1. Dimensional analysis Drag force on sphere depends on: The velocity U
The sphere diameter D
The viscosity m of the fluid
The density r of the fluid
Other characteristics of secondary importance
F = f ( U, D, m, r )
but dimensional analysis gives:
Dimensional Analysis
Jacob Y. Kazakia © 2002

2. Buckingham P Theorem Say we have :
The parameters can be grouped in n - m dimensionless groups
Where m is the number of independent dimensions needed to express the dimensions of all q ’s.
We then have a relationship:
Dimensional Analysis
Jacob Y. Kazakia © 2002

3. Ex1. Journal bearing
The load-carrying capacity W of a journal bearing depends on:
The diameter D
The length l
The clearance c
The angular speed w
The viscosity m of the lubricant l D c
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

4. Ex1 – cont.
3) Write the dimensions of all parameters in terms of the primary ones
D  L
l  L
c  L
 t –1
 M L –1 t –1
W  M L t -2
All 3 primary dimensions are needed.
r = 3 hence there must be
n – r = 6 – 3 = 3 dimensionless groups.
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 =  a = -1
t : - b - g =  b = 0
M : g =  g = 0
P1 = D-1 l
Dimensional Analysis
Jacob Y. Kazakia © 2002

5. Ex1 – cont -1. We found P 1 = l / D we similarly get P 2 = 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 =  a = -2
t : - b - g -2 =  b = -1
M : g +1 =  g = -1 or
P3 = D-2 w-1 m-1 W
P3 = W / (D2 w m )
6) Write the relation among the P ‘s :
Dimensional Analysis
Jacob Y. Kazakia © 2002

6. 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 =  a = 2
t : - b - g =  b = 1
M : g +1 =  g = -1 or
P4 = D2 w r / m (Reynolds number)
P4 = D2 w m-1 r
Resulting in:
Dimensional Analysis
Jacob Y. Kazakia © 2002

7. 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
We must find 6 – 3 = 3 groups.
[Q] = L3 / t
[m] = F t / L2
[r] = F t2 / L4
[g] = L / t2
[h] = L
[V] = L / t
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

8. 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 =  a = -1
t : - b + 2g +1 =  b = -1
F : g +1 =  g = -1
P2 = m / ( hVr )
Or equivalently
P2 = hVr / m
h a V b r g g  by inspection we get:
P3 = V2 / ( g h )
Result:
Dimensional Analysis
Jacob Y. Kazakia © 2002

9. 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
But this system has
no solution.
See next slide 
Dimensional Analysis
Jacob Y. Kazakia © 2002

10. Ex4. –cont. a - 3b - g - 1 = 0 - a - 2g -1 = 0 b + g +1 = 0
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

11. 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:
To attain dynamically similar model test we must have equal Reynolds numbers in both flow situations.
Dimensional Analysis
Jacob Y. Kazakia © 2002

12. 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
We must now find the pressure that will give this type of air density.
IDEAL GAS LAW:
(about 20 atmospheres)
The force:
Dimensional Analysis
Jacob Y. Kazakia © 2002