Water at 20C is flowing along a 5-m-long flat plate with a velocity of 50 m/s. Is the boundary-layer flow at this location laminar or turbulent? What will be the boundary layer thickness at a position 5 m from the leading edge? If each of the plate surfaces measures 500 m2, determine the drag force if both sides are exposed to the flow.

1. Boundary layer thickness = 0.039 m, 2. Flow is turbulent, 3. Total Drag Force = 1.5 x 10⁶ kg. m/s²

Explanation:

From the appendix of physical properties of gases and liquids, select the water property at a temperature of 20° C to obtain the values of density (ρ), and absolute viscosity (μ)

ρ = 998.2 kg/m³, μ = 993 x 10⁻⁶ Pa. s

Calculate the Reyonlds Number

Re = ρvx/μ, where v is the velocity of water and x is the distance at the leading edge

Substitute v = 50 m/s, x = 5m, ρ = 998.2kg/m³ and μ = 993 x 10⁻⁶ Pa. s

Re = 998.2 x 50 x 5/993 x 10⁻⁶ = 2.51 x 10⁸

Hence, the value of Re is greater than 3x 10⁶, thus flow is turbulent

The value of Re is in the turbulent region, then calculate the thickness of boundary layer, δ from the turbulent equation

δ/x = 0.376/Re (x) ^1/5

δ = (0.376). (5) / (2.51 x 10⁸) ^1/5 = 0.039m

Hence, the boundary layer thickness is 0.039 m

Calculate the local skin friction coefficient

C (fx) = 0.0576/Re (x) ^1/5

C (fx) = 0.0576 / (2.51 x 10⁸^1/5) = 0.0012

Calculate the total drag force on both sides of the surface

F (D) = [AC (fx) ρv² (∞) / 2] x 2 sides

F (D) = AC (fx) ρv² (∞), where A is the area of each plate and v (∞) is the velocity

F (D) = 500 x 0.0012 x 998.2 x 50 = 1.5 x 10⁶kg. m/s²

Hence, total drag force on both sides of the surface is 1.5 x 10⁶ kg. m/s²