250+ TOP MCQs on Measurement of Airspeed and Answers

Aircraft Performance Multiple Choice Questions on “Measurement of Airspeed”.

1. Which of the following is the correct formula for speed of sound?
a) (sqrt{gamma RT})
b) (sqrt{pRT})
c) (sqrt{rho Rt})
d) (sqrt{gamma R0T})
Answer: a
Clarification: The correct formula for speed of sound is (sqrt{gamma RT}) where R=characteristic gas constant, T=temperature and γ is the ratio of specific heat at constant pressure to that of specific heat at constant volume.

2. What is the speed of sound in air at an altitude where temperature is 299K?
a) 346.61K
b) 343.61K
c) 71.61 °C
d) 70.61 °C
Answer: a
Clarification: The answer is 346.61K. Given T=299K. We know that R=287J/kg-K, γ for air is 1.4.
From a=(sqrt{gamma RT})
a=(sqrt{1.4*287*299})
a=346.61K.

3. What is the relation between pressure and air speed in isentropic relations?
a) (frac{p_1}{p_2})=(Big{1+frac{gamma-1}{2}(frac{V1}{a1})^2Big}^frac{gamma}{gamma-1})
b) (frac{p_1}{p_2})=(Big{1+frac{gamma+1}{2}(frac{V1}{a1})^2Big}^frac{gamma}{gamma+1})
c) (frac{p_1}{p_2})=(Big{1+frac{gamma+1}{2}(frac{V1}{a1})^2Big}^frac{gamma}{gamma-1})
d) (frac{p_1}{p_2})=(Big{1+frac{gamma-1}{2}(frac{V1}{a1})^2Big}^frac{gamma}{gamma+1})
Answer: a
Clarification: The relation between pressure and air speed in isentropic relations is
(frac{p_1}{p_2})=(Big{1+frac{gamma-1}{2}(frac{V1}{a1})^2Big}^frac{gamma}{gamma-1}) where p1, p2 are pressures at two points, V1=velocity at one point, a1=speed of sound at point one and γ is the ratio of specific heat at constant pressure to that of specific heat at constant volume.

4. What is the pressure ratio of an aircraft moving in air at a mach number 1?
a) 1.893
b) 1.558
c) 1.444
d) 1.555
Answer: a
Clarification: The answer is 0.458. Given M=1. We know γ of air is 1.4. From (frac{p_1}{p_2}=Big{1+frac{gamma-1}{2}(M)^2Big}^frac{gamma}{gamma-1})
On substituting the values (frac{p_1}{p_2}=Big{1+frac{1.4-1}{2} 1^2Big}^frac{1.4}{1.4-1})
(frac{p_1}{p_2})=1.893.

5. What is the pressure ratio of an aircraft moving in air at a velocity 500m/s and speed of sound is 244 m/s?
a) 4.556
b) 3.327
c) 6.256
d) 2.565
Answer: b
Clarification: The answer is 0.458. Given V=500m/s, a=244m/s. We know γ of air is 1.4. From (frac{p_1}{p_2})=(Big{1+frac{gamma-1}{2}(frac{V1}{a1})^2Big}^frac{gamma}{gamma-1})
On substituting the values (frac{p_1}{p_2}=Big{1+frac{500}{244}Big}^frac{1.4}{1.4-1})
(frac{p_1}{p_2})=3.327.

6. Which of the following is the correct isentropic relation between pressure and temperature?
a) (frac{p_1}{p_2}=(frac{T_2}{T_1})^frac{gamma}{1-gamma})
b) (frac{p_1}{p_2}=(frac{T_2}{T_1})^frac{gamma-1}{gamma})
c) (frac{p_1}{p_2}=(frac{T_2}{T_1})^frac{gamma+1}{gamma})
d) (frac{p_1}{p_2}=(frac{T_2}{T_1})^frac{gamma}{gamma-1})
Answer: a
Clarification: (frac{p_1}{p_2}=(frac{T_2}{T_1})^frac{gamma}{1-gamma}) is the correct isentropic relation between pressure and temperature where p1, p2 are pressures, T1, T2 are temperatures and γ is the ratio of specific heat at constant pressure to that of specific heat at constant volume.

7. What is the speed of sound where density and pressure are 1.225kg/m3 and 101306N/m2?
a) 340.26m/s
b) 330.26m/s
c) 313 m/s
d) 325 m/s
Answer: a
Clarification: The answer is 340.26m/s. Given P=101306N/m2, ρ=1.225kg/m3 and we know that γ for air is 1.4. From the formula, a=(sqrt{frac{gamma P}{rho}})
a=(sqrt{frac{gamma*101306}{1.225}})
a=340.26m/s.

8. What is the formula for speed of sound in terms of pressure and density?
a) a=(sqrt{frac{gamma P}{rho}})
b) a=(sqrt{frac{gammarho}{P}})
c) a=(sqrt{frac{gamma RP}{rho}})
d) a=(sqrt{frac{gamma P}{rho R}})
Answer: a
Clarification: The formula for speed of sound in terms of pressure and density is given by a=(sqrt{frac{gamma P}{rho}}) where a=speed of sound, P=pressure, ρ=density and γ is the ratio of specific heat at constant pressure to that of specific heat at constant volume.

9. What is the relation between equivalent air speed and pressure ratio?
a) Ve=V(sqrt{sigma})
b) Ve=(frac{V}{sqrt{sigma}})
c) Ve=Vσ2
d) Ve=Vσ-2
Answer: a
Clarification: Ve=V(sqrt{sigma}) is the relation between equivalent air speed and pressure ratio where Ve is the equivalent air speed, V=velocity and σ is pressure ratio.

10. What is the equivalent air speed where velocity is 330m/s and pressure ratio is 8.447?
a) 959.1m/s
b) 1000m/s
c) 981m/s
d) 954m/s
Answer: a
Clarification: The answer is 959.1m/s. Given V=330m/s, σ=8.447. From the equation Ve=V(sqrt{sigma})
Ve=330(sqrt{8.447})
Ve=959.1m/s.

11. What is equivalent air speed?
a) The calibrated air speed corrected for scale-altitude error
b) The true air speed corrected for scale-altitude error
c) The indicated air speed corrected for scale-altitude error
d) The ground air speed corrected for scale-altitude error
Answer: a
Clarification: Equivalent air speed is the calibrated air speed corrected for scale-altitude error. The correction is done in calibrated equation of the airspeed indicator which is a function of calibrated air speed and height.

12. Which of the following is the full-law calibration equation?
a) pp-p=p0(Bigg[Big{1+frac{gamma-1}{2}left(frac{V_c}{a_0}right)^2Big}^{frac{gamma}{gamma-1}}-1Bigg])
b) pp-p=p0(Bigg[Big{1+frac{gamma+1}{2}left(frac{V_c}{a_0}right)^2Big}^{frac{gamma}{gamma-1}}-1Bigg])
c) p-p0=p0(Bigg[Big{1+frac{gamma-1}{2}left(frac{V_c}{a_0}right)^2Big}^{frac{gamma}{gamma-1}}-1Bigg])
d) pp-p=p0(Bigg[Big{1+frac{gamma-1}{2}left(frac{V_c}{a_0}right)^2Big}^{frac{gamma}{gamma+1}}-1Bigg])
Answer: a
Clarification: The full-law calibration equation is pp-p=p0(Bigg[Big{1+frac{gamma-1}{2}left(frac{V_c}{a_0}right)^2Big}^{frac{gamma}{gamma-1}}-1Bigg]) where pp, p, p0 are pressures, Vc is calibrated air speed and a0 is speed of sound and γ is the ratio of specific heat at constant pressure to that of specific heat at constant volume.

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