3.3 Second Law, Entropy, Exergy & Reversibility Analysis

3.3 Second Law, Entropy, Exergy & Reversibility Analysis

While the First Law of Thermodynamics establishes energy conservation, it places no constraint on the direction of energy transfers. Real physical processes occur spontaneously in one direction only. The Second Law of Thermodynamics identifies process directionality, asserts that energy has quality as well as quantity, and establishes theoretical performance limits for thermal power plants, heat pumps, and refrigeration systems.

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1. Classical Statements of the Second Law

Kelvin-Planck Statement

$\text{It is impossible for any device that operates on a thermodynamic cycle to receive heat from a single thermal reservoir}$ $\text{and produce a net amount of work.}$

This statement dictates that no heat engine can convert 100% of absorbed heat into useful work. Heat must be rejected to a low-temperature sink ($Q_L > 0$).

$\eta_{\text{th}} = \frac{W_{\text{net,out}}}{Q_H} = \frac{Q_H - Q_L}{Q_H} = 1 - \frac{Q_L}{Q_H} < 100\%$

Clausius Statement

$\text{It is impossible to construct a device that operates in a cycle and produces no effect other than standard work input}$ $\text{to transfer heat from a lower-temperature body to a higher-temperature body.}$

This asserts that heat cannot flow spontaneously from a cold medium to a warm medium without external work input ($W_{\text{in}} > 0$).

Coefficient of Performance (COP)

Refrigerators and heat pumps operate on cyclic refrigeration cycles:

$\text{COP}_{R} = \frac{\text{Desired Output}}{\text{Required Work}} = \frac{Q_L}{W_{\text{in}}} = \frac{Q_L}{Q_H - Q_L}$

$\text{COP}_{HP} = \frac{\text{Desired Output}}{\text{Required Work}} = \frac{Q_H}{W_{\text{in}}} = \frac{Q_H}{Q_H - Q_L} = \text{COP}_{R} + 1$

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2. The Carnot Cycle & Absolute Temperature Scale

The Carnot Cycle is a theoretical, totally reversible cycle composed of four reversible processes:

1. Reversible Isothermal Heat Addition ($T_H = \text{constant}$)
2. Reversible Adiabatic (Isentropic) Expansion
3. Reversible Isothermal Heat Rejection ($T_L = \text{constant}$)
4. Reversible Adiabatic (Isentropic) Compression

Carnot Principles

1. The efficiency of an irreversible heat engine is always less than the efficiency of a reversible engine operating between the same two reservoirs.
2. The efficiencies of all reversible heat engines operating between the same two reservoirs are identical.

Carnot Thermal Efficiency & COPs

For any reversible (Carnot) cycle, heat ratio equals absolute temperature ratio ($Q_L/Q_H = T_L/T_H$ in Kelvin):

$\eta_{\text{th,Carnot}} = 1 - \frac{T_L}{T_H}$

$\text{COP}_{R,\text{Carnot}} = \frac{T_L}{T_H - T_L}, \quad \text{COP}_{HP,\text{Carnot}} = \frac{T_H}{T_H - T_L}$

Crucial Rule: All temperatures in Carnot formulas MUST be expressed in absolute scale (Kelvin: $\text{K} = ^\circ\text{C} + 273.15$).

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3. Clausius Inequality & Definition of Entropy

The Clausius Inequality

For any cyclic process:

$\oint \frac{\delta Q}{T} \le 0$

• $\oint \frac{\delta Q}{T} = 0$ (Totally reversible cycle)
• $\oint \frac{\delta Q}{T} < 0$ (Irreversible cycle)
• $\oint \frac{\delta Q}{T} > 0$ (Impossible cycle)

Definition of Entropy ($S$)

Entropy is an extensive thermodynamic property defined by the differential relation:

$dS = \left( \frac{\delta Q}{T} \right)_{\text{rev}} \implies \Delta S = S_2 - S_1 = \int_1^2 \left( \frac{\delta Q}{T} \right)_{\text{rev}}$

The Increase of Entropy Principle

For any real, irreversible process in an isolated or combined system:

$S_{\text{gen}} \ge 0 \implies \Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} = S_{\text{gen}}$

• $S_{\text{gen}} = 0$: Reversible process
• $S_{\text{gen}} > 0$: Irreversible process
• $S_{\text{gen}} < 0$: Impossible process

Entropy Changes for Ideal Gases (Constant Specific Heats)

$\Delta s = s_2 - s_1 = c_v \ln\left(\frac{T_2}{T_1}\right) + R \ln\left(\frac{v_2}{v_1}\right)$

$\Delta s = s_2 - s_1 = c_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right)$

Isentropic ($s_2 = s_1$) Relations for Ideal Gases ($n = k = c_p/c_v$)

$\frac{T_2}{T_1} = \left( \frac{P_2}{P_1} \right)^{\frac{k-1}{k}} = \left( \frac{v_1}{v_2} \right)^{k-1}$

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4. Isentropic Efficiencies of Thermal Turbomachinery

Real adiabatic devices (turbines, compressors, pumps, nozzles) suffer from fluid friction, turbulence, and heat dissipation, causing entropy generation ($s_2 > s_1$). Isentropic efficiency ($\eta$) compares actual performance against an ideal reversible adiabatic process.

1. Isentropic Efficiency of a Turbine ($\eta_t$)

$\eta_t = \frac{\text{Actual Turbine Work}}{\text{Isentropic Turbine Work}} = \frac{w_a}{w_s} \approx \frac{h_1 - h_{2a}}{h_1 - h_{2s}}$

2. Isentropic Efficiency of a Compressor / Pump ($\eta_c$)

$\eta_c = \frac{\text{Isentropic Compressor Work}}{\text{Actual Compressor Work}} = \frac{w_s}{w_a} \approx \frac{h_{2s} - h_1}{h_{2a} - h_1}$

3. Isentropic Efficiency of a Nozzle ($\eta_n$)

$\eta_n = \frac{\text{Actual Exit Kinetic Energy}}{\text{Isentropic Exit Kinetic Energy}} = \frac{\mathbf{V}_{2a}^2}{\mathbf{V}_{2s}^2} \approx \frac{h_1 - h_{2a}}{h_1 - h_{2s}}$

  Turbine Expansion (T-s Diagram)    Compressor Compression (T-s Diagram)
  T ^                                T ^
    |   1                              |        2a (Actual)
    |  / \                             |       /|
    | /   \                            |      / | 2s (Isentropic)
    |/     \ 2a (Actual)               |     /  |
    |-------\                          |    1---|------>
    |        2s (Isentropic)           +------------------------> s
    +------------------------> s

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5. Exergy (Availability) & Irreversibility Analysis

Exergy (or Availability) is the maximum useful work potential that can be extracted from a system or fluid stream as it comes into complete thermodynamic equilibrium with its environment at the dead state ($P_0 = 101.325\text{ kPa}, T_0 = 298.15\text{ K} = 25^\circ\text{C}$).

Flow Exergy ($\psi$) per unit mass

$\psi = (h - h_0) - T_0 (s - s_0) + \frac{\mathbf{V}^2}{2} + g z$

Exergy Destruction & Gouy-Stodola Theorem

Energy cannot be destroyed, but exergy is destroyed whenever an irreversible process takes place. Exergy destruction ($X_{\text{destroyed}}$) or Irreversibility ($I$) is directly proportional to entropy generation:

$X_{\text{destroyed}} = I = T_0 S_{\text{gen}}$ $\dot{X}_{\text{destroyed}} = \dot{I} = T_0 \dot{S}_{\text{gen}}$

Second-Law Efficiency ($\eta_{\text{II}}$)

$\eta_{\text{II}} = \frac{\text{Thermal Efficiency Actual}}{\text{Thermal Efficiency Carnot}} = \frac{\text{Exergy Recovered}}{\text{Exergy Supplied}} = 1 - \frac{X_{\text{destroyed}}}{\text{Exergy Supplied}}$

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6. Step-by-Step Worked Numerical Calculation

Problem Statement

A heat engine operating on a steady flow process receives thermal energy from a high-temperature heat source at $T_H = 1000.0\text{ K}$ at a rate of $\dot{Q}H = 500.0\text{ kW}$. The heat engine rejects waste heat to the ambient atmosphere at $T_0 = T_L = 300.0\text{ K}$. The measured actual net electric power output of the heat engine is $\dot{W}{\text{net,out}} = 275.0\text{ kW}$.

Calculate:

1. The actual thermal efficiency ($\eta_{\text{th}}$) of the heat engine.
2. The maximum theoretical Carnot thermal efficiency ($\eta_{\text{th,Carnot}}$).
3. The rate of heat rejection ($\dot{Q}_L$) to the ambient air in kW.
4. The rate of entropy generation ($\dot{S}_{\text{gen}}$) in kW/K.
5. The rate of exergy destruction ($\dot{X}_{\text{destroyed}}$ or irreversibility $\dot{I}$) in kW.
6. The Second-Law Efficiency ($\eta_{\text{II}}$) of the heat engine.

Step-by-Step Solution

Step 1: Calculate actual thermal efficiency ($\eta_{\text{th}}$) $\eta_{\text{th}} = \frac{\dot{W}_{\text{net,out}}}{\dot{Q}_H} = \frac{275.0\text{ kW}}{500.0\text{ kW}} = 0.5500 \quad (55.00\%)$

Step 2: Calculate Carnot maximum thermal efficiency ($\eta_{\text{th,Carnot}}$) $\eta_{\text{th,Carnot}} = 1 - \frac{T_L}{T_H} = 1 - \frac{300.0\text{ K}}{1000.0\text{ K}} = 1 - 0.3000 = 0.7000 \quad (70.00\%)$

Step 3: Calculate rate of heat rejection ($\dot{Q}_L$) From First-Law rate energy balance: $\dot{Q}_H - \dot{Q}_L = \dot{W}_{\text{net,out}} \implies \dot{Q}_L = \dot{Q}_H - \dot{W}_{\text{net,out}}$ $\dot{Q}_L = 500.0\text{ kW} - 275.0\text{ kW} = 225.0\text{ kW}$

Step 4: Calculate rate of entropy generation ($\dot{S}_{\text{gen}}$) Apply entropy balance to the isolated combined system (heat engine + reservoirs): $\dot{S}_{\text{gen}} = \sum \frac{\dot{Q}_{\text{out}}}{T}_{\text{sink}} - \sum \frac{\dot{Q}_{\text{in}}}{T}_{\text{source}} = \frac{\dot{Q}_L}{T_L} - \frac{\dot{Q}_H}{T_H}$ $\dot{S}_{\text{gen}} = \frac{225.0\text{ kW}}{300.0\text{ K}} - \frac{500.0\text{ kW}}{1000.0\text{ K}} = 0.7500 - 0.5000 = 0.2500\text{ kW/K}$

Step 5: Calculate rate of exergy destruction ($\dot{X}_{\text{destroyed}}$) Using the Gouy-Stodola theorem ($T_0 = 300.0\text{ K}$): $\dot{X}_{\text{destroyed}} = T_0 \dot{S}_{\text{gen}} = 300.0\text{ K} \times 0.2500\text{ kW/K} = 75.0\text{ kW}$

Alternative Check using Available Energy: $\text{Max Available Work } \dot{W}_{\text{rev}} = \eta_{\text{th,Carnot}} \dot{Q}_H = 0.7000 \times 500.0 = 350.0\text{ kW}$ $\dot{X}_{\text{destroyed}} = \dot{W}_{\text{rev}} - \dot{W}_{\text{actual}} = 350.0\text{ kW} - 275.0\text{ kW} = 75.0\text{ kW}$

Step 6: Calculate Second-Law Efficiency ($\eta_{\text{II}}$) $\eta_{\text{II}} = \frac{\eta_{\text{th}}}{\eta_{\text{th,Carnot}}} = \frac{0.5500}{0.7000} = 0.7857 \quad (78.57\%)$

Final Summary of Results:

• Actual Thermal Efficiency $\eta_{\text{th}} = 55.00%$
• Carnot Maximum Efficiency $\eta_{\text{th,Carnot}} = 70.00%$
• Heat Rejection Rate $\dot{Q}_L = 225.0\text{ kW}$
• Entropy Generation Rate $\dot{S}_{\text{gen}} = 0.2500\text{ kW/K}$
• Exergy Destruction Rate $\dot{X}_{\text{destroyed}} = 75.0\text{ kW}$
• Second-Law Efficiency $\eta_{\text{II}} = 78.57%$