Ideal and real Brayton cycle for jet engines

Why it matters

Every gas-turbine engine — turbojet, turbofan, turboprop, turboshaft and the APU in the tail — runs on the Brayton cycle. The ideal cycle tells you that efficiency is set by pressure ratio and specific work by turbine inlet temperature; the real cycle, with component efficiencies and pressure losses, tells you why an engine with poor components may barely run at all. These two ideas drive the design of every compressor, combustor and turbine in the rest of this subject.

Key ideas

The four processes (closed-cycle idealisation).

• 1→2 compression in the compressor (isentropic in the ideal cycle).
• 2→3 heat addition at constant pressure in the combustor.
• 3→4 expansion in the turbine (and, in a jet engine, the nozzle) — isentropic in the ideal cycle.
• 4→1 heat rejection at constant pressure — in a real engine this is simply the hot exhaust mixing with the atmosphere; the open cycle is modelled as closed so cycle thermodynamics can be applied.

Air-standard assumptions. Working fluid is air, a calorically perfect gas with constant cp and γ (typically cp = 1005 J/(kg·K), γ = 1.4 for cold air; more accurate work uses hot-gas values of about 1150 J/(kg·K) and 1.33 after the combustor — take them from the question or data book). Fuel mass and kinetic energy changes inside the cycle are neglected.

What sets efficiency. For the ideal cycle, η depends only on the pressure ratio π: η = 1 − 1/π^((γ−1)/γ), equivalently 1 − T₁/T₂. Turbine inlet temperature does not appear.

What sets specific work. Net work per kg w = w_t − w_c depends on both π and the temperature ratio τ = T₃/T₁. For a given τ, w is zero at π = 1 and again at the π that makes T₂ = T₃; between them it peaks where T₂ = T₄, i.e. π^((γ−1)/γ) = √τ. A higher T₃ raises both the maximum work and the pressure ratio at which it occurs, so engines grow smaller and lighter for the same power.

The real cycle. Compression and expansion are not isentropic:

• Compressor isentropic efficiency η_c = (ideal work)/(actual work), so the actual temperature rise is larger than ideal.
• Turbine isentropic efficiency η_t = (actual work)/(ideal work), so less work is extracted.
• Stagnation pressure falls in the combustor (typically 2–6%) and in inlet and exhaust ducts.
• Turbine cooling air, bleed, and the change of cp and γ with temperature and composition also cost performance.

Back-work ratio. w_c/w_t is about 0.4–0.6 in gas turbines (much higher than steam plants), so a few points of component efficiency change net work a lot. With poor enough components the turbine cannot even drive the compressor — the reason early gas turbines failed. In the real cycle, efficiency also depends on T₃ and has its own optimum pressure ratio, higher than the one for maximum work.

Link to jet engines. In a turbojet the turbine extracts only the compressor work; the remaining enthalpy is expanded in the nozzle to make jet kinetic energy. The cycle's "net work" becomes the kinetic-energy gain used in the thermal efficiency of the propulsion topics.

Formulas

T₂/T₁ = π^((γ−1)/γ), T₃/T₄ = π^((γ−1)/γ) — isentropic compression and expansion; π = p₂/p₁ pressure ratio, T in K, γ ratio of specific heats.

w_c = cp·(T₂ − T₁), w_t = cp·(T₃ − T₄), q_in = cp·(T₃ − T₂) — per unit mass (J/kg), cp in J/(kg·K).

η_ideal = 1 − 1/π^((γ−1)/γ) = 1 − T₁/T₂ — ideal Brayton thermal efficiency.

π_opt^((γ−1)/γ) = √(T₃/T₁) — pressure ratio for maximum ideal net work; then w_max = cp·T₁·(√τ − 1)² with τ = T₃/T₁.

η_c = (T₂s − T₁)/(T₂ − T₁), η_t = (T₃ − T₄)/(T₃ − T₄s) — isentropic efficiencies; subscript s = isentropic end state.

η_real = (w_t − w_c)/q_in, BWR = w_c/w_t — real-cycle efficiency and back-work ratio.

P = ṁ·w_net — power (W) for mass flow ṁ (kg/s).

Worked examples

Example 1 (ideal cycle). Air enters at T₁ = 288 K; π = 10; T₃ = 1400 K; cp = 1005 J/(kg·K), γ = 1.4. Find T₂, T₄, net work, efficiency, and the pressure ratio for maximum work.

1. Exponent (γ − 1)/γ = 0.2857; π^0.2857 = 10^0.2857 = 1.9307.
2. T₂ = 288 × 1.9307 = 556.0 K; T₄ = 1400/1.9307 = 725.1 K.
3. w_c = 1005 × (556.0 − 288) = 269.4 kJ/kg; w_t = 1005 × (1400 − 725.1) = 678.2 kJ/kg.
4. w_net = 678.2 − 269.4 = 408.9 kJ/kg; q_in = 1005 × (1400 − 556.0) = 848.2 kJ/kg.
5. η = 408.9/848.2 = 0.482; check 1 − 1/1.9307 = 0.482.
6. Max-work pressure ratio: √(1400/288) = 2.2048 → π_opt = 2.2048^(1/0.2857) = 2.2048^3.5 = 15.9.

Answer: T₂ = 556 K, T₄ = 725 K, w_net ≈ 409 kJ/kg, η ≈ 0.482, π_opt ≈ 15.9 (giving w_max ≈ 420 kJ/kg).

Example 2 (GATE level, real cycle). Same conditions, but η_c = 0.85 and η_t = 0.90. Find net work, thermal efficiency and back-work ratio.

1. T₂ = T₁ + (T₂s − T₁)/η_c = 288 + 268.0/0.85 = 603.3 K.
2. T₄ = T₃ − η_t·(T₃ − T₄s) = 1400 − 0.90 × 674.9 = 792.6 K.
3. w_c = 1005 × 315.3 = 316.9 kJ/kg; w_t = 1005 × 607.4 = 610.4 kJ/kg.
4. w_net = 610.4 − 316.9 = 293.5 kJ/kg; q_in = 1005 × (1400 − 603.3) = 800.6 kJ/kg.
5. η = 293.5/800.6 = 0.367; BWR = 316.9/610.4 = 0.519.

Answer: w_net ≈ 293.5 kJ/kg, η ≈ 0.367, BWR ≈ 0.52. Component losses of 10–15% cut net work by 28% and efficiency from 48% to 37%.

Common mistakes

• Writing η = 1 − 1/π^(γ−1); the exponent is (γ − 1)/γ when π is a pressure ratio. The form 1 − 1/r^(γ−1) is for a volume (compression) ratio.
• Applying η_c the wrong way round: actual compressor work is larger than ideal (divide by η_c); actual turbine work is smaller (multiply by η_t).
• Assuming a higher T₃ raises ideal-cycle efficiency; it raises specific work, and raises efficiency only in the real cycle.
• Using kJ and J inconsistently with cp in J/(kg·K).
• Forgetting the combustor pressure drop when computing turbine pressure ratio in a real-cycle problem.

For GATE AE

Expect numericals on ideal efficiency from pressure ratio, compressor and turbine temperatures and works, net specific work and power, and the optimum pressure ratio for maximum work. Real-cycle questions add isentropic efficiencies and occasionally a combustor pressure loss or different cp for hot gas. Conceptual questions ask what pressure ratio and turbine temperature each control, and what the back-work ratio means.

Quick check

1. Ideal Brayton efficiency for π = 16, γ = 1.4?
2. In an ideal cycle, which variable alone sets efficiency?
3. What condition on temperatures gives maximum ideal net work?
4. If η_c falls, does compressor exit temperature rise or fall for the same pressure ratio?
5. Why must back-work ratio be well below 1 for a useful engine?

Answers: 1. 1 − 1/16^0.2857 = 0.547. 2. Pressure ratio (for a given γ). 3. T₂ = T₄, i.e. π^((γ−1)/γ) = √(T₃/T₁). 4. It rises. 5. Because net work = w_t(1 − BWR); as BWR approaches 1 nothing is left for thrust or shaft power.