Disco loud cracks have me puzzled.

[Tad]

Bob-PCP dieseling is keeping me up at night, trying to figure out the thermodynamics (never my strong point).

In springers it's simple.  The heat generated is equal to the energy used to compress the spring.  But calculating the heat generated in the PCP reservoir by rapid filling by scuba or detonation during firing is not so obvious (at least to me).  My old texts only deal with the equilibrium state, where there is no net change in temperature for the system.

Don't know if you're an engineering type, but if you have a theoretical handle on this, it would be much appreciated.

[rsterne]

Naw, not an engineer, just a curious sot trained as a chemist a zillion years ago.... The way I envision it is the air expanding out of the valve is doing so at around 1650 fps (the speed of air moelcules at 70*F).... They collide into the static air in the transfer port compressing it (and mixing with it, to be sure)....

Basically, the inrushing air works the same as if it was a piston (OK, maybe a flexible, porous piston), compressing the air that is already in the space.... No question, when you fill a gun from SCUBA it warms up.... and the faster you fill it, the hotter it gets.... I can only assume that when you fill slowly, the incoming air has more time to mix with the air already in the gun.... plus more time to lose heat to the gun itself....

Maybe lloyd-ss will chime in here.... he's got good ideas on this stuff....

Bob

[lloyd-ss]

Bob,

I believe it varies with the particular gun and valve.  There is both cooling and heating, and the net result will be localized heat gain or loss. When filling a gun, the valve in the out-going tank gets cold, and the receiving tank gets warm.  Heat was given up to the expanding air via the cold valve (the metal valve gave up heat), and then put back into the metal of the receiving tank.  Was there a net gain or loss?  Honestly, I do not know, but relying on the PV=nrT, there should have been no net change (maybe?). 
When firing a PCP, the air expands thru the valve, absorbing heat from the valve (making the valve cold---think CO2 valve) and is then re-compressed  behind the pellet.  The air will get hotter as it is re-compressed but I think the size of the transfer port ("expansion orifice") and how tightly the pellet is stuck in the breech will affect the final compressed temperature of the air behind the pellet. I think a stuck pellet would give the highest air temp, and therefore the possibility of dieseling.
This is almost like a refrigeration cycle, but without the liquid-vapor state changes.

This all goes counter to saying that the pee-wee pellets give the loud crack, but I think that is a totally different mechanism at work.

I am rusty on my thermo too (never did have  a good handle on it, LOL  ).

Just my 2 cents.
Lloyd

[Tad]

Thanks guys, didn't mean to hijack the original thread.

Bob-I see our problem.  I'm a chemist too.

Everybody has a feeling and a verbal description, but nothing quantitative.  I can easily calculate the heat rise in a spring scenario, but not the scuba tank scenario.

CO2 is a whole 'nother thing.  It gets cold because heat is absorbed to boil the liquid CO2.

The cooling of gases with expansion is due to the Joule Thomson effect, because real gases are not ideal.  An ideal gas would not get cold after expansion.  Real gases get hot or cold during expansion depending on there inversion temperature.

[JTank70]

Well,  I am certainly not a chemist or an engineer but it is still interesting to read you guys' thoughts on this stuff.

I got a good 100-150 more shots in and thankfully still no issue. So, whatever it was it has not happened again and I hope it never does.

Thanks for everyone's opinions and Ideas on this one.  They all helped me out and helped me to feel beter about the situation.

JT