Applying Bernoulli's Equation to a Siphon

In summary, the equation is used to calculate the flow rate of a siphon where the pressures at the surface of the water and at the exit point are different.
  • #1
bbar
4
0
I'm not sure if this is the right place to post this (because it isn't homework), but it could easily be a homework question.

This wiki page, http://en.wikipedia.org/wiki/Siphon#Explanation_using_Bernoulli.27s_equation", shows how to apply Bernoulli's equation to a siphon where both the surface of the water and the exit point of the siphon (point "c" in the diagram) are at the same pressure.

My question is this: how do you use the equation if the pressure at the surface of the water is different than the pressure at point c?

Thanks,
Brent
 
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  • #2
bbar said:
I'm not sure if this is the right place to post this (because it isn't homework), but it could easily be a homework question.

This wiki page, http://en.wikipedia.org/wiki/Siphon#Explanation_using_Bernoulli.27s_equation", shows how to apply Bernoulli's equation to a siphon where both the surface of the water and the exit point of the siphon (point "c" in the diagram) are at the same pressure.

My question is this: how do you use the equation if the pressure at the surface of the water is different than the pressure at point c?

Thanks,
Brent

Put the appropriate pressures into the Bernoulli equation. The wiki page that you cited had Patm as the pressure on both sides of the equation. It doesn't have to be the same pressure at both locales.
 
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  • #3
I understand now. Thanks a lot for you help.
 
  • #4
Another question. If I solve for VC and I'm ending up with a negative square root, does that just mean the siphon won't flow?

For example, the height of the starting point of my siphon is at, say, x=0. The end of the siphon is at x=-20m (20 meters below the start point). Also, the pressure at the starting point is standard (101 325 pascals) and the end point is at a pressure of 302 225 pascals. Let's just assume that [itex]\rho[/itex]=1.

So (02)/2 + (9.8)(0) + 101.325pa/1 = (VC2)/2 + (9.8m/s)(20m) + 302225pa

Solving for VC, I get [itex]\sqrt{}-402 192[/itex]

Is that right? Thanks again for all of your help.
 
  • #5


Hello Brent,

Thank you for your inquiry. I can provide you with some insights on applying Bernoulli's equation to a siphon.

Firstly, let's briefly review Bernoulli's equation. It states that the total energy of a fluid system remains constant along a streamline, and is expressed as:

P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2

Where P is pressure, ρ is density, v is velocity, g is the acceleration due to gravity, and h is height.

In the case of a siphon, the fluid (in this case, water) is flowing from a higher point to a lower point, driven by the difference in pressure between the two points. The equation is typically applied assuming that the pressure at the surface of the water (point "a" in the diagram) is equal to the pressure at the exit point of the siphon (point "c" in the diagram). This is because, in most cases, the siphon is open to the atmosphere at both points, and the pressure at these points is therefore equal.

However, if the pressure at the surface of the water is different than the pressure at point c, the equation can still be applied. In this case, the pressure term (P) in the equation will be different for each point. The other terms (ρ, v, g, and h) will remain the same, assuming that the fluid is the same and the heights are measured from the same reference point.

To determine the pressure at each point, you will need to consider the factors that can affect pressure, such as the depth of the water, the density of the fluid, and any external forces acting on the fluid.

In summary, Bernoulli's equation can still be applied to a siphon if the pressure at the surface of the water is different than the pressure at the exit point. You will just need to take into account the different pressure values at each point when solving the equation.

I hope this helps answer your question. If you have any further inquiries, please feel free to ask. Keep up the curiosity and enthusiasm for science!

Best regards,
 

1. What is Bernoulli's Equation?

Bernoulli's Equation is a fundamental principle in fluid dynamics that relates the pressure, velocity, and elevation of a fluid at any given point in a system.

2. How does Bernoulli's Equation apply to a siphon?

In a siphon, Bernoulli's Equation is used to explain the flow of a fluid from a higher elevation to a lower elevation without the use of pumps or external energy sources. It states that the sum of the pressure, kinetic energy, and potential energy of the fluid remains constant throughout the siphon.

3. What are the key components of Bernoulli's Equation?

The key components are pressure, velocity, and elevation. These variables are interrelated and any changes in one will affect the others.

4. How is Bernoulli's Equation applied in real-life situations?

Bernoulli's Equation has many practical applications, such as in the design of airplane wings, in the flow of water through pipes, and in the operation of carburetors in vehicles.

5. Are there any limitations to using Bernoulli's Equation in a siphon?

Yes, there are limitations to using Bernoulli's Equation in a siphon. It assumes that the fluid is incompressible, inviscid, and that there is no energy loss due to friction. In reality, there will always be some energy loss due to factors such as turbulence and viscosity.

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