7.2 Laws of Thermodynamics and Cyclic Processes

The First and Second Laws of Thermodynamics

The First Law of Thermodynamics is the principle of conservation of energy. For a closed system undergoing a change of state, the energy balance is: $\Delta U = Q - W$ where $\Delta U$ is the change in internal energy, $Q$ is the net heat transferred into the system, and $W$ is the net work done by the system on its surroundings. For a steady-state, steady-flow open system, the energy balance is expressed in terms of enthalpy ($h$), velocity ($V$), and elevation ($z$) in the FE Reference Handbook: $\dot{Q} - \dot{W}_s = \sum \dot{m}_{out} \left(h + \frac{V^2}{2} + gz\right)_{out} - \sum \dot{m}_{in} \left(h + \frac{V^2}{2} + gz\right)_{in}$ where $\dot{W}s$ is shaft work. For pumps, compressors, and turbines, kinetic and potential energy changes are typically negligible, simplifying the balance to $\dot{W}s = \dot{m}(h{in} - h{out})$ (assuming adiabatic operation).

The Second Law of Thermodynamics dictates the direction of natural processes. It asserts that the total entropy of an isolated system can never decrease over time. The change in entropy ($S$) for any process is: $\Delta S \ge \int \frac{dQ}{T}$ For a reversible process, this becomes $dS = dQ_{rev}/T$. For an ideal gas with constant specific heats, the change in specific entropy between two states is calculated using: $s_2 - s_1 = C_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right) = C_v \ln\left(\frac{T_2}{T_1}\right) + R \ln\left(\frac{v_2}{v_1}\right)$ For an isentropic process (reversible and adiabatic), $\Delta s = 0$. This leads to the isentropic relations for an ideal gas: $\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}} = \left(\frac{v_1}{v_2}\right)^{k-1}$ where $k = C_p / C_v$ is the specific heat ratio.

Thermodynamic Processes

Let's summarize the work and heat relationships for an ideal gas undergoing various processes:

• Isothermal ($T = \text{constant}$): Since internal energy of an ideal gas is a function of temperature only, $\Delta U = \Delta H = 0$. The First Law yields $Q = W = nRT \ln(V_2/V_1) = nRT \ln(P_1/P_2)$.
• Isobaric ($P = \text{constant}$): Work is $W = P(V_2 - V_1) = nR(T_2 - T_1)$. Heat transfer is $Q = \Delta H = n C_p (T_2 - T_1)$.
• Isochoric ($V = \text{constant}$): Since the volume is constant, boundary work $W = 0$. Heat transfer is $Q = \Delta U = n C_v (T_2 - T_1)$.
• Isentropic ($S = \text{constant}$, $Q = 0$): Work done by the gas is $W = -\Delta U = n C_v (T_1 - T_2) = \frac{P_1 V_1 - P_2 V_2}{k - 1}$.

Power and Refrigeration Cycles

Engineers use thermodynamic cycles to convert heat into work (power cycles) or to transfer heat from a cold region to a hot region (refrigeration and heat pump cycles).

The Carnot Cycle

The Carnot cycle represents the theoretical limit of thermodynamic efficiency. It consists of four reversible steps: isothermal expansion, isentropic expansion, isothermal compression, and isentropic compression.

• Thermal Efficiency ($\eta_{\text{Carnot}}$): For a power cycle operating between hot reservoir $T_H$ and cold reservoir $T_L$: $\eta_{\text{Carnot}} = 1 - \frac{T_L}{T_H}$
• Coefficient of Performance (COP): For refrigeration (R) and heat pump (HP) cycles: $COP_{R,\text{Carnot}} = \frac{T_L}{T_H - T_L}, \quad COP_{HP,\text{Carnot}} = \frac{T_H}{T_H - T_L}$ Note that all temperatures in these equations must be absolute (K or R).

The Rankine Cycle

The Rankine cycle is the model for steam power plants. It consists of:

1. 1 $\to$ 2 Isentropic Compression: Water is pressurized by a pump. Work input is $w_p = h_2 - h_1 \approx v_1(P_2 - P_1)$.
2. 2 $\to$ 3 Isobaric Heat Addition: High-pressure liquid is heated in a boiler to form superheated steam. Heat input is $q_{in} = h_3 - h_2$.
3. 3 $\to$ 4 Isentropic Expansion: Steam expands through a turbine to generate power. Work output is $w_t = h_3 - h_4$.
4. 4 $\to$ 1 Isobaric Heat Rejection: Wet steam is condensed back to liquid in a condenser. Heat rejected is $q_{out} = h_4 - h_1$.

The net thermal efficiency is $\eta_{\text{thermal}} = \frac{w_t - w_p}{q_{in}}$.

The Brayton Cycle

The Brayton cycle is the ideal cycle for gas turbine engines. It operates on a gas (usually air) and consists of isentropic compression (pump/compressor), isobaric heat addition (combustor), isentropic expansion (turbine), and isobaric heat rejection. The efficiency depends on the pressure ratio $r_p = P_2 / P_1$: $\eta_{\text{Brayton}} = 1 - \frac{1}{r_p^{(k-1)/k}}$

Vapor-Compression Refrigeration Cycle

This is the standard cycle for household refrigerators and air conditioners. It consists of:

1. 1 $\to$ 2 Isentropic Compression: Low-pressure vapor is compressed to high pressure. $w_c = h_2 - h_1$.
2. 2 $\to$ 3 Isobaric Heat Rejection: Vapor condenses into high-pressure liquid. $q_{out} = h_2 - h_3$.
3. 3 $\to$ 4 Isenthalpic Expansion: Liquid is throttled through an expansion valve. This is a constant-enthalpy process: $h_3 = h_4$.
4. 4 $\to$ 1 Isobaric Heat Addition: Liquid evaporates at low pressure, absorbing heat. $q_{in} = h_1 - h_4$.

The COP is $COP_R = \frac{h_1 - h_4}{h_2 - h_1}$.

Worked Example: Carnot Refrigerant Analysis

A vapor-compression refrigerator operates on a Carnot cycle between a cold space at $-10^\circ\text{C}$ and a hot reservoir at $35^\circ\text{C}$. Determine the maximum coefficient of performance ($COP_R$) and the minimum power input required to remove $5.0\text{ kW}$ of heat from the cold space.

Step 1: Convert temperatures to absolute scale (Kelvin). $T_L = -10 + 273.15 = 263.15 \text{ K}$ $T_H = 35 + 273.15 = 308.15 \text{ K}$

Step 2: Calculate the Carnot Coefficient of Performance ($COP_{R,\text{Carnot}}$). $COP_{R,\text{Carnot}} = \frac{T_L}{T_H - T_L} = \frac{263.15}{308.15 - 263.15} = \frac{263.15}{45.0} \approx 5.85$

Step 3: Calculate the minimum power input ($\dot{W}_{\text{in}}$). $COP_R = \frac{\dot{Q}_L}{\dot{W}_{\text{in}}} \implies \dot{W}_{\text{in}} = \frac{\dot{Q}_L}{COP_R}$ $\dot{W}_{\text{in}} = \frac{5.0 \text{ kW}}{5.85} \approx 0.855 \text{ kW} = 855 \text{ W}$

This represents the thermodynamic minimum work required. Any actual refrigerator operating between these temperatures will require more power because of irreversibilities (friction, pressure drops, non-isentropic compression).

Cycle Comparison Table