fluid_mechanics

344 Chapter 7 ■ Dimensional Analysis, Similitude, and Modeling
and, therefore, a 1, b 1 so that
ß 4 E Dg
This will always be true—you cannot affect the required number
of pi terms by using M, L, and T instead of F, L, and T, or vice
versa.
ß 4
which is the same as obtained using the FLT system. The other
pi terms would be the same, and the final result is the same; that is,
d
D f a h D , d D , E Dg b
(Ans)
DΔp
ρV 2
D 2 Δp
Vm
____ ρVD
μ
____ ρVD
μ
Once a correct set
of pi terms is obtained,
any other
set can be obtained
by manipulation of
the original set.
7.4.3 Uniqueness of Pi Terms
A little reflection on the process used to determine pi terms by the method of repeating variables
reveals that the specific pi terms obtained depend on the somewhat arbitrary selection of
repeating variables. For example, in the problem of studying the pressure drop in a pipe, we selected
D, V, and r as repeating variables. This led to the formulation of the problem in terms of
pi terms as
What if we had selected D, V, and m as repeating variables? A quick check will reveal that the pi
term involving ¢p / becomes
and the second pi term remains the same. Thus, we can express the final result as
Both results are correct, and both would lead to the same final equation for ¢p / . Note, however,
that the functions f and f 1 in Eqs. 7.5 and 7.6 will be different because the dependent pi terms
are different for the two relationships. As shown by the figure in the margin, the resulting graph
of dimensionless data will be different for the two formulations. However, when extracting the
physical variable, ¢p / , from the two results, the values will be the same.
We can conclude from this illustration that there is not a unique set of pi terms which
arises from a dimensional analysis. However, the required number of pi terms is fixed, and once
a correct set is determined, all other possible sets can be developed from this set by combinations
of products of powers of the original set. Thus, if we have a problem involving, say, three
pi terms,
we could always form a new set from this one by combining the pi terms. For example, we could
form a new pi term, ß¿ 2 , by letting
where a and b are arbitrary exponents. Then the relationship could be expressed as
or
¢p / D
rV 2
¢p / D 2
Vm
f a rVD
m b
¢p / D 2
Vm
f 1 a rVD
m b
ß 1 f1ß 2 , ß 3 2
ß¿ 2 ß a 2 ß b 3
ß 1 f 1 1ß¿ 2 , ß 3 2
ß 1 f 2 1ß 2 , ß¿ 2 2
All of these would be correct. It should be emphasized, however, that the required number of pi
terms cannot be reduced by this manipulation; only the form of the pi terms is altered. By using
(7.5)
(7.6)