This set of Aerodynamics Multiple Choice Questions & Answers (MCQs) focuses on “Linearized Pressure Coefficient”.

1. What is the relation between coefficient of pressure in terms of gamma and Mach number?

a) C_{p} = \(\frac {1}{γM_∞^2}\)) (1 – \(\frac {p}{p_∞}\))

b) C_{p} = \(\frac {2}{γM_∞^2}\)(\(\frac {p}{p_∞}\) – 1)

c) C_{p} = γM\(_∞^2\)(\(\frac {p}{p_∞}\))

d) C_{p} = \(\frac {γM_∞^2}{2} (\frac {p}{p_∞ – 1})\)

View Answer

Explanation: The relation between coefficient of pressure with gamma and Mach number are derived as follows:

Coefficient of pressure is given by C

_{p}= \(\frac {p – p_∞}{\frac {1}{2} ρ_∞ V_∞^{2}}\). Where, p

_{∞}, ρ

_{∞}, V

_{∞}are freestream pressure, density and velocity. We can manipulate the denominator by multiplying and diving it by γp

_{∞}. We get,

\(\frac {1}{2}\)ρ

_{∞}\(V_∞^2\) = \(\frac {1}{2}\frac {γp_∞}{γp_∞}\) ρ

_{∞}V

_{∞}

^{2}= \(\frac {γ}{2}\) p

_{∞}\(\frac {ρ_∞ V_∞^2}{γp_∞}\)

Since \(\frac {γp_∞}{ρ_∞}\) = a

^{2}the denominator becomes \(\frac {γ}{2}\)p

_{∞}\(\frac {V_∞^2}{a_∞^2} = \frac {γ}{2}\) p

_{∞}M\(_∞^2\) Since M = V/a.

Therefore the coefficient of pressure in terms of gamma and Mach number is:

C

_{p}= \(\frac {2}{γM_∞^2}\)(\(\frac {p}{p_∞}\) – 1)

2. Which of these is the relation for linearized pressure coefficient for two – dimensional bodies?

a) C_{p} = \(\frac {- 2u^{‘}}{V_∞}\)

b) C_{p} = \(\frac {- 2v^{‘}}{V_∞}\)

c) C_{p} = \(\frac {- 2w}{V_∞}\)

d) C_{p} = \(\frac {2u^{‘}}{V_∞}\)

View Answer

Explanation: The coefficient of pressure is given by

C

_{p}= \(\frac {2}{γM_∞^2} \big [ \big (1+ \frac {1}{2C_p T_∞}\)(V\(_∞^2\) – (u

^{‘2}+ v

^{‘2}+ w

^{‘2}))\()^{\frac {γ}{γ – 1}} – 1 \big ] \)

The velocity V\(_∞^2\) – (u

^{‘2}+ v

^{‘2}+ w

^{‘2}) is very very small which is equal to – (2u

^{‘}V

_{∞}+ u

^{‘2}+ v

^{‘2}+ w

^{‘2})

Since the perturbed velocity is very small. On binomial expansion, we arrive at

C

_{p}= –\(\frac {2u^{‘}}{V_∞}\) + (1 – M\(_∞^2\))\(\frac {u^{‘^{2}}}{V{_∞^{2}}} + \frac {v^{‘^{2}}+w^{‘^{2}}}{V_∞^{2}}\)

The values of \(\frac {u^{‘^2}}{V{_∞^2}} , \, \frac {v^{‘^{2}}+w^{‘^{2}}}{V_∞^{2}}\) are second order thus can be neglected. Therefore, linearized coefficient of pressure is given by C

_{p}= –\(\frac {- 2u^{‘}}{V_∞}\).

3. Which of these is the relation for linearized pressure coefficient for two – dimensional bodies?

a) C_{p} = \(\frac {-2u^{‘}}{V_∞} + \frac {v^{‘{^2}}+w^{‘^2}}{V_∞^{2}}\)

b) C_{p} = \(\frac {-2v^{‘}}{V{_∞^2}} \frac {v^{‘{^2}}+w^{‘^2}}{V_∞^{2}}\)

c) C_{p} = \(\frac {-2w}{V{_∞^2}} \frac {v^{‘{^2}}+w^{‘^2}}{V_∞^{2}}\)

d) C_{p} = –\(\frac {2u^{‘}}{V_∞}\) + (1 – M\(_∞^2\))\(\frac {u^{‘^{2}}}{V{_∞^{2}}} + \frac {v^{‘^{2}}+w^{‘^{2}}}{V_∞^{2}}\)

View Answer

Explanation: The non – linear coefficient of pressure is in the form of the relation below as obtained after binomial expansion.

C

_{p}= –\(\frac {2u^{‘}}{V_∞}\) + (1 – M\(_∞^2\))\(\frac {u^{‘^{2}}}{V{_∞^{2}}} + \frac {v^{‘^{2}}+w^{‘^{2}}}{V_∞^{2}}\)

For two dimensional terms, the second order terms can be neglected. But for three – dimensional the term \(\frac {v^{‘{^2}}+w^{‘^{2}}}{V_∞^{2}}\) has to be retained since it is not negligible. Apart from that \(\frac {u^{‘^2}}{V_∞^{2}}\) is neglected because the perturbed velocities have magnitudes that are very less than unity. This results in equation for linearized coefficient of pressure for three – dimensional bodies as C

_{p}= \(\frac {-2u^{‘}}{V_∞} + \frac {v^{‘{^2}}+w^{‘^2}}{V_∞^{2}}\).

4. What is the coefficient of pressure over an airfoil at supersonic flow at Mach 2 which is inclined to the freestream at 1.4 degrees?

a) 1.10

b) 1.92

c) 1.62

d) 2.81

View Answer

Explanation: Given, M\(_∞^2\) = 2, θ = 1.4 deg

For the supersonic flow, the linearized coefficient of pressure for small inclinations is given by:

C

_{p}= \(\frac {2θ}{\sqrt {M_∞^2 – 1}}\)

Substituting the values, we get

C

_{p}= \(\frac {2 × 1.4}{\sqrt {4 – 1}}\) = 1.62

5. Coefficient of pressure over the forward section of the hump in supersonic flow is negative.

a) True

b) False

View Answer

Explanation: The coefficient of pressure over a slender body in a supersonic flow, provided the inclination is small is given by

C

_{p}= \(\frac {2θ}{\sqrt {M_∞^2 – 1}}\)

The angle θ is positive when measured above the freestream horizontal and negative when measured below the freestream horizontal. Due to this reason, the coefficient of pressure in the forward portion of the hump is positive.

6. How does the coefficient of pressure vary for subsonic flow as the Mach number increases?

a) Increases

b) Decreases

c) Remains same

d) First increases, then decreases

View Answer

Explanation: The linearized coefficient of pressure for a subsonic flow varies with respect to the Mach number as follows:

C

_{p}∝ \(\frac {1}{\sqrt {1 – M_∞^2}}\)

According to this proportionality, as the Mach number increases, the coefficient of pressure also increases.

7. How does the coefficient of pressure vary for supersonic flow as the Mach number decreases?

a) Increases

b) Decreases

c) Remains same

d) First increases, then decreases

View Answer

Explanation: In the supersonic regime, according to the linearized coefficient of pressure is directly proportional to the local inclination and inversely proportional to \(\sqrt {M_∞^2 – 1}\). Due to this relation, as the Mach number decreases, the coefficient of pressure increases to a point where Mach number reaches transonic regime (M = 1) where the coefficient of pressure becomes infinity.

8. What is the coefficient of lift according to the linearized theory over a flat plate kept at an inclination of 3 degrees having a freestream Mach number of 2?

a) 0.01

b) 0.12

c) 0.85

d) 0.52

View Answer

Explanation: Given, M

_{∞}= 2, ∝ = 3 = 0.052 rad

According to the linearized theory, the coefficient of lift over a flat plate is given by:

c

_{l}= \(\frac {4∝}{\sqrt {M_∞^2 – 1}}\)

Substituting the values we get

c

_{l}= \(\frac {4 × 0.052}{\sqrt {4 – 1}}\) = 0.12

9. Linearized pressure distribution for higher deflection angle is inaccurate.

a) True

b) False

View Answer

Explanation: The linearized pressure distribution is usually inaccurate for higher deflection angles beyond 4 degrees. But when they are integrated to obtain the linearized coefficient of lift and drag, the inaccuracies are approximately compensated when adding the upper and lower surface.

**Sanfoundry Global Education & Learning Series – Aerodynamics.**

To practice all areas of Aerodynamics, __here is complete set of 1000+ Multiple Choice Questions and Answers__.

**Related Posts:**

- Apply for Aerospace Engineering Internship
- Check Aerospace Engineering Books
- Practice Aeronautical Engineering MCQs
- Practice Aerospace Engineering MCQs
- Apply for Aerodynamics Internship