Compression of gas spring 1

[BasWag]

Hi all,

I have a quetion about the compression of a gas spring like the one in the picture

.
We see that there are energy-losses (hysteresis) if we deflect the gas spring. So, for example, if we load the gas spring, this costs us 100J, but in unloading the gas spring 'only' returhs 80J. So the question is: where did the 20J go?

We initially thought that this energy is lost in fiction between the cylinder and the seal. However, the manufacturer suggested that this energy may be lost in heat generation due to the compression of the gas (nitrogen) inside the gas spring. I want to check this mathematically, but I am not sure whether I do this correctly.

Assume that we have a starting situation, in which:
Volume = V0 = 0.006 L
Pressure = P0 = 180 bar
Temperature = T0 = 293 K

Then we defect/compress the gas spring, to a situation in wich:
Volume = V1 = 0.0036 L
Pressure = P1 = ?
Temperature = T1 = ?

Now I want to know the amount of energy in Joules that is 'lost' in this compression.

What I did is:
Calculate P1 using P1=P0*(V0/V1)^gamma, in which gamma=cp/cv=1.4 --> P1 = 368 bar
Calculate T1 using T1=(P1*V1)/(P0*V0)*T0 --> T1 = 359 K
Calculate the amount of nitrogen gas inside the gas spring (in moles) using n=(P*V)/(R*T) --> n = 0.044 mol
Calculate the mass of the nitrogen gas inside the gas spring (in kg) using m=M*n --> m = 0.616*10^-3 kg
Calculate the energy that is 'lost' in heating the nitrogen during compression using Q=c*m*(T1-T0) --> Q = 42J

Is my calculation correct? How should I interpretate this result?
Deflecting/compressing the gas spring to its full stroke costs 60J. Does this mean that 42J out of this 60J is 'used' to heat the nitrogen during compression?
And is this energy 'lost', as in, does the gas spring return this energy in unloading?

Thank you in advance!

Bas

 

[IRstuff]

RV = nRT You compress something, it has to get hot; over time, the temperature eventually equilibrates to the environment, and that energy is now lost.

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[BasWag]

@IRstuff, I understand, but in the formula PV = nRT, P gets higher, but V gets lower (since we compress), right?

 

[EdStainless]

First make sure that you have the R for pure nitrogen.
Yes, P goes up and V goes down. And the product changes and hence the temperature changes.
n stays the same and so does R.
So P1V1/T1 = P2V2/T2

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed

 

[LittleInch]

Using your number also P has gone up by 2.04, but V has only deceased by 0.6. Therefore the PV has increased by 1.44

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[1503-44]

For a non-ideal, a real gas, PV = znRT
There will be a 20%+ error in volume at P2, if you do not use real gas EOS.

https://www.chemguide.co.uk/physical/kt/realgases.html

P1V1/T1 * z1 = P2V2/T2* z2
P1 V1 / T1 * Z1 = P2 V2 /T2 * Z2

z at 368 bar 360°K is 1.22
1.04 at 180 bar 293°K.

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[MortenA]

@OP, IRstuff has it right i believe,

When the air is heated from compression heat transfer to the walls of the piston starts - and that is not returned. The other stuff mention (compressability etc) should work "both ways" so losses from friction in the piston and heat loss when the compressed air is warmer should account for the difference. One way of check could be to do several compression decompression cycles rapidly. If the energy loss appears to be reducing then i would say it confirms this theory. It will not drop to zero because a part of it is friction and also because heat loss to the ambient atm starts when the piston itself is heated.

Best regards, Morten

--- Best regards, Morten Andersen

 

[1503-44]

Z increases the work and heat. Some will be reversed on decompression, but friction is not reversible and may increase the temperature of both the gas and the can, if that heat cannot escape the outside wall of the can at the rate at which they are heated. Its the same as an IC engine, but without the IC fire and exhaust system that takes a lot of the heat away. Or better, a deadheaded compressor (with no flow). They will heat quickly. Maybe you need some cooling fins now to avoid temperature rise in the system, seals and can. Meltdown? Depends on how fast you do the cycle vs cooling rate

A black swan to a turkey is a white swan to the butcher ... and to Boeing.