PTT 201/4 THERMODYNAMICSSEM 1 (2013/2014)

CHAPTER 6:Second Law of

Thermodynamics

Objectives• Introduce the second law of thermodynamics.

• Identify valid processes as those that satisfy both thefirst and second laws of thermodynamics.

• Discuss thermal energy reservoirs, reversible andirreversible processes, heat engines, refrigerators,and heat pumps.

• Describe the Kelvin-Planck and Clausius statementsof the second law of thermodynamics.

• Apply the second law to develop the absolutethermodynamic temperature scale.

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INTRODUCTION TO THE SECOND LAWThe 1st law of thermodynamics places no restriction on the direction of aprocess, but satisfying the 1st law does not ensure that the process can occur.Due to the lack the 1st law, the 2nd law of thermodynamics is introduced.

Transferringheat to a

paddle wheelwill not cause

A cup of hot coffee does notget hotter in a cooler room.

Transferringheat to a wirewill not

generateelectricity.

it to rotate.

These processescannot occur eventhough they are not

in violation of the firstlaw (energy isconserved).

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Work can always beconverted to heat directly and completely, but the reverse is not true (transferring heatto the water does not cause the shaft to rotate).

Processes occur in acertain direction, and notin the reverse direction.

A process must satisfy boththe first and second laws ofthermodynamics to proceed.

INTRODUCTION TO THE SECOND LAW

MAJOR USES OF THE SECOND LAW

1. The second law may be used to identify the direction of processes.

2. The second law also asserts that energy has quality as well as quantity.The first law is concerned with the quantity of energy and thetransformations of energy from one form to another with no regard to itsquality. The second law provides the necessary means to determine thequality as well as the degree of degradation of energy during a process.

3. The second law of thermodynamics is also used in determining thetheoretical limits for the performance of commonly used engineeringsystems, such as heat engines and refrigerators, as well as predicting thedegree of completion of chemical reactions.

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SECOND LAW OF THERMODYNAMICS

THERMAL ENERGY RESERVOIRS

A sourcesupplies

energy in theBodies with relatively large thermal form of heat,masses can be modeled as thermal and a sinkenergy reservoirs. absorbs it.

• A hypothetical body with a relatively large thermal energy capacity (mass x specific heat) that can supply or absorb finite amounts of heat without undergoing any change in temperature is called a thermal energy reservoir, or just a reservoir.

• In practice, large bodies of water such as oceans, lakes, and rivers as well as the atmospheric air can be modeled accurately as thermal energy reservoirs because of their large thermal energy storage capabilities or

thermal masses. 5

HEAT ENGINES

The devices that convert heat towork.

Part of the heat received by a heat engine is converted to work, while the rest is rejected to a sink.

1. They receive heat from a high-temperature source (solar energy,oil furnace, nuclear reactor,

etc.).2. They convert part of this heat towork (usually in the form of a

rotating shaft.)

3. They reject the remaining wasteheat to a low-temperature sink(the atmosphere, rivers, etc.).

4. They operate on a cycle.

Heat engines and other cyclicdevices usually involve a fluid toand from which heat is transferredwhile undergoing a cycle. This fluid is called the working fluid.

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A steam power plant

A portion of the work output ofa heat engine is consumedinternally to maintaincontinuous operation.

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Thermal efficiency

Schematic ofa heat engine.

Some heat engines perform betterthan others (convert more of theheat they receive to work).

Even the mostefficient heatengines reject

almost one-halfof the energy

they receive aswaste heat .

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Example 6-1:

Heat is transferred to heat engine from a furnace at a rate of80 MW. If the rate of waste heat rejection to a nearby riveris 50 MW, determine the net power output and thermal

efficiency for this heat engine?

Solution (refer to textbook p.280):

Assumption: Heat losses through the pipes and otherscomponents are negligible.

Answer:

i. Net power output = 30 MW

ii. Thermal efficiency = 0.375 or 37.5 %

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Example 6-2:

A car engine with a power output of 65 hp has a thermal Efficiency 24 percent. Determine the fuel consumption rateof this car if the fuel has a heating value of 44,000 kJ/kg (that is, 44,000 kJ of energy is released for each kg of fuel burned).

Solution (refer to textbook p.280):

Assumption: The power output of the car is constant.

Answer:The fuel consumption rate = 16.5 kg/h

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The Second Law ofThermodynamics:

Kelvin-Planck Statement

It is impossible for anydevice that operates on a

cycle to receive heat from asingle reservoir and producea net amount of work.

No heat engine can have a thermalefficiency of 100 percent, or as for a

power plant to operate, the working fluidmust exchange heat with the

environment as well as the furnace.The impossibility of having a 100%efficient heat engine is not due to

friction or other dissipative effects. It is alimitation that applies to both theidealized and the actual heat engines.

A heat engine that violates theKelvin-Planck statement of thesecond law.

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REFRIGERATORS AND HEAT PUMPSThe transfer of heat from a low-

temperature medium to a high-temperature one requires specialdevices called refrigerators.

Refrigerators, like heat engines,are cyclic devices.

The working fluid used in therefrigeration cycle is called arefrigerant.

The most frequently usedrefrigeration cycle is the vapor-compression refrigerationcycle.

In a home refrigerator, the freezercompartment where heat is absorbed bythe refrigerant serves as the evaporator,

Basic components of arefrigeration system and

and the coils usually behind therefrigerator where heat is dissipated to thekitchen air serve as the condenser.typical operating conditions.

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Coefficient of Performance

The efficiency of a refrigerator is expressedin terms of the coefficient of performance(COP).

The objective of a refrigerator is to removeheat (QL) from the refrigerated space.

The objective of a refrigerator is toremove QL from the cooled space.

Can the value of COPR begreater than unity? Yes, theamount of QL can be greater

than Wnet,in.13

The objectiveof a heat

pump is tosupply heatQH into thewarmer

space.

Heat Pumps

The work suppliedto a heat pump is

used to extractenergy from the

cold outdoors andcarry it into the

warm indoors.

for fixed values of QL and QH

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Example 6-3:

The food compartment of a refrigerator, as shown in figure, is maintained at 4° C by removing heat from it at a rate of360 kJ/min. If the required power input to therefrigerator is 2 kW, determine (a) the coefficient ofperformance of the refrigerator and (b) the rate of heatrejection to the room that houses the refrigerator?

Solution (refer to textbook p.285):

Assumption: Steady operating

conditions exist.

Answer:a) COPR = 3, 3 kJ of heat

removed for each kJ of worksupplied.

b) QH = QL + Wnet, in = 480 kJ/min 16

Example 6-4:

A heat pump is used to meet the heating requirements of ahouse and maintain it at 20° C. On a day when the outdoorair temperature drops to -2 ° C, the house estimated to loseheat at a rate of 80,000 kJ/h. If the heat pump under theseconditions has COP of 2.5, determine (a) the powerconsumed by the heat pump and (b) the rate at which heatis absorbed from the cold outdoor air?

Solution (refer to textbook p.286):

Assumption: Steady operating

conditions exist.

Answer:a) Wnet,in = 32,000 kJ/h.

b) QL = 48, 000 kJ/h17

The Second Law ofThermodynamics: Clausius

Statement

It is impossible to construct a device thatoperates in a cycle and produces no effectother than the transfer of heat from a lower-temperature body to a higher-temperaturebody.

It states that a refrigerator cannot operate unlessits compressor is driven by an external powersource, such as an electric motor.

This way, the net effect on the surroundingsinvolves the consumption of some energy in theform of work, in addition to the transfer of heatfrom a colder body to a warmer one. A refrigerator that violates

the Clausius statement ofthe second law.

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Equivalence of the Two Statements

Proof that theviolation of theKelvin-Planckstatement leadsto the violationof the Clausiusstatement.

The Kelvin-Planck and the Clausius statements are equivalentin their consequences, and either statement can be used asthe expression of the second law of thermodynamics.

Any device that violates the Kelvin-Planck statement alsoviolates the Clausius statement, and vice versa. 19

REVERSIBLE AND IRREVERSIBLE PROCESSES

Reversible process: A process that can be reversed without leaving any traceon the surroundings (actually do not occur in nature).

Irreversible process: A process that is not reversible.

• All the processes occurring in nature are irreversible.

• Why are we interested in reversible processes?• (1) they are easy to analyze and (2) they serve as

idealized models (theoretical limits) to whichactual processes can be compared.

• We try to approximate reversible processes. Why?

Two familiarreversible processes.

Reversible processes deliver the mostand consume the least work. 20

• The factors that cause a process to beirreversible are called irreversibilities .

Frictionrenders aprocess

irreversible.

• They include friction, unrestrained expansion,mixing of two fluids, heat transfer across a finite

temperature difference, electric resistance,inelastic deformation of solids, and chemicalreactions.

• The presence of any of these effects renders aprocess irreversible.

Irreversibilities

(a) Heattransfer througha temperaturedifference isIrreversible

irreversible, and(b) the reverseprocess is

impossible.

compressionand

expansionprocesses.

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Internally and Externally Reversible Processes•Internally reversible process: If no irreversibilities occur within the boundaries of

the system during the process.

• Externally reversible: If no irreversibilities occur outside the system boundaries.

•Totally reversible process: It involves no irreversibilities within the system or itssurroundings.

• A totally reversible process involves no heat transfer through a finite temperaturedifference, no nonquasi-equilibrium changes, and no friction or other dissipativeeffects.

A reversible processinvolves no internal andexternal irreversibilities. Totally and internally reversible heat

transfer processes. 22

THE CARNOT CYCLE

Reversible Isothermal Expansion (process 1-2, TH = constant)

Reversible Adiabatic Expansion (process 2-3, temperature drops from TH to TL)

Reversible Isothermal Compression (process 3-4, TL = constant)

Reversible Adiabatic Compression (process 4-1, temperature rises from TL to TH)

Execution of the Carnot cycle in a closed system.

P-V diagram of the Carnot cycle. P-V diagram of the reversed Carnot cycle.

The Reversed Carnot CycleThe Carnot heat-engine cycle is a totally reversible cycle.Therefore, all the processes that comprise it can be reversed, in which case it becomes the Carnot refrigeration cycle.

THE CARNOT PRINCIPLES

1. The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs.

2. The efficiencies of all reversible heat engines operating between the same two reservoirs are the same.

The Carnot principles. Proof of the first Carnot principle.

THE CARNOT PRINCIPLES

THE THERMODYNAMIC TEMPERATURESCALE

A temperature scale that isindependent of the properties ofthe substances that are used tomeasure temperature is called a

thermodynamic temperaturescale.

Such a temperature scale offersgreat conveniences in

thermodynamic calculations.

The arrangement ofheat engines used to

develop thethermodynamic

temperature scale.23

This temperature scale iscalled the Kelvin scale,and the temperatures onthis scale are calledabsolute temperatures.

For reversible cycles, theheat transfer ratio QH /QL

can be replaced by theabsolute temperature ratio

TH /TL.

A conceptual experimental setupto determine thermodynamic

temperatures on the Kelvinscale by measuring heattransfers QH and QL.

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