Nondimensionalization of the Wall Shear Formula John Grady

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

C 5. 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 • 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

Given Information d = 5 cm ε = 0. 25 mm Q (gal/min) 1. 5 3. 0 6. 0 9. 0 12. 0 14. 0 τw (Pa) 0. 05 0. 18 0. 37 0. 64 0. 86 1. 25

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

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

Pi Equations • Now use these parameters plus one other variable to find pi group by comparing exponents. • Π 1 = ρa Vb dc 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

Pi Groups • So, the first pi group is: → Π 1 = ρ1 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) = Cτ

Dimensionless Function • The three pi groups were used to find the dimensionless function for wall shear → Π 3 = fcn (Π 1, Π 2) → Cτ = fcn (Re, ε/d)

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

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.

Graph of Cτ vs. Re

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

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

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

• Questions? ?