Question

A pitot tube connected to a U-tube mercury manometer measures the speed of air flowing in the wind tunnel as shown in the figure below. The density of air is 1.23 kg m\(^{-3}\) while the density of water is 1000 kg m\(^{-3}\). For the manometer reading of \( h = 30 \) mm of mercury, the speed of air in the wind tunnel is _________ m s\(^{-1}\) (rounded off to 1 decimal place). 

 

Hint:

To calculate the airspeed using a pitot tube and manometer, use Bernoulli’s equation and the pressure difference indicated by the height of the mercury column.

Answer 1

The velocity of the air in the wind tunnel can be calculated using the Bernoulli equation and the manometer reading. The difference in pressure between the static pressure and the stagnation pressure in the pitot tube is balanced by the height of the mercury column in the manometer. Step 1: Formula to calculate velocity
The velocity is given by the equation: \[ v = \sqrt{2gh \left(\frac{\rho_m}{\rho}\right)} \] Where:
\( g \) is the acceleration due to gravity,
\( h \) is the height of the mercury column,
\( \rho_m \) is the density of mercury,
\( \rho \) is the density of air.
Step 2: Substitute values
Given that:
\( h = 30 \) mm = 0.03 m,
\( \rho_m = 13.6 \times 1000 \) kg/m\(^3\),
\( \rho = 1.23 \) kg/m\(^3\),
\( g = 10 \) m/s\(^2\),
Substitute into the formula: \[ v = \sqrt{2 \times 10 \times 0.03 \left(\frac{13.6 \times 1000}{1.23}\right)} = \sqrt{0.6 \times 11000} \approx \sqrt{6600} \approx 81.2 \, {m/s} \] Step 3: Conclusion
Thus, the speed of air in the wind tunnel is approximately \( \boxed{81.2} \, {m/s} \).