First slide

Fluid dynamics

Question

A liquid flows through a horizontal tube as shown in figure. The velocities of the liquid in the two sections, which have areas of cross-section $A_{1}$ and $A_{2}$ are $v_{1}$ and $v_{2}$ , respectively. The difference in the levels of the liquid in the two vertical tubes is h.Then

Moderate

A

$v_{2}^{2} - v_{1}^{2} = 2 gh$
B

$v_{2}^{2} + v_{1}^{2} = 2 gh$
C

$v_{2}^{2} - v_{1}^{2} = gh$
D

$v_{2}^{2} + v_{1}^{2} = gh$

Solution

Let $ρ$ be density of liquid flowing in the tube.
$A_{1} v_{1} = A_{2} v_{2}$
$\frac{v_{1}}{v_{2}} = \frac{A_{2}}{A_{1}}$
According to Bernoulli's equation for horizontal flow of liquid,
$P_{1} + \frac{1}{2} {ρv}_{1}^{2} = P_{2} + \frac{1}{2} {ρv}_{2}^{2}$
$P_{1} - P_{2} = \frac{1}{2} ({ρv}_{2}^{2} - {ρv}_{1}^{2})$
$hρg = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2}) (∵ P_{1} - P_{2} = hρg)$
$v_{2}^{2} - v_{1}^{2} = 2 hg$

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