Formula used
For an ideal Stirling cycle with perfect regeneration, the thermal efficiency equals the Carnot limit:
η = 1 − Tc/Th
- Th is the hot reservoir temperature (Kelvin).
- Tc is the cold reservoir temperature (Kelvin).
- η is efficiency (dimensionless).
If you provide heat input Qin, the tool also estimates
W = η·Qin and Qout = Qin − W.
These are idealized values for quick comparisons.
How to use this calculator
- Enter hot and cold reservoir temperatures, then choose their units.
- Select a display format for efficiency (percent, decimal, or both).
- Optional: enable heat input to estimate work and rejected heat.
- Click Calculate to see results above the form.
- Use Download CSV or Download PDF to export the last result.
Example data table
| Case | Hot temperature (K) | Cold temperature (K) | Ideal efficiency (%) | Example Qin (J) | Estimated work (J) |
|---|---|---|---|---|---|
| A | 900 | 300 | 66.667 | 1200 | 800 |
| B | 700 | 320 | 54.286 | 1000 | 542.86 |
| C | 1000 | 400 | 60.000 | 1500 | 900 |
Examples assume ideal regeneration and no mechanical losses.
Notes and limits
- Real engines are lower due to imperfect regeneration and friction.
- Always convert to Kelvin internally; the tool does this automatically.
- If temperatures are close, efficiency becomes small and sensitive.
Professional guide
1) What Stirling efficiency represents
This calculator reports the ideal thermal efficiency of a Stirling cycle when the regenerator is perfect. Under that assumption, the Stirling cycle reaches the same efficiency limit as a reversible heat engine operating between two reservoirs. The result is a fast benchmark for feasibility studies, comparisons, and sensitivity checks.
2) Temperature limits drive the ceiling
Efficiency depends only on the hot and cold reservoir temperatures: higher Th and lower Tc increase the limit. For example, 900 K and 300 K gives 66.667%, while 700 K and 320 K gives 54.286%. Small shifts matter: with Th=900 K, raising Tc from 300 K to 330 K drops the limit to 63.333%.
3) Why the ideal Stirling limit matches Carnot
The idealized Stirling cycle uses two isothermal processes and two constant-volume regenerative processes.
With perfect regeneration, internal heat shuttling does not require external heat exchange, so the only net external heat transfers
occur during isothermal expansion at Th and isothermal compression at Tc. Reversibility then yields
η = 1 − Tc/Th.
4) Using heat input to estimate work
When you enable heat input, the tool computes W = η·Qin. If η=0.60 and Qin=1500 J, the ideal work is
900 J and rejected heat is 600 J. These numbers are useful for sizing, but they ignore mechanical losses and imperfect heat transfer.
5) Real engines: typical performance ranges
Practical Stirling engines fall below the ideal limit due to pressure drops, finite heat exchanger effectiveness, leakage, and friction. A common engineering shorthand is to express real performance as a fraction of the Carnot limit. Many prototypes and commercial designs operate around 20–40% of the Carnot efficiency, depending on scale, materials, and operating frequency.
6) Regenerator effectiveness matters
The regenerator reduces external heat demand by storing heat between strokes. If regenerator effectiveness is below unity, additional external heat is required and efficiency falls. Improvements in matrix material, surface area, and flow distribution can raise effectiveness, but often increase pressure loss—so optimization balances heat recovery against pumping work.
7) Design sensitivity and safe inputs
Because the formula uses absolute temperature, always validate units. Negative Celsius values are allowed as long as the converted Kelvin value stays above zero. Also note that when Tc approaches Th, the calculated efficiency becomes small and sensitive, so measurement uncertainty can dominate early-stage comparisons.
8) How to interpret results for decisions
Use the calculator to set a best‑case target before detailed modeling. If the ideal limit is 55% and your application needs 25%, your realistic design goal might be 0.2–0.4 of Carnot, or roughly 11–22%. This helps decide whether to pursue higher temperature materials, better cooling, or alternative cycles.
FAQs
1) Is this efficiency the same as a real Stirling engine?
No. It is an ideal limit with perfect regeneration and no losses. Real engines are lower due to friction, leakage, pressure drops, and finite heat exchanger effectiveness.
2) Why must temperatures be in Kelvin internally?
The efficiency formula uses absolute temperature. Converting °C or °F to Kelvin ensures the ratio Tc/Th is physically meaningful and avoids invalid results near absolute zero.
3) What happens if the cold temperature is greater than the hot temperature?
The cycle cannot operate as a heat engine. The calculator flags this because it would imply negative or nonphysical efficiency for power production.
4) How is work output computed from heat input?
When enabled, the tool multiplies heat input by ideal efficiency: W = η·Qin. It then estimates rejected heat as Qout = Qin − W.
5) Can I use kWh, Wh, or BTU for heat input?
Yes. The calculator carries the selected energy unit through to work and rejected heat outputs, so comparisons remain consistent as long as you keep one unit per scenario.
6) Does higher hot temperature always improve efficiency?
In the ideal model, yes, because η increases as Th rises. In practice, higher temperatures can increase losses, material stress, and heat exchanger limits.
7) What is a good way to use this tool in design?
Use it as a ceiling estimate. Compare the ideal limit to your required efficiency, then apply a realistic fraction of Carnot (often 0.2–0.4) to set achievable performance targets.