Physics of Airtrack Operation, specifically, how an airtrack causes the glider float

adamxrt

thanks for the replies. that seems like the best logical structure to approach from, SC.

rcgldr. If i understand you correctly, pressure loss occurs when the air particle is still in the cutout of the hole. I was aware of that. There is then another second pressure loss as it passes out of the hole just above the surface of the track, again which i was aware of. But you are saying theres a sub loss after this? I dont fully understand how that comes to be? is it something to do with roughness of the glider surface?

rcgldr

The pressure of the air a short distance away from the edges of a slider is ambient. Assuming the slider is "floating", you have a pressure gradient than decreases from higher than ambient towards ambient as you get closer to the edge of a slider. The issue with the radial flow outwards and perpendicular to a hole is that the cross sectional area of the flow increases by the radius from the center of that hole (cross sectional area = 2 π x radius x height), so mass flow per unit area decreases with radius from a hole. This means you need sufficient mass flow as well as pressure to keep a slider "floating". I assume hole size and density of the holes are also factors.

adamxrt

to anyone reading

Can anyone postulate or think of a relationship or equation for an air gap which will output a friction value? Obviously i want friction low as possible.

adamxrt

This is really getting to me! I get a really a lower pressure inside the track than i have under the glider when i use the method my instructor gave me!!!!!!


Can i just use the density of air as 1.2 the whole time? or will it be different at different pressures. See im assuming incompressible flow!

heres what im doing roughly.



Using this i got a ridiculously small pressure around 250pa which isnt right. Im going to redo eveything AGAIN for the upteenth time and try and upload a picture of my working out. Am I on the right track? Or can anyone point out any flaws?


Can anyone help with equations? And if i have done anything stupid please point out vigourously, im not the most intelligent of people but im trying! Im not going to coast this year!!!

sophiecentaur

When you say that you get less pressure inside the track than under the slider, what do you actually mean? It seems to me that you must have calculated some things in the 'wrong order'.
There is bound to be a formula which gives an indication of the volume of air needed to produce a given pressure under the slider, based on squirting air through a slot (all the way round the edges). Using this value of pressure and the required volume of air, with another formula, you can then work out the pressure on the inside of the (covered) track holes which will achieve this flow. This pressure has to be greater than the pressure under the slider for air to flow in the direction you want. This internal pressure must be sustained whilst losing air from the covered holes plus all the uncovered ones- which tells you the total capacity of blower needed.

Whatever the formulae are, there is a logic that tells me that, if you apply them in the right order, you could never have less pressure inside than under the slider.

Stick at it. You'll get there in the end.

rcgldr

Yes, the required pressure won't be so much more than ambient that density will change enough to affect your calculations.

Getting back to the main problem, it would seem you need to determine the air flow rate between track and slider to achieve the required pressure to get the sliders to hover, then work your way back determining how much pressure (and minimal flow rate) is required to acheive that flow rate through the holes in the track.

Another potential issue is that between track and slider, the pressure will be greater and flow rate somewhat slower than the rest of the track.

sophiecentaur

They will appear as two 'parallel loads', as an electrical analogy - but at least they will appear as constant as the slider moves (if the slider is an integral number of hole spaces long). But the 'direct' path is only relevant when working out the blower capacity that's needed. (One thing at a time.)

I just wonder how much the tutor who set this problem actually knows just how open ended (no pun intended) this assignment is. Did he ever give anyone else the same assignment in the past, I wonder?

adamxrt

Im not sure he does, although its is to be expected, as final year projects are supposed to be very challenging.


rdgldr, I will attempt that, thought the problem is i cannot arrange bernousllis equation to fit/cant find the appropriate equation that describes the air motion inside the airtrack. During my education on thermodynamics, we only ever looked at pipe flow issues and head loss etc, nothing as abstract as this.


I also have another method to try. It as been suggested that i try using the "impulse of a jet " as im told the situation is very similar to that of air leaving a nozzle.
So, tonight, I belong to the girlfriend (necessary evil) and am off to a rugby match, but tomorrow night i have/ infact i will have time and wont be doing anything tomorow night so I will be back to continue this.


Also if anyone is interested, once the project gets fully under way i will continue this thread as a progress update on the build with pictures etc, so keep tuned if you are interested in momentum physics and collisions! I just need to solve my flow problems first!

sophiecentaur

I don't think the air flow along the pipe and inside the track is too significant. It's a massive chamber compared with the small exit holes and the pressure is not likely to vary much along its length.
The main, and very relevant, unknown is the flow of air under the slider and the friction there. All the other bits are much more straightforward.

This all seems a bit unfair unless you are allowed to do some practical measurements along the way. No serious engineering project would get away with a purely theoretical treatment before getting down to building something. It just wouldn't make commercial sense.

adamxrt

well guys, im very happy to say i have a solution for this problem. Please see below, im going to get it checked tomorrow.

=D

let me know if it looks dodgy...because it probably is! this is all my own work, apart from being given a structure to follow. All equations looked up and checked by myself.


Assuming incompressible flow. density of 1.2

Attached Files

Initial Airtrack calculation unchecked.doc (95.5 KB, 5 views)

adamxrt

Tutor liked the problem solution, though he says there are some minor mistakes, and my initial static velocity its far too big! Over all good news though, will redo and repost!

sophiecentaur

adam
I spotted a dodgy bit as soon as I saw your calculated air velocity (407m/s). The speed of sound in air is only 330m/s and that would imply some pretty extreme conditions in your track!! Sorry, but is supersonic flow likely?


The required (excess) pressure under the glider seems right @ 260Pa but I don't see how you get your next step at all (which gives you that unreal answer). That 'v²' equation springs up out of nowhere and what does it relate to? You seem to be assuming that air is flowing from a region of Atmospheric Pressure, in an uncontrolled way, into a region of 260Pa pressure. That, surely, is not what's happening. The pressure under the slider is Atmospheric Pressure PLUS 260Pa. In which case the air flow would, at least, be outwards.

You still seem to be tackling this the wrong way round. What is important is that there should be sufficient pressure under the glider as air leaks out. So, surely, you first need to be working out the airflow through the gap underneath the glider in order to maintain your 260Pa excess pressure. I suppose your equation could give you an airspeed under the edges of the slider which could be the right sort of value. The actual volume of air being moved per second would be the 'exit slot' CSA times this new velocity (area times velocity is volume flow rate). Here, you need to choose a ground clearance value.

Then you have to think about how many holes you have under the slider and the air flow required for each, with AP+260PA at their exit. Google Airflow through Orifice for some info. This is easier to find out about because it's 'gas stove engineering', which is a much more common field of study. Once you have done this, you can find just how much air needs to be pumped into the track, to maintain this pressure with a hundred or so little holes in it - same 'orifice' calculations.

Do you appreciate why I say it should be done this way round?

[Edit: I'm not sure your tutor has his eye on the ball if he doesn't spot that pressure error! Don't tell him I said so, though]

adamxrt

Yeah i appreciate what you are saying! I compiled that solution at 4am last night. I have no idea how I didnt pick up on the 407 being too big myself, put things into perspective, im unwell feel like crap, and had a nosebleed during one of my lectures earlier!! tutor was all over me about it aswell.


Though i am still embarassed about that value. No engineers should miss something as glaring as that ever!

he says my method seems to be right though? the v squared equation is rearranged from bernoullis equation. He himself suggested this 2 weeks ago, and i went away , rearranged it, came back and he said it was the right starting point. what i got wrong was i thought the 260 had to be taken away from atmosphere, when it is ALREADY the pressure difference (P-P(absolute)). SO

that V^2 is now, Sqrt(2*260/1.2), which gives me a V value of =20.817m/s!

And like you say, i then use Q=VA to find the airflow under the glider, this value will pretty much define everything...although thats what i was doing anyway in that solution? May I politley ask if you read the entire document, sophiecentaur? I do include this and also number of hole and ride height into the problem. Maybe it snot laid out that well.

But yes, i am using air flow through orifice equations too, that is where i got the impact of a jet and the mass flow rate equation later on in the document.

Im currently reworking the whole problem now, have class though very soon so ill not have it up until tonight maybe tomorrow.

sophiecentaur

Hi
No I didn't read it thoroughly after I came to your 'little gaff' haha. You're not the first to fail to make a reality check. Me too, frequently.

Can you take me through this jet equation? Even there, you end up with a velocity of 300m/s. Is that realistic? Wouldn't a real system with those speeds involved make a huge amount of noise? 150holes! But, if the volume of air in equals the volume of air out, it should end up moving somewhat faster through the holes.

I think that vsquared equation must be too simplified for finding the air flow. It assumes laminar flow and also that pressure is the same all over the bearing surface. This won't be true. There will be more pressure away from the edge (hence, less pressure where you are applying that volume flow calculation). The air in the centre will stay there because its pressure is only a bit higher than the air next to it. (That earlier practical comment of someones, about needing no holes up there confirms what I say. There will be a steep gradient of pressure, I think, between holes and edge. As a reality check, we wouldn't dream of putting the holes right near the edge, would we? So the pressure must be different all over the underside.

I have no idea what the right solution is but I can't even believe an exit speed of 20m/s is right, now I think of it. That's 72km/hr! It would blow your hat off - and, as for the air coming out of the holes at 300m/s, that would drill holes in your skin I think.
This is turning out to be quite hard ain't it?

btw, If you are shifting 0.012m³ at 3.42kPa, doesn't that represent over 4kW? Sorry but I think that's right.

cjl

20 m/s sounds believable to me. That's not that fast, and it will dissipate fairly quickly coming out of the holes (especially due to the small jet diameter). As for the power required? The change in volume during compression is only going to be about 3% (assuming standard sea level conditions), so as a really quick estimate, the change in volume will be 3% of 0.012m³/s, and the average gauge pressure will be about 1.8 kPa. This gives a power required of less than 1 watt. Admittedly, this is a very quick and dirty estimate, but it should at least give a decent order-of-magnitude guess.

(300 m/s definitely sounds too high though - I definitely agree about that)

sophiecentaur

It isn't the change in volume that counts, surely. You are pushing all the air out with that pressure difference. Your argument would imply that an incompressible fluid would take no power to pump across a pressure difference.

cjl

True enough. I was in a hurry, and accounted for the compression work but not the flow work. Compression work, as I said before, is less than a watt, so that can safely be neglected. Flow work is still only 41 watts though (0.012 m³/s * 3.42 kPa).