How much does a compressed air leak cost?

A 3 mm leak at 7 bar loses about 6.8 L/s of free air. With a compressor using 6.5 kW per m³/min, running 8,000 hours a year at 0.12 per kWh, that is about 2.6 kW and roughly 2,500 per year.

Compressed Air Leak Cost Calculator (bar or psi, mm or inch)

Enter the hole size and line pressure to see how much air, power and money a leak wastes in a year.

How the calculation works

Above roughly 0.9 bar gauge, air escaping through a hole reaches the speed of sound in the opening. The flow is then "choked": it depends only on the hole area and the absolute pressure upstream, not on the pressure outside. That makes the leak flow straightforward to estimate.

ṁ = C_d · A · P₀ · √(k / (R·T₀)) · (2 / (k+1))^{(k+1)/(2(k−1))}

Here ṁ is the mass flow in kg/s, C_d the discharge coefficient, A the hole area in m², P₀ the absolute line pressure in Pa, k = 1.4 for air, R = 287.05 J/(kg·K), and T₀ = 293.15 K (20 °C).

The mass flow is converted to free air at 1.01325 bar and 20 °C, which is how compressor capacity is rated. The wasted power is that flow multiplied by the compressor's specific power, and the annual energy is that power multiplied by the hours the system is pressurised.

kW wasted = free air flow (m³/min) × specific power (kW per m³/min)
Annual cost = kW wasted × hours × price per kWh

Worked example

A 3 mm sharp-edged leak on a 7 bar line, in a plant whose compressors use 6.5 kW per m³/min, pressurised 8,000 hours a year, paying 0.12 per kWh:

Absolute pressure: 7 + 1.013 = 8.013 bar, or 801,325 Pa.
Hole area: π × 0.003² ÷ 4 = 7.07 × 10⁻⁶ m².
Leak flow: 0.406 m³/min of free air, which is 6.8 L/s or 14.4 cfm.
Power wasted: 0.406 × 6.5 = 2.64 kW.
Energy: 2.64 × 8,000 = 21,130 kWh a year.
Cost: 21,130 × 0.12 = about 2,540 a year, from one hole you can barely see.

Reference table at 7 bar

Sharp-edged holes, 6.5 kW per m³/min, 8,000 hours a year, 0.12 per kWh.

Hole	L/s	cfm	kW	Cost per year
0.5 mm	0.19	0.4	0.07	70
1 mm	0.75	1.6	0.29	282
1.5 mm	1.69	3.6	0.66	634
2 mm	3.01	6.4	1.17	1,127
3 mm	6.77	14.4	2.64	2,536
4 mm	12.04	25.5	4.70	4,508
5 mm	18.81	39.9	7.34	7,044
6 mm	27.09	57.4	10.57	10,144

Flow rises with the square of the diameter: doubling the hole size multiplies the loss by four.

Assumptions and limits

The leak is treated as a single round hole. Cracks, worn seals and loose fittings have irregular shapes, so estimate an equivalent diameter or measure the flow with an ultrasonic leak detector.
Air at the leak is taken as 20 °C. Hotter air leaks slightly less mass for the same hole.
Specific power is treated as constant. A compressor running at part load in load/unload control often uses more energy per m³ than its full-load rating, so real savings can be higher, or lower if leak repairs do not reduce compressor running time.
Below about 0.9 bar gauge the flow is no longer choked and this formula overestimates the leak.

Questions

What discharge coefficient should I use?

Real leaks behave like sharp-edged holes, so 0.61 is the usual conservative choice. Use 0.97 only for smooth, well-rounded openings, such as an open blow-off nozzle.

How do I find my compressor's specific power?

Divide the compressor's total input power in kW by its free air delivery in m³/min at your operating pressure. Both values are usually on the manufacturer's data sheet. If you have several compressors, use the trim machine that actually responds to the leak demand.

How can I estimate total leakage in my plant?

On a non-production day with all air users off, run the compressors and record how much of the time they are loaded. Leakage as a percentage of capacity is roughly the loaded time divided by the total time. Many older systems lose a fifth or more of their air this way.

Is it worth fixing small leaks?

Usually yes. A 1 mm leak is inaudible in a running plant, but fifty of them on a 7 bar system add up to roughly 14 kW of compressor power, running all year.