Compressing air to an absolute pressure ratio r raises its temperature by the factor r to the power 0.2857 (the exponent (γ - 1) / γ for air, with γ = 1.4), so squeezing 20 °C air eightfold heats it to about 258 °C in the ideal case. That heat is why a compressor delivering 235.6 L/min at 7 bar gauge needs at least 827 W at the shaft and ideally 1,130 W, why two stages with a cooler between them save 15-20% of the power, and why a hot tank loses pressure as it cools.
Part of Pneumatic Systems
The gas laws turn compressor and tank questions into arithmetic: how hot the air gets, how much work it takes, and what pressure results when a tank warms or cools.
Boyle’s Law Covers the Slow Case and Adiabatic Compression the Fast One
All pressures here are absolute (gauge plus about 101,325 Pa) and all temperatures are in kelvin (°C + 273.15).
- Isothermal (very slow, perfectly cooled): p x V = constant. This is Boyle’s law, and it is the best a compressor could ever do.
- Adiabatic (fast, no heat escapes): p x V^γ = constant, with γ = 1.4 for air, a value recalled from engineering tables. A real compressor sits between the two but much nearer the adiabatic end.
For the adiabatic case the final temperature is T2 = T1 x r^k, where r = p2 / p1 and k = (γ - 1) / γ = 0.4 / 1.4 = 0.2857.
Compression Heating Makes Intercooling Necessary
With air entering at 20 °C (293.15 K):
| Pressure ratio r | Example | Ideal outlet temperature |
|---|---|---|
| 2 | 1 bar gauge | 84 °C |
| 2.83 | first stage of a two-stage unit to 7 bar gauge | 121 °C |
| 4 | first stage of a two-stage unit to 15 bar gauge | 162 °C |
| 7 | 6 bar gauge | 238 °C |
| 8 | 7 bar gauge | 258 °C |
| 16 | 15 bar gauge | 374 °C |
Check with r = 8: 8^0.2857 = e^(0.2857 x 2.0794) = 1.811, and 293.15 K x 1.811 = 531 K, which is 258 °C. The carrier article’s “roughly 230 °C” at “7 bar” matches r = 7 (238 °C), a ratio that corresponds to 7 bar absolute rather than 7 bar gauge. Real machines run cooler than these figures because heat leaks through the cylinder wall, but not by enough to relax the lubrication and burn hazards.
The Work of Compression Sets the Power Bill
Use the carrier’s compressor: bore 100 mm, stroke 100 mm, 300 RPM, single acting.
- Swept volume = π / 4 x 0.1² x 0.1 = 7.854 x 10⁻⁴ m³ (0.785 L)
- Free air per second = 7.854 x 10⁻⁴ x 300 / 60 = 0.003927 m³/s (235.6 L/min), before valve and clearance losses
- Isothermal power = p1 x Q x ln r = 101,325 x 0.003927 x ln 8 = 101,325 x 0.003927 x 2.0794 = 827 W
- Adiabatic power = [γ / (γ - 1)] x p1 x Q x (r^k - 1) = 3.5 x 101,325 x 0.003927 x 0.8114 = 1,130 W
- Ratio of the two: 827 / 1,130 = 73%
So even a perfect machine needs about 0.83 to 1.13 kW at the shaft, and real friction and valve losses add to that; the loss figures are not sourced here.
Where that power could come from. The corpus gives 785 W for a 20 L/s flow falling 5 m through an 80%-efficient overshot wheel (Water Wheel Types) and 50-75 W for one person cranking a chain pump (Chain Pump). The wheel cannot drive this compressor, and 1,130 / 75 = 15 people on cranks could only just do it. A compressor must be sized to the power on hand: at 785 W the ideal adiabatic limit is 785 / 1,130 x 235.6 = 164 L/min, and real output is lower.
Cooling Between Two Stages Saves Work
Split a ratio r into two equal stages of r^½ each, with air cooled back to the inlet temperature between them. The work, in units of p1 x Q, is the single-stage figure [γ / (γ - 1)] x (r^k - 1) compared with 2 x [γ / (γ - 1)] x (r^(k/2) - 1).
| Final pressure | Single stage | Two stages, equal ratio | Saving |
|---|---|---|---|
| 8 bar absolute (7 gauge) | 2.840 | 2.421 | 14.7% |
| 16 bar absolute (15 gauge) | 4.229 | 3.402 | 19.5% |
The carrier’s “15-20% less power” holds at 15 bar gauge and is slightly generous at 7. The equal-ratio split is the minimum-work split: the intermediate absolute pressure should be the square root of the product of inlet and final pressure, which for 16 bar is 4 bar absolute. At that split the first stage only reaches 162 °C instead of 374 °C. See Two-Stage with Intercooler.
A Tank’s Pressure Follows Its Temperature
At fixed volume, absolute pressure is proportional to absolute temperature.
- Cooling: a tank filled to 9 bar absolute while the air is at 60 °C (333.15 K) and then cooled to 20 °C reads 9 x 293.15 / 333.15 = 7.92 bar absolute, or 6.92 bar gauge. The gauge appears to show a leak when none exists.
- Heating: the same tank at 9 bar absolute heated to 150 °C (423.15 K), by a fire or hours in direct sun, reads 9 x 423.15 / 293.15 = 12.99 bar absolute, about 12 bar gauge. That is why a relief valve must vent at a pressure the tank can survive however hot it gets.
Expansion Cools the Air Just as Compression Heats It
Expanding adiabatically from 8 to 1 bar absolute drops 20 °C air to 293.15 x (1/8)^0.2857 = 162 K, which is -111 °C. That is the ideal limit and real tools stay far above it, because heat flows in from the tool body and expansion is not complete. It still explains why air-tool exhausts feel cold and why moisture in the air freezes in exhaust ports. See Moisture Removal.
Common Mistakes
- Putting gauge pressure into Boyle’s law. The law needs absolute pressure.
- Using Celsius in T2 = T1 x r^k. The ratio needs kelvin.
- Quoting a power without the power source. A compressor figure with no wheel, motor or crew behind it fails its own numbers.
- Treating the ideal-gas results as measured values. They are limits. Measure the real discharge temperature and delivery.
Safety
Hot discharge air and stored heat
The ideal outlet temperature at 7 bar gauge is 258 °C, and an unprotected discharge line or cylinder head burns on contact. Lubricating oil carried into the hot discharge line can bake into deposits, and compressor discharge lines have been the site of fires. Keep the discharge line short and clean, oil the compressor sparingly, and never place anything flammable against it. The same hot air cooling in a closed tank drops its pressure, while a hot tank near a fire gains pressure; the relief valve must be fitted and sized for the hottest case.
What’s Next
- Learn next: Piston Compressor — turning the arithmetic into a machine
- Combine with: Compressed Air — the same laws applied to storing wind or hydro energy