GATE 2025 ME – Question 49

Fluid Mechanics · Bernoulli's equation and fluid acceleration · 2 marks · Numerical answer

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).

Assume: Specific gravity of mercury = 13.6

Acceleration due to gravity = 10 m s$^{-2}$

A pitot tube in an air stream connected to a U-tube manometer filled with mercury, with the mercury level difference $h$ marked.

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Correct answer: 80.99 to 81.81

Explanation

The pressure difference across the pitot tube is $\Delta p = (\rho_{Hg} - \rho_{air})gh = (13600 - 1.23) \times 10 \times 0.03 = 4079.6$ Pa. By Bernoulli's equation the air speed is $V = \sqrt{\frac{2\Delta p}{\rho_{air}}} = \sqrt{\frac{2 \times 4079.6}{1.23}} = 81.4$ m/s.