Reciprocating compressor calculator
Calculate the compression of air and technical gases in a reciprocating compressor: from delivered volume flow, suction and discharge pressure and suction temperature you get mass flow, compression power, discharge temperature and the heat to be removed. Isothermal, isentropic and polytropic compression are available; for several stages the tool assumes intercooling back to suction temperature and shows the optimum intermediate pressures along with the savings against a single-stage machine, live with every input.
Calculation
Idealised compression with intercooling.
The calculation is reversible isothermal, isentropic or polytropic; between the stages the gas is cooled back to suction temperature. The discharge temperature is classified against empirical values for oil-lubricated reciprocating compressors: about 180 °C at the discharge valve plate is the usual upper limit, from roughly 200 °C the oil mist becomes critical. This is a classification, not a verification against a standard.
Model: ideal gas with temperature-independent heat capacities, reversible compression without friction, valve and flow losses, without clearance volume and leakage; with several stages complete intercooling back to suction temperature. The output is the internal compression power, not the coupling or motor power. A design and comparison tool, not a replacement for the manufacturer's performance data.
Results
Calculating …
Calculation in your browser, inputs go to our server only when you export or save.
Formulas and fundamentals
Mass flow and suction volume flow
The delivered volume flow V̇₂ refers to discharge pressure p₂ at suction temperature T₁, that is to the state after aftercooling. The ideal gas equation of state therefore gives an unambiguous mass flow, no matter how hot the gas leaves the compression:
Discharge temperature
Every reversible compression follows p·vᵐ = const with the process exponent m: isothermal m = 1, isentropic m = κ, polytropic m = n. This gives the temperature at the end of the stage:
Technical work and compression power
A compressor is an open, flow-through system. The governing quantity is therefore the technical work w_t = ∫v dp and not the boundary work of a closed system. Depending on the type of compression:
The compression power is the product of mass flow and specific technical work:
Multi-stage compression with intercooling
If the gas is cooled back to suction temperature between the stages, the total power is smallest when all stages share the same pressure ratio. For z stages the stage pressure ratio, the intermediate pressures and the total power follow directly:
For two stages the optimum intermediate pressure is exactly the geometric mean p = √(p₁·p₂).
Cooling duty and swept volume
If the gas leaves the compressor at suction temperature again after inter- and aftercooling, its enthalpy is unchanged. The first law for the open process then gives the total heat to be removed. From speed n_D and volumetric efficiency λ follows the required swept volume:
Sign convention: work and heat supplied to the gas count positive, heat removed is therefore negative.
Worked example
A reciprocating compressor delivers 110 m³/h of air from 0.96 bar to 16 bar at a suction temperature of 20 °C (R_s = 287.2 J/(kg·K), c_p = 1.004 kJ/(kg·K), κ = 1.4). From ṁ = p₂·V̇₂/(R_s·T₁) the mass flow is ṁ = 0.58 kg/s (2088 kg/h), and the suction volume flow is V̇₁ = ṁ·R_s·T₁/p₁ = 0.51 m³/s.
Compressed isothermally the temperature stays at 20 °C. The power is P = ṁ·R_s·T₁·ln(16/0.96) = 138 kW, and since the internal energy does not change, exactly this energy has to be removed as heat: Q̇ = −138 kW. This is the lower bound a cooled compression could reach.
Isentropically and in a single stage the temperature rises to T₂ = 293 K·(16/0.96)^0.2857 = 655 K, that is 382 °C, and the power grows to P = ṁ·c_p·(T₂−T₁) = 211 kW. Such a discharge temperature is unusable for an oil-lubricated reciprocating compressor.
In two stages with intercooling back to 20 °C the optimum intermediate pressure is p = √(0.96·16) bar = 3.92 bar. Each stage then only compresses with Π_St = 4.08, the temperature after each stage is 438 K = 165 °C and the total power drops to 169 kW. Compared with the single-stage machine that is 19.8 % less power at the same mass flow. Source of the figures: Langeheinecke/Jany/Thieleke, Thermodynamik für Ingenieure, 6th edition, examples 7.4 and 7.5.
Frequently asked questions
Why does multi-stage compression with intercooling save power?
The technical work is the integral ∫v dp. The hotter the gas gets during compression, the larger its specific volume v and the more expensive every further pressure increment becomes. Intercooling brings the gas back to suction temperature, and therefore to a small volume, before the next stage and thus lowers the work of that stage. In the example 0.96 bar to 16 bar the isentropic power drops from 211 kW in one stage to 169 kW in two stages, that is by 19.8 % at the same mass flow. The limiting case of infinitely many stages with complete intercooling is the isothermal compression at 138 kW.
Which type of compression should I choose?
Isothermal and isentropic are the two limiting cases: isothermal corresponds to ideally cooled compression and gives the smallest possible power, isentropic to uncooled compression and gives the largest. A real water-cooled reciprocating compressor lies in between and is calculated polytropically with n ≈ 1.25 to 1.35 for air. Uncooled or high-speed machines are closer to κ = 1.4. The exponent only describes the cooling during compression, not friction.
How hot may the gas get at the outlet?
For oil-lubricated reciprocating compressors the limit is not the material but the lubricant. As a rule of thumb about 180 °C at the discharge valve plate is the usual upper limit for mineral oil; from roughly 200 °C the entrained oil mist becomes critical because deposits and, in the extreme case, self-ignition can occur. If the limit is exceeded, an additional stage with intercooling, cooler suction air or a synthetic oil with higher temperature stability will help. The calculator classifies the discharge temperature against these empirical values; it does not verify against a standard.
What is the volumetric efficiency and how large is it?
The volumetric efficiency λ is the ratio of the actually aspirated volume flow to the geometric swept volume flow, so λ = V̇₁/(V_h·n_D). It is smaller than one because the gas in the clearance volume expands again on the return stroke, because the valves throttle the flow, because the suction air is heated by the hot walls and because piston rings and valves leak. Typical values are 0.7 to 0.9 and drop as the stage pressure ratio rises. Without speed and volumetric efficiency the calculator leaves the swept volume empty.
What does the entered volume flow refer to?
To the discharge pressure p₂ at suction temperature, that is to the state after aftercooling. This is the usual way of stating the delivered quantity and makes the mass flow unambiguous, no matter how hot the gas leaves the compression. The corresponding suction volume flow V̇₁, which governs frame size and swept volume, is output as a separate result.
Is the compression power already the motor power?
No. The calculator gives the internal compression power of the idealised process. For the coupling power the mechanical efficiency has to be added (roughly 0.85 to 0.95 for reciprocating compressors, depending on the drive train and belt drive), and for the electrical input the motor efficiency on top of that. Valve and flow losses as well as leakage are likewise not contained in the idealised process.
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