THEORETICAL ENERGY CALCULATION:

Permitted variance in air compressor performance data under PN2CPTC2, the Pneurop/Cagi Acceptance Code for Electrically Driven Packaged Displacement Air Compressor.

The acceptance code specifies that the following tolerances are permitted when carrying out performance tests on packaged air compressors.

Speed +/- 4%

Inlet Pressure (ambient) +/- 10%

Discharge Pressure +/- 3%

External coolant requirements +/- 10%

Coolant injection temp. +/- 10 deg C

In the first instance we shall simply carry out a few calculations on a theoretical 1000 cfm two stage inter-cooled oil-free packaged rotary air compressor with an interstage pressure of 26 psig, (Ö P2*P1) intercooling temperature of 90 deg F and a discharge pressure of 100 psig. The compression will not be isentropic as some cooling will take place during the compression cycle. To approximate actual compression conditions, we shall assume an adiabatic efficiency of 75% during compression of both stages. The actual atmospheric pressure is 14.7 psia, and the actual discharge pressure is 100 psig. Then we shall carry out the same calculations using the permitted tolerances on inlet pressure and discharge pressure.

Assume.

g = 1.4

J = 778

R = 53.3 ft lbf/lb deg R

Adiabatic efficiency 75%

m = mass of air flow in lb/min

Inlet temp = 80 deg F (540 deg R)

1st Stage

PV = mRT \ 144 * 14.7 * 1000 = m * 53.3 * 540 \ m = 73.55 lb/min

Cv = R/J(g-1) = 0.171 Btu/lb degF g = Cp/Cv

\ Cp = 0.171 * 1.4 = 0.239 Btu/lb degF

T2/T1 = (P2/P)**((g-1)/g))

\ T2 = 540 * (40.7/14.7)**((1.4-1/1.4) = 722 deg R

But the unit is only 75% isentropically efficient,

\ 0.75 = (722 - 540)/(T2- 540) = 782.7 deg R.

Work done/lb = Cp(T2 - T1)

\ Hp = mCp(T2 - T1) * J / 33000 = 73.55 * 0.239 (782.7 - 540) * 778 / 33000 = 100.58 hp

2nd Stage

T2/T1 = (P2/P)**((g-1)/g))

\ T2 = 550 * (114.7/40.7)**((1.4-1/1.4) = 739 deg R

But the unit is only 75% isentropically efficient,

\ 0.75 = (739 - 550)/(T2- 550) = 802 deg R.

Work done/lb = Cp(T2 - T1)

\ Hp = mCp(T2 - T1) * J / 33000 = 73.55 * 0.239 (802 - 550) * 778 / 33000 = 104.43 hp

Total package power ignoring motor losses, gearbox losses etc. = 205 hp

If we do the same calculations using the tolerances allowed by Pneurop/Cagi at the top end of the allowable conditions, i.e. absolute inlet pressure +10%, discharge pressure -3%, we get the following.

Atmospheric pressure is now 14.7psia. + 10% (i.e. 16.17)

Discharge pressure is now 100 psig - 3% (i.e. 97)

g = 1.4

J = 778

R = 53.3 ft lbf/lb deg R

Adiabatic efficiency 75%

m = mass of air flow in lb/min

Inlet temp = 80 deg F (540 deg R)

1st Stage

PV = mRT \ 144 * 16.17 * 1000 = m * 53.3 * 540 \ m = 80.9 lb/min

Cp = 0.239 Btu/lb degF

T2/T1 = (P2/P)**((g-1)/g))

\ T2 = 540 * (40.7/16.17)**((1.4-1/1.4) = 702.9 deg R

But the unit is only 75% isentropically efficient,

\ 0.75 = (702.9 - 540)/(T2- 540) = 757.2 deg R.

Work done/lb = Cp(T2 - T1)

\ Hp = mCp(T2 - T1) * J / 33000 = 80.9 * 0.239 (757.2 - 540) * 778 / 33000 = 99 hp

2nd Stage

T2/T1 = (P2/P)**((g-1)/g))

\ T2 = 550 * (111.7/40.7)**((1.4-1/1.4) = 733.9 deg R

But the unit is only 75% isentropically efficient,

\ 0.75 = (733.9 - 550)/(T2- 550) = 795 deg R.

Work done/lb = Cp(T2 - T1)

\ Hp = mCp(T2 - T1) * J / 33000 = 80.9 * 0.239 (795 - 550) * 778 / 33000 = 111.68 hp

Total package power ignoring motor losses, gearbox losses etc. = 210.6 hp

If we do the same calculations using the tolerances allowed by Pneurop/Cagi at the bottom end of the conditions, i.e. absolute inlet pressure -10%, discharge pressure +3%, we get the following.

Atmospheric pressure is now 14.7psia. - 10% (i.e. 13.23)

Discharge pressure is now 100 psig + 3% (i.e. 103)

g = 1.4

J = 778

R = 53.3 ft lbf/lb deg R

Adiabatic efficiency 75%

m = mass of air flow in lb/min

Inlet temp = 80 deg F (540 deg R)

1st Stage

PV = mRT \ 144 * 13.23 * 1000 = m * 53.3 * 540 \ m = 66.19 lb/min

Cp = 0.239 Btu/lb degF

T2/T1 = (P2/P)**((g-1)/g))

\ T2 = 540 * (40.7/13.23)**((1.4-1/1.4) = 744.4 deg R

But the unit is only 75% isentropically efficient,

\ 0.75 = (744.4 - 540)/(T2- 540) = 812.5 deg R.

Work done/lb = Cp(T2 - T1)

\ Hp = mCp(T2 - T1) * J / 33000 = 66.19 * 0.239 (812.5 - 540) * 778 / 33000 = 101.63 hp

2nd Stage

T2/T1 = (P2/P)**((g-1)/g))

\ T2 = 550 * (117.7/40.7)**((1.4-1/1.4) = 744.9 deg R

But the unit is only 75% isentropically efficient,

\ 0.75 = (744.9 - 550)/(T2- 550) = 809.9 deg R.

Work done/lb = Cp(T2 - T1)

\ Hp = mCp(T2 - T1) * J / 33000 = 66.19 * 0.239 (809.9 - 550) * 778 / 33000 = 96.93 hp

Total package power ignoring motor losses, gearbox losses etc. = 198.56 hp

Summary:

The machine has the following permutation of characteristics.

Case 1. Actual flow of 1000 cfm at 14.7 psia inlet with a discharge of 100 psig.

Mass flow rate 73.55 lb/min +/- 4%

Power absorbed. 205 hp +/- 5%

Case 2. Most efficient choice of pressure tolerance.

Mass flow rate 80.9 lb/min +/- 4%

Power absorbed. 210.6 hp +/- 5%

Case 3. Least efficient choice of pressure tolerance.

Mass flow rate 66.19 lb/min +/- 4%

Power absorbed. 198.56 hp +/- 5%

You can convert air mass flow rate back to a standard volume occupied by air at 14.504 psia (1 bara). This equates to 0.80 lb/cu ft. By using the magic words '1 bar absolute inlet pressure' in your sales leaflets, the above compressor will therefore produce anything from 827.37 cfm FAD up to 1011 cfm FAD, simply by working within the applicable pressure tolerances permitted under the Pneurop/Cagi test procedures.

Now we can apply the other tolerances, the volume flow rate can be overstated or understated by +/- 4%, independent of the pressure calculations. This widens the volume flow rate band even further, now being 794.27 cfm FAD at the bottom end and 1051.44 cfm FAD at the top. 

Now you can apply the specific energy tolerance of +/- 5%, this is independent of the other two calculations. The lowest power absorbed figure reduces to 188.63 hp, the highest power figure increases to 221.13 hp.

In terms of power conversion, you can basically choose your own figures, anything from:

3.81 cfm/hp (66.19 lb/min -4%, 198.56 hp + 5%)

up to

5.26 cfm/hp (80.9 lb/min + 4%, 210.6 hp -5%)

A working tolerance of around 30%. Don't lose sight of the fact that all of these calculations are for the same machine, just using the extremes of the permitted legal tolerances. Which figures do you think the compressor manufacturer will choose for their sales leaflets and presentations?