# Aerodynamics Questions and Answers – Mach Number Independence

This set of Aerodynamics Multiple Choice Questions & Answers (MCQs) focuses on “Mach Number Independence”.

1. At which flow regime does aerodynamic quantities such as coefficient of pressure, lift become independent of Mach number?
a) Subsonic
b) Supersonic
c) Transonic
d) Hypersonic

Explanation: For flows above Mach number 5 i.e. hypersonic flow, it is seen that the aerodynamic quantities such as coefficient of pressure, coefficient of life and wave drag become independent of the Mach number. This aspect was even described by Newton while formulating his Newtonian theory which states the coefficient of pressure is independent of Mach number at hypersonic speed.

2. Which of these equations represents non – dimensionalize continuity equation?
a) ∂$$\frac {(\overline {ρ} \, \overline {u})}{∂\overline {x}}$$ + ∂$$\frac {(\overline {ρ} \, \overline {v})}{∂\overline {y}}$$ + ∂$$\frac {(\overline {ρ} \, \overline {w})}{∂\overline {z}}$$ = 0
b) ρ u$$\frac {∂ \overline {u}}{∂\overline {x}}$$ + ρ u $$\frac {∂ \overline {v}}{∂\overline {y}}$$ + ρ u $$\frac {∂ \overline {w}}{∂\overline {z}}$$ = 0
c) ρ u$$\frac {∂ \overline {u}}{∂\overline {x}}$$ + ρ u $$\frac {∂ \overline {v}}{∂\overline {y}}$$ + ρ u $$\frac {∂ \overline {w}}{∂\overline {z}}$$ = –$$\frac {∂ \overline {p}}{∂\overline {x}}$$
d) ρ u$$\frac {∂ \overline {u}}{∂\overline {x}}$$ + ρ u $$\frac {∂ \overline {v}}{∂\overline {y}}$$ + ρ u $$\frac {∂ \overline {w}}{∂\overline {z}}$$ = –$$\frac {∂ \overline {p}}{∂y}$$

Explanation: In order to non – dimensionlize the continuity equation, we have to define the non – dimensional variables as follows:
x = $$\frac {x}{l}$$, y = $$\frac {y}{l}$$, z = $$\frac {z}{l}$$, p = $$\frac {p}{ρ_∞ V_∞^{2}}$$, ρ = $$\frac {ρ}{ρ_∞}$$
Where, x, y, z are velocity components
l is characteristic length
ρ is free stream density
V is free stream velocity
The continuity equation is given by
$$\frac {∂ρ}{∂t}$$ + ∂$$\frac {ρu}{∂x}$$ + ∂$$\frac {ρv}{∂y}$$ + ∂$$\frac {ρw}{∂z}$$ = 0
Since the flow is assumed to be steady, $$\frac {∂ρ}{∂t}$$ = 0. Substituting non – dimensional variables in the above equation we get
∂$$\frac {(\overline {ρ} \, \overline {u})}{∂\overline {x}}$$ + ∂$$\frac {(\overline {ρ} \, \overline {v})}{∂\overline {y}}$$ + ∂$$\frac {(\overline {ρ} \, \overline {w})}{∂\overline {z}}$$ = 0

3. Which boundary condition applied at the surface to non dimensionlize the governing equations?
a) V.n = 0
b) V × n = 0
c) V.(V × n) = 0
d) V × (V × n) = 0

Explanation: While non dimensionalising the governing equations for steady inviscid flow, the boundary condition applied is that the flow of tangent to the surface. This means that if V is the velocity vector and n is the unit normal vector at the surface, then for the flow to be tangent, V.n = 0.

4. Boundary condition V.n = 0 is applied at the surface to non dimensionlize the governing equations when fluid is being transferred.
a) True
b) False

Explanation: When there is no fluid transfer taking place between the surface, then in order to non dimensionalising the governing equations, we use the boundary condition V.n = 0. But, when there is transfer of fluid in or out of the surface then normal velocity vT has to be incorporated thus altering the boundary condition as V.n = vT.

5. For a flow with Mach number 4, P1 = 2.65 × 104 Pa what is the pressure behind the oblique shock wave having shock angle as 30 degrees?
a) 8.525 × 104 Pa
b) 11.925 × 104 Pa
c) 4.502 × 104 Pa
d) 15.278 × 104 Pa

Explanation: Given, P1 = 2.65 × 104 Pa, M1 = 4, β = 30°
The normal component of the Mach number upstream of the shockwave is given by:
Mn1 = M1sinβ = 4sin⁡(30) = 2
Using the normal shock table, for Mn1 = 2 we get the relation between the pressure upstream and downstream
$$\frac {P_2}{P_1}$$ = 4.5
Since P1 = 2.65 × 104, we get P2 = 4.5 × 2.65 × 104 = 11.925 × 104 Pa.

6. What happens to the pressure behind an oblique shock wave at very high Mach numbers?
a) p2 = $$\frac {2sin^2 β}{γ + 1}$$
b) p2 = $$\frac {γ – 1}{γ + 1}$$
c) p2 = 1 – $$\frac {2sin^2 β}{γ + 1}$$
d) p2 = $$\frac {2sin 2 β}{γ + 1}$$

Explanation: For the flow past an oblique shock, the pressure downstream of the shock wave is given by:
$$\frac {p_2}{p_∞}$$ = 1 + $$\frac {2γ}{γ + 1}$$(M$$_∞^{2}$$sin2β – 1)
We non dimensionalize this equation as
$$\frac {p_2}{p_∞} = \frac {\overline {p}_2 (ρ_∞ V_∞^{2})}{p_∞} = \frac {\overline {p}_2 V_∞^{2}}{RT_∞} =\frac {\overline {p}_2 γV_∞^{2}}{a_∞^{2}}$$ = p2γM$$_∞^2$$
Thus the non dimensionlised pressure behind the oblique shock is
p2 = $$\frac {1}{γM_∞^{2}} + \frac {2}{γ + 1}$$(sin2β –$$\frac {1}{M_∞^{2}}$$)
For a very high Mach number at hypersonic flow, we apply the limit M → ∞, resulting in
p2 = $$\frac {2sin^2 β}{γ + 1}$$

7. Mach number independence for conical cylinder is achieved at a lower Mach number compared to the sphere.
a) True
b) False

Explanation: The coefficient of drag over both conical cylinder and sphere becomes independent after a certain higher Mach number. This is known as Mach number independence. Although, this Mach number independence for sphere is achieve at a lower Mach number compared to the conical slender because for slender bodies, the Mach number independence occurs at a lower Mach number.

Sanfoundry Global Education & Learning Series – Aerodynamics.

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