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Manometer Problem

  1. Oct 18, 2014 #1
    1. The problem statement, all variables and given/known data
    The figure shows a schematic drawing of an open ended well-type manometer.
    Attached to one end of the manometer tube is a well (a reservoir with a higher crossectional
    area compared to that of the tube). The manometer is filled with a fluid with an SG = 2.95.
    The manometer tube has a diameter of 5mm and the well is cylindrical with a diameter of
    50mm.
    Initially, with the pressure on both sides of the manometer is atmospheric. The fluid level is
    consequently the same on both sides of the manometer. A gauge pressure of 2.1 psi is
    applied to the well side of the manometer.
    a) Calculate the difference in height of the fluid levels on each side of the manometer.
    b) Calculate the amount by which the fluid level on the well side is depressed. Calculate the
    amount by which the fluid level on the low pressure side rises.

    ysUdbqX.png

    2. Relevant equations

    ΔP = ρgh

    3. The attempt at a solution

    h = ΔP/ρg
    h = 2.1 psi to pas / 2950 kg/m^3 * 9.81
    2.1 psi to pas = 14478.99 Pa

    So, h = 14478.99/2950*9.81
    a) h = 0.5 meaning the difference in height is 0.5m?
    b) Not sure what exactly the answer requires. Does it require the height in respect to an algebraic representation of the height of the water initially?

    Cheers.
     
  2. jcsd
  3. Oct 19, 2014 #2

    haruspex

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    seems right, but you should provide an appropriate number of decimal places, even if the trailing digits are zeroes.
    It asks how much of the 0.5m difference is due to the level in the well going down and how much is due to the level in the tube going up. Hint: what do you think the ratio of those two quantities will be?
     
  4. Oct 19, 2014 #3
    Thank you for replying :) Well seeing as the narrow section is 5mm wide and the wide section is 50mm wide it could be stated that ten times the volume could fit in the wide section meaning the radio would be- 1n:10w (narrow, wide). So for 500mm

    50mm narrow to 450mm wide meaning that the fluid level on the well is depressed by 50mm and the fluid level on the other side is increased by 450mm?
     
  5. Oct 19, 2014 #4

    haruspex

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    You need to think about that again.
     
  6. Oct 21, 2014 #5
    Herpaderp, just realized I had to find the ratio of the who heights from v = 2pir^2 for each side. Thanks for the help, I am pretty sure I have that right now. I do apologize for asking another question a similar question has stumped me. Another question states that:

    A U-shaped tube is partially filled with water. One end of the tube is then capped
    so that the air in that end is trapped (initially at atmospheric pressure). The tube has a bore
    diameter of 10mm. The initial height of the entrapped air gap is 200mm. A gauge pressure of 1500 mmHg is then applied to the other end of the tube.

    And asks

    Explain how the function of this system is different in principle to the function of a
    closed ended manometer used to measure absolute pressure.

    Seeing as one end is closed and by knowing the height difference and density of the liquid the the pressure could be calculated. The only difference I could see if that the pressure measured would be at gauge but you could just subtract 1atm in pascals from the calculated result.
     
  7. Oct 21, 2014 #6

    haruspex

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    Are you suggesting that the trapped air will be at 1 atm throughout?
     
  8. Oct 21, 2014 #7
    One end of the tube is then capped so that the air in that end is trapped (initially at atmospheric pressure) hmm. That does imply that the pressure on that end will change and not be at an atmospheric pressure meaning the absolute pressure at the other end can not be calculated. I assume when it can be the open end of the tube is at 1atm and the other end is where the calculated pressure is.
     
  9. Oct 21, 2014 #8

    haruspex

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    It can be calculated, if you make certain assumptions about how the volume of the trapped air will change.
     
  10. Oct 21, 2014 #9
    I am not familiar with pressure in liquid solutions but I suppose it makes sense that the volume would change if the 1atm gas was sealed as the pressure would change and Pv = NRT meaning v = nRT/P (if the pressure increases then the volume decreases). You could state that the difference is that the volume of the sealed air would decrease meaning that a different calculation would have to be performed.
     
  11. Oct 21, 2014 #10

    haruspex

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    Yes, that's what I had in mind. But what assumption do you need to make (or extra measurement take) to apply that equation?
     
  12. Oct 22, 2014 #11
    I have just realized I have not done the second question correctly which was:

    A U-shaped tube is partially filled with water. One end of the tube is then capped
    so that the air in that end is trapped (initially at atmospheric pressure). The tube has a bore
    diameter of 10mm. The initial height of the entrapped air gap is 200mm. A gauge pressure of 1500 mmHg is then applied to the other end of the tube.

    Well the volume of the entrapped gas is 200mm * 2pi(10^2) = 125 663.706 millimeters^3 or 0.000125663706 m^3. The pressure was 1atm or 101 325 pascals. I have no idea what to do with that information.
     
  13. Oct 22, 2014 #12

    haruspex

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    You quoted the equation PV=nRT. If the pressure is increased to 1500mm Hg, what will happen to the volume? What do you need to assume does not change, in order to use that equation?
     
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