Experimental proof of Venturi Effect

In summary, the Venturi effect has been known for centuries and has been proven and reproven in various experiments. However, the increase in velocity inside a convergent nozzle in case of subsonic flow is a more specific aspect of the Venturi effect that the individual is interested in. They are looking for experimental proofs that show the increase in velocity is proportional to the decrease in cross-sectional area and that the velocity at the exit is close to the theoretical value as per the Venturi effect. While this information may not be readily available in modern literature, the individual may need to conduct their own experiments to satisfy their curiosity. The Venturi effect is a concept with broad applicability to mechanical engineering and is not limited to wind tunnel design and operation.
  • #71
In this video, the proof of Venturi Effect is much clear and ratio is far better.
 
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  • #72
T C said:
In this video, the proof of Venturi Effect is much clear and ratio is far better.
I don't see anything in that video about the Venturi Effect or the area/velocity ratio, with the possible exception of an out-of-focus/unreadable CFD snapshot. What results, exactly are you referring to? What ratio are you seeing?

Anyway, I wasn't trying to optimize accuracy, I was optimizing effort and cost. I'm quite certain you could produce better results than I did if you choose to put some effort into it, but your attempt was a far superior optimization of effort than mine was, so I doubt you will. I'm genuinely confused as to what you are after here (and I don't just mean this subject/thread, I mean your entire effort on PF).
 
  • #73
russ_watters said:
I don't see anything in that video about the Venturi Effect or the area/velocity ratio, with the possible exception of an out-of-focus/unreadable CFD snapshot. What results, exactly are you referring to? What ratio are you seeing?
Just look at the RPM with and without the flaps, that will be good example of velocity increase because RPM will increase in direct proportion to velocity. IMO, the best way get a proper venturi effect demonstration is using plastic or metal cones.
 
  • #74
T C said:
Just look at the RPM with and without the flaps, that will be good example of velocity increase because RPM will increase in direct proportion to velocity. IMO, the best way get a proper venturi effect demonstration is using plastic or metal cones.
It's just so disappointing that after all this effort trying to explain it to you that you think that has anything directly to do with the Venturi Effect. I feel like either you aren't trying at all or you are messing with us here.
 
  • #75
My main motto was to understand why my homemade experiment is unsuccessful and what is necessary to make it successful.
 
  • #76
T C said:
My main motto was to understand why my homemade experiment is unsuccessful and what is necessary to make it successful.
In that case you should re-read this thread from the start and put some real effort into learning the concepts. And for experimenting yourself, I showed you what could be done with a bigger fan, two boxes, a roll of duct tape and virtually no money or effort. If you put in even a small amount of effort and money you could surely do vastly better.

Or maybe better, you should probably enroll in an introductory fluid dynamics class, because you seem completely incapable of self-directed or forum-assisted learning. You may need an absolutely rigid structure that forces you to focus and doesn't allow you to wander off track. Here we just close threads and issue infractions. Maybe failing a test would help you focus.
 
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  • #77
T C said:
In case of a wind tunnel, IMO the main purpose is to create a flow at a specific speed. If a open air blower can generate more speed in comparison when fitted at the throat of a convergent nozzle shaped duct, why should one need that?

I know I'm a little late here, but to clear up the misconception:

The volumetric flow rate, as Russ points out, will nearly always be highest when the fan is out in the open. There might be some way to slightly exceed that with careful design, but really, open air is about as good as you'll get. However, flow velocity will be higher inside a well designed wind tunnel, with a converging nozzle, a test section, and then a diffuser with the fan at the exit of the diffuser. In addition, as boneh3ad correctly states, velocity is not the only goal of a wind tunnel - flow uniformity and straightness, and having an extremely controlled environment are also very important.
 
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  • #78
Is there any reason behind it? If so, what's that?
 
  • #79
T C said:
Is there any reason behind it? If so, what's that?
Conservation of energy, f=ma, etc.
 
  • #80
russ_watters said:
Conservation of energy, f=ma, etc.
Can you explain how the position of the fan (inlet or exhaust) can affect such laws?
 
  • #81
T C said:
Can you explain how the position of the fan (inlet or exhaust) can affect such laws?
It doesn't. The laws govern the flow and the addition of certain elements to the system affects the flow in keeping with the laws. Specifics on particular elements/fittings can be complicated, but much of that has already been discussed.
 
  • #82
So far, there is nothing in this thread that can clearly explain why a wind tunnel performs better when the blower is fitted and the exit instead of the the inlet.
 
  • #83
It doesn't. At least not necessarily. That's more related to keeping minimum turbulence and swirl in the test section, but from an energy and flow velocity standpoint, you could put the fan on either side (but you'd still always want a nozzle and diffuser to maximize performance, not just a fan in a tube).
 
  • #84
Nozzle is a must to get higher velocity at lesser power consumption. I can understand that.
 
  • #85
Both a nozzle and a diffuser. Without the diffuser, the pressure requirement to drive the nozzle is much higher, dramatically reducing the flow.
 
  • #86
cjl said:
Both a nozzle and a diffuser. Without the diffuser, the pressure requirement to drive the nozzle is much higher, dramatically reducing the flow.
Why?
 
  • #87
T C said:
Why?
Air doesn't like abrupt direction changes/transitions.
 
  • #88
T C said:
Why?
While Russ's statement is true, I don't think it really explains this very well. This really comes from 3 factors. First of all, you have to understand what a nozzle and diffuser do. A nozzle exchanges pressure for velocity, a diffuser does the opposite. So, at the entry of a nozzle, the pressure will be high and velocity low, while at the exit, the pressure will be lower and the velocity higher. On a diffuser, the entry will be at relatively high velocity and low pressure, and then as it slows the flow down, the pressure increases such that the exit conditions are a higher pressure and lower velocity.

The second important fact is that at any point where a wind tunnel is open to the atmosphere, the pressure must be equal to ambient pressure. This is because it is open to the atmosphere (admittedly, there are some minor caveats here, but we can largely ignore them for the moment).

Third, as was already discussed previously, fans flow more air (volumetrically) when the pressure change across the fan disk is as small as possible.

Now, taking those three things, let's think about a wind tunnel with only a nozzle. As I already said, the entrance of a nozzle must have a higher pressure than the exit. However, if the exit is open to the atmosphere after the test section, this means that the pressure just after the fan (just before the nozzle) must be higher than ambient, in order to drive the flow through the nozzle. This means you have to have a significant pressure gradient across the fan, reducing its flow. Having the fan on the exit doesn't help you that much either, since now the exact same logic still applies, except that the pressure just before the fan must be below ambient (since the entry to the nozzle is at ambient), again requiring a significant pressure jump across the fan.

However, now imagine a system with a nozzle and a diffuser. The air enters the nozzle, then loses pressure as it gains velocity. It then traverses the test section, and then in the diffuser, it loses velocity and the pressure rises again, and it is then exhausted to ambient. This means that the exit of the nozzle now doesn't have to be at ambient pressure - it can be below ambient pressure, since the flow will gain pressure in the diffuser. If we're still imagining a setup with a fan on the front, and you have the same pressure ratio across the nozzle, this means that less pressure is needed at the entrance to the nozzle than would be needed without the diffuser, since now the nozzle exit is actually below ambient pressure. Because of this, the pressure ratio across the fan can be much smaller, allowing it to flow more air (or, alternatively, to flow the same amount of air with less power).

This also still applies with an exhaust fan setup - basically, allowing the test section to operate below ambient pressure allows for a much smaller pressure change across the fan disk, improving operation substantially.
 
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  • #89
cjl said:
However, now imagine a system with a nozzle and a diffuser. The air enters the nozzle, then loses pressure as it gains velocity.
Seems like it would gain pressure in the forward direction, but maybe lose perpendicular pressure.
 
  • #90
paradisePhysicist said:
Seems like it would gain pressure in the forward direction, but maybe lose perpendicular pressure.
Define "perpendicular pressure." Also define "pressure in the forward direction." Given that pressure is a scalar quantity, I have no idea what you mean here.
 
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  • #91
boneh3ad said:
Define "perpendicular pressure." Also define "pressure in the forward direction." Given that pressure is a scalar quantity, I have no idea what you mean here.
Pressure equation is 2d (p=f/a) therefore it can have forward or perpendicular.

In the post I quoted there is a cone, nozzle where the water is flowing through, forward pressure I am referring to is the water streaming out the front of the cone, perpendicular is the pressure on the side walls of the cone.
 
  • #92
Pressure is isotropic though - that is to say, it is the same value in all directions. Yes, the standard definition of pressure is f/a, and that area needs to be in a particular direction, but one of the fun things about (static) pressure is that it acts equally in every direction.
 
  • #93
cjl said:
Pressure is isotropic though - that is to say, it is the same value in all directions. Yes, the standard definition of pressure is f/a, and that area needs to be in a particular direction, but one of the fun things about (static) pressure is that it acts equally in every direction.
Hmm, NASA says that dynamic pressure is of a moving gas, or in this case I'd assume liquid such as water.

"we can define a pressure force to be equal to the pressure (force/area) times the surface area in a direction perpendicular to the surface. If a gas is static and not flowing, the measured pressure is the same in all directions. But if the gas is moving, the measured pressure depends on the direction of motion. This leads to the definition of the dynamic pressure. "
https://www.grc.nasa.gov/WWW/k-12/rocket/dynpress.html

I of course am not a seasoned expert and new to all this but this is just my take.
 
  • #94
paradisePhysicist said:
Pressure equation is 2d (p=f/a) therefore it can have forward or perpendicular.

In the post I quoted there is a cone, nozzle where the water is flowing through, forward pressure I am referring to is the water streaming out the front of the cone, perpendicular is the pressure on the side walls of the cone.
Force has direction, but it does not inherit that direction from the pressure, but from the surface being acted on by pressure. In equation form,
[tex]\vec{F} = p\vec{A} = pA\hat{n}.[/tex]
Pressure is a scalar. Force is a vector and, when related to pressure, comes from a scalar pressure acting on a surface that has direction.

paradisePhysicist said:
Hmm, NASA says that dynamic pressure is of a moving gas, or in this case I'd assume liquid such as water.

"we can define a pressure force to be equal to the pressure (force/area) times the surface area in a direction perpendicular to the surface. If a gas is static and not flowing, the measured pressure is the same in all directions. But if the gas is moving, the measured pressure depends on the direction of motion. This leads to the definition of the dynamic pressure. "
https://www.grc.nasa.gov/WWW/k-12/rocket/dynpress.html

I of course am not a seasoned expert and new to all this but this is just my take.
Dynamic pressure (##q##) is effectively the difference between static (thermodynamic) pressure (##p##) and stagnation pressure (##p_0##). If a fluid is at rest, they are the same. If a fluid is moving, then it must be slowed down in order to reach stagnation, so ##p## rises and ##p_0>p##. The difference between them in an incompressible flow is ##q## and is the kinetic energy per unit volume in the fluid, or
[tex]q = \frac{1}{2}\rho v^2.[/tex]
This is Bernoulli's principle, ##p_0 = p + q##.
 
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  • #95
boneh3ad said:
Force has direction, but it does not inherit that direction from the pressure, but from the surface being acted on by pressure. In equation form,
[tex]\vec{F} = p\vec{A} = pA\hat{n}.[/tex]
Pressure is a scalar. Force is a vector and, when related to pressure, comes from a scalar pressure acting on a surface that has direction.Dynamic pressure (##q##) is effectively the difference between static (thermodynamic) pressure (##p##) and stagnation pressure (##p_0##). If a fluid is at rest, they are the same. If a fluid is moving, then it must be slowed down in order to reach stagnation, so ##p## rises and ##p_0>p##. The difference between them in an incompressible flow is ##q## and is the kinetic energy per unit volume in the fluid, or
[tex]q = \frac{1}{2}\rho v^2.[/tex]
This is Bernoulli's principle, ##p_0 = p + q##.
Basically I was saying that, if there is a tube of flowing water, the front is going to have more force and thus more pressure. An example would be a flowing river, if you are standing at the front of the river you are going to have much more pressure to deal with, but if you are sitting on the river bank and put your foot in the river the water pressure (the perpendicular pressure) will be much less.
 
  • #96
paradisePhysicist said:
Basically I was saying that, if there is a tube of flowing water, the front is going to have more force and thus more pressure. An example would be a flowing river, if you are standing at the front of the river you are going to have much more pressure to deal with, but if you are sitting on the river bank and put your foot in the river the water pressure (the perpendicular pressure) will be much less.

What you are describing is the concept of stagnation pressure. If a fluid (say, water) is flowing at a constant speed throughout, it has both static pressure and dynamic pressure. Static pressure is still a scalar and acts equally in all directions. If you were to non-intrusively measure the pressure in the flow, you would measure static pressure. It's the pressure felt by an object in the flow that is moving along with it and is how you would calculate the force on an immersed body.

Dynamic pressure is also a scalar quantity (it has the magnitude of velocity squared in it). This is the "extra" pressure you get when a fluid is moving, but the key here is that you never "feel" dynamic pressure.

What happens if you put an obstacle in the flow (such as yourself) is that the velocity of the fluid slows to 0 as it encounters you, which converts all of that dynamic pressure back into static pressure. At the very front end of you at the stagnation point, all dynamic pressure has been converted to static pressure. The static pressure at that point is therefore equal to the stagnation (or total) pressure, which is the constant in Bernoulli's principle. In essence, you still feel static pressure, but the static pressure has risen due to the change in flow conditions near your body.
 
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