Ideal gas law and question about cooliong ...

[vlaiv]

I'm a bit stuck and would appreciate some help on this one.

Don't know much about this part of physics and as a consequence - I found myself rather confused

While contemplating device that would flash freeze fruits and vegetables, I came up with following idea: Take old fridge (just as an insulated container - it won't do any cooling), put one atmosphere pressure release valve (or a bit above one atmosphere) on one side of it, and inlet on the other side of it.

Take 0.5L gas canister capable of storing 200 atm pressure and fill it up. Let it cool to some temperature like 5C. Connect it to inlet and release the gas. It should expand into fridge, and security valve should open up to release (ambient temp) air that was inside.

Common sense says that there will be drop in temperature when air is released from the canister. I'm trying to see what sort of difference in temperature I will get with this procedure (roughly) - hoping for about 40C delta T.

However when I try to fiddle around with ideal gas law - I get exactly 0 delta T - and that is what is confusing me. I'm probably not doing something right, but I can't figure out where I'm making mistake. Here is what happens:

P1 * V1 / T1 = P2 * V2 / T2

As far as I understand, if I fill 0.5L canister with air up to 200 atm pressure - it will contain 100L of air.

This means that 0.5 * 200 / T1 =  100 * 1 / T2

Or T1 = T2 - there will not be change in temperature when I release the gas from canister?

I can see that there is likely error in my reasoning that 0.5L canister will contain 100L of air at 200 atmospheres - but how do I go about calculating temperature drop?

[andrew s]

Have a look at the wikipedia  article "Joule-Thompson effect"

Regards Andrew 

[vlaiv]

Ah, yes, here is excerpt from wiki page on Joule-Thompson effect:

Quote

For an ideal gas, μ J T {\displaystyle \mu _{\mathrm {JT} }} is always equal to zero: ideal gases neither warm nor cool upon being expanded at constant enthalpy.

It also "explains" how to calculate this (no idea where to start with that):

https://en.wikipedia.org/wiki/Joule–Thomson_effect

[vlaiv]

1 minute ago, KevS said:

Vlad, try the following, might be right memory at my age  is problematical.

Can't remember the derivation off the top of my head, but will dig it out if required:

T(2) =P(2) V(2) T(1) / P(1) V(1)

K

Tried that one - and it is for ideal gas - which does not exhibit that behavior.

Both @andrew s and I posted "right solution" for this problem above. Well I still don't have a solution, but at least I know that it can't be derived from ideal gas law.

[vlaiv]

Best I could think of right now is to do some really crude approximation.

Problem is that every quantity depends on bunch of other quantities, and one ends up with really complicated differential equations.

Basis of my crude approximation would be:

- What sort of energy do I need to compress 100 liters of air to 200 atmospheres?

- What is heat capacity of air?

These of course are not constants (which makes things complicated) but let's do some approximations to at least see what sort of magnitude do we get as an answer.

If we assume isothermal compression, then work done is:

Let's say that we compress at 300 kelvins, and R (specific) of air is 287.058 J/kgK. 100L contains about 0.1276kg of air. Work done is: ~58KJ

(Not much energy needed in fact).

Now I'm going to just assume that heat capacity of air is something like 1KJ/kg K. This is really complex topic, but I've found some tables, and this value ranges in 0.7-1.5 depending on what you keep constant (constant volume or constant pressure) and vary other thing. I reckon 1KJ/kgK is a good approximation as any in this case.

So we have about 60KJ of energy, we have 0.1276Kg of air and we need to see how much kelvins can we get out of that

60 * 0.1276 / 1 = about 7.6K?

Not sure if this approximation is any good but it shows that this sort of thing will not get me delta T that I'm after.

 

[vlaiv]

Ok, this is deeply confusing (as is most of the physics if one does not know it )

Here is another approach - adiabatic expansion (probably closer to what will happen):

Taken from this page:

https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/Map%3A_University_Physics_II_-_Thermodynamics%2C_Electricity%2C_and_Magnetism_(OpenStax)/3%3A_The_First_Law_of_Thermodynamics/3.6%3A_Adiabatic_Processes_for_an_Ideal_Gas

We have following:

and hence this:

If we have 200 atm pressure and we quickly release gas to ambient pressure (1 atm) we will in fact have about 17 liters of gas out of 0.5 liter container (Cp/Cv for air ranges from 1.4 up to 1.58 at 100 atm, so I took it to be 1.5 for approximation).

And according to ideal gas law it will be at temperature of about 50K!

That sort of makes sense if PV=nRT then at the same pressure if temperature is five times lower, so it will be volume.

That is seriously different answer from above one.

[tonyowens_uk]

The whole area of heat flows in gas expansion is messy Vlad. First decide whether you want a practical engineering approach to refrigeration-in-a-box or a romp around the relevant thermodynamics. In my practice, the latter follows the former, not the reverse.

Joule-Thompson cooling does not work with near-ideal gas mixtures except in close proximity to the discharging jet - where significant but very localised chilling is obtained. There are lots of other similar options to look at, one of which is the Rank-Hilsch vortex tube. They are actually used occasionally in controls enclosure cooling. Use of a low cost phase changing material e.g. propane/butane mix (AP30 aerosol propellant) is another possibility. But the cooling power of a discharging liquid AP30 jet which flashes from liquid to vapour during passage through a nozzle from 3 bar gauge to ambient is not a lot. And bloody dangerous if there are ignition sources around. There is steam jet refrigeration, possible using water/steam as a refrigerant too, and employing a steam-jet ejector to create near vacuum pressure to evaporate and chill a mass of liquid water. Lots of possibilities. But compressed air, in itself, isnt a good place to start when trying to create useful cooling within a fridge-sized enclosure!

[saac]

For a 1st approximation I would tend to treat it as a reversible adiabatic expansion  so T2/T1 = (p2/p1) ^ (Ƴ-1)/ Ƴ .      Ƴ  (adiabatic heat capacity ratio) for air is approximately 1.4. 

T2 = (278) * (1/200) ^0.286

T2 = 61 K,   but air freezes at around  58K

200 atmo is too high!   In practice you would also need to allow for the efficiency of the nozzle used to throttle the gas so the final expanded temperature would be a little higher. 

Jim 

Edited October 26, 2019 by saac

[vlaiv]

Thanks all for input on the topic.

Luckily, I'm not trying to be very precise about all of that - just wanted approximation to see if there would be cooling with about 40K delta T.

From above two approaches - energy conservation / isothermal process and second one - adiabatic expansion, I think safe bet is that there will be some cooling in range of 8K to 200K delta T (although I'm well aware that this second figure is very unlikely).

It did show however that this topic is rather complicated and that there is no single formula that can be used to predict what will happen with good accuracy - it actually needs a bit more complicated simulation to get close. I think it is easier to actually test it out.

For all interested, here is what I had in mind:

I'll probably move to a new house (yet to be built - currently in phase of project and getting required permits) on piece of land that I purchased recently (yes there will be obsy too ) - and there is a lot of fruit there. That area is known for fruit production. Although you can buy frozen fruits and vegetables in supermarket at decent prices and there is no real commercial incentive to do that sort of thing - I fancied a way to do quick/deep freeze in home that does not involve expensive equipment and dangerous materials (like handling liquid nitrogen and such) - so I came up with above idea - use compressed air to quickly lower temperature. Grow your own food kind of approach (and store it for the winter). Had no idea if it would actually work, but recently I stumbled across these:

(I'm posting links to commercial items - but have no intention of promoting them in any way. If it is deemed against SGL policy - please remove them)

https://www.minidive.com/en/

And also a kick starter project that seems to offer similar thing:

Both of them have 0.5L bottles (that can be filled with hand pump up to 200 atm pressure). That is why I used those figure in above calculations to see if it would work.

In any case - I fancy idea of small scuba units which I can use on holidays , and if I could use it to do above as well - that would be nice (I have not found small pressure tanks anywhere else).

[saac]

Vlaiv I don't think it is as indeterminable as it may appear. Treating it as a simple adiabatic expansion will get you pretty close to the value in practice - nozzle considerations accepting (choking, throat velocity).  Nor is your target of -50 Celsius overly ambitious; for example a common CO2 fire extinguisher (55 bar , 3 litres) will easily get down to that range.  Discharge a CO2 fire extinguisher through a close fibre cloth and you will be able to collect lumps of dry ice at -78 Celsius - the horn of the extinguisher will itself drop by as much as 50 Celsius if the extinguisher is fully discharged. 

Out of interest what is your plan, are you looking to build the setup yourself - interesting project, good luck with it. 

Jim 

Edited October 27, 2019 by saac

[vlaiv]

3 hours ago, saac said:

Vlaiv I don't think it is as indeterminable as it may appear. Treating it as a simple adiabatic expansion will get you pretty close to the value in practice - nozzle considerations accepting (choking, throat velocity).  Nor is your target of -50 Celsius overly ambitious; for example a common CO2 fire extinguisher (55 bar , 3 litres) will easily get down to that range.  Discharge a CO2 fire extinguisher through a close fibre cloth and you will be able to collect lumps of dry ice at -78 Celsius - the horn of the extinguisher will itself drop by as much as 50 Celsius if the extinguisher is fully discharged. 

Out of interest what is your plan, are you looking to build the setup yourself - interesting project, good luck with it. 

Jim 

Yes - I thought of building one unit myself. Nothing much to it the way I envisioned it.

It will consist of old refrigerator - just casing and insulation, no need for other components so it can be broken unit. I'll need regular refrigerator and deep freeze to prepare and store goods. First step would be to take fruits / vegetables and spread them onto trays and get them cooled to regular refrigeration temperature - 5C or so. After that I load them into "fast freeze" unit. That is old fridge casing with some modifications:

- inlet that can be connected to high pressure tank - with nozzle

- maybe safety valve if pressure builds up (but on second though it might even create lower pressure inside if cool air starts mixing with hot air inside and there is drop in overall temperature in closed constant volume - pressure will go down, not sure if it will offset added air.

- Pressure gauge and thermometer - to monitor what is going on

After release of the compressed air if temperature drops enough it should start rising again. If insulation is good - maybe it will provide cold enough environment for deep freezing to be completed (I have no clue how long it will take - maybe 15 minutes or so?).

Next step is using regular freezer bags to package goods and storing them in regular freezer - ready to be used in winter months

Whole procedure will take maybe less than an hour, so I can do multiple rounds and store enough goods without much expense (except for electrical bill of compressor and refrigeration), but like I said - it's more about producing and storing own food than any economical gains.