Brayton cycle and gas turbines

The air-standard Brayton cycle: efficiency and net work versus pressure ratio and turbine inlet temperature, back-work ratio, isentropic efficiencies, regeneration, intercooling and reheat, with ideal and actual-cycle examples.

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Why it matters

Gas turbines power every jet aircraft, most naval and fast-patrol vessels, peaking power stations and the topping half of combined-cycle plants, and small ones drive APUs, microturbine generators and turbochargers. They are compact, light for their power and quick to start. The air-standard Brayton cycle explains their performance: why pressure ratio and turbine inlet temperature matter, why component efficiencies dominate, and when regeneration, intercooling and reheat help.

Key ideas

Open and closed cycles. A real gas turbine is open: air is drawn in, fuel burns in it, and the exhaust leaves. For analysis we use the air-standard model: air as an ideal gas of fixed composition, combustion replaced by heat addition from an external source, exhaust replaced by constant-pressure heat rejection. With constant specific heats this is the "cold air-standard" assumption.

Ideal Brayton (Joule) cycle. 1→2 isentropic compression; 2→3 heat addition at constant pressure (combustor); 3→4 isentropic expansion in the turbine; 4→1 heat rejection at constant pressure (to atmosphere or a cooler).

  • Ideal efficiency depends only on the pressure ratio r_p = p₂/p₁ and γ; it rises with r_p.
  • Net work per kg, however, depends on both r_p and the temperature ratio T₃/T₁. At a fixed T₃/T₁, net work is zero at r_p = 1 and again when T₂ reaches T₃, with a maximum in between. At that optimum T₂ = T₄ = √(T₁T₃).
  • Back-work ratio (compressor work ÷ turbine work) is high — 40–60 % — because the compressor handles a gas, not a liquid. This makes gas turbines very sensitive to component efficiencies: a few points lost in compressor or turbine efficiency can halve the net work.

Actual cycle. Isentropic efficiencies of compressor (η_c) and turbine (η_t), typically 0.80–0.90, and pressure losses in the combustor and ducts. The actual compressor exit is hotter, and the actual turbine exit is hotter, than the isentropic states.

Turbine inlet temperature (TIT). The key design parameter. Raising T₃ increases net work strongly, raises actual efficiency, and raises the optimum pressure ratio. It is limited by blade creep and oxidation, which is why modern machines use internal blade cooling, film cooling and thermal-barrier coatings to run gas well above the blade metal's limit.

Improvements.

  • Regeneration: a heat exchanger uses hot turbine exhaust to preheat compressed air before the combustor. It helps only when T₄ > T₂, so it suits low pressure ratios. Ideal regeneration gives η = 1 − (T₁/T₃) r_p^((γ−1)/γ), which falls as r_p rises — the opposite trend to the simple cycle.
  • Intercooling between compressor stages lowers compressor work and raises net work; on its own it lowers efficiency, because more fuel heat is needed, but it makes regeneration more effective.
  • Reheat between turbine stages raises turbine work; on its own it also lowers efficiency (hotter exhaust), but combined with regeneration it raises it. With many stages of intercooling, reheat and ideal regeneration, the cycle approaches the Ericsson cycle and Carnot efficiency.

Jet propulsion. In a turbojet the turbine produces only enough work to drive the compressor; the remaining enthalpy accelerates the gas in a nozzle, and thrust = ṁ (V_exit − V_flight) for a matched nozzle.

Formulas

r_p = p₂ / p₁ T₂ / T₁ = T₃ / T₄ = r_p^((γ−1)/γ) — isentropic compression and expansion. T in K, γ = c_p/c_v (1.4 for air). w_c = c_p (T₂ − T₁), w_t = c_p (T₃ − T₄), q_in = c_p (T₃ − T₂) (kJ/kg); c_p = 1.005 kJ/kg·K for air. η = 1 − 1 / r_p^((γ−1)/γ) = 1 − T₁ / T₂ — ideal simple Brayton cycle. r_p,opt = (T₃ / T₁)^(γ / (2(γ−1))) — pressure ratio for maximum net work (ideal cycle). w_net,max = c_p (√T₃ − √T₁)² η_c = (T₂s − T₁) / (T₂ − T₁), η_t = (T₃ − T₄) / (T₃ − T₄s) — isentropic efficiencies (s = isentropic state). Back-work ratio = w_c / w_t ε = (T₅ − T₂) / (T₄ − T₂) — regenerator effectiveness; T₅ = compressed air leaving the regenerator. Heat input becomes q_in = c_p (T₃ − T₅). η_regen,ideal = 1 − (T₁ / T₃) r_p^((γ−1)/γ) — ideal cycle with ε = 1. F = ṁ (V_e − V_a) — turbojet thrust (N) for a fully expanded nozzle; ṁ in kg/s, V in m/s.

Worked examples

Example 1 (standard, ideal cycle). Air enters the compressor of an ideal Brayton cycle at 100 kPa, 300 K. Pressure ratio 10, turbine inlet 1400 K. Take c_p = 1.005 kJ/kg·K, γ = 1.4. Find the net work, efficiency and back-work ratio.

  1. r_p^((γ−1)/γ) = 10^(0.2857) = 1.9307.
  2. T₂ = 300 × 1.9307 = 579.2 K; T₄ = 1400/1.9307 = 725.1 K.
  3. w_c = 1.005 × (579.2 − 300) = 280.6 kJ/kg; w_t = 1.005 × (1400 − 725.1) = 678.2 kJ/kg.
  4. w_net = 678.2 − 280.6 = 397.6 kJ/kg; q_in = 1.005 × (1400 − 579.2) = 824.9 kJ/kg.
  5. η = 397.6/824.9 = 0.482; check: 1 − 1/1.9307 = 0.482 ✓. Back-work ratio = 280.6/678.2 = 0.414.

w_net ≈ 398 kJ/kg, η ≈ 48.2 %, back-work ratio ≈ 0.41

Example 2 (GATE level, actual cycle with regenerator). Same cycle, but η_c = 0.85, η_t = 0.88. Find the actual net work and efficiency, then the efficiency with a regenerator of effectiveness 0.8.

  1. T₂ = T₁ + (T₂s − T₁)/η_c = 300 + 279.2/0.85 = 300 + 328.5 = 628.5 K.
  2. T₄ = T₃ − η_t (T₃ − T₄s) = 1400 − 0.88 × 674.9 = 1400 − 593.9 = 806.1 K.
  3. w_c = 1.005 × 328.5 = 330.1 kJ/kg; w_t = 1.005 × 593.9 = 596.9 kJ/kg; w_net = 266.7 kJ/kg.
  4. q_in = 1.005 × (1400 − 628.5) = 775.4 kJ/kg; η = 266.7/775.4 = 0.344. Back-work ratio = 330.1/596.9 = 0.553.
  5. Regenerator (T₄ = 806.1 K > T₂ = 628.5 K, so it helps): T₅ = T₂ + ε (T₄ − T₂) = 628.5 + 0.8 × 177.6 = 770.6 K.
  6. q_in = 1.005 × (1400 − 770.6) = 632.6 kJ/kg; work is unchanged, so η = 266.7/632.6 = 0.422.

Actual: w_net ≈ 267 kJ/kg, η ≈ 34.4 %; with regenerator η ≈ 42.2 %

Note how component inefficiencies cut net work by a third — the effect of the high back-work ratio.

Example 3 (optimum pressure ratio). For T₁ = 300 K and T₃ = 1400 K: r_p,opt = (1400/300)^(1.75) = 4.667^1.75 = 14.8, and w_net,max = 1.005 × (√1400 − √300)² = 1.005 × (37.417 − 17.321)² = 1.005 × 403.9 = 405.9 kJ/kg — only slightly more than at r_p = 10.

r_p,opt ≈ 14.8, w_net,max ≈ 406 kJ/kg

Common mistakes

  • Computing r_p^((γ−1)/γ) as r_p^((γ−1)) or r_p^(1/γ). For γ = 1.4 the exponent is 0.2857.
  • Applying an isentropic efficiency upside-down: compressor efficiency divides the ideal temperature rise; turbine efficiency multiplies the ideal drop.
  • Adding a regenerator when T₄ < T₂. It would then heat the exhaust and cool the compressed air.
  • Claiming intercooling or reheat alone always raises efficiency. They raise net work; efficiency gains need regeneration.
  • Using °C in temperature ratios.
  • Confusing the maximum-efficiency trend (keep raising r_p) with the maximum-work optimum.

For GATE ME

Expect ideal-cycle efficiency from r_p, temperatures at each state, net work and back-work ratio, optimum pressure ratio for maximum work, actual cycles with given isentropic efficiencies, regenerator effectiveness, and conceptual questions on intercooling, reheat and TIT. Practise the temperature chain T₁ → T₂ → T₃ → T₄ quickly and with consistent exponents; most errors are arithmetic.

Quick check

  1. Find the ideal Brayton efficiency for r_p = 8, γ = 1.4.
  2. At maximum net work in an ideal Brayton cycle, how are T₂ and T₄ related?
  3. When does a regenerator improve efficiency?
  4. Does intercooling alone raise net work, efficiency, or both?
  5. Why is a gas turbine more sensitive to compressor efficiency than a steam plant is to pump efficiency?

Answers: 1. 1 − 1/8^0.2857 = 1 − 1/1.811 ≈ 0.448 (44.8 %). 2. They are equal, both √(T₁T₃). 3. When the turbine exhaust is hotter than the compressor discharge (T₄ > T₂). 4. Net work only; efficiency falls unless regeneration is added. 5. The back-work ratio is 40–60 % compared with about 1–2 % for the pump.

Interview questions

All Thermal Engineering interview questions

Try answering each one aloud before you open it.

  1. 1.What is the Brayton cycle?Concept

    The Brayton cycle is a thermodynamic cycle that describes the workings of a constant-pressure heat engine, such as a gas turbine engine. It consists of four processes: isentropic compression, constant-pressure heat addition, isentropic expansion, and constant-pressure heat rejection. This cycle is commonly used in jet engines and power plants.

  2. 2.Explain the working principle of a gas turbine.Concept

    A gas turbine works on the principle of the Brayton cycle. Air is drawn into the compressor, where it is compressed to a high pressure. The compressed air is then mixed with fuel and ignited in the combustion chamber, producing high-temperature, high-pressure gases. These gases expand through the turbine, producing work that drives the compressor and generates power.

  3. 3.Why is the Brayton cycle preferred in jet engines?Application

    The Brayton cycle is preferred in jet engines because it operates efficiently at high speeds and altitudes. The cycle's continuous flow process allows for a steady thrust output, which is ideal for aircraft propulsion. Additionally, the high power-to-weight ratio of gas turbines makes them suitable for aviation applications.

  4. 4.What happens if the pressure ratio in a Brayton cycle is increased?Application

    For the ideal cycle the efficiency 1 − 1/r_p^((γ−1)/γ) rises steadily with pressure ratio, because heat is added at a higher mean temperature. Net work per kg does not keep rising: for a fixed turbine inlet temperature it peaks at r_p = (T₃/T₁)^(γ/(2(γ−1))) and then falls, since the compressor discharge approaches the turbine inlet temperature. In real machines with component losses the efficiency also peaks at a finite pressure ratio, and higher turbine inlet temperatures push both optima higher.

  5. 5.How does intercooling affect the performance of a gas turbine?Application

    Cooling the air between compressor stages brings the compression closer to isothermal, so compressor work falls and net work per kg rises. On its own, however, intercooling lowers the thermal efficiency, because the air leaves the compressor colder and the combustor must add more heat at a lower mean temperature. It pays off when combined with a regenerator, since the colder compressor discharge lets the regenerator recover more exhaust heat.

  6. 6.What is the role of a regenerator in a Brayton cycle?Application

    A regenerator is a heat exchanger that transfers heat from the hot turbine exhaust to the compressed air before it enters the combustor, so less fuel is needed to reach the same turbine inlet temperature and efficiency rises with no change in net work. It only works when the exhaust temperature exceeds the compressor discharge temperature, which is why it suits low pressure ratios. Its performance is described by effectiveness ε = (T₅ − T₂)/(T₄ − T₂).

  7. 7.Explain the concept of reheat in a gas turbine cycle.Concept

    Reheat expands the gas partly in a high-pressure turbine, adds heat again in a second combustor, and finishes the expansion in a low-pressure turbine. Because constant-pressure lines diverge on a T–s diagram, the total turbine work rises, so net work per kg increases. On its own, reheat lowers efficiency, because extra heat is added and the exhaust leaves hotter. Combined with regeneration, which recovers that hot exhaust, it raises efficiency.

  8. 8.Calculate the thermal efficiency of an ideal Brayton cycle with a pressure ratio of 8 and γ = 1.4.Numerical

    For the ideal Brayton cycle η = 1 − 1/r_p^((γ−1)/γ). The exponent is 0.4/1.4 = 0.2857, and 8^0.2857 = 1.811, so η = 1 − 1/1.811 = 0.448, or about 44.8 %. Note that it depends only on the pressure ratio and γ, not on the temperatures.

  9. 9.What are the limitations of the Brayton cycle in practical applications?Application

    In practical applications, the Brayton cycle faces limitations such as material constraints at high temperatures, which can affect turbine blade life and efficiency. Additionally, the cycle's efficiency is sensitive to pressure ratio and temperature limits, which are constrained by the design and materials used. Cooling and maintenance requirements also add to the complexity and cost.

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