Problem 3.9

Let subscript 1 correspond to sea level and subscript 2 correspond to an altitude of 12000 m.

Given: standard atmospheric conditions,
\( M_{1}=M_{2}=M \),
\( V_{1} - \) 120/3.6 m/s \( = V_{2} \).

To calculate: \( M=? \)

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

Writing \( V_{2} \) in terms of \( V_{1} \) and substituting the temperatures from standard atmospheric tables, \( T_{1}= \) 288.15 K, \( T_{2}= \) 216.65 K, we get,

\[ V_{1}\,\sqrt{\frac{T_{2}}{T_{1}}}=V_{1}-120/3.6\quad\implies\quad V_{1}=\frac{120/3.6}{1-\sqrt{\frac{T_{2}}{T_{1}}}}=\frac{120/3.6}{1-\sqrt{\frac{216.65}{288.15}}} \]

\[ V_{1}=250.818\,\text{m/s}\ . \]

Now the Mach number can be calculated using the velocity and sound-speed at sea-level as,

\[ M=\frac{V_{1}}{\sqrt{\gamma\,R\,T_{1}}}=\frac{250.818}{\sqrt{1.4\times287\times288.15}} \]

\[ \boxed{M=0.737131}\ . \]

Problem 3.9:

Solution: