Name: Introduction to Fluid Mechanics
Uploaded: 2019-07-19
Duration: 1 min 36 s
Description: A fluid can be at rest like water in a glass or in motion like water flowing in a pipe. This video explains fluid mechanics dealing with the behaviour of the...

Question

Consider the Venturi tube of figure. Let area $A$ equal $5 a .$ Suppose the pressure at A is $2.0 atm .$ Compute the values of velocity $v$ at $‘ A ’$ and velocity $v'$ at $‘ a ’$ that would make the pressure $p$ at $' a'$ equal to zero. Compute the corresponding volume flow rate if the diameter at $A$ is $5.0 cm .$ (The phenomenon at $a$ when $p$ falls to nearly zero is known as cavitation. The water vaporizes into small bubbles.) $(P_{atm} = 10^{5} N m^{- 2}, ρ = 1000 kg m^{- 3}) .$

Accepted Answer

Given,
Diameter at $5 a .$ is, $D = 5 cm$
Radius at $A$ is, $R = \frac{D}{2} = \frac{5}{2} = 2.5 cm$
Cross-sectional area at $' a'$ is, $A_{A} = π R^{2} = π \times 0 . 025^{2} = 1.96 \times 10^{- 3} m^{2}$
Now according to given question, Cross sectional area at $a$ is, $A_{a} = \frac{A_{A}}{5} = \frac{1.96 \times 10^{- 3}}{5} = 3.93 \times 10^{- 4} m^{2}$
Pressure at $A$ is, $P_{A} = 2 P_{atm} = 2 \times 10^{5} N m^{- 2}$
Pressure at $a$ is, $P_{a} = 0 N m^{- 2}$
Given system is in horizontal position, so change in potential energy of fluid is also zero, i.e. $h_{1} = h_{2} . . . (1)$

From Bernoulli's theorem, we have
$\frac{P_{A}}{ρ g} + \frac{v^{2}}{2 g} + h_{1} = \frac{P_{B}}{ρ g} + \frac{v'^{2}}{2 g} + h_{2}$
$\Rightarrow \frac{P_{A}}{ρ g} + \frac{v^{2}}{2 g} = \frac{P_{B}}{ρ g} + \frac{v'^{2}}{2 g}$ , from $(1)$

On substituting all the given values in above equation, we have
$\Rightarrow \frac{2 \times 10^{5}}{ρ} + \frac{v^{2}}{2} = 0 + \frac{v'^{2}}{2}$
$\Rightarrow \frac{v'^{2} - v^{2}}{2} = \frac{2 \times 10^{5}}{1000}$
$\Rightarrow v'^{2} - v^{2} = 400$

From equation of continuity, we have $Q = A_{A} v = A_{a} v'$ , where $Q$ represents volume flow rate through Venturi.
$\Rightarrow \frac{Q^{2}}{{A_{a}}^{2}} - \frac{Q^{2}}{{A_{A}}^{2}} = 400$
$\Rightarrow Q^{2} (\frac{1}{{(3.93 \times 10^{- 4})}^{2}} - \frac{1}{{(1.96 \times 10^{- 3})}^{2}}) = 400$
$\Rightarrow Q^{2} (6474629 - 260308) = 400$
$\Rightarrow Q^{2} = \frac{400}{6214321} = 6.436 \times 10^{- 5}$
$\Rightarrow Q = 8 \times 10^{- 3} m^{3} s^{- 1}$

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