Compressor Work
Shape gas processes and calculate boundary work.
Air cylinder
Closed-system boundary work: expansion work by the gas is positive; compression work by the gas is negative.
Process challenge
Pressure–volume process
The shaded strips represent closed-system boundary work, ∫p dV. Steady-flow compressor (shaft) work is calculated separately as −∫V dp.
The physics
For air, R = 0.287 kJ kg⁻¹ K⁻¹ and γ = 1.4. Perfect-gas equation: pV = mRT. With p in kPa and V in m³, pV is in kJ.
Isothermal process: n = 1, pV = constant, T₂ = T₁ and closed-system boundary work Wby = p₁V₁ ln(V₂/V₁).
Adiabatic reversible process: n = γ = 1.4, pVγ = constant, and Q = 0.
Polytropic process: pVn = constant. For n ≠ 1, closed-system boundary work Wby = (p₂V₂ − p₁V₁)/(1 − n).
Closed-system boundary work Wby is done by the gas: positive for expansion and negative for compression. Closed-system compression input work is −Wby.
For a reversible steady-flow compressor, the shaft work done by the fluid is Wshaft,by = −∫V dp. The positive compressor shaft input is Wshaft,in = ∫V dp. For p₂/p₁ > 1, isothermal compression requires the least shaft input and reversible adiabatic compression the most, for the same initial state and pressure ratio.
Steady-flow compressor input, n = 1: Wshaft,in = p₁V₁ ln(p₂/p₁). For n ≠ 1: Wshaft,in = [n/(n − 1)]p₁V₁[(p₂/p₁)(n−1)/n − 1].
Temperature ratio: T₂/T₁ = (p₂/p₁)(n−1)/n. Equivalently, T₂/T₁ = (V₁/V₂)n−1.
Assumptions: reversible, quasi-static perfect-gas processes, constant process index and negligible kinetic and potential energy changes. No mechanical losses are included. The steady-flow compressor comparison additionally assumes steady operation and an internally reversible flow process.