Questions: Entropy Balance and Irreversibility Analysis

Question 1

An engineer analyzes a heat exchanger that transfers 1,000 kW from a hot stream at 500°C to a cold stream at 100°C. She calculates S_gen > 0 for this device. What does this tell her about the process, and could she reduce S_gen without violating the second law?

• A. S_gen > 0 is the minimum possible; no heat exchanger transferring heat across a temperature difference can have lower entropy generation
• B. S_gen > 0 indicates irreversibility from heat transfer across a finite temperature difference; it could be reduced by minimizing the temperature difference (approaching reversibility), but cannot reach zero unless the difference is zero
• C. S_gen > 0 simply confirms heat is flowing from hot to cold, which is expected; this value is a fixed constant for any heat exchanger of this capacity
• D. S_gen > 0 means the heat exchanger violates the second law and must be redesigned

Answer: (B)

Entropy generation from heat transfer is proportional to the temperature difference across which it occurs and inversely proportional to both temperatures involved. A finite ΔT always produces S_gen > 0 — this is inherent to the irreversibility. In principle, S_gen could be reduced by bringing the streams' temperatures closer together (e.g., using a longer heat exchanger with more surface area), approaching the reversible limit of infinitesimal ΔT. But in practice, making ΔT → 0 requires infinite heat exchanger area, so there is always a design trade-off between entropy generation (efficiency loss) and equipment size (cost). S_gen cannot reach zero unless both streams are at the same temperature, which means no heat transfer occurs — the reversible limit is an ideal asymptote.

Question 2

An adiabatic turbine receives steam at high pressure and enthalpy and exhausts at lower pressure. The actual exit specific entropy s_out is measured to be slightly greater than the inlet specific entropy s_in. What does this indicate?

• A. A measurement error — entropy must decrease through a turbine because work is extracted from the fluid
• B. The turbine violates the second law, since entropy should be constant in an adiabatic device
• C. Normal real-world behavior: S_gen > 0 due to internal irreversibilities (friction, flow separation), so s_out > s_in as the entropy balance requires
• D. The turbine is operating isentropically, and the apparent increase is due to the reference state chosen for entropy calculation

Answer: (C)

For an adiabatic steady-state device, the entropy balance reduces to S_gen = ṁ(s_out − s_in). Since S_gen ≥ 0 always, we must have s_out ≥ s_in — exit specific entropy is at least as large as inlet specific entropy. An ideal (isentropic) turbine has S_gen = 0 and s_out = s_in exactly. A real turbine has S_gen > 0 from fluid friction, shock waves, and flow irreversibilities, so s_out > s_in. This increase in specific entropy through an adiabatic device is the quantitative signature of irreversibility and directly corresponds to reduced work output compared to the isentropic ideal — which is precisely what isentropic efficiency η_s measures.

Question 3

The entropy generation term S_gen in the entropy balance can be negative for a highly efficient, near-reversible process.

• T. True
• F. False

Answer: False

S_gen ≥ 0 is an absolute requirement of the second law — it can never be negative. A reversible process has S_gen = 0 exactly. Any real process with any irreversibility has S_gen > 0. A negative S_gen would imply spontaneous entropy destruction, which violates the second law. This is fundamental: unlike energy, entropy cannot be 'consumed' inside a system. It can be transferred out via heat (the Q̇/T_b term can be negative, carrying entropy out), but any entropy generated internally by irreversibilities stays and adds to the balance. The entropy balance equation has no negative-generation term precisely because second law forbids it.

Question 4

Transferring the same heat Q̇ at a lower boundary temperature T_b results in more entropy being transferred into or out of the system than transferring Q̇ at a higher T_b.

• T. True
• F. False

Answer: True

The entropy transfer associated with heat is Q̇/T_b — entropy and heat are related but not the same thing. At high temperature, a given amount of heat Q̇ carries less entropy (small Q̇/T_b). At low temperature, the same Q̇ carries more entropy (large Q̇/T_b). This is why heat rejection to a cold reservoir is thermodynamically 'costly' — more entropy must be exported at low temperature than was imported at high temperature, consistent with the second law. It also explains why heat pumps and refrigerators are inherently constrained: transferring heat from cold to hot requires adding entropy via work input to balance the entropy accounts.

Question 5

In the entropy balance equation, why must the heat transfer term be evaluated at the boundary temperature T_b where heat crosses the system boundary, rather than at an average internal system temperature?

Model answer: Entropy transfer by heat is Q̇/T, where T is the temperature at the exact location where heat crosses the boundary. If heat enters the system boundary at T_b = 300 K but the internal system temperature is 500 K, the entropy entering the system from outside is Q̇/300 — larger than Q̇/500. The difference is accounted for by S_gen: entropy is generated internally as heat flows from the boundary to the hotter interior. Using an average or internal temperature would misattribute this entropy generation as entropy transfer, corrupting the accounting. The entropy balance correctly separates entropy that crosses the boundary (via heat at T_b) from entropy created inside (S_gen). Only by using T_b can you isolate S_gen as a pure measure of internal irreversibility.

This is why high-temperature heat sources are thermodynamically valuable: a given Q̇ at high T_b delivers less entropy baggage into the system. Waste heat at low T_b carries proportionally more entropy for the same energy content, which is one reason low-temperature waste heat recovery has limited thermodynamic value — you're accepting a large entropy load with relatively little energy.