CE 230-Engineering Fluid Mechanics

Lecture # 40-43

Dimensional analysis and similitude

Why dimensional analysis?

Many problems in F.M. rely on experimental data

It is impractical to perform (cost, time) experimental study for every variation from original problem

Testing large structures is impossible so a model is used but how do we make sure it represents the prototype

Dimensional analysis helps in reducing the size of the problem

Suppose we are studying the pressure drop per unit length of pipe in a smooth long circular horizontal pipe assuming steady and incompressible flow

Pressure drop (ΔPl)is expected to depend on

Diameter, D Velocity, V Density, ρ Viscosity, μ

ΔPl = f(D,V, ρ, μ)

We should design an experimental study to understand and come up with the nature of the above relation

Since we have 5 variables lets keep 3 fixed and see the effect of the fourth on ΔPl

Say vary D

D

Del

ta P

l

Repeat for other variables

Vary V fixing others

V

Del

ta P

l

ETC

How many experiments we have to do?

Do you forecast any problem?

Buckingham PI theorem

The original variables can be replaced by a less number of non-dimensional terms (PI- terms)

no. π terms = no. Orignal v- no Basic units

Basic units

All variables can be expressed in terms of three basic unitsLength (L)Time (T)Mass (M)

Examples

Velocity , V, [L/T]

Density, ρ, [M/L3]

Viscosity, μ, [M/LT]

Pressure, P, [M/LT2]

Back to our problem

No O.V. = 5 No B.U. =3

Therefore no π-terms = 5-2

We will replace 5 variables by to two terms which is much easier to deal with.

Q: How do we develop the π-terms

Exponent method

1. List O.V.’s involvedΔpL, D, V, ρ, μ

2. Express each OV in basic unitsΔpL=[M/L2T2] D=[L] V=[L/T]

ρ= [M/L3] μ= [M/LT]

3. Determine # of π-terms 5-3 = 2

Exponent method continued

4. Select repeating variables from OVs = No Basic units used (here 3)

excluding primary variable (ΔpL) )

D, V, ρ

5. Form the π-terms as shown below

π1 = ΔpL Da Vb ρc

π2 = μ Dd Ve ρf

Exponent method continued

6. Determine the unknown exponents using the fact that each π-term is dimensionless

For π1 = ΔpL Da Vb ρc

units of π1 =[M/L2T2] [La] [Lb/Tb] [Mc/L3c]

Exponent for M 1+c=0 c=-1Exponent for T b=-2Exponent for L a=1

Exponent method continued

Therefore the firat non-dimensional parameter is:

π1 = ΔpL D/ρ V2

Similarly we can get

π2 = μ/ρ V D

Exponent method continued

The new relation now is : refore the firat non-dimensional parameter is:

π1 = f(π2 )

ΔpL D/ρ V2 = f(μ/ρ V D)Or

ΔpL =(ρ V2 /D) g(Re)

How do we know nature of g?

we need to vary the two variables only which is much easier than the original problem

D

Del

ta P

l

ΔpL D/ρ V2

Re