Centrifugal Vacuum: Is Mach 1.0 the Limit?

[blainiac]

I was thinking about how a person could create and maintain a vacuum at sea level. I was thinking of any supersonic objects (bullets, etc.), and that a vacuum (or partial) may be directly behind the object as it travels, as the air wouldn't have time to move into replace the void faster than the object. I'm sure this creates tremendous drag also.

Anyway, let's say we had a centrifugal fan. Ignoring any leaks, let's say we have a large one that was 1.0 meter in diameter. I say a large one, as we're trying to get the blade tips to the speed of sound. Also, let's say we prevented air from flowing in the normal sense by blocking the inlet, so no new air can flow into replace the air that's being moved unless it comes back.

Let's say we spin this fan at 5000 RPM somehow (blade speed close to the speed of sound).

So, if no new air could rush and replace the vacated area, would there be a vacuum in the space between the blades? I would think that as you approach Mach 1.0, that no new air could really enter the space... Is this true?

 

[Simon Bridge]

That's a cute idea - a similar one is to use such a fan to rotate the air in a chanmber so the centrifugal effect creates a radial air-pressure gradient. The pressure at the center would depend on the speed of the fan and the temperature of the air.

The place to start in understanding this sort of effect is to define what you mean by "vaccuum", which is your target. How low does the air pressure have to get?

The fan will produce turbulent air motion, as does the bullet, there is no actual vacuum in the sense of "the vacuum of space" or the kind of thing you can get just by pumping a lot of air out of a chamber.

 

[blainiac]

Thanks for the reply Simon.

My goal is to see if it is possible to lower the pressure on the outer side of the ring from atmospheric in the picture as much as possible. I originally was studying cavitation on metal surfaces in liquids, and thought something similar to the image attached would work in creating a constant cavity in water. These thoughts naturally lead to if it's possible to create a very low pressure gradient between blades in air at sea level, and if going faster than Mach 1.0 would indeed sustain this low pressure.

I also was thinking something of this design would become more efficient as it rotates faster. The 'blue' areas are low pressure areas.

 

[Simon Bridge]

Cavitation in water involves bubbles forming in the water - you have reasoned, by analogy, that the air equivalent to a bubble in water would be a vacuum, or very low pressure, pocket in air. I don't think the analogy holds.

Note, "a very low pressure gradient" would involve almost no change in pressure over the distances involved.
Presumably you hope for a pressure difference on the order of 1atmos - in a small distance. That would be a very high pressure gradient.

You were thinking that the design becomes more efficient as it rotates faster - but more efficient at doing what? How are you measuring efficiency?
The analogy with cavitation in liquids suggests you are thinking that the wheel takes less energy to maintain it's motion with higher speeds or something like that? This would require the turbulent layer of air around the blades to exact less drag with higher speeds.

I actually don't know. iirc deliberately interrupting air flow can reduce drag in airframes.

Getting above mach 1 should not be difficult - you could do the experiment quite cheaply.

 

[blainiac]

Thanks Simon,

Presumably you hope for a pressure difference on the order of 1atmos - in a small distance. That would be a very high pressure gradient.

Your presumption is correct! I instead meant a high pressure gradient... Thanks for catching that.

To consolidate my query, I'd like to know if the high speed rotation would indeed cause the air to vacate the space between the fins to create a relatively low pressure zone (up to 1 atmos if possible).

I believe air would rush to replace this vacation from above and below the fins as air doesn't move from the center in this design (as it is blocked from doing so). By efficiency, I was thinking there may be less drag when spinning very fast for some reason... I wish to disregard this aspect for now.

Getting above mach 1 should not be difficult - you could do the experiment quite cheaply.

Interesting... can you elaborate on this? Thank you kindly, I appreciate your insight and patience.

 

[Simon Bridge]

Air would try to circulate, yes.
You'll have lots of trouble with seals and containment, and lots of unhelpful turbulence.

I think a 1m diameter fan doing about 120rpm - the fins would be exceeding mach 1.
That's not a terribly fast rotation. You should be able to measure the drag for different speeds and plot a graph.
If 1m is too big for you, how about a 10cm fan doing 1200rpm?

 

[blainiac]

Very interesting, and I agree with the trouble with seals/containment/turbulence. As for the RPM, very interesting! I didn't think it could be so low. I attempted to calculate how radius, tip speed, and RPM were related, and got a much higher RPM. I'll try it again... I hope it is lower.

Mach 1 at STP is approximately:
SOS = 340.9 m/s (linear velocity wanted for tips at DIAMETER)
DIAMETER = 1.0 m
RPM = unknown

RPM*PI*DIAMETER = SOS (m/s)
RPM = SOS/PI*DIAMETER
RPM = SOS/3.14
RPM = 108.56 (for the SOS used)

WOW... That's a very low RPM for the tip speed desired. As for decreasing the size of the fan based on this information, I think that is a very good idea to have a 10 cm fan instead! It appears that the relationship between fan diameter and RPM required to get to Mach 1 is linear. Very interesting... Also measuring the drag for different speeds is a great idea. Once again, I appreciate your responses in helping me understand everything involved. :thumbs:

You are right, it would be inexpensive and fairly easy to play around with, much easier than I initially thought.

Edit: I believe I made an error... I changed the RPM to RPS, is this correct? I believe that's why I thought it may be too low initially.

RPM/60 = RPS
RPM = 108.56*60
RPM = 6513.6 required to reach tip speed of SOS

Hmm, I still hope I'm wrong.

 

[Simon Bridge]

Drat - I forgot the conversion from seconds to minutes: are correct.
Excuse - I'm battling a cold.

$60\text{rpm} = 1 \text{rps} = 2\pi \text{rad/s}$

velocity to diameter is: $2v=d\omega$

My 120rpm should be 7200rpm. A 10cm fan would need 72000rpm
That could be a bit harder to do... iirc A typical 80 mm, 30 CFM computer fan will spin at 2,600–3,000 rpm on 12 V DC power. ... vacuum cleaner and weed-trimmer motors can exceed 10000rpm when not loaded. ... ??
Looks doable.

 

[strive]

Well, this requires a lot of torque to spin at 340 m/s (outer edges), thus low cost electro-motors will not suffice (this thing with a max diameter of 120 mm and 20 mm long blades spun at 900 rps produces roughly 2.4 Nm of torque).

Also the low pressure region will be around 0.6 atm, while if you just block the inlet of a centrifugal compressor you can achieve somewhere around 0.3 atm (at half less torque).

However you could adapt this to lower the necessary torque and produce a small region with much lower pressure... it depends on what you need this for.

 

[blainiac]

Drat - I forgot the conversion from seconds to minutes: are correct.
Excuse - I'm battling a cold.

$60\text{rpm} = 1 \text{rps} = 2\pi \text{rad/s}$

velocity to diameter is: $2v=d\omega$

My 120rpm should be 7200rpm. A 10cm fan would need 72000rpm
That could be a bit harder to do... iirc A typical 80 mm, 30 CFM computer fan will spin at 2,600–3,000 rpm on 12 V DC power. ... vacuum cleaner and weed-trimmer motors can exceed 10000rpm when not loaded. ... ??
Looks doable.

No worries! I'm surprised you're able to comment with a cold! Very interesting by the way. I don't think it would be too difficult to test, especially since that RPM is attainable. The torque required may be more of an issue. Great video share.

Well, this requires a lot of torque to spin at 340 m/s (outer edges), thus low cost electro-motors will not suffice (this thing with a max diameter of 120 mm and 20 mm long blades spun at 900 rps produces roughly 2.4 Nm of torque).

Also the low pressure region will be around 0.6 atm, while if you just block the inlet of a centrifugal compressor you can achieve somewhere around 0.3 atm (at half less torque).

However you could adapt this to lower the necessary torque and produce a small region with much lower pressure... it depends on what you need this for.

Interesting, thanks for the explanation. 0.6 atm would be perfect for my application. I'm just wanting to create a pressure region very close to the wall that is between 0.0 (ideal, but unattainable it seems) and 0.75 atm for my particular interest. I believe I would need a relatively large area (larger than regular impeller inlets) with this relatively low pressure region to work with my test rig. With this design, I'm hoping to maximize this area, and the lower the pressure, all the better to test. I am not sure if any material can be used for the rotating fan at approximately 6,000 RPM at 1.0 m diameter without damaging itself.

I did have a few more questions though. If the device is spinning at 340 m/s at the outer edge, why is the low pressure region not lower than that? I understand that some of the air will circulate with the fan, but I would think that at great speeds, the air simply wouldn't be able to 'move' all the way into that region before being flung out. Is the answer similar to bullets traveling in air? It was mentioned above that there is turbulent air behind a bullet, but not really a vacuum... Is the air directly behind the bullet really able to drag along with it, even past the speed of sound?

 

[blainiac]

My brother has decided to assist me in building a very crude model to play with. Although the radius and RPM won't be nearly high enough for air, I will be testing it in water. I've read somewhere that cavitation can occur in water with velocities that are fairly low (15-25 m/s). With this, I will weld a wheel (from a vehicle) with approximately 30 'blades' onto it. I'm hoping to see cavitation in the spaces between the blades when an outboard motor is attached at shallow depth.

My hope is to see if only water vapor would exist near the middle of the wheel, between the blades to see how blade height, width, and spacing affect everything.

Edit: On another note (for air), I've been looking into what the centripetal acceleration of a fluid would be trying to 'fill a hole' a vacuum appeared in the atmosphere. Is there a maximum velocity towards a low pressure zone from atmospheric?

 

[Simon Bridge]

You realize that cavitation is not strictly what would happen in air right?
In water, the "cavity" is a bubble of vapour and the tendency to call them "voids" can be misleading.

 

[blainiac]

You realize that cavitation is not strictly what would happen in air right?
In water, the "cavity" is a bubble of vapour and the tendency to call them "voids" can be misleading.

Yes, I understand it can be misleading, and for that reason I should use terms such as gradient when referring to air (I've been reading about pressure gradient force, centrifuges, etc.)... In this case, for thought experiment's sake, I was imagining a sudden 'void' created in the atmosphere and the subsequent events. For the machine itself (in water), as I understand it, cavitation would occur if the pressure is lower than the vapor pressure, and will create a relatively low pressure vapor bubble as the fluid is liquid. I really should be careful with my words!

 

[Simon Bridge]

Yah - it can affect your thinking.
i.e. you won't get "bubbles" of low pressure in air like you do in water.

 

[blainiac]

Yah - it can affect your thinking.
i.e. you won't get "bubbles" of low pressure in air like you do in water.

That's because air really doesn't have a defined boundary being a gas and not a liquid, correct?

I see the fan (one operated in air) as having a relatively low pressure closest to the rim (between blades), and having a gradient going from that pressure to atmospheric pressure (or near) near the tips of the fan. I believe that's correct. I've been attempting to see how the centrifugal force moving the air radially outwards interacts with the centripetal force trying to keep the air between the blades... The accelerations involved seem extremely large...

For a 0.5 m radius fan with a blade tip speed of 340 m/s, I get:

a = 340^2/0.5 = 231.2 km/s^2 for the air.

Would atmospheric pressure around the fan (for a fan operated in air) be the centripetal force trying to keep the air between the blades? Would the centripetal acceleration in this case be anything comparable to the centrifugal acceleration?

 

[Simon Bridge]

Surface tension in the water is what gives it the ability to have bubbles - the vapour in the cavitation bubbles comes from a change in state in the surrounding water.

Air is a lot more loosely bound. To work the same thing by analogy, you'd need very low pressures so the much weaker binding forces for the air could maintain a bubble, and the state change (plasma) able to be contained in these bubbles. Basically this is not going to happen.Do not use centrifugal force in the same analysis with centripetal force - centrifugal is where you are rotating with the fan so the blades appear stationary. Centripetal force is what you use in an inertial frame. If in doubt - use inertial frames.

Consider a ball resting against one of the blades - the blade pushes it with a force perpedicular to the surface of contact. The ball tries to move in a straight line ... but the blade keeps contact. The result is the ball rolls towards the outer edge of the blade while the blade goes in a circle (the force making it do that is the force from the blade - there is no centrifugal force in this picture). When the ball reaches some sort of barrier to it's radial motion, the barrier provides a centripetal force on the ball to keep it going in a circle.

For air - pressure gradients take the place of the simple forces discussed here.

The centripetal force is an unbalanced force that results in an acceleration pointing to the center of the rotation.

 

[blainiac]

Thank you Simon for the details. I truly appreciate you taking your time with this thread, it means a lot!

Interesting point about not using the same analysis of centrifugal force with centripetal. I've been thinking that I could view them both similarly, thank you for that clarification!

I believe I will comment back to this thread once I've played with some more details and thinking.

 

[blainiac]

After building test models that are quite a bit different than earlier in the thread and much research, I came across a patent that aimed at doing something very similar to what my brother and I were attempting at this point; that is create a low pressure gradient on one side of a 'wing'. It looks like someone beat us to it in the '90s! :) The patent (now expired) is US Patent 5,328,333. It looks very simple, but is intriguing. The inventor even built a buggy pushed by a styrofoam version.

I was thinking of creating a test rig similar to the one described, but want to know the best way to make the pressure on the upper side of the disc as low possible. I was thinking that increasing the blade height relative to the disc would give sufficient time for the air to 'escape' before touching the disc's upper surface.

I believe I read somewhere that the root mean square velocity of air was around 500 m/s, which is very fast... Is there a way to determine how long blades would have to be to sufficiently evacuate air from between the blades? I was thinking that as long as the time it takes the air to go from the entrance (the top of the blade near the hoop) to the exit (the edge of the blade) is less than the time it takes an air particle to travel at 500 m/s from the top of the blade to the disc surface, everything would be fine. Any help for an ignoramus like me would be highly appreciated. :)

I was going to create another test where I have 90 0.3 m long blades welded on a metal tire rim that is 0.5 m in diameter. I was going to have a sheet metal disc on one side of the rim, so air will be forced to enter the device at the top. All of this on the shaft of a powerful boat motor rotating at around 4000 RPM.

To summarize my question, is there a way to determine what height the blades should be to minimize the pressure on the disc based on the diameter of the disc and the rotation velocity so that the least amount of air particles have 'time' to touch the disc before being flung out?

Attached is an image from the inventor's patent.

 

[strive]

I figure you meant the inner cylinder (the one that blades are attached to).
Speed of sound at ISA is roughly 340.3 m/s. (the root mean square doesn’t help you here)
I will refer to the distance between the inner cylinder and the max outer diameter as the depth of the blades, and to the distance between top and bottom disc as the height of the blade.
Divide the depth of the blades with speed of sound this tells you how long a particle of air requires to travel that distance if there is hard vacuum at the inner cylinder (of course there won’t be hard vacuum, and also as the air starts to move towards it the pressure gradient will decrease and thus also the velocity at which air is moving in will decrease).
Now take the max circumference (the path that the outer edge of the blade travels during 1 rotation) and multiply it with number of rotations per second.
Now take the same circumference and divide it with the number of the blades. Divide this with the result from the previous step and you get the time necessary for 1 blade to travel the distance between 2 blades. If this time is equal or shorter than the time obtained for an air particle to travel the distance of the blade depth you should obtain pressure around (or lower than) 0.1 Atm.

The height of the blades plays some role in this… but that is more complex. You will get somewhat higher pressure near the blade tips.

In case blades cannot reach velocities near or higher than 0.8*speed of sound you can simply multiply air density with a square of (blades velocity*sind(blades back sweep angle)) and divide this by 2. This is the total pressure of the air that is being flung out. Subtract this from atmospheric pressure and you’ll get the pressure at the inner cylinder.
Note that increasing the blades back sweep angle beyond 45° won’t work linearly (you’ll need a modificator).
Also note that as the air density inside decreases the total pressure decreases, thus the low pressure field will fluctuate.

None of this is accurate (and is highly simplified) it will only serve to shorten the testing period.

 

[blainiac]

Thank you very much for your helpful reply Strive. This is exactly what I've been trying to come up with, but haven't been able to. You're my hero. I'm about to go to work, but I'll pour over it once I'm back. Thanks again!

 

[strive]

No problem.
Just bear in mind that high rotational frequencies will incur high structural stress.

 

[blainiac]

Once again, thank you Strive.

I finally had some time to look over it, and I really appreciate the time you took. I had some questions naturally.

I used your equations and came up with some very nice results. I've been trying to think along those lines to answer some other questions. I was curious to look at the other 'entry' point as well, where the air can enter from above. Is there a reason the speed of sound is used and not the rms? I'm just curious. Naturally at first, I assumed the speed of sound was to be used, but other sources on the internet suggested the rms.

I've attached some pictures to illustrate what problems I'm trying to solve in addition to what you provided. The top-left picture is what the invention describes, and with that blade height, air doesn't have to travel far to reach the bottom before it is then flung out. With the lower picture, with increased blade height and less max circumference, air has to travel a longer distance down.

1) If the air has to travel longer vertically (takes longer to touch disc), would that help lower the pressure on the top side of the disc?

2) I'm assuming the blade depth must not be too shallow, otherwise the air will have less distance to travel to reach the inner circumference (where the cylinder is) as well?

3) Once air enters from above, would the air get dragged around with the blades on the fan for a ways, or does it move in a fairly straight line and eject quickly (where B occurs)?

I think that if I could answer these few questions, even simply, I would be extremely satisfied! In the images, I'm just concerned with the vertical pressure on the disc for simplicity and general inquiry (not the stresses of the inner cylinder at this point).

I really appreciate your feedback, more than you'll ever know! After this, I'll keep any questions to a minimum!

 

[strive]

Well speed of sound depends on the speed at which air molecules move (rms) but they are not held tightly together thus it takes some time for them to interact.

To answer your questions:
1.: well with given blade velocity you can really only get to a certain pressure differential. Increasing the blade height will help but only to a certain point. In order to calculate the exact pressure differential and best blade height you would have to approach this numerically.
If anyone has a better approach, please correct me.

2.: if the time that takes air to reach the inner cylinder is lower than the time necessary for 1 blade to cover the distance between 2 blades, you will have considerably lower average total pressure between the blades in comparison to max total pressure at the blade.
To properly describe this, it takes a few pictures and a page of text… so again this is an approximation.

3.: the air will move outward with the blade_velocity*sind(blade_back_sweep). Just calculate how long it will take it to exit and multiply that with angular velocity of the blades.

The stresses come from centripetal forces. Do not underestimate them if you intend to spin the blade tips at M˃1. Here is something to get you going in that direction: http://www.ewp.rpi.edu/hartford/users/papers/engr/ernesto/poworp/Project/4.%20Supporting_Material/Books/32669_04.pdf

Here the questions just start.

 

[blainiac]

Strive,

Thank you for taking your time clarifying the previous post and answering. It is much appreciated! Based on this information, I wrote a program to take inputs (RPM, max circumference, cylinder circumference, blade depth, disc area, etc., as well as some other important variables) and show results based on the information you've provided.

Although you stated that it is very approximated, I was impressed with the results. The inventor built two test rigs that were rotating at different shaft speeds with a powerful electric motor driving the fan. I plugged the inputs in based on his build specifications (eyeballed them from the videos) and compared his conclusions with the formulas, and was happy to see the results matched almost identical to the two tests he conducted. Very interesting!

Based on the information also, I can also see why blade height is not as important as I thought initially (although still important). It's interesting to see the difference the blade back-sweep has on the performance of the device.

You are correct in saying this is where the questions begin. The link is very helpful (still looking at the wealth of information there), I understand attempting to construct a device rotating that quickly will have high safety concerns, and will be of utmost importance. I am still thinking of the construction and materials best suited to this application that are within budget.

I'm starting to explore some of the information in the patent in more detail, especially the part where the inventor states that "as the foil reaches high speeds, thus disallowing air from replacing the discharged air, the drag due to the fins is reduced to a very low level because there is less and less air for the fins to strike. This low drag results in extremely high efficiency."

With your experience, would you have any further advice or insight into the device (safety is concern number one! I won't be standing in the plane of rotation any time soon!)? I am trying to think of modifications, but at this point design seems fairly straightforward (although building it without it flying apart and injuring someone would be another story!).

All in all, it is a very interesting/unique device. Once again thanks again for the useful information. It's invaluable to building a future fun project. I'll keep this thread updated with any building/questions.

 

[PressureFoiler]

Interesting discussion! Any progress since September 2014, blainiac? :)

 

[blainiac]

Interesting discussion! Any progress since September 2014, blainiac? :)

Yes, there has been a lot of progress. I will be building a large unit soon, but I have to wait until funds become available. Throughout the thread discussion, the formula below describes roughly how much thrust can be produced.

The thrust is in N, pressure is in Pa (N/m^2), rho is air density, theta is the angle for each blade from the root. I will post more when I can. I have written a Navier-Stokes fluid simulation to refine some aspects, but it's complicated. PM me for more details.

 

[PressureFoiler]

Yes, there has been a lot of progress. I will be building a large unit soon, but I have to wait until funds become available. Throughout the thread discussion, the formula below describes roughly how much thrust can be produced.

The thrust is in N, pressure is in Pa (N/m^2), rho is air density, theta is the angle for each blade from the root. I will post more when I can. I have written a Navier-Stokes fluid simulation to refine some aspects, but it's complicated. PM me for more details.

---
Blainiac,

Very nice! In your equation, is rho the density of air? Theta is the angle the fins make with the base (or inner diameter of fin connection point), yes? Wait, I see... You're using Bernoulli Navier-Stokes -- That is, you're assuming that the pressure will decrease based upon the increase in velocity of the air near the rim due to the fins striking the air, yes? Are you willing to question whether Bernoulli applies here? And are you willing to attempt to simulate this phenomenon using only Newton's laws instead? If so, perhaps maybe we can help each other... :)

Steve Quinn
Inventor - Rotating Thrust Producing Apparatus U.S. Patent 5,328,333 (expired -- the patent, not me... yet!)

 

[PressureFoiler]

Well speed of sound depends on the speed at which air molecules move (rms) but they are not held tightly together thus it takes some time for them to interact.

To answer your questions:
1.: well with given blade velocity you can really only get to a certain pressure differential. Increasing the blade height will help but only to a certain point. In order to calculate the exact pressure differential and best blade height you would have to approach this numerically.
If anyone has a better approach, please correct me.

2.: if the time that takes air to reach the inner cylinder is lower than the time necessary for 1 blade to cover the distance between 2 blades, you will have considerably lower average total pressure between the blades in comparison to max total pressure at the blade.
To properly describe this, it takes a few pictures and a page of text… so again this is an approximation.

3.: the air will move outward with the blade_velocity*sind(blade_back_sweep). Just calculate how long it will take it to exit and multiply that with angular velocity of the blades.

The stresses come from centripetal forces. Do not underestimate them if you intend to spin the blade tips at M˃1. Here is something to get you going in that direction: http://www.ewp.rpi.edu/hartford/users/papers/engr/ernesto/poworp/Project/4.%20Supporting_Material/Books/32669_04.pdf

Here the questions just start.

Strive states it well. Notice that he does NOT use Bernoulli's ... "Higher velocity over a surface equals lower pressure on that same surface". I believe his approach is right... But what do I know, I've been working this thing for over 25 years and still haven't got it off the ground. Help! :)