At two points 1 and 2 in a pipeline the velocities are V and 2V- respectively- Both the points are at the same elevation- The fluid density is - The flow can be assumed to be incompressible- inviscid- steady and irrotational- The difference in pressures P1 and P2 at points 1 and 2 is

Applying Bernoulli's equation, we get P_1ρ g+(V)^22g+Z=P_2ρ g+(2V)^22g+Z ⇒ P_1 P_2ρ g=4V^2 V^22g ⇒ P_1 P_2=32ρ V^2 ⇒ P_1 P_2=1.5ρ V^2 ...

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Fluid Mechanics

Bernoulli Equation-II

At two points...

Question

At two points 1 and 2 in a pipeline the velocities are V and 2V, respectively. Both the points are at the same elevation. The fluid density is $ρ$ . The flow can be assumed to be incompressible, inviscid, steady and irrotational. The difference in pressures P $_{1}$ and P $_{2}$ at points 1 and 2 is

0.5 ρ V^{2}

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1.5 ρ V^{2}

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2 ρ V^{2}

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3 ρ V^{2}

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Solution

The correct option is B $1.5 ρ V^{2}$
Applying Bernoulli's equation, we get
$\frac{P_{1}}{ρ g} + \frac{(V)^{2}}{2 g} + Z = \frac{P_{2}}{ρ g} + \frac{(2 V)^{2}}{2 g} + Z$
$\Rightarrow \frac{P_{1} - P_{2}}{ρ g} = \frac{4 V^{2} - V^{2}}{2 g}$
$\Rightarrow P_{1} - P_{2} = \frac{3}{2} ρ V^{2}$
$\Rightarrow P_{1} - P_{2} = 1.5 ρ V^{2}$

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At two points 1 and 2 in a pipeline the velocities are V and 2V, respectively. Both the points are at the same elevation. The fluid density is ρ. The flow can be assumed to be incompressible, inviscid, steady and irrotational. The difference in pressures P1 and P2 at points 1 and 2 is

The correct option is B 1.5ρV2Applying Bernoulli's equation, we get P1ρg+(V)22g+Z=P2ρg+(2V)22g+Z ⇒ P1−P2ρg=4V2−V22g ⇒ P1−P2=32ρV2 ⇒ P1−P2=1.5ρV2