Like Boiler Injector But Air-to-Air

When used with air, these are generally known as ejectors, eductors, or thermo-compressors. I'm not sure what your question is, but the air is not liquified (not usually, anyway). There is a pretty good article in Wikipedia.

In this concept, a thermo-compressor is used analogously to a boiler injector. It loads atmosphere into a pressure vessel in place of the usual mechanical compressor. Details are in my original post.

The usual objection to using the boiler injector as an analogy for loading an air tank with an air driven jet pump: the boiler injector works because of condensation, I think they call that a two-phase process. Compressed air is well-known for losing temperature quickly, such that an air-to-air jet pump might need 1000 psi drive to get 50 psi output, according to one engineer's off-the-cuff guesstimate. But it occurred to me that, because of the large temperature difference between input drive air and the same air in the reduced section of the nozzle, in the throat, that the air might condense, and there would be a liquid-driven process. Thus a closer analogy to the boiler injector.

Secondary inputs such as heating something somewhere or mechanically pumping air into the suction port of the injector might get results not appreciated by a superficial glance. Re-read the original post, I put a lot of thought into it, but there are a lot of blanks too and I could use some feedback on the concept.

Are you talking about one of these?

I have often thought of how they work as simply being a multi stage venturi booster/pump of sorts.

How is this relative to the Wright Bros flying experiments?

Yes, the picture is a boiler injector, first invented by Henri Giffard about 1858 when he was developing his airship. He was the first person to fly in a controllable machine, 50 years before the Wright Bros. Later he marketed the injector and made a fortune. It was considered impossible because of the paradox. You have a lower than ambient pressure in the injector, sucking water into the device, and yet you have the water being forced into a pressurized boiler. Sounds crazy but it's been standard equipment for over 150 years.

I'm going through a write-up on the boiler injector in a French engineering book by Poincaire, old book. He starts by pointing out that the variables at any point or cross-section of the injector will show the same total amount of energy coming out as what is going in, since there is no opportunity for energy to dissipate. He mentions the principle of conservation of energy. That's all the farther I've gotten, I am very slow to pick up technical discussions, but he says he has a crude way of showing how the paradox is not a problem.

Well obviously it works with steam, and my thought is that the only reason that something similar isn't done with air is that it isn't convenient. I just want to know if it should work, and if not, why not? If 1000 psi cold will make 50 psi, then why can't you heat the drive air and charge a 200 psi tank with it? Never mind the cost of the high pressure air, just assume there's a lot of it laying around, I am interested in whether it is even possible, and I can't see why not. Also interested in whether liquefaction of drive air would be the key to success, since that's how the boiler injector works.

What is the purpose of using air at 1000 PSI to pressurize a tank to 200 PSI with a injector nozzle?

What are you trying to gain here?

Overunity?

You wouldn't happen to be Smokie or DAS under a different user name now would you?

As far as why a injector nozzle works I am rather sure that the principle of how multi stage venturi action works and can transfer energy by trading first stage heat energy and particle velocity to make a greater outgoing pressure but at a lesser volume/flow rate and temperature in a second or third stage has been well studied and explained some time ago.

For basic reference. Venturi Effect

No I'm not someone else, I have only one handle here at this forum. Unlike steam, atmospheric pressure is generated by sunshine so you could bring up overunity if you wanted but I prefer not to, since it will only get me reamed and I was hoping to stay on topic. The main points are the ones I mentioned. Of course I'd be interested to know if there is a gain over conventional air compressors--the whole system is described in the original post--but first things first: is it even possible?

Thanks for noting that the boiler injector is multi-stage. I see it goes beyond the usual jet pump's convergent-divergent shape into a third cone or delivery section, which has an overflow attached to it. I will have to put some thought into this feature as it is no doubt not optional to the function of the boiler injector so my intended concept would have to do something similar or explain why not. If adapting the math and theory of the boiler injector to deal with air.

I understand that the venturi effect has been well explained, however, answers to the questions I brought up are not so easily found; how condensation works into it, I assume the latent heat energy and the density of an ultimately liquid drive have a lot to do with it. The boiler injector is a special case as jet pumps are concerned, but I will re-study venturi effect as you suggest.

Atmospheric pressure is caused by gravity and the mass of the elements that make up our atmosphere. Even if we had no sun we would still have an atmosphere.

I'm sure you know what you're talking about, so maybe you could tell me where I've gone wrong. The way I see it, if there were no sun, the elements making up our atmosphere such as oxygen and nitrogen would be frozen solid, mixed with the soil and rocks of the planet. No atmosphere. Then if the sun came into existence, the solid elements would thaw into liquid form, evaporate into gaseous form, and rise up from the ground to form an atmosphere. What have I left out?

If it was up to the sun we would not have an atmosphere. The solar winds it produces tend to blow atmospheres off of planets unless they have strong gravitational and magnetic fields.

The simple explanation. Why some planets have atmospheres and some don't.

More follow up reading. Atmospheres of the planets. Why We have an Atmosphere Why Mars lost most of its Atmosphere. History of Mars atmosphere.

By the way if you use these above link keywords in online searches you will probably find at least few weeks worth of reading material just on planetary atmospheric science.

So relating to your original post what exactly is your question or discussion of steam ejectors/injectors about anyway. So far most of you are posting makes little point or sense.

Thanks for the info on why planets have atmospheres.

I'm surprised you don't see the point of my post, but I'll try to restate it.

Since the 1850s, injectors have been used to replace mechanical water pumps for the purpose of charging boilers with fresh water. The reason they work is that the drive steam condenses, and the increased density of the resulting fluid, as well as the contribution of the latent heat of phase change energizing the resulting fluid, makes the output water able to put itself into the pressurized boiler. I think we can assume that without condensation, the device would not function.

So why isn't it being done with air, to replace mechanical compressors with injectors?

It is possible to put air into an air tank with an injector driven by water or steam.

Is it possible to put fresh air into an air tank with an injector driven by compressed air? Yes. I know a higher than tank pressure air drive jet would be required, I am assuming this air is available and would otherwise be thrown away. How do I determine how cold the air would get in the nozzle? How much pressure would be needed in the drive fluid? This relates to a point of curiosity: would the high pressure drive air condense due to pressure reduction in the nozzle? Because if so, then an article I am trying to study about "why the Giffard injector does not defy the laws of physics" might be adapted to apply to the analogous process with air.

If I were to reduce this discussion to one question, it is this: how to figure out for myself the temperature that will be reached in the throat of the injector at different drive pressures and temperatures.

Isn't it true that the air cannot flow faster than the speed of sound? Is that what you'd use as a limiting parameter?

I know the combined gas law pv/t = constant or pv^n = constant. Can this be adapted to a dynamics problem? I wouldn't know where to start. I'm thinking the Bernoulli equations might be more like it.

I'm a skeptic about other people's results, if a question is interesting enough, then I want to do the math myself, make a spreadsheet, and see what happens for myself. You're probably right but it would be interesting to see a summary of the process by which you reach your conclusion. Apparently you're saying there's no way an injector is going to lower the air temperature enough to condense the air, ever, under any circumstances. If so, I still would like to know how you know this.

No, it isn't true that the air cannot flow faster than sound. In fact, gas and steam ejectors use convergent-divergent nozzles so that the motive fluid is supersonic. In fact, the injector drawing posted by tcmtech shows a convergent-divergent nozzle.

I was misremembering something I had looked up once. Let me know if this is not true:

Mach one is max speed of fluid through the throat, the smallest part of a nozzle. Fluid won't flow faster than the speed of sound through an orifice. This is called choked flow. After choked flow, if the nozzle gets bigger again, the fluid speeds up. That's backwards to a subsonic throat where pressure goes down in the throat and back up in an enlarging section after that.

(If that is true, then my next question is, how does the air know it's flowing through an orifice and thus can't go over mach one? With me it's always why, why, why. That's why I ask questions.)

What brought this up: assertions have been made that it is impossible for air to condense momentarily in an injector due to the fact that pressure changes to velocity (pressure goes down) in a converging nozzle. No reasons have been given, but I want to know what to look up so I can continue my research into why there is no such thing as an air-to-air thermo-compressor that will cause the drive air to liquefy momentarily at some point of the process, no matter what the pressure, temperature, or flow conditions are. The key word is "why".

I am sure that, when pressure transforms to velocity in a nozzle, temperature will go down a lot. It seems to me that if the drive air is at a high enough pressure, then it has a lot of pressure to lose, therefore can reach a very low temperature. It's my understanding that temperature change due to compression and expansion is instantaneous, unlike pressure change due to radiators and heat exchangers and the like. What is the name of the natural phenomena that make it impossible for the drive air in an air-to-air injector reach -320 degrees F. or whatever is needed to liquefy air?

This is not the same as a process that liquefies air and puts it into a vessel to be transported somewhere and used for something. That sort of thing isn't relevant to my question. Air moves through the injector very quickly and if it gets cold enough in the right place, then it could liquefy for a split second. Like the boiler injector where the same thing happens, though for different reasons.

"Isn't it true that the air cannot flow faster than the speed of sound? Is that what you'd use as a limiting parameter?"

As stated before no not the least bit true. The nozzle on a common air hose can easily pass air at velocities well past supersonic. Also how do you think supersonic aircraft get moving faster than sound while using turbine engines that are basically just huge air compressors?

To me this whole thread is showing that you have a limited knowledge of the physics and properties of a lot of things and are making assumptions based on wrong or incomplete understandings of the related and modern understandings of the sciences involved which for us makes it hard to understand what you are asking implying or wanting to know more about.

Relating to making high pressure air cheap well I have to say that if you can make a 1000 PSI efficiently then you can make any pressure less than that even more efficiently without using the 1000 PSI step! As far as just regulating an air source of 1000 PSI down to 200 PSI thats what pressure regulators do and they do it at near 100% efficiency with no losses of the air itself.

Now relating to knowing things I tend to trust the data that has been provided by the countless number of scientists, physicists, and engineers who have studied these phenomena in extreme depth and have documented their work for the rest of the world to know what they found.

How do I know? Well I trust the people who say it is so!

"...if you can make a 1000 PSI efficiently then you can make any pressure less than that even more efficiently without using the 1000 PSI step..."

Notice I'm talking about putting a compressor inside a pressure vessel. This is not how we want to make all our compressed air, is it? It might be thermally very efficient, but what is the cost difference for special seals, space shuttle precision, etc?

The high pressure air has a specific task, which is to entrain vast amounts of fresh air and put it into a tank. The atmosphere is heated by the sun from 270 deg below zero celsius. That's a lot of free heat, upgraded from ambient to compressed air, thus made expandable, with a compression process that is extremely efficient. For every pound of air compressed to 1000 psi or whatever the drive pressure is, I'd expect many pounds of fresh air to be entrained. I would also provide mechanical pumping to the suction port so that the drive jet doesn't have to overcome the inertia of the air it is pumping.

I agree that there is no reason to make 1000 psi if there is no reason to.

What I'm suggesting is not analogous to a pressure regulator, except to the degree that 1000 psi air, regulated down to 200 psi, will get very cold and absorb ambient heat through any pipe or vessel wall that isn't insulated. The analogy I'm asking about has to do with the boiler injector and reaching condensation temperature in a nozzle.

I will continue to be curious about the "why" behind anyone's assertions. Opinions based on industry hearsay do not constitute science, whether they come from an engineer or not. The progress of science was not completed the day anyone got their degree.

I think that, since you want to know in details the 'Why' of many assertions that have been stated here, you need to take specific courses that relate to the topic you are interested in. This forum is not geared to provide extended lectures in the different disciplines required to understand and explain these assertions.

The different explanations or assertions given are to guide you to where you might find reasons to support or dismiss some of your ideas or questions.

for the Air Liquefaction domain, you need to search / study the tables available that give the different conditions required for the liquefying of Air or any other Gas. If you want to know how these tables were obtained, then you need to go to the theories and lab testing etc... You might get some facts and guidance here...

Actually I didn't ask for Why in details. I specified that I needed a summary of related phenomena or just the names of any phenomena to study, and I asked if Bernoulli equations were the right direction. This is in regards to how I can learn "How cold will air get moving through a converging nozzle when it starts from different pressures and temperatures." I am painfully aware that I will not easily understand the right wikipedia article when I do finally find it, or the right chapter in the right textbook when I find that. That's my problem. If someone would be kind enough to tell me what the chapter is likely to be called, that would be very nice.

I would be very surprised to find what I'm looking for in a book on liquefying gases, because I am not planning to collect, store or transport liquid air. Just take advantage of its transient existence for a split second. However, I will look for a section on nozzle flow in books and articles on liquid air. I will look hard. Can't go to school, too old etc. Tried that several times, got sent back against my will to the same prerequisite elementary algebra every single time, and got nothing out of that but a student loan to pay off the rest of my life. And several easy A's in algebra.

Compressed air in converging nozzle. Gets cold. How cold. The name of a relevant principle to study. Thanks!

Start looking in

en.wikipedia.org/wiki/Vortex_tube

and

www.newmantools.com/vortex.htm

This is about cooling using compressed air ....

you can also google for " Ranque-hilsch effect or vortex tube ".

Sorry, I do not have anything yet on how the ventury system you are contemplating will cool the air, and how to calculate the amount of cooling produced.

You aren't going to let me off without knowing where I think I can get away with making high pressure air for cheap. It's no secret and I don't mind telling you, but this has no bearing on my question, which is about whether a jet can condense air if the right pressure and temperature is introduced into the drive nozzle. The condensation question is a "have to know because I'm curious" and I think I should have a clue but I don't.

This will address your other question, where does the high pressure air come from.

When air is compressed, the work of compression is all lost as heat. To prevent this, steps would have to be taken to conserve the heat. There is good reason to dispel as much of the heat as possible in conventional compressors, because it will approach closer to isothermal compression which is cheaper than trying to compress hot air. If the air is getting hot, it is trying to expand while you are trying to compress it. So lose the heat during compression and there will be more air left over in the cooled tank, for the amount of work done. Otherwise extra work will have to be done to compress hot air and all the heat will be lost anyway after the air is delivered hot out of the compressor.

Less temperature rise per pressure unit is generated at higher pressure ranges, which is interesting though not a key point.

If a booster compressor were running inside the tank or inside a pressure vessel connected to the tank, then the heat that dissipates from it could be conserved in place; it would raise the pressure of the tank. So because of this, the cost of compression could be much lower than status quo engineering expects it to be. I don't know why this isn't done, except it would be a nuisance and most compressors are supposed to be conveniences, not anti-conveniences. Generally, the cost of running the compressor is just the cost of doing business, and the basic theory of air compressors is considered a closed case.

This compressor would take in tank air and boost it through one or two stages to 1000 or so psi, for use by the injector. All the compression work would stay in the tank or be returned to the tank by the injector.

This scenario is presented without apology by someone who does not believe in perpetual motion! If I'm wrong, just give specific reasons, not generalities.

There are examples of extremely efficient processes in engineering.

For example, the boiler injector system is nearly 100% efficient since residual energy in the drive fluid, which is not usable to entrain the suction fluid, returns to the boiler still ready for use. The 3% efficiency of entrainment doesn't mean much since the 96% efficiency of the system as a whole is a matter of simple logic. The injector is not God; it can't destroy energy!

Another example is resistance heaters, which are said to be 100% efficient, but in reality they are 100% INefficient. Their purpose is to waste electricity and they do this perfectly, by changing it all into resistance in the wire, which is heat. But the system as a whole isn't 100% efficient, because you have to generate the electricity first.

So when you talk about efficiency, you have to be specific. WHICH efficiency? Which output divided by which input? Generalizations do not reach me, I am a skeptic about peoples' opinions. The way some people overgeneralize the 2nd law, you'd think that tornados should be impossible.

So I'm claiming that engineering has missed the boat by not putting the compressor in the tank. No doubt you will disagree, and I'm eager to know your exact reasons, since I can't think of any.

Is it possible to use a high pressure air supply to raise a low pressure air source to an intermediate pressure? The short answer is yes.

However, I suspect you are trying to come up with a device which compresses air using only a moderate level of heat, similar to the feedwater injector which uses fire to pump water up to the boiler pressure. If so, that's a completely different situation. In order to do that, you would need liquid air available. Liquid water is freely available, but the cost to liquify air would make such a system impractical.

Your answer supports my interest in the question. My thesis is that liquid air would be required to make the device worth bothering with, if the device were practical. My question is, what is the topic I should research in order to learn why the air won't, and can't, liquefy in the nozzle? Bernoulli equation? Or...?

Thanks for the Penberthy catalog. I think they've been making boiler injectors since the dawn of time when they were invented, or shortly thereafter.

I do have a treatise on why the boiler injector paradox doesn't violate the law. It's in French and it's old, but I'm chipping away at it. The paradox is that it makes vacuum and a split second later it's piercing its way into a high pressure boiler.

Here's a hint I got from someone else today: "Re: liquid air and latent heat of vapourisation...note water has a latent heat of 2000 J/kg and air only 200 (I think)."

If this is true, and I will check on the accuracy of it, then for compressed air's latent heat to have the same energy as steam's latent heat, there would have to be ten times the mass of air compared to steam? I'll have to research this question. It still doesn't address the question of what temperatures can be reached in an air nozzle and how to compute it.

Thanks everyone for taking the time to add to the discussion. I'll let you know what I find out.

On Condensation: The Air flow into the Ventury, from the atmosphere, is due to a pressure drop at the restriction. This air expansion will absorb some energy and cools the inflowing air. But the most important energy absorbing action will be from the evaporation of any water contained in the air, as humidity.

To condense the Air to liquid form will require a much lower temperature than can be reached by the Ventury. Also, the entrained air will re-heat when it mixes with the main stream air. the impact will be negligeable due to the big difference in volumes.

Just a tought.

Thanks. So it sounds like a little humidity will contribute heat energy as fast as a lot of air. I hadn't thought of that.

I'm wondering how to determine what temperature will be reached in the nozzle. I know about pv/t = constant, but that is statics isn't it? I don't know how to proceed, to find out by doing math, how low the temperature will go.

Does the Bernoulli equation deal with this? I think as the nozzle constricts, the air speeds up, its pressure goes down, and its temperature goes down. But I don't know how to express the dynamics mathematically.

Thanks for your opinion.

i got rusted, for approximation you may use air values for n in the pv-t or some tables are made available in the back of a thermodynamics book using enthalpies for gases.

You may also use, 1st law and 2nd law analogy for thise using tables.

it's been a while, knots get tangled sometimes

i bet you're doing ms in UP i guess. Just like my bro before

"It still doesn't address the question of what temperatures can be reached in an air nozzle and how to compute it."

If you know your starting air temp and pressure and the ending temp or pressure you can calculate for the missing value.

Standard formula applies.