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What is the surface boundary condition for a thin airfoil at a subsonic flow? (Where shape of the airfoil is represented as y = f(x))(a) \(\frac {∂ϕ}{∂x}\) = V∞ \(\frac {df}{dx}\)(b) \(\frac {∂ϕ}{∂y} = \frac {df}{dy}\)(c) \(\frac {∂ϕ}{∂x}\) = – V\(_∞^2 \frac {df}{dx}\)(d) \(\frac {∂ϕ}{∂x} = \frac {dV_∞}{dx}\)I got this question in exam.My question comes from Linearized Subsonic Flow in division Linearized and Conical Flows of Aerodynamics

Answer»

The correct choice is (a) \(\frac {∂ϕ}{∂x}\) = V∞ \(\frac {df}{DX}\)

The explanation: For an airfoil with x – COMPONENT of velocity as V∞ + u^‘ and y – component of the velocity as v^‘, the SURFACE boundary condition is

\(\frac {df}{dx} = \frac {v^{‘}}{V_∞ + u^{‘}}\) = tanθ

Since it is a thin airfoil, the perturbation vector u^‘ is very SMALL in comparison to the freestream velocity V∞, RESULTING in \(\frac {df}{dx} = \frac {v^{‘}}{V_∞}\) = θ (Where tanθ ~ θ for small angles). Expressing the perturbation v^‘ in terms of velocity potential we get

v^‘ = \(\frac {∂ϕ}{∂x}\)

Substituting this in the above equation:

\(\frac {df}{dx} = \frac {\frac {∂ϕ}{∂x}}{V_∞}\) = θ

\(\frac {∂ϕ}{∂x}\) = V∞ \(\frac {df}{dx}\)



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