The loss in sound is miniscule in comparison to the energy present, and the concept of a 'refrigeration' loss is really only compression losses, as
that energy is actually lost during the compression of the gas as heat.
The long and short of it is that the energy is stored as numbers of air molecules, all trying to get away from each other. The total amount of molecules into the regulator and the total count out is the same.
The regulator makes it happen at specific rate. If you use 1 cubic foot of air from the tank, it doesn't matter if it happened at 140 PSI or at 70 PSI - the amount of air remains the same, and the only difference is that it takes twice as long to remove it from the tank.
The air passing through the tool at half pressure gives it half of the effective force, at half the rate, making it take twice as long to use the same amount of energy.
To elaborate - Charge up an air compressor from zero to 140PSI (or whatever value you like). Note how long it takes to charge the tank. This could be expressed in Wh of input if you felt like doing the math.
Discharge the tank with the regulator set wide open, into a non-restrictive discharge (cut-off hose?). Time how long it takes to get down to half of the 'full' PSI dictated above.
Recharge the compressor again, back to that full PSI value. Note how long it takes to charge the tank. Once again, get your watthour measurement.
Discharge the tank at 50% of the full pressure PSI setting, down to the 50% level in the tanks, again through your cut-off hose. Once again, note how long it takes to discharge the tank.
Recharge the compressor back up to full. Note that the compressor runs the same amount of time to recover the same amount of air, regardless of what pressure you use it at.
IE, negligible losses in the regulator.
Steve
