Design and Analysis of
Multi-Factored Experiments
Dimensional Analysis
L. M. Lye
DOE Course
1
Dimensional Analysis
• Dimensional analysis is a mathematical method
which is of considerable value in problems which
occur in engineering.
• Dimensional analysis is essentially a means of
utilizing partial knowledge of a problem when the
details are too obscure to permit an exact analysis.
• Dimensional Analysis is also known as Partial
Analysis – problem is partially solve and still
require experimentation to obtain a functional
relationship.
L. M. Lye
DOE Course
2
Basic Dimensions
• Variables used in engineering are usually
expressed in terms of a limited number of basic
dimensions namely: Mass (M), Length (L),
Time (T), and sometimes Temperature (q).
• E.g. Velocity, V= length/time or LT-1
•
Force, F = mass x accel (N) = MLT-2
•
Viscosity, m = N.s/m2 = ML-1T-1
•
Flow rate, Q = m3/s = L3T-1
• Hence most physical quantities we deal with
have only 3 basic dimensions: M, L, T.
L. M. Lye
DOE Course
3
Uses of Dimensional Analysis
• Four main uses of DA:
– Checking the consistency of units, making sure that the
LHS has the same units as the RHS of an equation. i.e
dimensional homogeneity. E.g. E (ML2T-2)= mc2
(ML2T-2)  apples + apples = apples.
– Determining the units of empirical coefficients – e.g. V
= C R1/2S1/2 What’s the units of Chezy’s C?
– Understanding complex phenomenon by expressing
functional relationships in terms of dimensionless
parameters to reduce the dimensionality of the problem
and to simplify the analysis. This helps to establish the
form of an equation relating a number of variables.
– To assist in the analysis of experimental results.
L. M. Lye
DOE Course
4
General Ideas
• If any physical quantity, J, is considered, it will be
possible to reduce it to some function of the three
fundamental dimensions, M, L, and T.
• i.e. J = f [M, L, T]
• If the magnitude of J is compared for two similar
systems, then:
• J”/J’ = f [M, L, T]”/ f [M, L, T]’= lJ
• Evidently, this ratio must be dimensionless.
• This is true if the function is in the form of a
product: J = K [Ma Lb Tc]
L. M. Lye
DOE Course
5
•
•
•
•
•
General ideas (continued)
Where K is a numeric value, and a, b, and c are
powers or indices who magnitudes have to be
determined.
E.g. J = velocity, then K=1, a=0, b=1, c= -1, since
velocity has dimensions LT-1.
It can be argued that dimensional relationships are
arbitrary, since the magnitudes depend on the choice
of units (feet, metres, pounds, kilograms, etc.)
For this reason, an equation which is a statement of
a physical law is often used in a dimensionless
form.
Dimensionless equations are completely general,
and are therefore frequently the basis for
representation of experimental data.
L. M. Lye
DOE Course
6
Methods of Dimensional Analysis
• There are many methods of dimensional analysis.
The techniques have been progressively refined
over the years.
• Rayleigh’s Indicial method – oldest method
• Buckingham p theorem – most famous method
• Hunsaker and Rightmire’s method – quick method
• Matrix Method – modern method using a
computer for matrix inversion.
• All methods are absolutely dependent on the
correct identification of all the factors which
govern the physical events being analyzed.
L. M. Lye
DOE Course
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• The omission of a single factor may give quite
misleading results.
• Hence, dimensional analysis is not unlike DOE.
Choice of factors that influence the response must
be carefully chosen.
• DOE can be used for factors that cannot be
expressed in physical quantities e.g. categorical
variables (colours, gender, material types, etc.)
• However, DOE can be combined with DA to
obtain functional relationships in an efficient way.
• More on this later.
L. M. Lye
DOE Course
8
Rayleigh’s method
• Consider a problem involving a scale model test of
a hydraulic machine. The thrust force F, velocity v,
viscosity m and density r of the fluid is given
including a typical size of the system, L, is also
given.
• Two questions must be posed, namely:
– How to analyze or plot the data in the most informative
way, and
– How to relate the performance of the model to that of
the working prototype.
L. M. Lye
DOE Course
9
Solution
• Let’s postulate that the force F is related to
the other given quantities:
[1]
F = f [r, v, m, L]
• The form of the function is completely
unknown, but it has been proposed earlier
that:
– The function must be in the form of a power
product
– There must be a dimensional balance between
both sides of the equation
L. M. Lye
DOE Course
10
• From [1], the equation maybe rewritten as:
[2]
F = K [ra vb mc, Ld]
• So obviously, for dimensional homogeneity, the
dimensions on the LHS must equal those on the
RHS.
• Expressing each quantity in [2] in terms of its
dimensions,
• MLT-2 = K[ (ML-3)a (LT-1)b (ML-1T-1)c Ld]
• Equating the indices for M, L, and T,
• M: 1 = a + c
• L: 1 = -3a + b - c + d
• T: -2 = -b - c or 2 = b + c
L. M. Lye
DOE Course
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• Thus we have 3 equations but 4 unknowns,
so a complete solution is not attainable.
• We can only obtain a partial solution.
• Let’s express all indices in terms of say c.
• a=1–c
• b=2–c
• d=2–c
• Substituting for a, b, and d in [2]:
• F = K [r1-c v2-c mc L2-c]
L. M. Lye
DOE Course
12
• Or,
c

 m 
2
2
 
F  K  r v L 
r vL  




• Since the function represents a product, it may be
restated as:
 r vL
F  r v L K 
 m
2 2
L. M. Lye
DOE Course



c
13
• Or as:
 r vL
 K 
2 2
rv L
 m
F



c
• Where: K and c are unknown and must be
obtained by experimentation.
• Key points to note from the above equation:
– Two groups have emerged from the analysis. If you
check, it will be found that both groups are
dimensionless.
– Dimensionless groups are independent of units and of
scale and are therefore equally applicable to the model
or to the prototype
L. M. Lye
DOE Course
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– Both groups represent ratios of forces: thrust
force/inertial force, and inertial force/viscous
force.
– All three fundamental dimensions are present.
Therefore, if the model is to truly represent the
prototype, then both model and prototype must
conform to the law of dynamic similarity, i.e. the
magnitude of each dimensionless group must be
the same for the model as for the prototype.
– The dimensionless groupings of variables are not
unique. Different dimensionless groups can
emerge.
– The Raleigh indicial method is okay as long as
the number of variables is small.
L. M. Lye
DOE Course
15
Another example
• The velocity of propagation of a pressure wave
through a liquid can be expected to depend on the
elasticity of the liquid represented by the bulk
modulus K, and its mass density r. Establish by D.
A. the form of the possible relationship.
• Assume: u = C Ka rb
• U = velocity = L T-1, r = M L-3, K = M L-1 T-2
• L T-1 = Ma L-a T-2a x Mb L-3b
• M: 0 = a + b
• L: 1 = -a – 3b
• T: -1 = - 2a
L. M. Lye
DOE Course
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• Therefore: a = ½ , b = -½, and a possible equation
is:
u C
K
r
• Rayleigh’s method is not always so straightforward.
Consider the situation of flow over a U-notched
weir.
• Q = f(r, m, H, g)
• [Q] = [C ra mb Hc gd]
[ ] => dimensions of
• Using the M, L, T system,
• [L3 T-1] = [ML-3]a [M L-1 T-1]b [L]c [L T-2]d
• M: 0 = a + b
(1)
• L: 3 = -3a –b +c + d
(2)
• T: -1 = - b – 2d
(3)
L. M. Lye
DOE Course
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• We have only 3 equations, but there are 4
unknowns. Need to express a, b, c, in terms
of d.
• b = 1 – 2d
• a = -b = 2d -1
• c = 3 + 3a + b – d = 1 + 3d
• Q = C r(2d-1) m(1-2d) H(1+3d) g(d)
2
3
mH  r H g

 C
r  m 2
L. M. Lye




d
or
DOE Course
 r 2H 3g 
 

2
mH
 m

rQ
18
Buckingham P theorem (1915)
• This is perhaps the mother of all dimensional
analysis methods. Many other methods were built
upon this method.
• Buckingham proposed that:
– If a physical phenomenon was a function of m quantities
and n fundamental dimensions, dimensional analysis
would produce (m-n) P groups;
– Each P should be a function of n governing variables plus
one more quantity;
– The governing quantities must include all fundamental
dimensions;
– The governing quantities must not combine among
themselves to form a dimensionless group;
– As each P is dimensionless, the final function must be
dimensionless, and therefore dimensionally:
–
f [P1, P2, …, Pm-n] = M0 L0 T0
L. M. Lye
DOE Course
19
Example 1
• Using the previous problem, there were 5
quantities (F, r, v, m, L) and 3 dimensions (M, L,
T), from which we can derive 2 groups.
• First choose 3 governing variables that together
must contain M, L, and T. Let’s choose: r, v, L.
• Combine them with one other variable to get P1:
P1 = ra vb Lc F = M0 L0 T0
• Therefore:
• (ML-3)a (LT-1)b Lc (MLT-2) = M0L0T0
L. M. Lye
DOE Course
20
Example 1 (continued)
•
•
•
•
•
•
•
Equating indices:
M: a+1 = 0, therefore a = -1
T: -b – 2 = 0, therefore b = -2
L: -3a + b + c + 1 = 0
Substituting for a and b, c = -2.
So, P1 = F/rv2L2
Repeat the process with the other variable:
P2 = ra vb Lc m = M0 L0 T0 = m/rvL, as before
L. M. Lye
DOE Course
21
Example 2
• The head loss per unit length (Dh/L) of pipe
in turbulent flow through a smooth pipe
depends on v, D, r, g, and m. Determine the
general form of the equation.
• F(Dh/L, v, D, r, m, g) = 0; 6 variables, 3
dimensions, 3 p terms.
• That is: p1 = f(p2, p3).
• Choose 3 repeating variables: v, D, and r
x
x
x
p1  v 1 D 1 r 1 m
L. M. Lye
DOE Course
22

 LT
•
•
•
•

 1 x1
 L  ML
x1

 3 x1
1
ML T
1
M: z1 + 1 = 0
L: x1 + y1 – 3z1 – 1 = 0
T: -x1 – 1 = 0
Therefore: x1 = -1, z1 = -1, and y1 = -1.
1
p1  v D
L. M. Lye
1
r
1
m or
m
r vD
DOE Course
or
r vD
m
 Re
23
• Final solution:
Dh

  Re, Fr
L
2

• Where Re = Reynold’s number, and Fr =
v
Froude number =
gD
L. M. Lye
DOE Course
24
Rightmire and Hunsaker’s method (1947)
• This method is easier to use and quicker
than Buckingham’s method. The method is
similar the Buckingham’s method by use of
repeating variables, but express them in
term of the variables themselves.
• e.g. D to represent Length, [L] = D
•
v to represent Time, [T] = L/v = D/v
•
r to represent Mass, [M] = rL3 = rD3
L. M. Lye
DOE Course
25
Example
 p1: g = L T-2 = v2D D-2 = v2 D-1,
2
v
• therefore p1 =
gD
 p2: m = M
L-1
T-1
=
rD v
3
D
• Therefore: p2 =
 p3 =
L. M. Lye
Dh
r vD
m
2
 r vD
 Re
, as before.
L
DOE Course
26
Another example
• A spherical drop of liquid of diameter D
oscillates under the influence of its surface
tension. Investigate the frequency of
oscillation f.
• F(f,s, D, r) = 0
• Answer:
L. M. Lye
f  K
s
rD
3
DOE Course
27
Matrix method
• Best for problems with many variables. Can be
solved using a matrix inversion routine.
• Consider a problem with 7 variables in 4
dimensions (M, L, T, q)
• First form the dimensional matrix:
a b
c
d
e
f
g
M
L
T
q
L. M. Lye
A
DOE Course
B
28
• We need to transform the above matrix to:
a b c d
a 1 0 0 0
b 0
1
0
0
c 0
0
1
0
e
f
g
d 0 0 0 1
• I = A-1A
D = A-1B
• A-1 = adj A/ |A|
• Whatever operation was done to get the unit
matrix on the left must be also done on the right to
get D.
L. M. Lye
DOE Course
29
• Consider the earlier problem of the head
loss per unit length.
• F(Dh/L, v, g, D, r, m) = 0
r
M 1
v
0
D
0
0
1
Dh/L
0
L -3
1
1
1
-1
0
0 -1
0
-2 -1
0
T
A
L. M. Lye
m
g
B
DOE Course
Inverse of A can
be obtained by
Matlab, Minitab
or by hand
30
A
D=
1
1

 0

 3
 0

2

  1
0
0
1
1
1
1
0 

1

1 
0

0

0 
1

D = 0
 3
0
0
1
0  0

1 1

1    2
1
1
1
0

0

0 
The three p terms are:
m Dh 
 gD
 2 , r vD , L 
v

The matrix method can obtain the p terms all in one go
instead of one term at a time.
L. M. Lye
DOE Course
31
Comments about Dimensional Analysis
• Most important but most difficult problem in
applying DA to any problem is the selection of the
variables involved. There is no easy way of
identifying the correct variables without specialized
knowledge about the phenomenon being
investigated.
• If you select too many variables, you get too many
p terms and may require much additional
experimentation to eliminate them.
• If important variables are omitted, then an incorrect
result will be obtained, which may prove to be
costly and difficult to ascertain.
L. M. Lye
DOE Course
32
•
•
•
•
Type of variables:
Geometry – length, angles, diameter, or area.
Material properties – r, m, elasticity, etc.
External effects – any variable that tends to
produce a change in the system e.g. forces,
pressures, velocities, gravity, etc.
• You need to keep the number of variables to a
minimum, and that they are independent. E.g. D
and Area need not be included together because
one is derived from the other. Therefore, heavy
thinking is required in variable selection  similar
to DOE.
L. M. Lye
DOE Course
33
Points to consider in the selection of variables:
•
•
•
•
Clearly define the problem. What is the main
response variable of interest? That is, what is
Y?
Consider the basic laws that govern the
phenomenon. Even a crude theory may be
useful.
Start the selection process by grouping the
variables in the 3 broad classes: geometry,
material properties, and external effects.
Consider other variables that may not fall into
one of the three categories, e.g. time,
temperature, colour, equipment, etc.
L. M. Lye
DOE Course
34
•
•
•
•
Be sure to include all quantities that enter the
problem even though some of them may be held
constant e.g. g. For D.A. it is the dimensions of
the quantities that are important – not specific
values.
Make sure that all variables are independent –
look for relationships among subsets of the
variables (same as DOE).
Remember that after a dimensional analysis,
you still need to carry out the experiment to
relate the dimensionless groups. Hence DOE
may be needed unless you have only 1 or 2 p
terms.
Remember the first lecture?
L. M. Lye
DOE Course
35
More on p terms
• Specific p terms obtained depend on the
somewhat arbitrary selection of repeating
variables. For example, if we choose:
 m, D, g instead of r, D, v, we would end up
with a different set of p terms. Both results
would be correct, and both would lead to
the same final equation for Dh/L, however,
the function relating the different p terms
would be different.
L. M. Lye
DOE Course
36
• Hence, there is not a unique set of p terms
which arises from a dimensional analysis.
However the required number of p terms is
fixed, and once a correct set is determined
all other possible sets can be developed
from this set by combination of products of
the original set.
• e.g. if we have problem involving 3 p terms,
p1 = f(p2, p3)
• we can combine the p terms to give a new p
term: p 2'  p 2a p 3b
L. M. Lye
DOE Course
37
• Then the relationship could be expressed as:

p1   p 2 ,p 3
'

or even as

p1   p
'
2 ,p 3

• All these would be correct; however, the
required number of p terms cannot be
reduced by this manipulation; only the form
of the p terms is altered.
p1 
D pD
rv
L. M. Lye
2
;
p2 
r vD
m
p 1  p 1p 2 
1
DOE Course
D pD r vD
rv
2
m

D pD
mv
38
2
• Which form of p terms is best? There is no
simple answer. Best to keep it simple. Some
p terms are well-known dimensionless
numbers like Reynolds, Froude, Mach,
Weber, Cauchy, Euler, etc.
• In pipe flow problems, the Reynold’s number
is prominent while in open channel and ocean
engineering problems, the Froude number is
more relevant.
• So it all depends on the field of investigation.
L. M. Lye
DOE Course
39
Correlation of Experimental Data
• One of the most important uses of
dimensional analysis is an aid in the efficient
handling, interpretation, and correlation of
experimental data.
• As noted earlier, DA cannot provide a
complete answer and a specific relationship
among the p terms cannot be determined
without experimentation. The degree of
difficulty obviously depends on the number
of p terms.
L. M. Lye
DOE Course
40
• Problems with:
• 1 p term  p = C where C = a constant
• 2 p terms  p1 = p2)  simple
regression problem
• > 2 p terms  p1 = (p2, p3)  multiple
regression problem
• With more and more p terms, a DOE
approach may be needed and may require
the use of RSM if relationship is nonlinear.
L. M. Lye
DOE Course
41
Example
• The pressure drop per unit length, Dp/L for the flow
of blood through a horizontal small diameter tube is
a function of flow rate Q, diameter D, and the blood
viscosity m. For a series of test with D = 2 mm and
m = 0.004 Ns/m2, the following data were obtained
for Dp measured over a length of 300 mm.
• Q (m3/s): 3.6 4.9 6.3
Dp (N/m2): 1.1 1.5 1.9
7.9
2.4
9.8 (x 10-6)
3.0 (x 104)
• Perform a DA for this problem and make use of the
data to determine a general relationship between Dp
and Q, one that is valid for other values of D, L and
m.
L. M. Lye
DOE Course
42
Solution
• 4 variables, F (Dp/L, D, Q, m) = 0,
• i.e. 4 – 3 = 1 p term
• From D.A. (try this yourself), we get:
(D p / L ) D
4
= constant. Try this yourself
mQ
• Substituting the values used in the
experiment,
(D p / L ) D
mQ
L. M. Lye
4
=
D p / 0 . 3 ( 0 . 002 )
0 . 004  Q
4
 1 . 33  10
DOE Course
8
Dp
Q
43
• Using the data obtained from the experiment,
(D p / L ) D
mQ
4
= [ 40.6, 40.7, 40.1 40.4 40.7]
• Average for constant C = 40.5, hence:
Dp
L
L. M. Lye
 40 . 5
mQ
D
DOE Course
4
44
Example 2
• A liquid flows with a velocity v through a hole in
the side of a tank. Assume that v = f(h, g, r, s).
Where h is the height of water above the hole, r is
the density of the fluid, and s the surface tension.
The density r is 1000 kg/m3, and s = 0.074 N/m.
The data obtained by changing h and measuring v
are:
• V (m/s) 3.13 4.43 5.42 6.25 7.00
• h (m) 0.50 1.00 1.50 2.00 2.50
• Plot the data using appropriate dimensional
variables. Could any of the original variables have
been omitted?
L. M. Lye
DOE Course
45
Solution
• 5 variables, F(v, h, g, r, s) = 0, 2 p terms.
• From dimensional analysis, v
 r gh 2 
 

gh

s
rgh2/s: 3.31 13.3 28.8 53.0 82.9
v/(gh)1/2: 1.41 1.41 1.41 1.41 1.41
Plotting the data will show that v is



gh
r gh
2
Independent of s which means that r and s
can be omitted. Of course this is well-known if
one were to apply the Bernoulli equation to solve
the problem.
L. M. Lye
DOE Course
46
References
• Chapter 2, 3, 4 notes from course website.
• Thomas Szirtes (1998): Applied Dimensional
Analysis and Modeling, McGraw Hill, 790
pages.
• Most books in physics and fluid mechanics.
• See by Islam and Lye (2007) on combined
use of DA and DOE.
L. M. Lye
DOE Course
47