Roman Inverted Siphons and Fluid Dynamics

  • #1

Homework Statement

"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 ____.

Homework Equations

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

Continuity Equation
A1V1 = A2V2

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 canceled 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
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?
  • #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?
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  • #4
(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)
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.
  • #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.
  • #6
Remember that there is already ambient pressure at the top of the pipe.
  • #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 canceled 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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