Problem Statement
Calculate the pressure and density of air at a height of 4000 m from sea-level where the pressure and temperature of the air are 10.143 N/cm² and 15°C respectively. The temperature lapse rate is given as 0.0065°C/m. Take the density of air at sea-level equal to 1.285 kg/m³.
Given Data
- Height, \(Z = 4000 \, \text{m}\)
- Sea-level Pressure, \(p_0 = 10.143 \, \text{N/cm}^2\)
- Sea-level Temperature, \(t_0 = 15^\circ\text{C}\)
- Temperature Lapse Rate, \(L = 0.0065 \, \text{K/m}\)
- Sea-level Density, \(\rho_0 = 1.285 \, \text{kg/m}^3\)
Solution
1. Convert Initial Conditions to SI Units
2. Calculate Gas Constant (R) and Polytropic Index (k)
First, find the specific gas constant for air from the sea-level conditions.
For a polytropic atmosphere with a constant lapse rate, we can find the index \(k\).
3. Calculate Pressure at 4000 m (\(p\))
Using the pressure-altitude relationship for a polytropic atmosphere:
4. Calculate Temperature and Density at 4000 m
Temperature (\(T\)):
Density (\(\rho\)):
Pressure at 4000 m: \( p \approx 60,350 \, \text{N/m}^2 \) or \( 6.035 \, \text{N/cm}^2 \)
Density at 4000 m: \( \rho \approx 0.84 \, \text{kg/m}^3 \)
Explanation of the Model
This problem uses a more realistic model of the Earth’s atmosphere than one with constant density. Here, we assume the temperature decreases linearly with altitude, a concept known as the temperature lapse rate (\(L\)). This is a good approximation for the troposphere (the lowest layer of the atmosphere).
This linear temperature change corresponds to a polytropic process for the air, where \(p/\rho^k = \text{constant}\). The calculation involves finding the specific gas constant (\(R\)) for air based on sea-level conditions, then using the lapse rate to determine the polytropic index (\(k\)). With these constants, we can use the barometric formula to find the pressure at any altitude, and then the ideal gas law to find the density.
Physical Meaning
The results show a significant drop in both pressure and density at an altitude of 4000 m (approx. 13,120 ft).
- The pressure drops to about 59% of its sea-level value (from 101,430 N/m² to 60,350 N/m²). This is because there is 4000 m less of the “air column” above, reducing the weight and thus the pressure.
- The density drops to about 65% of its sea-level value (from 1.285 kg/m³ to 0.84 kg/m³). This happens for two reasons: the lower pressure allows the air to expand, and the lower temperature causes it to contract. The pressure effect is dominant, leading to a significant overall decrease in density.
This is why it is harder to breathe at high altitudes and why aircraft require pressurized cabins. The air is simply “thinner”—there are fewer air molecules in a given volume.
