Problem 3.7

An airplane can fly at a speed of km/h at sea level where the temperature is ^oC. If the airplane flies at the same Mach number at an altitude where the temperature is ^oC, find the speed at which the airplane is flying at this altitude.

Given:
\( V_1 = \) 800 km/h = 800/3.6 m/s,
\( T_1 = \)15^oC = 288 K,
\( M_{1}=M_{2} \),
\( T_2 = \)-44^oC = 229 K.

To calculate: \( V_2=? \)

The problem description is schematically shown in Fig. 1.

Fig. 1: Schematic of the problem description

The airplane flies at the same Mach number at sea level and higher altitude, therefore,

\[ M_{1}=M_{2} \] \[ \frac{V_{1}}{a_{1}}=\frac{V_{2}}{a_{2}} \] \[ \frac{V_{1}}{\sqrt{\gamma\,R\,T_{1}}}=\frac{V_{2}}{\sqrt{\gamma\,R\,T_{2}}} \]

\[ V_{2}=\frac{V_{1}\sqrt{\gamma\,R\,T_{2}}}{\sqrt{\gamma\,R\,T_{1}}} = \frac{800\times\sqrt{1.4\times287\times229}}{\sqrt{1.4\times287\times288}} \]

\[ \boxed{V_{2}= 713.36\, \text{km/h}}\ . \]

Problem 3.7:

Solution: