Answer:
4.192 N
Explanation:
Step 1: Identify the given parameters
Velocity of airflow = 45m/s
air temperature = 20⁰
plate length and width = 1m and 0.5m respectively.
Step 2: calculate drag force due to shear stress, [tex]F_{s}[/tex]
[tex]F_{s} = C_{f} \frac{1}{2} (\rho{U_{o}}WL)[/tex]
Note: The density and kinematic viscosity of air at 20⁰ at 1 atm, is 1.2 kg/m³ and 1.5 X 10⁻⁵ N.s/m²
⇒The Reynolds number ([tex]R_{eL}[/tex]) based on the length of the plate is
[tex]R_{eL} =\frac{VXL}{U}[/tex]
[tex]R_{eL} =\frac{45X1}{1.5 X 10^{-5}}[/tex]
[tex]R_{eL}[/tex] = 3 X10⁶ (flow is turbulent, Re ≥ 500,000)
⇒The average shear stress coefficient ([tex]C_{f}[/tex]) on the "tripped" side of the plate is
[tex]C_{f} = \frac{0.074}{(R_{e})^\frac{1}{5}}[/tex]
[tex]C_{f} = \frac{0.074}{(3 X10^6)^\frac{1}{5}}[/tex]
[tex]C_{f}[/tex] = 0.0038
⇒The average shear stress coefficient ([tex]C_{f}[/tex]) on the "untripped" side of the plate is
[tex]C_{f} = \frac{0.523}{in^2(0.06XR_{e})} -\frac{1520}{R_{e}}[/tex]
[tex]C_{f} = \frac{0.523}{in^2(0.06 X 3X10^6)} -\frac{1520}{3X10^6}[/tex]
[tex]C_{f}[/tex] = 0.0031
The total drag force = [tex]\frac{1}{2}(1.2 X 45^2 X 1 X 0.5 (0.0038 +0.0031)[/tex]
The total drag force is 4.192 N
