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Bernoulli's equation misunderstanding

  1. Aug 12, 2014 #1
    1. The problem statement, all variables and given/known data

    A pump forces water at a constant flow rate through a pipe whose cross-sectional area, A, gradually decreases. At the exit point, A has decreased to 1/3 its value at the beginning of the pipe. If y= 60 cm(the distance from point 1 to the exit, where point 1 is where y=0) and the flow speed of the water just after it leaves the pump, which is point 1, is 1m/s what is the gauge pressure at point 1?

    I have the solution, but I don't know why a certain part of the solution is the way it is. More specifically, why is P at point 1 - Patm = to the gauge pressure. I know that the gauge pressure= pressure total - pressure at surface.

    2. Relevant equations
    Here is the solution.

    P1 + 1/2density*v1^2 = Patm + 1/2density*v2^2+density*gravity*y2(60cm)

    Then for some reason, which I do not know why P1 is the total pressure and we subtract
    P1-Patm = density*gravity*y2+1/2density*(3v1)^2-1/2density*v1^2

    The answer for the gauge pressure is 10^4 Pascals.

    Thanks Physics Forums!


    3. The attempt at a solution
     
  2. jcsd
  3. Aug 12, 2014 #2

    Simon Bridge

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    A pressure gauge compares the pressure on each side of a membrane.
    So if one end is attached, say, to a tire, and the other is open to the air, then $$P_{gauge}=P_{tire}-P_{air}$$

    ... I don't understand the question.
    I don't understand this description ... where is point 1? At the exit or at y=0 at both?
    Is the pipe horizontal or vertical?
    Is the narrow end the exit?

    Anyway - the key to understanding gauge pressure is to check where both ends of the gauge are connected.
     
  4. Aug 13, 2014 #3
    Hi Simon, sorry for the confusion. Here's an attempt to clarify my question, the pipe starts off horizontal, and the p1 is the horizontal reference where it is equal to 0. The pipe exit is the narrow end, and it 60m from the horizontal.

    I'm still having some trouble seeing the gauge pressure, but I think the total pressure is due to the pump. Thank you Simon for all your help!
     
  5. Aug 13, 2014 #4

    Simon Bridge

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    Where what is equal to 0?
    The height? The horizontal distances measurement (distances to the left are negative)? The pressure? The diameter?

    Anything that is "from the horizontal" would normally be measured starting perpendicularly to the horizontal ... so is this a height? Consider what an angle "from the horizontal" means.

    If you cannot describe a situation clearly, then you will have trouble working out the physics.

    The gauge pressure does not depend on what causes the pressure.
    The "gauge pressure" is simply the pressure reading on the pressure gauge.
    The pressure gauge simply shows you the difference in pressure between the two ends.
    So if P1 is the pressure at p1, and Px is the pressure at position x, then the gauge will show you
    Pg=Px-P1.
     
  6. Aug 13, 2014 #5
    Ahh I understand now. Lastly, can you get a negative pressure gauge, or should I always make pressure gauge positive?
     
  7. Aug 14, 2014 #6

    Simon Bridge

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    You can get a negative gauge pressure - if the probe pressure is lower than the reference pressure.
    It is just like using a voltmeter - it's just that common gauges are built to have the reference end in the air (by your hand say) by default so you only see the end you poke into something, like a tire.
     
  8. Aug 14, 2014 #7
    Thanks Simon! This helped a lot!
     
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