Calculate on the stagnation point on the nose of the plane.

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written 2.6 years ago by

prajapatijaimin • 3.2k

Solution:

Speed of aeroplane, $ V=1000 \mathrm{~km} / \mathrm{h}=\frac{1000 \times 1000}{60 \times 60}=277.77 \mathrm{~m} / \mathrm{s} \\ $

Pressure of air, $ \quad p_{0}=78.5 \mathrm{kN} / \mathrm{m}^{2} \\ $

Temperature of air, $ T_{0}=-8+273=265 \mathrm{~K} \\ $

For air : $ R=287 \mathrm{~J} / \mathrm{kg} \mathrm{K}, \gamma=1.4 \\ $

The sonic velocity for adiabatic flow is given by,

$$ \begin{aligned} C_{0} &=\sqrt{\gamma R T_{0}}=\sqrt{1.4 \times 287 \times 265}=326.31 \mathrm{~m} / \mathrm{s} \\\\ \therefore \text { Mach number, } \quad M_{0} &=\frac{V_{0}}{C_{0}}=\frac{277.77}{326.31}=0.851 \end{aligned} \\ $$

(i) Stagnation pressure, $ \mathbf{p}_{\mathbf{s}}: \\ $

The stagnation pressure, $\left(p_{s}\right)$ is given by the relation,

$$ \begin{aligned} p_{s} &=p_{0}\left[1+\left(\frac{\gamma-1}{2}\right) M_{0}^{2}\right]^{\frac{\gamma}{\gamma-1}} \\\\ p_{s} &=78.5\left[1+\left(\frac{1.4-1}{2}\right) \times 0.851^{2}\right]^{\frac{1.4}{1.4-1}} \\\\ &=78.5(1.145)^{3.5}=126.1 \mathbf{k N} / \mathbf{m}^{2} \quad \text { (Ans.) } \end{aligned} \\ $$

(ii) Stagnation temperature, $ \mathbf{T}_{\mathbf{s}}: \\ $

The stagnation temperature is given by,

$$ \begin{aligned} T_{s} &=T_{0}\left[1+\left(\frac{\gamma-1}{2}\right) M_{0}^{2}\right] \\\\ &=265\left[1+\frac{1.4-1} {2} \times 0.851^{2}\right]=303.4 \mathrm{~K} \text { or } \mathbf{3 0 . 4}{ }^{\circ} \mathrm{C} \quad \text { (Ans.) } \end{aligned} \\ $$

(iii) Stagnation density, $ \rho_{\mathbf{s}} : \\ $

The stagnation density $\left(\rho_{s}\right)$ is given by,

$$ \begin{aligned} &\frac{p_{s}}{\rho_{s}}=R T_{s} \text { or } \rho_{s}=\frac{p_{s}}{R T_{s}} \\\\ &\rho_{s}=\frac{126.1 \times 10^{3}}{287 \times 303.4}=1.448 \mathrm{~kg} / \mathrm{m}^{3} \quad \text { (Ans.) } \end{aligned} \\ $$

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