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Center of pressure

Example: Calculate the location of center of pressure of a airfoil section at an angle of attack of \({{\rm{5}}^{\rm{o}}}\), whose coefficient of lift, \({{\rm{c}}_{\rm{l}}}\) is 0.80 and \({{\rm{c}}_{{\rm{m,c/4}}}}\) is -0.08.Consider incompressible flow over the airfoil.
Solution:-
Location of center of pressure is given as \[{x_{cp}} = \left( {\frac{c}{4}} \right) - \frac{{M_{c/4}^\prime }}{{{L^{^\prime }}}}\]
Where\({x_{cp}}\)  is the distance of center of pressure from the leading edge of airfoil, and 'c' is the chord length .\(M_{c/4}^\prime \) is moment per unit span about quarter-chord point and \({L^\prime }\) is the lift per unit span;\[\begin{array}{l}{x_{cp}} = \frac{c}{4} - \frac{{M_{c/4}^\prime }}{{{L^\prime }}}\\\frac{{{x_{cp}}}}{c} = \frac{1}{4} - \frac{{\left( {M_{c/4}^\prime /{q_\infty }{c^2}} \right)}}{{\left( {{L^\prime }/{q_\infty }c} \right)}}\end{array}\]\[\begin{array}{l} = \frac{1}{4} - \left( {\frac{{{c_{m,c/4}}}}{{{c_l}}}} \right)\\ = \frac{1}{4} - \left( {\frac{{ - 0.08}}{{0.80}}} \right)\\ = \frac{1}{4} + 0.1\\ = 0.25 + 0.1\\ = 0.35\end{array}\]Therefore, location of center of pressure is at 0.35 with respect to chord length.

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