Question

Which of these represent compatibility equation along C+ characteristic line?

(a) dθ = tan⁡(θ – μ)

(b) dθ = \(\sqrt {M^2 + 1} \frac {dV}{V}\)

(c) dθ = \(\sqrt {M^2 – 1} \frac {dV}{V}\)

(d) dθ = tan⁡(θ – μ)\(\frac {dV}{V}\)

The question was asked in examination.

I need to ask this question from Determination of Compatibility Equations topic in division Numerical Techniques for Steady Supersonic Flow of Aerodynamics

Answer 1

Right choice is (c) dθ = \(\sqrt {M^2 – 1} \frac {dV}{V}\)

Explanation: If we set the numerator of the combination of momentum, continuity and energy equation represented by Cramer’s rule as zero, we get

(1 – \(\frac {u^{2}}{a^{2}}\))dudy + (1 – \(\frac {v^{2}}{a^{2}}\))dxdv = 0

On rearranging the terms we get

(1 – \(\frac {v^{2}}{a^{2}}\))dxdv = – (1 – \(\frac {u^{2}}{a^{2}}\))dudy

\(\frac {dv}{du} = \frac { – (1 – \frac {u^{2}}{a^{2}})dy}{(1 – \frac {v^{2}}{a^{2}})dx}\)

Substituting the value of characteristic curve in the above equation –

\(\frac {dy}{dx_{char}} =  \frac {- \frac {uv}{a^{2}} ± \sqrt {(\frac {u^2 + v^{2}}{a^{2}} ) – 1}}{1 – \frac {u^{2}}{a^{2}}}\)

We get, \(\frac {dv}{du} = \frac  {- (1 – \frac {u^{2}}{a^{2}})}{(1 – \frac {v^{2}}{a^{2} })} \big [ \frac { – \frac {uv}{a^{2}} ± \sqrt {(\frac {u^{2} + v^{2}}{a^{2}} ) – 1}}{1 – \frac {u^{2}}{a^{2}}} \big ] = \frac {\frac {uv}{a^{2}} ∓ \sqrt {(\frac {u^{2} + v^{2}}{a^{2}} ) – 1}}{1 – \frac {v^{2}}{a^{2}}}\)

u and v are the x and y – component of velocity V. u = Vcosθ and v = Vsinθ. On substituting these values we get

\(\frac {d(Vsinθ)}{d(Vcosθ)} = \frac {M^2 cosθsinθ ∓ \sqrt {M^2 – 1} }{1 – M^2 sin^2 θ}\)

dθ = ∓ \(\sqrt {M^2 – 1}\frac {dV}{V}\)

This the characteristic line along the C+ characteristic line is .dθ = + \(\sqrt {M^2 – 1}\frac {dV}{V}\).