14.1 First & Second Laws of Thermodynamics and Pure Substance Properties

14.1 First & Second Laws of Thermodynamics and Pure Substance Properties

Core FE Exam Principle: Thermodynamics focuses on the principles of energy conservation, heat-work transformations, and directional entropy constraints. On the NCEES FE Other Disciplines exam, success requires applying energy balances to closed systems and steady-flow control volumes, evaluating thermodynamic properties across phase regimes, and enforcing Second Law limits.

The First Law of Thermodynamics: Energy Conservation

The First Law of Thermodynamics states that energy cannot be created or destroyed; it can only change form or be transferred across system boundaries as heat or work.

Closed Systems (Fixed Mass / Control Mass)

For a system with constant mass where no mass crosses the system boundary, the First Law energy balance over a time interval or process is written as:

$$Q - W = \Delta E = \Delta U + \Delta KE + \Delta PE$$

where:

• (Q) = Net heat transfer added to the system ($\text{kJ}$ or $\text{Btu}$); (Q > 0) for heat entering, (Q < 0) for heat leaving.
• (W) = Net work done by the system on its surroundings ($\text{kJ}$ or $\text{Btu}$); (W > 0) for work done by system, (W < 0) for work done on system.
• $\Delta U = m (u_2 - u_1)$ = Change in internal thermal energy.
• $\Delta KE = \frac{1}{2} m (V_2^2 - V_1^2)$ = Change in kinetic energy.
• $\Delta PE = m g (z_2 - z_1)$ = Change in potential energy.

For stationary closed systems where changes in kinetic and potential energy are negligible ($\Delta KE \approx 0$, $\Delta PE \approx 0$):

$$Q - W = \Delta U = m(u_2 - u_1)$$

On a per-unit-mass basis:

$$q - w = \Delta u = u_2 - u_1$$

Enthalpy Definition

Enthalpy ((h)) is a compound thermodynamic property defined specifically to streamline energy balance evaluations involving flowing fluids:

$$h = u + P v \quad [\text{kJ/kg} \text{ or } \text{Btu/lbm}]$$

where (u) is specific internal energy, (P) is absolute pressure, and (v) is specific volume. The term (P v) represents the flow work required to push mass across a control volume surface.

Open Systems: Steady-Flow Energy Equation (SFEE)

Most industrial thermodynamic devices—such as turbines, pumps, compressors, nozzles, diffusers, throttling valves, and heat exchangers—operate as control volumes under steady-flow conditions. Under steady flow, mass entering the control volume equals mass leaving, and energy accumulation within the control volume is zero.

Steady-State Conservation of Mass

$$\sum \dot{m}_{in} = \sum \dot{m}_{out} \implies \dot{m} = \rho_1 A_1 V_1 = \rho_2 A_2 V_2 = \frac{A_1 V_1}{v_1} = \frac{A_2 V_2}{v_2}$$

Steady-Flow Energy Balance (Single Inlet, Single Outlet)

$$\dot{Q} - \dot{W} = \dot{m} \left[ \left( h_2 - h_1 \right) + \frac{V_2^2 - V_1^2}{2} + g\left( z_2 - z_1 \right) \right]$$

Dividing by mass flow rate $\dot{m}$ yields the per-unit-mass Steady-Flow Energy Equation:

$$q - w = \left( h_2 - h_1 \right) + \frac{V_2^2 - V_1^2}{2} + g\left( z_2 - z_1 \right)$$

Pure Substance Phase Behavior and Property Evaluation

A pure substance has a fixed chemical composition throughout. Water ($\text{H}_2\text{O}$), refrigerant R-134a, and nitrogen are common pure substances analyzed on the FE exam.

Phase Regimes on Thermodynamic Diagrams ($P-v$, $T-v$, $P-T$)

1. Compressed/Subcooled Liquid: Water at a temperature below the saturation temperature for a given pressure (or pressure above saturation pressure for a given temperature). Properties are approximated using saturated liquid values at local temperature: $v \approx v_f(T)$, $u \approx u_f(T)$, $h \approx h_f(T) + v_f(P - P_{sat})$.
2. Saturated Liquid (Point (f)): A liquid on the verge of vaporizing.
3. Saturated Liquid-Vapor Mixture: A two-phase mixture of liquid and vapor coexisting at saturation pressure $P_{sat}$ and saturation temperature $T_{sat}$.
4. Saturated Vapor (Point (g)): A vapor on the verge of condensing.
5. Superheated Vapor: A vapor at a temperature above the saturation temperature for a given pressure.

Vapor Quality ((x)) and Mixture Property Equations

In the two-phase saturated mixture region, temperature and pressure are dependent properties (specifying (P) automatically fixes $T_{sat}$). To fix the thermodynamic state, a second independent property—vapor quality ((x))—is required:

$$x = \frac{m_{vapor}}{m_{total}} = \frac{m_g}{m_f + m_g} \quad (0 \le x \le 1)$$

Any specific extensive property (y) (where (y) represents specific volume (v), internal energy (u), enthalpy (h), or entropy (s)) of a two-phase mixture is calculated as:

$$y = (1 - x) y_f + x y_g = y_f + x (y_g - y_f) = y_f + x y_{fg}$$ $$\text{Specific Volume: } v = v_f + x v_{fg} \implies x = \frac{v - v_f}{v_{fg}}$$ $$\text{Specific Enthalpy: } h = h_f + x h_{fg} \implies x = \frac{h - h_f}{h_{fg}}$$ $$\text{Specific Internal Energy: } u = u_f + x u_{fg}$$ $$\text{Specific Entropy: } s = s_f + x s_{fg}$$

where $y_{fg} = y_g - y_f$ represents the property change during vaporization at constant temperature and pressure.

The Second Law of Thermodynamics, Entropy, and Carnot Limits

While the First Law enforces energy conservation, the Second Law of Thermodynamics establishes the natural direction of energy flow and sets theoretical limits on cycle performance.

Classical Statements of the Second Law

• Kelvin-Planck Statement: It is impossible for any device that operates on a thermodynamic cycle to receive heat from a single thermal reservoir and deliver a net amount of work. (No heat engine can have a thermal efficiency of 100%, $\eta_{th} < 100\%$).
• Clausius Statement: It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body. (Refrigerators and heat pumps require net work input, $W_{in} > 0$).

Thermal Engine Efficiency and Carnot Cycle Limit

The thermal efficiency of any heat engine operating between a high-temperature reservoir at $T_H$ and a low-temperature reservoir at $T_L$ is:

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

The Carnot cycle is a reversible cycle that achieves the maximum possible theoretical efficiency between two thermal reservoirs operating at absolute temperatures $T_H$ and $T_L$ (in Kelvin or Rankine):

$$\left(\frac{Q_L}{Q_H}\right)_{rev} = \frac{T_L}{T_H} \implies \eta_{Carnot} = 1 - \frac{T_L}{T_H}$$

Exam Warning: Absolute temperatures (Kelvin $K = {}^\circ\text{C} + 273.15$ or Rankine $R = {}^\circ\text{F} + 459.67$) must be used in all Second Law and Carnot calculations.

Entropy and Clausius Inequality

Entropy ((S)) is a thermodynamic measure of molecular disorder or energy dispersion. The Clausius inequality states:

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

For a process between two states, the entropy change is:

$$\Delta S = S_2 - S_1 = \int_1^2 \frac{\delta Q_{rev}}{T} + S_{gen}$$

where $S_{gen}$ is the entropy generation:

• $S_{gen} = 0$: Reversible process (ideal, frictionless).
• $S_{gen} > 0$: Irreversible real process.
• $S_{gen} < 0$: Impossible process.

Comprehensive Worked Engineering Example

Problem Statement

Superheated steam enters an adiabatic steam turbine operating at steady state at $P_1 = 3.0 \text{ MPa}$ and $T_1 = 400^\circ\text{C}$. The steam expands through the turbine and exits at a saturated mixture pressure of $P_2 = 50 \text{ kPa}$ with a vapor quality of $x_2 = 0.92$. The mass flow rate of steam is $\dot{m} = 8.5 \text{ kg/s}$. Kinetic and potential energy changes across the turbine are negligible.

Steam Table Properties:

• At $P_1 = 3.0 \text{ MPa}$, $T_1 = 400^\circ\text{C}$: $h_1 = 3231.7 \text{ kJ/kg}$.
• At $P_2 = 50 \text{ kPa}$ ($T_{sat} = 81.32^\circ\text{C}$): $h_f = 340.54 \text{ kJ/kg}$, $h_{fg} = 2304.7 \text{ kJ/kg}$.

Calculate:

1. The exit specific enthalpy of the steam $h_2$.
2. The power output produced by the turbine $\dot{W}_{out}$ in megawatts (MW).
3. The Carnot thermal efficiency of an ideal heat engine operating between the turbine inlet temperature $T_1$ and condenser exit temperature $T_2 = T_{sat}(50\text{ kPa})$.

Step-by-Step Solution

Step 1: Evaluate Exit Specific Enthalpy $h_2$

Using the vapor quality mixture relation at state 2:

$$h_2 = h_f + x_2 h_{fg}$$ $$h_2 = 340.54 \text{ kJ/kg} + 0.92 \times (2304.7 \text{ kJ/kg})$$ $$h_2 = 340.54 + 2120.32 = 2460.86 \text{ kJ/kg}$$

Step 2: Compute Turbine Power Output $\dot{W}_{out}$

Apply the Steady-Flow Energy Balance (SFEE) to the adiabatic control volume ($\dot{Q} = 0$, $\Delta KE = 0$, $\Delta PE = 0$):

$$\dot{Q} - \dot{W}_{out} = \dot{m}(h_2 - h_1)$$ $$-\dot{W}_{out} = \dot{m}(h_2 - h_1) \implies \dot{W}_{out} = \dot{m}(h_1 - h_2)$$ $$\dot{W}_{out} = 8.5 \text{ kg/s} \times (3231.7 \text{ kJ/kg} - 2460.86 \text{ kJ/kg})$$ $$\dot{W}_{out} = 8.5 \times 770.84 \text{ kJ/kg} = 6552.14 \text{ kW} = 6.552 \text{ MW}$$

Step 3: Compute Carnot Efficiency Between Temperatures $T_1$ and $T_2$

Convert temperatures from Celsius to Kelvin:

$$T_H = T_1 = 400 + 273.15 = 673.15 \text{ K}$$ $$T_L = T_2 = 81.32 + 273.15 = 354.47 \text{ K}$$ $$\eta_{Carnot} = 1 - \frac{T_L}{T_H} = 1 - \frac{354.47}{673.15} = 1 - 0.5266 = 0.4734 \text{ or } 47.34\%$$

Final Answer: $h_2 = 2460.9 \text{ kJ/kg}$, power output $\dot{W}_{out} = 6.55 \text{ MW}$, and Carnot efficiency $\eta_{Carnot} = 47.3\%$.