Basics of Fluid Mechanics, 2014a

9.2. BUCKINGHAM–π–THEOREM 289
Equation (9.VI.a) leads to three equations as
⎛ ⎞
U
ρ
{}}{ {}}{
L
= ⎜
M
⎟
t ⎝ L 2 ⎠
a ⎛
⎜
⎝
⎞
σ
{}}{
M
⎟
t 2 ⎠
b ⎛
⎞
λ
{}}{
⎝ L ⎠
c
(9.VI.b)
Mass,M a+ b = 0
Length,L −2a + c = 1
time,t −2b = −1
The solution of equation set (9.VI.c) results in
√ σ
U =
λρ
(9.VI.c)
(9.VI.d)
Hence reduction of the surface tension by half will reduce the disturbance velocity by
1/ √ 2.
End Solution
Example 9.7:
Eckert number represent the amount of dissipation. Alternative number represents the
dissipation, could be constructed as
( dU
μ
dl
Diss =
ρU 2
l
U
) 2
( ) 2 dU
μ l
dl
=
ρU 3
(9.VII.a)
Show that this number is dimensionless. What is the physical interpretation it could
have? Flow is achieved steady state for a very long two dimensional channel where
the upper surface is moving at speed, U up , and lower is fix. The flow is pure Couette
flow i.e. a linear velocity. Developed an expression for dissipation number using the
information provided.
Solution
The nominator and denominator have to have the same units.
μ
{}}{
M
✓Lt
M
( dU
dl ) 2
{ }} {
✚✚L 2
t 2 ✚✚L 2
ρ U
l {}}{ {}}{
3
{}}{
M ✚✚L 3
✓L =
✚✚L 3 t 3
(9.VII.b)
= M t3 t 3