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Manometer problem

  1. Oct 11, 2006 #1
    Dear all,
    would anyone be able to help me with the following problem

    In the manometer shown in the figure, the reservoir diameters are 50 mm and the tube diameters are 10 mm. The upper fluid is an oil of density 850 kg/m3, and the lower fluid (hatched) is water. Initially the levels of oil in the two reservoirs are the same, and the levels of water in the two tubes are the same. If a pressure differential of 200 Pa is applied to the air above the oil in the reservoirs, what is then the difference in water levels?
    PICTURE SEE ATTACHED FILE

    Basically, I tried to calculate the change in height on the top rhs and then calculated the height difference at the water on the right hand side since the relation

    h1 = h2 (d/D)^2

    holds. due to the symmetry h on the lhs should be the same and hence dh=2h
    which gives me the wrong result.

    Can anyone please give me a hint? THat would be amazing! Thanks a lot!
     

    Attached Files:

  2. jcsd
  3. Oct 11, 2006 #2

    OlderDan

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    You have not defined h, h1, h2, d, or D and they are not in your picture. You have not said anything about how you calculated the change in the oil levels. Be more explicit, and show what you did to get your answer.
     
  4. Oct 11, 2006 #3

    andrevdh

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    Set up 2 equations for the pressure on the two sides of the manometer.

    One equation on top of the oil in the right arm - with the applied pressure differential. Call the height of the oil in the left arm [itex]h_o[/itex].

    Set up another equation at the top of the water level in the right arm of the manometer. Call the water height in the left arm [itex]h_w[/itex]. Call the height of the oil in the right arm a and in the left arm b.

    Find a relation between a,b,ho and hw. More exact you want a relation for a - b.
     
    Last edited: Oct 11, 2006
  5. Oct 11, 2006 #4
    Please see attached the definitions.

    pressure at C = pressure at D

    p(C) = p(D)

    p(D) = 1000g(x-h1) + h(2)*g*1000 + 200Pa

    p(C) = 850g*25h(1) + (h(2)-25h(1))g1000 + (x + h(1))*1000g

    now p(C) = p(D)

    and solve for h(1):

    200 - 1000gh(1) = 1000gh(1) - 2500gh(1) + 2150gh(1)

    h(1) = 1.09mm

    therefore the difference at the water levels is

    h(1)*25*2 = 54.5 mm

    which is wrong

    correct answer: 111mmm


    Can anyone see the mistake? Thanks so much!

    [Edit: sorry just saw the second reply with the definitions]
     

    Attached Files:

    Last edited: Oct 11, 2006
  6. Oct 11, 2006 #5

    andrevdh

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    Note slight change to previous post.
     
  7. Oct 11, 2006 #6

    andrevdh

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    Were are C and D located in the manometer diagram?
     
  8. Oct 11, 2006 #7
    C and D are at the equilibrium levels on the rhs and lhs respectively.
    Basically I though that the level of the water below the equilibrium level on the rhs must be (50/10)^2=25 times the height difference h1 at the large diameter on the top (see attached file in the above post)
     
    Last edited: Oct 11, 2006
  9. Oct 11, 2006 #8

    andrevdh

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    Note change inpost #3 again (sorry it was a bit confusing). I do not understand why you multiply the heights with the radia of the reserviour for the oil. That will make the pressure equations invalid!
     
  10. Oct 11, 2006 #9
    sorry, I'm confused. What do you mean by height of the oil exactly, i.e. from which point to which?
    Also shall I call the height of the oil on the left arm h0 or a (you ay two different things in your post).

    I thought that it was 25 h1, since the volume change should be the same. Since the density is constant and only the diameter changes I thought that

    h1 (top, rhs) = 5^2*h1 = h2(bottom rhs)

    But anyway, I shall try your attempt - if you could let me know the exact definitions first that would be awesome and I will give it a go!
    Thanks a lot!
     
  11. Oct 11, 2006 #10

    andrevdh

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    Are you trying to set up the equations that I suggested?
     
  12. Oct 11, 2006 #11

    andrevdh

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    The oil level in the left arm is a height [itex]h_o[/itex] above that in the right arm.

    The water level in the left arm is a height [itex]h_w[/itex] above that in the right arm.

    Both these heights are of cause with the applied pressure differential.
     
  13. Oct 11, 2006 #12

    andrevdh

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    The height of the oil above the water in the left arm is b and in the right arm a - again with the pressure differential in effect.
     
  14. Oct 12, 2006 #13
    Thank you. This is what I did:

    relation 0:

    b + h(w) = a + h(0)

    equation 1:

    200 = h(0)*850*g

    This h(0) = 0.024

    equation 2:

    850g*a = h(w)*g*1000 + b*850*g

    (a-b)*850g = h(w)*9810
    a-b = h(w)*1.18


    PUT IN RELATIONSHIP 0:

    a-b = h(w) - h(0)
    0.18h(w) = -h(0)
    h(w) = -133mm


    The correct result is 111mm - so is that just a matter of rounding or did I do something wrong? Also, why did I not have to use the given pipe diameters?
    Again, cheers for your help!
     
  15. Oct 12, 2006 #14

    andrevdh

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    850g*a = h(w)*g*1000 + b*850*g

    You forgot the pressure on top.

    It also seems I misled you a bit...

    The pressure in manometer is the same at the bottom of the manometer. As you move up in the two arms the pressures can be compared on the right and the left if you stay in the same fluid. So the equation above is fine (with the correction), but the comparison on top of the oil surface in the two arms is invalid. As soon as one move into different media up in the liquid the pressure cannot be compared any more due to the different densities of the media.

    How to solve the problem then?

    The volume of water moved in the tube will be the same as the volume of oil moved in the reserviour. This gives you another relation between [itex]h_o[/itex] and [itex]h_w[/itex].
     
    Last edited: Oct 12, 2006
  16. Oct 12, 2006 #15

    OlderDan

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    You do need the diameters because they affect the total depth of oil on each side. The pressure difference between the top and bottom surface of the oil is proportional to the total depth (resevior plus tube)

    andrevdh has made the crucial point I think you are missing.

    The pressures are equal on both sides at any level that has water on both sides. The pressure difference between the top and bottom surface of oil on each side is proportional to the total depth of oil. The total oil depths are different on the two sides, and the left side has an additional depth of water.

    You could use the top of the water on the right as a reference level where you know the pressures in both tubes are equal.

    On the left side, this reference pressure is
    Pleft = pressure change from depth of water + pressure change from depth of oil + pressure of gas on left

    On the right side it is just
    Pright = pressure change from depth of oil + pressure of gas on right

    Pleft=Pright

    I think you understand everything else you need.
     
    Last edited: Oct 12, 2006
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