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Fluid Mechanics

Bernoulli Equation-II

Air flows at ...

Question

Air flows at the rate of $1.5 m^{3} / s$ through a horizontal pipe with a gradually reducing cross-section as shown in the figure. The two cross-sections of the pipe have diameters of 400 mm and 200 mm. Take the air density as 1.2 $k g / m^{3}$ and assume inviscid incompressible flow. The change in pressure $(p_{2} - p_{1}$ ) (in kPa) between sections 1 and 2 is

-1.28

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2.56

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-2.13

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1.28

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Solution

The correct option is A -1.28
$\frac{p_{1}}{ρ g} + \frac{V_{1}^{2}}{2 g} + Z_{1} = \frac{p_{2}}{ρ g} + \frac{V_{2}^{2}}{2 g} + Z_{2}$
[where, Z = constant]
$A_{1} = \frac{π}{4} \times {0.4}^{2} = 0.1256 m^{2}$
$A_{2} = \frac{π}{4} \times {0.2}^{2} = 0.0314 m^{2}$
$\frac{p_{2} - p_{1}}{ρ g} = \frac{V_{1}^{2} - V_{2}^{2}}{2 g}$
$\frac{p_{2} - p_{1}}{1.2} = \frac{1}{2} [\frac{Q^{2}}{A_{1}^{2}} - \frac{Q^{2}}{A_{2}^{2}}]$
$p_{2} - p_{1} = 0.6 \times (1.5)^{2} [\frac{1}{A_{1}^{2}} - \frac{1}{A_{2}^{2}}]$
$p_{2} - p_{1} = 1.35 [\frac{1}{0.01577} - \frac{1}{0.000985}]$
=1.35[63.41-1015.22]
=-1284.94 Pa=-1.28 kPa

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The correct option is A -1.28p1ρg+V212g+Z1=p2ρg+V222g+Z2 [where, Z = constant] A1=π4×0.42=0.1256 m2 A2=π4×0.22=0.0314 m2 p2−p1ρg=V21−V222g p2−p11.2=12[Q2A21−Q2A22] p2−p1=0.6×(1.5)2[1A21−1A22] p2−p1=1.35[10.01577−10.000985] =1.35[63.41-1015.22] =-1284.94 Pa=-1.28 kPa