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Ideal Brayton cycle: diagrams and efficiency
Short answer
The ideal Brayton cycle has two isentropic and two constant‑pressure processes. Efficiency derivation: η = 1 − q_out/q_in = 1 − (T₄−T₁)/(T₃−T₂). Using isentropic ratios T₂/T₁ = T₃/T₄ = r_p^((γ−1)/γ) yields η = 1 − 1/r_p^((γ−1)/γ).
Sketch an ideal Brayton cycle on P-v and T-s diagrams. Also derive an expression for its efficiency in terms of pressure ratio.
The ideal Brayton cycle is the air-standard cycle of a gas turbine. It has two isentropic and two constant-pressure processes.
Diagrams
P-v diagram T-s diagram
P T
| 2 ____________ 3 | 3
| \ \ | /|
| \ \ | / |
| '. '. | / |
| '. '. | 2 _.-' |
| '-._ '-._ | | 4
| '- 1 _________ 4 | | .-'
| | | _.-'
| | 1.-'
+----------------------> v +--------------------> s
1-2 isentropic compression 2-3 constant-P heat addition
3-4 isentropic expansion 4-1 constant-P heat rejection
Derivation of efficiency
Take air as an ideal gas with constant cp. Per kg of air:
- Heat added (2-3, constant P): q_in = cp (T₃ − T₂)
- Heat rejected (4-1, constant P): q_out = cp (T₄ − T₁)
Thermal efficiency:
η = w_net/q_in = 1 − q_out/q_in = 1 − (T₄ − T₁)/(T₃ − T₂)
For the isentropic processes 1-2 and 3-4, with P₂ = P₃ and P₁ = P₄, the pressure ratio is r_p = P₂/P₁ = P₃/P₄:
T₂/T₁ = (P₂/P₁)^((γ−1)/γ) = r_p^((γ−1)/γ) T₃/T₄ = (P₃/P₄)^((γ−1)/γ) = r_p^((γ−1)/γ)
Hence T₂/T₁ = T₃/T₄, which gives T₃/T₂ = T₄/T₁. Then
(T₄ − T₁)/(T₃ − T₂) = (T₁/T₂) × (T₄/T₁ − 1)/(T₃/T₂ − 1) = T₁/T₂
So
η = 1 − T₁/T₂
η_Brayton = 1 − 1/r_p^((γ−1)/γ)
The efficiency depends only on the pressure ratio and γ: it rises as r_p rises. For example, r_p = 10 and γ = 1.4 gives η = 1 − 1/1.931 = 48.2 %.
✓ Answer checked by Claude Opus 5.5 (high effort)
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