Water flows through a 0.30m radius pipe at the rate of 0.20m^2/s. The pressure in the pipe is atmospheric. The pipe slants downhill and feeds into a second pipe with a radius of 0.15m, positioned 0.60m lower. What is the gauge pressure in the second pipe?(adsbygoogle = window.adsbygoogle || []).push({});

So, what I've figured from the problem and what I've learned so far about flow continuity and Bernoulli's principle gave me this.

FR=A_{1}v_{1}

FR=A_{2}v_{2}

Flow rate = Area of a Cross-section * Velocity

So, I can find the velocity for Bernoulli's equation.

P_{1}+ [tex]\rho[/tex]gh_{1}+ 1/2[tex]\rho[/tex]v_{1}^{2}= P_{2}+ [tex]\rho[/tex]gh_{2}+ 1/2[tex]\rho[/tex]v_{2}^{2}

And since there's no change in height at the first pipe cross-section, [tex]\rho[/tex]gh_{1}= 0, right?

P_{1}+ 1/2[tex]\rho[/tex]v_{1}^{2}= P_{2}+ [tex]\rho[/tex]gh_{2}+ 1/2[tex]\rho[/tex]v_{2}^{2}

So, then I solve for P_{2}, but since it wants gauge pressure, I have to subtract atmospheric pressure, which means that I can remove P_{1}from the equation, because the problem statement says it equals atmospheric pressure, right?

1/2[tex]\rho[/tex]v_{1}^{2}- 1/2[tex]\rho[/tex]v_{2}^{2}- [tex]\rho[/tex]gh_{2}= P_{g}

All I'm really looking for here is a logic check for this much of it. Am I right?

All help is greatly appreciated.

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# Bernoulli's Equation and Fluid Mechanics?

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