Joule (Brayton) Cycle — Efficiency Compared to Carnot Between Same Temperature Limits How does the ideal Joule (Brayton) cycle efficiency compare to the Carnot efficiency when both operate between the same maximum and minimum temperatures?

Introduction:
The Carnot efficiency is the theoretical maximum for any heat engine operating between two thermal reservoirs at temperatures T_hot and T_cold. Realistic cycles like the Joule (Brayton) cycle have different process structures and therefore lower efficiencies than the Carnot limit for the same temperature bounds.

Given Data / Assumptions:

• Ideal Brayton cycle with two isentropic and two isobaric processes.
• Same maximum and minimum cycle temperatures as the reference Carnot engine.
• Working fluid treated as an ideal gas with constant specific heats.

Concept / Approach:
Carnot efficiency depends only on temperatures: eta_Carnot = 1 - T_cold/T_hot. The Brayton efficiency depends on pressure ratio and the ratio of specific heats, and even in the ideal reversible limit it is strictly less than Carnot for the same temperature limits because heat addition/rejection in Brayton occurs over a temperature range, not isothermally at the extremes.

Step-by-Step Solution:
State the benchmark: eta_Carnot is the upper bound between the given T_hot and T_cold.Identify Brayton processes: isentropic compression/expansion + isobaric heat addition/rejection.Because heat is exchanged over finite temperature ranges, entropy generation equivalents prevent matching Carnot performance.Therefore, Brayton efficiency is less than Carnot between the same temperature limits.

Verification / Alternative check:
Even at the optimal pressure ratio for a given T_max/T_min, Brayton's efficiency remains below the Carnot bound, confirming the strict inequality.

Why Other Options Are Wrong:
“Greater than” or “equal to” would violate the second law or misinterpret the differing process paths; “none of these” is unnecessary since “less than” is correct.

Common Pitfalls:
Comparing at equal pressure ratios rather than equal temperature limits; forgetting that Carnot requires isothermal heat transfer at the extremes, which Brayton does not provide.

Final Answer:
less than Carnot cycle