91Ó°ÊÓ

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,${\mathrm{ÒÏ}}_{air}=1.03kg/m^3$.

1. The density of alcohol,$\mathrm{ÒÏ}=810kg/m^3$.

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:

$\mathrm{ÒÏ}âˆ\P Ä\frac{1}{2}\mathrm{ÒÏ}{g}^{2}h+cons\tan t$

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

The airflow obeys Bernoulli’s principle.

So,

${\mathrm{ÒÏ}}_Aâˆ\P Ä\frac{1}{2}{Pig}_{ir}{g}_{h}hA={\mathrm{ÒÏ}}_Bâˆ\P Ä\frac{1}{2}{\mathrm{ÒÏ}}_{ir}g{h}_{air}$

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

Then,

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

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,

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

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

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

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

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