PS7_solution

As far as the pressure ratio is concerned an angle of

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Unformatted text preview: ncerned, an angle of attack greater than 5 degrees yields unacceptable values for the pressure ratio as given by the linear theory, with a percentage deviation of more than 5 percent. Table-2b (deg) (rad) .( .( )/ ( )/ ) 5 15 30 Beyond an angle of attack of five degrees, the linear theory gives unphysical values for the pressure coefficient over the upper surface of the airfoil. Up to an angle of attack of about five degrees, the percentage deviation is reasonable. PROBLEM 2a The drag coefficient is given by ( ) √ Where k2 and k3 are contributions from the camber and thickness of an airfoil. In this problem, k2=0, since both the airfoils have no camber. Symmetric diamond airfoil The term k3 is evaluated using the formula, ∫( ̂ ) ̂ ̂ ∫( ̂ ) ̂ ̂ Where ̂ and ̂ are the y and x co-ordinates of the airfoil normalized by the chord length c. ̂ ( ̂ ) is the thickness distribution which for a symmetric airfoil is the same as the upper surface co-ordinate ̂ ( ̂ ). For the symmetric diamond airfoil () ̂ ̂ ̂ ̂ ̂ ̂ Since the rear surface of the airfoil has the same magnitude of slope, but an opposite sense of orientation with respect to the free stream, ̂ ̂ ∫( ̂ ) ̂ ̂ ̂ ∫ () ̂ ∫( () The drag coefficient is therefore ( ) √ ( ) √ Symmetric airfoil The equation for the top surface of the airfoil is ( ) ( ̂) In terms of non-dimensional co-ordinates ̂ ) ̂ ̂ ̂ ∫( ̂ ̂ ) ̂ ̂ ∫( ̂ ) ̂ ∫( ̂) ̂ ∫( ̂) ̂ ̂ () ( ) √ ( ) √ PROBLEM 2b The drag coefficient for the symmetric diamond airfoil ( ) is smaller than that for the symmetric curved airfoil ( ). The lift coefficient for both the airfoils would be the same since the lift coefficient is determined solely by the angle of attack and is independent of the camber and thickness...
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