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(B) According to Bernoulli's principle for a fluid flowing out of a pipe, the pressure difference is converted into kinetic energy.Applying Bernoulli'

Vedclass · Accepted Answer

$[\frac{2(P_1 - P_2)}{\rho}]^{1/2}$

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(B) According to Bernoulli's principle for a fluid flowing out of a pipe, the pressure difference is converted into kinetic energy.Applying Bernoulli's equation between the inside of the pipe (point $1$) and the outside (point $2$):$P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2$Assuming the pipe is horizontal $(h_1 = h_2)$ and the velocity inside the pipe is negligible $(v_1 \approx 0)$ compared to the exit velocity $(v_2 = v)$:$P_1 = P_2 + \frac{1}{2}\rho v^2$Rearranging for $v$:$\frac{1}{2}\rho v^2 = P_1 - P_2$$v^2 = \frac{2(P_1 - P_2)}{\rho}$$v = [\frac{2(P_1 - P_2)}{\rho}]^{1/2}$

$A$ closed pipe containing liquid showed a pressure $P_1$ by gauge. When the valve was opened, the pressure was reduced to $P_2$. The speed of water flowing out of the pipe is ($\rho$ = density of water).

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