Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Experience the ease of finding reliable answers to your questions from a vast community of knowledgeable experts. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
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
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
