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Mercury fluid physics problem

  1. Dec 11, 2006 #1
    I really honestly believe that I worked this out correctly, and explained my logic.. but I just wanted to have this checked to be on the safe side. Thanks a lot for helping, I appreciate the time everyone puts into reading this ^^;;

    1. The problem statement, all variables and given/known data

    Mercury is poured into a U tube in which the cross sectional area of the right limb is three units larger than the left one. The level of mercury in the narrow limb is 30 cm from the upper end of the tube. How much will the mercury level rise in the right limb when the left limb is filled to the top with water?

    2. Relevant equations
    (p_w)(g)(h_w)(A_1) = (p_m)(g)(h_m)(A_2)
    Where m = mercury, w = water, p = density, g = 9.8 m/s^2, h = height, and A = area
    (h_m) = (p_w)(h_w)(A_1) / (p_m)(A_2)
    30 cm = .30 m
    Density of water = 1000 kg/m^3
    Density of mercury = 13,600 kg/m^3

    3. The attempt at a solution
    (h_m) = (1000 kg/m^3)(x)(.3m) / (13,600 kg/m^3)(3x) = .007m
    The x's cancel out.. unit left remaining is just m on top because you want height, which is in meters.
    Logically speaking, since the limb on the right is 3x as large the height of the mercury will not be as great as .30 m on the other side because the area is larger. So .007 m sounds like a reasonable number. Also considering the density of mercury is much larger than that of water, it makes sense that not a very large height at all is needed to hold all the mercury.:rolleyes:
     
    Last edited: Dec 11, 2006
  2. jcsd
  3. Dec 11, 2006 #2

    OlderDan

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    What would you say about the pressures on the two sides of the tube at the level of the bottom of the water column in the smaller side. Are they the same? If not, how are they different?
     
  4. Dec 11, 2006 #3
    I got 0.007537688 metres. Or 0.008m I guess, I worked it out by saying the weight of the displaced water equalled the weight of the displaced mercury. I'm assuming by upper end of the table you meant top of the tube?
     
  5. Dec 11, 2006 #4
    Oh, I made a typo.. it should say tube, not table.. sorry!

    At the bottom of the water column on the smaller side..
    Well if the whole left is eventually filled with water, and mercury is pushed to the right, I would assume that there is some sort of difference or at least interaction in pressure between the water and mercury
     
  6. Dec 11, 2006 #5
    Oh, and I didn't take into account 1.013 x 10^5 atm of pressure because both sides should experience it..
     
  7. Dec 11, 2006 #6

    OlderDan

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    Where does the following equation you posted earlier come from?

    This equation does not give you equal pressures, and finding a level where the pressures are equal is essential in this problem. Equilibrium is obtained when the pressure on both sides of the tube is equal at some level where both sides contain mercury. The highest point where that is true is at the top of the mercury on the water side. The additionl pressure from the water column on the one side must equal the additional pressure from the mercury above this level on the other side. I cannot see where you have equated those pressures in your analysis.
     
  8. Dec 11, 2006 #7

    OlderDan

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    That's correct. You do not need to consider the atmospheric pressure.
     
  9. Dec 11, 2006 #8
    I believe the equation I got from earlier was related to Pascal's principle... I got it directly from my class notes.

    I thought about using Bernoulli's equation..
    P_1 + .5pv_1^2 + pgy_1 = P2 + .5pv_2^2 + pgy_2 but bebcause I dont have a v (velocity) to use then I couldn't solve it.
    I know that a_1v_1 = a_2v_2, so perhaps I could use that to solve the for a missing velocity, though that would still give me an unknown variable..

    not to mention I dont have P_1 or P_2, so that's even more unknown variables.
     
    Last edited: Dec 11, 2006
  10. Dec 11, 2006 #9
    I understand up to this point.

    Why is this true? Mercury and water should have different pressures?
     
  11. Dec 11, 2006 #10

    OlderDan

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    The equation you wrote is equating two forces being applied to pistions of different areas. I don't know the context of your class notes, but what you need in this problem is to equate two pressures, not two forces. You start out with the top of the mercury at the same level in both sides of the tube, and you have 30cm of space above the mercury in the narrow side. Then you pour in some water until the narrow side is filled. As you do this, the mercury level in the narrow side drops by some distance we can call d, and the mercury on the wider side moves up a distance we can call D.

    In terms of d and D:
    How high is the column of water?
    How much higher is the mercury in the wider side than the mercury in the narrow side?
    What is the relationship between D and d?

    The pressure at the top of the mercury in the narrow side must be the same as at a point at the same level in the wider side. (This is consistent with Bernoulli's equation, but since nothing is moving you don't need to pursue Bernoulli's equation to do the problem.) In the narrow side, the pressure at this level is due to the column of water above it (and the atmosphere). In the wider side, the pressure at this level is due to the column of mercury above it (and the atmosphere). You have to equate the pressure due to the mercury column on one side with the pressure due to the water column on the other side.

    If you answer the questions above, you will have the heights of these two columns in terms of the information given in the problem and the variables d and D. Which of these is the answer to the question being asked in the problem? Equate the pressures, relate these variables, and solve the problem.
     
  12. Dec 12, 2006 #11
    argh.. I had all these equations written out neatly using bold to emphasize my text according to your guidance with quotes but my internet window randomly shut off on me and I lost all the text! I'm so upset...

    I said that, in terms of pressure, P_1 + p_wgd = P_2 + p_mgD. They should be proportional, in that as one height decreases the other should increase, and if one pressure decreases the other pressure should increase
    The column of water, or d, therefore, must be d = (P_2 - P_1 + pmgD)/(p_wg)
    The mercury on the wider side , after water is added, should be similar..
    D = (P_1 - P_2 + p_wgd)/(p_mg)
    As D increases, d should decrease and vice versa.

    In terms of pressure, P_1 (the pressure on the left) = P_2 + p_mgD - p_wgd and P_2 = P_1 + p_wgd - p_mgD

    Am I going about this right before I continue?
     
  13. Dec 12, 2006 #12
    And congrats on your award ^^ You well deserve it
     
  14. Dec 12, 2006 #13

    OlderDan

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    Make a careful side by side sketch of your tube before and after the water is poured in. Before the water is added, the mercury levels are the same on both sides because the only thing above the mercury is the atmosphere. The narrow side has 30cm of empty tube above the mercury.

    When the water is poured in, the pressure at one level is equal on both sides of the tube only at levels that have mercury on both sides. Extend the line from the no-water sketch across the with-water sketch. The mercury level in the wider side has moved above the line by D and the mercury level in the narrow side has moved below the line by d. Draw a horizontal line at the top of the mercury on the narrow side. The height of the water column above this line is not 30cm because the mercury level has dropped. What is the height? In the wider side, the height of the mercury column above this line is not D because D is the distance from the line in the left sketch, not the new line in the right sketch.

    You have to get these distances right before you can solve the problem. There is a simple relationship between D and d, but you are not seeing it. The sketch should help.

    Thanks for the congrats. Glad I can be of some help.
     
  15. Dec 12, 2006 #14
    Okay.. I think the drawing should help

    "The height of the water column above this line is not 30 cm because the mercury level has dropped. What is this height?"

    The height should be d - 30 cm.

    "the wider side, the height of the mercury column above this line is not D because D is the distance from the line in the left sketch, not the new line in the right sketch."
    The height should be 30 cm + D

    Are these correct now? I drew the diagram and extended the lines and the two equations I just typed seem to be consistant. Now should I compare the pgh's on each side?
     
  16. Dec 12, 2006 #15

    OlderDan

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    Sorry, but no they are not right. The water column is more than 30cm high when the narrow side is filled, and D has no direct connection to the 30cm. D is whatever it has to be to contain the mercury that was pushed out of the narrow side into the wider side.
     
  17. Dec 12, 2006 #16
    I'll try again then..
    seeing as how water in the left will be lower than before, that means its height must have increased, in that case it's d + 30
    Therefore, when the water on the right rises, it should be (1/3)(d+30) because its area is 3x as large. I know you said that 30 cm has no direct connection to D, but the d + 30 is the height increased on the left side. If it increases that much, D must increase .333 times that much.
     
  18. Dec 12, 2006 #17

    OlderDan

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    You are getting close. d + 30cm is correct for the narrow side, but the distance the mercury level moved in the narrow side is d, not d + 30. So D is?
     
  19. Dec 12, 2006 #18
    You mean in the latter part of the sentence the mercury level moved in the wider side is d, not d+30? Let me think why.. both sides originally started at the same place.. if the mercury level goes down d distance (d+30 from original) D has no "30" to start with.. oh okay, now it's clear. In that case, it's only d like you said. D is therefore d/3.
     
  20. Dec 12, 2006 #19

    OlderDan

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    OK good. Now you know the height of the water above the "new line" at the bottom of the water level is d + 30cm, and you know D = d/3. At the level of this new line the pressure is the same on both sides. The pressure produced by the d + 30cm of water must be the same as the pressure pruduced by the mercury above this line in the sider side. What is the height of the mercury in the wider side relative to this line?
     
  21. Dec 12, 2006 #20
    On the right side, the pressure is pgD/3..so D must be 3P/pg ?
     
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