Mechanics of Fluids

174 Physical similarity and dimensional analysis
Step 5 Form a dimensionless group involving F and the three repeating
variables.
As a starting point, assume
N1 =
F
d α u β ϱ γ
where α, β and γ are to be evaluated.
For N1 to be dimensionless the indices must satisfy [MLT −2 ] =
[L] α [LT −1 ] β [ML −3 ] γ
Equating the indices of [L] we obtain
1 = α + β − 3γ
Equating the indices of [T] we obtain
−2 =−β
Equating the indices of [M] we obtain
1 = γ
Solving β = 2, γ = 1 and hence α = 2, yielding
N1 = F
d 2 u 2 ϱ
which is conventionally written in the form
N1 = F
ϱu 2 d 2
We next form a dimensionless group involving µ and the three
repeating variables. Using a similar analysis to that set out previously,
we write
Hence
N2 =
µ
d α u β ϱ γ
[ML −1 T −1 ]=[L] α [LT −1 ] β [ML −3 ] γ
which is solved, as before, to yield
Step 6 Hence
N1 = φ1(N2) or
N2 = µ
duϱ
F
ϱu2 � �
µ
= φ1
d2 duϱ
Finally, since µ/(duϱ) is recognised as the reciprocal of Reynolds
number, we can write the relation between the non-dimensional
quantities in the final form
F
ϱu2 = φ2
d2 � �
ϱud
= φ2(Re)
µ