A closed pipe containing liquid showed a pressure P 1 by gauge. When the valve was opened, pressure was reduced to P 2 . The speed of water flowing out of the pipe is ( ϱ =density of water)

Question

A closed pipe containing liquid showed a pressure P 1 ​ by gauge. When the valve was opened, pressure was reduced to P 2 ​ . The speed of water flowing out of the pipe is ( ϱ =density of water)

TestHub · Accepted Answer

Using Bernoulli's equation for horizontal flow:

$P_{1} + \frac{1}{2} ϱ v_{1}^{2} = P_{2} + \frac{1}{2} ϱ v_{2}^{2}$

Initially, the valve is closed, so the velocity of water $v_{1} = 0$ .

When the valve is opened, the pressure becomes $P_{2}$ and the velocity becomes $v_{2} = v$ .

Substituting these values into the equation:

$P_{1} + 0 = P_{2} + \frac{1}{2} ϱ v^{2}$

$P_{1} - P_{2} = \frac{1}{2} ϱ v^{2}$

$v^{2} = \frac{2 ( P _{1} - P _{2} )}{ϱ}$

$v = [\frac{2 ( P _{1} - P _{2} )}{ϱ}]^{\frac{1}{2}}$

Answer: $[\frac{2 ( P _{1} - P _{2} )}{ϱ}]^{\frac{1}{2}}$

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