Water is flowing through a horizontal tube. The pressure of the liquid in the portion where velocity is $2 m / s$ is $2 m$ of $H g$ . What will be the pressure in the portion where velocity is $4 m / s$

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Solution

According to Bernoulli,s principle
$P_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g h_{1}$ $= P_{2} + \frac{1}{2} ρ v_{2}^{2} + ρ g h_{2}$

$\Rightarrow$ $2 m H g + \frac{1}{2} ρ (2)^{2}$ $= P_{2} + \frac{1}{2} ρ (4)^{2}$

$\Rightarrow P_{2} = \frac{1}{2} \times 10^{3} (2^{2} - 4^{2}) + 2 m H g$

$\Rightarrow P_{2} = - 6000 + 2 H g$
$\Rightarrow - \frac{6000}{101325} (0.76) m H g + 2 m H g$

$\Rightarrow P_{2} = 1.955$ m of Hg

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Water is flowing through a horizontal tube. The pressure of the liquid in the portion where velocity is 2 m/s is 2 m of Hg. What will be the pressure in the portion where velocity is 4 m/s

According to Bernoulli,s principle P1+12ρv21+ρgh1 =P2+12ρv22+ρgh2⇒ 2mHg+12ρ(2)2 =P2+12ρ(4)2⇒P2=12×103(22−42)+2mHg⇒P2=−6000+2Hg⇒−6000101325(0.76)mHg+2mHg⇒P2=1.955 m of Hg

Water is flowing through a horizontal tube. The pressure of the liquid in the portion where velocity is $2 m / s$ is $2 m$ of $H g$ . What will be the pressure in the portion where velocity is $4 m / s$