A gas flowing through an orifice is throttled (causing turbulence and heating), and expanded (causing cooling). Thus, it is subject to energy conversions which reduce the amount of energy available to do work. The rate at which available energy is lost can be termed the pneumatic power, which is a function of the pressures, Lohm rate of the orifice, and the flow. For nitrogen, the relationship is shown on the graph below.

When the flow rate and pressure ratio is known, the resulting power consumption can be determined from the graph. If flow is not known, it can be readily calculated from the Lohm rate using the gas Lohm Law. Simply enter the graph at the appropriate pressure ratio (X-axis), and read vertically to the line corresponding to the applicable flow rate. The resulting power may then be read horizontally across on the Y-axis. Note that pressure ratio is the ratio of the absolute pressures.

For more precise calculations, or to extend the range of the pneumatic power graph, the following formulas may be used for nitrogen or air.

*Swipe to the right for more table information*

P_{1} = Supply Pressure (psia)
P_{2} = Exhaust Pressure (psia)
L = Lohm Rate |

Note that due to compressor inefficiencies, more energy will be needed to compress the air than will be expended when it flows through an orifice.

**Example:** For a 500 Lohm restrictor flowing nitrogen at 750 psia exhausting to 75 psia, the flow can be easily calculated from the gas Lohm Law.

Next, determine the pressure ratio, P_{1}/P_{2}, which in this example is 750/75 = 10. Then, from the graph above:

Pneumatic power = 2.5 HP

### Pneumatic Power Dissipation