Roman Inverted Siphons and Fluid Dynamics

In summary, the pressure in the pipe at the bottom of the valley is 2 atmospheres. This is found by using the Bernoulli's equation and converting all pressure units to pascals. By canceling out the velocity terms and substituting the known variables of density and height, we can solve for the pressure at the bottom of the valley.
  • #1
NexPhil
4
0

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!
 
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  • #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?
 
Last edited:
  • #4
NexPhil said:
(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
 

Related to Roman Inverted Siphons and Fluid Dynamics

1. How did the ancient Romans use inverted siphons for fluid transportation?

The ancient Romans used inverted siphons, also known as reverse siphons, to transport water over long distances without the need for pumps. These siphons were made of lead pipes and were placed underground to carry water from higher elevations to lower elevations. The siphons relied on gravity to create a vacuum that pulled the water through the pipes, enabling the water to flow continuously.

2. How did the design of Roman inverted siphons contribute to their efficiency?

The design of Roman inverted siphons played a crucial role in their efficiency. The siphons were constructed with a downward slope that allowed water to flow continuously through the pipes. They also had air vents at the highest points to release trapped air and maintain the vacuum. The use of lead pipes also reduced friction, further improving the flow of water.

3. What is the role of fluid dynamics in understanding the functioning of Roman inverted siphons?

Fluid dynamics is the study of how fluids, such as water, move and interact with their surroundings. It plays a crucial role in understanding the functioning of Roman inverted siphons. The siphons rely on the principles of fluid dynamics, such as Bernoulli's principle and the conservation of energy, to create a vacuum and maintain the continuous flow of water.

4. How did the construction of Roman inverted siphons impact their society?

The construction of Roman inverted siphons had a significant impact on their society. These siphons allowed the Romans to transport water over long distances, enabling them to supply water to their cities, towns, and farms. This led to the development of advanced agricultural practices, improved sanitation, and better living conditions, ultimately contributing to the growth and prosperity of the Roman Empire.

5. What lessons can we learn from the use of Roman inverted siphons in modern engineering and fluid dynamics?

The use of Roman inverted siphons in modern engineering and fluid dynamics provides valuable lessons. These ancient structures demonstrate the effectiveness of using gravity and natural forces to transport fluids, rather than relying on expensive and energy-consuming pumps. They also highlight the importance of understanding fluid dynamics in designing efficient and sustainable fluid transportation systems.

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