Difficulty: Easy
Correct Answer: Disagree
Explanation:
Introduction:
Engineers frequently idealize processes as isothermal (constant temperature) or isentropic (constant entropy). The question tests whether these labels inherently imply “reversible non-flow,” which is often assumed in classroom examples but is not universally true in practice.
Given Data / Assumptions:
Concept / Approach:
Isothermal simply means constant temperature. A process can be isothermal yet irreversible (e.g., finite temperature difference heat transfer, throttling of an ideal gas that keeps T roughly constant). Isentropic means constant entropy; in engineering, “isentropic” typically implies an ideal reversible adiabatic model, but real devices only approximate it and can be open-flow (e.g., turbines and compressors). Therefore, neither term inherently enforces “reversible non-flow” across all contexts.
Step-by-Step Solution:
Identify the logical claim: both process types “are reversible non-flow.”Counterexample 1 (isothermal): Gas expansion against a finite pressure drop with heat exchange at finite temperature difference is isothermal but irreversible.Counterexample 2 (isentropic): Turbine expansion is modeled as isentropic in an open steady-flow system, not a non-flow system; real expansions are not perfectly reversible.
Verification / Alternative check:
Textbook cycle analyses use isentropic efficiencies precisely because real devices deviate from reversibility. Similarly, isothermal compression in practice requires intercooling with irreversibilities, not perfect reversibility.
Why Other Options Are Wrong:
“Agree” would mistakenly equate the property labels with a specific mode (non-flow) and idealization (reversibility) that is not inherently required.
Common Pitfalls:
Assuming “isentropic” automatically equals “reversible adiabatic in a closed system”; forgetting that isothermal/isentropic can describe both closed and open systems.
Final Answer:
Disagree
Discussion & Comments