To determine the air temperature, air pressure, and air density at an altitude of 30,000 feet, we can use the standard atmospheric model.
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Convert altitude from feet to meters: \[ 30,000 \text{ feet} \times 0.3048 \text{ m/ft} = 9,144 \text{ m} \]
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Determine the temperature: In the standard atmosphere, for altitudes between 11,000 m and 20,000 m, the temperature decreases linearly. However, since 30,000 feet (or 9,144 m) is below 11,000 m, we can use the standard temperature for the lower troposphere.
The temperature at sea level is approximately 288.15 K (15°C). The lapse rate in the troposphere is approximately 6.49 K/km up to 11,000 m.
For the altitude of 9,144 m: \[ T = T_0 - L \times h \] Where:
- \( T_0 = 288.15 \text{ K} \) (sea level temperature)
- \( L = 6.49 \text{ K/km} \times \frac{1 \text{ km}}{1000 \text{ m}} = 0.00649 \text{ K/m} \)
- \( h = 9,144 \text{ m} \)
\[ T = 288.15 - (0.00649 \times 9144) \approx 288.15 - 59.4 \approx 228.75 \text{ K} \]
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Determine the pressure: From the standard atmosphere model, the pressure can be approximated for altitudes below 11,000 m using: \[ P = P_0 \left(1 - \frac{L \cdot h}{T_0}\right)^{\frac{g_0 \cdot M}{R \cdot L}} \]
Where:
- \( P_0 = 101325 \text{ Pa} \) (pressure at sea level)
- \( g_0 = 9.80665 \text{ m/s}^2 \) (acceleration due to gravity)
- \( M = 0.0289644 \text{ kg/mol} \) (molar mass of Earth's air)
- \( R = 8.31447 \text{ J/(mol·K)} \) (universal gas constant)
Substituting values: \[ P = 101325 \left(1 - \frac{0.00649 \cdot 9144}{288.15}\right)^{\frac{9.80665 \cdot 0.0289644}{8.31447 \cdot 0.00649}} \]
First calculate the term inside the brackets: \[ \frac{0.00649 \cdot 9144}{288.15} \approx 0.208 \] Then calculate: \[ P \approx 101325 \left(1 - 0.208\right)^{5.255} \approx 101325 \times (0.792)^{5.255} \approx 101325 \times 0.3713 \approx 37670 \text{ Pa} \]
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Determine the air density: Using the ideal gas law: \[ \rho = \frac{P}{R_d \cdot T} \] Where:
- \( R_d = 287.05 \text{ J/(kg·K)} \) is the specific gas constant for dry air
Substituting values: \[ \rho = \frac{37670}{287.05 \cdot 228.75} \approx \frac{37670}{6562.427} \approx 5.74 \text{ kg/m}^3 \]
Summary of Results
- Temperature \( T \approx 228.75 \text{ K} \)
- Pressure \( P \approx 37,670 \text{ Pa} \)
- Density \( \rho \approx 5.74 \text{ kg/m}^3 \)