fluid_mechanics

7.3 Determination of Pi Terms 337
Step 5
Step 6
Step 7
Step 8
variables to form a pi term. All of the required reference dimensions must be included
within the group of repeating variables, and each repeating variable must be dimensionally
independent of the others 1i.e., the dimensions of one repeating variable cannot be
reproduced by some combination of products of powers of the remaining repeating variables2.
This means that the repeating variables cannot themselves be combined to form
a dimensionless product.
For any given problem we usually are interested in determining how one particular
variable is influenced by the other variables. We would consider this variable to be
the dependent variable, and we would want this to appear in only one pi term. Thus, do
not choose the dependent variable as one of the repeating variables, since the repeating
variables will generally appear in more than one pi term.
Form a pi term by multiplying one of the nonrepeating variables by the product of
the repeating variables, each raised to an exponent that will make the combination
a
dimensionless. Essentially each pi term will be of the form u i u i b
1 u i c
2 u i
3 where u i is one of
the nonrepeating variables; u 1 , u 2 , and u 3 are the repeating variables; and the exponents
a i , b i , and c i are determined so that the combination is dimensionless.
Repeat Step 5 for each of the remaining nonrepeating variables. The resulting set of
pi terms will correspond to the required number obtained from Step 3. If not, check your
work—you have made a mistake!
Check all the resulting pi terms to make sure they are dimensionless. It is easy to make
a mistake in forming the pi terms. However, this can be checked by simply substituting
the dimensions of the variables into the pi terms to confirm that they are all dimensionless.
One good way to do this is to express the variables in terms of M, L, and T if the
basic dimensions F, L, and T were used initially, or vice versa, and then check to make
sure the pi terms are dimensionless.
Express the final form as a relationship among the pi terms, and think about what
it means. Typically the final form can be written as
By using dimensional
analysis, the
original problem is
simplified and defined
with pi terms.
D V
(1) (2)
ρ, μ
Δp = (p 1
– p 2 )/
ß 1 f1ß 2 , ß 3 , . . . , ß kr 2
where ß 1 would contain the dependent variable in the numerator. It should be emphasized
that if you started out with the correct list of variables 1and the other steps were completed
correctly2, then the relationship in terms of the pi terms can be used to describe the problem.
You need only work with the pi terms—not with the individual variables. However, it
should be clearly noted that this is as far as we can go with the dimensional analysis; that
is, the actual functional relationship among the pi terms must be determined by experiment.
To illustrate these various steps we will again consider the problem discussed earlier in this
chapter which was concerned with the steady flow of an incompressible Newtonian fluid through
a long, smooth-walled, horizontal circular pipe. We are interested in the pressure drop per unit
length, ¢p / , along the pipe as illustrated by the figure in the margin. First (Step 1) we must list
all of the pertinent variables that are involved based on the experimenter’s knowledge of the problem.
In this problem we assume that
¢p / f 1D, r, m, V2
where D is the pipe diameter, r and m are the fluid density and viscosity, respectively, and V is the
mean velocity.
Next 1Step 22 we express all the variables in terms of basic dimensions. Using F, L, and T
as basic dimensions it follows that
¢p / FL 3
D L
r FL 4 T 2
m FL 2 T
V LT 1