Calculate Brayton cycle efficiency, compressor pressure ratio, or specific heat ratio from any two inputs in an ideal gas turbine cycle.
Brayton Cycle Efficiency Formula
The ideal Brayton cycle thermal efficiency depends on the compressor pressure ratio and the specific heat ratio of the working gas. The calculator uses the air-standard Brayton cycle relation:
To solve for compressor pressure ratio:
To solve for specific heat ratio:
- η = Brayton cycle thermal efficiency, entered or displayed as a percent
- rp = compressor pressure ratio, equal to P2 / P1
- k = specific heat ratio, equal to Cp / Cv
- ln = natural logarithm
Choose the quantity to solve, enter the two displayed inputs, then select Calculate. For inverse k, pressure ratio must exceed 1 and efficiency must be greater than zero but below 100 × (1 − 1/rp). Pressure ratio 1 with zero efficiency does not determine a unique k.
These formulas assume constant specific heats, isentropic compression and expansion, and constant-pressure heat addition and rejection. Pressure ratio 1 is a degenerate zero-work limit. Actual gas turbines usually have lower efficiency because of compressor losses, turbine losses, pressure drops, heat losses, and non-ideal combustion.
Typical Specific Heat Ratio Values
The specific heat ratio depends on gas, temperature and state. The table gives approximate room-temperature ideal-gas values, not fixed values throughout a hot turbine cycle. For many Brayton cycle homework calculations using air, k = 1.4 is commonly used.
| Gas or Working Fluid | Typical k Value | Common Use |
|---|---|---|
| Air | 1.4 | Standard Brayton cycle problems |
| Nitrogen | About 1.4 | Approximation for air-like behavior |
| Carbon dioxide | About 1.3 | Gas calculations where CO2 is the working fluid |
| Helium | About 1.66 | Monatomic gas cycle examples |
Ideal Brayton Efficiency for Air
The table below shows ideal cycle efficiency values for air using k = 1.4.
| Pressure Ratio rp | Ideal Efficiency |
|---|---|
| 2 | 17.97% |
| 5 | 36.86% |
| 10 | 48.21% |
| 15 | 53.87% |
| 20 | 57.51% |
Example Brayton Cycle Efficiency Calculations
Example 1: Find efficiency from pressure ratio and k
Suppose the compressor pressure ratio is 10 and the working fluid is air with k = 1.4.
The ideal Brayton cycle efficiency is 48.2053%.
Example 2: Find pressure ratio from efficiency and k
Suppose the desired ideal efficiency is 50% and k = 1.4.
The required compressor pressure ratio is about 11.3137.
Brayton Cycle Efficiency FAQ
Why does Brayton cycle efficiency increase when pressure ratio increases?
In the ideal Brayton cycle, a higher compressor pressure ratio increases the temperature ratio during isentropic compression and expansion. In the efficiency formula, increasing rp makes the term 1 / rp(k-1)/k smaller, so the efficiency becomes larger.
Why must k be greater than 1?
The specific heat ratio is defined as k = Cp / Cv. For gases used in this type of thermodynamic calculation, Cp is greater than Cv, so k is greater than 1. If k is 1 or less, the ideal Brayton cycle efficiency formula is not valid.
Is this the same as actual gas turbine efficiency?
No. This is the ideal air-standard Brayton cycle efficiency. Actual gas turbine efficiency is affected by compressor efficiency, turbine efficiency, combustor pressure loss, mechanical losses, heat transfer, and real gas behavior. Use this result as an ideal cycle value, not as the guaranteed efficiency of a real machine.
