Difficulty: Medium
Correct Answer: all of these
Explanation:
Introduction / Context:
Heat-transfer irreversibility often arises from finite temperature differences between a system and an external reservoir. A regenerator exchanges heat internally between hot and cold streams, potentially eliminating those finite differences and thereby making the heat transfer reversible. This idea underlies ideal Stirling (isochoric regeneration) and Ericsson (isobaric regeneration) cycles, and it can be generalized conceptually.
Given Data / Assumptions:
Concept / Approach:
If the heat required during a process leg is supplied or removed internally at the same temperature as the system at each instant, then the temperature difference driving heat transfer is infinitesimal, so entropy generation due to heat transfer is zero. This logic applies not only to constant-pressure or constant-volume legs but also to appropriately guided polytropic paths, provided the regenerator can match the system temperature path on the opposite leg. Thus, regeneration offers a mechanism to make these heat-transfer legs internally reversible in the ideal limit.
Step-by-Step Solution:
Verification / Alternative check:
Stirling cycle (isothermal plus two isochoric regenerative legs) and Ericsson cycle (isothermal plus two isobaric regenerative legs) are textbook demonstrations. Polytropic recuperation is used industrially (recuperated Brayton) to reduce exergy destruction; the idealized limit corresponds to reversible internal heat exchange.
Why Other Options Are Wrong:
Choosing only one leg ignores the generality of the regenerative concept. 'None of these' contradicts well-established ideal cycles that are fully reversible with perfect regeneration.
Common Pitfalls:
Assuming regeneration eliminates all irreversibilities automatically—mechanical friction and pressure drops must also be minimized. Also, real regenerators have finite effectiveness, so practical processes remain slightly irreversible.
Final Answer:
all of these
Discussion & Comments