Introduction to the Second Law of Thermodynamics (on board)

Heat engine Thermal efficiency A cyclic heat engine

: a simple steam power plant Examples of heat engines

Refrigerators and heat pumps Coefficient of performance Heat pump is a device, which operating in a cycle, maintains a body at a temperature higher than the surroundings. Refrigerator is a device, which operating in a cycle, maintains a body at a temperature lower than the temperarature of its surroundings.

Example of a refrigerator A vapor compression refrigeration system

Introduction to the Second Law of Thermodynamics

Introducing the second law A process should satisfy the first law in order to occur. However, satisfying first law alone does not guarantee that the process will take place.

Examples of impossible processes that do not violate first law One more: A cup of coffee does not get hotter in a cooler room by absorbing heat from environment. Transferring heat to a resistance will not generate electrical energy Transferring heat to this paddle-wheel device will not cause the paddle-wheel to rotate and raise the mass through the pulley. Heat

Work was completely converted into heat in Joule’s experiment Q=W

Example of an heat engine: a simple steam power plant Some definitions (on board/discussion) Thermal energy reservoirs (source and sink) Heat engines Efficiency of a heat engine

Introducing the second law A process should satisfy the first law in order to occur. However, satisfying first law alone does not guarantee that the process will take place.

Examples of impossible processes that do not violate first law One more: A cup of coffee does not get hotter in a cooler room by absorbing heat from environment. Transferring heat to a resistance will not generate electrical energy Transferring heat to this paddle-wheel device will not cause the paddle-wheel to rotate and raise the mass through the pulley. Heat

Statements of the second law Two equivalent ways the second law can be stated are due to: Kelvin and Planck (“The Kelvin-Planck statement”) Clausius (“The Clausius statement”). The direction in which processes actually occur can be judged by taking the help of these two statements. Either of this statements can be used to detect impossible inventions and impossible processes.

Outline of our course of progression on second law DEDUCTION BASED ON either of the Kelvin-Planck and Clausiusstatements will give us the ability to C) state second law as an inequality involving engines/refrigerators in contact with more than one reservoirs B) Assign temperature values from a non-empirical perspective and find the most efficient refrigerators/engines Following step 1, the property entropy will be defined to allow another more mathematical statement of the second law and another way to judge the actual direction of processes.

The Kelvin Planck Statement of the Second Law It is impossible for any device that operates in a cycle to receive heat from a single reservoir and produce a net amount of work. Equivalently: “no heat engine can have a thermal efficiency of 100%”. “For a power plant to operate, the working fluid must exchange heat with the environment as well as the furnace.”

A heat engine that violates the Kelvin Planck statement You need more than one reservoir to convert heat to work by a cyclic engine (a cold reservoir, is needed to dump the heat which could not be converted to work).

Stating the Kelvin Planck statement analytically The Kelvin-Planck statement do not forbid cyclic devices operating with a single reservoir, but insists that such a cyclic device should receive work. So, according to the Kelvin Planck statement

Clausius statement of the second law of thermodynamics 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. A refrigerator is not a self-acting device: energy (electrical work to the motor driving the compressor) has to be provided from the surroundings to run a refrigerator.

A refrigerator that violates the Clausius statement

Equivalence of Kelvin-Planck and Clausius statements Violation of Clausius statementViolation of Kelvin-Planck statement Violation of Kelvin-Planck statementViolation of Clausius statement

Violation of KP  Violation of Clausius TH Q1 Q2 HE!+R R Q1 Q1 TC The net heat exchange of the cyclic device (HE+R) with the hot reservoir=Q2-Q TH Q =Q1+W =Q1+Q W W HE! TC

Violation of Clausius  Violation of KP HE+R!+TH W Q Q W R! HE Q-Q1 Q Q1 TC TH TC KP statement requires the device in contact with the single reservoir (here at Tc) to be a cyclic device. Because nothing happens to the TH reservoir (Qin=Qout=Q). the combined device (HE+R!+TH)is a cyclic device.

Violation of Clausius  Violation of KP (Alternative) TH Q Q Q and Q can be fed directly to H from R W W HE+R! HE R! Q-Q1 Q Q1 TC TC TH can be eliminated

Equivalence of Kelvin-Planck and Clausius statements Violation of Clausius statementViolation of Kelvin-Planck statement Violation of Kelvin-Planck statementViolation of Clausius statement

Perpetual motion machines (PMM) Any device that violates the first or the second law of thermodynamics is called a perpetual motion machine. Violates the First law: “perpetual machine of the first kind”: produces more energy than supplied. Violates the Second law: “perpetual motion machine of the second kind”: Allows the efficiency of cyclic heat engines to equal 100%.

Example of a PMM1 OK Not OK! Produces net energy output without energy input.

Identifying PMM2 by Kelvin-Planck/Clausius statement Not OK! Violates KP OK A PMM2 according to Kelvin-Planck statement is a device that: Operates in a cycle. Accepts heat from a single reservoir (surroundings). Develops a net work output. Example: A power plant with no condenser

Identifying PMM2 by Clausius statement tH tC A PMM2 according to Clausius statement is a device whose operation has the sole effect of transfer of heat from a low termperature to high temperature body.

How to make the most efficient heat engine Second law: no heat engine can have an efficiency of 100%. So, what is the maximum efficiency? It turns out (shown later) that maximum efficiency is realized when a heat engine runs on a cycle consisting of certain “idealized processes”.

Reversible process vacuum Reversible processes can be reversed leaving no trace on the surroundings. If the original process and its reverse is combined into a cycle, after the cycle is executed, both the system and surroundings will return to their original state. If the surroundings can be considered as a single thermal energy reservoir, no net heat and work exchange between the system and surroundings occurs during this cycle. Examples: Pendulum swinging in vacuum (can be studied in mechanical co-orddinates alone) Reversible work (slow or “quasiequilibrium expansion”) Reversible heat transfer (on board) Combinations thereof

Irreversible processes Processes that are not reversible are irreversible. After an irreversible process is executed, it is impossible to restore both the system and the surroundings to the original state. All “natural” or “spontaneous” processes are irreversible.

Irreversibilities Factors that render a process irreversible are irreversibilities. Examples: Friction Unrestrained expansion, fast expansion/contraction Heat transfer through a finite temperature difference Electric current flow through a resistance Inelastic deformation Mixing of matter with different compositions/states chemical reaction

Characteristics of reversible and irreversible processes Irreversible process In the intermediate stages the system is not in thermodynamic equilibrium. Fast. Driving forces (DT, DP etc. ) between the system and the surroundings and within parts of the system have finite magnitude. Dissipative mechanisms are present. Reversible process Passes through a succession of thermodynamic equilibrium states. Infinitely slow. Driving forces (DT, DP etc.) between the system and the surroundings and within parts of the system are infinitesimal in magnitude. Dissipative mechanisms (work done on the system incompletely converting to KE/PE change of the system) such as friction, Joule heating, inelastic deformation should be absent.

To show that heat transfer through a finite temperature difference is an irreversible process tH tH Q1-Q Q1 W=Q1-Q H Q W=Q1-Q Q Violation of Kelvin Planck statement tC Note: Heat transfer through an infinitesimal temperature difference is a reversible process.

Irreversible processes Processes that are not reversible are irreversible. After an irreversible process is executed, it is impossible to restore both the system and the surroundings to the original state. All “natural” or “spontaneous” processes are irreversible.

Irreversibilities Factors that render a process irreversible are irreversibilities. Examples: Friction Unrestrained expansion, fast expansion/contraction Heat transfer through a finite temperature difference Electric current flow through a resistance Inelastic deformation Mixing of matter with different compositions/states chemical reaction

How to conduct a reversible process? To conduct a process reversibly, at every stage of the there should be negligible “driving forces” from “property differentials between system and surroundings” such as DT, DP, D(composition), so that the system is “not driven” out of thermodynamic equilibrium. Reversible processes are therefore very slow. Example: Reversible heat transfer (on board) Reversible expansion/contraction (discussed with respect to quasi-equilibrium process) The system stays infinitesimally close to thermodynamic equilibrium during a reversible process. In practice, a thermodynamic process can at most approach reversibility With DT0, DP0 etc.

Reversible expansion/compression W=nw One small weight is removed at a time and the gas expands from a a volume Vito a volume Vf (see also discussion on quasi-equilibrium process). patm p, V

Usefulness of reversible processes: a demonstration W=nw Find work done by the system on surroundings when: Process 1: One small weight is removed at a time and the gas expands from a volume Vito a volume Vf. Process 2: All of the weights are removed at once from the piston at t=0 (an irreversible process) expands from a volume Vito a volume Vf patm Here, in order to keep the end states same; both processes are carried out isothermally. pi , Vi Note the careful choice of system boundary. In this diagram, p=patm is not the pressure of the system. Initial state: (pi=patm+W/A,Vi) Final state: (pf=patm,Vf) p=patm p p=patm p V V

Internal and external irreversibilities Internal irreversibility: Irreversibility located within the system boundaries. External irreversibility: Irreversibility located outside the system boundary; usually in the part of surroundings immediately adjacent to the system boundary. Internally reversible process: An idealization of a process in which no internal irreversibilities are present. (Totally) reversible process: A process with no internal and external irreversibilities. Another example: Example: thermal energy reservoirs undergo internally reversible processes (add to definition) Interpretation depends on choice of system boundary. Internally reversible process proceed through a succession of equilibrium states =quasi-equilibrium process

Reversible and irreversible processes between two equilibrium states p Internally Reversible v The path of an irreversible process cannot be shown on a property diagram, since intermediate states are not equilibrium states. The dotted line (shape does not matter) is just a convention to represent irreversible processes.

To show that a process is irreversible A process can be shown to be irreversible if it does not conform to the definition of a reversible process Non-zero energy exchange with the surroundings is required to return the system to initial state.

Example of irreversibility due to lack of equilibrium: unrestrained expansion of a gas 800 kPa 0 kPa A B A membrane separates a gas in chamber A from vacuum in chamber B. The membrane is ruptured and the gas expands Into chamber B until pressure equilibrium is established. The process is so fast and the container is insulated enough such that negligible heat transfer takes place between the gas and the surroundings during this process. At the end of the unrestrained expansion process, the gas (system) has the same internal energy, as it had initially.

To show that unrestrained expansion is an irreversible process 0 kPa 800 kPa 400 kPa System (gas) has been restored. Qout 800 kPa 0 kPa Converting Qout back completely to work by a cyclic device is impossible according to second law; hence the surroundings cannot be restored. Vacuum pump Win

Introduction to the Second Law of Thermodynamics (on board)