Question

Which of these is the relation for linearized pressure coefficient for two – dimensional bodies?

(a) Cp = \(\frac {-2u^{‘}}{V_∞} + \frac {v^{‘{^2}}+w^{‘^2}}{V_∞^{2}}\)

(b) Cp = \(\frac {-2v^{‘}}{V{_∞^2}} \frac {v^{‘{^2}}+w^{‘^2}}{V_∞^{2}}\)

(c) Cp = \(\frac {-2w}{V{_∞^2}} \frac {v^{‘{^2}}+w^{‘^2}}{V_∞^{2}}\)

(d) Cp =  –\(\frac {2u^{‘}}{V_∞}\) + (1 – M\(_∞^2\))\(\frac {u^{‘^{2}}}{V{_∞^{2}}} + \frac {v^{‘^{2}}+w^{‘^{2}}}{V_∞^{2}}\)

I got this question in an interview for job.

My question is based upon Linearized Pressure Coefficient topic in portion Linearized and Conical Flows of Aerodynamics

Answer 1

Right choice is (a) Cp = \(\frac {-2u^{‘}}{V_∞} + \frac {v^{‘{^2}}+w^{‘^2}}{V_∞^{2}}\)

Easy explanation: The non – linear coefficient of pressure is in the form of the relation below as obtained after binomial expansion.

Cp =  –\(\frac {2u^{‘}}{V_∞}\) + (1 – M\(_∞^2\))\(\frac {u^{‘^{2}}}{V{_∞^{2}}} + \frac {v^{‘^{2}}+w^{‘^{2}}}{V_∞^{2}}\)

For two dimensional terms, the second order terms can be neglected. But for three – dimensional the term \(\frac {v^{‘{^2}}+w^{‘^{2}}}{V_∞^{2}}\)  has to be retained since it is not negligible. Apart from that \(\frac {u^{‘^2}}{V_∞^{2}}\)  is neglected because the perturbed velocities have magnitudes that are very less than unity. This results in equation for linearized coefficient of pressure for three – dimensional bodies as Cp = \(\frac {-2u^{‘}}{V_∞} + \frac {v^{‘{^2}}+w^{‘^2}}{V_∞^{2}}\).