Question

A commercial aircraft flies at a Mach number of 0.67 at an altitude where the air temperature is -19.15 C. Assume ISA conditions.

Determine the altitude (in metres) this aircraft flies at.

Calculate the equivalent airspeed (in feet per second) of this aircraft.

Answer 1

We are given the Mach number and air temperature. We can use the equation for the speed of sound in air in terms of temperature in ISA conditions:

a = (331.3 + 0.606T) m/s

where a is the speed of sound in m/s and T is the air temperature in degrees Celsius. At -19.15 C, we have:

a = (331.3 + 0.606(-19.15)) m/s
= 320.8 m/s

The Mach number is defined as the ratio of the speed of the aircraft to the speed of sound, so we can write:

M = v/a

where M is the Mach number and v is the speed of the aircraft.

Solving for v, we get:

v = Ma = 0.67 x 320.8 = 215.0 m/s

To find the altitude, we need to use the standard atmosphere table. At ISA conditions, the speed of sound is 340.3 m/s at sea level and decreases with altitude. We can use the formula for the variation of temperature with altitude:

T = T0 - Lh

where T0 is the temperature at sea level (15 C), L is the lapse rate (-6.5 C/km), h is the altitude in km, and T is the temperature at the given altitude.

Solving for h, we get:

h = (T0 - T)/L = (15 - (-19.15))/(-6.5) = 5.98 km

Converting to metres, we get:

h = 5.98 km x 1000 m/km = 5980 m

Therefore, the aircraft is flying at an altitude of 5980 m.

To find the equivalent airspeed, we use the formula:

EAS = TAS/√(ρ/ρ0)

where EAS is the equivalent airspeed, TAS is the true airspeed in m/s, ρ is the air density at altitude, and ρ0 is the air density at sea level. We can assume a standard value of ρ0 = 1.225 kg/m3.

To find ρ, we use the standard atmosphere table to find the pressure and temperature at the given altitude. We can assume a value of 101325 Pa for the sea level pressure.

At 5980 m, the temperature is given by:

T = T0 - Lh = 15 - 6.5 x 5.98 = -21.27 C

The pressure can be found from the formula:

p/p0 = (T/T0)^(-g/RL)

where p is the pressure at altitude, p0 is the sea level pressure, g is the acceleration due to gravity (9.81 m/s2), R is the specific gas constant for air (287 J/kg K), and L is the lapse rate (-6.5 C/km).

Using the given values, we get:

p/p0 = (T/T0)^(-g/RL) = (253.88/288.15)^(-9.81/(-6.5 x 287)) = 0.300

Therefore, the pressure at 5980 m is:

p = 0.300 x 101325 = 30398 Pa

The air density can be found from the ideal gas law:

ρ = p/(RT)

where R is the specific gas constant for air and T is the temperature in Kelvin. Converting the temperature to Kelvin, we get:

T = -21.27 + 273.15 = 251.88 K

Substituting the values, we get:

ρ = 30398/(287 x 251.88) = 0.375 kg/m3

Now we can find the equivalent airspeed:

EAS = TAS/√(ρ/ρ0) = 215.0/√(0.375/1.225) = 275.8 m/s

Converting to feet per second, we get:

EAS = 275.8 x 3.281 = 905.0 ft/s

Therefore, the aircraft is flying at an equivalent airspeed of 905.0 ft/s.