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Pressure in a manometer

  1. Jan 28, 2014 #1

    Maylis

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    Hello,

    In this problem, and with problems dealing with pressures in manometers in general, I'm wondering if to calculate the pressure at any given point, you just look at what is above that point. For example, at Point C, there is 12 feet of mercury and 12.65 feet of oil on top of it, as well as the atmosphere. Does that mean that to find the pressure at point C, I just add in the ρgh from the 12 ft of mercury and the other ρgh from the oil as well as the P_atm?

    For the pressure at point A, I know that its filled with water up to that point, so the water must be counteracting that 12 feet of mercury and 12.65 feet of oil, but where do I include the actual water pressure itself in here?
     

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  3. Jan 28, 2014 #2

    SteamKing

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    The pressures at points B and C must be the same. Using the dimensions from the diagram, you can work back to find the pressure at point A.

    A quick note about your calculations: fresh water has a weight of 62.4 lbf / cu. ft., not lbm. Multiplication of the amount of water displaced by g = 32.2 ft/s^2 is not necessary.
     
  4. Jan 28, 2014 #3

    Maylis

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    Is my calculation incorrect then? I would then say the pressure of water is equal to the components on the right side i.e. the mercury, oil, and atmosphere.
     
  5. Jan 28, 2014 #4

    SteamKing

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    Numerically, the calculation appears OK, because you have multiplied by 32.2 and divided by 32.2. I was pointing out that your assumed units for the weight density of the water was already in lbf rather than lbm.
     
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