First slide

Fluid dynamics

Question

A pitot tube is shown in figure. Wind blows in the direction shown. Air at inlet A is brought to rest, whereas its speed just outside of opening B is unchanged. The U tube contains mercury of density $ρ_{m}$ . Find the speed of wind with respect to Pitot tube. Neglect the height difference between A and B and take the density of air as $ρ_{a}$

Difficult

A

$\sqrt{\frac{2 (ρ_{m} + ρ_{a}) g h}{ρ_{a}}}$
B

$\sqrt{\frac{(ρ_{m} - ρ_{a}) g h}{ρ_{a}}}$
C

$\sqrt{\frac{2 (ρ_{m} - ρ_{a}) g h}{ρ_{a}}}$
D

$\sqrt{\frac{(ρ_{m} + ρ_{a}) g h}{ρ_{a}}}$

Solution

Bernoulli's equation between A and B gives
$\frac{P_{A}}{ρ_{a}} = \frac{P_{B}}{ρ_{a}} + \frac{v^{2}}{2} \Rightarrow v^{2} = 2 [\frac{P_{A} - P_{B}}{ρ_{a}}] - - - - (1)$
Also equating pressures at horizontal level of E
$\begin{array}{l} P_{A} + ρ_{a} g y + ρ_{a} g h = P_{B} + ρ_{a} g y' + ρ_{a} g y + ρ_{m} g h \\ \Rightarrow P_{A} + ρ_{a} g h = P_{B} + ρ_{m} g h [∵ y^{'} = 0] \\ \Rightarrow P_{A} - P_{B} = (ρ_{m} - ρ_{a}) g h . \\ \Rightarrow v^{2} = \frac{2 (ρ_{m} - ρ_{a}) g h}{ρ_{a}} \\ \Rightarrow v = \sqrt{\frac{2 (ρ_{m} - ρ_{a}) g h}{ρ_{a}}} F r o m (1) \end{array}$

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