12.2 Entropy, Availability, Exergy & Properties of Pure Substances

11.2 Entropy, Availability, Exergy & Properties of Pure Substances

In thermal power engineering and mining utility analysis, understanding how energy degrades due to irreversibilities is critical. While the First Law tracks energy quantity, Entropy and Exergy evaluate energy quality, degradation, and maximum useful work output.

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1. Clausius Inequality & Definition of Entropy

The Clausius Inequality

For any thermodynamic cycle, whether reversible or irreversible, operating with thermal reservoirs:

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

• $\oint \frac{\delta Q}{T} = 0$: The cycle is internally reversible.
• $\oint \frac{\delta Q}{T} < 0$: The cycle is irreversible (real-world process with friction, throttling, or unrestrained expansion).
• $\oint \frac{\delta Q}{T} > 0$: The cycle is impossible (violates the Second Law).

               CLAUSIUS CYCLE INTEGRAL CRITERIA
                    /-----------------------\
                   |   oint (dQ / T) <= 0   |
                    \-----------------------/
                                |
        +-----------------------+-----------------------+
        |                                               |
        v                                               v
  oint (dQ/T) = 0                               oint (dQ/T) < 0
Reversible Engine Cycle                      Irreversible Engine Cycle
(Carnot Ideal Limit)                         (Real Thermal Machines)

Mathematical Definition of Entropy

Because the cyclic integral of $\left(\frac{\delta Q}{T}\right)_{rev}$ is identically zero for all reversible paths, it represents an exact differential of a state property called Entropy ($S$):

$$dS = \left(\frac{\delta Q}{T}\right)_{rev}$$

For any real (irreversible) process between states 1 and 2:

$$dS \ge \frac{\delta Q}{T} \implies dS = \frac{\delta Q}{T} + \delta S_{gen}$$

Where $\delta S_{gen}$ is the entropy generation within system boundaries:

• $\delta S_{gen} > 0$ for irreversible processes.
• $\delta S_{gen} = 0$ for reversible processes.
• $\delta S_{gen} < 0$ is thermodynamically impossible.

Principle of Increase of Entropy

For an isolated system that exchanges neither mass nor heat with surroundings ($\delta Q = 0$):

$$\Delta S_{isolated} \ge 0$$

Considering the system and its surroundings collectively as an isolated universe:

$$\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings} = S_{gen} \ge 0$$

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2. Gibbs $T\,ds$ Equations & Entropy Changes for Ideal Gases

From the First Law ($\delta Q = du + P\,dv$) and definition of entropy ($T\,ds = \delta Q_{rev}$):

$$T \, ds = du + P \, dv \qquad \text{(First } T\,ds \text{ equation)}$$

Using $h = u + Pv \implies dh = du + P\,dv + v\,dP$, substituting $du + P\,dv = dh - v\,dP$:

$$T \, ds = dh - v \, dP \qquad \text{(Second } T\,ds \text{ equation)}$$

Entropy Change Formulas for Ideal Gases ($P v = R T, du = c_v dT, dh = c_p dT$)

$$\Delta s = s_2 - s_1 = c_v \ln\left(\frac{T_2}{T_1}\right) + R \ln\left(\frac{v_2}{v_1}\right)$$ $$\Delta s = s_2 - s_1 = c_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right)$$ $$\Delta s = s_2 - s_1 = c_v \ln\left(\frac{P_2}{P_1}\right) + c_p \ln\left(\frac{v_2}{v_1}\right)$$

Special Case: For an isentropic process ($s_2 = s_1, \Delta s = 0$):

$$c_p \ln\left(\frac{T_2}{T_1}\right) = R \ln\left(\frac{P_2}{P_1}\right) \implies \frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma - 1}{\gamma}}$$

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3. Availability, Exergy & Irreversibility

Available vs. Unavailable Energy

When heat $Q_1$ is transferred from a high-temperature reservoir at $T_1$ to an engine rejecting heat to ambient surroundings at $T_0$:

• Available Energy ($AE$): Maximum theoretical work output $= W_{max} = Q_1 \left(1 - \frac{T_0}{T_1}\right)$.
• Unavailable Energy ($UAE$): Minimum heat rejected to environment $= Q_1 - AE = T_0 \Delta S$.

                      TOTAL HEAT SUPPLIED (Q_1)
+---------------------------------------+-----------------------------+
|        Available Energy (AE)          |   Unavailable Energy (UAE)  |
|     W_max = Q_1 * (1 - T_0 / T_1)     |       UAE = T_0 * Delta S   |
+---------------------------------------+-----------------------------+

Exergy Formulations

Exergy is the maximum theoretical useful work obtainable as a system comes into complete mechanical, thermal, and chemical equilibrium with the reference environment (the dead state: $P_0 = 101.325\text{ kPa}, T_0 = 298.15\text{ K}, V_0, U_0, S_0$).

1. Closed System (Non-Flow) Exergy ($\Phi$):

   $$\Phi = (U - U_0) + P_0(V - V_0) - T_0(S - S_0)$$ $$\text{Specific: } \phi = (u - u_0) + P_0(v - v_0) - T_0(s - s_0)$$

2. Open System (Flow) Exergy / Stream Availability ($\psi$):

   $$\psi = (h - h_0) - T_0(s - s_0) + \frac{V^2}{2000} + \frac{g z}{1000}$$

Gouy-Stodola Theorem & Irreversibility ($I$)

The rate of exergy destruction (irreversibility or lost work) is directly proportional to the total entropy generation rate of the universe:

$$I = W_{rev} - W_{actual} = T_0 \cdot S_{gen} = T_0 \cdot \Delta S_{universe}$$

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

Unlike First Law thermal efficiency (which measures energy quantity conversion), Second Law efficiency measures performance relative to the theoretical reversible ideal:

$$\eta_{II} = \frac{\text{Exergy Recovered / Output}}{\text{Exergy Supplied / Input}} = \frac{W_{actual}}{W_{reversible}} = \frac{\eta_{th}}{\eta_{Carnot}}$$

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4. Properties of Pure Substances & Phase Diagrams

A pure substance is homogeneous and chemically invariable throughout (e.g., water/steam, nitrogen, gaseous combustion products in single phase).

                TEMPERATURE-ENTROPY (T-s) STEAM DOME
      T ^
        |                  Critical Point (CP)
        |                        /\
        |      Saturated        /  \        Saturated
        |     Liquid Line      /    \      Vapour Line
        |      (x = 0)        / Wet  \       (x = 1)
        |                    / Region \ 
        |     Compressed    /  (x)     \   Superheated
        |       Liquid     /            \    Vapour
        |                 /              \
        +----------------+----------------+----------------> s

Key Reference Points for Pure Water

• Triple Point: All three phases (ice, liquid water, steam) coexist in thermodynamic equilibrium.
  • $T_{tp} = 0.01^{\circ}\text{C} = 273.16\text{ K}$
  • $P_{tp} = 0.6117\text{ kPa} = 0.006117\text{ bar} = 4.587\text{ mm Hg}$
  • Degrees of freedom ($F$) by Gibbs Phase Rule ($F = C - P + 2 = 1 - 3 + 2 = 0$).
• Critical Point: Meniscus between liquid and vapor disappears; latent heat of vaporization $h_{fg} = 0$.
  • $P_{cr} = 22.064\text{ MPa} = 220.64\text{ bar}$
  • $T_{cr} = 373.95^{\circ}\text{C} = 647.10\text{ K}$
  • Specific volume $v_{cr} = 0.003155\text{ m}^3/\text{kg}$

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5. Dryness Fraction (Steam Quality) & Property Formulations

In the two-phase wet mixture region:

$$\text{Dryness Fraction (Quality) } x = \frac{m_v}{m_v + m_l} = \frac{m_v}{m_{total}} \qquad (0 \le x \le 1)$$

• $x = 0$: Saturated liquid state (subscript $f$).
• $x = 1$: Saturated dry vapour state (subscript $g$).

Thermodynamic Relations for Wet Steam (Given $x$ at $P_{sat}, T_{sat}$)

Superheated Steam Formulations ($T_{sup} > T_{sat}$)

$$h = h_g + c_{pv} (T_{sup} - T_{sat})$$ $$s = s_g + c_{pv} \ln\left(\frac{T_{sup}}{T_{sat}}\right)$$

(Where $c_{pv} \approx 2.1\text{ kJ/kg}\cdot\text{K}$ is the average specific heat of superheated steam).

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6. Measurement of Steam Dryness Fraction (Calorimeters)

                      CALORIMETER COMPARISON
+--------------------------------+----------------------------------------+
| Calorimeter Type               | Working Principle & Measurement Range  |
+--------------------------------+----------------------------------------+
| Barrel / Tank Calorimeter      | Direct water mass condensation & heat  |
|                                | balance; prone to radiation loss.      |
+--------------------------------+----------------------------------------+
| Separating Calorimeter         | Mechanical separation of moisture      |
|                                | droplets by inertia; good for very wet |
|                                | steam ($x < 0.85$).                    |
+--------------------------------+----------------------------------------+
| Throttling Calorimeter         | Isenthalpic throttling ($h_1 = h_2$)   |
|                                | into superheated region; requires      |
|                                | $x > 0.95$ and superheated exit state. |
+--------------------------------+----------------------------------------+
| Combined Separating &          | Separating stage catches bulk water    |
| Throttling Calorimeter         | ($x_1$), throttling stage measures dry |
|                                | fraction ($x_2$); $x = x_1 \cdot x_2$. |
+--------------------------------+----------------------------------------+

Governing Equation for Throttling Calorimeter

$$h_1 = h_2 \implies h_{f1} + x_1 h_{fg1} = h_{g2} + c_{pv}(T_{sup2} - T_{sat2})$$ $$x_1 = \frac{h_{g2} + c_{pv}(T_{sup2} - T_{sat2}) - h_{f1}}{h_{fg1}}$$

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7. Worked Numerical Examples

Example 1: Entropy Generation in Heat Exchanger

Problem: In a counter-flow steam condenser, $20\text{ kg/s}$ of exhaust steam condenses isothermally at $T_{sat} = 50^{\circ}\text{C}$ ($323.15\text{ K}$), releasing $2380\text{ kJ/kg}$ of latent heat. Cooling water enters at $20^{\circ}\text{C}$ ($293.15\text{ K}$) and leaves at $35^{\circ}\text{C}$ ($308.15\text{ K}$). Assume specific heat of water $c_p = 4.184\text{ kJ/kg}\cdot\text{K}$. Calculate the rate of entropy generation ($\dot{S}_{gen}$).

Solution:

1. Rate of heat rejection by steam:

   $$\dot{Q} = \dot{m}_s \times h_{fg} = 20 \times 2380 = 47600\text{ kW}$$

2. Rate of entropy change of steam:

   $$\Delta \dot{S}_{steam} = \frac{-\dot{Q}}{T_{sat}} = \frac{-47600}{323.15} = -147.30\text{ kW/K}$$

3. Cooling water mass flow rate ($\dot{m}_w$):

   $$\dot{m}_w = \frac{\dot{Q}}{c_p (T_{w2} - T_{w1})} = \frac{47600}{4.184 \times (35 - 20)} = \frac{47600}{62.76} = 758.44\text{ kg/s}$$

4. Rate of entropy change of cooling water:

   $$\Delta \dot{S}_{water} = \dot{m}_w c_p \ln\left(\frac{T_{w2}}{T_{w1}}\right) = 758.44 \times 4.184 \times \ln\left(\frac{308.15}{293.15}\right)$$ $$\Delta \dot{S}_{water} = 3173.32 \times \ln(1.05117) = 3173.32 \times 0.04989 = +158.32\text{ kW/K}$$

5. Total entropy generation rate:

   $$\dot{S}_{gen} = \Delta \dot{S}_{steam} + \Delta \dot{S}_{water} = -147.30 + 158.32 = +11.02\text{ kW/K}$$

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Example 2: Stream Flow Exergy & Turbine Work

Problem: Steam enters an adiabatic turbine at $P_1 = 3\text{ MPa}, T_1 = 400^{\circ}\text{C}$ ($h_1 = 3231.7\text{ kJ/kg}, s_1 = 6.9235\text{ kJ/kg}\cdot\text{K}$) and exits at $P_2 = 10\text{ kPa}$ with quality $x_2 = 0.90$ ($h_2 = 2344.7\text{ kJ/kg}, s_2 = 7.4002\text{ kJ/kg}\cdot\text{K}$). The ambient temperature is $T_0 = 25^{\circ}\text{C} = 298.15\text{ K}$. Calculate:

1. Actual turbine work output per kg ($w_{act}$).
2. Maximum reversible work per kg ($w_{rev}$).
3. Exergy destroyed per kg ($i$).

Solution:

1. Actual work output:

   $$w_{act} = h_1 - h_2 = 3231.7 - 2344.7 = 887.0\text{ kJ/kg}$$

2. Reversible work output ($w_{rev} = \psi_1 - \psi_2$):

   $$w_{rev} = (h_1 - h_2) - T_0(s_1 - s_2) = 887.0 - 298.15 \times (6.9235 - 7.4002)$$ $$w_{rev} = 887.0 - 298.15 \times (-0.4767) = 887.0 + 142.13 = 1029.13\text{ kJ/kg}$$

3. Exergy destroyed (Irreversibility):

   $$i = w_{rev} - w_{act} = 1029.13 - 887.0 = 142.13\text{ kJ/kg}$$

   Direct Verification via Gouy-Stodola: $i = T_0 s_{gen} = 298.15 \times (7.4002 - 6.9235) = 142.13\text{ kJ/kg}$.