Calculator Inputs
Formula Used
The ideal Brayton (air-standard) thermal efficiency depends on pressure ratio and heat capacity ratio:
ηideal = 1 − 1 / rₚ(γ−1)/γ
For the real cycle, compressor and turbine isentropic efficiencies adjust temperature changes:
- Compressor: T₂s = T₁ · rₚ(γ−1)/γ, then T₂ = T₁ + (T₂s − T₁)/ηᶜ
- Turbine: T₄s = T₃ / rₚ(γ−1)/γ, then T₄ = T₃ − ηᵗ (T₃ − T₄s)
Thermal efficiency uses the temperature differences (cₚ cancels):
ηreal = 1 − (T₄ − T₁) / (T₃ − T₂)
Assumption: turbine expansion ratio equals compressor pressure ratio.
How to Use This Calculator
- Select Real for practical estimates, or Ideal for the air-standard limit.
- Enter the pressure ratio and γ. These drive the baseline cycle behavior.
- In Real mode, choose your temperature unit, then provide T₁ and T₃.
- Set ηᶜ and ηᵗ to represent component losses.
- Click Calculate. Results appear above the form for quick review.
- Use Download CSV or Download PDF to save outputs.
Example Data Table
| Mode | rₚ | γ | T₁ (K) | T₃ (K) | ηᶜ | ηᵗ | ηideal (%) | ηreal (%) |
|---|---|---|---|---|---|---|---|---|
| Ideal | 10 | 1.40 | — | — | — | — | 48.2 | — |
| Real | 10 | 1.40 | 300 | 1400 | 0.85 | 0.88 | 48.2 | 31–38 |
The real-cycle range depends strongly on component efficiencies and temperatures.
Brayton Cycle Efficiency in Practice
1) Cycle overview
The Brayton cycle models a gas turbine with compression, heat addition, expansion, and heat rejection. Thermal efficiency is the ratio of net work output to heat input.
2) Inputs this calculator uses
Ideal efficiency depends on pressure ratio rp and γ. Real mode adds T1, T3, compressor efficiency ηc, and turbine efficiency ηt, then estimates T2 and T4 from isentropic relations.
3) Pressure ratio trends
Higher rp generally increases ideal efficiency because the compressor raises temperature more for a given inlet state. Many industrial simple-cycle units operate around rp ≈ 6 to 15, while aero-derivative cores can exceed 25. Very high rp can reduce net work if T3 is limited.
4) Temperature limits and materials
Turbine inlet temperature T3 is a dominant lever for real-cycle performance. Modern machines may run roughly 1200 K to 1700 K depending on cooling and materials. Compressor inlet temperature T1 is often near ambient (about 288 K to 310 K), and higher T1 typically lowers efficiency.
5) Component efficiencies drive the gap
Isentropic efficiencies translate aerodynamic and mechanical losses into temperature penalties. Typical compressor values are about 0.80 to 0.90, and turbines often fall around 0.85 to 0.92. Lower ηc raises T2 (more work required), while lower ηt raises T4 (less work produced), both reducing thermal efficiency.
6) Back work ratio and net work
Gas turbines spend a large fraction of turbine work driving the compressor. The back work ratio (compressor work divided by turbine work) often lands around 0.4 to 0.7 in simple cycles. A lower ratio generally means more net specific work and better part-load flexibility.
7) Common enhancements beyond the simple cycle
Intercooling reduces compressor work, reheat increases turbine work, and regeneration recovers exhaust heat to preheat the compressed air. These features can raise efficiency or specific work, but they add pressure losses and hardware complexity. This calculator focuses on the simple-cycle baseline that is often used for screening and comparisons.
8) Interpreting results from this page
Use the ideal efficiency as an upper bound for a given rp and γ. In real mode, sanity-check that T2 remains below T3 and that the computed heat input (T3 - T2) stays positive. Export the table to document assumptions and compare scenarios consistently.
FAQs
1) What does the pressure ratio represent?
It is the compressor outlet pressure divided by compressor inlet pressure (p2/p1). Higher ratios usually improve the ideal efficiency, but practical gains depend on temperature limits and component efficiencies.
2) Why does γ affect the ideal efficiency?
γ controls the isentropic temperature change during compression and expansion. For air, γ is often near 1.4 at moderate temperatures, but it can decrease at high temperatures or with different working fluids.
3) Why is the real efficiency lower than the ideal value?
Real compressors and turbines are not perfectly isentropic. Losses increase the compressor temperature rise and reduce the turbine temperature drop, which reduces net work and increases the fraction of heat rejected.
4) Do I need cp to compute efficiency?
No. In this simplified model, cp cancels in the efficiency ratio. cp is only used if you enable optional outputs like specific net work (kJ/kg) and heat rate (kJ/kWh).
5) What happens if T3 is not much higher than T1?
The cycle has limited heat input and limited expansion work, so efficiency and net work typically drop. In extreme cases, the computed heat input can become non-positive, indicating an unrealistic input combination.
6) Does this tool include pressure losses and cooling flows?
No. It assumes the turbine expansion ratio equals the compressor pressure ratio and neglects combustor pressure loss and turbine cooling air. Those effects further reduce real performance, so treat outputs as screening-level estimates.
7) How can I improve efficiency in real systems?
Common paths include increasing turbine inlet temperature, improving component efficiencies, using recuperation at lower pressure ratios, or moving to combined-cycle operation that recovers exhaust heat in a bottoming steam cycle.