Free tool

Compressed air leak cost calculator

Enter the number and size of leaks, line pressure and operating hours to get the air lost, the power the leaks draw, the annual kWh wasted and what it costs.

Leak and system data

Currency Symbol used in the results

How to enter the leaks?

Number and size of leaks Known leak flow

Number of leaksleaks Leaks assumed to be the same size Hole diametermm Equivalent hole size, e.g. from an ultrasonic leak survey Hole type DOE method: table value × 0.61 (sharp) or × 0.97 (rounded). The DOE worked example assumes sharp-edged holes. Line pressurebar (g) Gauge pressure where the leak is. With a known flow it is only used for the leak table and can be left empty. Operating hours per yearh/year Hours the system is under pressure Compressor specific powerkW/(m³/min) If unknown, use 6.4 (the US DOE example value) Electricity price/kWh Unit price on your bill, all charges included

Result

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Loss per leak at this pressure

Hole diameter	Free air	Power	Annual cost

Method and check

The leak is modelled as compressible flow through an orifice (ideal gas, γ = 1.4, line air at 20 °C). Above about 0.9 bar(g) the flow is choked:

ṁ = Cd · A · p₀ · √(γ / (R · T₀)) · (2 / (γ + 1))^((γ + 1) / (2(γ − 1)))

Mass flow is converted to free air at 1.01325 bar absolute and 20 °C. At lower pressures the subsonic orifice equation is used.

Calibration and check: the Compressed Air Challenge table in the US DOE tip sheet "Minimize Compressed Air Leaks" lists 0.40, 1.55, 6.31, 25.22, 100.9 and 227 cfm for 1/64, 1/32, 1/16, 1/8, 1/4 and 3/8 inch holes at 100 psig (6.9 bar g). These are 0.97 times the ideal orifice flow; with that factor the model reproduces all five pressure rows of the table (70, 80, 90, 100 and 125 psig) within 0.6% for 1/16 inch and larger holes and within 3.3% for the two smallest, which the table gives to two digits. The table says to multiply by 0.61 for sharp-edged and 0.97 for well-rounded holes, so the effective coefficients in the tool are 0.97 × 0.61 = 0.592 (sharp-edged, default), 0.97 × 0.97 = 0.941 (well-rounded) and 0.97 (uncorrected table value).

Example: 10 sharp-edged leaks of 3 mm on a 7 bar(g) line lose 3.94 m³/min (65.7 l/s, 139 cfm) of free air. At 6.4 kW per m³/min that is 25.2 kW, or 151,374 kWh over 6,000 hours a year; at $0.12/kWh, about $18,165 a year.

Energy: power = free air flow × compressor specific power. The US DOE example uses about 18 kW per 100 cfm (≈ 6.4 kW per m³/min), which is the default.

Results are estimates; an exact figure needs a leak survey and compressor measurements. Replace the electricity price with the unit price on your own bill.

Sources

US Department of Energy, Compressed Air Tip Sheet: Minimize Compressed Air Leaks; Compressed Air Challenge leak table
Compressible flow: isentropic flow through an orifice

Seeing leaks all the time

When compressor current, flow and line pressure are monitored, the air used while the plant stands still, which is the leak load, shows up directly. See our energy monitoring and machine data collection pages. All tools are on the tools page.

Frequently asked questions

Common questions

How do you calculate the cost of a compressed air leak?

First find the free air flowing out of the leak from hole size, line pressure and discharge coefficient. Multiply that flow by the compressor specific power (kW per m³/min) to get the power the leak draws, by operating hours to get annual kWh, and by the electricity price to get the annual cost.

What method does the calculator use?

The leak is modelled as compressible flow through an orifice. Above about 0.9 bar(g) the flow is choked and proportional to absolute pressure. The model is calibrated to the US DOE / Compressed Air Challenge leak table, whose values are 0.97 times the ideal orifice flow: at 100 psig a 1/16 inch (1.59 mm) hole is listed at 6.31 cfm and the model gives 6.31 cfm. Across the five pressure rows (70–125 psig) the model is within 0.6% of the table for 1/16 inch and larger holes and within 3.3% for the two smallest.

Which hole type should I choose?

The same source says to multiply the table values by 0.97 for well-rounded holes and by 0.61 for sharp-edged holes. Sharp-edged (table × 0.61) is the default because DOE’s own worked example uses it, and it is the lower of the two corrected estimates. The real hole shape is hard to know, so the result is an estimate. "DOE table value" gives the uncorrected table figure and can serve as an upper bound.

What if I do not know my compressor specific power?

The US DOE example uses about 18 kW per 100 cfm, which is about 6.4 kW per m³/min. For your own compressor, use the specific power on the manufacturer data sheet (CAGI data sheet).

How accurate is the result?

It is an estimate. Real leaks have irregular shapes and pressure varies along the line. For an exact figure you need an ultrasonic leak survey and a measurement of compressor load and unload time.