Engineering task and calculation objective
The QVER module determines the specific compression work for the compression of gases according to the Energietechnische Arbeitsmappe (14th edition), a standard German energy engineering reference. From the isentropic exponent, compressibility factor, gas constant or molar mass, suction temperature and pressure ratio p2/p1, it calculates the work per unit mass flow — both for isentropic compression and for the limiting case of isothermal compression.
Being able to calculate the specific compression work is the first step in any compressor design: multiplied by the mass flow and divided by the efficiencies, it yields the power requirement of reciprocating, screw and turbo compressors. Applications range from compressed air stations and process gas compressors to natural gas and hydrogen compression in plant engineering.
Comparing the two limiting cases also reveals the savings potential of intercooling: isothermal compression represents the theoretical minimum work, while real cooled multistage compression lies between the isothermal and isentropic values.
Standard and calculation basis: Energietechnische Arbeitsmappe 14. Auflage: 1995
Calculation workflow
- Define the gas data: The inputs are the isentropic exponent κ, compressibility factor Z, molar mass or specific gas constant of the gas, and the suction temperature. For gas mixtures, these values can be imported from property modules such as Gasmix.
- Specify the pressure ratio: The pressure ratio p2/p1 formed from suction and discharge pressure (absolute pressures!) defines the compression duty. For multistage machines with intercooling, the stage pressure ratio is considered.
- Isentropic compression work: The specific work of reversible adiabatic compression follows from κ/(κ−1)·Z·R·T1·[(p2/p1)^((κ−1)/κ) − 1]. The compressibility factor corrects for the deviation from ideal gas behavior at the suction state.
- Isothermal compression work for comparison: For fully cooled compression the result is Z·R·T1·ln(p2/p1). The difference between the two values shows the maximum work that could be saved by intercooling.
- From work to power: The specific work, multiplied by the mass flow and divided by the isentropic or isothermal efficiency and the mechanical efficiency, yields the power requirement of the machine.
Input quantities
| Quantity | Symbol | Unit |
|---|---|---|
| Isentropic exponent | k | – |
| Compressibility | Z | – |
| Gas constant | R | J/(kg·K) |
| Molar mass | M∙ | kg/kmol |
| Temperature | T | K |
| Pressure ratio (p2/p1) | dp | – |
Calculated results
| Quantity | Symbol | Unit |
|---|---|---|
| spec. compression power | Ys | J/kg |
| spec. compression power (isothermal) | Yt | J/kg |
Worked example
Air (κ = 1.4; M = 28.96 kg/kmol; Z = 1.0) starting from 20 °C is compressed to three times the suction pressure (p2/p1 = 3). This worked example determines the specific isentropic compression work and, for comparison, the isothermal compression work.
Given values
| Isentropic exponent κ | 1.4 |
| Compressibility factor Z | 1.0 |
| Molar mass M | 28.96 kg/kmol |
| Suction temperature T1 | 20 °C = 293.15 K |
| Pressure ratio p2/p1 | 3 |
Solution
Specific gas constant
R = Ru / M = 8,314.46 / 28.96 = 287.1 J/(kg·K)
Isentropic compression work
ws = κ/(κ−1) · Z · R · T1 · [(p2/p1)(κ−1)/κ − 1]
With (κ−1)/κ = 0.2857 and 30.2857 = 1.3687:
ws = 3.5 · 1.0 · 287.1 · 293.15 · 0.3687 J/kg
ws ≈ 108.6 kJ/kg
Isothermal compression work
wisoth = Z · R · T1 · ln(p2/p1) = 1.0 · 287.1 · 293.15 · ln 3 J/kg
wisoth ≈ 92.5 kJ/kg
Isothermal compression requires about 15% less work — the theoretical savings potential of cooling at this pressure ratio.
Result
| Isentropic specific work ws | 108.6 kJ/kg |
| Isothermal specific work | 92.5 kJ/kg |
All values are illustrative. The applicable standard and project-specific boundary conditions remain authoritative.
Frequently asked questions
When do I calculate isentropically, when isothermally?
Isentropic compression is the reference model for uncooled machines such as turbo compressor stages; real machines are benchmarked against it via the isentropic efficiency. Isothermal compression is the theoretical best case with complete cooling and serves as the reference for strongly cooled reciprocating and screw compressors (isothermal efficiency). Real multistage compression with intercooling lies between the two limiting cases.
Why does the discharge temperature rise so sharply at high pressure ratios?
For isentropic compression, T2 = T1·(p2/p1)^((κ−1)/κ). At a pressure ratio of 3, air (κ = 1.4) already heats up from 20 °C to a good 120 °C. In practice the discharge temperature limits the stage pressure ratio — because of lubricating oil coking, material limits, and, for gases such as hydrogen or natural gas, also for safety reasons. That is why compression is done in multiple stages with intercooling.
What role does the compressibility factor Z play?
Z corrects the suction volume of real gases relative to the ideal gas. For air at moderate pressures Z ≈ 1 and is negligible; at high pressures, low temperatures, or for gases near saturation (CO2, natural gas at high pressure) Z deviates significantly from 1 and enters the compression work in direct proportion.
Is the calculated specific work already the coupling power?
No. The formulas give the reversible reference work. The actual coupling power is obtained only after dividing by the isentropic (or isothermal) efficiency of the machine — typically 0.7 to 0.85 — and the mechanical efficiency, each multiplied by the mass flow.