Recent content by sugarandfat

Thought about that, but the cooling of the air is done by the water in the tank, which is being circulated for free by the air bubbling through it. There's not really anything needs doing... Measuring coldth with a volume of water is a good idea, might do that.

Ok, cool, that all makes sense. So I guess the whole thing is a bit too real-worldy to accurately calculate. I can accept that. So now the question is, and this may be slightly more engineering than physics, what would be the best way forward for me to optimise the cooling effect? What's the...

Wow, nice reply, thanks. > 436K ie 163 C Seems a bit high... At 5 bar and 2.5 litres I measured the temp increase inside the bottle at something around 10 C. > 3.13 bar I'm using a bike pump with a pressure gauge to pressurise the bottle. Bringing it up to 5 bar and cooling the air, the...

> assuming an adiabatic process is not realistic. I'm not so much wanting to know what temperature it will ultimately end up as the amount of coldth in it, so I can see what kind of minimum temp is possible in an insulated container. > When and with what thermometer was the temperature of the...

As it is the air is only expanding freely, so no significant work. But a few people now have suggested I'll get a greater drop in temperature if I have it working on something; spinning a turbine or similar. There's not really any work needs to be done in the system, so I'd likely have the...

So it's not reversible and all bets are off?

Maybe the molar mass is off for some reason?

It can't possibly be -88 C, that's the boiling point of CO2... I did a quick test on this, at 5 bar, a 1.2 litre bottle, and 19 C room temp, the air came out at 10.5 C. At 7 bar and 16 C ambient it was 5.5 C. It wasn't exactly an optimised rig, but close ish.

Ok, now I get the volume as 6.31385 L PV=nRT 1 bar . 6.31385 L = .41571 mol . R . T = 185.091 K (kelvins) / -88.059 °C This can't be right..

Something's not right. P1.V1.γ = P2.V2.γ 5 bar . 2 L . 1.4 = 1 bar . V2 . 1.4 gives V2 as 10 litres. to calculate molar mass: PV=nRT 5 bar . 2 L = n . R . 293.15 K gives n as 0.41571 mol final temp: PV=nRT 1 bar . 19 L = .41571 mol . R . T = 293.15 K or 20 C, which is the...

Well, they should be the same, no? It'll start to warm up on contact with the outside air, but what will the temperature of the air leaving the bottle initially be when it has come back down to 1 bar?

Hi, I'm prototyping a non electrical refrigeration system and need some help with a calculation. I have a 2 litre bottle. I pump in air until the internal pressure is 5 bar. I allow the heat increase to dissipate back to the ambient 20 C. I release the air. What will the temperature of...