1. CE 230-Engineering Fluid Mechanics Lecture # 40-43 Dimensional analysis and similitude

2. 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

3. 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

4. Pressure drop (ΔP l )is expected to depend on Diameter, D Velocity, V Density, ρ Viscosity, μ ΔP l = f(D,V, ρ, μ)

5. 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 ΔP l Say vary D

6. Repeat for other variables Vary V fixing others

7. ETC How many experiments we have to do? Do you forecast any problem?

8. 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

9. Basic units All variables can be expressed in terms of three basic units Length (L) Time (T) Mass (M)

10. Examples Velocity, V, [L/T] Density, ρ, [M/L 3 ] Viscosity, μ, [M/LT] Pressure, P, [M/LT 2 ]

11. 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

12. Exponent method 1. List O.V.’s involved Δp L, D, V, ρ, μ 2. Express each OV in basic units Δp L =[M/L 2 T 2 ] D=[L]V=[L/T] ρ= [M/L 3 ] μ= [M/LT] 3. Determine # of π-terms 5-3 = 2

13. Exponent method continued 4. Select repeating variables from OVs = No Basic units used (here 3) excluding primary variable (Δp L) ) D, V, ρ 5. Form the π-terms as shown below π 1 = Δp L D a V b ρ c π 2 = μ D d V e ρ f

14. Exponent method continued 6. Determine the unknown exponents using the fact that each π-term is dimensionless For π 1 = Δp L D a V b ρ c units of π 1 =[M/L 2 T 2 ] [L a ] [L b /T b ] [M c /L 3c ] Exponent for M 1+c=0 c=-1 Exponent for Tb=-2 Exponent for L a=1

15. Exponent method continued Therefore the firat non-dimensional parameter is: π 1 = Δp L D/ρ V 2 Similarly we can get π 2 = μ/ ρ V D

16. Exponent method continued The new relation now is : refore the firat non-dimensional parameter is: π 1 = f(π 2 ) Δp L D/ρ V 2 = f( μ/ ρ V D) Or Δp L =( ρ V 2 /D) g( Re )

17. How do we know nature of g? we need to vary the two variables only which is much easier than the original problem ΔpL D/ρ V2 Re