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Brayton cycle; Atkinson and Stirling cycles

Short answer

Brayton efficiency: η = 1 – 1/rp^((γ-1)/γ); for rp=8, η=44.8%. Atkinson cycle: expansion ratio > compression ratio, constant-pressure heat rejection, higher efficiency but lower power density; achieved by late intake valve closing. Stirling cycle: closed external-heat cycle with regenerator; ideal efficiency equals Carnot; any heat source; clean but large and slow.

Derive the thermal efficiency of the air-standard Brayton cycle in terms of the pressure ratio and calculate it for a pressure ratio of 8. Write short notes on the Atkinson cycle and the Stirling cycle.

Brayton cycle efficiency

The Brayton (Joule) cycle is the ideal cycle of a gas turbine. It has isentropic compression 1-2 in a compressor, constant-pressure heat addition 2-3 in a combustor, isentropic expansion 3-4 in a turbine and constant-pressure heat rejection 4-1.

 Air --> [Compressor] --> [Combustor] --> [Turbine] --> Exhaust
             1-2             2-3             3-4        4-1
              \_______________shaft__________/ --> load
Qin=cp(T3−T2),Qout=cp(T4−T1)η=1−T4−T1T3−T2\begin{aligned} Q_{in} &= c_p (T_3 - T_2), \qquad Q_{out} = c_p (T_4 - T_1) \\ \eta &= 1 - \frac{T_4 - T_1}{T_3 - T_2} \end{aligned}

Since p2=p3p_2 = p_3 and p1=p4p_1 = p_4, T2/T1=T3/T4=rp(γ−1)/γT_2/T_1 = T_3/T_4 = r_p^{(\gamma-1)/\gamma}. Hence T4/T1=T3/T2T_4/T_1 = T_3/T_2 and

ηBrayton=1−T1T2=1−1rp(γ−1)/γ\eta_{Brayton} = 1 - \frac{T_1}{T_2} = 1 - \frac{1}{r_p^{(\gamma-1)/\gamma}}

For rp=8r_p = 8, γ=1.4\gamma = 1.4:

η=1−8−0.2857=1−0.552=0.448\eta = 1 - 8^{-0.2857} = 1 - 0.552 = 0.448

Answer: η=44.8%\eta = 44.8\%. It depends only on pressure ratio, not on the maximum temperature.

Atkinson cycle

  • Same as the Otto cycle (isentropic compression, constant-volume heat addition) but the expansion goes on until the gas reaches atmospheric pressure. The expansion ratio is larger than the compression ratio.
  • Heat is rejected at constant pressure.
  • It extracts more work from the same fuel, so the efficiency is higher than Otto's.
  • It is achieved with a long expansion stroke (original linkage design) or, in modern engines, by closing the inlet valve late (Miller/Atkinson-type timing). The cost is a lower power density, so it is used in hybrid cars.

Stirling cycle

  • A closed cycle with an external heat source and a regenerator. Processes: isothermal compression, constant-volume heating (regenerator), isothermal expansion, constant-volume cooling (regenerator).
  • With a perfect regenerator its efficiency equals the Carnot efficiency, 1−TL/TH1 - T_L/T_H.
  • Heat is supplied from outside, so any fuel or solar heat can be used and combustion is clean and quiet.
  • Drawbacks: large size, slow response and sealing problems, so it is rare in vehicles.

✓ Answer checked by Claude Opus 5.5 (high effort)

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