Compressed Air and Gas Flow

Compressibility, speed of sound and choked flow, pressure drop in air lines, sizing compressed-air pipework and leaks.

When is a gas compressible?

Liquids hold their density almost constant. Gases do not: the density of air rises in proportion to the absolute pressure and falls with temperature, ρ = p ÷ (R T). In slow flow with small pressure changes, a gas can be treated as incompressible and all the equations in this guide hold. The rule of thumb:

  • treat the gas as incompressible if the velocity is below about Mach 0.3 (about 103 m/s in air at 20 °C) and the pressure drop is below about 10 percent of the inlet absolute pressure;
  • otherwise use the compressible equations (isentropic flow for nozzles and short pipes, isothermal flow for long pipes, with density worked out at the local pressure).

Air at 20 °C and 1 atm has a density of 1.204 kg/m³; at 7 bar gauge (8 bar absolute) it is 9.52 kg/m³, eight times as dense. A standard volume (std, free air, FAD, Nm³) is the volume the gas would have at standard conditions (usually 1 atm and 0 °C or 20 °C): always state which, since compressors are rated in free air delivered and pipes in actual flow.

Speed of sound and Mach number

a = √(k R T)    Ma = v ÷ a

In air at 20 °C a = 343 m/s; at 100 °C 387 m/s. A pressure wave in a gas moves at this speed, and so does the information that the downstream pressure has changed: when the flow reaches Mach 1 in a nozzle the downstream pressure can no longer reach upstream, and the flow chokes. For air at Mach 0.6 the stagnation pressure is 1.276 times the static pressure and the stagnation temperature 1.072 times the static.

Choked flow through a nozzle or leak

When gas expands through a restriction, the speed at the narrowest point rises as the downstream pressure falls, until it reaches Mach 1 at the critical pressure ratio p₂ ÷ p₁ = (2 ÷ (k + 1))k ÷ (k − 1), which is 0.528 for air. Below that ratio (that is, when the upstream absolute pressure is more than about 1.9 times the downstream), lowering the downstream pressure further does not increase the flow: it is choked, and the mass flow depends only on the upstream pressure and temperature and the area.

Choked flow: mass flux against pressure ratio00.20.40.5280.60.810.00.20.40.60.8pressure ratio p₂ ÷ p₁mass flux (relative)critical ratio 0.528choked: flow cannot rise furthersubsonicFor air (k = 1.4) the throat reaches the speed of sound when p₂ ÷ p₁ falls to (2 ÷ (k + 1))^(k ÷ (k − 1))= 0.528. Lowering the pressure further does not increase the flow.
Mass flux through a nozzle against the pressure ratio. Below 0.528 the flow is choked.

ṁ = Cd A p₀ √(k ÷ (R T₀)) × (2 ÷ (k + 1))(k + 1) ÷ (2 (k − 1))

Worked example: a 5 mm hole in a compressed-air line

  • Upstream 7 bar gauge (8.01 bar absolute), 20 °C, discharging to atmosphere: the pressure ratio is 0.126, below 0.528, so the flow is choked.
  • Hole area 19.6 mm², Cd = 0.62: ṁ = 0.62 × 0.00001963 × 801 325 × √(1.4 ÷ (287 × 293.15)) × 0.5787 = 23.0 g/s.
  • At the standard density of 1.204 kg/m³ that is 19.1 L/s of free air (1.15 m³/min).

What a leak costs

Compressed air is the most expensive utility in many plants: it takes about 6 to 7 kW of compressor power for each m³/min of free air at 7 bar, about 10 percent of which turns into useful work, and a typical plant loses 20 to 30 percent of its air to leaks.

Hole diameterFree air lost at 7 bar gSame in m³/hYearly cost at 8 000 h and $0.12/kWh
1 mm0.8 L/s2.8 m³/h$286
3 mm6.9 L/s24.8 m³/h$2 577
5 mm19.1 L/s68.8 m³/h$7 159
10 mm76.5 L/s275.3 m³/h$28 636

The flow goes with the area, so the cost rises with the square of the diameter: a 10 mm hole wastes four times as much as a 5 mm one. The 5 mm leak above is 7.5 kW of compressor power. The Compressed Air Consumption Calculator adds up the demand of cylinders and tools, and finds the compressor size.

Pressure drop in an air line

In a long pipe, the gas expands as its pressure falls, so it speeds up along the pipe. For a moderate drop (below about 10 percent of the inlet absolute pressure) use the Darcy–Weisbach equation of Pipe Friction and the Moody Diagram with the density and velocity at the inlet; for bigger drops use the isothermal flow equation, p₁² − p₂² = f (L ÷ D) ṁ² R T ÷ A² (with ṁ the mass flow).

Worked example: 100 L/s of free air in 100 m of steel pipe at 7 bar g

The flow of 100 L/s free air is 12.6 L/s of actual air at 8 bar absolute (the volume shrinks by the pressure ratio). Density at the inlet 9.52 kg/m³, viscosity 18.1 µPa·s, roughness 0.045 mm.

PipeInside diameterVelocityRefDrop per 100 m
DN3235.1 mm13.1 m/s241 3620.02190.507 bar
DN4040.9 mm9.6 m/s207 1350.02140.231 bar
DN5052.5 mm5.8 m/s161 3680.02080.064 bar
DN6562.7 mm4.1 m/s135 1600.02050.026 bar
DN8077.9 mm2.7 m/s108 7240.02040.009 bar

The DN50 pipe drops 0.064 bar (0.8 percent of the inlet), at 5.8 m/s. The DN32 pipe is far too small: the drop is many times higher and the velocity too fast. Diameter matters enormously: halving it raises the drop by about 30 times.

Sizing compressed-air pipework

  • Total pressure loss from compressor to the furthest point should be under about 0.3 bar (4 psi), split between the main header (0.1 bar), the branches and the final hose and filter. Each lost bar is about 7 percent more compressor energy.
  • Velocity: 6 m/s or less in headers keeps the drop and noise down; 10 m/s or more is too much except for short drops.
  • Loop the main (a ring) so that air reaches a point from two directions, halving the effective flow and the drop.
  • Slope the main towards drain points, take branches off the top, and fit a drain to every low point: condensate is the other great source of trouble.
  • Use the right size for the end use: a 1/4 inch hose or an undersized quick coupler can drop 0.5 bar by itself at the flow a tool needs.
  • Pressure setting: do not raise the compressor pressure to make up for drop: reduce the drop instead, and set the pressure at the highest-demand device plus the losses.

Receivers

An air receiver stores compressed air so that the compressor can cycle and short bursts of demand do not pull the pressure down. Its volume follows from the free-air demand Q, the time t that it must supply it and the allowed pressure drop Δp: V = Q t patm ÷ Δp. To supply 100 L/s of free air for 30 s with a 1 bar drop needs V = 0.100 × 30 × 1.013 ÷ 1.0 = 3.04 m³ (3 040 liters). Receivers are pressure vessels and need a relief valve, a drain and an inspection regime.

Related

FAQ

Why does the flow stop increasing when I lower the downstream pressure?

Because the flow has choked: the speed at the narrowest section is the speed of sound, and pressure changes downstream can not travel upstream against it. The flow can only rise by raising the upstream pressure or enlarging the area.

What is the difference between scfm, Nm³/h and FAD?

Standard cubic feet per minute, normal cubic meters per hour and free air delivered all count the gas as the volume it would have at a reference pressure and temperature. They differ by the reference conditions (for example 20 °C or 0 °C), a few percent. Always check which one a datasheet uses.

How do I find leaks?

With an ultrasonic leak detector or soapy water, during a quiet shift. Fix them by tightening or replacing fittings and hoses; the biggest losses are often at the points of use (couplings, hoses, regulators, drain valves left open).

Does pipe material matter?

Friction does not differ much between steel, aluminum, copper and plastic; stainless and aluminum systems stay clean and avoid rust particles in tools. Plastic pipe must be rated for compressed air and for the compressor oil.

References

  • White, F. M., Fluid Mechanics, McGraw-Hill (compressible flow).
  • Çengel, Y. A. and Cimbala, J. M., Fluid Mechanics (compressible flow, choked flow).
  • Compressed Air and Gas Institute, Compressed Air and Gas Handbook; US Department of Energy, Improving Compressed Air System Performance: A Sourcebook for Industry.
  • ISO 8573 (compressed-air quality), ISO 1217 (displacement compressors, acceptance tests).

Last reviewed: October 2026. Energy and cost numbers are typical.