12.1 Fundamental Concepts, First & Second Laws of Thermodynamics

11.1 Fundamental Concepts, First & Second Laws of Thermodynamics

Engineering thermodynamics is the foundational science governing energy transformations, heat-work interactions, and thermal power generation. In Coal India Limited (CIL) operations, thermodynamic principles underpin the design, testing, and operation of heavy mining machinery, captive thermal power stations, reciprocating air compressors for underground ventilation, and turbo-machinery.

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1. Thermodynamic Systems, Surroundings & Boundaries

A thermodynamic system is defined as a quantity of matter or a region in space chosen for thermodynamic analysis. Everything external to the system is termed the surroundings. The real or imaginary surface separating the system from its surroundings is the boundary (which may be fixed or moving, real or imaginary).

+-------------------------------------------------------------------------+
|                         SYSTEM CLASSIFICATION                           |
+-------------------+-----------------+-----------------+-----------------+
| System Type       | Mass Transfer   | Energy Transfer | Typical CIL     |
|                   | Across Boundary | Across Boundary | Application     |
+-------------------+-----------------+-----------------+-----------------+
| Open System       | Yes             | Yes             | Steam turbine,  |
| (Control Volume)  |                 | (Heat & Work)   | boiler, pump,   |
|                   |                 |                 | compressor      |
+-------------------+-----------------+-----------------+-----------------+
| Closed System     | No              | Yes             | Piston-cylinder |
| (Control Mass)    |                 | (Heat & Work)   | gas expansion,  |
|                   |                 |                 | sealed vessel   |
+-------------------+-----------------+-----------------+-----------------+
| Isolated System   | No              | No              | Perfectly       |
|                   |                 | (Neither Heat   | insulated rigid |
|                   |                 | nor Work)       | tank, Universe  |
+-------------------+-----------------+-----------------+-----------------+

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2. Properties, States, Processes & Equilibrium

Intensive vs. Extensive Properties

• Intensive Properties: Independent of the mass or size of the system. Examples: Pressure ($P$), Temperature ($T$), Density ($\rho$), Specific volume ($v = V/m$), Specific internal energy ($u = U/m$), Specific enthalpy ($h = H/m$).
• Extensive Properties: Directly proportional to the mass or extent of the system. Examples: Total Volume ($V$), Total Mass ($m$), Total Internal Energy ($U$), Total Enthalpy ($H$), Total Entropy ($S$).
• Rule of Thumb: The ratio of two extensive properties yields an intensive property (e.g., $m/V = \rho$, $H/m = h$).

Thermodynamic Equilibrium

A system is in thermodynamic equilibrium if and only if it simultaneously satisfies four equilibrium conditions:

1. Thermal Equilibrium: Uniform temperature throughout the system ($dT/dx = 0$), with no temperature gradient relative to surroundings.
2. Mechanical Equilibrium: No unbalanced forces within the system or at the boundary; pressure is uniform throughout (neglecting gravity hydrostatic head in gases).
3. Chemical Equilibrium: No chemical reactions or spontaneous mass diffusion; chemical potential remains uniform across all phases.
4. Phase Equilibrium: The mass of each phase reaches dynamic steady state without net phase transformation.

Quasi-Static (Quasi-Equilibrium) Process

A process executed so slowly that the system departs only infinitesimally from thermodynamic equilibrium at every instant. The locus of these infinite succession of equilibrium states can be plotted as a continuous line on property diagrams ($P-v$, $T-s$). In the absence of dissipative effects (friction, unrestrained expansion, electrical resistance), a quasi-static process is internally reversible.

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3. Zeroth Law of Thermodynamics & Temperature Measurement

Zeroth Law Statement: If two bodies ($A$ and $B$) are each in thermal equilibrium with a third body ($C$), they are in thermal equilibrium with each other.

The Zeroth Law forms the scientific basis of temperature measurement and thermometry, establishing temperature as an intensive state property.

$$\text{If } T_A = T_C \quad \text{and} \quad T_B = T_C \implies T_A = T_B$$

Thermometric Properties & Scales

Thermometers utilize a thermometric property ($X$) that varies monotonically with temperature ($T = aX + b$):

Reference Point: Since 1954, the international thermodynamic temperature scale relies on a single fixed point: the Triple Point of Water, assigned an exact value of $273.16\text{ K} = 0.01^{\circ}\text{C}$ at $P = 0.6117\text{ kPa}$ ($4.587\text{ mm Hg}$). Absolute zero corresponds to $0\text{ K} = -273.15^{\circ}\text{C}$.

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4. First Law of Thermodynamics for Closed Systems

The First Law of Thermodynamics expresses the principle of conservation of energy:

For a Closed System Undergoing a Cyclic Process

The cyclic integral of heat transfer is equal to the cyclic integral of work transfer:

$$\oint \delta Q = \oint \delta W \qquad \text{(in consistent energy units, Joules)}$$

For a Closed System Undergoing a Non-Cyclic Process

$$\delta Q = dE + \delta W = dU + d(KE) + d(PE) + \delta W$$

For a stationary closed system ($d(KE) = 0, d(PE) = 0$):

$$\delta Q = dU + \delta W$$

Integrating from state 1 to state 2:

$$Q_{1-2} = (U_2 - U_1) + W_{1-2} = \Delta U + W_{1-2}$$

Nature of Work, Heat & Internal Energy

• Work ($\delta W$) and Heat ($\delta Q$): Path functions, boundary phenomena, transient quantities, inexact differentials (represented by $\delta W$ and $\delta Q$).
• Internal Energy ($U$): Point function, state property, exact differential ($dU$). For any cyclic process, $\oint dU = 0$.

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5. Quasi-Static Boundary Work ($P\,dV$) for Ideal Gas Processes

For a frictionless, quasi-static process in a closed piston-cylinder system, boundary work is:

$$W_{1-2} = \int_{V_1}^{V_2} P \, dV$$

               PRESSURE-VOLUME (P-v) PROCESS COMPARISON
    P ^
      | 1
      | *-------- Isobaric (n = 0)
      | |\  `.
      | | \   `.  Isothermal (n = 1)
      | |  \    `. 
      | |   \     `* Polytropic (1 < n < gamma)
      | |    \      `.
      | |     \       `* Reversible Adiabatic / Isentropic (n = gamma)
      | |      \        `.
      | |       \         `.
      | +--------+----------+------> V
     (Isochoric: n = infinity, vertical line)

Comprehensive Process Comparison Table (Ideal Gas: $PV = mRT$)

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6. Steady Flow Energy Equation (SFEE) for Open Systems

For a steady-state, steady-flow control volume with single inlet (1) and single exit (2):

$$\dot{m}\left[h_1 + \frac{V_1^2}{2000} + \frac{g z_1}{1000}\right] + \dot{Q} = \dot{m}\left[h_2 + \frac{V_2^2}{2000} + \frac{g z_2}{1000}\right] + \dot{W}_{cv}$$

Where:

• $h$ = specific enthalpy ($\text{kJ/kg}$)
• $V$ = fluid velocity ($\text{m/s}$), hence kinetic energy term is in $\text{kJ/kg}$
• $z$ = elevation head ($\text{m}$), $g = 9.81\text{ m/s}^2$, potential energy term is in $\text{kJ/kg}$
• $\dot{Q}$ = heat transfer rate ($\text{kW}$), $\dot{W}_{cv}$ = shaft work rate ($\text{kW}$), $\dot{m}$ = mass flow rate ($\text{kg/s}$)

SFEE Applied to Essential Thermal Engineering Devices

1. Steam / Gas Turbine: Well insulated ($\dot{Q} \approx 0$), $\Delta KE \approx 0, \Delta PE \approx 0$:

   $$W_T = \dot{m}(h_1 - h_2)$$

2. Centrifugal Compressor / Water Pump: Work input required, adiabatic ($\dot{Q} \approx 0$):

   $$W_C = \dot{m}(h_2 - h_1)$$

3. Nozzle (Expander): Accelerates fluid at the expense of pressure drop; $\dot{W} = 0, \dot{Q} = 0, \Delta PE = 0$:

   $$V_2 = \sqrt{V_1^2 + 2000(h_1 - h_2)} \approx \sqrt{2000(h_1 - h_2)} \quad \text{(when } V_1 \ll V_2\text{)}$$

4. Diffuser: Decelerates high-velocity fluid to increase static pressure; $\dot{W} = 0, \dot{Q} = 0$:

   $$h_2 - h_1 = \frac{V_1^2 - V_2^2}{2000}$$

5. Throttling Device (Expansion Valve, Porous Plug, Capillary Tube): Fluid flows through a narrow constriction; $\dot{W} = 0, \dot{Q} = 0, \Delta KE \approx 0, \Delta PE \approx 0$:

   $$h_1 = h_2 \qquad \text{(Strictly Isenthalpic Process)}$$
   • Joule-Thomson Coefficient: $\mu_J = \left(\frac{\partial T}{\partial P}\right)_h$
   • If $\mu_J > 0$, throttling causes a temperature drop (cooling, below inversion temperature).
   • If $\mu_J < 0$, throttling causes a temperature rise (heating).
   • For an ideal gas, enthalpy is solely a function of temperature ($h = f(T)$), so $h_1 = h_2 \implies T_1 = T_2$ and $\mu_J = 0$.

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7. Second Law of Thermodynamics & Thermal Reservoirs

While the First Law establishes energy conservation, it does not specify the direction of spontaneous processes. The Second Law provides directional criteria and defines absolute thermodynamic efficiency limits.

Classical Statements of the Second Law

+-------------------------------------------------------------------------+
|                   SECOND LAW CLASSICAL STATEMENTS                       |
+------------------------------------+------------------------------------+
| Kelvin-Planck Statement            | Clausius Statement                 |
| (Applies to Heat Engines)          | (Applies to Refrigerators / Pumps) |
+------------------------------------+------------------------------------+
| It is impossible for any device    | It is impossible to construct a    |
| that operates on a cycle to        | device that operates in a cycle    |
| receive heat from a single thermal | and produces no effect other than  |
| reservoir and produce a net amount | the transfer of heat from a cooler |
| of work.                           | body to a hotter body.             |
|                                    |                                    |
| Consequence: No heat engine can    | Consequence: Work input is always  |
| have 100% thermal efficiency.      | mandatory for refrigeration.       |
+------------------------------------+------------------------------------+

Equivalence: A violation of the Kelvin-Planck statement leads directly to a violation of the Clausius statement, and vice versa.

Perpetual Motion Machines (PMM)

• PMM-1 (First Kind): A hypothetical machine that produces work continuously without absorbing any energy from the surroundings. Violates the First Law of Thermodynamics.
• PMM-2 (Second Kind): A hypothetical machine that converts all absorbed heat completely into equivalent work while operating in a cycle with a single thermal reservoir (100% thermal efficiency). Violates the Second Law of Thermodynamics.
• PMM-3 (Third Kind): A device that operates with zero friction/dissipation indefinitely. Violates practical mechanical realities.

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8. Carnot Cycle, Carnot Theorem & Reversible Heat Engine / Heat Pump COP

The Carnot cycle is a totally reversible cycle operating between a high-temperature source ($T_H$) and low-temperature sink ($T_L$). It consists of 4 reversible processes:

1. Process 1-2: Reversible isothermal heat addition at $T_H$ ($Q_H = T_H \Delta S$).
2. Process 2-3: Reversible adiabatic (isentropic) expansion from $T_H$ to $T_L$.
3. Process 3-4: Reversible isothermal heat rejection at $T_L$ ($Q_L = T_L \Delta S$).
4. Process 4-1: Reversible adiabatic (isentropic) compression from $T_L$ to $T_H$.

             CARNOT ENGINE (HE)           CARNOT REFRIGERATOR / HEAT PUMP
                Source [ T_H ]                     Source [ T_H ]
                      |                                  ^
                      | Q_H                              | Q_H
                      v                                  |
                +-----------+                      +-----------+
                |   (HE)    | ---> W_net           |   (REF)   | <--- W_in
                +-----------+                      +-----------+
                      |                                  ^
                      | Q_L                              | Q_L
                      v                                  |
                 Sink [ T_L ]                       Sink [ T_L ]

Performance Formulations

1. Thermal Efficiency of Carnot Heat Engine:

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

2. Coefficient of Performance of Reversible Refrigerator:

   $$\text{COP}_{ref} = \frac{\text{Desired Effect (Cooling)}}{\text{Work Input}} = \frac{Q_L}{W_{in}} = \frac{Q_L}{Q_H - Q_L} = \frac{T_L}{T_H - T_L}$$

3. Coefficient of Performance of Reversible Heat Pump:

   $$\text{COP}_{HP} = \frac{\text{Desired Effect (Heating)}}{\text{Work Input}} = \frac{Q_H}{W_{in}} = \frac{Q_H}{Q_H - Q_L} = \frac{T_H}{T_H - T_L}$$

4. Fundamental COP Relationship:

   $$\text{COP}_{HP} = \text{COP}_{ref} + 1 = \frac{1}{\eta_{th, Carnot}}$$

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9. Worked Numerical Examples

Example 1: Polytropic Compression Work and Heat Transfer

Problem: A mass of $0.5\text{ kg}$ of air ($R = 0.287\text{ kJ/kg}\cdot\text{K}$, $\gamma = 1.4$) is compressed quasi-statically in a cylinder from $P_1 = 100\text{ kPa}, T_1 = 300\text{ K}$ to $P_2 = 800\text{ kPa}$ following the polytropic law $P V^{1.3} = C$. Calculate:

1. Final temperature $T_2$.
2. Boundary work transfer ($W_{1-2}$).
3. Heat transfer ($Q_{1-2}$).

Solution:

1. For polytropic process with index $n = 1.3$:

   $$\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{n-1}{n}} = \left(\frac{800}{100}\right)^{\frac{1.3-1}{1.3}} = (8)^{\frac{0.3}{1.3}} = 8^{0.23077} \approx 1.616$$ $$T_2 = 300 \times 1.616 = 484.8\text{ K}$$

2. Work transfer ($W_{1-2}$):

   $$W_{1-2} = \frac{m R (T_1 - T_2)}{n - 1} = \frac{0.5 \times 0.287 \times (300 - 484.8)}{1.3 - 1} = \frac{0.1435 \times (-184.8)}{0.3} = -88.4\text{ kJ}$$

   (Negative sign indicates work is done on the gas during compression).

3. Heat transfer ($Q_{1-2}$):

   $$Q_{1-2} = \left(\frac{\gamma - n}{\gamma - 1}\right) W_{1-2} = \left(\frac{1.4 - 1.3}{1.4 - 1.0}\right) (-88.4) = \left(\frac{0.1}{0.4}\right) (-88.4) = -22.1\text{ kJ}$$

   (Negative sign indicates heat is rejected from the system to surroundings).

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Example 2: SFEE for a Steam Nozzle

Problem: Dry saturated steam enters an insulated convergent-divergent nozzle at $P_1 = 2\text{ MPa}$ ($h_1 = 2799.5\text{ kJ/kg}$) with an inlet velocity of $50\text{ m/s}$. The exit enthalpy is measured as $h_2 = 2520.0\text{ kJ/kg}$. Determine the exit velocity $V_2$.

Solution: Applying SFEE with $\dot{Q} = 0, \dot{W}_{cv} = 0, \Delta z = 0$:

$$h_1 + \frac{V_1^2}{2000} = h_2 + \frac{V_2^2}{2000}$$ $$\frac{V_2^2}{2000} = (h_1 - h_2) + \frac{V_1^2}{2000} = (2799.5 - 2520.0) + \frac{50^2}{2000} = 279.5 + 1.25 = 280.75\text{ kJ/kg}$$ $$V_2 = \sqrt{2000 \times 280.75} = \sqrt{561500} \approx 749.33\text{ m/s}$$

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Example 3: Carnot Heat Engine Driving a Refrigerator

Problem: A reversible Carnot heat engine operates between a source at $T_1 = 800\text{ K}$ and a sink at $T_2 = 400\text{ K}$. All the power output of this engine is used to drive a reversible refrigerator operating between $T_3 = 260\text{ K}$ and $T_2 = 400\text{ K}$. If the heat supplied to the engine is $Q_1 = 2000\text{ kJ}$, find the heat extracted by the refrigerator ($Q_3$) from the refrigerated space.

Solution:

1. Engine thermal efficiency:

   $$\eta_E = 1 - \frac{T_2}{T_1} = 1 - \frac{400}{800} = 0.50$$ $$W_{net} = \eta_E \times Q_1 = 0.50 \times 2000 = 1000\text{ kJ}$$

2. Refrigerator COP:

   $$\text{COP}_{ref} = \frac{T_3}{T_2 - T_3} = \frac{260}{400 - 260} = \frac{260}{140} = 1.857$$

3. Cooling effect ($Q_3$):

   $$Q_3 = \text{COP}_{ref} \times W_{in} = 1.857 \times 1000 = 1857.14\text{ kJ}$$