Answer:
1. Boundary layer thickness = 0.039 m, 2. Flow is turbulent, 3. Total Drag Force = 1.5 x 10⁶ kg.m/s²
Step-by-step 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