Water $ρ = 1000 k g / m^{3}$ flows through a venturimeter with inlet diameter 80 mm and throat diameter 40 mm. The inlet and throat gauge pressures are measured to be 400 kPa and 130 kPa respectively. Assuming the venturimeter to be horizontal and neglecting friction, the inlet velocity (in m/s) is______
6

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Solution

The correct option is A 6
Given data:
$ρ = 1000 k g / m^{3}$
$d_{1} = 80 m m = 0.08 m$
$d_{2} = 40 m m = 0.04 m$

$p_{1} = 400 k P a = 400 \times 10^{3} P a$
$p_{2} = 130 k P a = 130 \times 10^{3} P a$

Applying Bernoulli's equation between sections 1 and 2.

$\frac{p_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} + z_{1} \frac{p_{2}}{ρ g} + \frac{v_{2}^{2}}{2 g} + z_{2}$

where $z_{1} = z_{1}$ for horizontal pipe

$∴ \frac{p_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} = \frac{p_{2}}{ρ g} + \frac{v_{2}^{2}}{2 g}$

$o r \frac{p_{1} - p_{2}}{ρ g} = \frac{v_{2}^{2} - v_{1}^{2}}{2 g}$

$o r \frac{p_{1} - p_{2}}{ρ} = \frac{v_{2}^{2} - v_{2}^{2}}{2 g} - - - (i)$

Applying continuity equation between sections 1 and 2.

$A_{1} V_{1} = A_{2} V_{2}$

$\frac{π}{4} d_{1}^{2} V_{1} = \frac{π}{2} d_{2}^{2} V_{2}$

$o r d_{1}^{2} V_{1} = d_{2}^{2} V_{2}$

$o r V_{2} = {(\frac{d_{1}}{d_{2}})}^{2} V_{1} = {(\frac{0.08}{0.04})}^{2} V_{1}$

$V_{2} = 4 V_{1}$

Substituting $V_{2} = 4 V_{1}$ in Eq.(i),we get

$\frac{400 \times 10^{3} - 130 \times 10^{3}}{1000} = \frac{(4 V_{1})^{2} - V_{1}^{2}}{2}$

$540 = 16 V_{1}^{2} - V_{1}^{2}$
$540 = 15 V_{1}^{2}$
$o r V_{1}^{2} = 36 o r V_{1} = 6 m / s$

Method II:

Given: $P_{1} = 400 k P a, = 400 \times 10^{3} P a$
$P_{2} = 130 k P a = 130 \times 10^{3} P a$
$z_{1} = z_{2} (V e n t u r i m e t e r i s h o r i z o n t a l)$

Discharge of venturimeter,

$Q = \frac{a_{1} a_{2} \sqrt{2 g h}}{\sqrt{a_{1}^{2} - a_{2}^{2}}}$

$h = (\frac{P_{1}}{ρ g} + z_{1}) - (\frac{P_{2}}{ρ g} + z_{2})$

$= (\frac{400 \times 10^{3}}{9810} - \frac{130 \times 10^{3}}{9810})$

=27.52 m
Inlet velocity $V_{1} = \frac{Q}{a_{1}} = \frac{a_{2} \sqrt{2 g h}}{\sqrt{a_{1}^{2} - a_{2}^{2}}}$

$= \frac{\frac{π}{4} (0.04)^{3} \sqrt{2 \times 9.81 \times 27.52}}{\sqrt{{[\frac{π}{4} (0.08)]}^{2} {[\frac{π}{4} (0.04)^{2}]}^{2}}}$

=6 m/s

POINTS TO REMEMBER
$\cdot$ Here, we have taken, discharge of venturimeter, Q as the ideal discharge, thats's why we applied:

$Q = \frac{a_{1} a_{2} \sqrt{2 g h}}{\sqrt{a_{1}^{2} - a_{2}^{2}}}$

$\cdot$ If we take actual discharge, then
$Q_{a c t u a l} = C_{D} Q$
where, $C_{D} = c o e f f i c i e n t o f d i s c h a r g e$
= 0.94 to 0.98

$\cdot$ Then in the problem,
$V_{a c t u a l} = 0.98 \times 6 = 5.88 m / s$

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