What this calculator does
Calculates isentropic outlet temperature, corrects work for compressor efficiency and converts shaft power to electrical input using motor efficiency.
Inputs
Results
Worked example: 0.5 kg/s of air from 1 to 8 bar
A plant air duty, calculated single-stage — and why the answer rules itself out.
Take 0.5 kg/s of air drawn at 20 °C and 1 bar absolute and delivered at 8 bar absolute (7 bar gauge, the standard plant-air pressure), with an isentropic efficiency of 75% and a 94%-efficient motor. For air, γ = 1.4 and R = 287 J/kg·K, so cₚ = γR/(γ − 1) = 1,004.5 J/kg·K.
The isentropic discharge temperature follows from the pressure ratio alone:
T₂ₛ = T₁ · r(γ−1)/γ = 293.15 × 80.2857 = 531.0 K = 257.9 °CReal compression is irreversible, so the actual temperature rise is the ideal rise divided by the efficiency, and the specific work follows from it:
ΔTact = (531.0 − 293.15) / 0.75 = 317.2 K → T₂ = 610.3 K = 337.2 °C w = cₚ · ΔTact = 1,004.5 × 317.2 = 318.6 kJ/kg Pshaft = ṁ · w = 0.5 × 318.6 = 159.3 kW Pelec = 159.3 / 0.94 = 169.5 kWRead the temperature, not just the power. 337 °C at the discharge flange is not a machine that gets built. Mineral lubricating oil degrades rapidly above roughly 200 °C, and in air service a hot oil film in the discharge line is a recognised fire and explosion hazard — which is why reciprocating air compressors are commonly limited to around 150–180 °C per stage, and oil-injected screws run far cooler because the injected oil absorbs the heat of compression as it is produced. The 159 kW is arithmetically correct and describes a machine that cannot exist in single-stage form.
The discharge temperature, not the power, is what decides the number of stages. That is the most useful thing this calculator tells you, and it is in the result panel above.
Where the second stage pays for itself
Intercooling both solves the temperature problem and reduces the power.
Splitting the compression and cooling the gas back to inlet temperature in between does two things at once. It halves the temperature rise per stage, and because the second stage then compresses denser, cooler gas, it reduces the total work. For equal work split the optimum intermediate pressure is the geometric mean:
pi = √(p₁ · p₂) = √(1 × 8) = 2.83 bar → stage ratio 2.83 eachRunning the same calculation on one stage of 2.83 and doubling it, with perfect intercooling back to 20 °C:
| Arrangement | Ratio per stage | Discharge temp. | Specific work | Shaft power |
|---|---|---|---|---|
| Single stage | 8.00 | 337 °C | 318.6 kJ/kg | 159.3 kW |
| Two stages, intercooled | 2.83 | 155 °C | 271.6 kJ/kg | 135.8 kW |
The discharge temperature falls from 337 °C to 155 °C, inside the practical band, and the shaft power falls by 14.7% — roughly 23 kW continuously. On a machine running 6,000 hours a year that is about 140 MWh of electricity, which is why plant air above a ratio of about 4 to 5 is almost always staged, and why the economics rarely need arguing.
The intercooler has to remove the heat of the first stage, about 68 kW here, and the aftercooler removes the second. Both duties, and the condensate they produce, are part of the installation that the compressor power figure does not include.
Cross-check against reality. At inlet conditions 0.5 kg/s of air is 25.2 m³/min of free air delivery, so the single-stage result is 6.3 kW per m³/min of shaft power and the two-stage result 5.4. Published specific powers for good industrial screw compressors at 7 bar g sit in roughly the same band, which is a reasonable sign the model is not far wrong for air at this pressure. Compare like with like though — vendor figures are usually package input power including the motor and drive, so compare them to the 6.7 kW per m³/min electrical figure rather than the shaft figure.
Isentropic or polytropic? The efficiency you enter matters
The two definitions are not interchangeable, and vendors quote the other one.
This calculator takes an isentropic efficiency: actual work compared with the work of a reversible adiabatic compression between the same pressures. That is the natural definition for a single stage and it is what most textbooks introduce first.
Turbomachinery vendors, and ASME PTC 10 and API 617, work in polytropic efficiency instead — the small-stage efficiency, applied to an infinitesimal pressure rise and integrated. The two are related through the polytropic exponent:
n / (n − 1) = [γ / (γ − 1)] · ηₚ for air at ηₚ = 0.78: n = 1.578The reason this matters is the reheat effect. Each stage delivers gas slightly hotter than ideal, and the next stage then has to compress that hotter, less dense gas. The loss compounds, so for a machine of fixed aerodynamic quality the isentropic efficiency falls as pressure ratio rises while the polytropic efficiency stays roughly constant. Polytropic efficiency therefore describes the machine; isentropic efficiency describes the machine and the duty together.
Consequence for this page: if you have a vendor's polytropic efficiency, do not type it into the isentropic field at high pressure ratio — at a ratio of 8 the isentropic value is several points lower, and using the polytropic number under-predicts both the power and the discharge temperature. At ratios below about 2 the two converge and the distinction stops mattering.
Typical efficiency ranges
| Machine type | Isentropic efficiency | Usual pressure-ratio range |
|---|---|---|
| Reciprocating, per stage | 75 – 90% | up to 4 – 5 |
| Oil-injected screw | 70 – 80% | up to 8 – 14 |
| Oil-free screw, two-stage | 70 – 78% | up to 10 |
| Centrifugal, per stage | 75 – 85% | 1.2 – 3 |
| Axial, per stage | 85 – 92% | 1.1 – 1.4 |
These are indicative ranges for orientation. Part-load behaviour differs sharply between types and is often what dominates annual energy use: a fixed-speed screw on load/unload control can consume 25–35% of full-load power while delivering nothing.
Where the ideal-gas model stops working
Everything this page assumes, and when each assumption breaks.
The calculation above rests on four assumptions. For air at moderate pressure all four hold well enough that the answer is useful. Each of them fails somewhere.
1. The gas is ideal (Z = 1)
Real behaviour enters through the compressibility factor:
pv = Z R T Z → 1 as p → 0 and far from the critical pointFor air at 8 bar and ambient temperature Z is about 0.998 and the error is invisible. For methane at 70 bar Z is around 0.85; for propane, CO₂ or a refrigerant near saturation it departs far further, and the specific work can be wrong by tens of percent. The rule of thumb is reduced pressure pr = p/pc: below about 0.1 the ideal-gas result is good, above 0.3 it is not. Heavy hydrocarbons and any gas approaching its saturation line need a real-gas equation of state and an enthalpy-based calculation, not this one.
2. Specific heats are constant
γ and cₚ are evaluated once, at inlet. Across a 317 K temperature rise they are not actually constant — for air cₚ increases by a few percent by 340 °C. Using the inlet value slightly under-predicts the work. Evaluating properties at the mean temperature is a simple improvement where it matters.
3. Compression is adiabatic
No heat is assumed to leave the gas during compression. Oil-injected screws break this assumption deliberately and substantially — the injected oil removes heat as it is generated, so the real process is closer to polytropic with n well below γ, the discharge temperature lands near 80–100 °C rather than the adiabatic value, and the work is lower than this page predicts. For that machine type treat the result as a conservative upper bound.
4. The gas is dry
Atmospheric air is not. Water vapour affects the effective gas constant, and more importantly it condenses in the intercooler and aftercooler, which is where the condensate load, the drain traps and the dryer sizing come from. None of that appears here.
What else the result does not include
- Volumetric efficiency and clearance. A reciprocating machine does not deliver its swept volume. Clearance re-expansion reduces delivered flow, increasingly so at high ratio, which affects the machine size rather than the specific work.
- Mechanical and drive losses. Bearings, seals, gearboxes and the couplings between them. The motor efficiency input covers the motor only.
- Pressure losses. Inlet filter, silencer, valves, intercoolers, aftercooler and piping all raise the pressure ratio the machine must actually produce above the ratio you entered. A 0.2 bar inlet loss and a 0.3 bar discharge loss turn a nominal 8:1 into something closer to 9:1.
- Surge, stonewall and turndown. A dynamic machine has an operating map, not a point. Whether the duty sits safely inside it is a selection question this calculation cannot answer.
- Multi-stage with unequal split, sidestreams or recycle. The geometric-mean rule assumes equal ratios, perfect intercooling and no extraction between stages.
Use it for: scoping a motor size, judging whether one stage or two, checking a vendor's figure is plausible, and seeing how discharge temperature moves with pressure ratio. Confirm with: real-gas software and vendor performance curves before anything is specified or purchased.
Sources
- ISO 1217, Displacement compressors — Acceptance tests — the basis on which free air delivery and specific power are measured and quoted.
- ASME PTC 10, Performance Test Code on Compressors and Exhausters — polytropic head and efficiency definitions.
- API Standard 617, Axial and Centrifugal Compressors and Expander-compressors; API Standard 618, Reciprocating Compressors for Petroleum, Chemical and Gas Industry Services — including discharge temperature limits.
- R. N. Brown, Compressors: Selection and Sizing — staging, real-gas treatment and machine selection.
- M. J. Moran and H. N. Shapiro, Fundamentals of Engineering Thermodynamics — isentropic and polytropic compression, reheat effect.
- Gas property data: γ and R for air at ambient conditions; compressibility factors from generalised correlations or an equation of state appropriate to the gas.
Frequently Asked Questions
Practical questions about assumptions, inputs and limitations.
The calculator requires mass flow. Convert actual or standard volumetric flow using density at the defined reference state.
It is a screening estimate for an ideal gas with constant properties. Actual discharge temperature depends on cooling, efficiency and gas properties.
No. Convert both pressures to absolute units before calculating pressure ratio.
Because a single stage at a high pressure ratio genuinely produces that temperature. Air taken from 1 to 8 bar absolute at 75% isentropic efficiency reaches about 337 C. That is above the point at which lubricating oil degrades, and in air service it is a recognised fire hazard, which is why reciprocating air compressors are typically limited to roughly 150 to 180 C per stage. A high calculated temperature is the calculator telling you the duty needs more than one stage.
For equal work split and perfect intercooling, use the geometric mean of the inlet and discharge pressures. Compressing from 1 to 8 bar absolute gives an intermediate pressure of the square root of 8, about 2.83 bar, and a ratio of 2.83 in each stage. For this example that drops discharge temperature from 337 C to 155 C and reduces shaft power by about 15%.
Isentropic, which is what this page assumes. Vendors and ASME PTC 10 normally quote polytropic efficiency, and the two are not interchangeable at high pressure ratio: because of the reheat effect, isentropic efficiency falls as ratio rises while polytropic efficiency stays roughly constant. Entering a polytropic figure in the isentropic field at a ratio of 8 under-predicts both the power and the discharge temperature.
For air and other light gases at modest pressure, yes. Judge it on reduced pressure, the operating pressure divided by the critical pressure: below about 0.1 the ideal-gas result is good, above about 0.3 it is not. Air at 8 bar has a compressibility factor near 0.998; methane at 70 bar is near 0.85, and heavy hydrocarbons, CO2 or any gas near its saturation line need a real-gas equation of state.
Because that machine is not adiabatic. Injected oil removes heat during compression rather than after it, so the real discharge temperature lands near 80 to 100 C instead of the adiabatic value and the actual work is lower than this page predicts. For oil-injected screws treat the result as a conservative upper bound.
Absolute, for both inlet and outlet. Pressure ratio is a ratio of absolute pressures, and using gauge values is one of the most common errors with this calculation. Standard plant air at 7 bar gauge is 8 bar absolute, and entering 7 and 1 instead of 8 and 1 understates the power by about 8% and the discharge temperature by around 20 K.