444
Chapter 10
dimensions L t
\D a
has dimensions L21"1
,
r]
has dimensions mass (M) L'11'1
, and p has dimensions M L 3.
Thus, we get 3 linear constraints on the exponents, which can be written
«1
f
2
1
1 - 1 - 3 '
-1
= -1
-1
0 - 1
0
Y
(2
)
0
0
0
0
1
1
5
s
This leaves 3 degrees of freedom for the six variables, which implies that the correlation can be
reformulated in terms of 3 dimensionless groups. (These groups were denoted “n,” in the original
formulation by Buckingham, thus the name of the theorem.) By choosing to express the other exponents in
terms of a and p, we get
y - p - 1
(3)
S ~ l-cu- p
(4)
e = -(1-ai
-
p)
(5)
This is dimensionally consistent. However, it does not result in the most often used dimensionless groups in
Table 10.6. To arrive at the more familiar dimensionless groups we set
at = I-a
(6)
This gives:
and thus
S —
a - p
£ - fi-a
(7)
(
8
)
(9)
which can be rearranged into:
1
-----
OC
n
a
ubd bPt
1
1_____I
.
D a P
TJ
(
10
)
or
Sh cc ScaR efi
(12)
where
db
is used as the characteristic length and
ub
as the characteristic velocity in the dimensionless groups.
In the current example a high velocity of the bubble relative to its surrounding liquid was assumed._______
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