Questions: Maximum Available Work: Carnot and Reversible Processes

Question 1

What does first-law analysis alone reveal about a throttle valve, and what does second-law analysis add?

• A. First law shows the throttle does useful work; second law confirms this by calculating entropy generation
• B. First law shows no useful work is done (isenthalpic); second law reveals that all pressure exergy is destroyed as lost work
• C. First law quantifies the heat released during throttling; second law determines the entropy change of the surroundings
• D. Both laws independently predict zero useful work, confirming the throttle is thermodynamically efficient

Answer: (B)

For an adiabatic throttle, h₁ = h₂ — the first law says nothing useful happened (no work out, no heat transferred). But the throttle's pressure drop through a constriction is highly irreversible, generating entropy S_gen > 0. By W_lost = T₀ × S_gen, the entire exergy of the pressure difference is destroyed. The first law masks this destruction by reporting only the enthalpy balance; the second law exposes what was lost. This gap between first- and second-law analysis is exactly what exergy accounting is designed to reveal.

Question 2

A turbine and a throttle valve are installed on parallel branches of the same pipeline, each producing an identical pressure drop. Which statement best describes the thermodynamic difference between them?

• A. The throttle extracts shaft work from the flow; the turbine generates entropy — they are thermodynamically equivalent
• B. The turbine converts pressure exergy into useful shaft work; the throttle destroys that same exergy as irreversibility
• C. Both devices destroy the same exergy, but the turbine additionally generates waste heat
• D. The turbine is more irreversible because rotating machinery generates more entropy than a simple constriction

Answer: (B)

The turbine performs a near-reversible expansion: pressure exergy drives shaft rotation, and W_actual ≈ W_max. The throttle performs the same pressure reduction irreversibly: S_gen > 0 and W_lost = T₀ × S_gen destroys the equivalent exergy with zero useful output. The physical process differs radically (ordered shaft work vs. chaotic dissipation) even though both achieve the same pressure drop. Replacing throttles with expanders is a standard engineering efficiency measure precisely because it recovers this otherwise-destroyed work.

Question 3

The first law of thermodynamics is sufficient to determine the maximum useful work extractable from a steady-flow process.

• T. True
• F. False

Answer: False

The first law gives only the enthalpy drop (h₁ − h₂) — it cannot distinguish between reversible and irreversible processes and treats all energy changes identically regardless of quality. The maximum work formula W_max = (h₁ − h₂) − T₀(s₁ − s₂) requires the second-law correction T₀(s₁ − s₂) to account for entropy changes. When entropy increases across a device (irreversibility present), the second term reduces W_max below the enthalpy drop. Using the first law alone overestimates the maximum work whenever entropy is generated.

Question 4

Lost work in a real process equals the ambient temperature T₀ multiplied by the entropy generated within that process.

• T. True
• F. False

Answer: True

W_lost = W_rev − W_actual = T₀ × S_gen. This is the quantitative form of the second law for engineering analysis: every irreversibility mechanism (heat transfer across finite ΔT, viscous friction, shock waves, mixing) generates entropy, and each unit of entropy generated destroys exactly T₀ joules of work potential. The formula transforms 'irreversibility is bad' from a qualitative statement into an exact accounting of what was lost and why — identifying the irreversibility sources that are worth addressing in design.

Question 5

A plant engineer proposes replacing a pressure-reducing throttle valve on a high-pressure steam line with a small steam turbine. Use the maximum work theorem to explain what thermodynamic improvement this achieves.

Model answer: The throttle is isenthalpic (h₁ = h₂) but generates significant entropy (S_gen > 0), destroying W_lost = T₀ × S_gen of work potential from the pressure difference — with zero useful output. The turbine performs a nearly reversible expansion over the same pressure drop, extracting shaft work W_actual ≈ (h₁ − h₂) − T₀(s₁ − s₂). The improvement equals approximately T₀ × S_gen_throttle — the work that the throttle was previously converting entirely into entropy is now recovered as useful power. The thermodynamic case is clear whenever the pressure drop carries significant exergy; economic justification depends on the scale of the installation.

This throttle-vs-expander comparison is the canonical illustration of the maximum work theorem. It shows that 'nothing happened' (first law: isenthalpic) and 'something was irreversibly destroyed' (second law: S_gen > 0) can coexist — and that second-law analysis is what makes the engineering opportunity visible.