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Thermodynamics · Chapter 7 of 10 · Advanced

Entropy

Entropy turns the second law into a number. It can be transferred with heat and generated by irreversibility, never destroyed, and the ideal of constant entropy sets the target every real turbine and compressor is measured against.

Readiness check

This chapter uses logarithms and the second law. Tick only what you can do closed-notes.

Evaluate a natural logarithm of a ratio.
State the Carnot efficiency and what reversibility means.
Read entropy from a steam table.
Use c_p, c_v, R, and k for an ideal gas.
Apply a turbine energy balance ẇ = h₁ − h₂.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsRevisit reversibility and Carnot in Chapter 6.

3 or more weak itemsReview the steam tables in Chapter 3.

The core idea

Entropy is a property that increases whenever a real process runs. Heat carries entropy in or out as Q/T, and irreversibility generates more; a perfect, reversible, adiabatic process keeps entropy constant.

dS = δQ/T (reversible)S_gen ≥ 0Δs = c_p ln(T₂/T₁) − R ln(P₂/P₁)

The Clausius inequality says that around any cycle the sum of δQ/T is at most zero, which defines a new property, entropy. For a process, the entropy change equals the entropy carried in with heat plus the entropy generated inside, and the generated part can never be negative. An adiabatic, reversible process generates nothing and carries nothing, so it is isentropic, a vertical line on a T-s diagram. Real devices slant to the right because they generate entropy.

The skill works when: you use the isentropic ideal as a target and compare the real process to it.

The skill breaks down when: entropy is expected to fall in an isolated system, or isentropic is confused with adiabatic alone.

The concept. On a T-s diagram an ideal turbine expansion is a vertical drop at constant entropy (1 to 2s). A real expansion generates entropy, so it slants right to 2a, ending warmer and doing less work.

The skills, taught in order

Five skills define entropy, compute it for an ideal gas and from tables, use the isentropic relations, and turn the isentropic ideal into a device efficiency.

7.1 Entropy and the Clausius inequality

Around any cycle, the cyclic sum of δQ/T is less than or equal to zero, with equality only for a reversible cycle. This guarantees that entropy S, defined by dS = δQ/T for a reversible path, is a property. For a process, ΔS = (entropy transfer with heat) + S_gen, and S_gen ≥ 0 always.

7.2 Entropy change of an ideal gas

For an ideal gas with constant specific heats, Δs = c_p ln(T₂/T₁) − R ln(P₂/P₁), or equivalently Δs = c_v ln(T₂/T₁) + R ln(v₂/v₁). The first form is the handiest when pressures are known. Entropy of a pure substance is read directly from the tables, just like enthalpy.

7.3 Isentropic processes and relations

Set Δs = 0 for an ideal gas and the property relations collapse into the isentropic relations: T₂/T₁ = (P₂/P₁)^(k−1)/k = (v₁/v₂)^k−1, and Pv^k = constant. These give the exit state of an ideal compressor or turbine before any efficiency is applied.

Quantity	Isentropic relation (ideal gas)
Temperature and pressure	T₂/T₁ = (P₂/P₁)^(k−1)/k
Temperature and volume	T₂/T₁ = (v₁/v₂)^k−1
Pressure and volume	Pv^k = constant

7.4 The T-s diagram

On a temperature-entropy diagram, the area under a reversible path is the heat transferred, and a vertical line is an isentropic process. Reading cycles on a T-s plot makes the rejected heat and the effect of irreversibility visible at a glance.

7.5 Isentropic efficiency

Real adiabatic devices fall short of the isentropic ideal. For a turbine the isentropic efficiency is η_T = (h₁ − h₂a)/(h₁ − h₂s), the actual work over the ideal work. For a compressor it inverts to η_C = (h₂s − h₁)/(h₂a − h₁), the ideal work over the actual, because a real compressor needs more.

Engineering connection: isentropic efficiency is how the ideal cycles of Chapters 9 and 10 are corrected to predict real turbine, pump, and compressor performance.

Worked example 1: isentropic compression of air

Air at 100 kPa and 300 K is compressed isentropically to 800 kPa. Using k = 1.4, c_p = 1.005, and R = 0.287 kJ/kg·K, find the exit temperature and the compressor work per kilogram, and confirm the entropy change is zero.

Figure 1. An ideal compressor raises pressure with no entropy change, so the exit temperature follows the isentropic relation and the work is the enthalpy rise.

ProblemFind the exit temperature, the work, and the entropy change for the isentropic compression in Figure 1.
Given / findP₁ = 100 kPa, T₁ = 300 K, P₂ = 800 kPa, k = 1.4, c_p = 1.005, R = 0.287 kJ/kg·K. Find T₂, w, Δs.
AssumptionsAir is an ideal gas with constant specific heats; the process is reversible and adiabatic, hence isentropic.
ModelThe isentropic relation gives T₂ from the pressure ratio; the adiabatic compressor work is the enthalpy rise c_p(T₂ − T₁).
EquationsT₂ = T₁ (P₂/P₁)^(k−1)/kw_in = c_p(T₂ − T₁)Δs = c_p ln(T₂/T₁) − R ln(P₂/P₁)
SolveT₂ = 300 × 8^0.2857 = 300 × 1.8114 = 543 K. w_in = 1.005 × (543 − 300) = 245 kJ/kg. Δs = 1.005 ln(1.811) − 0.287 ln(8) = 0.597 − 0.597 = 0.
CheckThe two entropy terms cancel, confirming the process really is isentropic. Compressing eightfold heats the air from 300 K to 543 K, the familiar reason compressed air is hot.
ConclusionThe isentropic relation fixes the ideal exit state, and the enthalpy rise is the minimum work needed. A real compressor would need more, captured by its isentropic efficiency.

Result. T₂ = 543 K, w_in = 245 kJ/kg, and Δs = 0.

Worked example 2: isentropic efficiency of a steam turbine

Steam enters a turbine at 3 MPa and 350 °C (h₁ = 3115.3 kJ/kg, s₁ = 6.7427 kJ/kg·K) and expands to 10 kPa. The turbine has an isentropic efficiency of 85 percent. Using the 10 kPa saturation data (s_f = 0.6492, s_fg = 7.5010 kJ/kg·K, h_f = 191.81, h_fg = 2392.82 kJ/kg), find the actual work and the actual exit enthalpy.

Figure 2. The ideal expansion drops to 2s at constant entropy; the real one ends at 2a with more entropy and less work. Efficiency is the ratio of the two enthalpy drops.

ProblemFind the actual work and exit enthalpy for the turbine in Figure 2.
Given / findh₁ = 3115.3 kJ/kg, s₁ = 6.7427 kJ/kg·K, exit 10 kPa, η_T = 0.85, with the saturation data given. Find w_actual and h₂a.
AssumptionsAdiabatic; the ideal exit is isentropic (s₂s = s₁); negligible kinetic and potential changes.
ModelFind the ideal exit from s₂s = s₁ (quality, then h₂s), take the ideal work, scale by efficiency, then back out the real exit enthalpy.
Equationsx_2s = (s₁ − s_f)/s_fgh_2s = h_f + x_2s h_fgw_actual = η_T(h₁ − h_2s)
Solvex_2s = (6.7427 − 0.6492)/7.5010 = 0.8124. h_2s = 191.81 + 0.8124 × 2392.82 = 2135.6 kJ/kg. Ideal work = 3115.3 − 2135.6 = 979.6 kJ/kg, so w_actual = 0.85 × 979.6 = 832.7 kJ/kg. Then h₂a = 3115.3 − 832.7 = 2282.6 kJ/kg.
CheckThe real exit enthalpy is higher than the ideal 2135.6 kJ/kg, meaning warmer, wetter steam and less work, exactly the effect of generated entropy.
ConclusionIsentropic efficiency bridges the ideal and the real: the perfect expansion sets the target, and the efficiency tells you how much of it the machine actually delivers.

Result. w_actual = 832.7 kJ/kg and h₂a = 2282.6 kJ/kg.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Adiabatic equals isentropic	Real adiabatic device treated as isentropic	"Is the process reversible as well as adiabatic?"	Isentropic needs both; real adiabatic devices generate entropy.
Expecting entropy to fall	Negative S_gen in an isolated system	"Is any entropy carried in by heat?"	Entropy generation is never negative; total entropy of an isolated system rises.
Temperatures in °C in the log	Entropy change comes out wrong	"Are T₁ and T₂ in kelvin?"	Use absolute temperature inside ln(T₂/T₁).
Inverting the efficiency	Turbine and compressor efficiencies swapped	"Is the device producing or consuming work?"	Turbine: actual over ideal; compressor: ideal over actual.

Practice ladder

Level 1 · Direct skill

Air is heated at constant pressure from 300 K to 600 K. Find its specific entropy change (c_p = 1.005 kJ/kg·K).

Show answer

At constant pressure, Δs = c_p ln(T₂/T₁) = 1.005 × ln(2) = 1.005 × 0.6931 = 0.697 kJ/kg·K. Entropy rises because heat is added.

Level 2 · Mixed concept

The air compressor of Worked Example 1 instead goes to 1000 kPa. What is the new isentropic exit temperature?

Show answer

T₂ = 300 × (1000/100)^0.2857 = 300 × 10^0.2857 = 300 × 1.931 = 579 K. A higher pressure ratio gives a higher exit temperature.

Level 3 · Independent problem

A compressor takes air from 290 K and ideally would reach 450 K, but its isentropic efficiency is 0.80. Find the actual exit temperature (c_p = 1.005 kJ/kg·K).

Show answer

Ideal work = c_p(450 − 290) = 160.8 kJ/kg. Actual work = ideal/η = 160.8/0.80 = 201 kJ/kg. ΔT_actual = 201/1.005 = 200 K, so T_actual = 290 + 200 = 490 K. A real compressor runs hotter than the ideal.

Level 4 · Transfer to real engineering

Explain, using entropy generation, why a throttle (which keeps enthalpy constant) is an inherently wasteful way to drop pressure compared with a turbine.

What good work looks like

A throttle generates entropy and extracts no work, while a turbine drops pressure closer to isentropically and captures work; the throttle destroys the work potential that the turbine recovers.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that my entropy change uses kelvin and the right sign on the pressure term."

"Give me four adiabatic devices; I will say which are close to isentropic."

"What is the exit temperature?" Applying the isentropic relation yourself is the skill.

"Compute the turbine efficiency." Forming the enthalpy-drop ratio is the point.

Portfolio task

Take a real turbine or compressor with a published efficiency. Find the ideal isentropic exit state, scale by the efficiency, and report the real work and exit state on a T-s sketch.

Must include: an isentropic exit state, an efficiency applied correctly, and a T-s sketch showing the rightward slant.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. What does the Clausius inequality establish?

That the cyclic sum of δQ/T is at most zero, which makes entropy a property.

2. Write the ideal-gas entropy change with pressures.

Δs = c_p ln(T₂/T₁) − R ln(P₂/P₁).

3. Give the isentropic T-P relation.

T₂/T₁ = (P₂/P₁)^(k−1)/k.

4. Define turbine isentropic efficiency.

η_T = actual work over ideal work = (h₁ − h₂a)/(h₁ − h₂s).

5. Why is isentropic not the same as adiabatic?

Isentropic means adiabatic and reversible; a real adiabatic device still generates entropy.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive the isentropic compression and turbine efficiency from a blank page.

+3 daysCompute entropy changes for two new processes.

+7 daysCarry entropy into exergy, Chapter 8.

+30 daysReuse isentropic efficiency in the cycles of Chapters 9 and 10.

Textbook mapping

Item	Mapping
Primary source	Borgnakke and Sonntag, Fundamentals of Thermodynamics, Chapter 8 (Entropy) and Chapter 9 (Second-Law Analysis for a Control Volume)
Cross-reference	Çengel and Boles, Ch. 7 · Moran, Ch. 6
Core topics	7.1 Clausius inequality · 7.2 Ideal-gas entropy · 7.3 Isentropic relations · 7.4 T-s diagram · 7.5 Isentropic efficiency
Engineering connection	Isentropic efficiency corrects every ideal cycle to real machine performance.
Read next	Chapter 8: Exergy.