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If a cone with half angle 30.2 degrees is kept in a flow at Mach number 3.5, then what is the value of Mach number downstream of the shockwave?(a) 1.110(b) 2.482(c) 1.648(d) 3.45I had been asked this question during an internship interview.Question is taken from Physical Aspects of Conical Flow topic in portion Linearized and Conical Flows of Aerodynamics

Answer»

Correct ANSWER is (c) 1.648

Explanation: Given, M1 = 3.5,θ = 30.2°

From the θ – β – M curve,

For M1 = 3.5 and θ = 30.2°, the value of β is 48°

Normal component of M1 is Mn1 = M1 sinβ = 3.5 × sin48 = 2.60

From the normal SHOCK table (gas table), for Mn1 = 2.6, we get

\(\FRAC {P_{02}}{P_{01}}\) = 0.4601 and Mn2 = 0.5039

Thus the Mach number downstream of the shock wave is

M2 = \(\frac {M_{n2}}{sin⁡(β – θ)} = \frac {0.5039}{sin⁡(48 – 30.2)}\) = 1.648



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