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Finding pressure

  1. Jan 14, 2016 #1
    1. The problem statement, all variables and given/known data
    i couldn't understand why the pressure P1 is given by P atm - ρ water (gh2) - ρ(oil)(gh1) + ρ(mercury)(gh3) .Where is the point P1 ? It's not indicated in the diagram ...
    2. Relevant equations


    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Jan 14, 2016 #2

    gneill

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    Staff: Mentor

    It looks like P1 is meant to be the pressure at the water/air interface in the tank. So, it's the same as the air pressure in the tank.
     
  4. Jan 14, 2016 #3
    isn't the air pressure at water surface = atmospheric pressure , which is 85.6Pa ?
     
  5. Jan 14, 2016 #4

    gneill

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    Staff: Mentor

    Nope. That would be the pressure at the open end of the "S" tube.

    The tank is a sealed enclosure except for the tube, and the enclosed air can have a different pressure than atmospheric.
     
  6. Jan 14, 2016 #5
    ok , can you explain why the pressure at P1 is found by subtracting the ρ water (gh2) and ρ(oil)(gh1) and plus ρ(mercury)(gh3) ??
     
  7. Jan 14, 2016 #6

    gneill

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    Staff: Mentor

    Sure. The calculation follows a path from the open air at the open end of the S-tube to a point where the pressure is that of the water/air interface. In the system, horizontal lines through the fluids are lines of equal pressure. Look at the path described in orange:
    Fig2.PNG
    Edit: Modified picture to describe a pressure change path from atmosphere to tank water surface with labels so points with equal pressures can be seen.
     
    Last edited: Jan 14, 2016
  8. Jan 14, 2016 #7
    If looks to me like they have the h2 and the h1 incorrectly switched in the equation.

    Chet
     
  9. Jan 14, 2016 #8

    gneill

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    Staff: Mentor

    I think you are right. I hadn't written out and simplified the full path equation before, now I have and I agree with you. Thanks for catching that.

    I've modified the diagram in my previous post to reflect the full path. Points with the same letter labels are at the same pressure. The OP should be able to follow the path summing the pressure changes, then simplify the result.
     
  10. Jan 14, 2016 #9
    He also needs to recognize that moving upward lowers the pressure and moving downward increases the pressure. I'm wondering if he is aware of this.
     
  11. Jan 14, 2016 #10
    so P1 = Patm +ρ (water)gh2 - ρ(oil)g(h1+h2) + ρ(oil)gh1 - ρ(water)g(h1+h2) + ρ(mercury)gh3
    P1 = Patm - ρ(water)gh1 - ρ(oil)gh2 + ρ(mercury)gh3 ???
     
  12. Jan 14, 2016 #11
    Yes. That's correct.
     
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