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# lectures-chapter10 - 650:460 Aerodynamics Chapter 10 Prof...

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Unformatted text preview: 650:460 Aerodynamics Chapter 10 Prof. Doyle Knight Tel: 732 445 4464, Email: [email protected] Office hours: Tues and Thur, 4:30 pm - 6:00 pm and by appointment Fall 2009 1 Supersonic Flows Around Thin Airfoils Linear Theory Assumptions • Inviscid • Steady • Supersonic • Thin airfoil Fig. 10.1 2 Supersonic Flows Around Thin Airfoils Linear Theory • Atributed to Jakob Ackeret. • Diploma in Mechanical Engineering from ETH Zurich (1920) • Worked with Ludwig Prandtl in G¨ ottingen (1921-1927) • PhD from ETH Zurich (1927) • Coined the term “Mach number” after Ernst Mach Jakob Ackeret (1898-1981) 3 Supersonic Flows Around Thin Airfoils Linear Theory Euler’s equation for compressible flow dp =- ρ UdU Assuming the changes dU are small compared to the freestream velocity U ∞ , we can integrate holding ρ U ≈ ρ ∞ U ∞ to obtain p- p ∞ =- ρ ∞ U ∞ ( U- U ∞ ) From Eqn (8.55) dU U = d θ √ M 2- 1 where d θ is the change in the angle of the velocity vector. 4 Supersonic Flows Around Thin Airfoils Linear Theory Defining the flow angle θ ∞ = 0 ( i.e. , measuring flow angles relative to the flow upstream), U ∞- U U ∞ = θ radicalbig M 2 ∞- 1 where we have used the approximation that the changes in velocity are small compared to U ∞ . A positive θ ( i.e. , a counter-clockwise rotation of the flow) results in a compression wave ( i.e. , a reduction in velocity and increase in pressure). Combining with the equation for the change in pressure yields p- p ∞ = ρ ∞ U 2 ∞ θ radicalbig M 2 ∞- 1 5 Supersonic Flows Around Thin Airfoils Linear Theory Thus, defining the pressure coefficient c p = p- p ∞ 1 2 ρ ∞ U 2 ∞ then c p = 2 θ radicalbig M 2 ∞- 1 On the airfoil surface, the tangent (see figure) is θ u = + dz u dx- α θ l =- dz l dx + α where z u ( x ) and z l ( x ) are the upper and lower surface of the airfoil 6 Supersonic Flows Around Thin Airfoils Lift The elemental lift on segment ABCD is dl = p l ds l cos θ l- p u ds u cos θ u Since ds cos θ = dx , dl = ( p l- p u ) dx and thus dc l = ( c p l- c p u ) d parenleftBig x c parenrightBig Fig. 10.4 7 Supersonic Flows Around Thin Airfoils Lift From previous c p l- c p u = 2 radicalbig M 2 ∞- 1 parenleftbigg 2 α- dz l dx- dz u dx parenrightbigg Therefore the lift coefficient is c l = integraldisplay 1 o dc l d parenleftBig x c parenrightBig Now integraldisplay 1 o dz l dx d parenleftBig x c parenrightBig = 0 since the leading and trailing edges are on the z = 0 axis. A similar result holds for dz u / dx ....
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lectures-chapter10 - 650:460 Aerodynamics Chapter 10 Prof...

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