Dimensional Analysis Jacob Y. Kazakia © 2002 1

Dimensional analysisDrag force on sphere depends on:a) The velocity Ub) The sphere diameter Dc) The viscosity of the fluidd) The density of the fluide) Other characteristics of

secondary importance

U

F = f ( U, D, , ) but dimensional analysis gives:

number

Reynolds

pressuredynamic

2

area

2 μ

UDρf

Uρ21

4Dπ

F

Dimensional Analysis Jacob Y. Kazakia © 2002 2

Buckingham TheoremSay we have :

0)q,.q,.........q,q,g(q

or

),......qq,q,f(qq

n1n321

1n4321

The parameters can be grouped in n - m dimensionless groups

mn321 ,.......ΠΠ,Π,Π

Where m is the number of independent dimensions needed to express the dimensions of all q ’s.

We then have a relationship:

0).,.........,,(G mn321

Dimensional Analysis Jacob Y. Kazakia © 2002 3

Ex1. Journal bearing

l

D

c

The load-carrying capacity W of a journal bearing depends on:

a) The diameter Db) The length lc) The clearance cd) The angular speed e) The viscosity 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 4

Ex1 – cont.3) Write the dimensions of all parameters in terms of the primary ones

D Ll Lc 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, ,

5) Find the groups: D l L t M L t L = L 0 t 0 M 0

L: t : - M :

1 = D-1 l

Dimensional Analysis Jacob Y. Kazakia © 2002 5

Ex1 – cont -1.

third group: D W L t M L t M L t -2 = L 0 t 0 M 0

L: t : - M :

3 = D-2 -1 W

We found 1 = l / D we similarly get 2 = c / D

or 3 = W / (D2 )

6) Write the relation among the ‘s :

D

c

D

l,f

μωD

W2

Dimensional Analysis Jacob Y. Kazakia © 2002 6

Ex2. Extension of ex1

D L t M L t M L-3 = L 0 t 0 M 0

L: t : - M :

4 = D2 -1 or 4 = D2 (Reynolds number)

Resulting in:

2

2,,f

μωD

W D

D

c

D

l

Suppose we had in the list of the parameters in the previous example. We would then have an additional ( nondimensional group)

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 g , h , and V where V is the belt speed. Predict the dependence of Q on the other variables.

[Q] = L3 / t[] = F t / L2

[] = F t2 / L4

[g] = L / t2

[h] = L [V] = L / t

There are six parameters. The primary dimensions are : F, L, and t ( 3 of them) The repeating parameters will be: h, V,

We must find 6 – 3 = 3 groups.

h V Q L L t F t L L3 t= L 0 t 0 F 0 or by inspection we obtain: 1 = Q / ( V h2 )

Dimensional Analysis Jacob Y. Kazakia © 2002 8

Ex3. – cont.

L: t : - F :

2 = / ( hV )Or equivalently2 = hV /

h V L L t F t L F tL= L 0 t 0 F0

h V g by inspection we get: 3 = V / ( g h )

Result:

)

gh

V,

μ

hVρf(

Vh

Q

numberFroude

2

numberReynolds

2

Dimensional Analysis Jacob Y. Kazakia © 2002 9

Ex4. Choice of repeating parametersA fluid of density and viscosity flows with speed V under pressure p. Determine a relation between the four parameters

[V] = L / t[] = M / L3

[p] = M / (L t2 )[] = M / ( L t )

V p L t M L M L t-2 M Lt -1= L 0 t 0 F0

L: t : - M : +

But this system has no solution.See next slide

Dimensional Analysis Jacob Y. Kazakia © 2002 10

Ex4. –cont.- +

1

1

1

110

201

131

1

2

1

110

330

131

1

1

1

000

110

131

Inconsistent system.No solution is possible

But if we chose out repeated variables differently: V, then everything is O.K.

V p L t M L M L t- M Lt -2= L 0 t 0 F0

Produces: = -2 , = -1, = 0 and hence p / V2 it is now easy to see why the previous choice did not work.

Dimensional Analysis Jacob Y. Kazakia © 2002 11

Model flowPrototype torpedo:D = 533 mm , Length = 6.7 ftSpeed in water: 28 m/sec

Model: Scale 1/5 in wind tunnelMax wind speed : 110 m/secTemperature : 20 0CForce 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, and we get by the earlier method:

μ

DVρf

DVρ

F22

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 p = 1 x 10-3 Pa secAir at 20 0C M = 1.8 x 10–5 Pa sec

33P

M

M

P

M

PPM m

kg22.9

1

0.018

1

5

110

28

m

kg999

μ

μ

D

D

V

Vρρ

We must now find the pressure that will give this type of air density.IDEAL GAS LAW:

MPa1.93K293Kkg

mN287

m

kg22.9RTρp

3MMM

(about 20 atmospheres)

kN43.71

5

110

28

22.9

999N618

D

D

V

V

ρ

ρFF

2

2

2

2M

2P

2M

2P

M

PMP

The force: