Difficulty: Easy
Correct Answer: Rankine efficiency
Explanation:
Introduction / Context:
Steam-plant performance is summarized by several efficiencies. It is important to distinguish component efficiencies (nozzle, blade, stage, internal/isentropic turbine) from cycle-level (thermal) efficiency. The question asks for the name of the ratio of the ideal (isentropic) turbine enthalpy drop to the total heat added in the boiler—an expression that mirrors the ideal Rankine thermal efficiency when pump work is small.
Given Data / Assumptions:
Concept / Approach:
For the ideal Rankine cycle, thermal efficiency is eta_th ≈ (h1 − h2s)/(h1 − h_fw), i.e., the isentropic enthalpy drop across the turbine divided by the boiler heat input. This is what many textbooks call the Rankine efficiency for the idealized case. Component factors such as stage efficiency or internal efficiency compare actual to isentropic within the turbine only and do not involve boiler heat input in the denominator. Reheat factor is the ratio of the sum of isentropic drops in stages to the overall isentropic drop—again a different concept.
Step-by-Step Solution:
Verification / Alternative check:
Energy-balance diagrams for the ideal Rankine cycle show turbine work equaling the isentropic drop; with pump work very small, the ratio above matches eta_th used for cycle benchmarking.
Why Other Options Are Wrong:
Reheat factor relates cumulative stage drops, not boiler heat. Stage efficiency compares a single stage output to its isentropic drop. Internal efficiency compares the actual turbine work to the isentropic drop. Nozzle efficiency is local to nozzles only.
Common Pitfalls:
Confusing cycle efficiency with turbine internal efficiency; overlooking pump work (often negligible but not zero in precise calculations).
Final Answer:
Rankine efficiency
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