In Example 5.4 we calculated the area ratio for Ma
Thus the area of the throat must be
=
3 as 4,23.
The stagnation pressure is obtained fiom Equation 5.17
The stagnation temperature is
The mass flow rate is
The power requirement of such a supersonic wind tunnel is often in the
order of megawatts. Cooling down the air would cause the sonic velocity
to be lowered thereby reducing the mass flow rate and saving mechanical
pumping power.
-
--
5.6 Compressible flow with friction and constant area
-Fanno flow-
-
In this section, we will consider compressible flow with fiiction in
insulated ducts of constant area. A process under these conditions is often
referred to as Fannoflow. An application of this theory would be in long
pipelines where gas or compressed air is conveyed and where the heat
flow from the surroundings is negligible.
Consider a pipe in which a gas is flowing. Obviously, there will be a
pressure drop in the flow direction to overcome the fiiction losses. The
density would thus decrease along the pipe due to the pressure drop.
However, the product pV would stay the same due to the constant area
and mass flow rate.
In addition, due to the no heat flow assumption, the gas would get hotter
due to friction. This would be another reason why the gas would be
expanding, becoming less dense.
The result is thus that the velocity of the gas would be increasing along
the pipeline. The maximum speed at which the flow could occur, would
be the speed of sound, i.e. Ma = 1. We denote this position with an
asterisk *. Thus & is the distance from position 1 up to the point where
sonic conditions are reached. This point might not be reached in fact, it
could actually be an imaginary point. The fact remains, ij'the pipe were
long enough, then at the point * the speed would be sonic velocity. Refer
figure 5.8. It follows that if we have a pipe with certain conditions at its
entrance, position 1, and we designate the conditions at its exit as
conditions 2, then we can say the actual length of the pipe is simply
Figrire 5.8 The concept of L *
Note, when refening to the entrance and exit of a pipe it does not imply
the pipe is open to the atmosphere, it merely means the pipe begins and
ends there. It could be connected to a compressor, a machine, or
whatever.
Remembering that the speed of sound is dependent on the temperature,
the velocity and temperature are changing along the pipe, and you would
realize that this is quite a complicated problem.
i.
The derivations of the relevant equations are beyond the scope of this
book (but not very difficult); we only present the equations applicable to
solve the problems. Suffice to point out that for the derivations one would
make use of the nonnal fluid flow and thermodynamic laws.
where f is Fanning'sfiictionfactor, Ma is the Mach nun~berat the
entrance ( o r any reference poinr) of the pipe and L* is the length of the
pipe up to the point where Ma = 1.
The ratio of pressure at the entrance to pressure at the point where the
sonic velocity is reached (* conditions), is given by
1
For the temperature ratio,
The velocity ratio
The stagnation pressure ratio
These equations were used to compile Table A8 in the appendix. We
explain the use of these equations in the following example.
Example 5.4:
A PVC pipe of internal diameter 50 mtn is 150 m long and conveys air at
100 kPa and 20'C. If the entrance velocity is 30 mds, determine the
conditions at its exit. Assume no heat transfer is taking place but do
account for friction.
Solution:
From the appendix tables we get for air at this temperature:
p= 1 8 , 2 2 ~ 1 0pas.
-~
From the gas law we have
The sonic velocity at the inlet is
/LL-
--
-1
Thus
-
a =
-
=
I
"
I
m
- - --
30
341,l
Ma,= -= 0,087 = 0.09
-
From the Fanno flow tables we obtain at Ma, 0,09
We now have to establish the f value. This can be obtained from a Moody
diagram or from formulas like the Churchill-Usagi formulas. Calculating
the Remolds number as
On the Moody chart for smooth pipes and for a Reynolds number of 10'
we getf = 0,0045.
Substituting thisf value with the dimensions of the acttial pipe we get
Using the relationship of equation 5.43 we have
From the Fanno flow tables we read
for
4 / L * = 29,5 = 27,932
-
D
-
then T,
T*
= 1,1946
P*
= 7,2866
We now use our 'trick' to get
pz = 2P..- P
p *,
P * PI
= (7,2866) (1 2 , : 6 2 ) 0 0 ~lo3 = 59,91 kPa +
I
Notes:
1. We see the temperature didn't change much, so the
viscosity would not have changed much and we were justified in
assuming the fiction factor f would not change much over the length of
the pipe.
2. For clarity reasons we didn't interpolate when using the
tables. For greater accuracy one should, or use the formulas.
5.7 Compressible frictionless flow with heat transfer
- Rayleigh flow
In this section we &nsider a constant area duct with compressible
fictionless flow to which heat is added or removed - also known as
Rayleighflow. This theory could be applied in cases where heat is added
directly to the fluid through an exothermic chemical reaction like
combustion, or where heat is added or removed fiom the fluid through
convection heat transfer as in heat exchangers or furnaces. The length of
piping is relatively short so fiction can be ignored, or the heat that is
transferred is much more than that developed by friction.
Consider a small control element of the flow of a fluid in a pipe as in
figure 5.9. The amount of heat that is added per unit mass of fluid over
the small length dr is denoted by dq. Over the whole length of pipe the
amount of heat transferred is q.
Figure 5.9 Control volume in a compressiblefluid with heat addition.
As shown in figure 5.9, the conditions before and after the control volume
change by a small amount. Remember, the main difference to your
previous studies is that this flow is compressible and as such the velocity
and density change all along the pipe. It is rather difficult to predict the
effect of the heat addition due to the complexity of the process, keeping
in mind because the process is ~ctionless,a high initial pressure would
have the effect of accelerating the flow, and remembering also heating
would generally cause expansion but the sonic velocity would also
increase, and not by a simple relationship. We therefore have to
investigate the equations which describe the process to be able too predict
the outcome. The analysis is rather involved, so we suffice by giving the
equations and the resulting conditions.
Applying the continuity equation, the momentum equation, the gas law,
the sonic velocity and Mach number definition, we can anive at a
formula that relates the static pressure at a given point, where the Mach
number is Ma, to the static pressure at the point where the Mach number
is I , designating this point with an asterisk *. Thus,
Similarly for the static temperature ratio
The velocity ratio is inversely proportional to the density ratio because
the mass flow rate ( m = p A V ) stays constant. Thus
Using the stagnation relationships of paragraph 1.3, a total or stagnation
temperature ratio can be derived, giving
The total pressure ratio is
When heat energy, q in wattsper kg/s of the flowingfluid Le. J/kg is
added to the fluid, the total enthalpy is increased. Because of the
relationship equation (5.4), h = C,T we can write
If heat is removed the total temperature would be reduced. Note, the heat
transfer is not only affecting the static temperature but also the dynamic
temperature, thus the total temperature.
The result of heat transfer would be the following:
Heat addition
Subsonic - this would cause the flow to achieve sonic conditions
eventually, but any additional heat would just cause expansion and
the mass flow rate would actually decrease.
Supersonic - flow would eventually also be Mach number unity
but this would have meant a deceleration.
Heat removal
Subsonic - this would have the effect of reducing the Mach number
Supersonic -the flow would be increasing in Mach number.
-
a
This is also sdown in figure 5.10 where T-s diagram is constructed and
shows the e&ct of heating and cooling on the Mach number.
Figure 5.10 m e T-s relationship during Rayleigh flow (Fom Fox)
Example 5.5
At the inlet to a constant area duct the Mach number is 0,2 and the static
pressure and static temperature 95 kPa and 30'C respectively. Heat is
added to the air flowing in the duct at a rate of 110 kJkg. Determine the
properties of the air at the end of the duct if the fiiction can be ignored.
Solution:
The duct can be modelled as in the following sketch.
1
y,
= 95
kPa
T, = 30°C
Ma, = 0,2
From the isentropic tables in the Appendix we read for MI = 0,2
273 + 30 5= 0.9921s To,= - 305,4 K
To,
0,9921
From the Rayleigh tables in the Appendix we read for M I= 0,2
The stagnation heat at point 2 can be obtained by the heat added
Expressing the stagnation temperature ratio in terms of the other ratios we
get
Now search for this value in the tables, we find it at Ma2 = 0,24
At this Mach number we also find
We use it to obtain
and
The velocities at the entrance and exit are
V, = m
~
a=-/, ,
x 0,2 = 69,sm l s +
Note how the properties have changed - only the pressure decreased, the
other properties increased.
5.8 Compressible flow at constant temperature
When gas flows at low velocities in a long pipeline through which heat
transfer can readily occur, we may regard the conditions as approaching
isothermal - constant temperature. Friction can also be taken into account
and the friction factor f stays constant - the Reynolds number would stay
ppqqtqpt becausepvstays constant for constant diameter, and the
viscosity stays constant because it is mainly dependent on the
temperature, which is constant.
The equation for these conditions is given below and example 5.6 shows
how to use it.
where point 1 and 2 are at the entrance and exit respectively and L is the
length of the pipe
Example 5.6
Compressed air is conveyed from a compressor to a water drill 200 m
away. The pipe is smooth and is 40 rnm diameter and at the entrance the
pressure is 700 kPa and the volumetric flow rate 0, 21 m3/s. Calculate the
pressure at the outlet. The temperature is assumed to be constant at 50°C.
Take R = 287 J/kg.K, the dynamic viscosity of air as 20 x 10' Pas.
Solution:
We obtain the f value by first getting the Reynolds number.
velocity
density
PVD - 7 , 5 5 ~ 1 6 ~ 0 , 0=2,41x10~
4
Reynolds number Re = P
20x10-~
From the Moody Chart in the Appendix or the Churchill-Usagi formulas
we get
f = 0,0037
Equation 5.55 is
As the equation is implicit for p2 we have to solve by iteration
The RHS is
Now guess a value for pl for the LHS
Tryp2 = 650xld PO
Try p2
=
625 x I @ Pa a LHS = -49.53~1o9
This is accurate enough and we accept it as the answer.
Problems
5.1 Calculate the speed of sound in helium at 50mC.
5.2 An aircraft travels at 250 m/s in air at a temperature of 5 'C and 70
kPa. Determine the Mach number.
5.3 A jet aircraft is travelling at 1000 kmlh at an altitude where the
temperature is -52 'C and the pressure 23 kPa. Is it travelling faster than
the speed of sound? Calculate the stagnation temperature, pressure and
density at this velocity.
5.4 A bullet is fired fiom a rifle and travels at 950 m/s through still air at
25 'C and pressure 100 kPa. Is the speed supersonic? Use the tables and
calculate the stagnation temperature, pressure and density before the
shock wave. Interpolate if necessary.
5.5 By using a computer spreadsheet, create graphs depicting the
stagnation temperature on a blunt object travelling in an atmosphere at
temperature 25 'C at speeds fiom 100 m/s up to 1000 m/s.
5.6 A blunt bullet is fired fiom a rifle and travels at 900 m/s through still
air at 25 'C and pressure 90 kPa. Is the speed supersonic? Calculate the
stagnation temperature, pressure and density before the shock wave. Also,
calculate the Mach number, static temperature, static and stagnation
pressure, and static and stagnation density after the normal shock wave.
Calculate the relative velocity of the bullet in the air.
5.7 In the previous problem, if the CDvalue of the bullet is 1,O and the
bullet has a diameter of 7,6 mm, a mass of 10 grams, determine the
deceleration it experiences. (Use static density).
5.8 Create a graph depicting static density to stagnation density for the
Mach number range fkom 0 to 10.
5.9 The mass flow rate of air in a 100 mm diameter pipe is to be
determined by a venturi meter with a throat diameter of 50 mm. The
gauge pressure at the inlet to the meter is 33 kPa and at the throat 5 kPa
Atmospheric pressure is 700 mm mercury. The air temperature at the inlet
to the meter is 15'C. Assuming isentropic flow and a discharge
coefficientof 0,97, calculate the mass flow rate and the air velocity in the
pipe.
5.10 An orifice plate installed in a pipe of 300 mm diameter is to be used
to calculate the mass flow rate of air into a turbo engine's intake. The
mass flow rate is expected to be in the region of 5 kgls at atmospheric
conditions of 100 kPa and 20'C. Assume the discharge coefficient of the
orifice plate is 0,6. Assume you would want to obtain a reading of about
100 mm mercury on a manometer connected in the "corner taps" way.
Determine (by trial and error) a suitable diameter for the orifice.
5.1 1 Create a graph by computer spreadsheet for the conditions of
problem 5.10 showing the mass flow rate versus the measurements fkom
the manometer for a usell range. State your assumptions clearly.
5.12 A commercial steel pipe is 625 m long and 75 mm diameter. Oxygen
is pumped into the pipe at 20 mls, and pressure 275 kPa. The inlet
temperature is 15'C. Determine the Mach number at the pipe exit.
5.13 Air flows into a 300 mm diameter cast iron pipe at a velocity of 3,75
m/s. The pressure and temperature is 200 kPa and 25'C respectively.
What length of pipe will cause the Mach number at the exit to be unity?
Also determine the exit pressure.
5.14 A pipe is made of a drawn metal and is 300 m long. Air flows into
the pipe at inlet conditions of 20 mls, 450 kPa and 5'C. The air must be
delivered at a Mach number of about 0,s. Select a suitable diameter and
calculate the exit pressure and temperature and velocity.
Air enters a smooth walled pipe of diameter 25 mm at a velocity of
90 mls, 200 kPa and 300 K. The length of the pipe is 5 m Determine the
conditions at the exit.
5.15
5.16 Air enters a 12 mm commercial steel pipe with a velocity of 450 m/s
a pressure of 210 kPa and temperature of -5'C. Determine the length
required for the pipe to be choked at the exit and the pressure there.
5.17 Air at -15'C and 15 kPa enters a duct at a velocity of 100 mls. Heat
is added to the fluid at a rate of 70 Wkg. Determine the pressure and
temperature of the air at exit. Assume specific heat at constant pressure to
be 1004 J/kg.K.
5.18 Air enters a duct at a velocity of 30 mls and it is desired to determine
how much heat must be added such that the Mach number at the exit will
be unity. Inlet conditions are 25'C and 100 kPa. Determine the exit
conditions of pressure and temperature and the heat transfer rate.
5.19 A duct with oxygen flowing in it is cooled at a rate of 140 kJ per kg
oxygen. Friction can be ignored. Determine the pressure, temperature and
Mach number and velocity at the exit. Inlet conditions are 30'C and 90
kPa and velocity 1330 mls.
5.20 A heated duct conveys air at a rate of 125 mls measured at the
entrance. The inlet temperature and pressure is O'C and 70 P a . The heat
is added at a rate of 140 kJkg of air. Calculate the mass flow rate if the
diameter of the pipe is 50 mm. Determine the conditions at the exit temperature, pressure, Mach number and velocity. Do these conditions
give the same mass flow rate?
5.21 Use a computer spreadsheet program and compile gas tables for
methane (y = 1,3 1) for Rayleigh flow for Mach numbers fkom 0 to 3 with
intervals of 0,2.
1
I
-
5.22 Air flows through a pipe of 150 mm diameter and length 300 m.The
mass flow rate is 4,5 kgls. The pressure at entry is 500 kPa and at the exit
125 kPa Assuming the flow to be isothermal at 60'C and the friction
factor to be constant, determine the iiiction factor, the heat transfer rate to
the air in the pipe in watts and the Mach number at the exit.
-
---
Appendices Figure A1
-
Dynamic viscosity of common fluids
vs. temperature
Temperature. C
-
Table A1
4
Properties of water
P (kglm3)
Dynamic
Viscosity
P (Pa4
Surface
Tension
o (Nlm)
Vapor
Pressure
(Pa)
0
999,9
1,787~10"
75,6 x105
O,6105x1O3
5
1000,O
1,519 xlo"
74,9 xlo"
0,8722 xlo3
10
999,7
1,307 xlo-'
74,2 xl0"
1,228 xlo3
20
998,2
1,002 xlo5
72,8 xlo"
2,338 xlo3
30
995,7
7,975 x10"1
71,2 X I O - ~
4,243 xlo3
40
992,2
6,529 xlo"
69,6 x10"
7,376 xlo3
50
988,l
5,468 x ~ o - ~ 67,9 x10"
12,33 xlo3
60
983,2
4,665 x10"
66 2 xlo"
19,92 xlo3
70
977,8
4,042 x10-~" 64,4 &lo-'
31,16 xlo3
80
971,8
3,547 xlo"
62,6 x10"
47,34 xlo3
90
965,3
3,147 xlo"
60,8 x10"
70,lO xlo3
100
958,4
2,818 xlo5
58,9 x10"
101,3 xlo3
Temperature
Density
("c)
I
IS(
.
-
-
-- -
Table A2 Wroperties of Air at standard
atmospheric pressure
Temperature
Density
("C)
P (kg/m3)
-40
-20
1,514
1,395
0
5
10
15
1,292
1,269
1,247
1,225
20
25
30
40
50
60
70
80
90
100
200
300
400
500
1000
1,204
1,184
1,165
1,127
1,109
1,060
1,029
0,9996
0,9721
0,9461
0,7461
0,6159
0,5243
0,4565
0,2772
Dynamic
Viscosity
P (Pa.s)
Specific heat
ratio
15,7x104
16,3x104
17,lxlod
17,3xl0"
17,6x10"
18,Ox10"
1,401
1,401
1,401
1,401
1,401
1,401
18,2x10"
18,5xl0"
18,6xlo4
18,7x10-~
19,5x10"
19,7xl0'
20,3xlod
20,7xl0"
21,4xlo4
1,401
1,401
1,400
1,400
1,400
1,399
1.399
1,399
1,398
1,397
1,390
1.379
1,368
1,357
1,321
21,7x10'
25,3x10"
29,8xl0"
33,2x10"
36,4x10"
50,4xlod
-
A
I
-
Table A3 The International standard'Atmosphere
Altitude
Pressure
Temperature
Density
(m)
-1000
(Pa)
113,9x103
("C)
21,5
P (kg/m3)
1,35
Dynamic
Viscosity
P (Pas)
18,2x1O4
0
101,3 xlo3
15,O
1,23
18,9x104
1000
89,88
83
1,11
17,6x104
2000
79,50
2,o
1,Ol
17,3x104
3000
70,12 xlo3
-43
0,909
16,9x104
4000
61,66 xlo3
-11,O
0,8 19
16,6x104
5000
54,05 xlo3
-17,5
0,736
16,3x106
6000
47,22 xlo3
-24,O
0,660
16,0x10"
7000
41,ll xlo3
-30,5
0,590
15,6x104
8000
35,65 xlo3
-36,9
0,526
15,2x104
9000
30,80 xlo3
-43,4
0,467
14,9x104
10 000
26,50 xlo3
-49,9
0,4 14
14,6 x10"
15 000
12,l 1 X
-56,5
0,195
14,2 x10"
20 000
55,29 xlo3
-56,s
0,0889
14,2 xl0"
25 000
25,49 xlo3
-51,6
0,0401
14,5 x10"
30 000
11,97 xlo3
-46,6
0,0184
14,8 x10"
40 000
0,287 xlo3
-22,8
0,00400
16,O xlo4
50 000
0,0798 xlo3
-2,5
0,00103
17,O xlo"
60 000
0,0220 XI o3
-26,l
0,000310
15,8 X I 0"
70 000
0,00522 x1 o3
-53,6
0,0000828
14,4 x10"
80 000
0,00105 x103
-743
0,0000185
13,2 x10"
~
O
~
Table A4 Physical properties of common
liquids at specified temperatures
Liquid and temperature
Density
Dynamic
Viscosity
P (kgfm3) P (Pass)
Surface
Tension
o (Nlm)
Benzene 20 'C
879
0,65 x
28,9 x10'~
Ethyl alcohol 20 'C
799
1 , 2 ~o1 - ~
22x10~~
0,96x 1o - ~
26x10~~
Carbon tetrachloride 20 'C 1590
Glycerin 20 'C
1260
620x1o5
63x10'~
Kerosene 20 'C
814
1,9x104
29x 1 o5
Mercury 20 'C
13 550
1,5x10"
480x10"
Meriam red oil 20 'C
Sea water 10 'C
Oils 38 'C
SAE 1OW
SAE 1OW -30
-SAE OW
1
-
-.Iable A5
-
--
.I
- I
---
rn
Ph
rys
at standard atmospheric pressure
(101,3 kPa) and temperature (15'C)
1
Gas
Density
Kinematic' R Gas
Viscosity constant
Cp
Y
(J1kg.K) ( J 4 . K )
P (kg/m3)
v (m2/s)
Air
1,22
14,6x104
287
1004
1.40
Carbon
dioxide
1,85
7,84x104
189
841
1,3
Helium
0,169
114x10"
2077
5187
1,66
Hydrogen
0,085
101x10"
4127
14223
I ,41
Methane
0,678
15,9x104
518
2208
1,31
Nitrogen
1,18
14,5x104
297
1041
1,40
Oxygen
1,35
15,0x10"
260
916
1,40
-
Table A6
Isentropic flow for gas with y = 1,4
(cont)
Table A6
Ma
P~PO
TITO
isentropic
AIA*
0.27
0.9506
0.9856
2.2385
0.28
0.9470
0.9846
2.1656
0.29
0.9433
0.9835
2.0979
0.30
0.9395
0.9823
2.0351
0.31
0.9355
0.9811
1.9765
0.32
0.9315
0.9799
1.9219
0.33
0.9274
0.9787
1.8707
0.34
0.9231
0.9774
1.8229
0.35
0.9188
0.9761
1.7780
0.36
0.9143
0.9747
1.7358
0.37
0.9098
0.9733
1.6961
0.38
0.9052
0.9719
1.6587
0.39
0.9004
0.9705
1.6234
0.40
0,8956
0.9690
1.5901
0.41
0.8907
0.9675
1.5587
0.42
0.8857
0.9659
1.5289
0.43
0.8807
0.9643
1.5007
0.44
0.8755
0.9627
1.4740
0.45
0.8703
0.9611
1.4487
0.46
0.8650
0.9594
1.4246
0.47
0.8596
0.9577
1.4018
0.49
0.8486
0.9542
1.3595
0.51
0.8374
0.9506
1.3212
0.53
0.8259
0.9468
1.2865
0.55
0.8142
0.9430
1.2549
0.57
0.8022
0.9390
1.2263
0.59
0.7901
0.9349
1.2003
0.61
0.7778
0.9307
1.I767
Table A6
Ma
(cont) isentropic
P~PO
TITO
AIA*
1.12
1.14
1.16
0.4568
0.4455
0.4343
0.7994
0.7937
0.7879
1.0113
1 .0153
1.0198
1.18
0.4232
0.7822
1.0248
1.20
1.22
1.24
0.4124
0.4017
0.3912
0.7764
1.0304
0.7706
0.7648
1.0366
1.0432
1.26
0.3809
0.7590
1.0504
1.28
1.30
1.32
1.34
1.36
1.38
1.40
0.3708
0.36091
0.35119
0.34166
0.7532
0.7474
0.7416
1.0581
1.0663
1.0750
0.7358
0.7300
0.7242
0.7184
1.0842
1.0940
1.1042
1.1149
1.42
1.44
1.46
1.48
0.7126
0.7069
0.7011
0.6954
1.1262
1.1379
1.1501
1.1629
1.50
1.52
1.54
1.56
0.30549
0.29693
0.28856
0.28039
0.27240
0.26461
0.25700
0.24957
0.6897
0.6840
0.6783
0.6726
1.1762
1.1899
1.2042
1.2190
1.58
0.24233
0.6670
1.2344
1.60
1.62
1.64
0.23527
0.22839
0.22168
0.6614
0.6558
0.6502
1.2502
1.2666
1.2836
1.66
0.21515
0.6447
1.3010
0.33233
0.32319
0.31424
Table A6
Ma
(cont)
PIPO
isentropic
T ~ o
AIA*
1.68
1.70
0.20879
0.20259
0.6392
1.3190
0.6337
1.3376
1.72
1.74
1.76
1.78
1.80
0.19656
0.19070
0.18499
0.17944
0.17404
0.6283
0.6229
0.6175
0.6121
0.6068
1.3567
1.3764
1.3967
1.4175
1.4390
1.82
1.84
1.86
1.88
0.16879
0.16369
0.15873
0.15392
0.6015
0.5963
0.5910
0.5859
1.4610
1.4836
1.5069
1.5308
1.90
1.92
1.94
1.96
1.98
2.00
2.05
2.10
2.15
2.20
0.14924
0.14470
0.5807
1.5553
0.14028
0.13600
0.13184
0.12780
0.11823
0.10935
0.10113
0.09352
0.5756
0.5705
0.5655
0.5605
0.5556
0.5433
0.5313
0.5196
0.5081
1.5804
1.6062
1.6326
1.6597
1.6875
1.7600
1.8369
1.9185
2.0050
2.25
2.30
0.08648
0.07997
0.4969
0.4859
2.0964
2.1931
2.35
0.07396
0.4752
2.2953
2.40
2.45
2.50
0.06840
0.06327
0.05853
0.4647
0.4544
0.4444
2.4031
2.5168
2.6367
2.55
0.05415
0.4347
2.7630
(cont)
Table A6
Ma
PIPO
isentropic
T ~ o
AIA*
2.60
0.05012
0.4252
2.65
2.70
0.04639
0.04295
0.4159
0.4068
3.0359
3.1830
2.75
0.3980
3.3377
2.80
0.03978
0.03685
0.3894
3.5001
2.85
0.03415
0.3810
3.6707
2.90
2.95
0.03165
0.3729
3.8498
0.02935
0.02722
0.02023
0.01880
0.01748
0.01625
0.01512
0.3649
0.3571
4.0376
4.2346
0.3281
0.3213
5.1210
5.3691
0.3147
0.3082
0.3019
0.2958
5.6286
5.9000
6.1837
6.4801
0.01062
0.00990
0.00924
0.2899
0.2841
0.2784
0.2729
0.2675
0.2623
6.7896
7.1128
7.4501
7.8020
8.1691
8.5517
3.85
0.00863
0.00806
0.2572
0.2522
8.9506
9.3661
3.90
0.00753
0.2474
9.7990
3.95
4.00
0.00704
0.00659
0.00616
0.00577
0.2427
0.2381
10.2496
10.7188
0.2336
0.2293
11.2069
11.7147
3.00
3.20
3.25
3.30
3.35
3.40
3.45
3.50
3.55
3.60
3.65
3.70
3.75
3.80
4.05
4.10
0.01408
0.01311
0.01221
0.01138
2.8960
(cont)
Table A6
Ma
PIPO
7.20
7.40
0.00020
0.00017
0.0880
0.0837
7.60
0.00014
0.0797
1 18.0799
133.5200
150.5849
7.80
8.00
0.00012
0.00010
0.0759
0.0725
169.4030
190.1094
8.20
0.00009
0.00007
0.0692
0.0662
212.8461
237.7622
0.00006
0.00005
0.00005
0.00004
0.00004
0.00003
0.00003
0.00002
0.0633
0.0607
0.0581
0.0558
0.0536
265.0142
294.7661
327.1893
362.4632
400.7753
0.0515
0.0495
0.0476
442.3210
487.3042
535.9375
8.40
8.60
8.80
9.00
9.20
9.40
9.60
9.80
10.00
Tmo
isentropic
AIA*
Table A7 Normal Shock Tables for a Gas with v=1.4
Ma1
Ma2
pap1
TZT1
pap1
1.OO
1.OOOO
1.OOOO
1.OOOO
1.OOOO
1.OOOO
0.5283
1.01
0.9901
1.0235
1.0066
1.0167
1.0000
0.5221
1.02
0.9805
1.0471
1.0132
1.0334
1.0000
0.5160
1.03
0.9712
1.0711
1.0198
1.0502
1.0000
0.5100
1.04
0.9620
1.0952
1.0263
1.0671
0.9999
0.5039
1.05
0.9531
1. I 196
1.0328
1.0840
0.9999
0.4979
1.06
0.9444
1.1442
1.0393
1.1009
0.9998
0.4920
1.07
0.9360
1.1691
1.0458
1.1179
0.9996
0.4861
1.08
0.9277
1.1941
1.0522
1.1349
0.9994
0.4803
1.09
0.9196
1.2195
1.0586
1.1520
0.9992
0.4746
1.1
0.9118
1.2450
1.0649
1.1691
0.9989
0.4689
1.I
1
0.9041
1.2708
1.0713
1.1862
0.9986
0.4632
1.12
0.8966
1.2968
1.0776
1.2034
0.9982
0.4576
1.I3
0.8892
1.3231
1.0840
1.2206
0.9978
0.4521
1.14
0.8820
1.3495
1.0903
1.2378
0.9973
0.4467
1.15
0.8750
1.3763
1.0966
1.2550
0.9967
0.4413
1.16
0.8682
1.4032
1.1029
1.2723
0.9961
0.4360
1.17
0.8615
1.4304
1.1092
1.2896
0.9953
0.4307
1.I8
0.8549
1.4578
1. I 154
1.3069
0.9946
0.4255
1.I9
0.8485
1.4855
1.I217
1.3243
0.9937
0.4204
1.2
0.8422
1.5133
1.1280
1.3416
0.9928
0.4154
1.21
0.8360
1.5415
1.1343
1.3590
0.9918
0.4104
1.22
0.8300
1.5698
1,1405
1.3764
0.9907
0.4055
p02lp01
pllp02
Table
normal shock
Ma1
T2lT1
~21~1
1.24
1.1531
1.4112
1.25
1.1594
1.4286
1.26
1.1657
1.4460
1.27
1.1720
1.4634
1.28
1.1783
1.4808
1.29
1.I846
1.4983
1.3
1.1909
1.5157
1.31
972
1.I
1.5331
1.32
1.2035
15505
1.33
1.2099
1.5680
1.34
1.2162
1.5854
1.35
1.2226
1.6028
1.36
1.2290
1.6202
1.37
1.2354
1.6376
1.38
1.2418
1.6549
1.39
1.2482
1.6723
1.4
1.2547
1.6897
1.41
1.2612
1.7070
1.42
1.2676
1.7243
1.43
1.2741
1.7416
1.44
1.2807
1.7589
1.45
1.2872
1.7761
1.46
1.2938
1.7934
1.47
1.3003
1.8106
1.48
1.3069
1.a278
Table
normal shock
Ma1
T21T1
1.49
1.5
1.51
1.52
1.53
1.3136
1.3202
1.3269
1.54
1 3470
1.3538
13 0 6
1.3674
1.3742
1.3811
1.3880
1.3949
1.4018
1.4088
1.4158
1.4228
1.4299
1.4369
1.4440
1.4512
1.4583
1.4655
1.4727
1.4800
1.4873
1.4946
1.5019
1.55
1.56
1.57
1.58
1.59
1.6
1.61
1.62
1.63
1.64
1.65
1.66
1.67
1.68
1.69
1.7
1.71
1.72
1.73
1.74
1.75
1.76
1.3336
1.3403
~21~1
1.8449
1.8621
1.8792
1 ,8963
1.9133
1.9303
1.9473
1.9643
1.9812
1.9981
2.0149
2.0317
2.0485
2.0653
2.0820
2.0986
2.1152
2.1318
2.1484
2.1649
2.1813
2.1977
2.2141
2.2304
2.2467
2.2629
2.2791
2.2952
Table
normal shock
TWTl
1.5093
1.5167
1.5241
1.5316
1.5391
1.5466
1.5541
1.5617
1 5693
1.5770
1.5847
1.5924
1.6001
1.6079
1.6157
1.6236
1.6314
1.6394
1.6473
1.6553
1.6633
1.6713
1.6794
1.6875
1.7038
1.7203
1.7369
1.7536
P2/~1
2.3113
2.3273
2.3433
2.3592
2.3751
2.3909
2.4067
2.4224
2.4381
2.4537
2.4693
2.4848
2.5003
2.5157
2.5310
2.5463
2.5616
2.5767
2.5919
2.6069
2.6220
2.6369
2.6518
2.6667
2.6962
2.7255
2.7545
2.7833
Table
(cont) normal shock
~ 2 1 ~ 1 T21T1
4.9783
5.0768
1.7705
5.1762
1.8046
~21~1
2.8119
2.8402
2.8683
5.2765
5.3778
5.4800
5.5831
5.6872
5.7922
1.8219
1 ,8393
1 .a569
1 .a746
1 .a924
1.9104
2.8962
2.9238
2.9512
2.9784
3.0053
3.0319
5.8981
6.0050
1.9285
1.9468
3.0584
3.0845
6.1128
1.9652
3.1105
6.2215
6.3312
6.4418
1.9838
3.1362
3.1617
1.7875
2.0025
2.0213
3.1869
2.0403
2.0595
2.0788
3.2119
3.2367
3.2612
7.2421
7.3602
2.0982
2.1178
2.1375
2.1574
2.1774
3.2855
3.3095
3.3333
3.3569
3.3803
7.4792
7.5991
2.1976
2.2179
7.7200
7.8418
2.2383
2.2590
3.4034
3.4263
3.4490
7.9645
2.2797
6.5533
6.6658
6.7792
6.8935
7.0088
7.1250
3.4714
3.4937
Table
(cont)
normal shock
Table
(cont)
Ma1
pap1
T21T1
3.44
13.6392
3.2337
3.48
3.52
13.9621
14.2888
3.2878
3.3425
3.56
3.6
3.3978
3.4537
3.64
14.6192
14.9533
15.2912
3.68
15.6328
3.72
3.76
3.8
3.84
15.9781
16.3272
16.6800
17.0365
3.5674
3.6252
3.6836
3.7426
3.88
3.92
17.3968
17.7608
3.96
4
18.1285
18.5000
4.04
4.08
4.12
4.16
4.2
4.24
4.28
4.32
18.8752
19.2541
19.6368
20.0232
20.4133
20.8072
21.2048
21.6061
4.1096
4.1729
4.2368
4.3014
4.3666
4.4324
4.36
4.4
4.44
22.0112
22.4200
22.8325
4.6334
4.7017
4.7706
4.48
23.2488
4.8401
4.52
23.6688
4.9102
normal shock
3.5103
3.8022
3.8625
3.9233
3.9848
4.0469
4.4988
4.5658
~ 2 1 ~ 1
4.2179
4.2467
4.2749
4.3026
4.3296
4.3561
4.3821
4.4075
4.4324
4.4568
4.4807
4.5041
4.5270
4.5494
4.5714
4.5930
4.6141
4.6348
4.6550
4.6749
4.6944
4.7135
4.7322
4.7505
4.7685
4.7861
4.8034
4.8203
Table
normal
shock
Ma1
T2lTl
~2/Pl
4.56
4.9810
4.6
5.0523
4.8369
4.8532
4.64
5.1243
4.8692
4.68
4.72
5.1969
5.2701
4.8849
4.9002
4.76
5.3440
4.9153
4.9301
4.8
5.4184
4.84
4.88
4.92
4.96
5
5.4935
5.5692
5.6455
5.7224
4.9446
5.8000
5.0000
5.1
5.2
5.3
5.4
5.9966
6.1971
5.0326
5.5
6.8218
5.1489
5.6
5.7
5.8
7.0378
7.2577
5.1749
5.1998
7.4814
7.7091
7.9406
5.2236
5.2464
5.2683
6.1
8.1760
5.2893
6.2
6.3
8.4153
8.6584
5.3094
5.3287
6.4
6.5
8.9055
5.3473
5.3651
5.9
6
6.6
6.4014
6.6097
9.1564
9.4113
4.9589
4.9728
4.9865
5.0637
5.0934
5.1218
5.3822
Table
(cont)
normal shock
Ma1
T21T1
6.7
6.8
9.6700
9.9326
6.9
7
10.1990
7.1
10.7436
11.0218
11.3038
7.2
7.3
7.4
7.5
7.6
7.7
7.8
7.9
8
8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
8.9
9
9.1
9.2
9.3
9.4
10.4694
1 1 ,5897
1 1.8795
12.1732
12.4707
12.7722
13.0775
13.3867
13.6998
14.0168
14.3377
14.6625
14.9911
15.3237
15.6601
16.0004
16.3446
16.6927
17.0447
17.4006
17.7603
18.1240
~21~1
5.3987
5.4145
5.4298
5.4444
5.4586
5.4722
5.4853
5.4980
5.5102
5.5220
5.5334
5.5443
5.5550
5.5652
5.5751
5.5847
5.5940
5.6030
5.6117
5.6201
5.6282
5.6361
5.6437
5.6512
5.6584
5.6653
5.6721
5.6787
Table
A7
(cant) normal shock
Table A8
Ma
-
TIT*
Fanno flow friction flow for gases with y = 1.4
VW
4fL*ID
pOlpO*
@P*
w
0
00
109.5434
57.8738
0.01 10
7134.4045
1.1999
54.7701
28.9421
0.0219
1778.4499
1.1998
36.5116
19.3005
0.0329
787.0814
0.04
1.1996
27.3817
14.4815
0.0438
440.3522
0.05
1.1994
21,9034
11.5914
0.0548
280.0203
0.06
1.1991
18.2508
9.6659
0.0657
193.0311
0.07
1.1988
15.6416
8.2915
0.0766
140.6550
0.08
1.1985
13.6843
7.2616
0.0876
106.7182
0.09
1.1981
12.1618
6.4613
0.0985
83.4961
0
1.2
m
0.01
1.2000
0.02
0.03
0.1
1.1976
10.9435
5.8218
0.1094
66.9216
0.11
1.I971
9.9466
5.2992
0.1204
54.6879
0.12
1.1966
9.1156
4.8643
0.1313
45.4080
38.2070
0.13
1.1960
8.4123
4.4969
0.1422
0.14
1.1953
7.8093
4.1824
0.1531
32.5113
0.1639
27.9320
0.15
1.I946
7.2866
3.9103
0.16
1.1939
6.8291
3.6727
0.1748
24.1978
21.1152
0.17
1.1931
6.4253
3.4635
0.1857
0.18
1.1923
6.0662
3.2779
0.1965
18.5427
0.19
1.1914
5.7448
3.1123
0.2074
16.3752
0.2
1.1905
5.4554
2.9635
0.2182
14.5333
0.21
1.1895
5.1936
2.8293
0.2290
12.9560
0.22
1.1885
4.9554
2.7076
0.2398
11.5961
0.23
1.1874
4.7378
2.5968
0.2506
10.4161
0.24
1.1863
4.5383
2.4956
0.2614
9.3865
0.25
1.1852
4.3546
2.4027
0.2722
8.4834
0.26
1.I840
4.1851
2.3173
0.2829
7.6876
0.27
1.1828
4.0279
2.2385
0.2936
6.9832
0.28
1.1815
3.8820
2.1656
0.3043
6.3572
0.29
1.1801
3.7460
2.0979
0.3150
5.7989
h
Table AB(cont.)
s Fanno
with
friction
PIP*
vw
3.6191
3.5002
3.3887
3.2840
3.1853
3.0922
3.0042
2.9209
2.8420
2.7671
2.6958
2.6280
2.5634
2.5017
2.4428
2.3865
2.3326
2.2809
2.2313
2.1838
2.1381
2.0942
2.0519
2.0112
1.9719
1.9341
1.8975
1 .a623
1 .a282
1.7952
0.3257
0.3364
0.3470
0.3576
0.3682
0.3788
0.3893
0.3999
0.4104
0.4209
0.4313
0.4418
0.4522
0.4626
0.4729
0.4833
0.4936
0.5038
0.5141
0.5243
0.5345
0.5447
0.5548
0.5649
0.5750
0.5851
0.5951
0.6051
0.6150
0.6249
-
Table 'A8(cont.)
Fanno
TIT.
plp*
-
Ma
=
+ith
friction
y = 1,4
L
pOIpO*
0.6348
0.6447
0.6545
0.6643
0.6740
0.6837
0.6934
0.7031
0.7127
0.7223
0.7318
0.7413
0.7508
0.7602
0.7696
0.7789
0.7883
0.7975
0.8068
0.8160
0.8251
0.8343
0.8433
0.8524
0.8614
0.8704
0.8793
0.8882
0.8970
0.9058
0.4908
0.4527
0.4172
0.3841
0.3533
0.3246
0.2979
0.2730
0.2498
0.2282
0.2081
0.1895
0.1721
0.1561
0.1411
0.1273
0.1145
0.1026
0.0917
0.0816
0.0723
0.0638
0.0559
0.0488
0.0423
0.0363
0.0310
0.0261
0.0218
0.0179
9
m
.
A
.
-
Table A8(cont.)
-
Fanno m
with
friction
Ma
Tn"
PIP*
pOIpO*
VW
0.9
1.0327
1.1291
1.0089
0.9146
0.91
1.0295
1.1150
1.0071
0.9233
0.92
1.0263
1.1011
1.0056
0.9320
0.93
1.0230
1.0876
1.0043
0.9407
0.94
1.0198
1.0743
1.0031
0.9493
0.95
1.0165
1.0613
1.0021
0.9578
0.9663
0.96
1.0132
1.0485
1.0014
0.97
1.0099
1.0360
1.0008
0.9748
0.98
1.0066
1.0238
1.0003
0.9833
0.99
I.0033
1.0118
1.0001
0.9916
1
1.oooo
1.moo
1.0000
1.oooo
1.01
0.9967
0.9884
1.0001
1.0083
1.02
0.9933
0.9771
1.0003
1.0166
1.03
0.9900
0.9660
1.0007
1.0248
1.04
0.9866
0.9551
1.0013
1.0330
1.05
0.9832
0.9443
1.0020
1.0411
1.06
0.9798
0.9338
1.0029
1.0492
1.07
0.9764
0.9235
1.0039
1.0573
1.08
0.9730
0.9133
1.0051
1.0653
1.0733
1.0812
I.09
0.9696
0.9034
1.0064
I. I
0.9662
0.8936
1.0079
1.12
0.9593
0.8745
1.0113
1.0970
1.14
0.9524
0.8561
1,0153
1.1126
1.16
0.9455
0.8383
1.0198
1.1280
1.18
0.9386
0.8210
1.0248
1.1432
1.2
0.9317
0.8044
1.0304
1.1583
1.22
0.9247
0.7882
1.0366
1.1732
1.24
0.9178
0.7726
1.0432
1.1879
1.2025
1.2169
1.26
0.9108
0.7574
1.0504
1.28
0.9038
0.7427
1.0581
- A8(cont.)--
- -
Fanno 'with
friction
-
y = I,4
plp*
pOIpO*
V W
4fL*ID
0.8969
0.7285
1.0663
1.2311
0.0648
Tn"
1.32
0.8899
0.7147
1.0750
1.2452
0.0716
1.34
0.8829
0.7012
1.0842
1.2591
0.0785
1.36
0.8760
0.6882
1.0940
1.2729
0.0855
1.38
0.8690
0.6755
1.1042
1.2864
0.0926
1.4
0.8621
0.6632
1. I 149
1.2999
0.0997
1.42
0.8551
0.6512
1.1262
1.3131
0.1069
1.44
0.8482
0.6396
1.1379
1.3262
0.1142
1.46
0.8413
0.6282
1.1501
1.3392
0.1215
1.48
0.8344
0.6172
1.1629
1.3520
0.1288
1.5
0.8276
0.6065
1.1762
1.3646
0.1361
1.52
0.8207
0.5960
1.1899
1.3770
0.1433
1.54
0.8139
0.5858
1.2042
1.3894
0.1506
1.56
0.8071
0.5759
1.2190
1.4015
0.1579
1.58
0.8004
0.5662
1.2344
1.4135
0.1651
1.6
1.62
0.7937
0.7869
0.5568
0.5476
1.2502
1.2666
1.4254
1.4371
0.1724
0.1795
1.64
0.7803
0.5386
1.2836
1.4487
0.1867
1.66
0.7736
0.5299
1.3010
1.4601
0.1938
1.68
0.7670
0.5213
1.3190
1.4713
0.2008
1.7
0.7605
0.5130
1.3376
1.4825
0.2078
1.72
0.7539
0.5048
1.3567
1.4935
0.2147
1.74
0.7474
0.4969
1.3764
1.5043
0.2216
1.76
0.7410
0.4891
1.3967
1.5150
0.2284
1.78
0.7345
0.4815
1.4175
1.5256
0.2352
1.8
0.7282
0.4741
1.4390
1.5360
0.2419
1.82
0.7218
0.4668
1.4610
1.5463
0.2485
1.84
0.7155
0.4597
1.4836
1.5564
0.2551
1.86
0.7093
0.4528
1.5069
1,5664
0.2616
1.88
0.7030
0.4460
1.5308
1.5763
0.2680
Fanno
with
friction
PIP*
pOIpO*
VN"
0.4394
1.5553
1.5861
0.4329
1,5804
1.5957
0.4265
1.6062
1.6052
0.4203
I,6326
1.6146
0.4142
1.6597
1.6239
0.4082
1.6875
1.6330
0.3939
1.7600
1.6553
0.3802
1,8369
1.6769
0.3673
1.9185
1.6977
0.3549
2.0050
1.7179
0.3432
2.0964
1.7374
0.3320
2.1931
1.7563
0.3213
2.2953
1.7745
0.3111
2.4031
1.7922
0.3014
2.5168
1,8092
0.2921
2.6367
1.8257
0.2832
2.7630
1.8417
0.2747
2.8960
1.8571
0.2666
3.0359
1.8721
0.2588
3.1830
1.8865
0.2513
3.3377
1.go05
0.2441
3.5001
1.9140
0.2373
3.6707
1.9271
0.2307
3.8498
1.9398
0.2243
4.0376
1.9521
0.2182
4.2346
1.9640
0.1961
5.1210
2.0079
0.1770
6.1837
2.0466
0.1606
7.4501
2.0808
0.1462
8.9506
2.1111
Table A8(cont.)
Ma
4
4.5
5
5.5
6
6.5
7
7.5
8
8.5
9
9.5
10
Fanno
TIT*
PIP*
0.2857
0.2376
0.2000
0.1702
0.1463
0.1270
0.1111
0.0980
0.0870
0.0777
0.0698
0.0630
0.0571
0.1336
0.1083
0.0894
0.0750
0.0638
0.0548
0.0476
0.0417
0.0369
0.0328
0.0293
0.0264
0.0239
with
pOIpO*
10.719
16.562
25.000
36.869
53.180
75.134
104.14
141.84
190.11
251.09
327.19
421.13
535.94
friction
VN"
2.1381
2.1936
2.2361
2.2691
2.2953
2.3163
2.3333
2.3474
2.3591
2.3689
2.3772
2.3843
2.3905
-
Table A9 Rayleiah flow frictionless with heat transfer v = 1.4
Ma
0
0.01
TIT"
0.0000
0.0006
TOITO*
0.0000
0.0005
plp*
2.4000
2.3997
pOIpO*
VN"
1.2679
1.2678
0.0000
0.0002
Table A9 (cont.) frictionless
TIT*
TOITO*
Ma
with heat
transfer
poipo*
1.2177
0.25
0.3044
0.2568
PIP*
2.2069
0.26
0.3250
0.2745
2.1925
1.2140
0.27
0.3457
0.2923
2.1 777
1.2102
0.28
0.3667
0.3104
2.1626
1.2064
0.29
0.3877
0.3285
2.1472
1.2025
0.3
0.4089
0.3469
2.1314
1.I985
0.31
0.4300
0.3653
2.1 154
1.1945
0.32
0.4512
0.3837
2.0991
1.1904
0.33
0.4723
0.4021
2.0825
1.A863
0.34
0.4933
0.4206
2.0657
1.1822
0.35
0.5141
0.4389
2.0487
1.1779
0.36
0.5348
0.4572
2.0314
1.1737
0.37
0.5553
0.4754
2.0140
1.1695
0.38
0.5755
0.4935
1.9964
1.1652
0.39
0.5955
0.51 13
1.9787
1.1609
0.4
0.6151
0.5290
1.9608
1.I566
0.41
0.6345
0.5465
1.9428
1.1523
0.42
0.6535
0.5638
1.9247
1.I480
0.43
0.6721
0.5808
1.go65
1.1437
1.8882
1.I394
1.8699
1.1351
1.8515
1.I308
1.8331
1.1266
1.8147
1.I224
1.7962
1.1182
Table A9 (cont.) frictionless
with heat
transfer
pOIpO*
Ma
TIT*
TOITO*
0.5
0.7901
0.6914
PIP*
1.7778
0.51
0.8051
0.7058
1.7594
1.lo99
0.52
0.8196
0.7199
1.7409
1.1059
0.53
0.8335
0.7336
1.7226
1.1019
0.54
0.8469
0.7470
1.7043
1.0979
0.55
0.8599
0.7599
1.6860
1.0940
0.56
0.8723
0.7725
1.6678
1.0901
0.57
0.8842
0.7847
1.6496
1.0863
0.58
0.8955
0.7965
1.6316
1.0826
0.59
0.9064
0.8079
1.6136
1.0789
0.6
0.9167
0.8189
1.5957
1.0753
0.61
0.9265
0.8296
1.5780
1.0717
0.62
0.9358
0.8398
1.5603
1.0682
0.63
0.9447
0.8497
15428
1.0648
0.64
0.9530
0.8592
1.5253
1.0615
1.1141
-- -
-
1 J".'
Table A9 (cont.) frictionless
TIT*
TOITO*
0.9401
0.9455
0.9505
0.9553
0.9598
0.9639
0.9679
0.9715
0.9749
0.9781
0.9810
0.9836
0.9861
0.9883
0.9903
0.9921
0.9937
0.9951
0.9963
0.9973
0.9981
0.9988
0.9993
0.9997
0.9999
with heat
transfer
Table A9 (cont.) frictionless
with heal
transfer
Ma
TIT
TOITO"
PIP*
pOIpO*
1
1.oooo
1.oooo
1.oooo
1.oooo
1.02
0.9930
0.9997
0.9770
1.0002
1.04
0.9855
0.9989
0.9546
1.0008
1.06
0.9776
0.9977
0.9327
1.0017
1.08
0.9691
0.9960
0.91 15
1.0031
1.1
0.9603
0.9939
0.8909
1.0049
1.12
0.9512
0.991 5
0.8708
1.0070
1.14
0.9417
0.9887
0.8512
1.0095
1.16
0.9320
0.9856
0.8322
1.0124
1.18
0.9220
0.9823
0.8137
1.0157
1.2
0.9118
0.9787
0.7958
1.0194
1.22
0.901 5
0.9749
0.7783
1.0235
1.24
0.891 1
0.9709
0.7613
1.0279
1.26
0.8805
0.9668
0.7447
1.0328
1.28
0.8699
0.9624
0.7287
1.0380
1.3
0.8592
0.9580
0.7130
1.0437
1.32
0.8484
0.9534
0.6978
1.0497
1.34
0.8377
0.9487
0.6830
1.0561
1.36
0.8269
0.9440
0.6686
1.0629
1.38
0.8161
0.9391
0.6546
1.0701
1.4
0.8054
0.9343
0.641 0
1.0777
1.42
0.7947
0.9293
0.6278
1.0856
1.44
0.7840
0.9243
0.6149
1.0940
1.46
0.7735
0.9193
0.6024
1.lo28
1.48
0.7629
0.9143
0.5902
1.1120
I!
I
I
j
I
1
i
I
I
I
i
!
-
Table A9 (cont.) frictionless
with heat
transfer
PIP*
0.5783
pOIpO*
1.1215
Ma
1.5
0.7525
TORO*
0.9093
1.52
1.54
0.7422
0.7319
0.9042
0.8992
0.5668
1.1315
0.5555
1.1419
1.56
1.58
0.7217
0.7117
0.8942
0.8892
0.5446
1.1527
0.5339
1.1640
1.6
1.62
0.7017
0.691 9
0.8842
0.8792
0.5236
1.64
0.6822
0.8743
0.5036
1.66
0.6726
0.8694
0.4940
1.68
0.6631
0.8645
0.4847
1.7
0.6538
0.8597
0.4756
1.72
0.6445
0.8549
0.4668
1.74
0.6355
0.8502
0.4581
1.76
0.6265
0.8455
0.4497
1.78
0.6176
0.8409
0.4415
1.8
0.6089
0.8363
0.4335
1.82
0.6004
0.8317
0.4257
1.84
0.5919
0.8273
0.4181
1.86
0.5836
0.8228
0.4107
1.88
0.5754
0.8185
0.4035
1.9
0.5673
0.8141
0.3964
1.92
0.5594
0.8099
0.3895
1.94
0.5516
0.8057
0.3828
1.96
0.5439
0.8015
0.3763
2
0.5289
0.7934
0.3636
TTT*
0.5135
Table A9 (cont.) frictionless
with heat
transfer
pOlpO*
Ma
TIT"
TOITO*
2.05
0.5109
0.7835
PIP*
0.3487
2.1
0.4936
0.7741
0.3345
1.6162
2.15
0.4770
0.7649
0.3212
1.6780
2.2
0.4611
0.7561
0.3086
1.7434
2.25
0.4458
0.7477
0.2968
1.8128
2.3
0.4312
0.7395
0.2855
1.8860
2.35
0.4172
0.7317
0.2749
1.9634
2.4
0.4038
0.7242
0.2648
20451
2.45
0.3910
0.7170
0.2552
2.1311
2.5
0.3787
0.7101
0.2462
2.221 8
2.55
0.3669
0.7034
0.2375
2.3173
2.6
0.3556
0.6970
0.2294
2.4177
2.65
0.3448
0.6908
0.2216
2.5233
2.7
0.3344
0.6849
0.2142
2.6343
2.75
0.3244
0.6793
0.2071
2.7508
2.8
0.3149
0.6738
0.2004
2.8731
2.85
0.3057
0.6685
0.1940
3.0014
2.9
0.2969
0.6635
0.1 879
3.1359
2.95
0.2884
0.6586
0.1820
3.2768
3
0.2803
0.6540
0.1765
3.4245
3.2
0.2508
0.6370
0.1 565
4.0871
3.4
0.2255
0.6224
0.1397
4.8783
3.6
0.2037
0.6097
0.1 254
5.8173
3.8
0.1848
0.5987
0.1 131
6.9256
4
0.1683
0.5891
0.1026
8.2268
1.5579
Table A9 (cont.) frictionless
with heat
transfer
pOlpO*
4.2
TIT"
0.1539
TOITO*
0.5807
PIP*
0.0934
9.7473
4.4
0.1412
0.5732
0.0854
11.5155
4.6
0.1300
0.5666
0.0784
13.5629
4.8
0.1200
0.5608
0.0722
15.9234
5
0.1111
0.5556
0.0667
18.6339
5.2
0.1032
0.5509
0.0618
21.7344
5.4
0.0960
0.5467
0.0574
25.2679
5.6
0.0896
0.5429
0.0534
29.2806
5.8
0.0838
0.5394
0.0499
33.8223
6
0.0785
0.5363
0.0467
38.9459
6.2
0.0737
0.5335
0.0438
44.7084
6.4
0.0693
0.5309
0.041 1
51.1700
6.6
0.0653
0.5285
0.0387
58.3953
6.8
0.0616
0.5264
0.0365
66.4524
7
0.0583
0.5244
0.0345
75.4138
7.2
0.0552
0.5225
0.0326
85.3562
7.4
0.0523
0.5208
0.0309
96.3605
7.6
0.0496
0.5193
0.0293
108.5124
7.8
0.0472
0.5178
0.0278
121.go17
8
0.0449
0.5165
0.0265
136.6235
8.2
0.0428
0.5152
0.0252
152.7774
8.4
0.0408
0.5140
0.0241
170.4680
8.6
0.0390
0.51 30
0.0230
189.8050
8.8
0.0373
0.51 19
0.0219
210.9036
9
0.0356
0.51 10
0.0210
233.8840
Ma
Table A9 (cont.) frictionless
Ma
TIT"
9.2
with heat
transfer
plp*
pOIpO*
VN*
0.0341
TOITO*
0.5101
0.0201
258.8719
1.6999
9.4
0.0327
0.5092
0.0192
285.9989
1.7005
9.6
0.0314
0.5085
0.0185
315.4021
1.7011
9.8
0.0301
0.5077
0.0177
347.2245
1.7016
10
0.0290
0.5070
0.0170
381.6149
1.7021
References and Further reading
Douglas, Gasiorek ,Swafiield., Fluid Mechanics. Longman 1985
Fox R. W., and McDonald, A.T., Introduction to Fluid Mechanics. New
York: Wiley 1985.
Hucho, W. H., Aerodynamics of Road Vehicles. Butterworth-Heinemann,
1987.
Janna, W. S., Introduction to FluidMechanics. PWS-Kent 1993
Munson, B. R., and Young, D.E.F., and Okiishi, T. H., Fundamentals of
Fluid Mechanics. New York: 1985 Wiley.
Scior-Rylski, A. J., Road Vehicle Aerodynamics. Pentech Press, London,
1975.
White F. A., FluidMechanics. Mcgraw-Hill2003
-
172
- -
- I
Answers to Problems
-
1.1-1.3 computer problems
1.4 AB = 0,012 mA31sBC=-0,015mA3/s
DC=-0,012mA31s BD=0,048mA3/s
CA=-0,028mA31s
1.5-1.6 not given
1.7 3,38 kPa
1.8 derivation
2.1-2.13 derivations
2.14
2.196 mls
2.15 velocity ratio = 1.054
2.16 1175
2.17 234.4 kN
2.18 194.4 mls 12.99 mls
2.19 for Re similarity 0,028 m 371 mls x N
2.20 0.977 rnls
2.21 378 kPa
3.1-3.4 computer graphs
3.5 derivations
3.6 0,0482 N
3.7 0,106 m
3.8 theorv
3.9 28,23 rnls
3.11 0.005 m 0.0903 Pa 0,0233 m 0,4905 Pa
3.12 15,35 N
3.13 615 600 tonnes, 63 km
3.14 329,3 N
232,9 N
3.15 194 Nm
3.16 17,96 N
3.17 2257 N
3341 N
3.18 2,336 MW 5548 MW 5,2 mls
3.19 7,2 MW
0.375 mm
3.20 378 kW
3.21 30 mm
123 mls
4.1 derivation
4.2 126,lN
4.3 996,3 N
85 mls
4.4 27,74 kN
4.5 3146 N
32814 N
4.618,46N
15N
4.7 108 N 3 kW
4.8 derivation
4.9 0,924 mls
4.10 037 mls
4.11 3,8 mls
4.12 37,gdegrees
4.13 3756 N
3984 N
4.14 795 kNm
4.15 5,5 Hz
516 N
4.16 3,63 rnls 0,0042 Hz 1,78 Hz
4.17 0,271
0,040
4.18 216,7 kN 0 47,2 kN '3,6 8,6 19,7kN
4.19 open
4.20 R102 000
4.21 35 kmlh
5.1 1055 mls
5.2 0,748
5.3 40,29 kPa 259 K 0,541 kglmA3
5.4 747 K 24,94 kPa 11,64 kglmA3
5.5 graphs
5.6 701,2 K 1,799 MPa 8,938 kglmA3
0,5038 667,3 K 6953 kPa 826,9 kPa
3,631 kglmA3 4,109 kg/mA3 260,9 mls
5.7 560 m/sA2
5.8 graphs
5.9 133 kglmA3
5.10
5.11 graph
5.12 0,13
5.13 103 km 1,826 kPa
5.14 0,0265m 246,6 K 31,8 kPa 251,2 rnls
5.15 300 K 1858 kPa
5.16 0,038 m 308 kPa
5.17 324,4 K 14,4 kPa
5.18 7.57 MJlkg 6529 K 42,l kPa
5.19 85 K 24,4 kPa 1373 kPa
5.20 395,4 K 60.4 kPa 209,8 mls
5.21 computer problem
5.22 0,0051 0,532 79,4 kW