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 Q (gal/min) 1.53.06.09.012.014.0 τ w (Pa)0.050.180.370.640.861.25 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. Π 1 = ρ a V b d c μ -1 = (ML -3 ) a (LT -1 ) b (L) c (ML -1 T -1 ) -1 = M 0 L 0 T 0 This leads to a = 1, b = 1, and c = 1

8. Pi Groups So, the first pi group is: → Π 1 = ρ 1 V 1 d 1 μ -1 = (ρVd) / μ = Re The other pi groups were found to be: → Π 2 = ε / d → Π 3 = τ / (ρV 2 ) = C τ

9. Dimensionless Function The three pi groups were used to find the dimensionless function for wall shear → Π 3 = fcn (Π 1,Π 2 ) → C τ = 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, C τ was calculated for each Reynolds number (ε/d was not used since it was constant for all values of Q) Finally, C τ vs. Re was plotted using Excel.

12. Graph of C τ 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 → R 2 = 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??