Question

A pitot tube (Figure) on a high-altitude aircraft measures a differential pressure of $\mathbf{180}\mathbf{}\mathit{P}\mathit{a}$. What is the aircraft’s airspeed if the density of the air is $\mathbf{0}\mathbf{.}\mathbf{031}\mathit{k}\mathit{g}\mathbf{/}{\mathit{m}}^{\mathbf{3}}\mathbf{}\mathbf{?}$

Short answer

The plane’s airspeed is$1.1\times {10}^{2}m/s$.

Step-by-step solution

Given information

1. The pressure difference measured by the pitot tube is$180Pa$.

2. The density of air$0.031kg/m^3$.

Determining the concept of Bernoulli’s equation

Using Bernoulli’s equation, find the expression for the airspeed of the plane. Then using this expression, find the airspeed of the plane for the given values. According to Bernoulli’s equation, as the speed of a moving fluid increases, the pressure within the fluid decreases.

The formula is as follows:

$p\forall +\frac{1}{2}\rho g^{2}h+\text{constant}$

Where, p is pressure, v is velocity, h is height, g is the acceleration due to gravity, h is height, and $\rho$is density.

Determining the plane’s airspeed

The airflow obeys Bernoulli’s principle.

So,

$$\begin{gathered} p_A+\frac{1}{2}{\text{ρ}}_{air}^2{\text{ghtρ}}_{air}={\rho }_B+\frac{1}{2}{\rho }_{air}^{2}g{h}_{air} \\ {p}_A-p_B=\frac{1}{2}{}_{air}\left({}^2\right) \\ \Delta p=\frac{1}{2}{}_{air}({}^2) \end{gathered}$$

Thus, rearranging,

$$\begin{gathered} v=\sqrt{\frac{2∆p}{{\rho }_{air}}} \\ =\sqrt{\frac{2\times 180Pa}{0.031kg/m^3}} \\ =107.76m/s \\ \approx 1.1\times {10}^{2}m/s \end{gathered}$$

Hence, the plane’s airspeed is $1.1\times {10}^{2}m/s$.