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## Main Question or Discussion Point

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?

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

FR=A

Flow rate = Area of a Cross-section * Velocity

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

P

And since there's no change in height at the first pipe cross-section, [tex]\rho[/tex]gh

P

So, then I solve for P

1/2[tex]\rho[/tex]v

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

All help is greatly appreciated.

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.