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Aerospace
Compressor specific work from total pressure ratio
Shaft work per kilogram of gas to compress it through total pressure ratio πc, from the inlet total temperature, constant cp and a total-to-total isentropic efficiency ηc.
- Ref
- AER-48
- Rev
- v1
- Field
- Aerospace
- Result
- wc [J/kg]
Adiabatic compression to a total pressure ratio πc needs the isentropic work cpTt,in[πc^((γ−1)/γ) − 1], and real compressors need more by the factor 1/ηc, the total-to-total isentropic efficiency. The result is the shaft work per kilogram of gas, before mechanical losses. Higher inlet temperature costs more work for the same pressure ratio, which is why cooling the intake helps.
Shaft work per kilogram of gas to compress it through total pressure ratio πc, from the inlet total temperature, constant cp and a total-to-total isentropic efficiency ηc. It assumes adiabatic compression, and it excludes mechanical losses.
Calculator
- Ideal (isentropic) work146,500 J/kg
- Outlet total temperature, Tt,in + w/cp471.5 K
- Power for 20 kg/s of air3,446 kW
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- Compressing the air to 4 times the inlet pressure needs 172.3 kJ/kg and heats it to 471.53 K.
- The inefficiency adds 25,845 J/kg, 17.65 % on top of the ideal work.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (J/kg) | Local step | Change in result |
|---|---|---|---|---|---|---|
| πc Total pressure ratio | 1 to 40 | 0 | 662,600 | 662,600 | +10% | +8.44 % |
| γ Heat-capacity ratio | 1.1 to 1.67 | 47,620 | 263,800 | 216,200 | +10% | +28.8 % |
| Tt,in Inlet total temperature | 200 to 500 K | 114,900 | 287,200 | 172,300 | +10% | +10 % |
| cp Specific heat at constant pressure | 900 to 1,300 J/(kg·K) | 154,400 | 223,000 | 68,610 | +10% | +10 % |
| ηc Isentropic efficiency | 0.7 to 0.95 | 154,200 | 209,200 | 55,060 | +10% | −9.09 % |
Sweep of Total pressure ratio
| Total pressure ratio | Specific work (J/kg) |
|---|---|
| 1 | 0 |
| 5.875 | 233,500 |
| 10.75 | 344,300 |
| 15.63 | 423,000 |
| 20.5 | 485,800 |
| 25.38 | 538,600 |
| 30.25 | 584,600 |
| 35.13 | 625,500 |
| 40 | 662,600 |
Sweep one variable
Variables and how each one contributes
Specific work
What the equation gives you. Everything else feeds it.
Specific heat at constant pressure
The specific heat at constant pressure, in J/(kg·K). Raises the work in direct proportion. Raising it by 10 % (to 1,105) moves the result by +10 %, from 172,300 to 189,500 J/kg.
Inlet total temperature
The inlet total temperature, in K. Raises the work in direct proportion. Raising it by 10 % (to 330) moves the result by +10 %, from 172,300 to 189,500 J/kg.
Heat-capacity ratio
The heat-capacity ratio cp/cv of the gas, about 1.4 for air. A lower γ (hotter, more complex gas) changes the compressible-flow exponents and the sizes of the shock and expansion effects. Raising it by 10 % (to 1.54) moves the result by +28.8 %, from 172,300 to 221,900 J/kg.
Total pressure ratio
The total pressure ratio, outlet over inlet, at least 1. Raises the work, slowly at first and then ever more slowly in proportion. Raising it by 10 % (to 4.4) moves the result by +8.44 %, from 172,300 to 186,800 J/kg.
Isentropic efficiency
The total-to-total isentropic efficiency of the compressor. A lower efficiency raises the work in inverse proportion. Raising it by 10 % (to 0.935) moves the result by −9.09 %, from 172,300 to 156,600 J/kg.
Worked example with the values above
- Temperature rise term: πc^((γ−1)/γ) − 1 = 4^0.28571 − 1 = 0.485994.
- Ideal work: cp Tt × term = 1,005 × 300 × 0.485994 = 146,454.4 J/kg.
- wc = ideal work / ηc = 172,299.3 J/kg.
Assumptions and limits
- Constant specific heats in each stream, adiabatic components and total-to-total efficiencies.
- Steady operation at one operating point, with no bleed and no leakage unless the page says so.
Common mistakes
- Mixing total and static states.
- Using a total-to-static efficiency where a total-to-total one is needed.
- Forgetting the fuel added to the turbine mass flow.
Statistics and uncertainty
Constant-cp cycle calculations are rough: cp rises from 1005 J/(kg·K) in cold air to about 1150 J/(kg·K) in hot gas, which shifts results by 5 % to 10 %. Component efficiencies from maps are good to 1 % to 2 % at design and worse off design.
Where it is used
- Matching compressors, combustors and turbines and finding fuel flow and power.
- Sizing engines and checking cycle designs.
Maths you need
- Isentropic efficiency: ideal work divided by actual work (or the reverse for a turbine)
- Total temperature: the temperature the gas would reach at rest
- Brayton cycle: compression, heat addition, expansion