Formula Used
This calculator models a compressor using ideal-gas relationships and steady-flow energy. It estimates gas power and shaft power from temperature rise and efficiencies.
- Pressure ratio: r = P₂ / P₁
- Isentropic outlet temperature: T₂s = T₁ · r^((k−1)/k)
- Actual outlet temperature: T₂ = T₁ + (T₂s − T₁)/ηᵢₛ
- Polytropic option: T₂ = T₁ · r^((n−1)/n)
- Specific work: w = cₚ · (T₂ − T₁)
- Gas power: P_gas = ṁ · w
- Shaft power: P_shaft = P_gas / ηₘ
- Automatic cₚ: cₚ = kR/(k−1) (using R in kJ/(kg·K))
Note: Use absolute pressures and consistent gas properties for best results.
How to Use This Calculator
- Enter the mass flow rate and choose its unit.
- Enter inlet temperature and select the correct unit.
- Enter inlet and outlet absolute pressures with units.
- Select a model: isentropic uses k, polytropic uses n.
- Keep cp on auto, or enter it directly.
- Provide efficiencies as fractions or percentages.
- Click Calculate to view results above the form.
- Use CSV or PDF buttons to export the results.
Example Data Table
| ṁ (kg/s) | T₁ (K) | P₁ (kPa) | P₂ (kPa) | k | ηᵢₛ | ηₘ | Approx. Shaft Power (kW) |
|---|---|---|---|---|---|---|---|
| 0.75 | 300 | 101.325 | 600 | 1.4 | 0.80 | 0.95 | ~240 |
| 0.30 | 290 | 100 | 300 | 1.4 | 0.78 | 0.97 | ~55 |
| 1.20 | 310 | 120 | 900 | 1.33 | 0.82 | 0.96 | ~520 |
These values are illustrative and will vary with gas properties.
Compressor Power Article
1) What compressor power represents
Compressor power is the shaft energy rate needed to raise a gas from inlet pressure to discharge pressure at a given mass flow. This calculator estimates ideal thermodynamic power, then applies efficiency and mechanical losses to report realistic shaft and electrical power in kW or hp.
2) Key pressure and temperature inputs
Inlet pressure and outlet pressure define the pressure ratio, a primary driver of work. Inlet temperature sets the specific volume and the enthalpy rise. For air service near ambient, inlet temperatures of 288–310 K are common. Hot suction conditions materially increase power.
3) Isentropic vs polytropic compression
For an ideal isentropic step, the temperature rise follows the isentropic exponent k. Real machines behave closer to polytropic compression, especially with multiple stages and intercooling. If you know a polytropic exponent n, the calculator can use it to better match measured performance.
4) Mass flow, units, and density checks
Mass flow is entered directly, so volumetric conversions require density at inlet conditions. A quick sanity check: 1.0 m³/min of air at 101.3 kPa and 300 K is roughly 0.02 kg/s. If your result differs greatly, recheck pressure units and temperature scale. For dry air, cp near 1005 J/kg·K is often used.
5) Efficiency and loss modeling
Isentropic efficiency ηis typically ranges from 0.65–0.85 for small to medium compressors, while large optimized units can be higher. Mechanical efficiency ηm often sits around 0.90–0.98 depending on bearings, gearing, and coupling losses. Lower efficiencies increase required shaft power proportionally.
6) Estimating electrical input power
Electrical input depends on motor efficiency and drive losses. For many industrial motors, rated efficiencies around 0.90–0.96 are typical, and VFD losses may add 1–3%. If you include motor efficiency, the calculator reports both shaft power and approximate electrical power, supporting breaker and cable sizing.
7) Typical power scaling with pressure ratio
As pressure ratio increases, power rises faster than linearly because discharge temperature climbs. For air with k ≈ 1.4, moving from a 2:1 to a 4:1 ratio can increase ideal work by well over 50% at constant mass flow. Multi‑stage compression with intercooling reduces temperature rise and can cut power.
8) Practical application notes
Use consistent absolute pressures; gauge values must be converted to absolute by adding atmospheric pressure. Confirm allowable discharge temperature, lubrication limits, and cooling capacity. Treat results as engineering estimates and validate against manufacturer curves for final selection.
FAQs
What is the difference between ideal and shaft power?
Ideal power comes from thermodynamic compression work only. Shaft power divides ideal power by isentropic efficiency and then accounts for mechanical losses. Electrical power further divides by motor and drive efficiency.
Should I enter gauge or absolute pressure?
Enter absolute pressures. If you have gauge values, add local atmospheric pressure to convert. Using gauge directly will understate the pressure ratio and produce an unrealistically low power estimate.
Which exponent should I use: k or n?
Use k for an isentropic estimate when no test data exists. Use a polytropic exponent n when you have vendor curves or measured temperatures, because n can capture heat transfer and staging effects.
Why does power increase strongly with pressure ratio?
Higher ratios raise discharge temperature and the required enthalpy rise. The work scales with a power of the pressure ratio, not linearly, so doubling discharge pressure can add much more than double the power.
What typical efficiencies are reasonable?
For many industrial air compressors, ηis of 0.70–0.85 is common, and mechanical efficiency of 0.90–0.98 is typical. Use manufacturer data whenever possible for final sizing.
How do I convert volumetric flow to mass flow?
Compute inlet density from pressure and temperature, then multiply by inlet volumetric flow. For air near 101.3 kPa and 300 K, 1.0 m³/min is roughly 0.02 kg/s.
Does this include intercooling or multi‑stage compression?
The calculator models a single equivalent compression step. Intercooling and multiple stages can reduce power significantly at high ratios. For staged systems, use an effective exponent or compare against vendor stage data.