Input interpretation
Blasius boundary layer thickness
Equation
δ = 4.9 sqrt((η x)/(U_∞ ρ)) | δ | Blasius boundary layer thickness x | distance along plate U_∞ | free stream velocity η | dynamic viscosity ρ | fluid density (valid for laminar flow over flat plate)
Input values
distance along plate | 5 meters free stream velocity | 10 m/s (meters per second) dynamic viscosity | 0.001003 Pa s (pascal seconds) fluid density | 0.9982 g/cm^3 (grams per cubic centimeter)
Results
Blasius boundary layer thickness | 3.473 mm (millimeters) = 0.1367 inches = 0.3473 cm (centimeters) = 0.003473 meters
Possible intermediate steps
Calculate the Blasius boundary layer thickness using the following information: known variables | | x | distance along plate | 5 m U_∞ | free stream velocity | 10 m/s η | dynamic viscosity | 0.001003 Pa s ρ | fluid density | 0.9982 g/cm^3 Convert known variables into appropriate units using the following: 1 Pa s = 1000 g/(m s): 1 g/cm^3 = 1×10^6 g/m^3: known variables | | x | distance along plate | 5 m U_∞ | free stream velocity | 10 m/s η | dynamic viscosity | 1.003 g/(m s) ρ | fluid density | 998230 g/m^3 The relevant equation that relates Blasius boundary layer thickness (δ), distance along plate (x), free stream velocity (U_∞), dynamic viscosity (η), and fluid density (ρ) is: δ = 4.9 sqrt((η x)/(U_∞ ρ)) Substitute known variables into the equation: known variables | | x | distance along plate | 5 m U_∞ | free stream velocity | 10 m/s η | dynamic viscosity | 1.003 g/(m s) ρ | fluid density | 998230 g/m^3 | : δ = 4.9 sqrt((1.003 g/(m s)×5 m)/(10 m/s×998230 g/m^3)) Separate the numerical part, 4.9 sqrt((1.003×5)/(10×998230)), from the unit part, sqrt((g/(m s) m)/(m/s×g/m^3)) = m: δ = 4.9 sqrt((1.003×5)/(10×998230)) m Evaluate 4.9 sqrt((1.003×5)/(10×998230)): δ = 0.0034731 m Convert 0.0034731 m into mm (millimeters) using the following: 1 m = 1000 mm: Answer: | | δ = 3.473 mm