Respuesta :
Solution :
Given :
Rectangular wingspan
Length,L = 17.5 m
Chord, c = 3 m
Free stream velocity of flow, [tex]$V_{\infty}$[/tex] = 200 m/s
Given that the flow is laminar.
[tex]$Re_L=\frac{\rho V L}{\mu _{\infty}}$[/tex]
[tex]$=\frac{1.225 \times 200 \times 3}{1.789 \times 10^{-5}}$[/tex]
[tex]$= 4.10 \times 10^7$[/tex]
So boundary layer thickness,
[tex]$\delta_{L} = \frac{5.2 L}{\sqrt{Re_L}}$[/tex]
[tex]$\delta_{L} = \frac{5.2 \times 3}{\sqrt{4.1 \times 10^7}}$[/tex]
= 0.0024 m
The dynamic pressure, [tex]$q_{\infty} =\frac{1}{2} \rho V^2_{\infty}$[/tex]
[tex]$ =\frac{1}{2} \times 1.225 \times 200^2$[/tex]
[tex]$=2.45 \times 10^4 \ N/m^2$[/tex]
The skin friction drag co-efficient is given by
[tex]$C_f = \frac{1.328}{\sqrt{Re_L}}$[/tex]
[tex]$=\frac{1.328}{\sqrt{4.1 \times 10^7}}$[/tex]
= 0.00021
[tex]$D_{skinfriction} = \frac{1}{2} \rho V^2_{\infty}S C_f$[/tex]
[tex]$=\frac{1}{2} \times 1.225 \times 200^2 \times 17.5 \times 3 \times 0.00021$[/tex]
= 270 N
Therefore the net drag = 270 x 2
= 540 N
