1. 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
Explanation: The answer is 0.458. Given V=500m/s, a=244m/s. We know γ of air is 1.4. From \(\frac{p_1}{p_2}\)=\(\Big\{1+\frac{\gamma-1}{2}(\frac{V1}{a1})^2\Big\}^\frac{\gamma}{\gamma-1}\)
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.
2. 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) \(\frac{p_1}{p_2}=(\frac{T_2}{T_1})^\frac{\gamma}{\gamma-1}\)
Explanation: \(\frac{p_1}{p_2}=(\frac{T_2}{T_1})^\frac{\gamma}{1-\gamma}\) 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.
3. 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
Explanation: 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.
4. What is the formula for speed of sound in terms of pressure and density?
a) a=\(\sqrt{\frac{\gamma P}{\rho}}\)
b) a=\(\sqrt{\frac{\gamma\rho}{P}}\)
c) a=\(\sqrt{\frac{\gamma RP}{\rho}}\)
d) a=\(\sqrt{\frac{\gamma P}{\rho R}}\)
Explanation: 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.
5. 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
Explanation: 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.
6. 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
Explanation: 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.
7. 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
Explanation: 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.
8. Which of the following is the full-law calibration equation?
a) pp-p=p0\(\Bigg[\Big\{1+\frac{\gamma-1}{2}\left(\frac{V_c}{a_0}\right)^2\Big\}^{\frac{\gamma}{\gamma-1}}-1\Bigg]\)
b) pp-p=p0\(\Bigg[\Big\{1+\frac{\gamma+1}{2}\left(\frac{V_c}{a_0}\right)^2\Big\}^{\frac{\gamma}{\gamma-1}}-1\Bigg]\)
c) p-p0=p0\(\Bigg[\Big\{1+\frac{\gamma-1}{2}\left(\frac{V_c}{a_0}\right)^2\Big\}^{\frac{\gamma}{\gamma-1}}-1\Bigg]\)
d) pp-p=p0\(\Bigg[\Big\{1+\frac{\gamma-1}{2}\left(\frac{V_c}{a_0}\right)^2\Big\}^{\frac{\gamma}{\gamma+1}}-1\Bigg]\)
Explanation: The full-law calibration equation is pp-p=p0\(\Bigg[\Big\{1+\frac{\gamma-1}{2}\left(\frac{V_c}{a_0}\right)^2\Big\}^{\frac{\gamma}{\gamma-1}}-1\Bigg]\) 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.
9. What is the formula for mach number?
a) M=\(\frac{V}{a}\)
b) M=\(\frac{a}{V}\)
c) V=\(\frac{a}{M}\)
d) a=\(\frac{M}{V}\)
Explanation: The formula for mach number is M=\(\frac{V}{a}\) where M=mach number, V=velocity, a=speed of sound. There is no unit for mach number as it is a ratio of speed of the object to the speed of sound in the surrounding air.
10. What is the mach number of an aircraft flying with a speed of 450m/s at a temperature of 290 K?
a) 1.318
b) 2.045
c) 3.214
d) 0.235
Explanation: The answer is 1.318. Given T=290K , V= 450m/s. We know γ of air=1.4 and R=287 J/kg-K.
From the formula-M=\(\frac{V}{a}\) where a=\(\sqrt{\gamma RT}\)
On substituting the values to find “a”,
we have a=\(\sqrt{1.2*287*290}\)
a=341.35m/s.
Now substituting the value of “a” in the formula=\(\frac{V}{a}\),
we get M=\(\frac{450}{341.35}\)
M=1.318.