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Work, Fluid, Pressure

  1. Apr 16, 2012 #1
    1. The problem statement, all variables and given/known data
    A plastic cylindrical tube, weight-less, area = 10 cm^2, length = 2m, upper-closed, vertically immersed in mercury, completely filled by mercury.
    The upper side is at the same level of mercury, the outer pressure is 760 torr. Calculate the work needed to raise the tube of 1,52 m.
    Mercury density: 13595kg/m^3.

    2. The attempt at a solution
    I have realised that I need to win the weight force and the outer pressure force. My result is: 550 J (I have added weight force work to the outer pressure work).
    My book says the result is 116 J.

    Can you help me? Who is wrong?
    Thank you very much
     
  2. jcsd
  3. Apr 16, 2012 #2
    Let me fully understand this problem. The 2 m cylinder of mercury is held in the mercury reservoir in a vertical position, then raised until 1.52 m is out of the mercury. Both ends of the filled plastic are sealed. Or is only the top end sealed and the mercury is free to flow out as it is raised from the reservoir?
     
  4. Apr 16, 2012 #3
    OK, based on the answer you provide, the bottom of the tube is open.

    Hint:

    Break the problem into two parts. First determine how much work is done if the cylinder were sealed at both ends. That is a calculus problem.

    Next, consider how much extra work was done by the mercury not coming out of the bottom of the cylinder as it is raised. 760 torr is atmospheric pressure so you can figure out how much mercury remains in the tube. It will have a perfect vacuum at its top. Determine how much work was saved by the tube not being completely full of mercury. Subtract it from your first answer and you'll get the 116 J.
     
  5. Apr 16, 2012 #4
    I'm sorry I didn't understand the second part.

    This is my attempt using your hint (consider that may be a language problem :D )
    - Weight force (with full cylinder):[itex] mg=10^-3*13595*2*9.8=266,46 N[/itex]
    - Pressure force: [itex]10^5*10^-3=100 N[/itex]
    - Work:[itex](266,46+100)*1.52=557 J [/itex]

    -Weight force (remaining mercury): [itex]10^-3*0.48*13595*9.8=64 N[/itex]
    - Pressure force: 100
    - Work: [itex](64+100)*1.52=250 N[/itex]

    Difference between full cylinder work and saved work: 307 J and not 116J
    What's wrong? Thank you very very much
     
  6. Apr 17, 2012 #5
    Let's start from the beginning. Your weight force of 266 N is correct. But let's take an alternate approach.

    Work is integral(F*dh). Assuming the volume of the tube is negligible, the only work done is the work of raising the mercury in the tube above the pool surface (due to Archimedes Principle). Writing this mathematically you have:

    W = integral(rho*g*A*h*dh) from 0 to 1.52 m where

    rho is density, kg/m^3
    g is gravity, m/sec^2
    A is cross sectional area, m^2
    h is column of mercury above pool surface, meters

    The above will give you the work done if the mercury were not exiting the bottom of the tube and forming a vacuum in the top. Therefore the actual work done must be less than this because the mercury does not rise with the tube because some exits the open lower end.

    First thing to do is calculate the volume of the evacuated space in the tube. That is where atmospheric pressure is considered. If you raise the tube 1.52 meters, is a full vacuum created? And if so, how much empty volume is above the mercury in the tube? Once you determine that, figure out how much work is needed to raise that weight of mercury that amount and subtract it from your result from the integration above.
     
  7. Apr 17, 2012 #6
    Hi, the first work is rho*g*A*(h^2/2)=154 J. Isn't it?

    Is the second work integral of "rho*g*A*h dh" from 0 to 0.48 m?

    thanks a lot!
     
  8. Apr 17, 2012 #7
    "Hi, the first work is rho*g*A*(h^2/2)=154 J. Isn't it?"

    That is correct.

    For the second part I did not use an integration but you could. The mercury in the tube has only risen 0.76 m because that produces full vacuum. In order for the center of gravity (assuming no mercury exits) to rise from the 0.76 level to the 1.14 meter level (halfway between 0.76 and 1.52 m), it must rise another 0.38 m. The weight is 13595*9.81*0.76/1000 = 101 N. Multiplying that by 0.38 m gives 38.4 J. Subtract that from the 154 J and you get the answer. Or you can integrate from 0 to 0.76 m if you choose.
     
  9. Apr 17, 2012 #8
    OK, but why do you choose the halfway between 0,76 and 1.52? What's the logic behind?
     
  10. Apr 17, 2012 #9
    That is how far the center of mass of the cylinder of mercury was raised to get from the surface of the actual column to its position if the column were 1.52 m. In other words you have 0.76 m more mercury raised in the initial calculation. The distance is from the surface to the center of mass of the extra amount.
     
  11. Apr 17, 2012 #10
    As an additional explanation, suppose you were asked to determine how much work is needed to raise a baloon full of water lying on a table to some height. The baloon will change shape depending on how you pick it up. The work done is the change in elevation of its centers of gravity multiplied by its weight. The center of gravity changes its position due to the baloon deforming but the work is the difference in elevation of the centers of gravity multiplied by the weight.
     
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