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Liquid Level question

  1. Aug 8, 2012 #1
    So I work for a drilling company that stores oil in large cylindrical tank facilities. The total volume of each tank is 38,500 bbls.

    In each specific tank facility, a radar tank gauge sits atop a 10" pipe inside the main tank, measuring the level of the liquid inside the pipe. Usually, a vent hole is drilled in this 10" pipe to allow gas to be released as the fluid level rises, however in one of the facilities, no hole was drilled in the top of the pipe, which allowed pressure to build inside of the pipe.

    We noticed the error when the tank was filled all the way to the top (a height known to be 44'), however the radar gauge only showed a level of 32'.

    The gauge now shows 15', and my boss wants to know what the volume of the liquid in the tank currently is.

    Is this possible to solve with the given information? I also know that when the height of the liquid is 10', there are 8,750 bbls in the tank.
     
  2. jcsd
  3. Aug 8, 2012 #2

    mfb

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    Like in the attached image?

    With internal air pressure [itex]p_{in}[/itex], outside atmospheric pressure [itex]p_{out}[/itex], the density of oil [itex]\rho[/itex] and gravitational acceleration g:
    [tex]p_{in}-p_{out}=12ft \cdot \rho \cdot g[/tex]

    Assuming the measurement with the full tank was slow enough to have the same air temperature inside and outside, pressure is proportional to inverse volume and therefore inverse height between ceiling and oil level. [tex]\frac{p_{in}}{p_{out}} = \frac{h_{total}}{h_{total}-32ft}[/tex]

    Solve for htotal...
    [itex](h-32)p_i = h p_o[/itex] => [itex](h-32)(p_o+12\rho g) = h p_o[/itex] => [itex](12 \rho g) h = 32(p_o + 12 \rho g)[/itex]
    => [itex]h_{total} = 32 ft + \frac{32 p_{out}}{12 \rho g}[/itex]

    Using pout=10^5 Pa, [itex]\rho \approx 1g/cm^3[/itex], g=10m/s^2, this corresponds to about htotal=120ft. You have quite a large volume of air in this pipe.


    Assuming the volume of air inside did not change and the temperature is the same, it is possible to do the same backwards with the new pressure p':
    [tex]\frac{p'_{in}}{p_{out}} = \frac{h_{total}}{h_{total}-15ft}[/tex]

    The pressure difference now corresponds to the unknown height difference x between the oil levels:
    [tex]x = \frac{p'_{in}-p_{out}}{\rho \cdot g}[/tex]

    With the new variable [itex]a=\frac{p_{out}}{\rho g}[/itex] (~33ft), the total formula for the height of the oil gets handy:
    [tex]h = 15ft + a \frac{15ft}{\frac{32}{12}a+32ft-15ft}[/tex]
    This can be used for other values of the radar gauge in the obvious way (just replace the 15 everywhere). For 32 ft, you get h=44ft, as expected, and for 0 you get 0.

    Therefore, my prediction is a height difference of ~4.7 ft, for a total height of ~20 ft, corresponding to 17500bbls. If you can check this somehow, I would be interested in the result ;). Note that the calculation has some assumptions inside which may or may not be true, so do not trust the result.
    What about drilling the missing hole? ;)
     
  4. Aug 8, 2012 #3
    Thank you! Yes, that's the first thing I suggested, why not just drill the vent hole?

    If you were to use ρ≈1g/cm3 and g=10m/s^2, wouldn't you first need to convert the h to SI units as well? Just a thought. Other than that, looks good. I've been stuck trying to use Boyle's law and getting nowhere.

    I'll keep you updated if I am able to figure out the actual height and volume of our tank.

    BTW, was there supposed to be an attachment to go with your post, because I am not seeing one?
     
  5. Aug 8, 2012 #4
    Nevermind, works both ways!
     
  6. Aug 8, 2012 #5
    Wait wait wait, if you enter the constants you assumed ( pout=10^5 Pa, ρ≈1g/cm3, g=10m/s^2), into the derived formula htotal=32ft+32pout12ρg, the result is around 58, not 120.

    However, when I work it using metric units, it comes out to about 34.97, or about 120ft! What?
     
  7. Aug 9, 2012 #6

    mfb

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    Strange, the attachment was there before. Next try.

    I converted the result of the fraction p/(rho g)=10m in ft as all other lengths were given in ft.
     

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  8. Aug 9, 2012 #7
    That image is right.

    I talked to my boss this morning. He said that all the calculations made sense, but assured me there was an error somewhere as the total height of the pipe is actually 47 ft (not 120).

    I told him that this information would have been useful to know before. He also informed me that the density of the oil we are using is actually .8156 g/cm^3.
     
  9. Aug 9, 2012 #8

    mfb

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    Does it have a variable cross-section somewhere? Or additional air volume close to the top? This would be treated as "height", even if it is not directly a height. If not, I think the assumption of constant air mass in the tube or something else is wrong.
     
  10. Aug 9, 2012 #9
    I don't think there is additional air volume. I'm with you though, unless there is not some unknown vent, I'm not sure how it is possible that the height is less than 120.

    Are you sure that pin/pout=htota/(lhtotal−32ft) is correct? Where did you get that from?
     
  11. Aug 9, 2012 #10

    mfb

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    Ideal gas law: pV=NkT. Assuming the same temperature and the same amount of air before and after filling (and approximating air as ideal gas), the right side is constant. htotal represents the volume, divided by the (constant) cross-section of the measuring pipe.
     
  12. Aug 9, 2012 #11
    Ah, so just Boyle's law. PV=k... or P1V1 = P2V2.

    So simply put you are saying (P_out)(h_total)= (P_inside)(h_total - h_reading)? And the two times you were comparing were before any oil was in the tank, and when it was completely filled?

    It makes sense, somehow we're missing something though because the actual height of the gauge is not 120ft.
     
  13. Aug 9, 2012 #12
    Would this make sense, kind of modified the method, but used many of the same ideas.

    Given: htotal = 47 ft
    ρoil = 815.6 kg/m3
    Temperature is constant
    g=9.8 m/s2
    Pout = 10^5 Pa
    hreading1=32ft
    hreading2=15ft

    pin/pout =((htotal))/((htotal-hreading1))

    Pin = (47ft/(47ft-15ft))(10^5 Pa) = 313,333.3 Pa

    Assuming the volume of air inside did not change and the temperature is the same,

    p'in/pin=((htotal-hreading1))/((htotal-hreading2))
    p’in=((47ft-32ft)/(47ft-15ft)) (313,333.3 Pa) = 146,875 Pa
    I used p’in and pin because the gas is never completely drained, so the air trapped in the pipe should stay constant.

    The pressure difference now corresponds to the unknown height difference x between the oil levels:

    hoil- hreading =((p'in-pout))/((ρ⋅g))
    hoil=((146,875 Pa – 10^5 Pa))/(( 815.6 kg/m3*9.8 m/s2)) (3.28 ft/m)+ 15 ft
    hoil = 34.2 ft = 29,956 bbls
     
  14. Aug 9, 2012 #13

    mfb

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    That value is wrong, you cannot get such an overpressure with 12 ft of oil level difference.
    If that thing is a perfect, closed cylinder with a total height of 47ft and the other values are as given in the first post, some air went out or something really weird happened.
     
  15. Aug 9, 2012 #14
    Whoops mistyped that. I meant:

    Pin = (47ft/(47ft-32ft))(10^5 Pa) = 313,333.3 Pa

    But that doesn't change the over-pressure from the oil difference.
     
  16. Aug 10, 2012 #15

    mfb

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    Actually I did not look at the formula at all, just at the result, and this is wrong. The issue here is the inconsistency between 47ft total height and the first measurement of the liquid level (difference).
     
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