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Roman Inverted Siphons and Fluid Dynamics

  1. Jan 24, 2015 #1
    1. The problem statement, all variables and given/known data
    "Inverted siphons were used by the Romans to cross some valleys. If the height difference between the top of the valley and the bottom is 10 meters, the pressure in the pipe at the bottom of the valley will be ____.

    2. Relevant equations
    Bernoulli's Equation
    P1+(pv^2)/2 + pgh1 = constant

    Continuity Equation
    A1V1 = A2V2

    3. The attempt at a solution
    I know that the density of water is 1000 kg/m3, g = 9.81 m/s2, and the height is 10 meters. I substituted these numbers into the Bernoulli's equation. However, based on the continuity equation, I cancelled the velocity in the Bernoulli's equation. The pressure on top should be 1 atmospheres due to air pressure.

    (1)+(1000)/2+(1000)(9.81)(10) = P2+(1000)/2+(1000)(9.81)(10)
    501 + 98100 = P2 + 98100
    P2 = 501

    At this point, it is obvious that is not the answer. The answer was 2 atmospheres.

    Thanks for reading!
  2. jcsd
  3. Jan 24, 2015 #2


    User Avatar

    Staff: Mentor

    Can you explain your terms in your implementation of the Bernoulli's equation?

    (1)+(1000)/2+(1000)(9.81)(10) = P2+(1000)/2+(1000)(9.81)(10)

    What are the units involved in each term? Are they consistent?
  4. Jan 24, 2015 #3
    (1 atmosphere)+(1000 kg/m3)/2 + (1000 kg/m3 (9.81 m/s2)(10 meters) = P2 + (1000 kg/m3)/2 + (1000 kg/m3)(9.81 m/s2)(10 meters)

    I actually wanted to know if the bernoulli's equation is appropriate for this problem or not. I am only given the height difference and asked to find the pressure at the bottom. Also the mass and velocity should be the same throughout the pipe, should it not?
    Last edited: Jan 24, 2015
  5. Jan 24, 2015 #4


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    Staff: Mentor

    So on the left you have a pressure in atmospheres, a density in kg/m3, and another pressure in Pascals. The terms on the right hand side are equally disparate. That's a big problem.

    You can't cancel a part of a term (velocity in this case) from each side of an equation. You can only cancel whole terms. If the velocity is the same so that the terms as a whole are the same on each side, cancel the whole terms.

    Make sure that your pressure terms are using the same units.
  6. Jan 25, 2015 #5
    I reread the problem again and decided to use a different approach using the pressure formula since I'm only given the height, but know density of water and acceleration of gravity.

    pressure = density of water * acceleration of gravity * height
    pressure = 1000*9.81*10
    pressure = 9810 Pa
    9810 Pa ≈ 1 atmosphere

    However, that still does not round to the appropriate answer unless I double the height.
  7. Jan 25, 2015 #6


    User Avatar

    Staff: Mentor

    Remember that there is already ambient pressure at the top of the pipe.
  8. Jan 25, 2015 #7
    Thank you again for helping. Here's what I have done so far:
    I converted all the pressure units from atmospheres to Pascals and redo the problem using the Bernoulli's equation.

    1 atm = 101,325 Pascals
    I cancelled the whole terms of the (pv^2)/2 from the equation and substituted the known variables.

    P1 + pgh1 = P2+ pgh2

    (101,325 Pascals)+(1,000 kg/m3)(9.81 m/s2)(10 meters) = P2 + (1,000 kg/m3)(9.81 m/s2)(0 meters)

    (101,325 Pascals) + (98,100 kg m2/s2) = P2 + 0
    199, 425 Pascals) = P2
    Convert pascals to atm
    199,425 Pascals = 1.968 ≈ 2 atm
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