The pressure leads from Pitot-static tube mounted on an aircraft were connected to a pressure gauge in the cockpit. The dial of the pressure gauge is calibrated to readthe aircraft speed in m/s. The calibration is done on the ground by applying a known pressure across the gauge and calculating the equivalent velocity using incompressible Bernoulli’s equation and assuming that the density is $1.224 kg/m3$.

The gauge having been calibrated in this way the aircraft is flown at $9200 m$, where the density is $0.454 kg/m3 $and ambient pressure is $30 kN/m2$. The gauge indicates a velocity of $152 m/s$. What is the true speed of the aircraft ?

(UPSC exam)

Solution:

Bernoulli's equation for an incompressible flow is given by,

$$ p+\frac{\rho V^{2}}{2}=\text { constant } \\ $$

The stagnation pressure $\left(p_{s}\right)$ created at Pitot-static tube,

$$ p_{s}=p_{0}+\frac{\rho_{0} V_{0}^{2}}{2} \text { (neglecting compressibility effects) }...(1) \ $$ $$ p_{0}=30 \mathrm{kN} / \mathrm{m}^{2}, V_{0}=152 \mathrm{~m} / \mathrm{s}, \rho_{0}=1.224 \mathrm{~kg} / \mathrm{m}^{3} \ $$

Here,

$$ \therefore \quad p_{s}=30+\frac{1.224 \times 152^{2}}{2} \times 10^{-3}=44.139 \mathrm{kN} / \mathrm{m}^{2} \\\\ $$

Neglecting compressibility effect, the speed of the aircraft when,

$ \rho_{0}=0.454 \mathrm{~kg} / \mathrm{m}^{3} \text { is given by [using eqn. (i)], } \\ $

$$or,$$

$$ \therefore \quad V_{0}=249.57 \mathrm{~m} / \mathrm{s} \\ $$

Sonic velocity,

$$ \quad C_{0}=\sqrt{\gamma R T_{0}}=\sqrt{\gamma \frac{p_{0}}{\rho_{0}}}=\sqrt{1.4 \times \frac{30 \times 10^{3}}{0.454}}=304.16 \mathrm{~m} / \mathrm{s} \\ $$

Mach number,

$$ \quad M=\frac{V_{0}}{C_{0}}=\frac{249.57}{304.16}=0.82 \\ $$

Compressibility correction factor,

$$ =\left(1+\frac{M_{0}^{2}}{4}\right) $$

neglecting the terms containing higher powers of $M_{0}, \\$ $ \begin{aligned} &=\left(1+\frac{0.82}{4}\right)=1.168 \\ \therefore \quad \text { True speed of aircraft } &=\frac{249.57}{\sqrt{1168}}=230.9 \mathrm{~m} / \mathrm{s} \end{aligned} \\ $

$ \text { Hence true speed of aircraft }=\mathbf{2 3 0 . 9} \mathrm{m} / \mathbf{s} \text { (Ans.) } $