Free Air Delivery Measurement Method

In such a case you work wit the mass of air not with the volume. At the end you compute back to air flow. Your problem is more complex due to the fact that the compressor works against a variable pressure so that the leaks are variable in time. You may as 1st simplification assume leaks proportional to pressure in the reservoir.

http://www.em-ea.org/Guide%20Books/book-4/4.8.Compressors.pdf

http://www.retscreen.net

Volume of free air in the tank is 2 m3 when it kept at atomospheric pressure (1.03 kg/cm2 absolute or 0 gauge). If you add 2 m3 more free air in to the tank, its pressure becomes 2.06 kg/cm2 absolute or 1.03 kg/cm2 gauge assuming that temperature is unchanged. Hope you can do rest of the maths.

The equation of state for a gas is v [m^3/kg] = p [N/m^2]/(R* T[°K]) with R as the universal gas constant and T the absolute temperature.

The air mass in the reservoir can be computed and using same equation in volume at inlet conditions. The rest you can compute on your own.

v = p/RT ?

I thought the relation was PV = CT , where C is a constant, reference....

http://en.wikipedia.org/wiki/Ideal_gas_law#Empirical

2kL = 2000 litres??

If the ambient were 30 Celsius, close off reservoir valve at the end of delivery, then wait until the receiver reservoir temperature is ambient (reservoir pressure stops falling).

Then calculate the original volume at ambient as...

2kL x (settled pressure in reservoir)/ambient air pressure

Deduct 2kL, because the reservoir started full of air, to get the amount delivered.

If it is a piston pump, I would expect it to draw in the same volume on each rotation and the volume ratio to vary little with ambient temperature.

If you want, for example, to correct mass (of a given volume V) at 20 Celsius, M20, to 30 Celsius at same pressure, then...

M30 = M20 x (273.2 + 20)/(273.2 +30)

- noting it will be less dense at 30 Celsius.