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.
To determine the altitude at which the aircraft flies, we need to use the International Standard Atmosphere (ISA) conditions. According to ISA, the temperature decreases at a rate of 1.98 degrees Celsius per 1,000 feet.
First, let's convert the temperature to Kelvin:
-19.15 degrees Celsius = 253.00 Kelvin
Next, we need to find the standard temperature at sea level, which is 15 degrees Celsius or 288.15 Kelvin.
Now, we can use the temperature difference to find the altitude:
Temperature difference = standard temperature at sea level - temperature at altitude
Temperature difference = 288.15 K - 253.00 K
Temperature difference = 35.15 K
The standard lapse rate is 1.98 degrees Celsius per 1,000 feet, which is equivalent to 6.49 Kelvin per 1,000 feet.
Altitude = Temperature difference / Lapse rate
Altitude = 35.15 K / 6.49 K/1,000 ft
Converting feet to meters:
1 foot = 0.3048 meters
Altitude = 35.15 K / (6.49 K/1,000 ft) * (1,000 ft / 0.3048 m)
Altitude = 35.15 K / 6.49 K * 0.3048
Altitude ≈ 5154 meters
Therefore, the aircraft flies at an altitude of approximately 5154 meters.
To calculate the equivalent airspeed (EAS), we need to use the formula:
EAS = True airspeed × sqrt(ρ0 / ρ)
where:
ρ0 = density at sea level under ISA conditions
ρ = density at the given altitude
The ratio of density can be computed using the formula:
ρ/ρ0 = (T0 / T) * (P / P0)
where:
T0 = temperature at sea level under ISA conditions
T = temperature at the given altitude
P0 = pressure at sea level under ISA conditions
P = pressure at the given altitude
Standard values for T0, P0, and ρ0 are available in the International Standard Atmosphere (ISA) tables.
Using the ISA tables, we look up the standard values at sea level:
T0 = 288.15 K
P0 = 101325 Pa
Now, let's find the pressure at the given altitude. The pressure decreases at a rate of approximately 11.3 Pa per foot.
Pressure difference = 11.3 Pa/ft × (altitude in feet)
Converting to meters:
1 foot = 0.3048 meters
Pressure difference = 11.3 Pa/ft × (altitude in meters / 0.3048 m/ft)
Pressure difference ≈ 11.3 × altitude
Therefore, the pressure at the given altitude is:
P = P0 - Pressure difference
P ≈ 101325 Pa - 11.3 × altitude
Now, let's calculate the density ratio:
ρ/ρ0 = (T0 / T) * (P / P0)
ρ/ρ0 = (288.15 K / T) * (P / 101325 Pa)
Finally, we can calculate the equivalent airspeed:
EAS = True airspeed × sqrt(ρ0 / ρ)
Without the true airspeed given, we cannot calculate the equivalent airspeed. Please provide the true airspeed to compute the EAS.
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.