This set of Gas Dynamics Multiple Choice Questions & Answers (MCQs) focuses on “Mach Reflection”.

1. What is the result of the interaction between a normal shock and right running oblique shock?

a) Formation of higher strength normal shock

b) Formation of lower strength normal shock

c) It gives rise to a reflected left running oblique shock

d) It gives rise to a reflected right running oblique shock

View Answer

Explanation: Intersection of a normal shock and right running oblique shock gives rise to a left running reflected shock in order to bring the flow to its original direction. The reflected oblique shock is of lesser strength than the right running oblique shock.

2. Considering Mach reflections, what is the relation between Mach numbers of different regions? Consider upstream of 1^{st} shock as region 1, upstream of reflected shock as region 2 and downstream of the reflected shock as region 3.

a) M_{1} > M_{2} > M_{3}

b) M_{1} < M_{2} < M_{3}

c) M_{1} < M_{2} > M_{3}

d) M_{1} = M_{2} > M_{3}

View Answer

Explanation: When Mach reflections are considered, there is no simple oblique shock wave but the reflection of waves and the strength of the reflected shock reduces due to deflection involved in the process. This leads to the reduction in flow Mach number between different regions.

3. What scenario describes the shock system when M_{2} is less than the detachment Mach number for the wall deflection required?

a) All the shocks become curved

b) Straight shock system with downstream flow parallel to the original flow direction

c) All shocks become curved and the downstream flow is parallel to the original flow direction

d) Straight shock system with flow unparallel to the original flow direction

View Answer

Explanation: The reflected shock wave must have less strength as compared to the incident shock wave because of the deflection involved in the process. When the M

_{2}is less than the detachment Mach number for a given wall deflection, the entire picture of the flow field changes, all the shocks become curved, and the flow behind the shock may not be parallel to the wall.

4. When two oblique shocks of opposite family intersect with a normal shock bridging them, what phenomenon is observed?

a) Strong shock solution is observed

b) Detached curved shock is observed behind and ahead of the normal shock

c) Straight shock system can be seen

d) Detached curved shock is observed ahead of the normal shock and straight shock is observed ahead of the normal shock

View Answer

Explanation: A strong shock solution is physically impossible for the scenario, but detached shock is part of the strong shock solution. The curved shock system is observed when two oblique shocks of the opposite family intersect.

5. If there is a supersonic flow over a blunt edge, then the detached shock is observed. Where is the oblique shock observed in the detached shock?

a) On the blunt edge point

b) Just above the blunt edge point

c) Far away from the edge

d) Near the trailing edge of the section

View Answer

Explanation: In a detached shock, the normal shock is observed on the blunt edge and then slowly loses strength and becomes an oblique shock just above the edge point. Finally the detached shock becomes a Mach line far away from the edge.

6. What is true about the flow properties above and below the discontinuity?

a) The pressure on both sides are equal

b) The flow direction of both sides are different

c) The density and temperature are the same on both the sides

d) The pressure on both sides are different

View Answer

Explanation: On either side of the discontinuity the flow direction needs to be same. The pressure is also same on both sides as the flow is not turning.

7. What is the shape of the pressure curve on the upper surface of a thin diamond wedge airfoil at zero angle of attack?

a) Pressure increases instantly and becomes constant, and then suddenly becomes negative and becomes constant

b) Pressure decreases linearly in a straight line

c) Pressure decrease follows a parabolic path

d) Pressure increases and becomes constant

View Answer

Explanation: According to shock expansion theory, shock wave interactions determine the aerodynamic loads on the surface of the airfoil. Oblique shock and expansion waves decrease and increase the pressure respectively.

8. What is the shape of the pressure curve on the upper surface of a circular arc airfoil at zero angle of attack?

a) Pressure increases and becomes constant, and then becomes negative instantly

b) Pressure first increases instantly and then decreases linearly in a straight line

c) Pressure decrease follows a parabolic path and then becomes atmospheric pressure again

d) Pressure increases and becomes constant till the trailing edge of the airfoil

View Answer

Explanation: According to shock expansion theory, shock wave interactions determine the aerodynamic loads on the surface of the airfoil. Detached shock is formed on the blunt edge which increases the pressure and then it is linearly decreased along the upper length of the airfoil.

9. What is the shape of the pressure curve on the surface of a flat plate at some angle of attack?

a) Pressure increases and becomes constant on the top surface and on the bottom surface

b) Pressure decreases linearly in a straight line thought the length of the edge of the plate

c) The pressure curve is of square shape with unequal pressure value on top and bottom surfaces

d) Pressure increases and becomes constant

View Answer

Explanation: According to shock expansion theory, shock wave interactions determine the aerodynamic loads on the surface of the airfoil. We can see that when a flat plate is at an angle of attack, the pressure on either side of the airfoil is unequal and hence the shape of the pressure curve is square.

10. What is the lift and drag value per unit span when a thin diamond wedge airfoil of thickness t and pressure above the top surface is p_{2} and p_{3} at zero angle of attack?

a) Drag = 0 and Lift = (p_{2} – p_{3})t

b) Drag = 0 and Lift = (p_{2} + p_{3})t

c) Lift = 0 and Drag = (p_{2} – p_{3})t

d) Lift = 0 and Drag = (p_{3} – p_{2})t

View Answer

Explanation: Since the angle of attack is zero, the pressure on the top and bottom surface is equal which leads to zero lift. For drag, it is given by, Drag = 2(p

_{2}t – p

_{3}t) × \(\frac {1}{2}\) = (p

_{2}– p

_{3})t.

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