Thermodynamics: Heat Engines, Heat Pumps, and Entropy

Introduction to Thermodynamic Cycles and Heat Engines

A thermodynamic cycle is defined as a loop on a pressure-volume ($P$-$V$) diagram. In these cycles, a system (typically an ideal gas) is subjected to various processes but eventually returns to its initial thermodynamic state, characterized by the same variables: number of moles ($n$), volume ($V$), pressure ($P$), and temperature ($T$).

• Heat Engines: Devices that convert thermal energy into mechanical work through a cyclical process. Examples include:

  • Diesel Engine: A high-efficiency internal combustion engine.

  • Otto Cycle: An ideal cycle that mimics the performance of gasoline engines.

• Heat Pumps: Devices that move thermal energy in an unnatural direction (from cold to hot) using work as an input.

The Physics of Heat Engines

Heat engines are governed by the first and second laws of thermodynamics. While energy is conserved, not all heat energy can be converted into usable work.

• Work per Cycle: On a $P$-$V$ diagram, the work done per cycle is given by the area enclosed by the loop.

  • $W_{\text{tote}} = \pm \text{Area}$

  • A counterclockwise (CCW) cycle results in a positive total work (work done on the gas).

  • A clockwise (CW) cycle results in a negative total work, characteristics of a heat engine.

• Usable Output Work ($W_{\text{eng}}$): This represents the mechanical work extracted from the engine per cycle, such as moving a car or lifting a crane. It is mathematically the negative of the work done on the gas during a clockwise cycle:

  • $W_{\text{eng}} = -W_{\text{tote}}$

  • $W_{\text{eng}}$ is a positive quantity.

• Expansion vs. Compression Work: In a heat engine, the expansion work (which is negative) is greater in magnitude than the compression work (which is positive) per cycle, resulting in a net loss of energy by the gas to the environment as work.

Energy Reservoirs and Schematics

• Temperature (Energy) Reservoir: A system so large that adding or removing heat does not significantly change its temperature (e.g., the ocean).

  • $T_{\text{initial}} \approx T_{\text{final}}$

• Heat Engine Schematic:

  • Hot Reservoir ($T_H$): Provides the fuel heat ($Q_H$). For the gas, this is an energy gain (Q_H > 0).

  • Cold Reservoir ($T_C$): Receives waste heat ($Q_C$). For the gas, this is an energy loss (Q_C < 0).

  • Work Out ($W_{\text{eng}}$): The usable energy produced per cycle.

The First Law for Heat Engines

For an ideal gas in a complete cycle, the change in thermal energy ($\Delta E_{\text{thermal}}$) is zero because the temperature returns to its starting point ($\Delta T = 0$).

• First Law Equation:

  • $0 = Q_{\text{total}} + W_{\text{tote}}$

  • $0 = (Q_H + Q_C) - W_{\text{eng}}$

• Energy Balance:

  • $Q_H - |Q_C| = W_{\text{eng}}$

  • This implies that energy entering the system ($Q_H$) must equal energy leaving the system ($|Q_C| + W_{\text{eng}}$).

Numerical Example: Heat Engine Waste

Consider an engine where:

• Fuel heat ($Q_H$) = $10\,\text{Joules}$

• Usable work ($W_{\text{eng}}$) = $7\,\text{Joules}$

• To find waste heat ($Q_C$):

  • $|Q_C| = Q_H - W_{\text{eng}}$

  • $|Q_C| = 10\,\text{J} - 7\,\text{J} = 3\,\text{J}$

  • Since it is a loss, $Q_C = -3\,\text{J}$.

Thermal Efficiency ($e$)

Efficiency is the ratio of what we want to the price we pay.

• Formula:

  • $e = \frac{W_{\text{eng}}}{Q_H}$

  • $e = \frac{Q_H - |Q_C|}{Q_H} = 1 - \frac{|Q_C|}{Q_H}$

• Constraints: Efficiency is always a fraction (e < 1) and is usually expressed as a percentage. It can never be $100\%$ ($1.0$) because the Second Law of Thermodynamics dictates that waste heat ($Q_C$) is never zero.

• Typical Values:

  • Automobile Engines: $20\times - 25\%$ ($e \approx 0.20 - 0.25$).

  • Diesel Engines: $30\times - 45\%$ ($e \approx 0.30 - 0.45$).

The Second Law of Thermodynamics

There are two primary ways to state the Second Law of Thermodynamics conceptually in the context of cycles:

1. Kelvin-Planck Statement: It is impossible to use heat energy from a temperature reservoir and produce an equal amount of work in a cycle. This means conversion cannot be $100\%$ efficient; waste heat must exist.

2. Clausius Statement: It is impossible to move heat from a cold temperature reservoir to a hot temperature reservoir in a cycle without the input of energy through work. This applies to heat pumps.

Heat Pumps: Coolers and Heaters

A heat pump uses a reversed (counterclockwise) cycle to pump heat from a cold reservoir to a hot reservoir. This is an "unnatural" flow facilitated by work input.

• Work Motor ($W_{\text{motor}}$): Energy put into the gas by a compressor. It is positive.

• Signs:

  • Q_C > 0 (Heat gain from the cold reservoir).

  • Q_H < 0 (Heat loss to the hot reservoir).

• First Law for Heat Pumps:

  • $|Q_H| = W_{\text{motor}} + Q_C$

• Coefficient of Performance ($\kappa$):

  • $\kappa$ for Cooling: $\\kappa_C = \frac{Q_C}{W_{\text{motor}}}$ (What you want is to remove heat from the cold interior).

  • $\kappa$ for Heating: $\\kappa_H = \frac{|Q_H|}{W_{\text{motor}}}$ (What you want is heat delivered to the hot interior).

Reversibility and Irreversibility

• Reversible Process: A process where no net heat is transferred to the surroundings upon completing a cycle. These are idealizations and do not exist in real life.

• Irreversible Process: All real-life processes are irreversible and leave the surroundings hotter.

• Example: Adiabatic Free Expansion:

  • A gas expands into a vacuum through a membrane.

  • $Q = 0, W = 0, \Delta T = 0$.

  • To return the gas to its initial state, one must compress it (increasing temperature) and then remove heat. Because heat must be transferred to the surroundings to complete the cycle, the process is irreversible.

The Carnot Cycle

The Carnot Cycle is a mathematical idealization consisting of four reversible processes and represents the most efficient engine possible between two temperature reservoirs.

• Structure on a $P$-$V$ Diagram:

  1. Isothermal Expansion (at $T_H$): Fuel heat ($Q_H$) enters the system.

  2. Adiabatic Expansion: Temperature drops from $T_H$ to $T_C$; no heat exchange.

  3. Isothermal Compression (at $T_C$): Waste heat ($Q_C$) leaves the system.

  4. Adiabatic Compression: Temperature rises from $T_C$ back to $T_H$; no heat exchange.

• Carnot Theorem: No engine run between two reservoirs can be more efficient than a Carnot engine. Real engines are always less efficient than the Carnot value because real processes are irreversible.

• Carnot Formulas (Using Absolute Temperatures in Kelvin):

  • $e_{\text{Carnot}} = 1 - \frac{T_C}{T_H}$

  • $\kappa_{\text{Carnot, Heater}} = \frac{T_H}{T_H - T_C}$

  • $\kappa_{\text{Carnot, Cooler}} = \frac{T_C}{T_H - T_C}$

Detailed Example: Monatomic Ideal Gas in a Carnot Cycle

Given:

• $n = 2.34\,\text{moles}$

• Monatomic gas: $C_V = \frac{3}{2}R$

• $T_H = 720\,\text{K}$

• State A: $P_A = 1400\,\text{kPa}, V_A = 10\,\text{L}$

• State D: $V_D = 15\,\text{L}$

• State C: $V_C = 24\,\text{L}$

Variable Calculations:

• Using Adiabatic relationship ($P_A V_A^{\gamma} = P_D V_D^{\gamma}$) where $\gamma = \frac{5}{3}$:

  • $P_D = 1400 \times \left(\frac{10}{15}\right)^{5/3} = 712\,\text{kPa}$

• Using temp-vol relationship ($T_A V_A^{\gamma-1} = T_D V_D^{\gamma-1}$):

  • $T_D = 720 \times \left(\frac{10}{15}\right)^{2/3} = 549\,\text{K}$ (This is $T_C$).

• Isothermal calculation for $P_C$:

  • $P_C = 712 \times \left(\frac{15}{24}\right) = 445\,\text{kPa}$.

• Finding $V_B$ for State B (Adiabatic with C):

  • $V_B = 24 \times \left(\frac{549}{720}\right)^{3/2} = 16\,\text{L}$.

• Finding $P_B$:

  • $P_B = 445 \times \left(\frac{24}{16}\right)^{5/3} = 875\,\text{kPa}$.

Work and Heat Calculations:

• Isothermal Expansion (A to B):

  • $W_{AB} = nRT_H \ln\left(\frac{V_A}{V_B}\right) = 2.34 \times 8.31 \times 720 \times \ln\left(\frac{10}{16}\right) = -6583\,\text{J}$

  • $Q_H = -W_{AB} = 6583\,\text{J}$

• Isothermal Compression (C to D):

  • $W_{CD} = nRT_C \ln\left(\frac{V_C}{V_D}\right) = 2.34 \times 8.31 \times 549 \times \ln\left(\frac{24}{15}\right) = 5020\,\text{J}$

  • $Q_C = -W_{CD} = -5020\,\text{J}$

• Adiabatic Steps: The total work from the two adiabatic steps ($W_{BC} + W_{DA}$) is exactly zero because $W_{BC} = -4990\,\text{J}$ and $W_{DA} = 4990\,\text{J}$.

• Final Efficiency:

  • $W_{\text{eng}} = |(-6583 + 5020)| = 1560\,\text{J}$

  • $e = \frac{1560}{6583} = 0.237 = 23.7\%$

  • Using Temperature: $e = 1 - \frac{549}{720} = 23.7\%$

Entropy ($S$)

Entropy is a thermodynamic quantity representing the state of disorder. Like potential energy, the change in entropy ($\Delta S$) is more significant than absolute values.

• Definition:

  • $\Delta S = \int_i^f \frac{dQ}{T}$

• Units: Joules per Kelvin ($\text{J/K}$).

• Calculations:

  • For isothermal processes (where $T$ is constant):

    • $\Delta S = \frac{Q}{T}$

  • In an isothermal expansion, $Q$ is positive, so $\Delta S$ is positive.

• Entropy and the Second Law:

  • For all real-life (irreversible) processes, the entropy of the universe must increase:

    • \Delta S_{\text{universe}} > 0

  • Reversible processes (idealized) result in $\Delta S_{\text{universe}} = 0$.

  • Local entropy can decrease (e.g., human beings creating ordered structures), but this is always accompanied by a larger increase in entropy elsewhere in the universe.