Brayton cycle with regeneration, reheat and intercooling

The ideal and real Brayton cycle, efficiency, net work, optimum pressure ratio and back-work ratio, isentropic efficiencies, and the effects of regeneration, intercooling and reheat.

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Why it matters

The Brayton (Joule) cycle is the ideal cycle of every gas turbine — turbojets, turbofans, turboprops, helicopter turboshafts and the auxiliary power unit. Its analysis explains why engine designers chase high pressure ratios and turbine inlet temperatures, why ground and marine gas turbines use regenerators, intercoolers and reheat, and why aero engines usually do not.

Key ideas

Ideal simple Brayton cycle (air standard, steady flow):

  • 1→2 isentropic compression in the compressor (pressure ratio rp = p₂/p₁)
  • 2→3 constant-pressure heat addition in the combustor
  • 3→4 isentropic expansion in the turbine
  • 4→1 constant-pressure heat rejection (to the atmosphere in an open cycle)

Since both isentropic processes share the same pressure ratio, T₂/T₁ = T₃/T₄ = rp^((γ−1)/γ) and the efficiency depends on rp only: η = 1 − 1/rp^((γ−1)/γ).

Net work and the optimum pressure ratio. For fixed T₁ and turbine inlet temperature T₃, net work is zero at rp = 1 and again when T₂ reaches T₃; in between there is a maximum at rp,opt = (T₃/T₁)^(γ/(2(γ−1))), where T₂ = T₄ = √(T₁·T₃). Raising T₃ (better blade materials and cooling) increases both specific work and the useful range of rp.

Back-work ratio = w_C/w_T. In gas turbines it is high (often 40–60%) because the gas is compressed, not pumped as a liquid. Small component inefficiencies therefore cut net work sharply.

Real cycle. Compressor and turbine are adiabatic but irreversible, with isentropic efficiencies η_C = (T₂s − T₁)/(T₂ − T₁) and η_T = (T₃ − T₄)/(T₃ − T₄s); pressure losses in the combustor and ducts are often ignored in GATE problems.

Regeneration. When the turbine exhaust T₄ is hotter than the compressor exit T₂, a heat exchanger (regenerator) preheats the compressed air with exhaust heat, reducing the fuel needed. Net work is unchanged, heat input falls, so efficiency rises. Effectiveness ε = (T₅ − T₂)/(T₄ − T₂). For an ideal regenerator (ε = 1) and ideal machines, η = 1 − (T₁/T₃)·rp^((γ−1)/γ) — efficiency now falls as rp rises, and regeneration is useless once T₂ ≥ T₄. That is why low-pressure-ratio ground turbines use regenerators and high-pressure-ratio aero engines do not (weight and volume also rule them out in aircraft).

Intercooling (cooling between compressor stages) reduces compressor work because the work of a steady-flow compressor ∝ ∫v dp and cooler air has smaller v. Reheat (combustion between turbine stages) increases turbine work. Each on its own raises net work but lowers the efficiency of the simple cycle, because heat is then added at lower average temperature (reheat) or rejected and resupplied (intercooling). Combined with regeneration they raise efficiency, and with many stages the cycle approaches the Ericsson cycle, whose efficiency equals Carnot's. Ideal two-stage intercooling or reheat uses equal pressure ratios per stage, p_i = √(p_low·p_high). The afterburner of a military jet is a form of reheat used for thrust, not efficiency.

Formulas

T₂/T₁ = T₃/T₄ = rp^((γ−1)/γ) (ideal) η = 1 − 1/rp^((γ−1)/γ) (ideal simple cycle) w_net = cp·[(T₃ − T₄) − (T₂ − T₁)] rp,opt = (T₃/T₁)^(γ/(2(γ−1))) (maximum specific work) η_C = (T₂s − T₁)/(T₂ − T₁); η_T = (T₃ − T₄)/(T₃ − T₄s) ε = (T₅ − T₂)/(T₄ − T₂) (regenerator effectiveness) η = 1 − (T₁/T₃)·rp^((γ−1)/γ) (ideal cycle, ideal regenerator) p_i = √(p₁·p₂) (optimum intermediate pressure, two stages)

  • rp: compressor pressure ratio; γ = 1.4, cp = 1.005 kJ/(kg·K) for air
  • T₁: compressor inlet, T₃: turbine inlet (maximum) temperature, K
  • T₅: compressed-air temperature leaving the regenerator, K
  • w: specific work, kJ/kg

Worked examples

Example 1 (standard). An ideal Brayton cycle has rp = 10, T₁ = 300 K and T₃ = 1400 K. Find the efficiency, the net work and the back-work ratio.

  1. rp^((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 × 279.2 = 280.6 kJ/kg; w_T = 1.005 × 674.9 = 678.2 kJ/kg
  4. w_net = 397.6 kJ/kg; q_in = 1.005 × (1400 − 579.2) = 824.9 kJ/kg
  5. η = 397.6/824.9 = 0.482 (= 1 − 1/1.9307 ✓); back-work ratio = 280.6/678.2 = 0.414

Answer: η = 48.2%, w_net = 397.6 kJ/kg, back-work ratio 41.4%

Example 2 (GATE level). A regenerative gas turbine: rp = 8, T₁ = 300 K, T₃ = 1300 K, η_C = 0.85, η_T = 0.90, regenerator effectiveness 0.80. Find the net work and the efficiency with and without the regenerator.

  1. rp^((γ−1)/γ) = 8^0.2857 = 1.8114
  2. Compressor: T₂s = 543.4 K; T₂ = T₁ + (T₂s − T₁)/η_C = 300 + 243.4/0.85 = 586.4 K
  3. Turbine: T₄s = 1300/1.8114 = 717.7 K; T₄ = T₃ − η_T·(T₃ − T₄s) = 1300 − 0.9 × 582.3 = 775.9 K
  4. w_C = 1.005 × 286.4 = 287.8 kJ/kg; w_T = 1.005 × 524.1 = 526.7 kJ/kg; w_net = 238.9 kJ/kg
  5. Regenerator: T₅ = T₂ + ε·(T₄ − T₂) = 586.4 + 0.8 × 189.5 = 738.0 K
  6. q_in = 1.005 × (1300 − 738.0) = 564.8 kJ/kg → η = 238.9/564.8 = 0.423
  7. Without regenerator: q_in = 1.005 × (1300 − 586.4) = 717.2 kJ/kg → η = 0.333

Answer: w_net = 238.9 kJ/kg; η = 42.3% with the regenerator, 33.3% without

Common mistakes

  • Using the pressure ratio exponent (γ−1)/γ backwards, or using the volume-ratio form from piston cycles.
  • Applying the isentropic efficiency the wrong way round: actual compressor work is larger (divide by η_C), actual turbine work smaller (multiply by η_T).
  • Assuming reheat or intercooling alone increases efficiency; without regeneration they increase work but reduce efficiency.
  • Adding a regenerator when T₂ > T₄ — heat would flow the wrong way.
  • Forgetting that regeneration changes heat input, not net work.

For GATE AE

This cycle links directly to the propulsion syllabus, so it is a frequent source of questions. Expect efficiency and net-work numericals, optimum pressure ratio for maximum work, the effect of component efficiencies, regenerator effectiveness, and conceptual questions on why reheat or intercooling alone lowers efficiency. Practise drawing the T–s diagram with every state number before calculating.

Quick check

  1. Find the ideal Brayton efficiency for rp = 10 (γ = 1.4).
  2. What is the optimum pressure ratio for maximum work when T₃/T₁ = 4?
  3. Does regeneration change the net work of an ideal cycle?
  4. With an ideal regenerator, does efficiency rise or fall as rp increases?
  5. What is the main purpose of reheat?

Answers: 1. 48.2%. 2. 4^1.75 ≈ 11.3. 3. No — it reduces heat input only. 4. It falls. 5. To increase turbine (net) work.

Interview questions

All Engineering Thermodynamics interview questions

Try answering each one aloud before you open it.

  1. 1.What is the Brayton cycle and how does it work?Concept

    The Brayton cycle is a thermodynamic cycle that describes the workings of a constant-pressure heat engine, such as a jet engine or a gas turbine. It consists of four processes: isentropic compression, constant-pressure heat addition, isentropic expansion, and constant-pressure heat rejection. The cycle is typically represented on a pressure-volume (P-V) or temperature-entropy (T-S) diagram.

  2. 2.Explain the role of regeneration in the Brayton cycle.Concept

    Regeneration in the Brayton cycle involves using a heat exchanger to transfer heat from the exhaust gases to the compressed air before it enters the combustion chamber. This process increases the thermal efficiency of the cycle by reducing the amount of fuel needed to reach the desired turbine inlet temperature. It effectively recycles some of the waste heat, improving overall efficiency.

  3. 3.What is the purpose of reheat in the Brayton cycle?Concept

    Reheat splits the expansion into stages and adds heat (burns more fuel) between them at constant pressure, so the second turbine starts again at a high temperature. Because turbine work per unit pressure ratio grows with temperature (w ∝ ∫v dp), the total turbine work and the net specific work increase. On its own, however, reheat lowers the thermal efficiency of the simple cycle, because the extra heat is added at a lower average temperature; it improves efficiency only when combined with regeneration, which recovers the hotter exhaust.

  4. 4.Describe the process of intercooling in the Brayton cycle.Concept

    Air is compressed in stages and cooled at constant pressure in a heat exchanger between them, ideally back to the inlet temperature. Steady-flow compression work is ∫v dp, so cooler, denser air needs less work for the remaining pressure rise; for two ideal stages the work is minimised with equal pressure ratios, p_i = √(p₁p₂). Intercooling increases the net work, but on its own it lowers cycle efficiency because the air leaves the compressor colder and needs more fuel; with regeneration it raises efficiency.

  5. 5.What happens to the efficiency of a Brayton cycle if regeneration is not used?Application

    If regeneration is not used in a Brayton cycle, the efficiency of the cycle decreases. Without regeneration, more fuel is required to reach the desired turbine inlet temperature because the heat from the exhaust gases is not utilized. This results in higher fuel consumption and lower thermal efficiency.

  6. 6.How does reheat affect the specific work output of a Brayton cycle?Application

    Reheat increases the specific work output: after partial expansion the gas is heated back to near the turbine inlet temperature, and the remaining expansion then takes place at higher temperature and specific volume, giving more work for the same pressure drop. The price is extra fuel and a higher exhaust temperature, so without a regenerator the cycle efficiency falls. In aero engines the afterburner is a form of reheat used to boost thrust at the cost of fuel consumption.

  7. 7.Calculate the thermal efficiency of a simple Brayton cycle with a pressure ratio of 10 and a specific heat ratio (γ) of 1.4.Numerical

    For the ideal air-standard Brayton cycle η = 1 − 1/rp^((γ−1)/γ). The exponent is 0.4/1.4 = 0.2857, so 10^0.2857 = 1.931 and η = 1 − 1/1.931 = 1 − 0.518 = 0.482, i.e. about 48.2%. The efficiency depends only on the pressure ratio, not on the turbine inlet temperature, which instead governs the net work.

  8. 8.An ideal Brayton cycle with two-stage intercooling, two-stage reheat and an ideal regenerator has a compressor inlet temperature of 300 K, a turbine inlet temperature of 1500 K and an overall pressure ratio of 12. Estimate its efficiency.Numerical

    Use equal stage pressure ratios √12 = 3.464, so each stage has a temperature ratio 3.464^(0.4/1.4) = 1.426. Each compressor stage heats air from 300 K to 427.9 K and each turbine stage cools gas from 1500 K to 1051.8 K, so w_C = 2cp(127.9) and w_T = 2cp(448.2); with ideal regeneration the heat input equals the turbine work, giving η = 1 − w_C/w_T = 1 − (300/1500) × 1.426 ≈ 0.715, about 71.5%. With more stages it approaches the Carnot (Ericsson) limit 1 − 300/1500 = 80%.

  9. 9.What are the limitations of using reheat and intercooling in the Brayton cycle?Application

    The limitations of using reheat and intercooling in the Brayton cycle include increased complexity and cost of the system due to additional components like heat exchangers and reheaters. These modifications also require more space and maintenance. Additionally, the benefits of reheat and intercooling may diminish at lower pressure ratios or when the cycle is not operating near its design point.

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