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Thermodynamics - Double U Tube Manometer with 2 Closed Ends

  1. Aug 31, 2009 #1
    Thermodynamics -- Double U Tube Manometer with 2 Closed Ends

    1. The problem statement, all variables and given/known data
    GIVEN:
    Fresh water and sea water in parallel horizontal pipelines are connected to each other by a double U tube manometer.

    Density of sea water is 1035 kg/m^3
    Density of air is 1.2 kg/m^3
    Standard temp and pressure

    A link to the relevant diagram is here:
    https://engineering.purdue.edu/ME200/Fall%202009/SpecialProblems/SpecialProblems3-5.pdf" [Broken]

    FIND:
    a) Pressure difference between the pipelines
    b) Can the air column be ignored in your analysis?
    c) If the air is replaced with oil of specific gravity 0.72, then what is the pressure difference in kPa between the two pipelines? Can the oil column be neglected?

    2. Relevant equations
    P-Patm = rho * g * h of liquid

    3. The attempt at a solution
    The problem I am running into is that each end of the manometer is closed so I'm not sure how to find an initial pressure. If one end were open to atmospheric pressure I think I could find the pressure difference and work my way back, if that makes sense. However since both ends are closed "bulbs" of sea and fresh water, I'm not really sure where to start.

    Right now I have labeled the point where the fresh water meets the mercury as point a, the point in the mercury at that same level as point b, the point where the air meets the sea water as point d, and the point at that same level as point c. I am thinking that I can then use the rho*g*l equation somehow, I'm just unclear as to how since Patm isn't evident. Pressure at point a and point be should be equal though, and pressure at point c and point d should be equal.

    I am thinking that there is some sort of equivalency I can use or something... I have the entire manometer hookup as my system but perhaps that is where I am going wrong.

    I am sorry I don't have much worked out, if anyone can even give me some hints as to a starting point I would appreciate it greatly.

    Thanks!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 31, 2009 #2

    Mapes

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    Re: Thermodynamics -- Double U Tube Manometer with 2 Closed Ends

    Hi ynotbathe, welcome to PF. You don't need an absolute pressure, because you're only asked to find a pressure difference. Go ahead and assign a pressure [itex]p_0[/itex] to any of the points you mentioned; it's going to disappear in the final analysis. Although you've given one version of the [itex]\rho g h[/itex] equation, there's a more general version that doesn't necessitate using [itex]P_\mathrm{atm}[/itex]. Does this help?
     
  4. Sep 1, 2009 #3
    Re: Thermodynamics -- Double U Tube Manometer with 2 Closed Ends

    could you explain how you went about doing this problem? im a little lost
     
  5. Sep 2, 2009 #4
    Re: Thermodynamics -- Double U Tube Manometer with 2 Closed Ends

    This is what I did, but I'm pretty sure it's wrong:

    Pa = Pb = arbitrary constant = rho*g*h = 13570 kg/m^3 (density of mercury) * 9.8 m/s^2 (gravitational constant) * .1 m (difference in height) = 13298.6 Pa

    Pc = Pd = arbitrary constant = rho*g*h = 1.2 kg/m^3 (density of air) * 9.8 m/s^2 (gravitational constant) * .4 m (difference in height) = 4.707 Pa

    In addition to not thinking these numbers are right, I'm not really sure where to go from here. Any hints or suggestions to what I'm doing wrong?

    Thanks!
     
  6. Sep 2, 2009 #5

    Mapes

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    Re: Thermodynamics -- Double U Tube Manometer with 2 Closed Ends

    The [itex]\rho g h[/itex] equation is for finding the difference in pressure between two points. Lacking an absolute reference, as is the case here, you can only find a pressure relative to an arbitrary point. So pick an arbitrary point -- say, one of the two pipes -- and work your way to the other pipe.
     
  7. Oct 3, 2010 #6
    Re: Thermodynamics -- Double U Tube Manometer with 2 Closed Ends

    u can try this method. Start with one of the enclosed bulb (eg; P1) add the "rho"gh terms when u go down the tube and subtract the terms when you go up the tube until you reach the other enclosed bulb. The effect of air air column is usually neglected in most problem as the density is small and so, the "rho"gh term for the air column is assumed to be zero.Try this link, http://www.fuzzyturtle.net/TDEC/tdec202/hw01.pdf" [Broken]
     
    Last edited by a moderator: May 5, 2017
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