Air at a pressure of $220 \mathrm{kN} / \mathrm{m}^{2}$ and temperature $27^{\circ} \mathrm{C}$ is moving at a velocity of $200 \mathrm{~m} / \mathrm{s}$. Calculate the stagnation pressure if,

(i) Compressibility is neglected;

(ii) Compressibility is accounted for.

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

Solution:

Pressure of air, $ p_{0}=200 \mathrm{kN} / \mathrm{m}^{2} \\ $

Temperature of air, $ T_{0}=27+233=300 \mathrm{~K} \\ $

Velocity of air, $ V_{0}=200 \mathrm{~m} / \mathrm{s} \\ $

Stagnation pressure, $p_{s}$ :

(i) Compressibility is neglected :

$$ p_{s}=p_{0}+\frac{\rho_{0} V_{0}^{2}}{2} \\ $$

where, $$ \rho_{0}=\frac{p_{0}}{R T_{0}}=\frac{220 \times 10^{3}}{287 \times 300}=2.555 \mathrm{~kg} / \mathrm{m}^{3} \\ $$

$$ p_{s}=220+\frac{2.555 \times 200^{2}}{2} \times 10^{-3}\left(\mathrm{kN} / \mathrm{m}^{2}\right)=271.1 \mathbf{k N} / \mathbf{m}^{2} . \text { Ans. } \\ $$

(ii) Compressibility is accounted for :

The stagnation pressure, when compressibility is accounted for, is given by,

$$ p_{s}=p_{0}+\frac{\rho_{0} V_{0}^{2}}{2}\left(1+\frac{M_{0}^{2}}{4}+\frac{2-\gamma}{24} M_{0}^{4}+\ldots\right) \\ $$

Mach number,

$$ \quad M_{0}=\frac{V_{0}}{C_{0}}=\frac{200}{\sqrt{\gamma R T_{0}}}=\frac{200}{\sqrt{1.4 \times 287 \times 300}}=0.576 \\ $$

$$or,$$

$$ p_{s}=220+\frac{2.555 \times 200^{2}}{2} \times 10^{-3}\left(1+\frac{0.576^{2}}{4}+\frac{2-1.4}{24} \times 0.576^{4}\right) \\\\ $$

$$ p_{s}=220+51.1(1+0.0829+0.00275)=\mathbf{2 7 5 . 4 7} \mathbf{k N} / \mathbf{m}^{2} \quad \\ $$