To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.
The altitude is,
[tex]z = 20km[/tex]
And the velocity can be written as,
[tex]V = 3530km/h (\frac{1000m}{1km})(\frac{1h}{3600s})[/tex]
[tex]V = 980.55m/s[/tex]
From the properties of standard atmosphere at altitude z = 20km temperature is
[tex]T = 216.66K[/tex]
[tex]k = 1.4[/tex]
[tex]R = 287 J/kg[/tex]
Velocity of sound at this altitude is
[tex]a = \sqrt{kRT}[/tex]
[tex]a = \sqrt{(1.4)(287)(216.66)}[/tex]
[tex]a = 295.049m/s[/tex]
Then the Mach number
[tex]Ma = \frac{V}{a}[/tex]
[tex]Ma = \frac{980.55}{296.049}[/tex]
[tex]Ma = 3.312[/tex]
So front stagnation temperature
[tex]T_0 = T(1+\frac{k-1}{2}Ma^2)[/tex]
[tex]T_0 = (216.66)(1+\frac{1.4-1}{2}*3.312^2)[/tex]
[tex]T_0 = 689.87K[/tex]
Therefore the temperature at its front stagnation point is 689.87K
