Question

What is the relation between coefficient of pressure in terms of gamma and Mach number?

(a) Cp = \(\frac {1}{γM_∞^2}\)) (1 – \(\frac {p}{p_∞}\))

(b) Cp = \(\frac {2}{γM_∞^2}\)(\(\frac {p}{p_∞}\)  – 1)

(c) Cp = γM\(_∞^2\)(\(\frac {p}{p_∞}\))

(d) Cp = \(\frac {γM_∞^2}{2} (\frac {p}{p_∞ – 1})\)

The question was posed to me during a job interview.

My query is from Linearized Pressure Coefficient in division Linearized and Conical Flows of Aerodynamics

Answer 1

The correct answer is (b) Cp = \(\frac {2}{γM_∞^2}\)(\(\frac {p}{p_∞}\)  – 1)

Easiest explanation: The relation between coefficient of pressure with gamma and Mach number are derived as follows:

Coefficient of pressure is given by Cp = \(\frac {p – p_∞}{\frac {1}{2} ρ_∞ V_∞^{2}}\). Where, p∞, ρ∞, V∞ are freestream pressure, density and velocity. We can manipulate the denominator by multiplying and diving it by γp∞. We get,

\(\frac {1}{2}\)ρ∞\(V_∞^2\) = \(\frac {1}{2}\frac {γp_∞}{γp_∞}\) ρ∞V∞^2 = \(\frac {γ}{2}\) p∞  \(\frac {ρ_∞ V_∞^2}{γp_∞}\)

Since \(\frac {γp_∞}{ρ_∞}\) = a^2 the denominator becomes \(\frac {γ}{2}\)p∞\(\frac {V_∞^2}{a_∞^2} = \frac {γ}{2}\) p∞ M\(_∞^2\) Since M = V/a.

Therefore the coefficient of pressure in terms of gamma and Mach number is:

Cp = \(\frac {2}{γM_∞^2}\)(\(\frac {p}{p_∞}\)  – 1)