The Second Law of Thermo

7/25/2019 The Second Law of Thermo

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e econ aw o ermo ynam cs

Ref. 1: Cengel & Boles, Chapter 6


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Objectives nro uce e secon aw o ermo ynamcs

Discuss: thermal energy reservoirs, reversible and irreversible

, , ,

Discuss: the KelvinPlanck and Clausius statements of the second

machines

devices Describe the Carnot c cle; examine the Carnot rinci les, idealized

Carnot heat engines, refrigerators, and heat pumps

Determine the expressions for the thermal efficiencies andcoe cens o per ormance or revers e ea engnes, eapumps, and refrigerators

Swinburne University of Technology2


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The second law of thermodynamicsThe first law places no restriction on the direction of a

rocess and satisf in the first law does not uarantee

that the process will occur

The second law of thermodynamics asserts that:

the energy hasqualityas well asquantity

A process can occur when and only when it satisfies both

the first and the second laws of thermod namics



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The second law of thermodynamics Physical processes in nature can proceed toward equilibrium

spontaneously:

Water flows down a waterfall

Heat flows from a high temperature to a low temperature

Q (heat transfer)

A spontaneous process may be reversed, but it will not reverse

mposs e

Swinburne University of Technologyse sponaneous4


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The second law of thermodynamicsThe second law also asserts that energy has a quality

engineersFor example, the energy stored in a hot container (higher

temperature) has higher quality (ability to work) in

comparison with the energy contained (at lower temperature)in the surroundings

The second law is also used in determining the

engineering systems, such as heat engines and

Swinburne University of Technologyre rgerao5


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Heat (thermal) reservoirAheat reservoiris a sufficiently large system in stable

e uilibrium to which and from which finite amounts of

heat can be transferred without any change in its

Exam les: Lakes rivers atmos here oceans

Aheat source ahigh temperature heat reservoir

Aheat sink a low temperature heat reservoir to

Swinburne University of Technologywhich heat is transferred6


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A work reservoir a sufficiently large system in stableequilibrium to which and from which finite amounts of work can

be transferred adiabatically without any change in its pressure

Thermodynamic cycle a series of processes that returns to

s orgna s aethe ro erties of the s stem at the end of the c cle are the

same as at its beginning

P P T T u u v v etcf i f i f i f i= = = =, , , , .



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Heat EnginesHeat engine a thermodynamic system operating in a

thermod namic c cle in which convert heat to work

Characteristics:They receive heat from a high-temperature source (nuclear

They convert part of this heat to work

They reject the remaining waste heat to a low-temperature

sink

They operate in a cycle



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Heat Engines The following figure illustrates a steam power plant as a heat

engine operating in a thermodynamic cycle

Source, THEnergy source

QinBoiler

Turbine

in

HeatEngine

WnetWoutWin Wnet = Wout - Win

Qout

Sink, TL

= -Ener sink lake, r iver

on enser

Qout


ne n ou


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Thermal Efficiency, thThermal efficiency index of performance of a work-producing

device or a heat engine and is defined by:

The ratio of the net work output (the desired result) to the heat input (the

costs to obtain the desired result)

th =Desired Result

Required Input

For a heat engine the desired result is the net work done and the input is

.

net outW ,th

inQ=

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w ere

10

W W W

Q Q

net out out in

in net

, =


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Thermal Efficiency, thth always less than 1 or less than 100 percent

The cycle thermal efficiency may be written as



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Thermal Efficiency, th The thermal efficiencies of work-producing devices are in

general low

Ordinary spark-ignition automobile engines, th ~ 20%-25%

~ -, th

Power plants th ~ 40%-60%

Is it possible to save the rejected heat Qout in a power cycle?The answer is NO. Without the coolin in condenser the c cle cannot

be completed.

Ever heat en ine must waste some ener b transferrin it to a lo -temperature reservoir in order to complete the cycle, even in idealized

cycle.



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Example 1

A steam power plant produces 50 MW of net work while burning fuel to

produce 150 MW of heat energy at the high temperature. Determine the

cycle thermal efficiency and the heat rejected by the cycle to the

surroundings.th

net out

HQ=

,

= 150 M

MW= = .150 0 333 or 33.3%

W Q Qnet out H L, =

= Wnet= 50 MW

MW MW

net out,

=

=

150 50


QL= ?


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Example 2

A 600 MW steam power plant, which is cooled by a nearby river, has a

thermal efficiency of 40 percent. Determine the rate of heat transfer to the

river water. Will the actual heat transfer rate be higher or lower than this

value? Why?



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Example 3

An automobile engine consumes fuel at a rate of 28 L/h and delivers 60kW of

power to the wheels. If the fuel has a heating value of 44,000 kJ/kg and a

density of 0.8 g/cm3, determine the efficiency of this engine.



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Second Law Statements

Kelvin-Planck statement :

heat from a single reservoir and produce a net amount of work

Source, TH

Wnet = Qin

n

EngineThermal efficiency

of 100%

no heat engine can have a thermal efficiency of 100%

Qout = 0



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Heat Pumps and RefrigeratorsA heat pump/refrigerator a thermodynamic system

o eratin in a thermod namic c cle that removes heat

from a low-temperature body and delivers heat to a

-

cannot occurs b itself; re uires external ener in

the form of work or heat from the surroundings

,used in the cycles are called refrigerant.



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Heat Pumps and Refrigerators Refrigerators and heat pumps are essentially the same devices;

they differ in their objectives only

Refrigerator is to maintain the refrigerated space at a low temperature

-

supplies the heat to a warmer medium

WARMWARM

house

QHdesired

output

environment

QH

HPWin

RWin

COLD

environment

L

COLD

QL output

Swinburne University of Technology18HEAT PUMP

REFRIGERATORVapor-compression refrigeration cycle


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Coefficient of Performance, COPThe index of performance of a refrigerator or heat pump

is ex ressed in terms of the coefficient of erformance

COP, the ratio of desired result to input.

=Desired Result

Required Input

This measure of performance may be larger than 1we want the COP to be as large as possible



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Coefficient of Performance, COP For refrigerator the desired result is the heat

supplied at the low temperature and the inputWARM

environment

is the net work into the device to make the

cycle operate. R Win

QH

COP QL=

QLdesired

output

net in, COLDrefrigerated space

REFRIGERATOR

Applying the first law to the cyclic refrigerator:

,W W Q Q

L H in cycle

in net in H L= = COP

Q QR

L

H L

=



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Coefficient of Performance, COP For a heat pump, the desired result is the heat

transferred to the higher temperatureand theWARM

house

input is the net work into the device to make the

cycle operate.HP

QHdesired

output

COP Q QH H= =

Win

QL

W Q Qnet in H L,COLD

environment

HEAT PUMP

Under the same operating conditions the COPHP and COPR are related by

COP COPHP R= + 1


HP


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Most existin heat um s use the cold outside air as the heat source in

winter (air-source HP).

In cold climates their efficiency drops considerably when temperatures

are below the freezing point.

In such cases, geothermal (ground-source) HP that use theground as the heat source can be used.

Air conditioners are basically refrigerators whose refrigerated space is aroom or a building instead of the food compartment.

e o a rergerator ecreases wt ecreasng rergeraton

temperature.

s no economca o rergerae o a ower emperaure an

needed.



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Heat Pump and Air Conditioner Ratings HP and AC are rated using the EER system (energy efficiency

rating)

the amount of heating (cooling) on a seasonal basis in Btu/hr per unit

The heat transfer rate is often given in terms of tons of heating

. , .

Relationship between EER andCOPR

EER = 3.412 COPR

Most air conditioners have an EER between 8 to 12 (COP of 2.3

to 3.5)



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Example 4

A food refrigerator is to provide a 15,000 kJ/h cooling effect while rejecting

22,000 kJ/h of heat. Calculate the COP of this refrigerator.

Reservoir

22,000 kJ/h

in

15,000 kJ/h

Reservoir



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Example 5 o .

is losing heat to the outside air through the walls and the windows at a rate of 60,000kJ/h while the energy generated within the house from people, lights and appliances

amounts to 4000 kJ/h. For a COP of 2.5, determine the required power input to the

heat pump.



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Second Law Statements

Clausius statement :

Source (TH)

operates in a cycle and produces no effectQL

lower-temperature body to higher-

Win = 0

In other words, a refrigerator will notL

an external power source Source (TL)



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Source (TH)Source (TH)

+

Refri erator

Win = 0

Refri erator

Wnet = QH

H

Heat engine

QLQL

T =

QL = 0

Source (TL)Source (TL)

The Kelvin-Planck and the Clausius statements are e uivalent in theirconsequences.

Any device violates the Kelvin-Planck statement also violates the Clausius


statement and vice versa


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Assume that the heat engine shown is violating the Kelvin-Planck

statement by absorbing heat from a single reservoir and producingan equal amount of work W.

The output of the engine drives a refrigerator that transfers anamount of heat QL from the low-T reservoir and an amount of heat

QH +QL to the high-T reservoir.

The combination of the heat engine and refrigerator acts like a

heat pump that transfers heat QL from the low-T reservoir without

any external energy input.

This is a violation of the Clausius statement of the second law.



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Perpetual-Motion MachinesAny device that violates the first or second law of

thermod namics is called a er etual-motion machine.

If the device violates the first law, it is a perpetual-

.

If the device violates the second law, it is a perpetual-

moton mac ne o t e secon n .



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Perpetual-Motion Machines

A perpetual-motion machine that violates

A perpetual-motion machine that

violates the first law (PMM1). Ref. 1

(PMM2). Ref. 1

Despite numerous attempts, no perpetual-motion machine is known to have

worked. If somethin sounds too ood to be true it robabl is.



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Swinburne University of Technology31http://en.wikipedia.org/wiki/Image:WaterScrewPerpetualMotion.png


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Reversible Processes

A reversible processa process that can be reversedwithout leavin an trace on the surroundin

A reversible process is a quasi-equilibrium, or quasi-

both system and surroundings are returned to their initial

states at the end of the reverse process

, ,without any space limitation



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Reversible Processes

Reversible processes are idealizations of actualrocesses.

We use reversible process concept because:

they are easy to analyze (since system passes througha series of equilibrium states)

they serve as limits (idealized models) to which the

ac ua processes can e compare

reversible rocesses deliver the most and consume the

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R ibl P


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Reversible Processes Internally reversible process

- , ,

place, can be reversed and leave no change in the system

when the process is reversed, the system passes through

exactly the same equilibrium states while returning to its

initial state

a quasi-equilibrium process, which, once having takenplace, can be reversed and in so doing leave no change in

the system or surroundings



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I ibl P


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Irreversible Process An irreversible process is a process that is not reversible

.

occur because of the following:

rc on

Unrestrained expansion of gases

Heat transfer through a finite temperature difference Mixin of two different substances

Hysteresis effects

Any deviation from a quasi-static process


L t M d R i


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Last Monday ReviewThe second law of thermodynamics

Kevin-Planck statement, Clausius statement

Reversible and irreversible processes


C t C l


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Carnot CycleThe efficiency of a heat-engine greatly depends on how

the individual rocesses that make u the c cle are

executed

The net work (or efficiency) can be maximized by using

reversible rocesses which rovide u er limits on the

performance of real cycles

coas a arno - n ro uce e concep

of cyclic operation and devised a reversible cycle that iscomposed of four reversible processes, two isothermals

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C t C l


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Carnot Cycle Process 1-2 Reversible

isothermal expansion

Reversible heat transfer from the

heat source at TH to the workingfluid which is also at TH. The fluid

expands slowly, doing work on the

surroun ngs

Process 2-3 Reversible

adiabatic expansion

during which the system does work.

As a result the workin fluid

Swinburne University of Technology39temperature decreases from T

Hto T

LRef 1

Carnot Cycle


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Carnot Cycle

Process 3-4 Reversibleisothermal compression:

The system is brought in contact

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The Second Law of Thermo