Find the Mach number when an aeroplane is flying at 900 km/ hour through still air having a pressure of 8 $N/cm^2$ and temperature -$15^0$ C. take k=1.4 and R= 287 J/kg K. Calculate the pressure, density and temperature of air at stagnation point on the nose of plane.

Given :

$ \begin{aligned} \text{v} =900 \space {Km}/{P h} &=\frac{900 \times 1000}{60 \times 60} \\ &= 250 \space {m} / {s} \\ \text {Pressure of a air }\left(P \right) &=8 \space {N} /{cm}^{2} \\&=8 \times 10^{4} \space{N} / {m}^{2} \\ T =15^{\circ}{C} &=15+273=288\space {K} \\ K &=1.4\\ \end{aligned} $

Now for Adiabatic process,

The velocity of sound

$ \begin {aligned} C &=\sqrt{K R T} \\ &=\sqrt{1.4 \times 287 \times 288} \\ &=340.17 {m} / {s} \\ \end{aligned} $

Mach number(M)

$ \begin {aligned} M &= \frac{v}{C}=\frac{250}{340.17} \\ M &=0.735 \\ \end{aligned} $

Stagnation Pressure (Ps)

$ \begin {array}{I} P_S &=P \left(1+\frac{k-1}{2} \times M^{2}\right)^{\frac{K}{K-1}} \\ &=8\times10^{4}\left(1+\frac{1.4-1}{2} \times 0.735^{2}\right)^{\frac{1.4}{1.4-1}} \\ &= 8.24 \times 10^{4} N/ m^{2} \\ \end{array} $

Stagnation Temperature (Ts)

$ \begin {aligned} T_{S} &=T\left(1+\frac{k-1}{2} \times M^{2}\right) \\ &=288\left(1+\frac{1.4-1}{2} \times 0.735^{2}\right) \\ &=319.12 \mathrm{K} \\ T_{S} &=46.12^{\circ} \mathrm{C} \end{aligned} $

Stagnation density $\left(\rho_{s}\right)$ $ \begin{aligned} \rho_{S} &=\frac{P_{s}}{R T_{s}}=\frac{8.24 \times 10^{4}}{287 \times 319.12} \\ &=0.8995 \mathrm{~kg} / \mathrm{m}^{3} \end{aligned} $