Questions: Power Cycle Analysis and Thermal Efficiency

Question 1

An engineer claims they can design a power cycle that achieves 100% thermal efficiency by making all internal processes perfectly reversible — no friction, no heat losses, perfectly isentropic compression and expansion. Is this claim valid?

• A. Yes — a perfectly reversible cycle has no irreversibilities, so all heat input becomes work
• B. No — even a perfectly reversible (Carnot) cycle must reject heat to the cold reservoir; the second law requires it, and efficiency is bounded by η = 1 − T_cold/T_hot < 1
• C. Yes — the Carnot cycle is 100% efficient by definition
• D. No — but only because real materials introduce irreversibilities that cannot be completely eliminated

Answer: (B)

The Carnot efficiency η = 1 − T_cold/T_hot is the efficiency of a perfectly reversible cycle, and it is less than 100% whenever T_cold > 0 K. The second law requires that any heat engine operating between a hot and cold reservoir must reject some heat — work cannot be extracted from heat with perfect efficiency. Making processes reversible removes internal irreversibilities but cannot change the fundamental constraint that some heat must flow to the cold reservoir. 100% efficiency would require either T_cold = 0 K or T_hot = ∞, both physically unachievable.

Question 2

A power plant engineer adds a regenerator that uses exhaust heat to preheat the working fluid before it enters the boiler. This modification:

• A. Increases efficiency beyond the Carnot limit for the same operating temperatures, since energy is being recycled
• B. Reduces the external heat input needed for the same net work output, improving efficiency — without exceeding the Carnot limit
• C. Increases efficiency by lowering the cold reservoir temperature
• D. Has no effect on efficiency — it merely moves heat around within the cycle

Answer: (B)

Regeneration uses heat that would otherwise be rejected to the cold reservoir to preheat the working fluid instead, reducing how much external fuel (Q_in) is needed to reach the target operating temperature. Since η = W_net / Q_in, reducing Q_in for the same W_net raises efficiency. Crucially, regeneration does NOT violate or exceed the Carnot limit — it is internal heat exchange that reduces external heat input, not a mechanism that extracts more work than thermodynamics permits. The Carnot limit depends only on T_hot and T_cold, which the regenerator doesn't change.

Question 3

According to the Carnot efficiency formula η = 1 − T_cold/T_hot, efficiency increases when the hot reservoir temperature rises or the cold reservoir temperature falls, regardless of the specific cycle design used.

• T. True
• F. False

Answer: True

This is a fundamental implication of the second law. The Carnot efficiency is determined entirely by the temperature ratio between the hot and cold reservoirs — it is the theoretical maximum for any cycle operating between those temperatures, regardless of whether the working fluid is steam, gas, or anything else, and regardless of the specific cycle design (Rankine, Brayton, Otto). This is why power plant engineers pursue superheat (raising T_hot), higher pressure ratios (allowing expansion to lower effective T_cold), and other strategies to push the operating temperatures further apart.

Question 4

Regeneration allows a well-designed power cycle to exceed the Carnot efficiency for its operating temperature limits.

• T. True
• F. False

Answer: False

No real or theoretical cycle can exceed the Carnot efficiency for a given pair of reservoir temperatures — this is a statement of the second law of thermodynamics. Regeneration improves efficiency by reducing external heat input for the same net work, bringing actual efficiency closer to the Carnot limit. It operates within thermodynamic constraints, not around them. The Carnot limit is an absolute ceiling set by the temperature ratio, not a practical target that better engineering can surpass.

Question 5

A power plant draws heat from a reservoir at 800 K and rejects heat to a cooling system at 300 K. Explain why even a perfectly designed (reversible) engine between these temperatures cannot convert more than ~62.5% of the heat input to work, and what this implies for real power plants operating between the same temperatures.

Model answer: The Carnot efficiency η = 1 − T_cold/T_hot = 1 − 300/800 = 0.625, or 62.5%. This is an absolute upper bound set by the second law: any engine operating between 800 K and 300 K must reject at least 37.5% of its heat input to the cold reservoir, regardless of how perfectly it is designed. The second law requires this heat rejection — there is no thermodynamic process that can extract all the energy from a temperature difference as work. Real power plants operating between these temperatures will achieve less than 62.5% efficiency because their processes are irreversible (friction, heat transfer across temperature gradients, turbine and compressor inefficiencies), typically achieving 35–45%. Improving real cycle efficiency means both minimizing irreversibilities AND operating between the widest possible temperature difference.

The Carnot efficiency gives engineers a benchmark: how close to the theoretical limit is our actual cycle? The gap between Carnot efficiency and actual efficiency quantifies the irreversibility penalty. A plant achieving 40% between 800 K and 300 K is operating at 40/62.5 = 64% of its theoretical maximum — there is room for improvement through reduced irreversibilities. But no amount of engineering improvement can close the gap to zero while T_cold remains above 0 K.