Find the lift per unit span and the circulation at an angle of attack of \(4^{\circ}\), for this airfoil.

A NACA 4421 airfoil having a chord length of \(2\,m\) is flying at an velocity of \(70\,m/s\) at an altitude of \(2\,km\). Find the lift per unit span and the circulation at an angle of attack of \(4^{\circ}\), for this airfoil.

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Lift is a force which is generated mostly from wings of an aircraft, moving through air. Lift per unit span is the lift which is produced per unit span of the wing. Span is the distance from one wing tip to the other. Lift per unit span is \[{L}’=\frac{1}{2}\rho _{\infty}V_{\infty}^{2}c\left ( 1 \right )c_{l}\]At, an altitude of \(2\,km\) density is \(1.0066\,kg/m^3\). For an angle of attack of \(4^{\circ}\), the coefficient of lift for this airfoil is approximately \(0.8\). Therefore, lift per unit span will be \[{L}’=\frac{1}{2}\rho _{\infty}V_{\infty}^{2}c\left ( 1 \right )c_{l}\]\[\Rightarrow {L}’=\frac{1}{2}\times 1.0066\times \left ( 70 \right )^{2}\times 2\left ( 1 \right )\times 0.8\]\[=3945.872\,N\]Circulation is the line integral of velocity field around a closed curve in a fluid flow. Lift per unit span in a two dimensional flow field is directly proportional to circulation. It is also called Kutta-Joukowski theorem. \[{L}’=\rho _{\infty}V_{\infty}\tau\] Therefore, circulation around the airfoil will be \[{L}’ = \rho _{\infty}V_{\infty}\tau\]\[\Rightarrow  3945.872 = 1.0066\times \left ( 70 \right )\times\tau\]\[\Rightarrow \tau = 56\,m^{2}/s\]

Answered on 2nd August 2021.
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