# Aerodynamics Questions and Answers – Critical Mach Number

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

1. What is a critical Mach number?
a) Freestream Mach number for which stagnation pressure is obtained
b) Freestream Mach number for which sonic flow is obtained
c) Free stream Mach number for which supersonic flow is obtained
d) Freestream Mach number for which hypersonic flow is obtained

Explanation: When the freestream Mach number is increased for a flow over an airfoil, the local Mach number on the top surface of the airfoil increases too. There is a point in when the freestream Mach number is increased to a point where sonic speed is achieved on the top surface. This is known as the critical Mach number.

2. The local Mach number at a point on the airfoil reaches 1 for critical Mach number.
a) True
b) False

Explanation: For a freestream Mach number, the local Mach number at the airfoil varies based on the pressure distribution. At the upper surface, there is a point with minimum pressure where the local Mach number is maximum. Thus for critical Mach number, this local Mach number at the upper surface of the airfoil is unity.

3. What is the coefficient of pressure at the minimum pressure point for an airfoil with critical Mach number as 0.6?
a) 1.53
b) 1.66
c) 1.42
d) 1.15

Explanation: Given, Mcrit = 0.6, γ = 1.4 (air)
The critical Mach number is calculated using the relation
(Cp)crit = $$\frac {2}{γM_{crit}^{2}} \bigg [ \frac {1 + \frac {1}{2}(γ – 1)M_{crit}^{2}}{1 + \frac {1}{2}(γ – 1)} \bigg ]^{\frac {γ}{γ – 1}}$$ – 1
On substituting the values, we get
(Cp)crit = $$\frac {2}{1.4×0.6^{2}} \bigg [ \frac {1 + \frac {(1.4 – 1)}{2} 0.6^{2}}{1 + \frac {(1.4 – 1)}{2}} \bigg ]^{\frac {1.4}{1.4 – 1}}$$ – 1
(Cp)crit = 3.97$$\big [ \frac {1.07}{1.20} \big ]$$3.5 – 1 = 1.66

4. What is the coefficient of pressure at minimum pressure point a function of?
a) Critical Mach number
b) Freestream Mach number
c) Chord/thickness ratio of airfoil
d) Length of the airfoil

Explanation: Coefficient of pressure at minimum pressure point which is present at the upper surface of the airfoil is given by the relation:
(Cp)crit = $$\frac {2}{γM_{crit}^{2}} \bigg [ \frac {1 + \frac {1}{2}(γ – 1)M_{crit}^{2}}{1 + \frac {1}{2}(γ – 1)} \bigg ]^{\frac {γ}{γ – 1}}$$ – 1
In this the value of gamma is constant depending on the medium. The only varying quantity is critical Mach number. Thus, the coefficient of pressure at minimum pressure point is a function of only critical Mach number.

5. Thicker the airfoil, higher is the critical Mach number.
a) True
b) False

Explanation: According to the Prandtl – Glauert rule, the incompressible pressure coefficient is given by:
Cp = $$\frac {C_{p_0}}{\sqrt {1 – M_∞^{2}}}$$
For airfoils which are thin, the flow has less expansion resulting in less magnitude of Cp0. This results in a higher critical Mach number value. On the other hand thick airfoils have higher magnitude of Cp0 because of strong flow expansion. This results in lower critical Mach number.

6. What is the relation between the drag divergence Mach number and critical Mach number?
a) Mdrag – divergence = Mcrit
b) Mdrag – divergence > Mcrit
c) Mdrag – divergence < Mcrit
d) Mdrag – divergence × Mcrit = 0

Explanation: Drag – divergence Mach number is greater than critical Mach number. At this Mach speed, there is a region of local supersonic flow followed by a shock wave. This results in pressure drag eventually causing boundary layer separation.

7. Which of these techniques is employed to delay the critical Mach number?
a) Increasing the thickness of airfoil
b) Swept wing
c) Increase camber
d) Decrease drag – divergence Mach number

Explanation: There are two ways employed to increase the critical Mach number thereby increasing drag – divergence Mach number. First is to reduce the thickness of the airfoil, and second is to add sweep to the wing.

8. Which equation is used to compute the critical Mach number of the airfoil?
a) (Cp)crit = $$\frac {2}{γM_{crit}^{2}} \bigg [ \frac {1 + \frac {1}{2}(γ – 1)M_{crit}^{2}}{1 + \frac {1}{2}(γ – 1)} \bigg ]^{\frac {γ}{γ – 1}}$$ – 1
b) (Cp)crit = $$\frac {2}{γM_{crit}^{2}} \bigg [ \frac {1 + \frac {1}{2}(γ – 1)M_{crit}^{2}}{1 + \frac {1}{2}(γ – 1)} \bigg ]^{\frac {γ}{γ + 1}}$$ + 1
c) (Cp)crit = γM$$_{crit}^{2} \bigg [ \frac {1 + \frac {1}{2}(γ – 1)}{1 + \frac {1}{2}(γ – 1)} \bigg ]^{\frac {γ}{γ – 1}}$$ – 1
d) (Cp)crit = γM$$_{crit}^{2} \bigg [ \frac {1 + \frac {1}{2}(γ – 1)}{1 + \frac {1}{2}(γ – 1)M_{crit}^{2}} \bigg ]^{\frac {γ}{γ – 1}}$$ – 1

Explanation: The formula for the coefficient of pressure for an isentropic flow is given by:
Cp = $$\frac {2}{γM_∞^{2}} \bigg ( \frac {p}{p_∞} – 1 \bigg )$$
For an isentropic flow, the ratio of pressure at a point to the freestream pressure is given by:
$$\frac {p}{p_∞} = \bigg [ \frac {1 + \frac {(γ – 1)}{2} M_∞^{2}}{1 + \frac {(γ – 1)}{2} M^{2}} \bigg ]^{\frac {γ}{γ – 1}}$$
Substituting this in the above equation we get
Cp = $$\frac {2}{γM_∞^{2}} \bigg [ \bigg ( \frac {1 + \frac {(γ – 1)}{2} M_∞^{2}}{1 + \frac {(γ – 1)}{2} M^{2}}\bigg ) ^{\frac {γ}{γ – 1}} – 1 \bigg ]$$
At critical Mach number, local Mach number M = 1 and freestream Mach number is equal to the critical Mach number. Substituting these we finally arrive at the relation:
(Cp)crit = $$\frac {2}{γM_{crit}^{2}} \bigg [ \frac {1 + \frac {1}{2}(γ – 1)M_{crit}^{2}}{1 + \frac {1}{2}(γ – 1)} \bigg ]^{\frac {γ}{γ – 1}}$$ – 1

9. What happens to the flow around the airfoil at upper critical Mach number?
a) The flow around airfoil becomes subsonic
b) The flow around airfoil becomes supersonic
c) The flow around airfoil becomes sonic
d) The flow around airfoil becomes hypersonic

Explanation: Critical Mach number is of two kinds – lower and upper critical Mach number. When the flow around the airfoil is at the upper critical Mach number, then the flow around the entire airfoil reaches supersonic speed.

10. Which drag is prominent after exceeding the critical Mach number?
a) Form drag
b) Wave drag
c) Pressure drag
d) Skin – friction drag

Explanation: When the flow exceeds critical Mach number, there is a formation of supersonic region which is followed by a shock wave. There is a pressure loss across the shock wave which results in large pressure drag.

Sanfoundry Global Education & Learning Series – Aerodynamics.

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