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Sagot :
An airplane flying across the sky experience drag force determined by the factors including the speed of flight, coefficient of skin friction and the reference surface area
The boundary layer thickness is approximately 0.233 cm
The total skin friction drag, is approximately 265 N
Reason:
First part:
Given parameters are;
Chord length, L = 3 m
Velocity of the plane, V = 200 m/s
Density of the air, ρ = 1.225 kg/m³
Viscosity of the air, μ = 1.81 × 10⁻⁵ kg/(m·s)
The Reynolds number is given as follows;
[tex]R_{eL} = \dfrac{\rho \times V \times L}{\mu}[/tex]
Therefore;
[tex]R_{eL} = \dfrac{1.255 \times 200 \times 3}{1.81 \times 10^{-5}} = 4.16022099448 \times 10^7 \approx 4.16 \times 10^7[/tex]
Boundary layer thickness, [tex]\delta_L[/tex], for laminar flow, is given as follows;
[tex]\dfrac{ \delta_L }{L}=\dfrac{5.0}{\sqrt{R_{eL} } }[/tex]
[tex]{ \delta_L }=\dfrac{5.0 \times L}{\sqrt{R_{eL} } }[/tex]
Which gives;
[tex]{ \delta_L }=\dfrac{5.0 \times 3}{\sqrt{4.16 \times 10^{7}} } \approx 2.33 \times 10^{-3 }[/tex]
The boundary layer thickness, [tex]\delta_L[/tex] ≈ 2.33 × 10⁻³ m = 0.233 cm
Second Part
The total skin friction is given as follows;
[tex]Dynamic \ pressure, q = \dfrac{1}{2} \cdot \rho \cdot V^2[/tex]
Therefore;
[tex]q = \dfrac{1}{2} \times 1.225 \times 200^2 = 24,500[/tex]
The dynamic pressure, q = 24,500 N/m²
Skin friction drag coefficient, [tex]C_D[/tex], is given as follows;
[tex]C_D = \dfrac{1.328}{\sqrt{R_{eL} } }[/tex]
Therefore;
[tex]C_D = \dfrac{1.328}{\sqrt{4.16 \times 10^7 } } \approx 2.06 \times 10^{-4}[/tex]
Skin friction drag, [tex]D_f[/tex], is given as follows;
[tex]D_f[/tex] = q × [tex]C_D[/tex] × A
Where;
A = The reference area
∴ [tex]D_f[/tex] = 24,500 N/m² × 2.06 × 10⁻⁴ × 3 m × 17.5 m = 264.9675 N ≈ 265 N
The total skin friction drag, [tex]D_f[/tex] ≈ 265 N
Learn more here:
https://brainly.com/question/12977939
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