Calculator Inputs
Example Data Table
| Case | Gas | Model | P1 | P2 | T1 | ṁ | Stages | ηc | Actual Work | Electrical Power |
|---|---|---|---|---|---|---|---|---|---|---|
| Reference Study | Air | Isentropic | 1 bar | 7 bar | 25 °C | 2 kg/s | 2 with perfect intercooling | 85% | 225.83 kJ/kg | 485.13 kW |
This sample uses mechanical efficiency of 98%, motor efficiency of 95%, 8000 operating hours, and an electricity rate of 0.12 per kWh.
Formula Used
Pressure ratio: rp = P2 / P1
Stage pressure ratio: rs = rp1 / N
Inlet density: ρ1 = P1 / (ZRT1)
Inlet volumetric flow: V̇1 = ṁ / ρ1
Isothermal specific work: w = ZRT1 ln(P2 / P1)
Isentropic discharge temperature: T2 = T1(P2 / P1)(k-1)/k
Isentropic specific work: w = (k / (k - 1)) ZRT1[(P2 / P1)(k-1)/k - 1]
Polytropic discharge temperature: T2 = T1(P2 / P1)(n-1)/n
Polytropic specific work: w = (n / (n - 1)) ZRT1[(P2 / P1)(n-1)/n - 1]
Actual specific work: wactual = wideal / ηc
Gas power: Pgas = ṁ × w
Shaft power: Pshaft = Pgas / ηmech
Electrical power: Pelec = Pshaft / ηmotor
Annual energy: E = Pelec,kW × operating hours
Annual cost: Cost = E × electricity rate
How to Use This Calculator
- Choose a gas preset or enter custom gas properties.
- Select isothermal, isentropic, or polytropic compression.
- Enter inlet pressure, outlet pressure, and inlet temperature.
- Provide mass flow rate, compressor efficiency, and stage count.
- Add optional mechanical and motor efficiencies for electrical demand.
- Enter yearly operating hours and your electricity rate.
- Click the calculate button to see work, power, temperature, and cost.
- Use the CSV or PDF buttons to export the result summary.
FAQs
1. What does compressor work mean?
Compressor work is the energy needed to raise a gas from inlet pressure to outlet pressure. It is often expressed as specific work in kJ/kg and then converted into power using mass flow rate.
2. Which model should I choose?
Use isothermal for ideal cooling, isentropic for adiabatic benchmark estimates, and polytropic when measured compressor behavior falls between those extremes. Polytropic often matches practical rotating compressor studies better.
3. Why is actual work higher than ideal work?
Real compressors have losses from friction, leakage, turbulence, heat transfer, and imperfect compression paths. Efficiency accounts for those losses, so actual work and required power are higher than ideal theoretical values.
4. What is the benefit of multistage compression?
Splitting compression into stages reduces stage pressure ratio and can lower required work when intercooling is used. It also helps control discharge temperature and improves equipment reliability at high pressure ratios.
5. Why does the calculator ask for gas constant and k?
Different gases respond differently during compression. The gas constant affects density and work, while the specific heat ratio affects temperature rise and isentropic calculations. Presets help fill typical values quickly.
6. What does the compressibility factor do?
The compressibility factor adjusts ideal-gas behavior for real-gas effects. A value near 1.0 means the gas behaves almost ideally. Deviations become more important at high pressures or unusual temperatures.
7. Can I estimate annual operating cost here?
Yes. Enter yearly operating hours and your electricity rate. The calculator estimates annual electrical energy use from calculated motor demand and multiplies it by the entered rate for a rough operating cost.
8. Is this suitable for final equipment design?
It is useful for screening, budgeting, and comparison studies. Final equipment design should also include vendor curves, suction conditions, gas composition, Reynolds effects, allowable temperature limits, and project safety margins.