Find the Mach number when an airplane is flying at 900 kmph through still air having a pressure of 8.0 N/cm^2 and temperature $-15^o$ C. Take k = 1.4 and R = 287 J/kg K. Calculate the pressure,temperature and density of the air at the stagnation point on the nose of the plane.

Stagnation Properties of a fluid is defined as "When a flowing fluid past an immersed body, and at a point on the boy if the resultant velocity becomes zero, then the values of pressure, temperature and density at that point are called as stagnation properties.

• That point is called as stagnation point.
• The values of pressure,temperature and density at that point represent as stagnation pressure, stagnation temperature and stagnation density.

Data : Speed of airplane $V = 900 kmph = \frac{900 \times 1000}{60 \times 60} = 250 m/s$

Pressure of air $P_1 = 8 N/cm^2 = 8 \times 10^4 N/m^2$

( $\because 1/cm^2 = 10^4 1/m^2$ or $1m^2 = 10^4 cm^2)$

Temperature, $t_1 = -15˚ c$

$\therefore T_1 = -15 + 273 = 258˚ k$

k = 1.4

Now, For adiabatic process, the velocity of sound $c_1 = \sqrt {KRT}$

$\therefore c_1 = \sqrt {1.4 \times 287 \times 258}$

$c_1 = 321.96 m/s$

Now, Mach number, $M_1 = \frac{v_1}{c_1} = \frac{250}{321.96}$

$\therefore M_1 = 0.776$

Stagnation pressure, Ps

$p_s = p_1 (1 + \frac{k-1}{2} \times m_1^2) ^\frac{k}{k-1}$

$ = 8 \times 10_4 ( 1 + \frac{1.4-1}{2} \times 0.776^2)$

$p_s = 11.91 \times 10^4 N/m^2$

Stagnation Temperature, T,

$T_s = T_1 (1 + \frac{k-1}{2} \times M_1^2)$

$= 258 \times ( 1 + \frac{1.4-1}{2} \times 0.776^2)$

Ts = 289.07˚ K

Ts in ˚C (unit)

$\therefore$ Ts = 289.07 - 273

Ts = 16.07 ˚C

Stagnation Density, $\rho_s$

$\rho_s = \frac{Ps}{RTs}$

As R = 287.J/kg K, the value of $p_s$ should be taken in $N/m^2$, so that the value of $\rho_s$ is obtained in $kg/m^3$

$\therefore \rho_s = \frac{11.91 \times 10^4}{287 \times 289.07} = 1.43 kg/m^3$