The Stirling Cycle: Isothermal Legs with Regeneration

1 · Predict

The Stirling cycle uses 2 isothermal legs and 2 isochoric legs, no adiabatic legs at all. Its efficiency can in principle match the Carnot value if a perfect "regenerator" recycles heat between the isochoric legs. Without an ideal regenerator, does the simple cycle efficiency you measure fall short of the Carnot bound?

2 · Set Up

  1. Open the stirling-cycle preset and press Reset. 1 mol of argon cycles between 600 K and 300 K using isothermal and isochoric legs only.
  2. Enable the per-leg work readout and the cycle-efficiency readout.
  3. For each leg of the cycle (1 through 4), read off the leg's target volume and the work done by the gas during that leg.

3 · Collect Data

Leg numberLeg's target volume V (m³)Work done by gas this leg (J)
1
2
3
4

Sketch the P-V diagram for all 4 legs, marking the 2 isothermal (curved) and 2 isochoric (vertical) segments.

4 · Analyze

  1. Confirm legs 2 and 4 (isochoric) show W = 0. Sum all 4 legs' work for the net cycle work, and compare the cycle's efficiency reading to the Carnot bound η_c = 1 − 300/600 = 0.5.
  2. This cycle's raw efficiency (without an ideal regenerator recovering the isochoric legs' heat) stays below 0.5. Explain, using the first law, why heat dumped during the isochoric-cooling leg represents energy the simple engine doesn't recover.

5 · Extend

  1. A real Stirling engine uses a regenerator — a mesh that stores heat released during isochoric cooling and returns it during isochoric heating. Explain why a perfect regenerator would let this cycle approach the Carnot efficiency even without adiabatic legs.
  2. Stirling engines are prized for running on any external heat source (solar, waste heat, even temperature differences in space) rather than requiring internal combustion. Why might a closed-cycle engine with external heating be attractive for those applications?

The Physics Behind This Experiment

Carnot Bound on Cycle Efficiency

No cycle operating between temperatures Tc and Th — Stirling included — can exceed the Carnot efficiency η_c = 1 − Tc/Th. A real (non-regenerated) Stirling cycle falls short of this bound because it dumps unrecovered heat during its isochoric-cooling leg.

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