Questions: Brayton Cycle Modifications: Intercooling and Reheating

Question 1

A two-stage Brayton cycle with intercooling is compared to a single-stage cycle with the same overall pressure ratio and inlet temperature. What is the primary thermodynamic benefit of intercooling?

• A. Intercooling increases turbine inlet temperature, allowing more work extraction
• B. Intercooling reduces compression work by keeping gas cooler during each compression stage, approaching isothermal compression
• C. Intercooling increases the pressure ratio achievable by the compressor
• D. Intercooling eliminates entropy generation in the compression process

Answer: (B)

Compressor work is proportional to the absolute temperature at the compressor inlet: w_c = c_p(T_out − T_in). By cooling the gas back toward inlet temperature between stages, intercooling ensures the second stage compresses cooler, denser gas — which requires less work to reach the same pressure ratio. This approaches the thermodynamic ideal of isothermal compression (pT = constant), the theoretical limit of minimum compression work. Intercooling does not affect turbine inlet temperature or pressure ratio capability.

Question 2

Why does the combination of intercooling and reheating specifically enable effective regeneration, while a simple Brayton cycle without these modifications cannot benefit as much from a regenerator?

• A. Intercooling and reheating reduce the mass flow rate, making heat exchange more practical
• B. Intercooling lowers the temperature of compressed air leaving the compressor, while reheating raises the turbine exhaust temperature — creating the temperature difference needed for the regenerator to transfer exhaust heat to the compressed air
• C. Intercooling and reheating together increase the pressure ratio, which is required for regeneration to work
• D. A regenerator requires two separate heat sources, which intercooling and reheating provide

Answer: (B)

In a simple Brayton cycle at high pressure ratio, the compressed air exits the compressor hotter than the turbine exhaust — so a heat exchanger between them would actually transfer heat the wrong way. Intercooling lowers the compressor outlet temperature, and reheating raises the turbine exhaust temperature. This reversal in relative temperatures makes the regenerator thermodynamically feasible: exhaust heat can now flow into the cooler compressed air before combustion, reducing the fuel needed.

Question 3

Adding intercooling to a Brayton cycle usually increases overall thermal efficiency, even without a regenerator.

• T. True
• F. False

Answer: False

Intercooling alone does NOT necessarily improve thermal efficiency — it may actually reduce it. While intercooling decreases compression work (improving net work output), it also reduces the average temperature at which heat is added (since the intercooled air enters the combustor at a lower temperature), which lowers cycle efficiency by the Carnot argument. The efficiency benefit of intercooling is only fully realized when paired with a regenerator that recovers the exhaust heat and compensates for the lower combustor inlet temperature.

Question 4

For a two-stage Brayton cycle with intercooling, the total compressor work is minimized when both stages operate at equal pressure ratios (each compresses by √r_p, where r_p is the overall pressure ratio).

• T. True
• F. False

Answer: True

The optimal interstage pressure that minimizes total two-stage compression work is the geometric mean of inlet and outlet pressures: P_i = √(P_1 × P_2). This means each stage has the same pressure ratio √r_p, and with perfect intercooling back to inlet temperature, each stage does identical work. Any deviation from equal pressure ratios increases total work. This result extends to N stages, where optimal work is achieved with equal pressure ratios of r_p^(1/N) per stage.

Question 5

Explain the fundamental thermodynamic reason why cooling gas between compressor stages reduces the total work required to achieve a given pressure ratio.

Model answer: Compression work equals the area under the process path on a p-v diagram (or equivalently, the enthalpy rise for steady-flow devices). For a fixed pressure ratio, compressing hot gas requires more work than compressing cool gas because hot gas has larger specific volume — you are pushing against a larger volume at each pressure increment. By cooling the gas between stages, you reset the specific volume to a lower value before the next pressure rise, so each subsequent compression stage acts on denser, easier-to-compress gas. In the limit of infinitely many stages with intercooling back to inlet temperature, this approaches isothermal compression, which is the theoretical minimum work for a given pressure ratio.

This is the key insight: compression work depends on the temperature (and thus specific volume) of the gas being compressed, not just the pressure ratio. Intercooling exploits this by reducing the specific volume before each compression stage.