Consider an airplane flying at a pressure altitude of 33,500 ft and a density altitude of 32,000 ft. Calculate the outside air temperature.
The expression for temperature of the air in terms density is,
\(T=\frac{p}{\rho R}\)
Here, \(\rho\) is the density of air, \(p\) is the pressure and \(R\) is the characteristic gas constant.
From appendix \(\mathrm{B}\), the value of pressure corresponding to pressure altitude of \(33,500 \mathrm{ft}\) is \(535.89 \mathrm{lb} / \mathrm{ft}^{2}\)
From appendix \(B\), the value of density corresponding to density altitude of \(32,000 \mathrm{ft}\) is \(8.2704 \times 10^{-4}\) slug \(/ \mathrm{ft}^{3}\)
Calculate the temperature of air as shown below:
Substitute \(535.89 \mathrm{lb} / \mathrm{ft}^{2}\) for \(p, 8.2704 \times 10^{-4}\) slug \(/ \mathrm{ft}^{3}\) for \(\rho\) and \(1716 \mathrm{ft} \cdot \mathrm{lb} / \mathrm{slug} / \mathrm{R}\) for \(R\) in the
equation \(T=\frac{p}{\rho R}\).
$$ \begin{aligned} T &=\frac{535.89 \mathrm{lb} / \mathrm{ft}^{2}}{\left(8.2704 \times 10^{-4} \text { slug } / \mathrm{ft}^{3}\right)(1716 \mathrm{ft} \cdot \mathrm{lb} / \mathrm{slug} / \mathrm{R})} \\ &=378 \mathrm{R} \end{aligned} $$
Hence, the outside air temperature is \(378^{\circ} \mathrm{R}\).