Reynolds(1842-1912)

Ernst Mach (1838 – 1916)

SIGNIFICANT DIMENSIONLESS GROUPS SIGNIFICANT DIMENSIONLESS GROUPS IN FLUID MECHANICSIN FLUID MECHANICS

FORCES:FORCES:

(in Fluid Mechanics)

Inertia

Viscous VLAdy

dUA

Pressure 2LpAp

Gravity gLmg 3

Surface Tension L

Compressibility 2LEAE vv

DIMENSIONLESSGRUOPS

PHYSICAL MEANINGS OFPHYSICAL MEANINGS OF DIMENSIONLESS GROUPS DIMENSIONLESS GROUPS

Reynolds Number (Re)

Euler Number(Eu = Cp)

Force_Inertia

Force_essurePr

Froude Number(Fr) Force_Gravity

Force_Inertia

Weber Number(We)

Force_Tension_Surface

Force_Inertia

Mach Number(M)

ilityCompressib_to_due_Force

Force_Inertia

Inertia

Viscous VLAdy

dUA

Pressure 2LpAp

Gravity gLmg 3

Surface Tension L

Compressibility 2LEAE vv

FLOW SIMILARITY AND MODEL STUDYFLOW SIMILARITY AND MODEL STUDY

EXPERIMENTAL DATA

To calculate secondary data:

Must be ScaledScaled

- Forces- Moments, etc.

There must be similarity between MODEL and PROTOTYPE

- GEOMETRIC SIMILARITY:

- KINEMATIC SIMILARITY:

- DYNAMIC SIMILARITY:

- Similar in shape - Constant (linear) scales

- Similar flow kinematics - Constant scales (in magnitudes)

- All forces scaled constantly - Needs geometric & kinematic sim’s.

FLOW SIMILARITY AND MODEL STUDY (FLOW SIMILARITY AND MODEL STUDY (Cont’dCont’d))

For complete analysis All contributing forces must be presented: - Viscous force

- Pressure force

- Surface tension force

- etc.

Buckingham – theorem can be used

EXAMPLEEXAMPLE

Drag force analysis on a sphere: F = f (D, V, , )

VD

fDV

F122

From – theorem, we have:

F

, , V

D

Similarity in the ratio of drag to inertia forces between model & prototype

The flow will dynamically similar if:

prototypeelmodDV

F

DV

F

2222

prototypeelmod

VDVD

Also:

Similarity in the ratio of inertia to viscous forces between model & prototype

EXAMPLE PROBLEM 7.4EXAMPLE PROBLEM 7.4

Given: Sonar transducer model tested in a wind tunnel

a) VmFind: b) Fp

Solution:

VD

fDV

F22

prototypeelmod

VDVD

The test should be run at:

EXAMPLE PROBLEM 7.4EXAMPLE PROBLEM 7.4

m

mmm

DVRe

5.02 x 105

p

pp

prototype

p

DVVDRe

5.02 x 105

Therefore:

And:

Vm = 156 ft/sec 47.55 m/s

Finally:

Fp = 54.6 lbf