Air has a velocity of 1000 km/h at a pressure of 9.81 kN/m2 in vacuum and a temperature of 47°C. Compute its stagnation properties and the local Mach number. Take atmospheric pressure = 98.1 kN/m2, R = 287 J/kg K and γ = 1.4.

Solution:

Velocity of air, $ \quad V_{0}=1000 \mathrm{~km} / \mathrm{h}=\frac{1000 \times 1000}{60 \times 60}=277.78 \mathrm{~m} / \mathrm{s} \\ $

Temperature of air, $ \quad T_{0}=47+273=320 \mathrm{~K} \\ $

Atmospheric pressure, $ \quad p_{a t m}=98.1 \mathrm{kN} / \mathrm{m}^{2} \\ $

Pressure of air (static), $ \quad p_{0}=98.1-9.81=88.29 \mathrm{kN} / \mathrm{m}^{2} \\ $

Sonic velocity, $ \quad C_{0}=\sqrt{\gamma R T_{0}}=\sqrt{14 \times 287 \times 320}=358.6 \mathrm{~m} / \mathrm{s} \\ $

$\therefore$ Mach number, $ \quad M_{0}=\frac{V_{0}}{C_{0}}=\frac{277.78}{358.6}=0.7746 \\ $

Stagnation pressure, $ \mathbf{p}_{\mathrm{s}}: \\ $

The stagnation pressure is given by,

$$ p_{s}=p_{0}\left[1+\left(\frac{\gamma-1}{2}\right) M_{0}^{2}\right]^{\frac{\gamma}{\gamma-1}} \\ $$

$$or,$$

$$ p_{s}=88.29\left[1+\frac{1.4-1}{2} \times 0.7746^{2}\right]^{\frac{1.4}{1.4-1}} \\ $$

$$ =88.29(1.12)^{3.5}=131.27 \mathrm{kN} / \mathbf{m}^{2} \quad \text { (Ans.) } \\ $$

Stagnation temperature, $ \mathrm{T}_{\mathrm{s}}: \\ $

$$ \begin{aligned} &T_{s}=T_{0}\left[1+\left(\frac{\gamma-1}{2}\right) M_{0}^{2}\right] \\\\ &T_{s}=320\left[1+\frac{14-1}{2} \times 0.7746^{2}\right]=358.4 \mathrm{~K}^{2} \text { or } \mathbf{8 5 . 4}^{\circ} \mathbf{C} \quad \text { (Ans.) } \\ \end{aligned} $$

Stagnation density, $ \rho_{\mathrm{s}}: \\ $

$$ \rho_{s}=\frac{p_{s}}{R T_{s}}=\frac{131.27 \times 10^{3}}{287 \times 358.4}=1.276 \mathrm{~kg} / \mathrm{m}^{3} \\ $$

Compressibility factor at, $ \mathrm{M}=0.8 : \\ $

Compressibility factor,

$$ \begin{aligned} &=1+\frac{M_{0}^{2}}{4}+\frac{2-\gamma}{24} M_{0}^{4}+\ldots \\\\ &=1+\frac{0.8^{2}}{4}+\frac{2-1.4}{24} \times 0.8^{4}=1.1702 \end{aligned} \\ $$