Bernoulli's Equation and Fluid Mechanics?

  • Thread starter wmrunner24
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  • #1
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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=A1v1
FR=A2v2

Flow rate = Area of a Cross-section * Velocity

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

P1 + [tex]\rho[/tex]gh1 + 1/2[tex]\rho[/tex]v12 = P2 + [tex]\rho[/tex]gh2 + 1/2[tex]\rho[/tex]v22

And since there's no change in height at the first pipe cross-section, [tex]\rho[/tex]gh1= 0, right?

P1 + 1/2[tex]\rho[/tex]v12 = P2 + [tex]\rho[/tex]gh2 + 1/2[tex]\rho[/tex]v22

So, then I solve for P2, but since it wants gauge pressure, I have to subtract atmospheric pressure, which means that I can remove P1 from the equation, because the problem statement says it equals atmospheric pressure, right?

1/2[tex]\rho[/tex]v12 - 1/2[tex]\rho[/tex]v22 - [tex]\rho[/tex]gh2 = Pg

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

Answers and Replies

  • #2
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That looks correct. There is actually no reason to set h1=0 because the problem gave you the height difference between the two pipes.
 
  • #3
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Oh? I was under the impression that whenever you wrote something like [tex]\rho[/tex]gh1, it was assumed to be the change in height, like in gravitational potential energy. I said h1 was zero because as you moved left to right, at the first cross-section, there was no change yet so it was 0. But, if I understand what you're saying, I factor to get h1 - h2 together and replace it with [tex]\Delta[/tex]h, right? Which actually seems to make more sense...
 

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