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CE 230-Engineering Fluid Mechanics

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

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CE 230-Engineering Fluid Mechanics

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  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 (ΔPl)is expected to depend on • Diameter, D • Velocity, V • Density, ρ • Viscosity, μ ΔPl = 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 ΔPl 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/L3] • Viscosity, μ, [M/LT] • Pressure, P, [M/LT2]

  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 • List O.V.’s involved ΔpL, D, V, ρ, μ • Express each OV in basic units ΔpL=[M/L2T2] D=[L] V=[L/T] ρ= [M/L3] μ= [M/LT] • Determine # of π-terms 5-3 = 2

  13. Exponent method continued • Select repeating variables from OVs = No Basic units used (here 3) excluding primary variable (ΔpL) ) D, V, ρ • Form the π-terms as shown below π1 = ΔpLDa Vbρc π2 = μDd Veρf

  14. Exponent method continued • Determine the unknown exponents using the fact that each π-term is dimensionless For π1 = ΔpLDa Vbρc units ofπ1 =[M/L2T2] [La] [Lb/Tb] [Mc/L3c] Exponent for M 1+c=0 c=-1 Exponent for T b=-2 Exponent for L a=1

  15. Exponent method continued Therefore the firat non-dimensional parameter is: π1 = ΔpLD/ρ V2 Similarly we can get π2 = μ/ρ V D

  16. Exponent method continued The new relation now is : refore the firat non-dimensional parameter is: π1 = f(π2 ) ΔpLD/ρ V2 = f(μ/ρ V D) Or ΔpL =(ρ V2 /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

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