Introduction to the Second Law of Thermodynamics (on board)

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Heat engine

A cyclic heat engine

Net work output

Thermalefficiency

Examples of heat engines

: a simple steam power plant

Refrigerators and heat pumps

Refrigerator is a device, whichoperating in a cycle, maintains abody at a temperature lower thanthe temperarature of its surroundings.

Heat pump is a device, whichoperating in a cycle, maintains abody at a temperature higher thanthe surroundings.

Desired effect

RequirCO

Coefficientof performance

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 devicewill not cause the paddle-wheel to rotate and raise the mass through the pulley.

Work was completely converted into heat in Joule’s experiment

Some definitions (on board/discussion)

• Thermal energy reservoirs (source and sink)• Heat engines • Efficiency of a heat engine

Example of an heat engine: a simple steam power plant

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 devicewill not cause the paddle-wheel to rotate and raise the mass through the pulley.

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

1. DEDUCTION BASED ON either of the Kelvin-Planck and Clausius statements will give us the ability toC) 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

2. 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

1TER,cycle,net 0W

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 statementViolation of Kelvin-Planck statement

• Violation of Kelvin-Planck statementViolation of Clausius statement

The net heat exchange of the cyclicdevice (HE+R) with the hot reservoir=Q2-Q

Violation of KP Violation of Clausius

Violation of Clausius Violation of KP

HE+R!+TH

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.

TH can be eliminated and Q can be fed directly to H from R

Violation of Clausius Violation of KP (Alternative)

Equivalence of Kelvin-Planck and Clausius statements

• Violation of Clausius statementViolation of Kelvin-Planck statement

• Violation of Kelvin-Planck statementViolation 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

OKNot OK! Produces netenergy output without energy input.

( )out outQ W

,net out out inW W W

Identifying PMM2 by Kelvin-Planck/Clausius statement

• 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

OKNotOK!ViolatesKP

Identifying PMM2 by Clausius statement

• 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

• 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

vacuum

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

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.

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.

Characteristics of reversible and irreversible processes

To show that heat transfer through a finite temperature difference is an irreversible process

W=Q1-Q

W=Q1-QQ

Violation of Kelvin Planckstatement

Note: Heat transfer through an infinitesimal temperature difference is a reversibleprocess.

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

• 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)

How to conduct a reversible process?

The system stays infinitesimally close to thermodynamic equilibrium during a reversible process. In practice, a thermodynamic process can at most approach reversibility With

DT0, DP0 etc.

Reversible expansion/compression

One small weight is removed at a time and the gas expands from a a volume Vi

to a volume Vf (see also discussion on quasi-equilibrium process).

Usefulness of reversible processes: a demonstration

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 Vi to 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 Vi

to a volume Vf

pi , Vi

Initial state: (pi=patm+W/A,Vi)

Final state: (pf=patm,Vf)

rev irrev boundary boundary

irrev atm i revf

F dx P dVw

wVw p V

Here, in order to keep the end states same; both processes arecarried out isothermally.

p=patm

Note the careful choice of system boundary.In this diagram,p=patm is not the pressure of the system.

Internal and external irreversibilities

Internally reversible process proceed through a succession ofequilibrium states =quasi-equilibrium process

• 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.

Interpretation depends on choice ofsystem boundary.

Example: thermal energy reservoirsundergo internally reversible processes(add to definition)

Another example:

Reversible and irreversible processes between two equilibrium states

The path of an irreversible process cannot be shown on a property diagram, since intermediate states are notequilibrium states. The dotted line (shape does not matter) isjust a convention to represent irreversible processes.

Internally Reversible

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

A membrane separates a gas in chamber A from vacuum in chamber B. The membrane is ruptured and the gas expandsInto chamber B until pressure equilibrium is established. Theprocess is so fast and the container is insulated enough such that negligible heat transfer takes place betweenthe gas and the surroundings during this process.

800 kPa 0 kPaBA

At the end of the unrestrained expansion process, the gas (system) has the same internal energy, as it had initially.

800 kPa

To show that unrestrained expansion is an irreversible process

800 kPa 0 kPa

Vacuum pumpWin

400 kPa

Converting Qout back completely towork by a cyclic device is impossible according tosecond law; hence the surroundingscannot be restored.

System (gas) has been restored.

Introduction to the Second Law of Thermodynamics (on board)

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