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PTT 201/4 THERMODYNAMICS SEM 1 ( 2013/2014). CHAPTER 6: Second Law of Thermodynamics. Objectives. • Introduce the second law of thermodynamics. • Identify valid processes as those that satisfy both the first and second laws of thermodynamics.

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SEM 1 (2013/2014)

  • CHAPTER 6:

  • Second Law of Thermodynamics


• Introduce the second law of thermodynamics.

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

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

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

• Apply the second law to develop the absolute thermodynamic temperature scale.



The 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.


heat to a

paddle wheel

will not cause

it to rotate.

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

These processescannot occur eventhough they are not

Transferringheat to a wirewill not

in violation of the first


law (energy is




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

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

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



1. The second law may be used to identify the directionof 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 the

transformations 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 the theoretical limits for the performance of commonly used engineering systems, such as heat engines and refrigerators, as well as predicting the degree of completion of chemical reactions.



A source


energy in the

form of heat,

Bodies with relatively large thermal

masses can be modeled as thermal

and a sink

energy reservoirs.

absorbs it.

• A hypothetical body with a relatively large thermal energy capacity(mass xspecific heat) that can supplyor absorb finite amounts of heatwithout

undergoing any change in temperatureis called a thermal energyreservoir, or just a reservoir.

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

thermal masses.



The devices that convert heat to work.

1. They receive heat from a high-

temperature source (solar energy,

oil furnace, nuclear reactor, etc.).2. They convert part of this heat to

work (usually in the form of a

rotating shaft.)

3. They reject the remaining waste

heat to a low-temperature sink

(the atmosphere, rivers, etc.).

They operate on a cycle.

Heat engines and other cyclicdevices usually involve a fluid to

and from which heat is transferred

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

while undergoing acycle. This fluid

is called theworking fluid.


A steam power plant

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


Thermal efficiency

Schematic of

a heat engine.

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

Even the most efficient heat engines rejectalmost one-half

of the energy

they receive as waste heat .


Example 6-1:

Heat is transferred to heat engine from a furnace at a rate of

80 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.



Net power output = 30 MW


Thermal efficiency = 0.375 or 37.5 %


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.


The fuel consumption rate = 16.5 kg/h


The Second Law of


Kelvin-Planck Statement

It is impossible for any

device 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

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

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 the

idealized and the actual heat engines.



The 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 used

refrigeration cycle is the vapor-compression refrigerationcycle.

In a home refrigerator, the freezer

compartment where heat is absorbed by

the refrigerant serves as the evaporator,

and the coils usually behind the

Basic components of arefrigeration system and

refrigerator where heat is dissipated to thekitchen air serve as the condenser.

typical operating conditions.


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.

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

The objective of a refrigerator is toremove QLfrom the cooled space.

than Wnet,in.


Heat Pumps

The objectiveof a heat

pump is to

supply heatQHinto thewarmer


The work suppliedto a heat pump is

used to extract

energy from thecold outdoors and carry it into the

warm indoors.

for fixed values of QLand QH


Example 6-3:

The food compartment of a refrigerator, as shown in figure, is maintained at 4° C by removing heat from it at arate of360 kJ/min. If the required power input to the

refrigerator is 2 kW, determine (a) the coefficient of

performance 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:

COPR= 3, 3 kJ of heat removed for each kJ of worksupplied.

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


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 lose

heat at a rate of 80,000 kJ/h. If the heat pump under these

conditions has COP of 2.5, determine (a) the power

consumed 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/h


The Second Law of

Thermodynamics: Clausius


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 surroundings

involves 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 violatesthe Clausius statement ofthe second law.


Equivalence of the Two Statements

Proof that the

violation of theKelvin-Planckstatement leads

to the violation

of the Clausiusstatement.

The Kelvin-Planck and the Clausius statements are equivalent

in their consequences, and either statement can be used asthe expression of the second law of thermodynamics.

Any device that violates the Kelvin-Planck statement also

violates the Clausius statement, and vice versa.



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 familiar

reversible processes.

Reversible processes deliver the most

and consume the least work.


The factors that cause a process to be irreversible are called irreversibilities .

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

Frictionrenders aprocess

temperature difference, electric resistance,

inelastic deformation of solids, and chemicalreactions.


• The presence of any of these effects renders a process irreversible.


(a) Heat

transfer througha temperaturedifference is


irreversible, and


(b) the reverseprocess is





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 its


A totally reversible process involves no heat transfer through a finite temperature

difference, no nonquasi-equilibrium changes, and no friction or other dissipative


A reversible process

involves no internal and

Totally and internally reversible heat

external irreversibilities.

transfer processes.



Execution of the Carnot cycle in a closed system.

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

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

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

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

P-V diagram of the Carnot cycle.

P-V diagram of the reversed Carnot cycle.

The Reversed Carnot Cycle

The 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.

Proof of the first Carnot principle.

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

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


A temperature scale that is

independent of the properties ofthe substances that are used to

measure temperature is called a

thermodynamic temperaturescale.

Such a temperature scale offersgreat conveniences in

thermodynamic calculations.

The arrangement of

heat engines used to

develop the

thermodynamictemperature scale.


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

For reversible cycles, theheat transfer ratio QH /QLcan be replaced by the

A conceptual experimental setupto determine thermodynamic

temperatures on the Kelvin

absolute temperature ratio

scale by measuring heattransfers QHand QL.