Basics of Fluid Mechanics, 2014a

302 CHAPTER 9. DIMENSIONAL ANALYSIS
Solution
The coefficients a and b multiply different terms and therefore must have different units.
The results must be unitless thus a
L 0 = a
x
{}}{
2
[ ] 1
L 2 =⇒ a =
L 2
(9.XI.b)
From equation (9.XI.b) it clear the conversion of the first term is U x = ah 2 x. The
exponent appears a bit more complicated as
(
L 0 = b exp h x )
( x
= b exp (h) exp = b exp (h) exp (x) (9.XI.c)
h
h)
Hence defining
b = 1
(9.XI.d)
exp h
With the new coefficients for both terms and noticing that y = h −→ y =1now can
be written as
a
U x (y =1)
{}}{
= ah 2 x 2 +
U 0
b
{ }} {
b exp (h) exp (x) =a x 2 + b exp x
(9.XI.e)
Where a and b are the transformed coefficients in the dimensionless presentation.
End Solution
After the boundary conditions the initial condition can undergo the non–dimensional
process. The initial condition (9.21) utilizing the previous definitions transformed into
U x (x =0)
U 0x
= U 0y
U 0x
f(y) (9.28)
Notice the new dimensionless group of the velocity ratio as results of the boundary
condition. This dimensionless number was and cannot be obtained using the Buckingham’s
technique. The physical significance of this number is an indication to the
“penetration” of the initial (condition) velocity.
The main part of the analysis if conversion of the governing equation into a dimensionless
form uses previous definition with additional definitions. The dimensionless
time is defined as t = tU 0x /h. This definition based on the characteristic time of
h/U 0x . Thus, the derivative with respect to time is
∂U x
∂t
Ux
U
{}}{
0x
= ∂ ∂
U x U 0x
}{{}
t
tU 0x
h
= U 0x 2
h
h
U 0x
∂U x
∂t
(9.29)