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Fluid Pressure

  1. Feb 10, 2014 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    fluid pressure = fluid density * depth * g

    3. The attempt at a solution
    My understanding is that fluid pressure at any two points at the same depth is equal, for a fluid at rest. But because one end is open, the pressure at point C is less than B and equal to A? I don't understand intuitively how this works.

    Any help would be appreciated - thanks!
  2. jcsd
  3. Feb 10, 2014 #2

    Simon Bridge

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    Well at point A - what is exerting the pressure?
    What about at point D?

    Would this be a high or a low pressure compared with the rest?

    Remains to consider B and C ... what is exerting the pressure in each case?
    Is there anything else that would contribute to the pressure?
  4. Feb 11, 2014 #3
    At point A, we have atmospheric pressure, as does point D, so they are equal I think. B has the weight of the water molecules above it, so it experiences more pressure than A and D. I'm just not sure why C is < B when they are at the same depth. Is it because molecules at C are moving but not at B?
  5. Feb 11, 2014 #4

    Simon Bridge

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    That's the one - notice that B is placed conspicuously far away from the entrance to the pipe that C is in.

    You could also look at it like this: The pressure has to drop as you approach D or A.
    C is closer to the low pressure region than B is.
  6. Feb 11, 2014 #5
    What's happening is that the fluid that is approaching the outlet pipe has to get accelerated from the very low velocity away from the pipe (e.g., at point B ) to the higher velocity at point C. To accelerate it, you need an imbalance of force. This is supplied by a pressure variation, from higher pressure away from the pipe entrance to lower pressure within the pipe.
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