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Unformatted text preview: Wind tunnel test of an F18 fighter plane model. Testing of models is imperative in the design
of complex, expensive fluidsengineering devices. Such tests use the principles of dimensional
analysis and modeling from this chapter. (Courtesy of Mark E. Gibson/Visuals Unlimited)  v v 276  eText Main Menu  Textbook Table of Contents  Study Guide Chapter 5
Dimensional Analysis
and Similarity
Motivation. In this chapter we discuss the planning, presentation, and interpretation
of experimental data. We shall try to convince you that such data are best presented in
dimensionless form. Experiments which might result in tables of output, or even multiple volumes of tables, might be reduced to a single set of curves—or even a single
curve—when suitably nondimensionalized. The technique for doing this is dimensional
analysis.
Chapter 3 presented gross controlvolume balances of mass, momentum, and energy which led to estimates of global parameters: mass flow, force, torque, total heat
transfer. Chapter 4 presented infinitesimal balances which led to the basic partial differential equations of fluid flow and some particular solutions. These two chapters covered analytical techniques, which are limited to fairly simple geometries and welldefined boundary conditions. Probably onethird of fluidflow problems can be attacked
in this analytical or theoretical manner.
The other twothirds of all fluid problems are too complex, both geometrically and
physically, to be solved analytically. They must be tested by experiment. Their behavior is reported as experimental data. Such data are much more useful if they are expressed in compact, economic form. Graphs are especially useful, since tabulated data
cannot be absorbed, nor can the trends and rates of change be observed, by most engineering eyes. These are the motivations for dimensional analysis. The technique is
traditional in fluid mechanics and is useful in all engineering and physical sciences,
with notable uses also seen in the biological and social sciences.
Dimensional analysis can also be useful in theories, as a compact way to present an
analytical solution or output from a computer model. Here we concentrate on the presentation of experimental fluidmechanics data. 5.1 Introduction Basically, dimensional analysis is a method for reducing the number and complexity
of experimental variables which affect a given physical phenomenon, by using a sort
of compacting technique. If a phenomenon depends upon n dimensional variables, dimensional analysis will reduce the problem to only k dimensionless variables, where
the reduction n k 1, 2, 3, or 4, depending upon the problem complexity. Generally n k equals the number of different dimensions (sometimes called basic or pri  v v 277  eText Main Menu  Textbook Table of Contents  Study Guide Chapter 5 Dimensional Analysis and Similarity mary or fundamental dimensions) which govern the problem. In fluid mechanics, the
four basic dimensions are usually taken to be mass M, length L, time T, and temperature , or an MLT system for short. Sometimes one uses an FLT system, with force
F replacing mass.
Although its purpose is to reduce variables and group them in dimensionless form,
dimensional analysis has several side benefits. The first is enormous savings in time
and money. Suppose one knew that the force F on a particular body immersed in a
stream of fluid depended only on the body length L, stream velocity V, fluid density
, and fluid viscosity , that is,
F f(L, V, , ) (5.1) Suppose further that the geometry and flow conditions are so complicated that our integral theories (Chap. 3) and differential equations (Chap. 4) fail to yield the solution
for the force. Then we must find the function f(L, V, , ) experimentally.
Generally speaking, it takes about 10 experimental points to define a curve. To find
the effect of body length in Eq. (5.1), we have to run the experiment for 10 lengths L.
For each L we need 10 values of V, 10 values of , and 10 values of , making a grand
total of 104, or 10,000, experiments. At $50 per experiment—well, you see what we
are getting into. However, with dimensional analysis, we can immediately reduce
Eq. (5.1) to the equivalent form
F
V2L2
or CF g VL
(5.2) g(Re) i.e., the dimensionless force coefficient F/( V2L2) is a function only of the dimensionless Reynolds number VL / . We shall learn exactly how to make this reduction in
Secs. 5.2 and 5.3.
The function g is different mathematically from the original function f, but it contains all the same information. Nothing is lost in a dimensional analysis. And think of
the savings: We can establish g by running the experiment for only 10 values of the
single variable called the Reynolds number. We do not have to vary L, V, , or separately but only the grouping VL / . This we do merely by varying velocity V in, say,
a wind tunnel or drop test or water channel, and there is no need to build 10 different
bodies or find 100 different fluids with 10 densities and 10 viscosities. The cost is now
about $500, maybe less.
A second side benefit of dimensional analysis is that it helps our thinking and planning for an experiment or theory. It suggests dimensionless ways of writing equations
before we waste money on computer time to find solutions. It suggests variables which
can be discarded; sometimes dimensional analysis will immediately reject variables,
and at other times it groups them off to the side, where a few simple tests will show
them to be unimportant. Finally, dimensional analysis will often give a great deal of
insight into the form of the physical relationship we are trying to study.
A third benefit is that dimensional analysis provides scaling laws which can convert data from a cheap, small model to design information for an expensive, large prototype. We do not build a milliondollar airplane and see whether it has enough lift
force. We measure the lift on a small model and use a scaling law to predict the lift on  v v 278  eText Main Menu  Textbook Table of Contents  Study Guide 5.1 Introduction 279 the fullscale prototype airplane. There are rules we shall explain for finding scaling
laws. When the scaling law is valid, we say that a condition of similarity exists between the model and the prototype. In the simple case of Eq. (5.1), similarity is achieved
if the Reynolds number is the same for the model and prototype because the function
g then requires the force coefficient to be the same also:
If Rem Rep then CFm CFp (5.3) where subscripts m and p mean model and prototype, respectively. From the definition
of force coefficient, this means that
Fp
Fm p
m Vp
Vm 2 Lp
Lm 2 (5.4) for data taken where pVp Lp / p
mVm Lm / m. Equation (5.4) is a scaling law: If you
measure the model force at the model Reynolds number, the prototype force at the
same Reynolds number equals the model force times the density ratio times the velocity ratio squared times the length ratio squared. We shall give more examples later.
Do you understand these introductory explanations? Be careful; learning dimensional
analysis is like learning to play tennis: There are levels of the game. We can establish some
ground rules and do some fairly good work in this brief chapter, but dimensional analysis in the broad view has many subtleties and nuances which only time and practice and
maturity enable you to master. Although dimensional analysis has a firm physical and
mathematical foundation, considerable art and skill are needed to use it effectively.
EXAMPLE 5.1
A copepod is a water crustacean approximately 1 mm in diameter. We want to know the drag
force on the copepod when it moves slowly in fresh water. A scale model 100 times larger is
made and tested in glycerin at V 30 cm/s. The measured drag on the model is 1.3 N. For similar conditions, what are the velocity and drag of the actual copepod in water? Assume that
Eq. (5.1) applies and the temperature is 20°C. Solution
From Table A.3 the fluid properties are:
Water (prototype): p 0.001 kg/(m s) p 998 kg/m3 Glycerin (model): m 1.5 kg/(m s) m 1263 kg/m3 The length scales are Lm 100 mm and Lp 1 mm. We are given enough model data to compute the Reynolds number and force coefficient
Rem mVmLm
m CFm Fm
Vm2Lm2
m (1263 kg/m3)(0.3 m/s)(0.1 m)
1.5 kg/(m s)
1.3 N
(1263 kg/m3)(0.3 m/s)2(0.1 m)2 25.3
1.14  v v Both these numbers are dimensionless, as you can check. For conditions of similarity, the prototype Reynolds number must be the same, and Eq. (5.2) then requires the prototype force coefficient to be the same  eText Main Menu  Textbook Table of Contents  Study Guide 280 Chapter 5 Dimensional Analysis and Similarity Rep
or Rem Vp
CFp or 998Vp(0.001)
0.001 25.3 0.0253 m/s CFm 1.14
Fp 7.31 2.53 cm/s Ans. Fp
998(0.0253)2(0.001)2
10 7 N Ans. It would obviously be difficult to measure such a tiny drag force. Historically, the first person to write extensively about units and dimensional reasoning
in physical relations was Euler in 1765. Euler’s ideas were far ahead of his time, as were
those of Joseph Fourier, whose 1822 book Analytical Theory of Heat outlined what is now
called the principle of dimensional homogeneity and even developed some similarity rules
for heat flow. There were no further significant advances until Lord Rayleigh’s book in
1877, Theory of Sound, which proposed a “method of dimensions” and gave several examples of dimensional analysis. The final breakthrough which established the method as
we know it today is generally credited to E. Buckingham in 1914 [29], whose paper outlined what is now called the Buckingham pi theorem for describing dimensionless parameters (see Sec. 5.3). However, it is now known that a Frenchman, A. Vaschy, in 1892 and
a Russian, D. Riabouchinsky, in 1911 had independently published papers reporting results equivalent to the pi theorem. Following Buckingham’s paper, P. W. Bridgman published a classic book in 1922 [1], outlining the general theory of dimensional analysis. The
subject continues to be controversial because there is so much art and subtlety in using dimensional analysis. Thus, since Bridgman there have been at least 24 books published on
the subject [2 to 25]. There will probably be more, but seeing the whole list might make
some fledgling authors think twice. Nor is dimensional analysis limited to fluid mechanics or even engineering. Specialized books have been written on the application of dimensional analysis to metrology [26], astrophysics [27], economics [28], building scale
models [36], chemical processing pilot plants [37], social sciences [38], biomedical sciences [39], pharmacy [40], fractal geometry [41], and even the growth of plants [42]. 5.2 The Principle of
Dimensional Homogeneity In making the remarkable jump from the fivevariable Eq. (5.1) to the twovariable
Eq. (5.2), we were exploiting a rule which is almost a selfevident axiom in physics.
This rule, the principle of dimensional homogeneity (PDH), can be stated as follows:
If an equation truly expresses a proper relationship between variables in a physical
process, it will be dimensionally homogeneous; i.e., each of its additive terms will
have the same dimensions.
All the equations which are derived from the theory of mechanics are of this form. For
example, consider the relation which expresses the displacement of a falling body
S S0 V0t 1
2 gt2 (5.5)  v v Each term in this equation is a displacement, or length, and has dimensions {L}. The
equation is dimensionally homogeneous. Note also that any consistent set of units can
be used to calculate a result.  eText Main Menu  Textbook Table of Contents  Study Guide 5.2 The Principle of Dimensional Homogeneity 281 Consider Bernoulli’s equation for incompressible flow
p 12
V
2 gz const (5.6) Each term, including the constant, has dimensions of velocity squared, or {L2T 2}.
The equation is dimensionally homogeneous and gives proper results for any consistent set of units.
Students count on dimensional homogeneity and use it to check themselves when
they cannot quite remember an equation during an exam. For example, which is it:
S 1
2 gt2? or S 12
2 g t? (5.7) By checking the dimensions, we reject the second form and back up our faulty memory. We are exploiting the principle of dimensional homogeneity, and this chapter simply exploits it further.
Equations (5.5) and (5.6) also illustrate some other factors that often enter into a dimensional analysis:
Dimensional variables are the quantities which actually vary during a given case
and would be plotted against each other to show the data. In Eq. (5.5), they are
S and t; in Eq. (5.6) they are p, V, and z. All have dimensions, and all can be
nondimensionalized as a dimensionalanalysis technique.
Dimensional constants may vary from case to case but are held constant during a
given run. In Eq. (5.5) they are S0, V0, and g, and in Eq. (5.6) they are , g,
and C. They all have dimensions and conceivably could be nondimensionalized, but they are normally used to help nondimensionalize the variables in the
problem.
Pure constants have no dimensions and never did. They arise from mathematical
manipulations. In both Eqs. (5.5) and (5.6) they are 1 and the exponent 2, both
2
of which came from an integration: t dt 1 t2, V dV 1 V2. Other common
2
2
dimensionless constants are and e.
Note that integration and differentiation of an equation may change the dimensions
but not the homogeneity of the equation. For example, integrate or differentiate
Eq. (5.5):
S dt S0t
dS
dt 1
2 V0 V0t2 1
6 gt3 gt (5.8a)
(5.8b)  v v In the integrated form (5.8a) every term has dimensions of {LT}, while in the derivative form (5.8b) every term is a velocity {LT 1}.
Finally, there are some physical variables that are naturally dimensionless by virtue
of their definition as ratios of dimensional quantities. Some examples are strain (change
in length per unit length), Poisson’s ratio (ratio of transverse strain to longitudinal strain),
and specific gravity (ratio of density to standard water density). All angles are dimensionless (ratio of arc length to radius) and should be taken in radians for this reason.
The motive behind dimensional analysis is that any dimensionally homogeneous
equation can be written in an entirely equivalent nondimensional form which is more  eText Main Menu  Textbook Table of Contents  Study Guide 282 Chapter 5 Dimensional Analysis and Similarity compact. Usually there is more than one method of presenting one’s dimensionless data
or theory. Let us illustrate these concepts more thoroughly by using the fallingbody
relation (5.5) as an example. Ambiguity: The Choice of
Variables and Scaling Parameters1 Equation (5.5) is familiar and simple, yet illustrates most of the concepts of dimensional analysis. It contains five terms (S, S0, V0, t, g) which we may divide, in our thinking, into variables and parameters. The variables are the things which we wish to plot,
the basic output of the experiment or theory: in this case, S versus t. The parameters
are those quantities whose effect upon the variables we wish to know: in this case S0,
V0, and g. Almost any engineering study can be subdivided in this manner.
To nondimensionalize our results, we need to know how many dimensions are contained among our variables and parameters: in this case, only two, length {L} and time
{T}. Check each term to verify this:
{S} {S0} {L} {t} {T} {V0} {LT 1 } {g} {LT 2 } Among our parameters, we therefore select two to be scaling parameters, used to define dimensionless variables. What remains will be the “basic” parameter(s) whose effect we wish to show in our plot. These choices will not affect the content of our data,
only the form of their presentation. Clearly there is ambiguity in these choices, something that often vexes the beginning experimenter. But the ambiguity is deliberate. Its
purpose is to show a particular effect, and the choice is yours to make.
For the fallingbody problem, we select any two of the three parameters to be scaling parameters. Thus we have three options. Let us discuss and display them in turn.
Option 1: Scaling parameters S0 and V0: the effect of gravity g.
First use the scaling parameters (S0, V0) to define dimensionless (*) displacement
and time. There is only one suitable definition for each:2
S
S0 S* t* V0t
S0 (5.9) Substitute these variables into Eq. (5.5) and clean everything up until each term is dimensionless. The result is our first option:
S* 1 t* 1
t*2
2 gS0
2
V0 (5.10) This result is shown plotted in Fig. 5.1a. There is a single dimensionless parameter ,
which shows here the effect of gravity. It cannot show the direct effects of S0 and V0,
since these two are hidden in the ordinate and abscissa. We see that gravity increases
the parabolic rate of fall for t* 0, but not the initial slope at t* 0. We would learn
the same from fallingbody data, and the plot, within experimental accuracy, would
look like Fig. 5.1a.
1  v v I am indebted to Prof. Jacques Lewalle of Syracuse University for suggesting, outlining, and clarifying this entire discussion.
2
Make them proportional to S and t. Do not define dimensionless terms upside down: S0/S or S0/(V0t).
The plots will look funny, users of your data will be confused, and your supervisor will be angry. It is not
a good idea.  eText Main Menu  Textbook Table of Contents  Study Guide 5.2 The Principle of Dimensional Homogeneity
5 8 g S0
=2
V02
1 0.5 gS
V02 0 S ** = S
S0 g S0
=2
V02
1 6 0.5
0.2 S *= 4 3 283 0
4 2 2 1
0 0 3 2 0 1 Vt
t* = 0
S0 2
gt
t ** =
V0 (a) (b) 1 3 10 8 V0 S *** = S
S0 √gS0 Fig. 5.1 Three entirely equivalent
dimensionless presentations of the
fallingbody problem, Eq. (5.5): the
effect of (a) gravity, (b) initial displacement, and (c) initial velocity.
All plots contain the same information. =2 1 6
0.5
0
4 2 0 0 1 2 3 t *** = t √g / S0
(c) Option 2: Scaling parameters V0 and g: the effect of initial displacement S0.
Now use the new scaling parameters (V0, g) to define dimensionless (**) displacement and time. Again there is only one suitable definition:
S** Sg
2
V0 t** t g
V0 (5.11)  v v Substitute these variables into Eq. (5.5) and clean everything up again. The result is
our second option:
12
gS0
S**
t**
t**
(5.12)
2
2
V0  eText Main Menu  Textbook Table of Contents  Study Guide Chapter 5 Dimensional Analysis and Similarity This result is plotted in Fig. 5.1b. The same single parameter again appears and here
shows the effect of initial displacement, which merely moves the curves upward without changing their shape.
Option 3: Scaling parameters S0 and g: the effect of initial speed V0.
Finally use the scaling parameters (S0, g) to define dimensionless (***) displacement and time. Again there is only one suitable definition:
S
S0 S*** t*** t g
S0 1/2 (5.13) Substitute these variables into Eq. (5.5) and clean everything up as usual. The result
is our third and final option:
S*** 1 t*** 1
t***2
2 1 V0
gS0 (5.14) This final presentation is shown in Fig. 5.1c. Once again the parameter appears, but
we have redefined it upside down,
1/ , so that our display parameter V0 is in
the numerator and is linear. This is our free choice and simply improves the display.
Figure 5.1c shows that initial velocity increases the falling displacement and that the
increase is proportional to time.
Note that, in all three options, the same parameter appears but has a different
meaning: dimensionless gravity, initial displacement, and initial velocity. The graphs,
which contain exactly the same information, change their appearance to reflect these
differences.
Whereas the original problem, Eq. (5.5), involved five quantities, the dimensionless
presentations involve only three, having the form
S gS0
2
V0 fcn(t , ) (5.15) The reduction 5 3 2 should equal the number of fundamental dimensions involved
in the problem {L, T}. This idea led to the pi theorem (Sec. 5.3).
The choice of scaling variables is left to the user, and the resulting dimensionless
parameters have differing interpretations. For example, in the dimensionless dragforce
formulation, Eq. (5.2), it is now clear that the scaling parameters were , V, and L,
since they appear in both the drag coefficient and the Reynolds number. Equation (5.2)
can thus be interpreted as the variation of dimensionless force with dimensionless viscosity, with the scalingparameter effects mixed between CF and Re and therefore not
immediately evident.
Suppose that we wish to study drag force versus velocity. Then we would not use
V as a scaling parameter. We would use ( , , L) instead, and the final dimensionless
function would become
CF F
2 fcn(Re) Re VL (5.16) In plotting these data, we would not be able to discern the effect of or , since they
appear in both dimensionless groups. The grouping CF again would mean dimension  v v 284  eText Main Menu  Textbook Table of Contents  Study Guide 5.2 The Principle of Dimensional Homogeneity 285 less force, and Re is now interpreted as either dimensionless velocity or size.3 The plot
would be quite different compared to Eq. (5.2), although it contains exactly the same
information. The development of parameters such as CF and Re from the initial variables is the subject of the pi theorem (Sec. 5.3). Some Peculiar Engineering
Equations The foundation of the dimensionalanalysis method rests on two assumptions: (1) The
proposed physical relation is dimensionally homogeneous, and (2) all the relevant variables have been included in the proposed relation.
If a relevant variable is missing, dimensional analysis will fail, giving either algebraic difficulties or, worse, yielding a dimensionless formulation which does not resolve the process. A typical case is Manning’s openchannel formula, discussed in Example 1.4:
1.49 2/3 1/2
V
RS
(1)
n
Since V is velocity, R is a radius, and n and S are dimensionless, the formula is not dimensionally homogeneous. This should be a warning that (1) the formula changes if the
units of V and R change and (2) if valid, it represents a very special case. Equation (1)
in Example 1.4 (see above) predates the dimensionalanalysis technique and is valid
only for water in rough channels at moderate velocities and large radii in BG units.
Such dimensionally inhomogeneous formulas abound in the hydraulics literature.
Another example is the HazenWilliams formula [30] for volume flow of water through
a straight smooth pipe
dp 0.54
Q 61.9D2.63
(5.17)
dx
where D is diameter and dp/dx is the pressure gradient. Some of these formulas arise
because numbers have been inserted for fluid properties and other physical data into
perfectly legitimate homogeneous formulas. We shall not give the units of Eq. (5.17)
to avoid encouraging its use.
On the other hand, some formulas are “constructs” which cannot be made dimensionally homogeneous. The “variables” they relate cannot be analyzed by the dimensionalanalysis technique. Most of these formulas are raw empiricisms convenient to a
small group of specialists. Here are three examples:
B
S
0.0147DE 25,000
100 R (5.18) 140
130 API (5.19) 3.74
DE 0.26tR 172
tR (5.20) Equation (5.18) relates the Brinell hardness B of a metal to its Rockwell hardness R.
Equation (5.19) relates the specific gravity S of an oil to its density in degrees API.  v v 3
We were lucky to achieve a size effect because in this case L, a scaling parameter, did not appear in
the drag coefficient.  eText Main Menu  Textbook Table of Contents  Study Guide 286 Chapter 5 Dimensional Analysis and Similarity Equation (5.20) relates the viscosity of a liquid in DE, or degrees Engler, to its viscosity tR in Saybolt seconds. Such formulas have a certain usefulness when communicated between fellow specialists, but we cannot handle them here. Variables like
Brinell hardness and Saybolt viscosity are not suited to an MLT dimensional system. 5.3 The Pi Theorem There are several methods of reducing a number of dimensional variables into a smaller
number of dimensionless groups. The scheme given here was proposed in 1914 by
Buckingham [29] and is now called the Buckingham pi theorem. The name pi comes
from the mathematical notation , meaning a product of variables. The dimensionless
groups found from the theorem are power products denoted by 1, 2, 3, etc. The
method allows the pis to be found in sequential order without resorting to free exponents.
The first part of the pi theorem explains what reduction in variables to expect:
If a physical process satisfies the PDH and involves n dimensional variables, it can
be reduced to a relation between only k dimensionless variables or ’s. The reduction j n k equals the maximum number of variables which do not form a pi
among themselves and is always less than or equal to the number of dimensions describing the variables.
Take the specific case of force on an immersed body: Eq. (5.1) contains five variables
F, L, U, , and described by three dimensions {MLT}. Thus n 5 and j 3. Therefore it is a good guess that we can reduce the problem to k pis, with k n j 5
3 2. And this is exactly what we obtained: two dimensionless variables 1 CF and
Re. On rare occasions it may take more pis than this minimum (see Example
2
5.5).
The second part of the theorem shows how to find the pis one at a time:
Find the reduction j, then select j scaling variables which do not form a pi among
themselves.4 Each desired pi group will be a power product of these j variables plus
one additional variable which is assigned any convenient nonzero exponent. Each
pi group thus found is independent.
To be specific, suppose that the process involves five variables
1 f ( 2, 3, 4, 5) Suppose that there are three dimensions {MLT} and we search around and find that indeed j 3. Then k 5 3 2 and we expect, from the theorem, two and only two
pi groups. Pick out three convenient variables which do not form a pi, and suppose
these turn out to be 2, 3, and 4. Then the two pi groups are formed by power products of these three plus one additional variable, either 1 or 5:
1 ( 2)a( 3)b( 4)c 1 M0L0T0 2 ( 2)a( 3)b( 4)c 5 M0L0T0 Here we have arbitrarily chosen 1 and 5, the added variables, to have unit exponents.
Equating exponents of the various dimensions is guaranteed by the theorem to give
unique values of a, b, and c for each pi. And they are independent because only 1  v v 4  Make a clever choice here because all pis will contain these j variables in various groupings. eText Main Menu  Textbook Table of Contents  Study Guide 5.3 The Pi Theorem
Table 5.1 Dimensions of FluidMechanics Properties 287 Dimensions
Quantity Symbol Length
Area
Volume
Velocity
Acceleration
Speed of sound
Volume flow
Mass flow
Pressure, stress
Strain rate
Angle
Angular velocity
Viscosity
Kinematic viscosity
Surface tension
Force
Moment, torque
Power
Work, energy
Density
Temperature
Specific heat
Specific weight
Thermal conductivity
Expansion coefficient MLT L
A FLT L
L2
L3
LT 1
LT 2
LT 1
L3T 1
MT 1
ML 1T 2
T1
None
T1
ML 1T 1
L2T 1
MT 2
MLT 2
ML2T 2
ML2T –3
ML2T 2
ML–3
L2T 2
ML–2T
MLT –3 V
dV/dt
a
Q
m
˙
p,
˙ F
M
P
W, E
T
cp, c
k L
L2
L3
LT 1
LT 2
LT 1
L3T 1
FTL 1
FL 2
T1
None
T1
FTL 2
L2T 1
FL 1
F
FL
FLT 1
FL
FT2L–4
L2T
FL
FT 1
2
1 1 2 1 3
1 1 1 contains 1 and only 2 contains 5. It is a very neat system once you get used to the
procedure. We shall illustrate it with several examples.
Typically, six steps are involved:
1.
2.
3. 4.  v v 5.  List and count the n variables involved in the problem. If any important variables are missing, dimensional analysis will fail.
List the dimensions of each variable according to {MLT } or {FLT }. A list is
given in Table 5.1.
Find j. Initially guess j equal to the number of different dimensions present, and
look for j variables which do not form a pi product. If no luck, reduce j by 1
and look again. With practice, you will find j rapidly.
Select j scaling parameters which do not form a pi product. Make sure they
please you and have some generality if possible, because they will then appear
in every one of your pi groups. Pick density or velocity or length. Do not pick
surface tension, e.g., or you will form six different independent Webernumber
parameters and thoroughly annoy your colleagues.
Add one additional variable to your j repeating variables, and form a power
product. Algebraically find the exponents which make the product dimensionless. Try to arrange for your output or dependent variables (force, pressure drop,
torque, power) to appear in the numerator, and your plots will look better. Do eText Main Menu  Textbook Table of Contents  Study Guide Chapter 5 Dimensional Analysis and Similarity 6. this sequentially, adding one new variable each time, and you will find all
n j k desired pi products.
Write the final dimensionless function, and check your work to make sure all pi
groups are dimensionless. EXAMPLE 5.2
Repeat the development of Eq. (5.2) from Eq. (5.1), using the pi theorem. Solution
Write the function and count variables: Step 1 F f(L, U, , ) there are five variables (n 5) List dimensions of each variable. From Table 5.1 Step 2 F
{MLT L
2 } U {L} {LT 1 {ML 3} } {ML 1T 1 } Step 3 Find j. No variable contains the dimension , and so j is less than or equal to 3 (MLT). We inspect the list and see that L, U, and cannot form a pi group because only contains mass and
only U contains time. Therefore j does equal 3, and n j 5 3 2 k. The pi theorem
guarantees for this problem that there will be exactly two independent dimensionless groups. Step 4 Select repeating j variables. The group L, U, Step 5 Combine L, U, with one additional variable, in sequence, to find the two pi products.
First add force to find 1. You may select any exponent on this additional term as you please,
to place it in the numerator or denominator to any power. Since F is the output, or dependent,
variable, we select it to appear to the first power in the numerator:
1 LaUb cF (L)a(LT we found in step 3 will do fine. 1b ) (ML 3)c( MLT 2 ) M0L0T0 Equate exponents:
Length: a b 3c Time: 1 0 c 1 0 2 Mass: 0 b We can solve explicitly for
a 2 b 2 c 1 F
U2L2 CF Therefore
1 L 2U 2 1 F Ans. This is exactly the right pi group as in Eq. (5.2). By varying the exponent on F, we could have
found other equivalent groups such as UL 1/2/F1/2.  v v 288  eText Main Menu  Textbook Table of Contents  Study Guide 5.3 The Pi Theorem 289 Finally, add viscosity to L, U, and to find 2. Select any power you like for viscosity. By
hindsight and custom, we select the power 1 to place it in the denominator:
2 LaUb c 1 La(LT 1b ) (ML 3)c(ML 1T 1 ) 1 M0L0T0 Equate exponents:
Length: a b 3c 1 0 c 1 0 1 0 Mass:
Time: b from which we find
a
Therefore 2 b L1U1 c 1 1
UL 1 Re Ans. We know we are finished; this is the second and last pi group. The theorem guarantees that the
functional relationship must be of the equivalent form
F
U2L2 g UL Ans. which is exactly Eq. (5.2). EXAMPLE 5.3
Reduce the fallingbody relationship, Eq. (5.5), to a function of dimensionless variables. Why
are there three different formulations? Solution
Write the function and count variables S f (t, S0, V0, g) five variables (n 5) List the dimensions of each variable, from Table 5.1:
S
{L} t S0 {T} {L} V0
LT g
1 } {LT 2 }  v v There are only two primary dimensions (L, T), so that j 2. By inspection we can easily
find two variables which cannot be combined to form a pi, for example, V0 and g. Then j
2, and we expect 5 2 3 pi products. Select j variables among the parameters S0, V0, and
g. Avoid S and t since they are the dependent variables, which should not be repeated in pi
groups.
There are three different options for repeating variables among the group (S0, V0, g). Therefore we can obtain three different dimensionless formulations, just as we did informally with the
fallingbody equation in Sec. 5.2. Take each option in turn:  eText Main Menu  Textbook Table of Contents  Study Guide Chapter 5 Dimensional Analysis and Similarity
1. Choose S0 and V0 as repeating variables. Combine them in turn with (S, t, g):
ab
S1S 0V0 1 f cd
t1S 0V 0 2 e
g1S 0V 0 3 Set each power product equal to L0T0, and solve for the exponents (a, b, c, d, e, f ). Please
allow us to give the results here, and you may check the algebra as an exercise:
a 1 b 0 c S
S0 S* 1 1 d e V0t
S0 t* 2 1 1 f 2
Ans. gS0
2
V0 3 Thus, for option 1, we know that S* fcn(t*, ). We have found, by dimensional analysis, the same variables as in Eq. (5.10). But here there is no formula for the functional relation — we might have to experiment with falling bodies to establish Fig. 5.1a.
2. Choose V0 and g as repeating variables. Combine them in turn with (S, t, S0):
a
S1V 0 gb 1 c
t1V 0gd 2 e
S1V 0gf
0 3 Set each power product equal to L0T0, and solve for the exponents (a, b, c, d, e, f ). Once
more allow us to give the results here, and you may check the algebra as an exercise.
a 2 b 1 c Sg
2
V0 S** 1 1 d t** 2 1 e tg
V0 1 f 2
Ans. gS0
2
V0 3 Thus, for option 2, we now know that S** fcn(t**, ). We have found, by dimensional
analysis, the same groups as in Eq. (5.12). The data would plot as in Fig. 5.1b.
3. Finally choose S0 and g as repeating variables. Combine them in turn with (S, t, V0):
a
S1S 0gb 1 c
t1S0 gd 2 1e
V0S0g f 3 Set each power product equal to L0T0, and solve for the exponents (a, b, c, d, e, f ). One
more time allow us to give the results here, and you may check the algebra as an exercise:
a
1 1 b S*** 0
S
S0 1
2 c
2 t*** d 1
2
g
t
S0 1
2 e f 1
2
V0
gS0 3 Ans. Thus, for option 3, we now know that S*** fcn(t***,
1/
). We have found, by dimensional analysis, the same groups as in Eq. (5.14). The data would plot as in Fig. 5.1c.
Dimensional analysis here has yielded the same pi groups as the use of scaling parameters
with Eq. (5.5). Three different formulations appeared, because we could choose three different
pairs of repeating variables to complete the pi theorem. EXAMPLE 5.4
At low velocities (laminar flow), the volume flow Q through a smallbore tube is a function only
of the tube radius R, the fluid viscosity , and the pressure drop per unit tube length dp/dx. Using the pi theorem, find an appropriate dimensionless relationship.  v v 290  eText Main Menu  Textbook Table of Contents  Study Guide 5.3 The Pi Theorem 291 Solution
Write the given relation and count variables:
Q f R, , dp
dx four variables (n 4) Make a list of the dimensions of these variables from Table 5.1:
Q
{L3T R
1 } dp/dx
{ML 1T {L} 1 {ML 2T } 2 } There are three primary dimensions (M, L, T), hence j 3. By trial and error we determine that
R, , and dp/dx cannot be combined into a pi group. Then j 3, and n j 4 3 1. There
is only one pi group, which we find by combining Q in a power product with the other three:
1 Ra b dp c 1
Q
dx (L)a(ML 1T 1b ) (ML 2T 2c ) (L3T 1 ) M0L0T0
Equate exponents:
Mass: b c b 2c 3 0 Time: b 2c 1 0 Solving simultaneously, we obtain a 4, b Length: a 1 or 1 R 0 1, c
4 1 1. Then dp
dx Q
R 4(dp/dx) 1 Q
const Ans. Since there is only one pi group, it must equal a dimensionless constant. This is as far as dimensional
analysis can take us. The laminarflow theory of Sec. 6.4 shows that the value of the constant is /8. EXAMPLE 5.5
Assume that the tip deflection of a cantilever beam is a function of the tip load P, beam length
L, area moment of inertia I, and material modulus of elasticity E; that is,
f (P, L, I, E). Rewrite
this function in dimensionless form, and comment on its complexity and the peculiar value of j. Solution
List the variables and their dimensions:
P  v v {L}  eText Main Menu {MLT  L
2 } {L} I E 4 {ML 1T {L } Textbook Table of Contents  2 } Study Guide 292 Chapter 5 Dimensional Analysis and Similarity There are five variables (n 5) and three primary dimensions (M, L, T), hence j 3. But try as
we may, we cannot find any combination of three variables which does not form a pi group.
This is because {M} and {T} occur only in P and E and only in the same form, {MT 2}. Thus
we have encountered a special case of j 2, which is less than the number of dimensions (M,
L, T). To gain more insight into this peculiarity, you should rework the problem, using the (F,
L, T) system of dimensions.
With j 2, we select L and E as two variables which cannot form a pi group and then add
other variables to form the three desired pis:
LaEbI1 1 (L)a(ML 1T 2b ) (L4) from which, after equating exponents, we find that a
LaEbP1 2 from which we find a 2, b from which a
or 1, b 1, or LaEb 3 0, or (L)a(ML 1T 1 2 M0L0T0 4, b 2b ) (MLT 0, or 2 1 I/L4. Then M0L0T0 ) P/(EL2), and (L)a(ML 1T 2b ) (L) M0L0T0 /L. The proper dimensionless function is 3 L f P
I
,
EL2 L4 2, 1), Ans. 3 f( (1) This is a complex threevariable function, but dimensional analysis alone can take us no further.
We can “improve” Eq. (1) by taking advantage of some physical reasoning, as Langhaar
points out [8, p. 91]. For small elastic deflections, is proportional to load P and inversely
proportional to moment of inertia I. Since P and I occur separately in Eq. (1), this means
that 3 must be proportional to 2 and inversely proportional to 1. Thus, for these conditions, L
or (const) P L4
EL2 I (const) PL3
EI (2) This could not be predicted by a pure dimensional analysis. Strengthofmaterials theory predicts that the value of the constant is 1 .
3 5.4 Nondimensionalization of
the Basic Equations We could use the pitheorem method of the previous section to analyze problem after
problem after problem, finding the dimensionless parameters which govern in each
case. Textbooks on dimensional analysis [for example, 7] do this. An alternate and very
powerful technique is to attack the basic equations of flow from Chap. 4. Even though
these equations cannot be solved in general, they will reveal basic dimensionless parameters, e.g., Reynolds number, in their proper form and proper position, giving clues
to when they are negligible. The boundary conditions must also be nondimensionalized.
Let us briefly apply this technique to the incompressibleflow continuity and momentum equations with constant viscosity:  v v Continuity:  eText Main Menu V  0 Textbook Table of Contents (5.21a)  Study Guide 5.4 Nondimensionalization of the Basic Equations dV
2
g
p
V
dt
Typical boundary conditions for these two equations are Momentum: Fixed solid surface: V Inlet or outlet:
Free surface, z (5.21b) 0 Known V, p
: w d
dt p 293 pa (5.22)
1 (Rx Ry 1) We omit the energy equation (4.75) and assign its dimensionless form in the problems
(Probs. 5.42 and 5.45).
Equations (5.21) and (5.22) contain the three basic dimensions M, L, and T. All variables p, V, x, y, z, and t can be nondimensionalized by using density and two reference constants which might be characteristic of the particular fluid flow:
Reference velocity U Reference length L For example, U may be the inlet or upstream velocity and L the diameter of a body
immersed in the stream.
Now define all relevant dimensionless variables, denoting them by an asterisk:
V*
x
L x*
t* tU
L y* V
U
y
z*
L
p p* z
L (5.23) gz
U2 All these are fairly obvious except for p*, where we have slyly introduced the gravity
effect, assuming that z is up. This is a hindsight idea suggested by Bernoulli’s equation (3.77).
Since , U, and L are all constants, the derivatives in Eqs. (5.21) can all be handled
in dimensionless form with dimensional coefficients. For example,
u
x (Uu*)
(Lx*) U u*
L x* Substitute the variables from Eqs. (5.23) into Eqs. (5.21) and (5.22) and divide through
by the leading dimensional coefficient, in the same way as we handled Eq. (5.12). The
resulting dimensionless equations of motion are:
Continuity: * V*
dV*
dt* Momentum: 0 *p* UL (5.24a)
*2(V*) The dimensionless boundary conditions are:
Fixed solid surface:  v v Inlet or outlet:  eText Main Menu  V* 0 Known V*, p* Textbook Table of Contents  Study Guide (5.24b) 294 Chapter 5 Dimensional Analysis and Similarity Free surface, z* *:
p* w*
pa
U2 gL
z*
U2 d*
dt*
U2L (5.25)
1 (R*
x R* 1)
y These equations reveal a total of four dimensionless parameters, one in the momentum
equation and three in the freesurfacepressure boundary condition. Dimensionless Parameters In the continuity equation there are no parameters. The momentum equation contains
one, generally accepted as the most important parameter in fluid mechanics:
Reynolds number Re UL It is named after Osborne Reynolds (1842 – 1912), a British engineer who first proposed
it in 1883 (Ref. 4 of Chap. 6). The Reynolds number is always important, with or without a free surface, and can be neglected only in flow regions away from highvelocity
gradients, e.g., away from solid surfaces, jets, or wakes.
The noslip and inletexit boundary conditions contain no parameters. The freesurfacepressure condition contains three:
Euler number (pressure coefficient) Eu pa
U2 This is named after Leonhard Euler (1707 – 1783) and is rarely important unless the
pressure drops low enough to cause vapor formation (cavitation) in a liquid. The Euler
number is often written in terms of pressure differences: Eu
p/( U2). If p involves
vapor pressure p , it is called the cavitation number Ca (pa p )/( U2).
The second pressure parameter is much more important:
Froude number Fr U2
gL It is named after William Froude (1810 – 1879), a British naval architect who, with his
son Robert, developed the shipmodel towingtank concept and proposed similarity
rules for freesurface flows (ship resistance, surface waves, open channels). The Froude
number is the dominant effect in freesurface flows and is totally unimportant if there
is no free surface. Chapter 10 investigates Froude number effects in detail.
The final freesurface parameter is
Weber number We U2L  v v It is named after Moritz Weber (1871 – 1951) of the Polytechnic Institute of Berlin, who
developed the laws of similitude in their modern form. It was Weber who named Re
and Fr after Reynolds and Froude. The Weber number is important only if it is of order unity or less, which typically occurs when the surface curvature is comparable in
size to the liquid depth, e.g., in droplets, capillary flows, ripple waves, and very small
hydraulic models. If We is large, its effect may be neglected.  eText Main Menu  Textbook Table of Contents  Study Guide 5.4 Nondimensionalization of the Basic Equations 295 If there is no free surface, Fr, Eu, and We drop out entirely, except for the possibility of cavitation of a liquid at very small Eu. Thus, in lowspeed viscous flows with
no free surface, the Reynolds number is the only important dimensionless parameter. Compressibility Parameters In highspeed flow of a gas there are significant changes in pressure, density, and temperature which must be related by an equation of state such as the perfectgas law,
Eq. (1.10). These thermodynamic changes introduce two additional dimensionless parameters mentioned briefly in earlier chapters:
Mach number Ma U
a Specificheat ratio k cp
c The Mach number is named after Ernst Mach (1838 – 1916), an Austrian physicist. The
effect of k is only slight to moderate, but Ma exerts a strong effect on compressibleflow properties if it is greater than about 0.3. These effects are studied in Chap. 9. Oscillating Flows If the flow pattern is oscillating, a seventh parameter enters through the inlet boundary condition. For example, suppose that the inlet stream is of the form
u U cos t Nondimensionalization of this relation results in
u
U u* cos L
t*
U The argument of the cosine contains the new parameter
Strouhal number St L
U  v v The dimensionless forces and moments, friction, and heat transfer, etc., of such an oscillating flow would be a function of both Reynolds and Strouhal numbers. This parameter is named after V. Strouhal, a German physicist who experimented in 1878 with
wires singing in the wind.
Some flows which you might guess to be perfectly steady actually have an oscillatory pattern which is dependent on the Reynolds number. An example is the periodic
vortex shedding behind a blunt body immersed in a steady stream of velocity U. Figure 5.2a shows an array of alternating vortices shed from a circular cylinder immersed
in a steady crossflow. This regular, periodic shedding is called a Kármán vortex street,
after T. von Kármán, who explained it theoretically in 1912. The shedding occurs in
the range 102 Re 107, with an average Strouhal number d/(2 U) 0.21. Figure
5.2b shows measured shedding frequencies.
Resonance can occur if a vortex shedding frequency is near a body’s structuralvibration frequency. Electric transmission wires sing in the wind, undersea mooring
lines gallop at certain current speeds, and slender structures flutter at critical wind or
vehicle speeds. A striking example is the disastrous failure of the Tacoma Narrows suspension bridge in 1940, when windexcited vortex shedding caused resonance with the
natural torsional oscillations of the bridge.  eText Main Menu  Textbook Table of Contents  Study Guide 296 Chapter 5 Dimensional Analysis and Similarity (a)
0.4
Data spread St = ω d
2π U 0.3 0.2 0.1 Fig. 5.2 Vortex shedding from a
circular cylinder: (a) vortex street
behind a circular cylinder (from
Ref. 33, courtesy of U.S. Naval Research Laboratory); (b) experimental shedding frequencies (data
from Refs. 31 and 32). 0
10 10 2 10 3 10 4
ρ Ud
Re =
µ 10 5 10 6 107 (b) Other Dimensionless Parameters We have discussed seven important parameters in fluid mechanics, and there are others. Four additional parameters arise from nondimensionalization of the energy equation (4.75) and its boundary conditions. These four (Prandtl number, Eckert number,
Grashof number, and walltemperature ratio) are listed in Table 5.2 just in case you fail
to solve Prob. 5.42. Another important and rather sneaky parameter is the wallroughness ratio /L (in Table 5.2).5 Slight changes in surface roughness have a strik  v v 5
Roughness is easy to overlook because it is a slight geometric effect which does not appear in the
equations of motion.  eText Main Menu  Textbook Table of Contents  Study Guide 5.4 Nondimensionalization of the Basic Equations
Table 5.2 Dimensionless Groups in
Fluid Mechanics Parameter Qualitative ratio
of effects Definition 297 Importance UL Inertia
Viscosity Always Ma U
a Flow speed
Sound speed Compressible flow Froude number Fr U2
gL Inertia
Gravity Freesurface flow Weber number We Inertia
Surface tension Freesurface flow Cavitation number
(Euler number) Ca Pressure
Inertia Cavitation Prandtl number Pr cp
k Dissipation
Conduction Heat convection Eckert number Ec U2
cpT0 Kinetic energy
Enthalpy Dissipation Specificheat ratio k Enthalpy
Internal energy Compressible flow Strouhal number St Oscillation
Mean speed Oscillating flow Wall roughness
Body length Turbulent, rough walls Buoyancy
Viscosity Natural convection Wall temperature
Stream temperature Heat transfer Static pressure
Dynamic pressure Aerodynamics, hydrodynamics Lift force
Dynamic force Aerodynamics, hydrodynamics Drag force
Dynamic force Aerodynamics, hydrodynamics Reynolds number Re Mach number Roughness ratio U2L
p p
U2 cp
c
L
U L Grashof number
Temperature ratio Tw
T0 Pressure coefficient Cp Lift coefficient CL Drag coefficient CD TgL3 Gr 2 p
1
2 p
U2
L 1
2 U2A
D 1
2 U2A 2  v v ing effect in the turbulentflow or highReynoldsnumber range, as we shall see in Chap.
6 and in Fig. 5.3.
This book is primarily concerned with Reynolds, Mach, and Froudenumber effects, which dominate most flows. Note that we discovered all these parameters (except /L) simply by nondimensionalizing the basic equations without actually solving
them.
If the reader is not satiated with the 15 parameters given in Table 5.2, Ref. 34 contains a list of over 300 dimensionless parameters in use in engineering. See also
Ref. 35.  eText Main Menu  Textbook Table of Contents  Study Guide 298 Chapter 5 Dimensional Analysis and Similarity Dimensional analysis is fun, but does it work? Yes; if all important variables are included in the proposed function, the dimensionless function found by dimensional
analysis will collapse all the data onto a single curve or set of curves.
An example of the success of dimensional analysis is given in Fig. 5.3 for the measured drag on smooth cylinders and spheres. The flow is normal to the axis of the cylinder, which is extremely long, L/d → . The data are from many sources, for both liquids and gases, and include bodies from several meters in diameter down to fine wires
and balls less than 1 mm in size. Both curves in Fig. 5.3a are entirely experimental;
the analysis of immersed body drag is one of the weakest areas of modern fluidmechanics theory. Except for some isolated digitalcomputer calculations, there is no
theory for cylinder and sphere drag except creeping flow, Re 1.
The Reynolds number of both bodies is based upon diameter, hence the notation
Red. But the drag coefficients are defined differently: A Successful Application 5 4 Cylinder
length effect
Transition to turbulent
boundary layer (10 4 < Re < 10 5)
L/d CD
2
Cylinder (two – dimensional)
1 CD ∞
40
20
10
5
3
2
1 3 1.20
0.98
0.91
0.82
0.74
0.72
0.68
0.64 Sphere
0
10 10 2 10 3 10 4
ρ Ud
Red =
µ
(a) 10 5 10 6 10 7 1.5
1.0
CD 0.7  v v Fig. 5.3 The proof of practical dimensional analysis: drag coefficients of a cylinder and sphere:
(a) drag coefficient of a smooth
cylinder and sphere (data from
many sources); (b) increased
roughness causes earlier transition
to a turbulent boundary layer. 0.5
0.3
10 4 Cylinder: ε = 0.02
−
d
0.009
0.007
0.004
0.002
0.0005 _
L=∞
d Smooth
10 5
Red 10 6 (b)  eText Main Menu  Textbook Table of Contents  Study Guide 5.4 Nondimensionalization of the Basic Equations drag
1
2 CD
1
2 299 cylinder U2Ld
drag
U2 1 d2
4 (5.26)
sphere They both have a factor 1 as a traditional tribute to Bernoulli and Euler, and both are
2
based on the projected area, i.e., the area one sees when looking toward the body from
upstream. The usual definition of CD is thus
CD 1
2 drag
U2(projected area) (5.27) However, one should carefully check the definitions of CD, Re, etc., before using data
in the literature. Airfoils, e.g., use the planform area.
Figure 5.3a is for long, smooth cylinders. If wall roughness and cylinder length are
included as variables, we obtain from dimensional analysis a complex threeparameter
function
CD f Red, L
,
dd (5.28) To describe this function completely would require 1000 or more experiments. Therefore
it is customary to explore the length and roughness effects separately to establish trends.
The table with Fig. 5.3a shows the length effect with zero wall roughness. As length
decreases, the drag decreases by up to 50 percent. Physically, the pressure is “relieved”
at the ends as the flow is allowed to skirt around the tips instead of deflecting over and
under the body.
Figure 5.3b shows the effect of wall roughness for an infinitely long cylinder. The
sharp drop in drag occurs at lower Red as roughness causes an earlier transition to a
turbulent boundary layer on the surface of the body. Roughness has the same effect on
sphere drag, a fact which is exploited in sports by deliberate dimpling of golf balls to
give them less drag at their flight Red 105.
Figure 5.3 is a typical experimental study of a fluidmechanics problem, aided by
dimensional analysis. As time and money and demand allow, the complete threeparameter relation (5.28) could be filled out by further experiments. EXAMPLE 5.6
The capillary rise h of a liquid in a tube varies with tube diameter d, gravity g, fluid density ,
surface tension , and the contact angle . (a) Find a dimensionless statement of this relation.
(b) If h 3 cm in a given experiment, what will h be in a similar case if the diameter and surface tension are half as much, the density is twice as much, and the contact angle is the same? Solution
Part (a) Step 1 Write down the function and count variables:  v v h  eText Main Menu  f(d, g, , ,) Textbook Table of Contents n  6 variables Study Guide Chapter 5 Dimensional Analysis and Similarity Step 2 List the dimensions {FLT} from Table 5.2:
h Step 3 d {L} {L} g
{LT 2 {FT2L 4} } {FL 1} none Find j. Several groups of three form no pi: , , and g or , g, and d. Therefore j 3, and we
expect n j 6 3 3 dimensionless groups. One of these is obviously , which is already
dimensionless:
Ans. (a) 3 If we had carelessly chosen to search for it by using steps 4 and 5, we would still find Step 4 3 . Select j repeating variables which do not form a pi group: , g, d. Step 5 Add one additional variable in sequence to form the pis:
abc Add h: (FT 2L 4)a(LT gdh 1 2b ) (L)c(L) F 0L0T 0 Solve for
a b Finally add 0 c 1
h
d 00 Therefore g d 1h 1 Ans. (a) , again selecting its exponent to be 1:
abc gd 2 (FT 2L 4)a(LT 2b ) (L)c(FL 1) F0L0T0 Solve for
a Step 6 b 1
1 Therefore 2 c g 1d 2 2 Ans. (a) gd2 The complete dimensionless relation for this problem is thus
h
d F gd2 , Ans. (a) (1) This is as far as dimensional analysis goes. Theory, however, establishes that h is proportional
to . Since occurs only in the second parameter, we can slip it outside
h
d actual gd2 F1( ) Example 1.9 showed theoretically that F1( ) h gd or F1( ) 4 cos . 1
We are given h1 for certain conditions d1, 1, 1, and 1. If h1 3 cm, what is h2 for d2 2 d1, 2
1
2 1, and 2
2 1, 2
1? We know the functional relation, Eq. (1), must still hold at condition 2 Part (b) h2
d2 2
2
2gd 2 , F 2 But
2
2
2gd 2  v v 300  eText Main Menu  1
2 1 2 1g( 1 d1)2
2 1
2
1gd1 Textbook Table of Contents  Study Guide 5.5 Modeling and Its Pitfalls 301 Therefore, functionally,
h2
d2 F 1
2
1gd1 , 1 h1
d1 We are given a condition 2 which is exactly similar to condition 1, and therefore a scaling law
holds
h2 h1 d2
d1 (3 cm) 1
21 d
d1 1.5 cm Ans. (b) If the pi groups had not been exactly the same for both conditions, we would have had to know
more about the functional relation F to calculate h2. 5.5 Modeling and Its Pitfalls So far we have learned about dimensional homogeneity and the pitheorem method,
using power products, for converting a homogeneous physical relation to dimensionless form. This is straightforward mathematically, but there are certain engineering difficulties which need to be discussed.
First, we have more or less taken for granted that the variables which affect the
process can be listed and analyzed. Actually, selection of the important variables requires considerable judgment and experience. The engineer must decide, e.g., whether
viscosity can be neglected. Are there significant temperature effects? Is surface tension
important? What about wall roughness? Each pi group which is retained increases the
expense and effort required. Judgment in selecting variables will come through practice and maturity; this book should provide some of the necessary experience.
Once the variables are selected and the dimensional analysis is performed, the experimenter seeks to achieve similarity between the model tested and the prototype to
be designed. With sufficient testing, the model data will reveal the desired dimensionless function between variables
1 f( 2, 3, ... k) (5.29) With Eq. (5.29) available in chart, graphical, or analytical form, we are in a position
to ensure complete similarity between model and prototype. A formal statement would
be as follows:
Flow conditions for a model test are completely similar if all relevant dimensionless
parameters have the same corresponding values for the model and the prototype.
This follows mathematically from Eq. (5.29). If 2m
2p,
3m
3p, etc., Eq. (5.29)
guarantees that the desired output 1m will equal 1p. But this is easier said than done,
as we now discuss.
Instead of complete similarity, the engineering literature speaks of particular types
of similarity, the most common being geometric, kinematic, dynamic, and thermal. Let
us consider each separately.  v v Geometric Similarity Geometric similarity concerns the length dimension {L} and must be ensured before
any sensible model testing can proceed. A formal definition is as follows:  eText Main Menu  Textbook Table of Contents  Study Guide 302 Chapter 5 Dimensional Analysis and Similarity A model and prototype are geometrically similar if and only if all body dimensions
in all three coordinates have the same linearscale ratio.
Note that all length scales must be the same. It is as if you took a photograph of the prototype and reduced it or enlarged it until it fitted the size of the model. If the model is
to be made onetenth the prototype size, its length, width, and height must each be onetenth as large. Not only that, but also its entire shape must be onetenth as large, and
technically we speak of homologous points, which are points that have the same relative
location. For example, the nose of the prototype is homologous to the nose of the model.
The left wingtip of the prototype is homologous to the left wingtip of the model. Then
geometric similarity requires that all homologous points be related by the same linearscale ratio. This applies to the fluid geometry as well as the model geometry.
All angles are preserved in geometric similarity. All flow directions are preserved.
The orientations of model and prototype with respect to the surroundings must be
identical.
Figure 5.4 illustrates a prototype wing and a onetenthscale model. The model
lengths are all onetenth as large, but its angle of attack with respect to the free stream
is the same: 10° not 1°. All physical details on the model must be scaled, and some
are rather subtle and sometimes overlooked:
1.
2.
3. The model nose radius must be onetenth as large.
The model surface roughness must be onetenth as large.
If the prototype has a 5mm boundarylayer trip wire 1.5 m from the leading
edge, the model should have a 0.5mm trip wire 0.15 m from its leading edge.
If the prototype is constructed with protruding fasteners, the model should have
homologous protruding fasteners onetenth as large. 4. And so on. Any departure from these details is a violation of geometric similarity and
must be justified by experimental comparison to show that the prototype behavior was
not significantly affected by the discrepancy.
Models which appear similar in shape but which clearly violate geometric similarity should not be compared except at your own risk. Figure 5.5 illustrates this point. Homologous
points a * 40 m
1m a
10°
Vp 4m
10°
8m Vm v v Fig. 5.4 Geometric similarity in
model testing: (a) prototype;
(b) onetenthscale model.  (a)  0.1 m eText Main Menu  Textbook Table of Contents 0.8 m (b)  Study Guide * 5.5 Modeling and Its Pitfalls V2 V1
Huge
sphere V3 303 V4 Large
sphere Medium
sphere Tiny
sphere (a) V1 V2 V3 Large 4:1
ellipsoid Fig. 5.5 Geometric similarity and
dissimilarity of flows: (a) similar;
(b) dissimilar. Medium 3.5:1
ellipsoid Small 3:1
ellipsoid (b) The spheres in Fig. 5.5a are all geometrically similar and can be tested with a high expectation of success if the Reynolds number or Froude number, etc., is matched. But
the ellipsoids in Fig. 5.5b merely look similar. They actually have different linearscale
ratios and therefore cannot be compared in a rational manner, even though they may
have identical Reynolds and Froude numbers, etc. The data will not be the same for
these ellipsoids, and any attempt to “compare” them is a matter of rough engineering
judgment. Kinematic Similarity Kinematic similarity requires that the model and prototype have the same lengthscale
ratio and the same timescale ratio. The result is that the velocityscale ratio will be
the same for both. As Langhaar [8] states it:
The motions of two systems are kinematically similar if homologous particles lie at
homologous points at homologous times.
Lengthscale equivalence simply implies geometric similarity, but timescale equivalence may require additional dynamic considerations such as equivalence of the
Reynolds and Mach numbers.
One special case is incompressible frictionless flow with no free surface, as sketched in
Fig. 5.6a. These perfectfluid flows are kinematically similar with independent length and
time scales, and no additional parameters are necessary (see Chap. 8 for further details).
Frictionless flows with a free surface, as in Fig. 5.6b, are kinematically similar if
their Froude numbers are equal
Frm 2
Vm
gLm 2
Vp
gLp Frp (5.30) Note that the Froude number contains only length and time dimensions and hence is
a purely kinematic parameter which fixes the relation between length and time. From
Eq. (5.30), if the length scale is  v v Lm  eText Main Menu  Textbook Table of Contents Lp  (5.31) Study Guide 304 Chapter 5 Dimensional Analysis and Similarity V1p
V1m = β V1p Dm =
α Dp V∞ m = βV∞ p Dp V∞ p Model V2 m = βV2 p V2 p
Prototype
(a)
λp
Prototype
waves: Cp
Hp
Period Tp Fig. 5.6 Frictionless lowspeed
flows are kinematically similar:
(a) Flows with no free surface are
kinematically similar with independent length and timescale ratios;
(b) freesurface flows are kinematically similar with length and time
scales related by the Froude
number. Vp
λm = α λ p Hm = α Hp
Cm = C p √α
Model
waves:
Vm = V p √α Period Tm = T p √α
(b) where is a dimensionless ratio, the velocity scale is
1/ 2 Vm
Vp
Tm
Tp and the time scale is Lm
Lp Lm/Vm
Lp/Vp (5.32) (5.33) These Froudescaling kinematic relations are illustrated in Fig. 5.6b for wavemotion
modeling. If the waves are related by the length scale , then the wave period, propagation speed, and particle velocities are related by
.
If viscosity, surface tension, or compressibility is important, kinematic similarity is
dependent upon the achievement of dynamic similarity. Dynamic Similarity  v v Dynamic similarity exists when the model and the prototype have the same lengthscale ratio, timescale ratio, and forcescale (or massscale) ratio. Again geometric sim  eText Main Menu  Textbook Table of Contents  Study Guide 5.5 Modeling and Its Pitfalls 305 ilarity is a first requirement; without it, proceed no further. Then dynamic similarity
exists, simultaneous with kinematic similarity, if the model and prototype force and
pressure coefficients are identical. This is ensured if:
1. For compressible flow, the model and prototype Reynolds number and Mach
number and specificheat ratio are correspondingly equal.
2. For incompressible flow
a. With no free surface: model and prototype Reynolds numbers are equal.
b. With a free surface: model and prototype Reynolds number, Froude number, and
(if necessary) Weber number and cavitation number are correspondingly equal.
Mathematically, Newton’s law for any fluid particle requires that the sum of the pressure force, gravity force, and friction force equal the acceleration term, or inertia force,
Fp Fg Ff Fi The dynamicsimilarity laws listed above ensure that each of these forces will be in the
same ratio and have equivalent directions between model and prototype. Figure 5.7 shows
an example for flow through a sluice gate. The force polygons at homologous points have
exactly the same shape if the Reynolds and Froude numbers are equal (neglecting surface
tension and cavitation, of course). Kinematic similarity is also ensured by these model laws. Discrepancies in Water and Air
Testing The perfect dynamic similarity shown in Fig. 5.7 is more of a dream than a reality because true equivalence of Reynolds and Froude numbers can be achieved only by dramatic changes in fluid properties, whereas in fact most model testing is simply done
with water or air, the cheapest fluids available.
First consider hydraulic model testing with a free surface. Dynamic similarity requires equivalent Froude numbers, Eq. (5.30), and equivalent Reynolds numbers
VmLm
m VpLp (5.34) p Fpp
Fgp
Fip
Ffp Fim v v Fig. 5.7 Dynamic similarity in
sluicegate flow. Model and prototype yield identical homologous
force polygons if the Reynolds and
Froude numbers are the same corresponding values: (a) prototype;
(b) model.   eText Main Menu Fpm a a' (a)  Textbook Table of Contents (b)  Study Guide Fgm Ffm 306 Chapter 5 Dimensional Analysis and Similarity But both velocity and length are constrained by the Froude number, Eqs. (5.31) and
(5.32). Therefore, for a given lengthscale ratio , Eq. (5.34) is true only if
m
p Lm Vm
Lp Vp 3/2 (5.35) For example, for a onetenthscale model,
0.1 and 3/2 0.032. Since p is undoubtedly water, we need a fluid with only 0.032 times the kinematic viscosity of water to achieve dynamic similarity. Referring to Table 1.4, we see that this is impossible: Even mercury has only oneninth the kinematic viscosity of water, and a mercury
hydraulic model would be expensive and bad for your health. In practice, water is used
for both the model and the prototype, and the Reynoldsnumber similarity (5.34) is unavoidably violated. The Froude number is held constant since it is the dominant
parameter in freesurface flows. Typically the Reynolds number of the model flow is
too small by a factor of 10 to 1000. As shown in Fig. 5.8, the lowReynoldsnumber
model data are used to estimate by extrapolation the desired highReynoldsnumber
prototype data. As the figure indicates, there is obviously considerable uncertainty in
using such an extrapolation, but there is no other practical alternative in hydraulic model
testing.
Second, consider aerodynamic model testing in air with no free surface. The important parameters are the Reynolds number and the Mach number. Equation (5.34)
should be satisfied, plus the compressibility criterion
Vm
am Vp
ap (5.36) Elimination of Vm /Vp between (5.34) and (5.36) gives
Lm am
Lp ap m
p (5.37) Since the prototype is no doubt an air operation, we need a windtunnel fluid of low
viscosity and high speed of sound. Hydrogen is the only practical example, but clearly
it is too expensive and dangerous. Therefore wind tunnels normally operate with air as
the working fluid. Cooling and pressurizing the air will bring Eq. (5.37) into better Range
of Rem Range
of Re p
Powerlaw
extrapolation log CD  v v Fig. 5.8 Reynoldsnumber extrapolation, or scaling, of hydraulic data
with equal Froude numbers. Uncertainty
in prototype
data estimate Model
data: 105 106 107 108 log Re  eText Main Menu  Textbook Table of Contents  Study Guide 5.5 Modeling and Its Pitfalls 307 Fig. 5.9 Hydraulic model of a barrierbeach inlet at Little River,
South Carolina. Such models of necessity violate geometric similarity
and do not model the Reynolds
number of the prototype inlet.
(Courtesy of U.S. Army Engineer
Waterways Experiment Station). agreement but not enough to satisfy a lengthscale reduction of, say, onetenth. Therefore Reynoldsnumber scaling is also commonly violated in aerodynamic testing, and
an extrapolation like that in Fig. 5.8 is required here also.
Finally, a serious discrepancy of another type occurs in hydraulic models of natural
flow systems such as rivers, harbors, estuaries, and embayments. Such flows have large
horizontal dimensions and small relative vertical dimensions. If we were to scale an estuary model by a uniform linear length ratio of, say, 1 1000, the resulting model would
be only a few millimeters deep and dominated by entirely spurious surfacetension or Webernumber effects. Therefore such hydraulic models commonly violate geometric similarity by “distorting” the vertical scale by a factor of 10 or more. Figure 5.9
shows a hydraulic model of a barrierbeach inlet in South Carolina. The horizontal scale
reduction is 1 300, but the vertical scale is only 1 60. Since a deeper channel flows more
efficiently, the model channel bottom is deliberately roughened more than the natural channel to correct for the geometric discrepancy. Thus the friction effect of the discrepancy
can be corrected, but its effect on, say, dispersion of heat and mass is less well known. EXAMPLE 5.7  v v The pressure drop due to friction for flow in a long smooth pipe is a function of average flow
velocity, density, viscosity, and pipe length and diameter: p fcn(V, , , L, D). We wish to
know how p varies with V. (a) Use the pi theorem to rewrite this function in dimensionless
form. (b) Then plot this function, using the following data for three pipes and three fluids:  eText Main Menu  Textbook Table of Contents  Study Guide Chapter 5 Dimensional Analysis and Similarity D, cm L, m Q, m3/h 1.0
1.0
1.0
2.0
2.0
2.0
3.0
3.0
3.0 5.0
7.0
9.0
4.0
6.0
8.0
3.0
4.0
5.0 0.3
0.6
1.0
1.0
2.0
3.1
0.5
1.0
1.7 *V Q/A, A
†Gasoline.
‡Water.
§Mercury. p, Pa , kg/m3 4,680
22,300
70,800
2,080
10,500
30,400
540
2,480
9,600 680†
680†
680†
998‡
998‡
998‡
13,550§
13,550§
13,550§ , kg/(m s)
2.92 E4†
2.92 E4†
2.92 E4†
0.0010‡
0.0010‡
0.0010‡
1.56 E3§
1.56 E3§
1.56 E3§ V, m/s*
1.06
2.12
3.54
0.88
1.77
2.74
0.20
0.39
0.67 D2/4. (c) Suppose it is further known that p is proportional to L (which is quite true for long pipes
with wellrounded entrances). Use this information to simplify and improve the pitheorem formulation. Plot the dimensionless data in this improved manner and comment upon the results. Solution
There are six variables with three primary dimensions involved {MLT}. Therefore we expect that
j 6 3 3 pi groups. We are correct, for we can find three variables which do not form a pi
product, for example, ( , V, L). Carefully select three ( j) repeating variables, but not including
p or V, which we plan to plot versus each other. We select ( , , D), and the pi theorem guarantees that three independent powerproduct groups will occur:
a b Dc 1 or D2
1 d p 2 p e D fV g
3 VD
2 2 h D iL L
D 3 We have omitted the algebra of finding (a, b, c, d, e, f, g, h, i) by setting all exponents to zero
M 0, L0, T 0. Therefore we wish to plot the dimensionless relation
D2 p
2 fcn VD L
,
D Ans. (a) We plot 1 versus 2 with 3 as a parameter. There will be nine data points. For example, the
first row in the data above yields
D2 p
2 VD (680)(1.06)(0.01)
2.92 E4 (680)(0.01)2(4680)
(2.92 E4)2
24,700 L
D 3.73 E9 500 The nine data points are plotted as the open circles in Fig. 5.10. The values of L /D are listed
for each point, and we see a significant length effect. In fact, if we connect the only two points
which have the same L /D ( 200), we could see (and crossplot to verify) that p increases
linearly with L, as stated in the last part of the problem. Since L occurs only in 3 L /D, the
function 1 fcn( 2, 3) must reduce to 1 (L /D) fcn( 2), or simply a function involving
only two parameters:  v v 308  eText Main Menu  Textbook Table of Contents  Study Guide 5.5 Modeling and Its Pitfalls 309 1011
900
700 L = 200
D Fig. 5.10 Two different correlations
of the data in Example 5.7: Open
circles when plotting D2 p/ 2
versus ReD, L/D is a parameter;
once it is known that p is proportional to L, a replot (solid circles)
of D3 p/(L 2) versus ReD collapses into a single powerlaw
curve. 1010 Π1 133 300 500 200 400 100 109 108
Π1
7
Π3 10 0.155 ReD1.75 10 6
10 4 105 ReD D3
L p
2 VD fcn flow in a long pipe Ans. (c) We now modify each data point in Fig. 5.10 by dividing it by its L /D value. For example, for the
first row of data, D3 p/(L 2) (3.73 E9)/500 7.46 E6. We replot these new data points as
solid circles in Fig. 5.10. They correlate almost perfectly into a straightline powerlaw function:
D3
L p 0.155 2 VD 1.75 Ans. (c) All newtonian smooth pipe flows should correlate in this manner. This example is a variation of
the first completely successful dimensional analysis, pipeflow friction, performed by Prandtl’s
student Paul Blasius, who published a related plot in 1911. For this range of (turbulentflow)
Reynolds numbers, the pressure drop increases approximately as V1.75. EXAMPLE 5.8
The smoothsphere data plotted in Fig. 5.3a represent dimensionless drag versus dimensionless
viscosity, since ( , V, d) were selected as scaling or repeating variables. (a) Replot these data to
display the effect of dimensionless velocity on the drag. (b) Use your new figure to predict the
terminal (zeroacceleration) velocity of a 1cmdiameter steel ball (SG 7.86) falling through
water at 20C. Solution
To display the effect of velocity, we must not use V as a repeating variable. Instead we choose
( , , d) as our j variables to nondimensionalize Eq. (5.1), F fcn(d, V, , ). (See Example
5.2 for an alternate approach to this problem.) The pi groups form as follows:
a  v v 1  eText Main Menu  bc dF F
2 Textbook Table of Contents e
2  fg dV Vd Study Guide Ans. (a) 310 Chapter 5 Dimensional Analysis and Similarity That is, a 1, b
2, c 0, e 1, f
1, and g 1, by using our powerproduct techniques
of Examples 5.2 to 5.6. Therefore a plot of F/ 2 versus Re will display the direct effect of velocity on sphere drag. This replot is shown as Fig. 5.11. The drag increases rapidly with velocity up to transition, where there is a slight drop, after which it increases more quickly than ever.
If the force is known, we may predict the velocity from the figure.
For water at 20°C, take
998 kg/m3 and
0.001 kg/(m s). For steel, s
3
7.86 water 7840 kg/m . For terminal velocity, the drag equals the net weight of the sphere in
water. Thus
F ( Wnet w)g s 6 d3 (7840 998)(9.81) (0.01)3 6 0.0351 N Therefore the ordinate of Fig. 5.11 is known:
(998 kg/m3)(0.0351 N)
[0.001 kg/(m s)]2 F Falling steel sphere: 2 3.5 E7 From Fig. 5.11, at F/ 2 3.5 E7, a magnifying glass reveals that Red
estimate of the terminal fall velocity is
Vd 20,000 or V 20,000[0.001 kg/(m s)]
(998 kg/m3)(0.01 m) 2 E4. Then a crude 2.0 m
s 1011
1010
Transition:
109
108 µ2 ρF = π
CD Re2
8 107
106
10 5
10 4
103
102
10
1
0.1  v v Fig. 5.11 Crossplot of spheredrag
data from Fig. 5.3a to isolate diameter and velocity.  1 eText Main Menu 10  10 2 10 3
ρ Vd
Re =
µ 10 4 10 5 10 6 Textbook Table of Contents  Study Guide Ans. (b) Problems 311
Better accuracy could be obtained by expanding the scale of Fig. 5.11 in the region of the given
force coefficient. However, there is considerable uncertainty in published drag data for spheres,
so the predicted fall velocity is probably uncertain by at least 5 percent.
Note that we found the answer directly from Fig. 5.11. We could use Fig. 5.3a also but would
have to iterate between the ordinate and abscissa to obtain the final result, since V is contained
in both plotted variables. Summary Chapters 3 and 4 presented integral and differential methods of mathematical analysis
of fluid flow. This chapter introduces the third and final method: experimentation, as
supplemented by the technique of dimensional analysis. Tests and experiments are used
both to strengthen existing theories and to provide useful engineering results when theory is inadequate.
The chapter begins with a discussion of some familiar physical relations and how
they can be recast in dimensionless form because they satisfy the principle of dimensional homogeneity. A general technique, the pi theorem, is then presented for systematically finding a set of dimensionless parameters by grouping a list of variables
which govern any particular physical process. Alternately, direct application of dimensional analysis to the basic equations of fluid mechanics yields the fundamental
parameters governing flow patterns: Reynolds number, Froude number, Prandtl number, Mach number, and others.
It is shown that model testing in air and water often leads to scaling difficulties for
which compromises must be made. Many model tests do not achieve true dynamic similarity. The chapter ends by pointing out that classic dimensionless charts and data can
be manipulated and recast to provide direct solutions to problems that would otherwise
be quite cumbersome and laboriously iterative. Problems
Most of the problems herein are fairly straightforward. More difficult or openended assignments are labeled with an asterisk. Problems labeled with an EES icon, for example, Prob. 5.61, will benefit from the use of the Engineering Equation Solver (EES), while
problems labeled with a computer icon may require the use of a
computer. The standard endofchapter problems 5.1 to 5.91 (categorized in the problem list below) are followed by word problems
W5.1 to W5.10, fundamentals of engineering exam problems FE5.1
to FE5.10, comprehensive applied problems C5.1 to C5.4, and design projects D5.1 and D5.2.
Problem distribution 5.1
5.2
5.2
5.3 Topic Problems Introduction
Choosing proper scaling parameters
The principle of dimensional homogeneity
The pi theorem 5.1 – 5.6
5.7 – 5.9
5.10 – 5.17
5.18 – 5.41  v v Section  eText Main Menu  5.4
5.4
5.5
5.5
5.5 Nondimensionalizing the basic equations
Data for spheres and cylinders
Scaling of model data
Froude and Machnumber scaling
Inventive rescaling of the data 5.42 – 5.47
5.48 – 5.57
5.58 – 5.74
5.75 – 5.84
5.85 – 5.91 P5.1 For axial flow through a circular tube, the Reynolds number for transition to turbulence is approximately 2300 [see
Eq. (6.2)], based upon the diameter and average velocity. If
d 5 cm and the fluid is kerosine at 20°C, find the volume
flow rate in m3/h which causes transition.
P5.2 In flow past a thin flat body such as an airfoil, transition to
turbulence occurs at about Re 1 E6, based on the distance
x from the leading edge of the wing. If an airplane flies at
450 mi/h at 8km standard altitude and undergoes transition
at the 12 percent chord position, how long is its chord (wing
length from leading to trailing edge)? Textbook Table of Contents  Study Guide 312 Chapter 5 Dimensional Analysis and Similarity
P5.3 An airplane has a chord length L 1.2 m and flies at a
Mach number of 0.7 in the standard atmosphere. If its
Reynolds number, based on chord length, is 7 E6, how high
is it flying?
P5.4 When tested in water at 20°C flowing at 2 m/s, an 8cmdiameter sphere has a measured drag of 5 N. What will be the
velocity and drag force on a 1.5mdiameter weather balloon moored in sealevel standard air under dynamically
similar conditions?
P5.5 An automobile has a characteristic length and area of 8 ft
and 60 ft2, respectively. When tested in sealevel standard
air, it has the following measured drag force versus speed:
V, mi/h 20 40 31 115 P5.12 P5.13 60 Drag, lbf P5.11 249 The same car travels in Colorado at 65 mi/h at an altitude
of 3500 m. Using dimensional analysis, estimate (a) its
drag force and (b) the horsepower required to overcome air
drag.
*P5.6 SAE 10 oil at 20°C flows past an 8cmdiameter sphere. At
flow velocities of 1, 2, and 3 m/s, the measured sphere drag
forces are 1.5, 5.3, and 11.2 N, respectively. Estimate the
drag force if the same sphere is tested at a velocity of 15
m/s in glycerin at 20°C.
P5.7 A body is dropped on the moon (g 1.62 m/s2) with an
initial velocity of 12 m/s. By using option 2 variables,
Eq. (5.11), the ground impact occurs at t** 0.34 and
S** 0.84. Estimate (a) the initial displacement, (b) the final displacement, and (c) the time of impact.
P5.8 The Bernoulli equation (5.6) can be written in the form
p p0 1
2 V2 gz (1) where p0 is the “stagnation” pressure at zero velocity and
elevation. (a) State how many scaling variables are needed
to nondimensionalize this equation. (b) Suppose that we
wish to nondimensionalize Eq. (1) in order to plot dimensionless pressure versus velocity, with elevation as a parameter. Select the proper scaling variables and carry out and
plot the resulting dimensionless relation.
P5.9 Modify Prob. 5.8 as follows. Suppose that we wish to nondimensionalize Eq. (1) in order to plot dimensionless pressure
versus gravity, with velocity as a parameter. Select the
proper scaling variables and carry out and plot the resulting
dimensionless relation.
P5.10 Determine the dimension {MLT } of the following quantities: (d) u
x 2 2 (b) (p
1 p0) dA (c) cp T
xy u
dx dy dz
t  v v (a) u  eText Main Menu  P5.14 P5.15 P5.16 All quantities have their standard meanings; for example,
is density.
For a particle moving in a circle, its centripetal acceleration
takes the form a fcn(V, R), where V is its velocity and R
the radius of its path. By pure dimensional reasoning,
rewrite this function in algebraic form.
The velocity of sound a of a gas varies with pressure p and
density . Show by dimensional reasoning that the proper
form must be a (const)(p/ )1/2.
The speed of propagation C of a capillary wave in deep water is known to be a function only of density , wavelength
, and surface tension . Find the proper functional relationship, completing it with a dimensionless constant. For
a given density and wavelength, how does the propagation
speed change if the surface tension is doubled?
Consider laminar flow over a flat plate. The boundary layer
thickness grows with distance x down the plate and is also
a function of freestream velocity U, fluid viscosity , and
fluid density . Find the dimensionless parameters for this
problem, being sure to rearrange if neessary to agree with
the standard dimensionless groups in fluid mechanics, as
given in Table 5.2.
It is desired to measure the drag on an airplane whose velocity is 300 mi/h. Is it feasible to test a onetwentiethscale
model of the plane in a wind tunnel at the same pressure
and temperature to determine the prototype drag coefficient?
Convection heattransfer data are often reported as a heattransfer coefficient h, defined by
.
Q hA T
.
where Q heat flow, J/s
A surface area, m2
T temperature difference, K The dimensionless form of h, called the Stanton number, is
a combination of h, fluid density , specific heat cp, and
flow velocity V. Derive the Stanton number if it is proportional to h.
P5.17 In some heattransfer textbooks, e.g., J. P. Holman, Heat
Transfer, 5th ed., McGrawHill, 1981, p. 285, simplified formulas are given for the heattransfer coefficient from Prob.
5.16 for buoyant or natural convection over hot surfaces.
An example formula is
h 1.42 T
L 1/4 where L is the length of the hot surface. Comment on the
dimensional homogeneity of this formula. What might be
the SI units of constants 1.42 and 1 ? What parameters might
4
be missing or hidden?
P5.18 Under laminar conditions, the volume flow Q through a
small triangularsection pore of side length b and length L Textbook Table of Contents  Study Guide Problems 313
is a function of viscosity , pressure drop per unit length
p/L, and b. Using the pi theorem, rewrite this relation in
dimensionless form. How does the volume flow change if
the pore size b is doubled?
P5.19 The period of oscillation T of a water surface wave is assumed to be a function of density , wavelength , depth h,
gravity g, and surface tension . Rewrite this relationship
in dimensionless form. What results if is negligible? Hint:
Take , , and g as repeating variables.
P5.20 The power input P to a centrifugal pump is assumed to be
a function of the volume flow Q, impeller diameter D, rotational rate , and the density and viscosity of the
fluid. Rewrite this as a dimensionless relationship. Hint:
Take , , and D as repeating variables.
P5.21 In Example 5.1 we used the pi theorem to develop Eq. (5.2)
from Eq. (5.1). Instead of merely listing the primary dimensions of each variable, some workers list the powers of
each primary dimension for each variable in an array:
F
1
1
2 M
L
T L
0
1
0 U
0
1
1 1
3
0 v v  P5.27 P5.28 P5.29 H
,N
L Viscosity effects are negligible. Find appropriate pi groups
for this problem and rewrite the function above in dimensionless form.
P5.23 The period T of vibration of a beam is a function of its length
L, area moment of inertia I, modulus of elasticity E, density , and Poisson’s ratio . Rewrite this relation in dimensionless form. What further reduction can we make if
E and I can occur only in the product form EI? Hint: Take
L, , and E as repeating variables.
P5.24 The lift force F on a missile is a function of its length L,
velocity V, diameter D, angle of attack , density , viscosity , and speed of sound a of the air. Write out the dimensional matrix of this function and determine its rank.  P5.26 1
1
1 This array of exponents is called the dimensional matrix for
the given function. Show that the rank of this matrix (the
size of the largest nonzero determinant) is equal to j n
k, the desired reduction between original variables and the
pi groups. This is a general property of dimensional matrices, as noted by Buckingham [29].
P5.22 When freewheeling, the angular velocity
of a windmill
is found to be a function of the windmill diameter D, the
wind velocity V, the air density , the windmill height H as
compared to the atmospheric boundary layer height L, and
the number of blades N:
fcn D, V, , P5.25 eText Main Menu  P5.30 P5.31 (See Prob. 5.21 for an explanation of this concept.) Rewrite
the function in terms of pi groups.
When a viscous fluid is confined between two long concentric cylinders as in Fig. 4.17, the torque per unit length
T required to turn the inner cylinder at angular velocity
is a function of , cylinder radii a and b, and viscosity .
Find the equivalent dimensionless function. What happens
to the torque if both a and b are doubled?
A pendulum has an oscillation period T which is assumed
to depend upon its length L, bob mass m, angle of swing ,
and the acceleration of gravity. A pendulum 1 m long, with
a bob mass of 200 g, is tested on earth and found to have a
period of 2.04 s when swinging at 20°. (a) What is its period when it swings at 45°? A similarly constructed pendulum, with L 30 cm and m 100 g, is to swing on the
moon (g 1.62 m/s2) at
20°. (b) What will be its period?
In studying sand transport by ocean waves, A. Shields in
1936 postulated that the threshold waveinduced bottom
shear stress required to move particles depends upon gravity g, particle size d and density p, and water density and
viscosity . Find suitable dimensionless groups of this problem, which resulted in 1936 in the celebrated Shields sandtransport diagram.
A simply supported beam of diameter D, length L, and modulus of elasticity E is subjected to a fluid crossflow of velocity V, density , and viscosity . Its center deflection
is assumed to be a function of all these variables. (a) Rewrite
this proposed function in dimensionless form. (b) Suppose
it is known that is independent of , inversely proportional to E, and dependent only upon V 2, not and V separately. Simplify the dimensionless function accordingly.
Hint: Take L, , and V as repeating variables.
When fluid in a pipe is accelerated linearly from rest, it begins as laminar flow and then undergoes transition to turbulence at a time ttr which depends upon the pipe diameter
D, fluid acceleration a, density , and viscosity . Arrange
this into a dimensionless relation between ttr and D.
In forced convection, the heattransfer coefficient h, as defined in Prob. 5.16, is known to be a function of stream velocity U, body size L, and fluid properties , , cp, and k.
Rewrite this function in dimensionless form, and note by
name any parameters you recognize. Hint: Take L, , k, and
as repeating variables.
The heattransfer rate per unit area q to a body from a fluid
in natural or gravitational convection is a function of the
temperature difference T, gravity g, body length L, and
three fluid properties: kinematic viscosity , conductivity k,
and thermal expansion coefficient . Rewrite in dimensionless form if it is known that g and appear only as the
product g . Textbook Table of Contents  Study Guide 314 Chapter 5 Dimensional Analysis and Similarity P5.32 A weir is an obstruction in a channel flow which can be calibrated to measure the flow rate, as in Fig. P5.32. The volume flow Q varies with gravity g, weir width b into the paper, and upstream water height H above the weir crest. If it
is known that Q is proportional to b, use the pi theorem to
find a unique functional relationship Q(g, b, H). P5.37 H
Q P5.38 Weir P5.39 P5.32 EES P5.33 A spar buoy (see Prob. 2.113) has a period T of vertical (heave)
oscillation which depends upon the waterline crosssectional
area A, buoy mass m, and fluid specific weight . How does
the period change due to doubling of (a) the mass and (b) the
area? Instrument buoys should have long periods to avoid wave
resonance. Sketch a possible longperiod buoy design.
P5.34 To good approximation, the thermal conductivity k of a gas
(see Ref. 8 of Chap. 1) depends only upon the density ,
mean free path , gas constant R, and absolute temperature
T. For air at 20°C and 1 atm, k 0.026 W/(m K) and
6.5 E8 m. Use this information to determine k for hydrogen at 20°C and 1 atm if
1.2 E7 m.
P5.35 The torque M required to turn the coneplate viscometer in
Fig. P5.35 depends upon the radius R, rotation rate , fluid
viscosity , and cone angle . Rewrite this relation in dimensionless form. How does the relation simplify it if it is
known that M is proportional to ? P5.40 P5.41 P5.42 Ω R θ P5.43 S
t θ
Fluid P5.35 v v .
P5.36 The rate of heat loss, Qloss through a window or wall is a
function of the temperature difference between inside and
outside T, the window surface area A, and the R value of
the window which has units of (ft2 h °F)/Btu. (a) Using   eText Main Menu Buckingham pi theorem, find an expression for rate of heat
loss as a function of the other three parameters in the problem. (b) If the temperature difference T doubles, by what
factor does the rate of heat loss increase?
The pressure difference p across an explosion or blast wave
is a function of the distance r from the blast center, time t,
speed of sound a of the medium, and total energy E in the
blast. Rewrite this relation in dimensionless form (see Ref.
18, chap. 4, for further details of blastwave scaling). How
does p change if E is doubled?
The size d of droplets produced by a liquid spray nozzle is
thought to depend upon the nozzle diameter D, jet velocity
U, and the properties of the liquid , , and . Rewrite this
relation in dimensionless form. Hint: Take D, , and U as
repeating variables.
In turbulent flow past a flat surface, the velocity u near the
wall varies approximately logarithmically with distance y
from the wall and also depends upon viscosity , density
, and wall shear stress w. For a certain airflow at 20°C
and 1 atm, w 0.8 Pa and u 15 m/s at y 3.6 mm. Use
this information to estimate the velocity u at y 6 mm.
Reconsider the slantedplate surface tension problem (see
Fig. C1.1) as an exercise in dimensional analysis. Let the
capillary rise h be a function only of fluid properties, gravity, bottom width, and the two angles in Fig. C1.1. That is,
h fcn( , , g, L, , ). (a) Use the pi theorem to rewrite
this function in terms of dimensionless parameters. (b) Verify that the exact solution from Prob. C1.1 is consistent with
your result in part (a).
A certain axialflow turbine has an output torque M which is
proportional to the volume flow rate Q and also depends upon
the density , rotor diameter D, and rotation rate . How does
the torque change due to a doubling of (a) D and (b) ?
Nondimensionalize the energy equation (4.75) and its
boundary conditions (4.62), (4.63), and (4.70) by defining
T* T/T0, where T0 is the inlet temperature, assumed constant. Use other dimensionless variables as needed from Eqs.
(5.23). Isolate all dimensionless parameters you find, and
relate them to the list given in Table 5.2.
The differential equation of salt conservation for flowing
seawater is  u S
x S
y w S
z 2 S
x2 2 S
y2 2 S
z2 where is a (constant) coefficient of diffusion, with typical units of square meters per second, and S is the salinity
in parts per thousand. Nondimensionalize this equation and
discuss any parameters which appear.
P5.44 The differential energy equation for incompressible twodimensional flow through a “Darcytype” porous medium is
approximately Textbook Table of Contents  Study Guide Problems 315
p
x cp T
x 2 pT
yy cp k T
y2 0 where is the permeability of the porous medium. All other
symbols have their usual meanings. (a) What are the appropriate dimensions for ? (b) Nondimensionalize this
equation, using (L, U, , T0) as scaling constants, and discuss any dimensionless parameters which arise.
P5.45 In naturalconvection problems, the variation of density due
to the temperature difference T creates an important buoyancy term in the momentum equation (5.30). To firstorder
accuracy, the density variation would be
T),
0(1
where is the thermalexpansion coefficient. The momentum equation thus becomes
0 dV
dt (p 0gz) 2 T gk 0 V where we have assumed that z is up. Nondimensionalize this
equation, using Eqs. (5.23), and relate the parameters you
find to the list in Table 5.2.
P5.46 The differential equation for compressible inviscid flow of
a gas in the xy plane is
2 2
2 t t (u2 2 (u2 ) a2) x2
2 ( 2 a2) 2
2 y 2u xy 0 where is the velocity potential and a is the (variable) speed
of sound of the gas. Nondimensionalize this relation, using
a reference length L and the inlet speed of sound a0 as parameters for defining dimensionless variables.
P5.47 The differential equation for smallamplitude vibrations
y(x, t) of a simple beam is given by
2 A y 2 t 4 EI y
x4 0 beam material density
crosssectional area
area moment of inertia
Young’s modulus where
A
I
E  v v Use only the quantities , E, and A to nondimensionalize y,
x, and t, and rewrite the differential equation in dimensionless form. Do any parameters remain? Could they be removed by further manipulation of the variables?
P5.48 A smooth steel (SG 7.86) sphere is immersed in a stream
of ethanol at 20°C moving at 1.5 m/s. Estimate its drag in
N from Fig. 5.3a. What stream velocity would quadruple its
drag? Take D 2.5 cm.
P5.49 The sphere in Prob. 5.48 is dropped in gasoline at 20°C. Ignoring its acceleration phase, what will its terminal (constant) fall velocity be, from Fig. 5.3a?  eText Main Menu  P5.50 When a microorganism moves in a viscous fluid, it turns
out that fluid density has nearly negligible influence on the
drag force felt by the microorganism. Such flows are called
creeping flows. The only important parameters in the problem are the velocity of motion U, the viscosity of the fluid
, and the length scale of the body. Here assume the microorganism’s body diameter d as the appropriate length
scale. (a) Using the Buckingham pi theorem, generate an
expression for the drag force D as a function of the other
parameters in the problem. (b) The drag coefficient discussed in this chapter CD D/( 1 U2A) is not appropriate
2
for this kind of flow. Define instead a more appropriate drag
coefficient, and call it Cc (for creeping flow). (c) For a spherically shaped microorganism, the drag force can be calculated exactly from the equations of motion for creeping flow.
The result is D 3 Ud. Write expressions for both forms
of the drag coefficient, Cc and CD, for a sphere under conditions of creeping flow.
P5.51 A ship is towing a sonar array which approximates a submerged cylinder 1 ft in diameter and 30 ft long with its axis
normal to the direction of tow. If the tow speed is 12 kn
(1 kn 1.69 ft/s), estimate the horsepower required to tow
this cylinder. What will be the frequency of vortices shed
from the cylinder? Use Figs. 5.2 and 5.3.
P5.52 A 1indiameter telephone wire is mounted in air at 20°C
and has a natural vibration frequency of 12 Hz. What wind
velocity in ft/s will cause the wire to sing? At this condition what will the average drag force per unit wire
length be?
P5.53 Vortex shedding can be used to design a vortex flowmeter
(Fig. 6.32). A blunt rod stretched across the pipe sheds vortices whose frequency is read by the sensor downstream.
Suppose the pipe diameter is 5 cm and the rod is a cylinder
of diameter 8 mm. If the sensor reads 5400 counts per
minute, estimate the volume flow rate of water in m3/h. How
might the meter react to other liquids?
P5.54 A fishnet is made of 1mmdiameter strings knotted into
2 2 cm squares. Estimate the horsepower required to tow
300 ft2 of this netting at 3 kn in seawater at 20°C. The net
plane is normal to the flow direction.
P5.55 The radio antenna on a car begins to vibrate wildly at 500
Hz when the car is driven at 55 mi/h. Estimate the diameter of the antenna.
P5.56 A wooden flagpole, of diameter 5 in and height 30 ft, fractures at its base in hurricane winds at sea level. If the fracture stress is 3500 lbf/in2, estimate the wind velocity in
mi/h.
P5.57 The simply supported 1040 carbonsteel rod of Fig. P5.57
is subjected to a crossflow stream of air at 20°C and 1 atm.
For what stream velocity U will the rod center deflection be
approximately 1 cm? Textbook Table of Contents  Study Guide 316 Chapter 5 Dimensional Analysis and Similarity
tion speed for which the power will not exceed 300 W?
What will the pressure rise be for this condition?
*P5.63 The pressure drop per unit length p/L in smooth pipe flow
is known to be a function only of the average velocity V,
diameter D, and fluid properties and . The following data
were obtained for flow of water at 20°C in an 8cmdiameter pipe 50 m long: D = 1 cm, L = 60 cm
δ = 1 cm?
U Q, m3/s P5.58 For the steel rod of Prob. 5.57, at what airstream velocity
U will the rod begin to vibrate laterally in resonance in its
EES
first mode (a half sine wave)? Hint: Consult a vibration text
under “lateral beam vibration.”
P5.59 We wish to know the drag of a blimp which will move in
20°C air at 6 m/s. If a onethirtiethscale model is tested in
water at 20°C, what should the water velocity be? At this
velocity, if the measured water drag on the model is 2700 N,
what is the drag on the prototype blimp and the power required to propel it?
P5.60 A prototype water pump has an impeller diameter of 2 ft
and is designed to pump 12 ft3/s at 750 r/min. A 1ftdiameter model pump is tested in 20°C air at 1800 r/min, and
Reynoldsnumber effects are found to be negligible. For
similar conditions, what will the volume flow of the model
be in ft3/s? If the model pump requires 0.082 hp to drive it,
what horsepower is required for the prototype?
P5.61 If viscosity is neglected, typical pumpflow results from
Prob. 5.20 are shown in Fig. P5.61 for a model pump tested
EES
in water. The pressure rise decreases and the power required
increases with the dimensionless flow coefficient. Curvefit
expressions are given for the data. Suppose a similar pump
of 12cm diameter is built to move gasoline at 20°C and a
flow rate of 25 m3/h. If the pump rotation speed is 30 r/s,
find (a) the pressure rise and (b) the power required.
Pressur
e ris
e P ≈ 0.5 + 3Q
ρ Ω 3D 5
ΩD3 r we Po ∆p
Q
≈ 6.0 – 120
ρ Ω 2D 2
ΩD3 (( 2 Pump data
(Ω in r/s) 0.005 0.01 0.015 0.020 p, Pa P5.57 5800 20,300 42,100 70,800 Verify that these data are slightly outside the range of Fig.
5.10. What is a suitable powerlaw curve fit for the present
data? Use these data to estimate the pressure drop for flow
of kerosine at 20°C in a smooth pipe of diameter 5 cm and
length 200 m if the flow rate is 50 m3/h.
P5.64 The natural frequency of vibration of a mass M attached
to a rod, as in Fig. P5.64, depends only upon M and the
stiffness EI and length L of the rod. Tests with a 2kg mass
attached to a 1040 carbonsteel rod of diameter 12 mm and
length 40 cm reveal a natural frequency of 0.9 Hz. Use these
data to predict the natural frequency of a 1kg mass attached
to a 2024 aluminumalloy rod of the same size.
ω
M L Stiffness EI P5.64
P5.65 In turbulent flow near a flat wall, the local velocity u varies
only with distance y from the wall, wall shear stress w, and
fluid properties and . The following data were taken in
the University of Rhode Island wind tunnel for airflow,
0.0023 slug/ft3,
3.81 E7 slug/(ft s), and w 0.029
lbf/ft2: Q
= flow coefficient
ΩD3 P5.61  v v *P5.62 Modify Prob. 5.61 so that the rotation speed is unknown but
D 12 cm and Q 25 m3/h. What is the maximum rota  eText Main Menu  0.021 0.035 0.055 0.080 0.12 0.16 u, ft/s
0 y, in 50.6 54.2 57.6 59.7 63.5 65.9 (a) Plot these data in the form of dimensionless u versus dimensionless y, and suggest a suitable powerlaw curve fit.
(b) Suppose that the tunnel speed is increased until u 90
ft/s at y 0.11 in. Estimate the new wall shear stress, in
lbf/ft2. Textbook Table of Contents  Study Guide Problems 317  v v Use these data to predict the drag force of a similar 15in
P5.66 A torpedo 8 m below the surface in 20°C seawater cavitates
diamond placed at similar orientation in 20°C water flowat a speed of 21 m/s when atmospheric pressure is 101 kPa.
ing at 2.2 m/s.
If Reynoldsnumber and Froudenumber effects are negligible, at what speed will it cavitate when running at a depth P5.71 The pressure drop in a venturi meter (Fig. P3.165) varies
only with the fluid density, pipe approach velocity, and diof 20 m? At what depth should it be to avoid cavitation at
ameter ratio of the meter. A model venturi meter tested in
30 m/s?
water at 20°C shows a 5kPa drop when the approach veP5.67 A student needs to measure the drag on a prototype of charlocity is 4 m/s. A geometrically similar prototype meter is
acteristic dimension dp moving at velocity Up in air at stanused to measure gasoline at 20°C and a flow rate of
dard atmospheric conditions. He constructs a model of char9 m3/min. If the prototype pressure gage is most accurate
acteristic dimension dm, such that the ratio dp/dm is some
at 15 kPa, what should the upstream pipe diameter be?
factor f. He then measures the drag on the model at dynamically similar conditions (also with air at standard at P5.72 A onefifteenthscale model of a parachute has a drag of
450 lbf when tested at 20 ft/s in a water tunnel. If Reynoldsmospheric conditions). The student claims that the drag
number effects are negligible, estimate the terminal fall veforce on the prototype will be identical to that measured on
locity at 5000ft standard altitude of a parachutist using the
the model. Is this claim correct? Explain.
prototype if chute and chutist together weigh 160 lbf. NeP5.68 Consider flow over a very small object in a viscous fluid.
glect the drag coefficient of the woman.
Analysis of the equations of motion shows that the inertial terms are much smaller than the viscous and pressure P5.73 The yawing moment on a torpedo control surface is tested
on a oneeighthscale model in a water tunnel at 20 m/s, usterms. It turns out, then, that fluid density drops out of
ing Reynolds scaling. If the model measured moment is
the equations of motion. Such flows are called creeping
14 N m, what will the prototype moment be under similar
flows. The only important parameters in the problem are
conditions?
the velocity of motion U, the viscosity of the fluid , and
the length scale of the body. For threedimensional bod P5.74 A onetenthscale model of a supersonic wing tested at 700
m/s in air at 20°C and 1 atm shows a pitching moment of
ies, like spheres, creeping flow analysis yields very good
0.25 kN m. If Reynoldsnumber effects are negligible, what
results. It is uncertain, however, if such analysis can be
will the pitching moment of the prototype wing be if it is
applied to twodimensional bodies such as a circular
flying at the same Mach number at 8km standard altitude?
cylinder, since even though the diameter may be very
small, the length of the cylinder is infinite for a twodi P5.75 A onetwelfthscale model of an airplane is to be tested at
20°C in a pressurized wind tunnel. The prototype is to fly
mensional flow. Let us see if dimensional analysis can
at 240 m/s at 10km standard altitude. What should the tunhelp. (a) Using the Buckingham pi theorem, generate an
nel pressure be in atm to scale both the Mach number and
expression for the twodimensional drag D2D as a functhe Reynolds number accurately?
tion of the other parameters in the problem. Use cylinder
.
diameter d as the appropriate length scale. Be careful — *P5.76 A 2ftlong model of a ship is tested in a freshwater tow
tank. The measured drag may be split into “friction” drag
the twodimensional drag has dimensions of force per unit
(Reynolds scaling) and “wave” drag (Froude scaling). The
length rather than simply force. (b) Is your result physimodel data are as follows:
cally plausible? If not, explain why not. (c) It turns out
that fluid density cannot be neglected in analysis of
0.8
1.6
2.4
3.2
4.0
4.8
creeping flow over twodimensional bodies. Repeat the Tow speed, ft/s
dimensional analysis, this time with included as a pa Friction drag, lbf 0.016 0.057 0.122 0.208 0.315 0.441
rameter. Find the nondimensional relationship between
Wave drag, lbf
0.002
0.021
0.083
0.253
0.509
0.697
the parameters in this problem.
P5.69 A onesixteenthscale model of a weir (see Fig. P5.32) has
The prototype ship is 150 ft long. Estimate its total drag
a measured flow rate Q 2.1 ft3/s when the upstream wawhen cruising at 15 kn in seawater at 20°C.
ter height is H 6.3 in. If Q is proportional to weir width P5.77 A dam spillway is to be tested by using Froude scaling with
b, predict the prototype flow rate when Hproto 3.2 ft.
a onethirtiethscale model. The model flow has an average
P5.70 A diamondshaped body, of characteristic length 9 in, has
velocity of 0.6 m/s and a volume flow of 0.05 m3/s. What
the following measured drag forces when placed in a wind
will the velocity and flow of the prototype be? If the meaEES
tunnel at sealevel standard conditions:
sured force on a certain part of the model is 1.5 N, what
will the corresponding force on the prototype be?
V, ft/s
30
38
48
56
61
P5.78 A prototype spillway has a characteristic velocity of 3 m/s
F, lbf
1.25
1.95
3.02
4.05
4.81
and a characteristic length of 10 m. A small model is con  eText Main Menu  Textbook Table of Contents  Study Guide 318 P5.79 P5.80 P5.81 P5.82 P5.83 P5.84 Chapter 5 Dimensional Analysis and Similarity
structed by using Froude scaling. What is the minimum scale
ratio of the model which will ensure that its minimum Weber number is 100? Both flows use water at 20°C.
An East Coast estuary has a tidal period of 12.42 h (the
semidiurnal lunar tide) and tidal currents of approximately
80 cm/s. If a onefivehundredthscale model is constructed
with tides driven by a pump and storage apparatus, what
should the period of the model tides be and what model current speeds are expected?
A prototype ship is 35 m long and designed to cruise at
11 m/s (about 21 kn). Its drag is to be simulated by a 1mlong model pulled in a tow tank. For Froude scaling find
(a) the tow speed, (b) the ratio of prototype to model drag,
and (c) the ratio of prototype to model power.
An airplane, of overall length 55 ft, is designed to fly at 680
m/s at 8000m standard altitude. A onethirtiethscale model
is to be tested in a pressurized helium wind tunnel at 20°C.
What is the appropriate tunnel pressure in atm? Even at this
(high) pressure, exact dynamic similarity is not achieved. Why?
A prototype ship is 400 ft long and has a wetted area of
30,000 ft2. A oneeightiethscale model is tested in a tow
tank according to Froude scaling at speeds of 1.3, 2.0, and
2.7 kn (1 kn 1.689 ft/s). The measured friction drag of
the model at these speeds is 0.11, 0.24, and 0.41 lbf, respectively. What are the three prototype speeds? What is the
estimated prototype friction drag at these speeds if we correct for Reynoldsnumber discrepancy by extrapolation?
A onefortiethscale model of a ship’s propeller is tested in
a tow tank at 1200 r/min and exhibits a power output of 1.4
ft lbf/s. According to Froude scaling laws, what should the
revolutions per minute and horsepower output of the prototype propeller be under dynamically similar conditions?
A prototype oceanplatform piling is expected to encounter
currents of 150 cm/s and waves of 12s period and 3m P5.85
P5.86
P5.87 P5.88 P5.89 P5.90 P5.91 height. If a onefifteenthscale model is tested in a wave
channel, what current speed, wave period, and wave height
should be encountered by the model?
Solve Prob. 5.49, using the modified spheredrag plot of Fig.
5.11.
Solve Prob. 5.49 for glycerin at 20°C, using the modified
spheredrag plot of Fig. 5.11.
In Prob. 5.62 it was difficult to solve for because it appeared in both power and flow coefficients. Rescale the
problem, using the data of Fig. P5.61, to make a plot of dimensionless power versus dimensionless rotation speed. Enter this plot directly to solve Prob. 5.62 for .
Modify Prob. 5.62 as follows: Let
32 r/s and Q 24
m3/h for a geometrically similar pump. What is the maximum diameter if the power is not to exceed 340 W? Solve
this problem by rescaling the data of Fig. P5.61 to make a
plot of dimensionless power versus dimensionless diameter.
Enter this plot directly to find the desired diameter.
Knowing that p is proportional to L, rescale the data of
Example 5.7 to plot dimensionless p versus dimensionless
diameter. Use this plot to find the diameter required in the
first row of data in Example 5.7 if the pressure drop is increased to 10 kPa for the same flow rate, length, and fluid.
Knowing that p is proportional to L, rescale the data of
Example 5.7 to plot dimensionless p versus dimensionless viscosity. Use this plot to find the viscosity required
in the first row of data in Example 5.7 if the pressure drop
is increased to 10 kPa for the same flow rate, length, and
density.
Develop a plot of dimensionless p versus dimensionless
viscosity, as described in Prob. 5.90. Suppose that L 200
m, Q 60 m3/h, and the fluid is kerosine at 20°C. Use your
plot to determine the minimum pipe diameter for which the
pressure drop is no more than 220 kPa. Word Problems W5.2 W5.3
W5.4 In 98 percent of data analysis cases, the “reducing factor”
j, which lowers the number n of dimensional variables to
n j dimensionless groups, exactly equals the number of
relevant dimensions (M, L, T, ). In one case (Example
5.5) this was not so. Explain in words why this situation
happens.
Consider the following equation: 1 dollar bill 6 in. Is
this relation dimensionally inconsistent? Does it satisfy
the PDH? Why?
In making a dimensional analysis, what rules do you follow for choosing your scaling variables?
In an earlier edition, the writer asked the following question about Fig. 5.1: “Which of the three graphs is a more
effective presentation?” Why was this a dumb question?  v v W5.1  eText Main Menu  W5.5 W5.6 W5.7 This chapter discusses the difficulty of scaling Mach and
Reynolds numbers together (an airplane) and Froude and
Reynolds numbers together (a ship). Give an example of
a flow which would combine Mach and Froude numbers.
Would there be scaling problems for common fluids?
What is different about a very small model of a weir or
dam (Fig. P5.32) which would make the test results difficult to relate to the prototype?
What else are you studying this term? Give an example
of a popular equation or formula from another course
(thermodynamics, strength of materials, etc.) which does
not satisfy the principle of dimensional homogeneity. Explain what is wrong and whether it can be modified to be
homogeneous. Textbook Table of Contents  Study Guide Comprehensive Problems
W5.8 W5.9 Some colleges (e.g., Colorado State University) have environmental wind tunnels which can be used to study, e.g.,
wind flow over city buildings. What details of scaling
might be important in such studies?
If the model scale ratio is
Lm / Lp, as in Eq. (5.31),
and the Weber number is important, how must the model 319 and prototype surface tension be related to for dynamic
similarity?
W5.10 For a typical incompressible velocity potential analysis in
Chap. 4 we solve 2
0, subject to known values of
/ n on the boundaries. What dimensionless parameters
govern this type of motion? Fundamentals of Engineering Exam Problems
FE5.1 FE5.2 FE5.3 FE5.4 FE5.5 Given the parameters (U, L, g, , ) which affect a certain liquid flow problem, the ratio V 2/(Lg) is usually
known as the
(a) velocity head, (b) Bernoulli head, (c) Froude number,
(d ) kinetic energy, (e) impact energy
A ship 150 m long, designed to cruise at 18 kn, is to be
tested in a tow tank with a model 3 m long. The appropriate tow velocity is
(a) 0.19 m/s, (b) 0.35 m/s, (c) 1.31 m/s,
(d ) 2.55 m/s, (e) 8.35 m/s
A ship 150 m long, designed to cruise at 18 kn, is to be
tested in a tow tank with a model 3 m long. If the model
wave drag is 2.2 N, the estimated fullsize ship wave drag
is
(a) 5500 N, (b) 8700 N, (c) 38,900 N,
(d ) 61,800 N, (e) 275,000 N
A tidal estuary is dominated by the semidiurnal lunar tide,
with a period of 12.42 h. If a 1 500 model of the estuary
is tested, what should be the model tidal period?
(a) 4.0 s, (b) 1.5 min, (c) 17 min, (d ) 33 min, (e) 64 min
A football, meant to be thrown at 60 mi/h in sealevel air
(
1.22 kg /m3,
1.78 E5 N s /m2), is to be tested
using a onequarter scale model in a water tunnel (
998 kg/m3,
0.0010 N s/m2). For dynamic similarity, what is the proper model water velocity?
(a) 7.5 mi/h, (b) 15.0 mi/h, (c) 15.6 mi/h,
(d ) 16.5 mi/h, (e) 30 mi/h FE5.6 A football, meant to be thrown at 60 mi/h in sealevel air
(
1.22 kg/m3,
1.78 E5 N m2), is to be tested using a onequarter scale model in a water tunnel (
998
kg/m3,
0.0010 N s/m2). For dynamic similarity,
what is the ratio of prototype force to model force?
(a) 3.86 1, (b) 16 1, (c) 32 1, (d ) 56 1, (e) 64 1
FE5.7 Consider liquid flow of density , viscosity , and velocity U over a very small model spillway of length scale L,
such that the liquid surface tension coefficient is important. The quantity U2L / in this case is important and
is called the
(a) capillary rise, (b) Froude number, (c) Prandtl number, (d ) Weber number, (e) Bond number
FE5.8 If a stream flowing at velocity U past a body of length L
causes a force F on the body which depends only upon U,
L, and fluid viscosity , then F must be proportional to
(a) UL / , (b) U 2L2, (c) U/L, (d ) UL, (e) UL/
FE5.9 In supersonic wind tunnel testing, if different gases are
used, dynamic similarity requires that the model and prototype have the same Mach number and the same
(a) Euler number, (b) speed of sound, (c) stagnation enthalpy, (d) Froude number, (e) specific heat ratio
FE5.10 The Reynolds number for a 1ftdiameter sphere moving
at 2.3 mi/h through seawater (specific gravity 1.027, viscosity 1.07 E3 N s/m2) is approximately
(a) 300, (b) 3000, (c) 30,000, (d) 300,000, (e) 3,000,000 Comprehensive Problems
Estimating pipe wall friction is one of the most common
tasks in fluids engineering. For long circular rough pipes in
turbulent flow, wall shear w is a function of density , viscosity , average velocity V, pipe diameter d, and wall
roughness height . Thus, functionally, we can write w
fcn( , , V, d, ). (a) Using dimensional analysis, rewrite
this function in dimensionless form. (b) A certain pipe has
d 5 cm and
0.25 mm. For flow of water at 20°C,  v v C5.1  eText Main Menu  measurements show the following values of wall shear
stress:
Q, gal/min
w, Pa 1.5 3.0 6.0 9.0 12.0 14.0 0.05 0.18 0.37 0.64 0.86 1.25 Plot these data using the dimensionless form obtained in
part (a) and suggest a curvefit formula. Does your plot reveal the entire functional relation obtained in part (a)? Textbook Table of Contents  Study Guide 320
C5.2 Chapter 5 Dimensional Analysis and Similarity
When the fluid exiting a nozzle, as in Fig. P3.49, is a gas,
instead of water, compressibility may be important, especially if upstream pressure p1 is large and exit diameter d2
is small. In this case, the difference p1 p2 is no longer
controlling, and the gas mass flow m reaches a maximum
˙
value which depends upon p1 and d2 and also upon the absolute upstream temperature T1 and the gas constant R. Thus,
functionally, m fcn(p1, d2, T1, R). (a) Using dimensional
˙
analysis, rewrite this function in dimensionless form. (b) A
certain pipe has d2 1 mm. For flow of air, measurements
show the following values of mass flow through the nozzle: analysis. Let the vertical velocity be a function only of distance from the plate, fluid properties, gravity, and film
thickness. That is, w fcn(x, , , g, ). (a) Use the pi theorem to rewrite this function in terms of dimensionless parameters. (b) Verify that the exact solution from Prob. 4.80
is consistent with your result in part (a).
C5.4 The Taco Inc. model 4013 centrifugal pump has an impeller
of diameter D 12.95 in. When pumping 20°C water at
1160 r/min, the measured flow rate Q and pressure
rise p are given by the manufacturer as follows:
Q, gal/min T1, K 300 300 500 800 p1, kPa 200 250 300 300 0.037 0.046 0.055 0.043 0.034 200 300 400 500 600 700 36 35 34 32 29 23 300 ˙
m, kg/s C5.3 300 2 p, lb/in Plot these data in the dimensionless form obtained in part
(a). Does your plot reveal the entire functional relation obtained in part (a)?
Reconsider the fully developed draining vertical oilfilm
problem (see Fig. P4.80) as an exercise in dimensional (a) Assuming that p fcn( , Q, D, ), use the pi theorem to rewrite this function in terms of dimensionless parameters and then plot the given data in dimensionless form.
(b) It is desired to use the same pump, running at 900 r/min,
to pump 20°C gasoline at 400 gal/min. According to your
dimensionless correlation, what pressure rise p is expected, in lbf/in2? Design Projects
We are given laboratory data, taken by Prof. Robert Kirchhoff and his students at the University of Massachusetts,
for the spin rate of a 2cup anemometer. The anemometer
was made of pingpong balls (d 1.5 in) split in half, facing in opposite directions, and glued to thin ( 1 in) rods
4
pegged to a center axle. (See Fig. P7.91 for a sketch.) There
were four rods, of lengths
0.212, 0.322, 0.458, and
0.574 ft. The experimental data, for wind tunnel velocity U
and rotation rate , are as follows:
0.212
U, ft/s
18.95
22.20
25.90
29.94
38.45 0.322 , r/min U, ft/s
435.00
545.00
650.00
760.00
970.00 18.95
23.19
29.15
32.79
38.45 0.458 , r/min U, ft/s
225.00
290.00
370.00
425.00
495.00 20.10
26.77
31.37
36.05
39.03 0.574 , r/min U, ft/s
140.00
215.00
260.00
295.00
327.00 23.21
27.60
32.07
36.05
39.60 , r/min
115.00
145.00
175.00
195.00
215.00 Assume that the angular velocity of the device is a function of wind speed U, air density and viscosity , rod
length , and cup diameter d. For all data, assume air is
at 1 atm and 20°C. Define appropriate pi groups for this
problem, and plot the data above in this dimensionless
manner.  v v D5.1  eText Main Menu  Comment on the possible uncertainty of the results.
As a design application, suppose we are to use this
anemometer geometry for a largescale (d 30 cm) airport
wind anemometer. If wind speeds vary up to 25 m/s and
we desire an average rotation rate
120 r/min, what
should be the proper rod length? What are possible limitations of your design? Predict the expected (in r/min) of
your design as affected by wind speeds from 0 to 25 m/s.
D5.2 By analogy with the cylinderdrag data in Fig. 5.3b, spheres
also show a strong roughness effect on drag, at least in the
Reynolds number range 4 E4 ReD 3 E5, which accounts for the dimpling of golf balls to increase their distance traveled. Some experimental data for roughened
spheres [43] are given in Fig. D5.2. The figure also shows
typical golfball data. We see that some roughened spheres
are better than golf balls in some regions. For the present
study, let us neglect the ball’s spin, which causes the very
important sideforce or Magnus effect (See Fig. 8.11) and
assume that the ball is hit without spin and follows the equations of motion for plane motion (x, z):
mx
˙˙
where F cos
F CD 24 Textbook Table of Contents m˙˙
z
D2(x 2
˙  F sin
z 2)
˙ W
tan Study Guide ˙
1z
x
˙ References 321 0.6
Golf ball
0.5
Drag coefficient, CD The ball has a particular CD(ReD) curve from Fig. D5.2 and
is struck with an initial velocity V0 and angle 0. Take the
ball’s average mass to be 46 g and its diameter to be 4.3 cm.
Assuming sealevel air and a modest but finite range of initial conditions, integrate the equations of motion to compare the trajectory of “roughened spheres” to actual golfball calculations. Can the rough sphere outdrive a normal
golf ball for any conditions? What roughnesseffect differences occur between a lowimpact duffer and, say, Tiger
Woods? 0.4
0.3
900
0.2 10 5 1250 10 5
500 10 0.1 D 150 Rough spheres
5 10 Smooth sphere 5 0
2 104 105 106 4 106 Reynolds number, UD/ D5.2 References 2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17. P. W. Bridgman, Dimensional Analysis, Yale University Press,
New Haven, CT, 1922, rev. ed., 1931.
A. W. Porter, The Method of Dimensions, Methuen, London,
1933.
F. M. Lanchester, The Theory of Dimensions and Its Applications for Engineers, CrosbyLockwood, London, 1940.
R. EsnaultPelterie, L’Analyse dimensionelle, F. Rouge, Lausanne, Switzerland, 1946.
G. W. Stubbings, Dimensions in Engineering Theory, CrosbyLockwood, London, 1948.
G. Murphy, Similitude in Engineering, Ronald, New York,
1950.
H. E. Huntley, Dimensional Analysis, Rinehart, New York,
1951.
H. L. Langhaar, Dimensional Analysis and the Theory of
Models, Wiley, New York, 1951.
W. J. Duncan, Physical Similarity and Dimensional Analysis, Arnold, London, 1953.
C. M. Focken, Dimensional Methods and Their Applications,
Arnold, London, 1953.
L. I. Sedov, Similarity and Dimensional Methods in Mechanics, Academic, New York, 1959.
E. C. Ipsen, Units, Dimensions, and Dimensionless Numbers,
McGrawHill, New York, 1960.
E. E. Jupp, An Introduction to Dimensional Methods, CleaverHume, London, 1962.
R. Pankhurst, Dimensional Analysis and Scale Factors, Reinhold, New York, 1964.
S. J. Kline, Similitude and Approximation Theory, McGrawHill, New York, 1965.
B. S. Massey, Units, Dimensional Analysis, and Physical Similarity, Van Nostrand Reinhold, New York, 1971.
J. Zierep, Similarity Laws and Modeling, Dekker, New York,
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32. 33. W. E. Baker et al., Similarity Methods in Engineering Dynamics, Spartan, Rochelle Park, NJ, 1973.
E. S. Taylor, Dimensional Analysis for Engineers, Clarendon
Press, Oxford, England, 1974.
E. de St. Q. Isaacson and M. de St. Q. Isaacson, Dimensional
Methods in Engineering and Physics, Arnold, London, 1975.
P. LeCorbeiller, Dimensional Analysis, Irvington, New York,
1966.
V. J. Skoglund, Similitude — Theory and Applications, International, Scranton, PA, 1967.
M. S. Yalin, Theory of Hydraulic Models, Macmillan, London, 1971.
J. J. Sharp, Hydraulic Modeling, Butterworth, London, 1981.
G. I. Barenblatt, Dimensional Analysis, Gordon and Breach,
New York, 1987.
R. EsnaultPelterie, Dimensional Analysis and Metrology, F.
Rouge, Lausanne, Switzerland, 1950.
R. Kurth, Dimensional Analysis and Group Theory in Astrophysics, Pergamon, New York, 1972.
F. J. deJong, Dimensional Analysis for Economists, North
Holland, Amsterdam, 1967.
E. Buckingham, “On Physically Similar Systems: Illustrations of the Use of Dimensional Equations,” Phys. Rev., vol. 4,
no. 4, 1914, pp. 345 – 376.
“Flow of Fluids through Valves, Fittings, and Pipe,” Crane
Co. Tech. Pap. 410, Chicago, 1957.
A. Roshko, “On the Development of Turbulent Wakes from
Vortex Streets,” NACA Rep. 1191, 1954.
G. W. Jones, Jr., “Unsteady Lift Forces Generated by Vortex
Shedding about a Large, Stationary, Oscillating Cylinder at
High Reynolds Numbers,” ASME Symp. Unsteady Flow, 1968.
O. M. Griffin and S. E. Ramberg, “The Vortex Street Wakes
of Vibrating Cylinders,” J. Fluid Mech., vol. 66, pt. 3, 1974,
pp. 553 – 576. Textbook Table of Contents  Study Guide 322 Chapter 5 Dimensional Analysis and Similarity Encyclopedia of Science and Technology, 8th ed., McGrawHill, New York, 1997.
35. H. A. Becker, Dimensionless Parameters, Halstead Press (Wiley), New York, 1976.
36. D. J. Schuring, Scale Models in Engineering, Pergamon
Press, New York, 1977.
37. M. Zlokarnik, Dimensional Analysis and ScaleUp in Chemical Engineering, SpringerVerlag, New York, 1991.
38. W. G. Jacoby, Data Theory and Dimensional Analysis, Sage,
Newbury Park, CA, 1991.  v v 34.  eText Main Menu  39. B. Schepartz, Dimensional Analysis in the Biomedical Sciences, Thomas, Springfield, IL, 1980.
40. A. J. Smith, Dosage and Solution Calculations: The Dimensional Analysis Way, Mosby, St. Louis, MO, 1989.
41. J. B. Bassingthwaighte et al., Fractal Physiology, Oxford
Univ. Press, New York, 1994.
42. K. J. Niklas, Plant Allometry: The Scaling of Form and
Process, Univ. of Chicago Press, Chicago, 1994.
43. R. D. Blevins, Applied Fluid Dynamics Handbook, van Nostrand Reinhold, New York, 1984. Textbook Table of Contents  Study Guide ...
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This note was uploaded on 10/27/2009 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.
 Spring '08
 Sakar
 The Land

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