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Find the elevation

  1. Jun 30, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the elevation at point A. in the figure shown.

    2. Relevant equations

    P = pgh

    3. The attempt at a solution

    So i Converted 0.21kg/cm^2 to Kn/m^2 and i got 20.6kN/m^2 or 20.6kPa

    and -30cm Hg i got 40kN/m^2 or 40kPa
     

    Attached Files:

  2. jcsd
  3. Jun 30, 2014 #2

    haruspex

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    I'm not sure I understand the diagram. It appears to be two tanks side by side, connected by a pipe at the bottom and sealed at the top. Pressures in the airspaces in the tanks are given. Right?
    So, now calculate the pressures at 33m in each, then at 30m in each, etc.
     
  4. Jun 30, 2014 #3
    Yes the pressure in the airspaces in the tanks are given.
    I'll start with the left tank
    P = Patm + (0.82)(9.81)(38-30)
    is it correct?
     
  5. Jul 1, 2014 #4

    haruspex

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    If by Patm you mean the pressure of the air in that tank, that term is ok. But watch out for the units in the other term. What does "specific gravity" mean exactly?
     
  6. Jul 1, 2014 #5
    Specific gravity is the ratio of the density of a substance to the density. that's why it has no unit
    yes If Patm =14.7 psi = 101.325kPa = 101.325kN/m^2
    and
    P = 101.325 + (0.82)(9.81(38-30)
    P = 36.9714 kPa
    What i don't know is how to do I form an equation where i can get the height (h)
     
  7. Jul 1, 2014 #6

    haruspex

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    OK, but you need to specify the reference substance. Did you mean
    Specific gravity is the ratio of the density of a substance to the density of water.​
    ?
    The density of that substance should therefore figure in the formula. As it stands you have (spec grav) * (acceleration) * ( height), which has dimension L2/T2.

    If you continue to work out the pressure at each height, working down from the top, and put in an unknown h for the height you need, you eventually get to two expressions for the pressure right at at the bottom. They must be equal, of course.
     
  8. Jul 2, 2014 #7
    is this correct?
    (-40) + (0.82)(9.81)(38-30) + (1.5) (9.81) (h) - (9.81)(h+3) - 20.6 = P1
    and i'll perform another on the other side? or should I set P1 = 0?
     
  9. Jul 2, 2014 #8
    When I perform P1 = 0
    I get h = 5.235
     
  10. Jul 2, 2014 #9

    haruspex

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    Two problems.
    1. I had been ignoring the minus sign in "-30cm Hg". I don't understand how an absolute air pressure can be negative. These are sealed containers, so it shouldn't be relative to ambient pressure. Do you have any explanation for that?
    2. You are ignoring my comment about density and units. If the s.g. is 0.82 relative to water, what is the density in kg/m3?
     
  11. Jul 3, 2014 #10
    The minus sign is not an absolute pressure i think.
    The density of oil is not given. And also gasoline. In order to get their density you have to multiply their specific gravity to the density of water which is 9.81kg/m^3
     
  12. Jul 3, 2014 #11

    haruspex

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    No, that's where you are going wrong. The 9.81 is gravitational acceleration. That factor converts mass to weight. The density of water is much larger.
     
  13. Jul 4, 2014 #12
    Oh yeah sorry. It's not kg/m^3 ... It's 9.81 kN/m^3 forgot to change that sorry
     
  14. Jul 4, 2014 #13

    haruspex

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    Right, but answer my other question; what is the density of water in these units?
     
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