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Pressure and height

  1. Aug 23, 2011 #1
    1. The problem statement, all variables and given/known data
    question is in the diagram.
    basically im not sure how to use the specific gravity and convert it into pressure.

    2. Relevant equations

    P=p(rho)gh

    3. The attempt at a solution

    -
     

    Attached Files:

  2. jcsd
  3. Aug 23, 2011 #2

    tiny-tim

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    Hi TyErd! :smile:

    (have a rho: ρ :wink:)

    Use the specific gravity to find ρ, the density …

    ρ-of-oil = ρ-of-water times specific-gravity-of-oil :wink:

    (so specific gravity of water = 1)
     
  4. Aug 23, 2011 #3
    ohhh! so the density of any liquid is density of water x specific gravity of the liquid?
     
  5. Aug 23, 2011 #4
    in knowing that my working out is where density of water is 1000...: 80 + 1000(0.3)(9.81) - 1000(13.6)(9.81)(h) - 1000(0.72)(9.81)(0.75) = P_atmospheric. now what, i have two unknowns or do i assume P_atm is 101.325?
     
    Last edited: Aug 23, 2011
  6. Aug 23, 2011 #5
    even if i do use 101.325 the answer is still wrong... i don't know why.
     
  7. Aug 24, 2011 #6

    tiny-tim

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    Hi TyErd! :smile:

    (just got up :zzz: …)
    Yup! :biggrin:
    ah, the question says the gage pressure (or gauge pressure) is 80 KPa …

    gauge pressure is defined as actual pressure minus atmospheric pressure …

    so you can ignore the atmospheric pressure! :wink:
     
  8. Aug 24, 2011 #7
    oh yeah so then what does all of that equal? it should equal the pressure at the top of the oil column so would that just be the 0 because we can assume the mass of air negligible?
     
  9. Aug 24, 2011 #8

    tiny-tim

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    yes :smile:

    the sum of the mgh's has to equal the difference between the outside air and the measurer (80 KPa) …

    you can think of them as being an initial and final pressure, much like initial and final energy in a "solids" equation :wink:
    no, the mass of air is very heavy, it's because the 80 KPa is given as the difference from that :wink:
     
  10. Aug 24, 2011 #9
    okay so i did this: 80 + 1000(0.3)(9.81) - 1000(13.6)(9.81)(h) - 1000(0.72)(9.81)(0.75) =0, h=-0.017047 which is incorrect. where have i gone wrong?
     
  11. Aug 24, 2011 #10

    tiny-tim

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    80,000 ? o:)
     
  12. Aug 24, 2011 #11
    even then its incorrect. the answer is h=0.582m
     
  13. Aug 24, 2011 #12

    tiny-tim

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    I do get 0.582 :confused:

    Please show your full calculations.
     
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