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Finding gauge pressure when the faucet is turned off

  1. Nov 22, 2008 #1
    1. The problem statement, all variables and given/known data
    The main water line enters a house on the first floor. The line has a gauge pressure of 1.70 X 10^5 Pa.

    (a) A faucet on the second floor, 5.2 m above the first floor, is turned off. What is the gauge pressure at this faucet?
    Pa
    (b) How high above the water main could a faucet be before no water would flow from it, even if the faucet were open?


    2. Relevant equations
    P(gauge)=[tex]\rho[/tex]gh
    Pabs = Patm + Pgauge
    Pabs = 1.01 X 10^5Pa + [tex]\rho[/tex]gh
    density of water = 1000kg/m^3

    3. The attempt at a solution
    i tried to use both of these equations but it didn't work. do i need to use 1.70 X 10^5Pa when solving the problem?
    I tried the first equation and got 51012Pa (1000 X 9.81 X 5.2)
    i tried using the second one and got the wrong answer too.
    then i tried to find the density at the closed water fountain.. but i wasn't even
    sure how to use that information.
    which equation should i use to solve a?
    i'm sure i can solve b after finding this.
     
  2. jcsd
  3. Nov 22, 2008 #2

    Doc Al

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    Staff: Mentor

    How do you find the difference in pressure between two points in a fluid?
     
  4. Nov 22, 2008 #3
    i'll just use "p" for rho..

    i thought the difference might be pgh, and that pgh would be the gauge pressure.
    so i just multiplied density of water by gravity by height.. but that didn't work.
    i just don't know if there's supposed to be a different equation, since the faucet is
    closed and thus a closed valve (i'm not even sure it's a closed valve..)
     
  5. Nov 22, 2008 #4

    Doc Al

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    Staff: Mentor

    That's it.
    Try it again. Note that ρgh is the change in pressure.
     
  6. Nov 22, 2008 #5
    ok this time i tried the differences in gauge pressure,
    so i subtracted pgh from 1.70 X 10^5Pa, and got 118988Pa..
    would this be the right answer?
     
  7. Nov 22, 2008 #6

    Doc Al

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    Looks good.
     
  8. Nov 22, 2008 #7
    thank you!!
    that really helped a lot :)
     
  9. Nov 22, 2008 #8
    to find the height, would i use torricelli's theorum and then find V1 (velocity 1)?
     
  10. Nov 23, 2008 #9

    Doc Al

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    No. To find the height needed for part (b) you'll do the same sort of thing you did for part (a). In part (a) you were given a height and you found the pressure difference; for part (b) you have the pressure difference (you can figure it out) and need to find the height.

    What must the gauge pressure be for no water to flow? What's the pressure difference at that height compared to the pressure at the main level?
     
  11. Nov 23, 2008 #10
    would the gauge pressure have to be zero for no water to flow?
    and the pressure difference would be gauge pressure, what I calculated in (a), I think.
     
  12. Nov 23, 2008 #11

    Doc Al

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    Staff: Mentor

    Right!
    No. What you calculated in (a) was the pressure difference for a 5.2 m change in height.

    For (b) the pressure difference is between the main line pressure (given) and a gauge pressure of zero.
     
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