A pitot tube (Figure) is used to determine the airspeed of an airplane. It consists of an outer tube with a number of small holes B (four are shown) that allow air into the tube; that tube is connected to one arm of a U-tube. The other arm of the U-tube is connected to hole $\mathit{A}$at the front end of the device, which points in the direction the plane is headed. At $\mathit{A}$the air becomes stagnant so that ${\mathit{v}}_{\mathbf{A}}\mathbf{=}\mathbf{0}$. At $\mathit{B}$, however, the speed of the air presumably equals the airspeed $\mathit{v}$of the plane.

(a) Use Bernoulli’s equation to show that, $\mathit{v}\mathbf{=}\sqrt{\frac{\overbrace{\mathbf{p}}\mathbf{g}\mathbf{h}}{{\mathbf{\rho }}_{\mathbf{a}\mathbf{i}\mathbf{r}}}}$where $\mathit{\rho }$is the density of the liquid in the U-tube and $\mathit{h}$is the difference in the liquid levels in that tube.

(b) Suppose that the tube contains alcohol and the level difference $\mathit{h}$is $\mathbf{26}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}$. What is the plane’s speed relative to the air? The density of the air is $\mathbf{1}\mathbf{.}\mathbf{03}\mathbf{}\mathit{k}\mathit{g}\mathbf{/}{\mathit{m}}^{\mathbf{3}}$and that of alcohol is $\mathbf{810}\mathbf{}\mathit{k}\mathit{g}\mathbf{/}{\mathit{m}}^{\mathbf{3}}$.

1. The speed of air at the end A, $v_A=0m/s$.

2. At point B, the speed of air = the speed of the plane $=v$.

3. The pitot tube is horizontal.

4. The height difference in U-tube,$h=26.0cm$.

5. The density of air,${\rho }_{air}=1.03kg/m^3$.

1. The density of alcohol,$\rho =810kg/m^3$.

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

The equation is as follows:

$\rho \forall \frac{1}{2}\rho g^{2}h+cons\tan t$

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

The airflow obeys Bernoulli’s principle.

So,

${\rho }_A\forall \frac{1}{2}{Pig}_{ir}{g}_{h}hA={\rho }_B\forall \frac{1}{2}{\rho }_{ir}g{h}_{air}$

It is given that the pipe is horizontal and$v_A=0m/s$.

Then,

$∆p={\rho }_A-p_B=\frac{1}{2}\text{air}\left({}^2\right)$

The difference in pressure is also indicated by the height difference in the liquid columns of the U-tube, i.e.,h.

Thus,

$∆pgh$

Hence, by combining these two equations,

$$\begin{gathered} \frac{1}{2}{\rho }_{air}\left(v^2\right)=\rho gh \\ v=\sqrt{\frac{2\rho gh}{{\rho }_{air}}} \end{gathered}$$

Hence, using Bernoulli’s principle, it is proved that$v=\sqrt{\frac{2\rho gh}{{\rho }_{air}}}$.

Use the equation developed in part (a) to determine the plane’s speed relative to air as,

$$\begin{gathered} v=\sqrt{\frac{2\rho gh}{{\rho }_{air}}} \\ =\sqrt{\frac{2\times \left(810kg/m^3\right)\times \left(9.8m/s^2\right)\times \left(26.0\times {10}^{-2}m\right)}{1.03kg/m^3}} \\ =63.3m/3 \end{gathered}$$

Hence, the plane’s speed relative to the air is $63.3m/3$.