Question

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) 3.45

I 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 1

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