How does the siphon method work and why does the pipe width matter?

  • Thread starter Thread starter shalikadm
  • Start date Start date
  • Tags Tags
    Bernouli
AI Thread Summary
The discussion centers on the siphon method of water removal, specifically addressing the significance of pressure and pipe width. It is noted that if the pressure at point C equals the vapor pressure of water, the siphon ceases to function due to boiling. The diagram referenced illustrates that water from symmetrically placed holes will hit the ground at the same point, emphasizing the importance of flow dynamics. Concerns are raised about using a large diameter siphon with a smaller outflow tube, suggesting that height is more critical than width for maintaining flow. Additionally, if the pipe is too wide, bubbles can disrupt the flow, potentially causing the siphon to fail.
shalikadm
Messages
63
Reaction score
0
1. I found this diagram on book but there weren't any description.can someone tell me, what its trying to tell specially by those two red lines meeting the ground at the same place...?
2013_05_23_15_32_52.jpg

2.this is a diagram for siphon method of removing water. I have read somewhere that the siphon stops if the pressure at point c is equal to the vapour pressure of water. what is the reason for that.?
Pc = Po - hρg
2013_05_23_17_02_26.jpg
 
Last edited:
Physics news on Phys.org
Handwritten in a book, with no descriptions? Uh huh...

The first question would be asking which pair of colored arcs accurately represent the path of water pouring out of that structure.

For the question on siphons, if the pressure at C equals or is lower than the vapor pressure of water, it boils and no longer can flow up the siphon.
 
Well, I guess it is just trying to tell you about the water flow. There doesn't seem to be anything aphysical about the diagram at least at first glance and without making any measurements or calculations. The two red lines indicate that the water flow from points symmetrically placed about the center will hit the ground at the same location.
 
physwizard said:
The two red lines indicate that the water flow from points symmetrically placed about the center will hit the ground at the same location.
Yes...so what's the physics behind that ? water drains from symetrically placed holes from the center will catch the ground at the same place...?

russ_watters said:
The first question would be asking which pair of colored arcs accurately represent the path of water pouring out of the structure
all arcs are paths of water pouring.red color shows some special feature..
russ_watters said:
it boils and no longer can flow up the siphon.
further...siphon needs a continuous flow of water.water bubbles disturb the continuous flow...am i correct?
 
Last edited:
Apologies if this thread is not exactly germane to my issue but... I have been unable to find information and comment re. the following: If I were to establish a large diameter siphon and have the outflow diameter reduced substantially by attaching a much smaller (but contiguous) tube, would I then be able to have an uphill endpoint for my water flow because the total water mass (and therefore the force exerted) in the "downhill" tube would far exceed that in the smaller "uphill" tube?
 
For this purpose it is height alone that matters, not pipe width. You get the same amount of suction with a 1/8 inch reed on the outlet side as you do with a 36 inch pipe.

There is a concern that if the pipe is too wide, bubbles can rise up the outflow faster than water carries them down. If this occurs, the water can drain out of the outflow until the siphon no longer works at all.
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Thread 'Recovering Hamilton's Equations from Poisson brackets'
The issue : Let me start by copying and pasting the relevant passage from the text, thanks to modern day methods of computing. The trouble is, in equation (4.79), it completely ignores the partial derivative of ##q_i## with respect to time, i.e. it puts ##\partial q_i/\partial t=0##. But ##q_i## is a dynamical variable of ##t##, or ##q_i(t)##. In the derivation of Hamilton's equations from the Hamiltonian, viz. ##H = p_i \dot q_i-L##, nowhere did we assume that ##\partial q_i/\partial...
Back
Top