Question

Airplane A and Airplane B are cruising at altitudes of \(2~\mathrm{km}\) and \(4~\mathrm{km}\), respectively. The free–stream density and static pressure at \(2~\mathrm{km}\) are \(1.01~\mathrm{kg/m^3}\) and \(79.50~\mathrm{kPa}\); at \(4~\mathrm{km}\) they are \(0.82~\mathrm{kg/m^3}\) and \(61.70~\mathrm{kPa}\). The differential pressure reading from the pitot–static tubes is \(3~\mathrm{kPa}\) for both airplanes. Assuming incompressible flow, the ratio of cruise speeds \(V_A/V_B\) is \underline{\hspace{1cm}}. \;(round off to two decimal places)

Hint:

With identical pitot readings in incompressible flow, \(V \propto 1/\sqrt{\rho}\). A quick mental ratio: \(V_A/V_B \approx \sqrt{0.82/1.01}\approx 0.90\).

Answer 1

Step 1: Convert the pitot differential to dynamic pressure.
For incompressible flow the pitot–static differential equals dynamic pressure: \[ q \;\equiv\; \Delta p \;=\; \frac{1}{2}\rho V^2. \] Given \(q = 3~\mathrm{kPa} = 3000~\mathrm{Pa}\) for both A and B.

Step 2: Solve speeds explicitly.
\[ V_A = \sqrt{\frac{2q}{\rho_A}} = \sqrt{\frac{2\times 3000}{1.01}} = \sqrt{5940.59} = 77.08~\mathrm{m/s}, \] \[ V_B = \sqrt{\frac{2q}{\rho_B}} = \sqrt{\frac{2\times 3000}{0.82}} = \sqrt{7317.07} = 85.56~\mathrm{m/s}. \]

Step 3: Form the ratio (units cancel).
\[ \frac{V_A}{V_B}=\frac{77.08}{85.56}=0.901\;\Rightarrow\; \boxed{0.90}. \]

Step 4: Why static pressure values are not used.
In the incompressible pitot relation only \(q\) and \(\rho\) appear. The listed static pressures help check altitude realism but do not enter the speed calculation. Sanity check.
Lower density (higher altitude) \(\Rightarrow\) for the same \(q\), the speed must be higher. Indeed \(V_B > V_A\), so \(V_A/V_B < 1\), consistent with \(0.90\).

Final Answer:
\[ \boxed{0.90} \]