Dimensional analysis
U
F = f ( U, D, m, r )
but dimensional analysis gives:
Drag force on sphere depends on:
a) The velocity U
b) The sphere diameter D
c) The viscosity m of the fluid
d) The density r of the fluid
e) Other characteristics of
secondary importance
 ρ UD 

 f 
2
μ 
πD  1

2





ρU
2
Reynolds
4






number

 dynamic
F

area
Dimensional Analysis
Jacob Y. Kazakia © 2002

pressure
1
Buckingham P Theorem
Say we have :
q1  f(q2 , q 3 , q 4 ,......qn 1 )
or
g(q1 , q 2 , q 3 ,..........qn 1 , q n )  0
The parameters can be grouped in n - m dimensionless groups
Π1 , Π2 , Π3 ,.......Πn m
Where m is the number of independent dimensions needed to
express the dimensions of all q ’s.
We then have a relationship:
G ( P1, P2 , P3 ,..........Pn m )  0
Dimensional Analysis
Jacob Y. Kazakia © 2002
2
Ex1. Journal bearing
The load-carrying capacity W of a
journal bearing depends on:
D
a) The diameter D
c
b) The length l
c) The clearance c
d) The angular speed w
l
e) The viscosity m of the lubricant
We must now go through the following 6 steps:
1) The number of parameters n = 6 ( including W )
2) Select as primary dimensions: M, L, t ( mass, length, time)
Dimensional Analysis
Jacob Y. Kazakia © 2002
3
Ex1 – cont.
3) Write the dimensions of all parameters in terms of the primary ones
D L
lL
All 3 primary dimensions are needed.
c L
r = 3 hence there must be
w  t –1
n – r = 6 – 3 = 3 dimensionless groups.
m  M L –1 t –1
W  M L t -2
4) Select repeating parameters: D, w, m
5) Find the groups:
D a w b m g l  La tb M g Lg tg L = L 0 t 0 M 0
L: a  g + 1  0
t: -bg 0
M: g 0
Dimensional Analysis
 a  1
b0
g0
Jacob Y. Kazakia © 2002
P1 = D-1 l 4
Ex1 – cont -1.
We found
P 1= l / D
we similarly get
P2=c/D
third group:
D a w b m g W  La tb M g Lg tg M L t -2 = L 0 t 0 M 0
L: a  g + 1  0
t : - b  g 2  0
M : g +1  0
P3 = D-2 w1 m-1 W
or
P3 = W / (D2 w m )
6) Write the relation among the P ‘s :
Dimensional Analysis
 a  2
 b  1
 g  1
Jacob Y. Kazakia © 2002
W
l c
 f , 
2
D ωμ
D D
5
Ex2. Extension of ex1
Suppose we had r in the list of the parameters in the previous
example. We would then have an additional P ( nondimensional
group)
D a w b m g r  La tb M g Lg tg M L-3 = L 0 t 0 M 0
L: a  g  3  0
t: -bg 0
M : g +1  0
P4 = D2 w m-1 r
Resulting in:
Dimensional Analysis
or
a2
b1
 g  1
P4 = D2 w r / m (Reynolds number)
 l c D 2w r 
W

 f  , ,
2
D ωμ
m 
D D
Jacob Y. Kazakia © 2002
6
Ex3. Belt in viscous fld.
A continuous belt moving vertically through a bath of viscous liquid
drags a layer of thickness h along with it. The volume flow rate Q
of the liquid depends on m , r , g , h , and V where V is the belt
speed. Predict the dependence of Q on the other variables.
There are six parameters.
The primary dimensions are :
F, L, and t ( 3 of them)
The repeating parameters will be: h, V, r
[Q] = L3 / t
[m] = F t / L2
[r] = F t2 / L4
[g] = L / t2
[h] = L
[V] = L / t
We must find 6 – 3 = 3 groups.
h a V b r g Q  La Lb tb F g t2g L4g L3 t1 = L 0 t 0 F 0
or by inspection we obtain: P1 = Q / ( V h2 )
Dimensional Analysis
Jacob Y. Kazakia © 2002
7
Ex3. – cont.
h a V b r g m  La Lb tb F g t2g L4g F t L2 = L 0 t 0 F0
L: a + b  4g  2  0
t : - b + 2g +1  0
F : g +1  0
 a  1
 b  1
 g  1
h a V b r g g  by inspection we get:
Result:
Dimensional Analysis
P2 = m / ( hVr )
Or equivalently
P2 = hVr / m
P3 = V2 / ( g h )
2
Q
hVρ V
 f(
,
)
2
h V
μ
gh
 
Jacob Y. Kazakia © 2002
Reynolds Froude
number number
8
Ex4. Choice of
repeating parameters
A fluid of density r and viscosity m flows with speed V under
pressure p. Determine a relation between the four parameters
[V] = L / t
[r] = M / L3
[p] = M / (L t2 )
[m] = M / ( L t )
V a r b p g m  La ta M b L3b M g Lg t-2g M L1 t -1 = L 0 t 0 F0
L: a  3b  g  1  0
t : - a  2g 1  0
M : b + g +1  0
Dimensional Analysis
Jacob Y. Kazakia © 2002
But this system has
no solution.
See next slide 
9
Ex4. –cont.
a  3b  g  1  0
- a  2g 1  0
b + g +1  0
1
3
1 1
1  3 1 1
1
0
0
1
2 1
1 1
0 3 3 2
0
1
1 1
1  3 1 1
0
1
1 1
0
0
0 1
Inconsistent system.
No solution is possible
But if we chose out repeated variables differently: V, r, m
then everything is O.K.
V a r b m g p  La ta M b L3b M g Lg t-g M L1 t -2 = L 0 t 0 F0
Produces: a = -2 , b = -1, g = 0 and hence p / r V2
it is now easy to see why the previous choice did not work.
Dimensional Analysis
Jacob Y. Kazakia © 2002
10
Model flow
Prototype torpedo:
D = 533 mm , Length = 6.7 ft
Speed in water: 28 m/sec
Model: Scale 1/5 in wind tunnel
Max wind speed : 110 m/sec
Temperature : 20 0C
Force on model: 618 N
Find : a) required wind tunnel pressure for dynamically similar test
b) Expected drag force on prototype
We assume the parameters: F, V, D, r, m and we get by the earlier
method:
ρ V D
F

 f 
2 2
ρV D
 μ 
To attain dynamically similar model test we must have equal
Reynolds numbers in both flow situations.
Dimensional Analysis
Jacob Y. Kazakia © 2002
11
Model flow – cont.
Water at 20 0C mp = 1 x 10-3 Pa sec
Air at 20 0C
mM = 1.8 x 10–5 Pa sec
VP DP μ M
kg 28 5 0.018
kg
ρM  ρP
 999 3
 22.9 3
VM DM μ P
m 110 1 1
m
We must now find the pressure that will give this type of air density.
IDEAL GAS LAW:
kg
N m
p M  ρ M RTM  22.9 3 287
293K  1.93MPa
m
kg K
(about 20 atmospheres)
The force:
ρ P VP2 D2P
999 282 52
FP  FM
 618 N
 43.7 kN
2
2
2
ρ M VM DM
22.9110 1
Dimensional Analysis
Jacob Y. Kazakia © 2002
12