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Nondimensionalization of the Wall Shear Formula

Nondimensionalization of the Wall Shear Formula. John Grady BIEN 301 2/15/07. C5.1 Problem Statement . For long circular rough pipes in turbulent flow, wall shear stress can be written as a function of ρ , μ , ε , d, and V. Required .

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Nondimensionalization of the Wall Shear Formula

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  1. Nondimensionalization of the Wall Shear Formula John Grady BIEN 301 2/15/07

  2. C5.1 Problem Statement • For long circular rough pipes in turbulent flow, wall shear stress can be written as a function of ρ, μ, ε, d, and V

  3. Required • Rewrite the function for wall shear in dimensionless form • Use the formula found to plot the given data of volume flow and shear stress and find a curve fit for the data

  4. Given Information d = 5 cm ε = 0.25 mm

  5. Assumptions • Incompressible • Turbulent • Long circular rough pipe • Viscid liquid • Constant temperature

  6. Pi Equations • The function for wall shear contained 6 variables • The primary dimensions for these variables were found to include M,L,T • Therefore, we should use 3 scaling parameters • Scaling parameters were ρ, V, and d

  7. Pi Equations • Now use these parameters plus one other variable to find pi group by comparing exponents. • Π1 = ρa Vb dc μ-1 = (ML-3)a (LT-1)b (L)c (ML-1T-1) -1 = M0L0T0 • This leads to a = 1, b = 1, and c = 1

  8. Pi Groups • So, the first pi group is: • Π1 = ρ1 V1 d1μ-1 = (ρVd) / μ = Re • The other pi groups were found to be: • Π2 = ε / d • Π3 = τ / (ρV2) = Cτ

  9. Dimensionless Function • The three pi groups were used to find the dimensionless function for wall shear • Π3 = fcn (Π1,Π2) • Cτ = fcn (Re,ε/d)

  10. Data Conversion • The volume flow rates given were converted to SI units • The values for ρ and μ for water @ 20°C were looked up • The value for the average velocity, V, was found by using the following formula • V = Q / A

  11. Data Manipulation • Next, the values for V, d, ρ, and μ were used to calculate the Reynolds number for each Q given • Then, Cτ was calculated for each Reynolds number (ε/d was not used since it was constant for all values of Q) • Finally, Cτ vs. Re was plotted using Excel.

  12. Graph of Cτ vs. Re

  13. Analysis of Graph • A power curve fit was used to find a formula for shear stress as a function of the Reynolds number • τ = 3.6213(Re)-0.6417 • R2 = 0.9532

  14. Conclusion • It was determined that the formula for wall shear can be reduced to an equation using a single variable, Re • This can save a lot of time and money testing different flows

  15. Biomedical Application • An application of this problem could involve the flow of various fluids into a subject using an IV or catheter • Wall shear would need to be factored in to determine the correct flow rate of the fluid

  16. Questions??

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