Thermodynamics · Chapter 7
Thermodynamics · Chapter 7

Entropy

Entropy is a state property that converts the second law into a quantitative audit. It tracks energy dispersal, tests whether a proposed process is possible, and separates entropy transferred with heat or mass from entropy generated by irreversibility.

Learning objectives

  • Define entropy through a reversible comparison path without confusing entropy with heat.
  • Select closed-system or control-volume entropy balances and apply a consistent sign convention.
  • Determine entropy changes from phase-appropriate tables or justified ideal-gas and incompressible models.
  • Distinguish adiabatic, reversible, and isentropic processes and evaluate device efficiencies.
  • Verify units, energy conservation, and nonnegative entropy generation while judging property uncertainty.
7–1

Entropy

A property, not a substance

For an internally reversible comparison path, \(dS=\delta Q_{rev}/T\). The comparison path evaluates a state change; it does not make the actual path reversible. Entropy has units kJ/K, and specific entropy has units kJ/(kg\!\cdot\!K).

Clausius definition
\[\Delta S=\int_1^2\frac{\delta Q_{rev}}{T},\qquad \oint\frac{\delta Q}{T}\le0\]

Physical meaning

Entropy measures how broadly energy is distributed among accessible microscopic states. Engineering balances use it chiefly as an accounting property for irreversibility, not as a vague synonym for disorder.

7–2

The Increase of Entropy Principle

For system plus surroundings, \(S_{gen}\ge0\). Zero denotes a reversible ideal; a negative computed value exposes a sign, boundary-temperature, state, or property error.

Closed-system entropy balance
\[\Delta S_{sys}=\sum_k\int\frac{\delta Q_k}{T_{b,k}}+S_{gen}\]
EXAMPLE 1Cooling a calibration block
A sealed 6.00 kg aluminum calibration block cools from 420 K to 310 K while delivering heat to a laboratory. Using the supplied constant specific heat, determine the block entropy change.

Recognition strategy

A solid with modest pressure change signals the incompressible entropy relation. The requested property change depends only on end states, although the actual cooling path need not be reversible.

Given information

  • Closed system: 6.00 kg fixed aluminum mass; process is cooling, not a cycle.
  • Absolute temperatures are 420 K and 310 K; pressure effects, kinetic energy, and potential energy are neglected.
  • Supplied educational property: constant \(c=0.900\ \text{kJ/(kg}\!\cdot\!\text{K)}\); verify against authoritative material data for release work.
  • Heat into the block is positive and work out is positive; calculation uses a mass basis and an approximate incompressible model.

Model and interpretation

The boundary follows the block, so no mass crosses it. Entropy is a state property; use an internally reversible comparison path to evaluate the change, not to claim that actual finite-temperature cooling is reversible.

Step-by-step solution
1. Classify. Treat aluminum as an incompressible solid with constant supplied specific heat over the stated range.
2. Formulate. For this approximate property model, \(\Delta S=m c\ln(T_2/T_1)\), with both temperatures absolute.
3. Substitute. \(\Delta S=(6.00)(0.900)\ln(310/420)\ \text{kJ/K}\); the logarithm is dimensionless.
4. Calculate. Since \(\ln(310/420)=-0.3036\), the block change is \(\Delta S=-1.64\ \text{kJ/K}\).

Verification

Units: The product \(mc\) has kJ/K and the logarithm is dimensionless, so entropy units are kJ/K.

Signs and direction: Cooling gives a negative system entropy change; this does not violate the second law because the laboratory gains entropy.

Plausibility: The magnitude is below \(mc=5.40\ \text{kJ/K}\), consistent with a moderate fractional temperature decrease; total entropy generation must remain nonnegative.

Final result: \(\Delta S_{\text{block}}=-1.64\ \text{kJ/K}\) for the stated approximate property model.

Engineering interpretation

The negative value describes entropy leaving the block with heat, not entropy destruction. A full laboratory audit needs the boundary temperature history; only then can total entropy generation and irreversibility be quantified.

7–3

Entropy Change of Pure Substances

Locate each state by two independent intensive properties. In a saturated mixture only, \(s=s_f+x s_{fg}\) with \(0\le x\le1\). In compressed-liquid or superheated regions, use the matching authoritative table and interpolate between bracketing entries; never assign quality in a single phase.

7–4

Isentropic Processes

Isentropic means constant entropy. A reversible adiabatic process is isentropic, but an adiabatic real device generally generates entropy. Isentropic states are ideal comparisons at the same inlet state and specified exit pressure.

7–5

Property Diagrams Involving Entropy

On a T–s plot, reversible heat is the area under the path, while a vertical path is constant entropy. A cycle encloses net reversible heat, equal to net work by the first law.

sTT_HT_L1234
Figure 7–1  On a T–s diagram a Carnot cycle is a rectangle; the enclosed area is the net work.
7–6

What Is Entropy?

Entropy can remain constant while energy changes form, and system entropy can decrease when sufficient entropy leaves with heat or mass. Only the entropy generation term has an unconditional nonnegative sign.

7–7

The T ds Relations

Property relations
\[Tds=du+Pdv,\qquad Tds=dh-vdP\]

These exact simple-compressible property relations connect entropy with \(u\) for a closed mass and \(h\) for flow analysis. Their use does not require the actual process to be reversible.

7–8

Entropy Change of Liquids and Solids

With negligible specific-volume change and supplied constant heat capacity, \(s_2-s_1=c\ln(T_2/T_1)\). Use absolute temperature and replace the constant with an integral when c varies materially.

7–9

The Entropy Change of Ideal Gases

Constant-property ideal gas
\[s_2-s_1=c_p\ln\frac{T_2}{T_1}-R\ln\frac{P_2}{P_1}\]

At broad temperature ranges, obtain \(s^\circ(T)\) from an authoritative ideal-gas table and interpolate, then retain the pressure correction. Confirm ideal-gas validity before using either form.

EXAMPLE 2Diagnosing an air-line state change
A sealed flexible air pouch moves from 290 K and 120 kPa absolute to 360 K and 250 kPa absolute. With supplied constant ideal-gas properties, find its specific entropy change.

Recognition strategy

Two ideal-gas end states and a requested entropy change call for the temperature-pressure relation. The pouch is closed even though its boundary moves; pressure work affects energy, not the state-property calculation directly.

Given information

  • Closed system, 1 kg mass basis; process between two equilibrium states, not a cycle.
  • All pressures are absolute and temperatures are kelvin; kinetic and potential energy changes are neglected.
  • Supplied ideal-gas data: \(c_p=1.005\ \text{kJ/(kg}\!\cdot\!\text{K)}\), \(R=0.287\ \text{kJ/(kg}\!\cdot\!\text{K)}\).
  • Heat into and boundary work out are positive; constant properties are approximate and variable-property air data should be used for release work.

Model and interpretation

Use \(s_2-s_1=c_p\ln(T_2/T_1)-R\ln(P_2/P_1)\). This is an ideal-gas property relation, not an entropy balance, so it does not by itself determine heat or entropy generation.

Step-by-step solution
1. Identify states. Both independent pairs \((T,P)\) define single-phase ideal-gas states on an absolute basis.
2. Formulate. Write \(\Delta s=c_p\ln(T_2/T_1)-R\ln(P_2/P_1)\) before numerical substitution.
3. Evaluate terms. The thermal term is \(1.005\ln(360/290)=0.217\ \text{kJ/(kg}\!\cdot\!\text{K)}\).
4. Combine. The pressure term is \(0.287\ln(250/120)=0.21065\ \text{kJ/(kg}\!\cdot\!\text{K)}\), giving \(\Delta s=+0.00665\ \text{kJ/(kg}\!\cdot\!\text{K)}\).

Verification

Units: Both logarithms use ratios, and multiplying by a specific heat gives kJ/(kg·K) as required.

Signs and direction: Heating raises entropy while compression lowers it; their near cancellation correctly leaves a small positive result.

Plausibility: Substitution into the same relation in reverse changes only the sign, and no claim of zero entropy generation follows from this property change.

Final result: \(s_2-s_1\approx +0.0067\ \text{kJ/(kg}\!\cdot\!\text{K)}\), positive but close to zero.

Engineering interpretation

The pouch approaches an isentropic state-to-state change, yet it is not proved reversible or adiabatic. Measurement and property uncertainty may exceed this small difference, so professional work should use variable-property data and an uncertainty study.

7–10

Reversible Steady-Flow Work

For an internally reversible steady one-stream device, \(w_{rev}=-\int v\,dP-\Delta ke-\Delta pe\). Compression work falls when cooling keeps specific volume low; this relation is a benchmark, not an actual device guarantee.

7–11

Minimizing the Compressor Work

Intercooling and staging reduce gas specific volume during compression. Equal pressure ratios minimize ideal work for identical stages with complete intercooling, but pressure drop, cooler cost, condensation, and off-design maps change the professional optimum.

7–12

Isentropic Efficiencies of Steady-Flow Devices

Actual versus ideal components
\[\eta_t=\frac{h_1-h_{2a}}{h_1-h_{2s}},\qquad \eta_c=\frac{h_{2s}-h_1}{h_{2a}-h_1}\]

Actual and isentropic exit states share inlet state and exit pressure. Use enthalpy, not internal energy, for steady-flow devices. Efficiencies must come from supplied or vendor data, never assumption disguised as fact.

7–13

Entropy Balance

Steady control volume
\[0=\sum_i\dot m_i s_i-\sum_e\dot m_e s_e+\sum_k\frac{\dot Q_k}{T_{b,k}}+\dot S_{gen}\]

Entropy crosses with mass and heat but not work. Use boundary temperature, not bulk fluid temperature, for entropy transfer by heat.

EXAMPLE 3Entropy audit of an insulated exchanger
In an insulated steady heat exchanger, 2.00 kg/s of oil cools from 430 K to 350 K while 3.00 kg/s of water warms from 295 K. Find the water outlet temperature and entropy-generation rate.

Recognition strategy

An insulated two-stream control volume requires mass and energy balances first, then an entropy balance. Each liquid stays single phase, allowing supplied incompressible heat capacities to close the states.

Given information

  • Steady control volume encloses both streams; no mixing, shaft work, or external heat transfer occurs.
  • Supplied constants: \(c_o=2.10\) and \(c_w=4.18\ \text{kJ/(kg}\!\cdot\!\text{K)}\); temperatures are absolute.
  • Each stream conserves mass; pressure changes, kinetic energy, and potential energy are neglected.
  • Heat into the whole control volume and work out are positive; this is an approximate incompressible model with a mass-rate basis.

Model and interpretation

For the whole exchanger, external \(\dot Q=0\), but energy passes internally between streams. Enthalpy changes reduce to \(c\Delta T\); entropy changes reduce to \(c\ln(T_2/T_1)\).

Step-by-step solution
1. Energy balance. Set \(\dot m_oc_o(430-350)=\dot m_wc_w(T_{w2}-295)\) for steady operation.
2. Solve outlet state. The oil releases \(336\ \text{kW}\), hence \(T_{w2}=295+336/(3.00\times4.18)=321.794\ \text{K}\), reported as \(321.8\ \text{K}\).
3. Entropy balance. With no boundary heat, \(\dot S_{gen}=\dot m_oc_o\ln(350/430)+\dot m_wc_w\ln(321.794/295)\).
4. Evaluate audit. The oil term is \(-0.8646\ \text{kW/K}\), the water term is \(+1.0902\ \text{kW/K}\), so \(\dot S_{gen}=0.2256\ \text{kW/K}\), reported as \(0.226\ \text{kW/K}\).

Verification

Units: Energy terms are kJ/s = kW and entropy-rate terms are kJ/(s·K) = kW/K.

Signs and direction: Oil heat loss equals water heat gain, while positive entropy generation confirms the permitted heat-flow direction.

Plausibility: The water exit remains below the 350 K oil exit, avoiding a temperature cross; mass, energy, and entropy audits all close with \(\dot S_{gen}\ge0\).

Final result: \(T_{w2}=321.8\ \text{K}\) and \(\dot S_{gen}=0.226\ \text{kW/K}\).

Engineering interpretation

The positive generation measures degradation caused by finite temperature differences. Release work needs temperature-dependent vendor fluid properties, pressure-drop effects, heat loss, fouling, and instrument uncertainty rather than these supplied constants.

Problem-solving strategy

  1. Recognize whether the boundary encloses a fixed mass, one steady component, or an entire cycle.
  2. Identify every state with absolute pressure and temperature, phase, and an authoritative property source.
  3. Write mass, energy, entropy, and when relevant exergy balances before inserting component models.
  4. Separate ideal benchmark states from actual states and iterate when properties depend on the unknown temperature.
  5. Audit units, signs, cycle closure, entropy generation, and sensitivity to supplied data.

Chapter recap

  • Define the boundary and distinguish entropy transfer from entropy generation.
  • Select phase-valid properties and use absolute temperature in every logarithm or ratio.
  • Keep actual and isentropic states separate and require \(S_{gen}\ge0\).

Before moving on

You should be able to identify states, write closed and open entropy balances, calculate generation, and explain why an adiabatic real process need not be isentropic.