An aircraft flies at an altitude of 30,000 feet. Determine the air temperature (in [K]), air pressure (in [Pa]) and air density (in [kg/m3]) at this altitude, according to the standard atmosphere.
To determine the air temperature, air pressure, and air density at an altitude of 30,000 feet according to the standard atmosphere, we can follow these steps:
1. Convert the altitude from feet to meters:
30,000 feet * 0.3048 meters/foot = 9,144 meters
2. Use the standard temperature lapse rate to determine the temperature:
In the troposphere, the temperature decreases at an average rate of 6.5°C per kilometer (or 1.98°C per 1,000 feet).
The temperature at sea level is 288.15 K (15°C).
Therefore, the temperature at 9,144 meters can be calculated as follows:
Temperature = 288.15 K - (0.00198°C/ft * 9,144 ft)
= 288.15 K - 18.08°C
= 270.07 K
Thus, the air temperature at an altitude of 30,000 feet is approximately 270.07 K.
3. Use the barometric formula to calculate the air pressure:
The barometric formula relates the pressure, temperature, and altitude.
P = P0 * (1 - (L * h / T0))^((g * M) / (R * L))
Where:
P0 = Sea level standard atmospheric pressure = 101,325 Pa
L = Temperature lapse rate = 0.0065 K/m
h = Height or altitude above sea level = 9,144 m
T0 = Sea level standard temperature = 288.15 K
g = Acceleration due to gravity = 9.81 m/s^2
M = Molar mass of Earth's air = 0.02897 kg/mol
R = Ideal gas constant = 8.314 J/(mol K)
Plugging in the values, we get:
P = 101,325 Pa * (1 - (0.0065 K/m * 9,144 m / 288.15 K))^((9.81 m/s^2 * 0.02897 kg/mol) / (8.314 J/(mol K) * 0.0065 K/m))
P = 101,325 Pa * (1 - (0.0065 K/m * 9,144 m / 288.15 K))^(-3.486)
After solving this equation, we find that the air pressure at an altitude of 30,000 feet is approximately 22,323.47 Pa.
4. Use the ideal gas law to calculate air density:
The ideal gas law relates pressure, temperature, and density.
PV = nRT
Where:
P = Air pressure = 22,323.47 Pa
V = Volume (can be assumed to be constant)
n = Number of moles of air (can be approximated)
R = Ideal gas constant = 8.314 J/(mol K)
T = Temperature = 270.07 K
By rearranging the equation, we can solve for the air density, ρ:
ρ = (P / R) * (1 / T)
ρ = (22,323.47 Pa / 8.314 J/(mol K)) * (1 / 270.07 K)
After solving this equation, we find that the air density at an altitude of 30,000 feet is approximately 0.362 kg/m^3.
Therefore, according to the standard atmosphere, the air temperature at an altitude of 30,000 feet is approximately 270.07 K, the air pressure is approximately 22,323.47 Pa, and the air density is approximately 0.362 kg/m^3.