Chapter 7

DIMENSIONAL ANALYSIS
AND MODELING

Fluid Mechanics: Fundamentals and Applications, 2nd Edition

Yunus A. Cengel, John M. Cimbala
McGraw-Hill, 2010
1
 DIMENSIONAL ANALYSIS AND MODELING
7–1 Dimensions and Units

7–2 Dimensional Homogeneity

Nondimensionalization of Equations

7–3 Dimensional Analysis and Similarity

7–4 The Method of Repeating Variables and the

Buckingham Pi Theorem

7–5 Experimental Testing and Incomplete Similarity

Setup of an Experiment and Correlation of Experimental Data
Incomplete Similarity
Wind Tunnel Testing
Flows with Free Surfaces

2
 Objectives

Develop a better understanding of dimensions, units, and dimensional

homogeneity of equations

Understand the numerous benefits of dimensional analysis

Know how to use the method of repeating variables to identify non dimensional
parameters

Understand the concept of dynamic similarity and how to apply it to experimental

modeling

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 7–1 DIMENSIONS AND UNITS
Dimension: A measure of a physical quantity (without numerical values).
Unit: A way to assign a number to that dimension.
There are seven primary dimensions (also called fundamental or basic dimensions):
mass length time temperature
electric current amount of light amount of matter.
All non-primary dimensions can be formed by some combination of the seven primary
dimensions.

A dimension is a measure of a physical quantity without numerical values, while a unit is a way to
assign a number to the dimension. For example, length is a dimension, but centimeter is a unit.

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5
EXAMPLE 7–1: An engineer is studying how some insects
are able to walk on water. A fluid property of importance in
this problem is surface tension (ss), which has dimensions
of force per unit length. Write the dimensions of surface
tension in terms of primary dimensions.

Solution The primary dimensions of surface tension are to

be determined.
The water strider is an insect that
Analysis Force has dimensions of mass times acceleration,
can walk on water due to surface
or {mL/ t2}. Thus,
tension.

The usefulness of expressing the dimensions of a variable or constant in terms of

primary dimensions will become clearer in the discussion of the method of repeating
variables in Section 7–4.

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 7–2 DIMENSIONAL HOMOGENEITY
The law of dimensional homogeneity: Every additive term in an equation must have the
same dimensions.

You can’t add apples and oranges! Total energy of a system at state 1 and at state 2.

7
 An equation that is not dimensionally
homogeneous is a sure sign of an error.

8
EXAMPLE 7–2 Dimensional Homogeneity of the Bernoulli Equation
Probably the most well-known (and most misused) equation in fluid mechanics is the
Bernoulli equation (Fig. 7–6), discussed in Chap. 5. The standard form of the
Bernoulli equation for incompressible irrotational fluid flow is

Solution We are to verify that the primary dimensions of each additive term in Eq. 1 are the
same, and we are to determine the dimensions of constant C.
Analysis (a) Each term is written in terms of primary dimensions,

All three additive terms have the same dimensions.

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 (b) From the law of dimensional homogeneity, the constant must have the
same dimensions as the other additive terms in the equation. Thus,

If the dimensions of any of the terms were different from the others, it would
indicate that an error was made somewhere in the analysis.

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 DIMENSIONAL CONSISTENCY

Quantity Symbol SI Units Dimensions

--------------------------------------------------------------------------------------------------
Mass m, M kg [m]=M
Length L,x,y m [L]=L
Time t,T s [t]=T
Frequency f Hz=1/s [f]=1/T
Angular velocity  rad/s []=1/T
Rotational speed N rev/s [N]=1/T
Velocity u,v,q m/s [u]=L/T
Acceleration a,g m/s2 [a]=L/T2
Density  (rho) kg/m3 []=M/L3
Force F,D N=kg.m/s2 [F]=M.L/T2
Stress  (tau) Pa=N/m2 []=M/L.T2
Pressure p Pa [p]=M/L.T2
Kinematic viscosity  (nu) m2/s []=L2/T
Dynamic viscosity  (mu) Pa.s []=M/LT
Work W J=N.m [W]=M.L2/T2
Energy E J [J]=M.L2/T2
Power P W=J/s [P]=M.L2/T3
 DIMENSIONAL CONSISTENCY

Quantity Symbol SI Units Dimensions

-------------------------------------------------------------------------------------------------
Mass m, M kg [m]=M
Length L,x,y m [L]=L
Time t,T s [t]=T
Frequency f Hz=1/s [f]=1/T
Angular velocity  rad/s []=1/T
Rotational speed N rev/s [N]=1/T
Velocity u,v,q m/s [u]=L/T
Acceleration a,g m/s2 [a]=L/T2
Density  (rho) kg/m3 []=M/L3
Force F,D N=kg.m/s2 [F]=M.L/T2
Stress  (tau) Pa=N/m2 []=M/L.T2
Pressure p Pa [p]=M/L.T2
Kinematic viscosity  (nu) m2/s []=L2/T
Dynamic viscosity  (mu) Pa.s []=M/LT
Work W J=N.m [W]=M.L2/T2
Energy E J [J]=M.L2/T2
Power P W=J/s [P]=M.L2/T3
DIMENSIONAL CONSISTENCY

Quantity Symbol SI Units Dimensions

------------------------------------------------------------------------------------------------
Mass m, M kg [m]=M
Length L,x,y m [L]=L
Time t,T s [t]=T
Frequency f Hz=1/s [f]=1/T
Angular velocity  rad/s []=1/T
Rotational speed N rev/s [N]=1/T
Velocity u,v,q m/s [u]=L/T
Acceleration a,g m/s2 [a]=L/T2
Density  (rho) kg/m3 []=M/L3
Force F,D N=kg.m/s2 [F]=M.L/T2
Stress  (tau) Pa=N/m2 []=M/L.T2
Pressure p Pa [p]=M/L.T2
Kinematic viscosity  (nu) m2/s []=L2/T
Dynamic viscosity  (mu) Pa.s []=M/LT
Work W J=N.m [W]=M.L2/T2
Energy E J [J]=M.L2/T2
Power P W=J/s [P]=M.L2/T3
 DIMENSIONS

How many of the following dimensional equalities are incorrect?

(a) five (b) four (c) three (d) two (e) one

(1) Radius [R] = L

L
(2) Velocity [V] =
T
(3) Density [] = M
L3
ML
(4) Dynamic viscosity [] =
T
(5) Drag force [D] = ML
T2
(6) Surface tension [] = M
T2
 DIMENSIONS

How many of the following dimensional equalities are incorrect?

(a) five (b) four (c) three (d) two (e) one

(1) Radius [R] = L

L
(2) Velocity [V] =
T
(3) Density [] = M
L3
ML Pa.s  Ns  kgms  M
(4) Dynamic viscosity [] =
T m 2 m 2 s 2 LT
(5) Drag force [D] = ML
T2
(6) Surface tension [] = M
T2
 UNITS AND DIMENSIONS

The units of the quantities listed below are all correct. How many of
the corresponding dimensions are correct?
(a) six (b) five (c) four (d) three (e) two

(1) Mass m (kg) [m] = M

ML
(2) Force F (N) [F] =
T2
(3) Frequency f (Hz) [f] = L
T
(4) Stress  (Pa or N/m2) []= ML
T2 2
(5) Work W (J or N.m) [W] = ML2
T
2
(6) Power P (W or J/s or N.m/s) ML[P] =
2
T
M
(7) Surface tension  (N/m) []= 2
T
 UNITS AND DIMENSIONS
The units of the quantities listed below are all correct. How many of
the corresponding dimensions are correct?
(a) six (b) five (c) four (d) three (e) two

(1) Mass m (kg) [m] = M

ML
(2) Force F (N) [F] = 2
T
L
(3) Frequency f (Hz) [f] =
T
ML
(4) Stress  (Pa or N/m2) []= 2
T 2
ML
(5) Work W (J or N.m) [W] =
T2
ML2
(6) Power P (W or J/s or N.m/s) [P] =
T2
M
(7) Surface tension  (N/m) []= 2
T
 UNITS AND DIMENSIONS

Frequency f (Hz) f  1
T

Stress  (Pa or N/m2) 

F N kgm kg
 2 2 2 2    M
A m ms ms LT 2

J kgm2 kgm2 ML2

Power P (W or J/s or N.m/s)  2
 3 P  3
s s.s s T
 UNITS AND DIMENSIONS

DIMENSIONS AND UNITS MAY BE MULTIPLIED AND DIVIDED (CANCELLED)

Example 1 Dimensions and units of a/u a   L

u   L dimensions
T2 T
 a  L T 1 = [ f ] [FREQUENCY] m/s2 m/s units
 u   T 2 . L  T
m / s2
 1 / s  Hz
m/ s
Example 2 Dimensions and units of p/u2  p  M
   M u   L
LT 2 L3 T
Pa kg/m3 m/s

 p  M L3 T 2 Dimensionless
Pa

N / m2
1
 u 2   LT 2 . M . L2  1 kg / m . m / s  kg / m.s 
3 2 2 2
  (Non-Dimensional)
(because N = kg.m/s2)
 UNITS AND DIMENSIONS

milli= 10-3 kilo= 103

103 PREFIX RULE
micro= 10-6 Mega= 106
nano= 10-9 Giga= 109

Multiply Add
To Powers Indices
Divide Subtract

Work out dimensions from units or definitions & vice versa for units
 UNITS AND DIMENSIONS

Each term in a physical equation must have the same dimensions (and units)

v = u + at s = u.t +(1/2)at2
L L L L L
 2
.T L .T 1. 2 .T 2
T T T T T
p = .g.h T = 2 l/g

M M L T2
2
 3 . 2 .L T  1.1. L.
LT L T L
4
1 πR Δp
D  C D ρV 2 A Q 
2 8μ L
M
2 1. L4 .
ML M L L3 LT 2
 1. 1. 3 . 2 .L2 
T2 L T T M
1. .L
LT
Constants have dimension 1 not zero
 DIMENSIONS

Given the following

L ML
Radius [R] = L Velocity [V] = Drag force [D] =
T T2
L2
Density [] = M Kinematic viscosity [] =
L3 T
which of the following is correct?

(1)  D  (2) VR 

 V 2 R   1
      1

(a) both (b) neither (c) only the first (d) only the second
 DIMENSIONS

Given the following

L ML
Radius [R] = L Velocity [V] = Drag force [D] =
T T2
L2
Density [] = M Kinematic viscosity [] =
L3 T
which of the following is correct?

(1)  D  (2) VR 

 V 2 R   1     1
 

(a) both (b) neither (c) only the first (d) only the second
 Nondimensionalization of Equations
Nondimensional equation: If we divide each term in the equation by a collection of
variables and constants whose product has those same dimensions, the equation is
rendered nondimensional.
Normalized equation: If the nondimensional terms in the equation are of order unity, the
equation is called normalized.
Each term in a nondimensional equation is dimensionless.
Nondimensional parameters: In the process of nondimensionalizing an equation of
motion, nondimensional parameters often appear—most of which are named after a
notable scientist or engineer (e.g., the Reynolds number and the Froude number).
This process is referred to by some authors as inspectional analysis.

A nondimensionalized form of the Bernoulli equation is formed

by dividing each additive term by a pressure (here we use P).
Each resulting term is dimensionless (dimensions of {1}).

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Dimensional variables: Dimensional quantities that change or vary in
the problem. Examples: z (dimension of length) and t (dimension of
time).
Nondimensional (or dimensionless) variables: Quantities that
change or vary in the problem, but have no dimensions. Example:
Angle of rotation, measured in degrees or radians, dimensionless
units.
Dimensional constant: Gravitational constant g, while dimensional,
remains constant.
Parameters: Refer to the combined set of dimensional variables,
Object falling in a
nondimensional variables, and dimensional constants in the problem. vacuum. Vertical
Pure constants: The constant 1/2 and the exponent 2 in equation. velocity is drawn
Other common examples of pure constants are  and e. positively, so w < 0 for a
falling object.

25
To nondimensionalize an equation, we need to select scaling parameters, based on the
primary dimensions contained in the original equation.

Froude number

In a typical fluid flow problem, the scaling parameters

usually include a characteristic length L, a characteristic
velocity V, and a reference pressure difference P0  P.
Other parameters and fluid properties such as density,
viscosity, and gravitational acceleration enter the
problem as well.

26
 The two key advantages of nondimensionalization of an equation.

27
EXAMPLE 7–3: Your little brother’s high school physics class conducts experiments in a large
vertical pipe whose inside is kept under vacuum conditions. The students are able to remotely
release a steel ball at initial height z0 between 0 and 15 m (measured from the bottom of the pipe),
and with initial vertical speed w0 between 0 and 10 m/s. A computer coupled to a network of
photosensors along the pipe enables students to plot the trajectory of the steel ball (height z plotted
as a function of time t) for each test. The students are unfamiliar with dimensional analysis or
nondimensionalization techniques, and therefore conduct several “brute force” experiments to
determine how the trajectory is affected by initial conditions z0 and w0. First they hold w0 fixed at 4
m/s and conduct experiments at five different values of z0: 3, 6, 9, 12, and 15 m. The experimental
results are shown in Fig. 7–12a. Next, they hold z0 fixed at 10 m and conduct experiments at five
different values of w0: 2, 4, 6, 8, and 10 m/s. These results are shown in Fig. 7–12b. Later that
evening, your brother shows you the data and the trajectory plots and tells you that they plan to
conduct more experiments at different values of z0 and w0. You explain to him that by first
nondimensionalizing the data, the problem can be reduced to just one parameter, and no further
experiments are required. Prepare a nondimensional plot to prove your point and discuss.

Solution A nondimensional plot is to be generated from all the available trajectory data.
Specifically, we are to plot z* as a function of t*.

Assumptions The inside of the pipe is subjected to strong enough vacuum pressure that
aerodynamic drag on the ball is negligible.

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Figures 7-12a,b. Trajectories of a steel ball falling in a vacuum: Figure 7-13.Trajectories of a steel
(a) w0 fixed at 4 m/s, and (b) z0 fixed at 10 m. ball falling in a vacuum.

Equation is valid as is the nondimensionalization. As previously discussed, this problem combines

three of the original dimensional parameters (g, z0, and w0) into one non dimensional parameter,
the Froude number. After converting to the dimensionless variables, the 10 trajectories of Fig. 7–
12a and b are replotted in dimensionless format in Fig. 7–13. It is clear that all the trajectories are
of the same family, with the Froude number as the only remaining parameter. Fr2 varies from
about 0.041 to about 1.0 in these experiments. If any more experiments are to be conducted,
they should include combinations of z0 and w0 that produce Froude numbers outside of this
range. A large number of additional experiments would be unnecessary, since all the trajectories
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would be of the same family as those plotted in Fig. 7–13.
EXAMPLE 7–4. The gravitational constant at the surface of the moon is only about one-
sixth of that on earth. An astronaut on the moon throws a baseball at an initial speed of
21.0 m/s at a 5° angle above the horizon and at 2.0 m above the moon’s surface. (a)
Using the dimensionless data of Example 7–3 shown in Fig. 7–13, predict how long it
takes for the baseball to fall to the ground. (b) Do an exact calculation and compare the
result to that of part (a).
Solution Experimental data obtained on earth are to be used to
predict the time required for a baseball to fall to the ground on the
moon.
Assumptions 1 The horizontal velocity of the baseball is irrelevant.
2 The surface of the moon is perfectly flat near the astronaut. 3
There is no aerodynamic drag on the ball since there is no
atmosphere on the moon. 4 Moon gravity is one-sixth that of earth.

a) The Froude number is calculated based on the value Throwing a baseball on

of gmoon and the vertical component of initial speed, the moon

30
 (b) An exact calculation is obtained by setting z equal to zero in Eq. 7–5 and solving for time
t (using the quadratic formula),

If the Froude number had landed between two of the trajectories of Fig. 7–13,
interpolation would have been required. Since some of the numbers are precise to
only two significant digits, the small difference between the results of part (a) and
part (b) is of no concern. The final result is t=3.0 s to two significant digits.

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 In a general unsteady fluid flow problem with a free surface, the scaling parameters include a
characteristic length L, a characteristic velocity V, a characteristic frequency f, and a reference
pressure difference P0  P. Nondimensionalization of the differential equations of fluid flow produces
four dimensionless parameters: the Reynolds number, Froude number, Strouhal number, and Euler
number.

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 7–3 DIMENSIONAL ANALYSIS AND SIMILARITY
In most experiments, to save time and money, tests are performed on a geometrically
scaled model, rather than on the full-scale prototype. In such cases, care must be taken
to properly scale the results. We introduce here a powerful technique called dimensional
analysis.
The three primary purposes of dimensional analysis are
To generate nondimensional parameters that help in the design of experiments
(physical and/or numerical) and in the reporting of experimental results
To obtain scaling laws so that prototype performance can be predicted from model
performance
To predict trends in the relationship between parameters
The principle of similarity
(1) Geometric similarity—the model must be the same shape as the prototype, but may
be scaled by some constant scale factor.
(2) Kinematic similarity—the velocity at any point in the model flow must be
proportional (by a constant scale factor) to the velocity at the corresponding point in the
prototype flow.
(3) Dynamic similarity—When all forces in the model flow scale by a constant factor to
corresponding forces in the prototype flow (force-scale equivalence).

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 Kinematic similarity is achieved
when, at all locations, the speed
in the model flow is
proportional to that at
corresponding locations in the
prototype flow, and points in
the same direction.

In a general flow field, complete similarity between a model and prototype is

achieved only when there is geometric, kinematic, and dynamic similarity.

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 We let uppercase Greek letter Pi () denote a nondimensional parameter.
In a general dimensional analysis problem, there is one  that we call the
dependent , giving it the notation 1.
The parameter 1 is in general a function of several other ’s, which we call
independent ’s.

To ensure complete similarity, the model and prototype must be geometrically

similar, and all independent groups must match between model and prototype.

To achieve similarity

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 The Reynolds number Re is formed by the ratio of
density, characteristic speed, and characteristic
length to viscosity. Alternatively, it is the ratio of
characteristic speed and length to kinematic
viscosity, defined as  =/.

Geometric similarity between a prototype car of

The Reynolds number is the most well
length Lp and a model car of length Lm. known and useful dimensionless
parameter in all of fluid mechanics.

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EXAMPLE 7–5: The aerodynamic drag of a new sports car is to be predicted at a speed of 50.0
mi/h at an air temperature of 25°C. Automotive engineers build a onefifth scale model of the car to
test in a wind tunnel. It is winter and the wind tunnel is located in an unheated building; the
temperature of the wind tunnel air is only about 5°C. Determine how fast the engineers should run
the wind tunnel in order to achieve similarity between the model and the prototype.

Solution We are to utilize the concept of similarity to determine the speed of the wind tunnel.
Assumptions 1 Compressibility of the air is negligible (the validity of this approximation is
discussed later). 2 The wind tunnel walls are far enough away so as to not interfere with the
aerodynamic drag on the model car. 3 The model is geometrically similar to the prototype. 4
The wind tunnel has a moving belt to simulate the ground under the car. (The moving belt is
necessary in order to achieve kinematic similarity everywhere in the flow, in particular
underneath the car.)

A drag balance is a device used in a wind tunnel to measure

the aerodynamic drag of a body. When testing automobile
models, a moving belt is often added to the floor of the wind
tunnel to simulate the moving ground (from the car’s frame
of reference).

37
 This speed is quite high (about 100 m/s), and the wind tunnel may not be able to
run at that speed. Furthermore, the incompressible approximation may come into
question at this high speed

38
EXAMPLE 7–6: This example is a follow-up to Example
7–5. Suppose the engineers run the wind tunnel at 221
mi/h to achieve similarity between the model and the
prototype. The aerodynamic drag force on the model
car is measured with a drag balance. Several drag
readings are recorded, and the average drag force on
the model is 21.2 lbf. Predict the aerodynamic drag
force on the prototype (at 50 mi/h and 25°C).

By arranging the dimensional parameters as nondimensional ratios, the units cancel nicely even
though they are a mixture of SI and English units. Because both velocity and length are squared
in the equation for 1, the higher speed in the wind tunnel nearly compensates for the model’s
smaller size, and the drag force on the model is nearly the same as that on the prototype. In fact,
if the density and viscosity of the air in the wind tunnel were identical to those of the air flowing
over the prototype, the two drag forces would be identical as well.

39
40
If a water tunnel is used instead of a wind tunnel to test their one-fifth scale model, the
water tunnel speed required to achieve similarity is

One advantage of a water tunnel is that the required

water tunnel speed is much lower than that required
for a wind tunnel using the same size model
(221 mi/h for air and 16.1 mi/h for water).

Similarity can be achieved even when the model fluid is

different than the prototype fluid. Here a submarine model
is tested in a wind tunnel.

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7–4 THE METHOD OF REPEATING VARIABLES AND THE BUCKINGHAM PI THEOREM

How to generate the nondimensional

parameters, i.e., the ’s?

There are several methods that have been

developed for this purpose, but the most
popular (and simplest) method is the
method of repeating variables.

A concise summary of the six steps that

comprise the method of repeating variables.

42
Detailed description of the six steps that comprise the method of repeating variables*

Step 2 List the primary dimensions for each of the n parameters.

Step 3 Guess the reduction j. As a first guess, set j equal to the number of primary dimensions
represented in the problem. The expected number of ’s (k) is equal to n minus j, according to the
Buckingham Pi theorem
If at this step or during any subsequent step, the analysis does not work out, verify that you
have included enough parameters in step 1. Otherwise, go back and reduce j by one and try
again.
Step 4 Choose j repeating parameters that will be used to construct each . Since the
repeating parameters have the potential to appear in each , be sure to choose them wisely.
Step 5 Generate the ’s one at a time by grouping the j repeating parameters with one of the
remaining parameters, forcing the product to be dimensionless. In this way, construct all k ’s. By
convention the first , designated as 1, is the dependent (the one on the left side of
the list). Manipulate the ’s as necessary to achieve established dimensionless groups.
Step 6 Check that all the ’s are indeed dimensionless. Write the final functional relationship in the
form of Eq. 7–11.
43
 Step 1

Step 2

Step 3

Step 4

Setup for dimensional analysis of a ball falling in a vacuum.

Elevation z is a function of time t, initial vertical speed w0,
initial elevation z0, and gravitational constant g.

44
Step 5

45
 The  groups that result
from the method of
repeating variables are
guaranteed to be
dimensionless because
It is wise to choose
we force the overall
common parameters as
exponent of all seven
repeating parameters
primary dimensions to
since they may appear in
be zero.
each of your
dimensionless  groups.

The mathematical rules for adding and

subtracting exponents during
multiplication and division, respectively.

46
 Established nondimensional
parameters are usually named after
a notable scientist or engineer.

47
Step 6

The method of repeating variables cannot predict the

exact mathematical form of the equation.

A quick check of your

algebra is always wise.

48
49
50
51
52
EXAMPLE 7–7 Pressure in a Soap Bubble: Some children are playing with soap bubbles, and
you become curious as to the relationship between soap bubble radius and the pressure inside the
soap bubble (Fig. 7–29). You reason that the pressure inside the soap bubble must be greater than
atmospheric pressure, and that the shell of the soap bubble is under tension, much like the skin of
a balloon. You also know that the property surface tension must be important in this problem. Not
knowing any other physics, you decide to approach the problem using dimensional analysis.
Establish a relationship between pressure difference P Pinside Poutside, soap bubble radius R,
and the surface tension ss of the soap film.

Solution The pressure difference between the inside of a soap

bubble and the outside air is to be analyzed by the method of
repeating variables.
Assumptions 1 The soap bubble is neutrally buoyant in the air, and
gravity is not relevant. 2 No other variables or constants are
important in this problem.

Step 1 There are three variables and constants in this problem;

n = 3. They are listed in functional form, with the dependent
The pressure inside a soap bubble is variable listed as a function of the independent variables and
greater than that surrounding the constants:
soap bubble due to surface tension
in the soap film.

53
Step 2 The primary dimensions of each parameter are listed. The dimensions of surface
tension are obtained from Example 7–1, and those of pressure from Example 7–2.

Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in
the problem (m, L, and t).

k= n – j= 3 -3 = 0

At times like this, we need to first go back and make sure that we are not neglecting some
important variable or constant in the problem. Since we are confident that the pressure
difference should depend only on soap bubble radius and surface tension, we reduce the
value of j by one,

If this value of j is correct, k=n-j= 3- 2 = 1. Thus we expect one , which is more physically
realistic than zero ’s.

54
Step 5 We combine these repeating parameters into a product with the dependent
variable P to create the dependent π ,

Step 6 We write the final functional relationship. In the case at hand, there is only one ,
which is a function of nothing. This is possible only if the  is constant.

Discussion This is an example of how we can sometimes predict trends with dimensional analysis,
even without knowing much of the physics of the problem. For example, we know from our result
that if the radius of the soap bubble doubles, the pressure difference decreases by a factor of 2.
Similarly, if the value of surface tension doubles, P increases by a factor of 2. Dimensional analysis
cannot predict the value of the constant in Eq. 3; further analysis (or one experiment) reveals that
the constant is equal to 4 (Chap. 2).
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EXAMPLE 7–8: Some aeronautical engineers are designing an airplane and wish to
predict the lift produced by their new wing design. The chord length Lc of the wing is
1.12 m, and its planform area A (area viewed from the top when the wing is at zero
angle of attack) is 10.7 m2. The prototype is to fly at V = 52.0 m/s close to the ground
where T = 25°C. They build a one-tenth scale model of the wing to test in a pressurized
wind tunnel. The wind tunnel can be pressurized to a maximum of 5 atm. At what speed
and pressure should they run the wind tunnel in order to achieve dynamic similarity?

Solution We are to determine the speed and pressure at which to run the
wind tunnel in order to achieve dynamic similarity.
Assumptions 1 The prototype wing flies through the air at standard
atmospheric pressure. 2 The model is geometrically similar to the
prototype.

Step 1 List of relevant parameters:

Step 2

56
 Step 3

Step 4

Step 5

57
 Step 6

At 25°C, c is approximately 346 m/s, and the Mach number of the prototype airplane wing is Map =
52.0/346 = 0.150—subsonic.
A common rule of thumb is that for Mach numbers less than about 0.3, as is the fortunate case
here, compressibility effects are practically negligible.

This example illustrates one of the (frustrating) limitations of dimensional analysis; namely,
You may not always be able to match all the dependent ’s simultaneously in a model test.
Compromises must be made in which only the most important ’s are matched. In many
practical situations in fluid mechanics, the Reynolds number is not critical for dynamic
similarity provided that Re is high enough. If the Mach number of the prototype were
significantly larger than about 0.3, we would be wise to precisely match the Mach number
rather than the Reynolds number in order to ensure reasonable results. Furthermore, if a
different gas were used to test the model, we would also need to match the specific heat
ratio (k), since compressible flow behavior is strongly dependent on k.

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Solution We are to generate a nondimensional
relationship between shear stress and other
parameters.
Assumptions 1 The flow is fully developed. 2
The fluid is incompressible. 3 No other
parameters are significant in the problem.

59
 Step 1 There are 6 variables and constants. n=6. They are listed in functional form, with the
dependent variable listed as a function of the independent variables and constants:

Step 2 The primary dimensions of each parameter are listed. Note that shear stress is a force per
unit area, and thus has the same dimensions as pressure.

Step 3 As a first guess, j is set equal to 3, the number of primary dimensions represented in the
problem (m, L, and t).

If this value of j is correct, the expected

number of ’s is k = n-j=6-3 = 3.

Step 4: We choose three repeating parameters since j = 3. Following the guidelines of Table 7–3,
we cannot pick the dependent variable tw. We cannot choose both ε and D since their dimensions
are identical, and it would not be desirable to have μ or ε appear in all the π’s. The best choice of
repeating parameters is thus V, D, and ρ.

60
 Step 5 The dependent П is generated:

Although the Darcy friction factor for pipe

flows is most common, you should be aware Step 6 We write the final functional relationship as
of an alternative, less common friction
factor called the Fanning friction factor. The
relationship between the two is f = 4Cf .

The result applies to both laminar and turbulent fully developed pipe flow; it turns out, however, that the second
independent (roughness ratio /D) is not nearly as important in laminar pipe flow as in turbulent pipe flow. This
problem presents an interesting connection between geometric similarity and dimensional analysis. Namely, it is
necessary to match /D since it is an independent in the problem. From a different perspective, thinking of
roughness as a geometric property, it is necessary to match /D to ensure geometric similarity between two pipes.

61
To verify the validity of Eq. 1 of Example 7–9, we use computational fluid dynamics
(CFD) to predict the velocity profiles and the values of wall shear stress for two
physically different but dynamically similar pipe flows:

• Air at 300 K flowing at an average speed of 14.5 ft/s through a pipe of inner diameter
1.00 ft and average roughness height 0.0010 ft.

• Water at 300 K flowing at an average speed of 3.09 m/s through a pipe of inner
diameter 0.0300 m and average roughness height 0.030 mm.
The two pipes are clearly geometrically similar since they are both round pipes.
They have the same average roughness ratio (/D = 0.0010 in both cases).
We have carefully chosen the values of average speed and diameter such that the two
flows are also dynamically similar. Specifically, the other independent  (the Reynolds
number) also matches between the two flows.

62
 Normalized axial velocity profiles for fully
developed flow through a pipe as predicted by
CFD; profiles of air (circles) and water (crosses)
are shown on the same plot.

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7–5 EXPERIMENTAL TESTING, MODELING, AND INCOMPLETE SIMILARITY

One of the most useful applications of dimensional analysis is in designing physical

and/or numerical experiments, and in reporting the results of such experiments.
In this section we discuss both of these applications, and point out situations in which
complete dynamic similarity is not achievable.

Setup of an Experiment and Correlation of Experimental Data

Consider a problem in which there are five original parameters (one of which is the
dependent parameter).
A complete set of experiments (called a full factorial test matrix) is conducted.
This testing would require 54 = 625 experiments.
Assuming that three primary dimensions are represented in the problem, we can reduce
the number of parameters from five to two (k = 5  3 = 2 nondimensional groups), and
the number of independent parameters from four to one.
Thus, for the same resolution we would hen need to conduct a total of only 5 1 = 5
experiments.

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 For a two- problem, we plot dependent
dimensionless parameter (1) as a function of
independent dimensionless parameter (2). The
resulting plot can be (a) linear or (b) nonlinear.
In either case, regression and curve-fitting
techniques are available to determine the
relationship between the ’s.

If there are more than two ’s in the problem

(e.g., a three-  problem ora four-  problem),
we need to set up a test matrix to determine the
relationship between the dependent  and the
independent ’s. In many cases we discover that
one or more of the dependent ’s has negligible
effect and can be removed from the list of
necessary dimensionless parameters.

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 Incomplete Similarity

We have shown several examples in which the nondimensional groups are easily
obtained with paper and pencil through straightforward use of the method of repeating
variables.
In fact, after sufficient practice, you should be able to obtain the ’s with ease—
sometimes in your head or on the “back of an envelope.”
Unfortunately, it is often a much different story when we go to apply the results of
our dimensional analysis to experimental data.
The problem is that it is not always possible to match all the ’s of a model to the
corresponding ’s of the prototype, even if we are careful to achieve geometric similarity.
This situation is called incomplete similarity.
Fortunately, in some cases of incomplete similarity, we are still able to extrapolate
model test data to obtain reasonable full-scale predictions.

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 Wind Tunnel Testing
We illustrate incomplete similarity with the
problem of measuring the aerodynamic drag force
on a model truck in a wind tunnel.
One-sixteenth scale.
The model is geometrically similar to the
prototype.
The model truck is 0.991 m long. Wind tunnel has
a maximum speed of 70 m/s.
The wind tunnel test section is 1.0 m tall and 1.2 m
wide. Measurement of aerodynamic drag on a
model truck in a wind tunnel equipped with a
drag balance and a moving belt ground plane.

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 To match the Reynolds number between model and prototype, the wind tunnel should
be run at 429 m/s. This is impossible in this wind tunnel.

What do we do? There are several options:

(1) Use a bigger wind tunnel. Automobile manufacturers typically test with three-
eighths scale model cars and with one-eighth scale model trucks and buses in very large
wind tunnels.
(2) We could use a different fluid for the model tests. For example, water can achieve
higher Re numbers, but more expensive.
(3) We could pressurize the wind tunnel and/or adjust the air temperature to increase
the maximum Reynolds number capability.
(4) If all else fails, we could run the wind tunnel at several speeds near the maximum
speed, and then extrapolate our results to the full-scale Reynolds number.
Fortunately, it turns out that for many wind tunnel tests the last option is quite viable.

68
 The Langley full-scale wind tunnel (LFST) is large
enough that full-scale vehicles can be tested.

For many objects, the drag coefficient levels off at

Reynolds numbers above some threshold value. This
fortunate situation is called Reynolds number
independence. It enables us to extrapolate to
prototype Reynolds numbers that are outside of the
range of our experimental facility.

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EXAMPLE 7–10:A one-sixteenth scale model tractor-trailer truck (18-wheeler) is tested in a wind
tunnel. The model truck is 0.991 m long, 0.257 m tall, and 0.159 m wide. During the tests, the
moving ground belt speed is adjusted so as to always match the speed of the air moving through
the test section. Aerodynamic drag force FD is measured as a function of wind tunnel speed; the
experimental results are listed in Table 7–7. Plot the drag coefficient CD as a function of the
Reynolds number Re, where the area used for the calculation of CD is the frontal area of the model
truck (the area you see when you look at the model from upstream), and the length scale used for
calculation of Re is truck width W. Have we achieved dynamic similarity? Have we achieved
Reynolds number independence in our wind tunnel test? Estimate the aerodynamic drag force on
the prototype truck traveling on the highway at 26.8 m/s. Assume that both the wind tunnel air and
the air flowing over the prototype car are at 25°C and standard atmospheric pressure.

Solution We are to calculate and plot CD as a function

of Re for a given set of wind tunnel measurements and
determine if dynamic similarity and/or Reynolds
number independence have been achieved. Finally, we
are to estimate the aerodynamic drag force acting on
the prototype truck.
Assumptions 1 The model truck is geometrically
similar to the prototype truck. 2 The aerodynamic drag
on the strut(s) holding the model truck is negligible.

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 Have we achieved dynamic similarity? we have geometric similarity between model and
prototype, but the Reynolds number of the prototype truck is

Aerodynamic drag coefficient as a

function of the Reynolds number.
The values are calculated from
wind tunnel test data on a model
truck.

71
where the width of the prototype is specified as 16 times that of the model. Comparison
of above Equations reveals that the prototype Reynolds number is more than six times
larger than that of the model. Since we cannot match the independent ’s in the problem,
dynamic similarity has not been achieved.
Have we achieved Reynolds number independence? From CD Figure we see that Reynolds
number independence has indeed been achieved—at Re greater than about 5x105, CD
has leveled off to a value of about 0.76 (to two significant digits). Since we have achieved
Reynolds number independence, we can extrapolate to the full-scale prototype, assuming
that CD remains constant as Re is increased to that of the full-scale prototype.

We give our final result to two significant digits. More than that cannot be justified. As
always, we must exercise caution when performing an extrapolation, since we have no
guarantee that the extrapolated results are correct.

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 Flows with Free Surfaces

For the case of model testing of flows with free surfaces (boats and ships,
floods, river flows, aqueducts, hydroelectric dam spillways, interaction of waves
with piers, soil erosion, etc.), complications arise that preclude complete
similarity between model and prototype.
For example, if a model river is built to study flooding, the model is often several
hundred times smaller than the prototype due to limited lab space.
Researchers often use a distorted model in which the vertical scale of the model
(e.g., river depth) is exaggerated in comparison to the horizontal scale of the
model (e.g., river width).

In addition, the model riverbed slope is often made proportionally steeper than
that of the prototype.

These modifications result in incomplete similarity due to lack of geometric

similarity.
Model tests are still useful under these circumstances, but other tricks (like
deliberately roughening the model surfaces) and empirical corrections and
correlations are required to properly scale up the model data.

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 In many flows involving a liquid with a free surface,
both the Reynolds number and Froude number are
relevant nondimensional parameters. Since it is not
always possible to match both Re and Fr between
model and prototype, we are sometimes forced to
settle for incomplete similarity.

To ensure complete similarity we would need to use a liquid

whose kinematic viscosity satisfies this equation.
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 (a) (c)

A NACA 0024 airfoil being tested in a

towing tank
at Fr (a) 0.19,
(b) 0.37,
and (c) 0.55. In tests like this, the
Froude number is the most important
nondimensional parameter.

(b)
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EXAMPLE 7–10
In the late 1990s the U.S. Army Corps of Engineers designed an experiment to model the flow of the
Tennessee River downstream of the Kentucky Lock and Dam. Because of laboratory space
restrictions, they built a scale model with a length scale factor of Lm /Lp = 1/100. Suggest a liquid
that would be appropriate for the experiment.

Solution We are to suggest a liquid to use in an experiment involving a one-hundredth scale

model of a lock, dam, and river.
Assumptions 1 The model is geometrically similar to the prototype. 2 The model river is deep
enough that surface tension effects are not significant.

76
Thus, we need to find a liquid that has a viscosity of 1x109 m2/s. A quick glance
through the appendices yields no such liquid. Hot water has a lower kinematic
viscosity than cold water, but only by about a factor of 3. Liquid mercury has a very
small kinematic viscosity, but it is of order 10-7 m2/s—still two orders of magnitude too
large to satisfy Eq. 1. Even if liquid
mercury would work, it would be too expensive and too hazardous to use in such a
test. What do we do? The bottom line is that we cannot match both the Froude number
and the Reynolds number in this model test. In other words, it is impossible to achieve
complete similarity between model and prototype in this case. Instead, we do the best
job we can under conditions of incomplete similarity. Water is typically used in such
tests for convenience.

It turns out that for this kind of experiment, Froude

number matching is more critical than Reynolds
number matching. As discussed previously for wind
tunnel testing, Reynolds number independence is
achieved at high enough values of Re. Even if we are
unable to achieve Reynolds number independence, In many experiments involving free
we can often extrapolate our low Reynolds number surfaces, we cannot match both the
model data to predict full-scale Reynolds number Froude number and the Reynolds
number. However, we can often
behavior (Fig. 7–44). A high level of confidence in
extrapolate low Re model test data to
using this kind of extrapolation comes only after predict high Re prototype behavior.
much laboratory experience with similar problems.
77
We mention the importance of similarity in the production of Hollywood movies in which
model boats, trains, airplanes, buildings, monsters, etc., are blown up or burned.

Movie producers must pay attention to dynamic similarity in order to make the small-
scale fires and explosions appear as realistic as possible.

You may recall some low-budget movies where the special effects are unconvincing.

In most cases this is due to lack of dynamic similarity between the small model and the
full-scale prototype.

If the model’s Froude number and/or Reynolds number differ too much from those of the
prototype, the special effects don’t look right, even to the untrained eye.

78
 Summary

• Dimensions and units

• Dimensional homegeneity
– Nondimensionalization of Equations
– Vapor Pressure and Cavitation
• Dimensional analysis and similarity
• The method of repeating variables and the Buckingham
pi theorem
• Experimental testing, modeling and, incomplete
similarity
– Setup of an Experiment and Correlation of Experimental
Data
– Incomplete Similarity
– Wind Tunnel Testing
– Flows with Free Surfaces

79