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Fluid Dynamics of two water reservoirs

  1. Feb 13, 2012 #1
    1. The problem statement, all variables and given/known data

    The two water reservoirs shown in the figure are open to the atmosphere, and the water has density 1000 kg/m3. The manometer contains incompressible mercury with a density of 13,600 kg/m3. What is the difference in elevation h if the manometer reading m is 25.0 cm?

    p12.13.jpg

    2. Relevant equations

    Bernoulli's Equation

    Atmospheric pressure + 1/2(density of fluid)(speed^2) + (density)(gravity)(change in height) = constant

    Continuity Equation

    (density)(speed)(area) = constant [all fluids]

    (speed)(area) = constant [incompressible fluids]

    3. The attempt at a solution

    I have no idea.
     
  2. jcsd
  3. Feb 13, 2012 #2
    "I have no idea."

    You must show some ideas else we cannot assist you. We are not permitted to solve it for you - forum rules.

    Hint: Think about how pressure relates to depth of fluids.
     
  4. Feb 13, 2012 #3
    I'm going to guess I have to start with the mercury. It's not open to the atmosphere so I should be able to ignore atmospheric pressure at that point.

    Going by depth the pressure on the left side of the mercury "u" should be (density)(gravity)(depth)= 33354 Pa?

    I could be going off here, I'm not really familiar with manometres.

    Edt:

    Googled a manometre, turns out that's the pressure difference that I've calculated.

    Would I be able to solve for h if I divide out gravity and the density of water?

    Doing so yields 3.4m as h, and the answer is actually 3.15m. So that's out. I think I need to go back and factor in atmospheric pressure.

    Subtracting atmospheric pressure from Δp gives me 3.38m, closer, but not the answer and I doubt that's what I should be doing anyway.

    Δp=p2-p1 I assume, and p1 must just be atmospheric pressure since it's at the top of the water. p2 should just be atmospheric pressure + ρgh, but that just rings back to 3.4m. So no, I'm missing something.

    Well they're both on the top of the water, so that line of thinking is flawed. I'd make a guess to say the pressure difference might refer to the difference between the top of the water and the depth h in one side, but I've already tried that and it didn't work out.
     
    Last edited: Feb 13, 2012
  5. Feb 14, 2012 #4
    You need to write an equation for the pressures at two points. Draw three horizontal lines.

    Line 1 is at the top of the lower reservoir's surface. It should extend to the left leg of the manometer. Line 2 is at the water mercury interface in the right leg of the manometer and it should also extend to the left leg of manometer. Line 3 is at the mercury water interface on the left leg and it should extend to the right leg. Label the distance from line 2 to line 1 as d. In your picture h should now be the distance from line 1 to the top of the upper reservoir. The distance between lines 2 and 3 should be 25 cm.

    With line 3 the pressure must be equal where it crosses both legs of the manometer. If the pressures are not equal, the fluid would be moving (Bernoulli's Eqn.)

    Now write equations for the pressures where line 3 crosses the manometer. Equate them. Solve.
     
  6. Feb 14, 2012 #5
    So I'm pretty sure that you're saying that pressure at the top of the left side and the top of the right side of the mercury are equal.

    But I'm not sure how that would help me solve for h or for d.
     
  7. Feb 14, 2012 #6
    Pressures along line 3 are equal.

    Write an expression for the pressure in the left leg at line 3 at the interface. Do the same for where line 3 crosses the right leg. Equate them. You have introduced a new variable, d, but you'll see what happens when you solve for h.
     
  8. Feb 14, 2012 #7
    Pressure along line 3 = (13600kg/m^3)(9.81m/s)(0.25m) ?

    I'm not sure what you're asking me to equate this to.
     
  9. Feb 14, 2012 #8
    The pressure consists of all the liquid above the point. I'm not asking for pressure along the line, I'm asking for pressure where the line crosses the leg of the U-tube. On the left leg you have three components making up the pressure.

    You have 25 cm of hg.
    You have a depth of water extending from line 2 to the surface of lower reservoir called d. d is an unknown.
    You have a depth of water from surface of right reservoir to left reservoir surface. That depth is h, another unknown.

    The pressure at the hg-water interface in the left leg is the sum of the above.


    Now look at where this line, line 3, crosses the right leg of the U-tube. Write another equation for the pressure at that point. It will have two components. Since the pressures where line 3 crosses the U-tube must be equal, equate them and solve for h.
     
  10. Feb 14, 2012 #9
    So (25cm)(gravity)(density of Hg) + d(gravity)(water density) + h(gravity)(water density) = pressure at water-Hg interface in the left tube = pressure at equal point in right tube?
     
  11. Feb 14, 2012 #10
    Yes, you'll see that the d cancels out.
     
  12. Feb 14, 2012 #11
    How do I generate an equation for the point in the right tube?
     
  13. Feb 14, 2012 #12
    Sum the pressures of the sections of fluids that are above it. Same method as you did with left leg.
     
  14. Feb 14, 2012 #13
    (25cm)(gravity)(density of Hg) + d(gravity)(water density) + h(gravity)(water density) =

    d(gravity)(water density) + h(gravity)(water density) ?

    or maybe

    (25cm)(gravity)(density of Hg) + d(gravity)(water density) + h(gravity)(water density) =

    d(gravity)(water density) + h(gravity)(water density) + (25cm)(gravity)(water density) ?

    Should there be a part about atmospheric pressure on the right?
     
  15. Feb 14, 2012 #14
    No, the last term on the RHS is wrong. What is the column of liquid above the intersection of line 3 and the surface of the reservoir on the right? You got part of it correct. It consists of two sections.
     
  16. Feb 14, 2012 #15
    Atmospheric pressure is on both reservoirs so it would cancel out.

    What about the column of Hg on the right side of the manometer?
     
  17. Feb 14, 2012 #16
    (25cm)(gravity)(density of Hg) + d(gravity)(water density) + h(gravity)(water density) =

    d(gravity)(water density) + h(gravity)(water density)

    The term "h(gravity)(water density)" is wrong. It's the last term on the second line of your equation.

    Going up the right leg you have "d(gravity)(water density)" and a mercury (Hg) component.
     
  18. Feb 14, 2012 #17
    (25cm)(gravity)(density of Hg) + d(gravity)(water density) + h(gravity)(water density) =

    d(gravity)(water density) + (25cm)(gravity)(density of Hg)?

    Wouldn't that just cancel out?

    Just looking at it makes me think that I should use

    (25cm)(gravity)(density of water) + d(gravity)(water density) + h(gravity)(water density) =

    d(gravity)(water density) + (25cm)(gravity)(density of Hg)

    Basically the pressure on the left side being caused by 25cm of water as opposed to 25cm of mercury

    Yep, that works out.

    Divide gravity out of everything, cancel the d term, then solve for h=3.15m.

    -------------------------------------------------------------------------------------

    (.25m)(1000kg/m^3) + h(1000kg/m^3) = (.25m)(13600kg/m^3)

    h(1000kg/m^3)= 3150kg/m^2

    h = 3.15m
     
    Last edited: Feb 14, 2012
  19. Feb 14, 2012 #18
    Yes, I misread your LHS from the previous page and indicated it was ok when it was not. I missed that.

    Your second equation is correct. Problem done.
     
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