A compressed air system at a customer’s plant. The compressor cycled on every 3 minutes even though no machine was running. I put a ultrasonic leak detector on the lines. It found 12 leaks. The worst was a 2 mm hole in a 6 mm tube elbow. That single leak was consuming about 30 L/min of free air. At 8000 operating hours a year, that’s enough wasted air to cost $400/year in electricity. This is how I calculate leak rates.
The orifice equation
A leak through a small hole (or a loose fitting) follows the orifice equation for compressible flow. For a hole at supply pressure (6 bar gauge = 7 bar absolute), the flow is sonic (choked) because the pressure ratio across the hole is more than 2:1. The flow rate is:
Q = 198 × d² × P
Where Q is in L/min (free air, at standard conditions), d is the hole diameter in mm, and P is the absolute pressure in bar. For a 2 mm hole at 7 bar absolute: Q = 198 × 4 × 7 = 5,544 L/min. Wait — that’s enormous. That can’t be right. Let me reconsider. The constant depends on units. Let me use a simpler rule of thumb.
The practical rule
For compressed air at 6 bar gauge, a 1 mm diameter hole leaks about 6 L/min. A 2 mm hole leaks 24 L/min. A 3 mm hole leaks 54 L/min. These are rough numbers but they’re what I use in the field.
| Hole size (mm) | Leak rate (L/min free air) | Annual cost (8000 h, $0.05/1000L) |
|---|---|---|
| 0.5 | 1.5 | $3.50 |
| 1.0 | 6 | $14 |
| 2.0 | 24 | $58 |
| 3.0 | 54 | $130 |
| 5.0 | 150 | $360 |
The 2 mm hole was 24 L/min. The customer had 12 leaks averaging 1 mm each — that’s 72 L/min total. At 8000 hours, that’s 72 × 60 × 8000 = 34,560,000 L/year = 34,560 m³/year. At $0.05 per 1000 L, that’s $1,728/year. Fixing the leaks cost $200 in fittings and one hour of labor. It paid for itself in 5 weeks.
How I find leaks
Ultrasonic leak detector: A handheld device that listens for the high-frequency hiss of escaping air. I walk the line. Every fitting, every valve exhaust, every tube connection. The detector gives a signal strength. I mark the loud ones.
Soap solution: For suspected leaks, I spray soapy water. Bubbles form where the air is escaping. This is slower but definitive. I use it after the detector has narrowed it down.
Shutoff test: At night (no production), I close the main supply valve and watch the pressure gauge. If it drops from 6 bar to 4 bar in 10 minutes, the system has significant leaks. The drop rate tells me the total leak rate. For a 100 L receiver, a 2 bar drop in 10 minutes is about 20 L/min leak rate.
The common leak locations
- Push-to-connect fittings: The most common. A tube cut at an angle leaks slowly. A tube not fully inserted leaks. Fix: recut square, reinsert.
- Valve exhaust ports: A valve that’s slightly worn exhausts continuously. Fix: rebuild or replace the valve.
- Quick couplings: Unused couplings with no plug leak. Fix: install plugs on unused ports.
- Hose fittings: Pipe thread fittings that aren’t sealed with Teflon tape leak. Fix: re-seal.
- Cylinder rod seals: A worn rod seal bleeds air slowly. Fix: re-pack the cylinder.
The compressor sizing consequence
If the system has 100 L/min of leaks, the compressor must produce that air just to maintain pressure. A 10 kW compressor producing 1000 L/min is running at 10% capacity just feeding leaks. That’s wasted electricity. Fixing the leaks doesn’t just save air — it lets the compressor unload more often. The payback is often under 6 months.
The number I estimate: a 1 mm hole leaks 6 L/min at 6 bar. Walk the system with an ultrasonic detector. Fix everything over 1 mm. The annual cost of a 2 mm hole is about $60 — fix it. The plant that cycled its compressor every 3 minutes wasn’t undersized; it was feeding 12 leaks. Air leaks don’t show up on a pressure gauge — they show up on the electric bill.
