First slide

Fluid dynamics

Question

The vessel of area of cross-section A has liquid to a height H. There is a hole at the bottom of vessel having area of cross-section a. The time taken to decrease the level from H₁ to H₂ will be

Very difficult

A

$\frac{A}{a} \sqrt{(\frac{2}{g})} [\sqrt{H_{1}} - \sqrt{H_{2}}]$
B

$\sqrt{(2 gh)}$
C

$\sqrt{[2 gh (H_{1} - H_{2})]}$
D

$\frac{A}{a} \sqrt{(\frac{g}{2})} [\sqrt{H_{1}} - \sqrt{H_{2}}]$

Solution

The average velocity of efflux
$v = \frac{\sqrt{2 {gH}_{1}} + \sqrt{2 {gH}_{2}}}{2}$
let t be the time taken to empty the tank from level H₁ to H₂. Then
$\frac{\sqrt{2 {gH}_{1}} + \sqrt{2 {gH}_{2}}}{2} \times a \times t = A [H_{1} - H_{2}] or t = \frac{A}{a} \times \sqrt{(\frac{2}{g})} \times [\frac{H_{1} - H_{2}}{\sqrt{H_{1}} + \sqrt{H_{2}}}] = \frac{A}{a} \times \sqrt{(\frac{2}{g})} \times [\frac{(H_{1} - H_{2}) \times (\sqrt{H_{1}} - \sqrt{H_{2}})}{(\sqrt{H_{1}} + \sqrt{H_{2}}) (\sqrt{H_{1}} - \sqrt{H_{2}})}] = \frac{A}{a} \times \sqrt{(\frac{2}{g}) \times (\sqrt{H_{1}} - \sqrt{H_{2}})}$
Alternate method. Let I be the height of the water level at any instant. Now, the rate of decrease of water level is
-dh/ dt. Therefore,
$- A (\frac{dh}{dt}) = av = a \sqrt{(2 gh)} or - (\frac{dh}{dt}) = \frac{a}{A} \sqrt{(2 gh)} Integrating this expression, we get - \int_{H_{1}}^{H_{2}} \frac{dh}{\sqrt{h}} = \frac{a}{A} \sqrt{(2 g)} \int_{0}^{t} dt - 2 [\sqrt{h}]_{H_{1}}^{H_{2}} = \frac{a}{A} \sqrt{(2 g)} \cdot t 2 [\sqrt{H_{1}} - \sqrt{H_{2}}] = \frac{a}{A} \sqrt{(2 g)} \cdot t ∴ t = \frac{A}{a} \sqrt{(\frac{2}{g})} [\sqrt{H_{1}} - \sqrt{H_{2}}]$

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