2.
Deen 1.3 (Use the whisk wire diameter, viscosity and surface tension as recurring variables)

Π
2
=
L
a
2
M
LT
⎛
⎝
⎜
⎞
⎠
⎟
b
2
M
T
2
⎛
⎝
⎜
⎞
⎠
⎟
c
2
L
T
⎛
⎝
⎜
⎞
⎠
⎟
.
For the first dimensionless group we find that
L
a
1
−
b
1
+
3
=
0
M
b
1
+
c
1
=
0
T
−
b
1
−
2c
1
=
0
so that
b
1
=
c
1
=
0 and a
1
=
−
3
yielding
Π
1
=
V
D
3
.
For the second group we find that
L
a
2
−
b
2
+
1
=
0
M
b
2
+
c
2
=
0
T
−
b
2
−
2c
2
−
1
=
0
.
Then
c
2
=
−
1, b
2
=
1 and a
2
=
0
.
So the second group is the Capillary number in Table 1.6,
Π
2
=
μ
U
γ
3.
Deen 1.8
Hints:
For part (a), find k dimensionless groups as ratios of the internal dimensions to the pipe
diameter D, before beginning the usual procedure to find the
P
groups. One of the remaining
groups should correspond to one of the well-known groups in Table 1.6—which one would you
expect to be relevant?
P
Solution
3
1

pressure drop
D
P and the viscosity
μ
.
Of the latter two, the pressure drop is chosen because it is the
measurable quantity in the experiments, and is of primary interest in the design.
It is not necessary
to choose the viscosity as a non-recurring variable, but that choice is suggested by the statement in
part (c) that the pressure drop becomes independent of viscosity at high Reynolds number.
If the
viscosity only appears in one dimensionless group, then that observation can be modeled by using
equations that render one dimensionless group irrelevant when Re>>1.
The recurring quantities are
then D,
r
, and U.
As suggested in the hint, k dimensionless groups can be found immediately as ratios of the internal
dimensions to D, or
Π
i
=
d
i
D
for 1
≤
i
≤
k
.
The two remaining groups have the form
Π
k
+
1
=
D
a
1
ρ
b
1
U
c
1
μ
and
Π
k
+
2
=
D
a
2
ρ
b
2
U
c
2
Δ
P
.
Substituting dimensions,
Π
k
+
1
=
L
a
1
M
L
3
⎛
⎝
⎜
⎞
⎠
⎟
b
1
L
T
⎛
⎝
⎜
⎞
⎠
⎟
c
1
M
LT
and
Π
k
+
2
=
L
a
2
M
L
3
⎛
⎝
⎜
⎞
⎠
⎟
b
2
L
T
⎛
⎝
⎜
⎞
⎠
⎟
c
2
M
LT
2
.
Finding the coefficients shows that
Π
k
+
1
=
μ
ρ
UD
=
1
Re
and
Π
k
+
2
=
Δ
P
ρ
U
2
.