Second

Law

ofT

hermodynam

ics&

Entropy

Dr.

Md.

Zahurul

Haq

Pro

fessorD

epartm

ent

ofM

echanica

lEngin

eering

Bangla

desh

University

ofEngin

eering

&Tech

nolo

gy

(BU

ET

)D

haka

-1000,Bangla

desh

http

://za

huru

l.buet.a

c.bd/

ME

6101:

Cla

ssicalT

herm

odynam

ics

http://zahurul.buet.ac.bd/ME6101/

©D

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odynam

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ME6101

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56

Overview

1H

eatEngines

2T

heSecond

Law

ofT

hermodynam

ics

3Entropy

4Second-L

awof

Therm

odynam

icsfor

CV

System

s

5Entropy

ofa

Pure

Substance

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

2/

56

Heat

Engin

es

Som

eO

bservationsin

Work

&H

eatConversions

T104

Work

canalw

aysbe

converted

toheat

directly

and

com

pletely,

but

the

reverseis

not

true.

T054

©D

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d.

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lH

aq

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ET

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Law

ofT

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odynam

ics

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56

Heat

Engin

es

T055

©D

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lH

aq

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Law

ofT

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odynam

ics

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56

Heat

Engin

es

Therm

alReservoir

T136

Ath

erm

alre

serv

oir

isa

closedsystem

with

thefollow

ingcharacteristics:

•Tem

perature

remains

uniformand

constantduring

apro

cess.

•Changes

within

thetherm

alreservoir

are

internallyreversible.

•H

eattransfer

toor

froma

thermal

reservoironly

resultsin

anincrease

or

decreasein

theinternal

energyof

the

reservoir.

Atherm

alreservoir

isan

idealizationw

hichin

practicecan

be

closely

approximated.

Large

bodies

ofwater,

suchas

oceans

andlakes,

andthe

atmosphere

behave

essentiallyas

thermal

reservoirs.

©D

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d.

Zahuru

lH

aq

(BU

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)Second

Law

ofT

herm

odynam

ics

ME6101

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56

Heat

Engin

es

Heat

Engine:

Classifi

cations

T058

©D

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d.

Zahuru

lH

aq

(BU

ET

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Law

ofT

herm

odynam

ics

ME6101

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56

Heat

Engin

es

(Heat)

Engine

T106

T105

T107

©D

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d.

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Law

ofT

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odynam

ics

ME6101

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56

Heat

Engin

es

T109

T108

Therm

alEffi

ciency,ηth

=W

net,out

Qin

=1−

Qout

Qin

=1−

QL

QH

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

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56

Heat

Engin

es

Refrigerator/A

ir-conditioner

T110

T142

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

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56

Heat

Engin

es

T144

©D

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d.

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aq

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odynam

ics

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/56

Heat

Engin

es

T111

T138

Coeffi

cientof

Perform

ance,COPR=

Desire

dO

utp

ut

Require

dIn

put=

QL

Wnet,in

©D

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d.

Zahuru

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aq

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Law

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odynam

ics

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/56

Heat

Engin

es

Heat

Pum

p

T112

T137

Coeffi

cientof

Perform

ance,COPHP=

Desire

dO

utp

ut

Require

dIn

put=

QH

Wnet,in

COPHP=

QH

Wnet,in

=Q

L+W

net,in

Wnet,in

=COPR+

1

©D

r.M

d.

Zahuru

lH

aq

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odynam

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/56

Heat

Engin

es

Reversible

Engines

T035

T036

•A

reversib

lepro

cess

fora

systemis

defined

asa

process

thatonce

havingtaken

placecan

be

reversedand

inso

doingleave

nochange

in

eitherthe

systemor

thesurrounding.

•A

reversiblepow

ercycle

canbe

changedto

areversible

refrigeration

cycleby

justreversing

allthe

heatand

work

flow

quantities.

©D

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d.

Zahuru

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aq

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odynam

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/56

The

Second

Law

ofT

herm

odynam

icsK

elvin-Planck

(KP)

Statem

ent

Kelvin-P

lanck(K

P)

statement

Itis

impossible

toconstruct

adevice

thatw

illop

eratein

acycle

and

produce

noeff

ectother

thanthe

raisingof

aweight

andthe

exchangeof

heatw

itha

singlereservoir.

T028

Wnet≦

0for

singlereservoir

©D

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odynam

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/56

The

Second

Law

ofT

herm

odynam

ics

T755

Anon-co

ntin

uous

process

that

converts

heat

towork

with

100%

efficien

cy.

T756

Pro

cess

(a)

vio

late

sth

eSeco

nd

Law

ofT

herm

odynam

ics.

©D

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/56

The

Second

Law

ofT

herm

odynam

ics

Clausius

Statem

ent

Clausius

statement

Itis

impossible

toconstruct

adevice

thatop

eratesin

acycle

andpro

duces

noeff

ectother

thanthe

transferof

heatfrom

aco

olerbody

toa

hotter

body.

T029

©D

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The

Second

Law

ofT

herm

odynam

ics

Equivalence

ofStatem

ents

T030

Vio

lation

ofClau

sius

(C)

statemen

t⇒

violatio

nofK

evlin-P

lanck

(KP)

statemen

t.

©D

r.M

d.

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aq

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odynam

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The

Second

Law

ofT

herm

odynam

ics

T113

Vio

lation

ofK

evlin-P

lanck

(KP)

statemen

t⇒

violatio

nofClau

sius

(C)

statemen

t.

©D

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Zahuru

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The

Second

Law

ofT

herm

odynam

ics

3O

bservationsof

Two

Statem

ents

1B

oth

arenegativ

esta

tem

ents;

negativestatem

entsare

impossible

toprove

directly.Every

relevantexp

eriment

thathas

been

conducted,

eitherdirectly

orindirectly,

verifies

thesecond

law,and

noexp

eriment

hasever

been

conductedthat

contradictsthe

secondlaw

.T

hebasis

of

thesecond

lawis

thereforeexp

erimental

evidence.

2B

oth

state

ments

areequiv

ale

nt.

Two

statements

areequivalent

if

thetruth

ofeither

statement

implies

thetruth

ofthe

otheror

ifthe

violationof

eitherstatem

entim

pliesthe

violationof

theother.

3B

oth

state

ments

state

the

impossib

ilityofPerp

etu

alM

otio

n

Mach

ine

of2nd

Kin

d(P

MM

2).

©D

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/56

The

Second

Law

ofT

herm

odynam

ics

Perp

etualM

otionM

achines

1A

perp

etual-motion

machine

ofthe

first

kind(P

MM

1)would

create

work

fromnothing

orcreate

energy,thus

violatingthe

first

law.

2A

perp

etual-motion

machine

ofthe

secondkind

(PM

M2)

would

extractheat

froma

sourceand

thenconvert

thisheat

completely

into

otherform

sof

energy,thus

violatingthe

secondlaw

.

T140

Aperp

etual-m

otio

nm

achin

eofth

eseco

nd

kind.

©D

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Zahuru

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aq

(BU

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odynam

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/56

The

Second

Law

ofT

herm

odynam

ics

Carnot’s

Principles

1It

isim

possible

toconstruct

an

enginethat

operates

betw

eentw

o

givenreservoirs

andis

more

efficient

thana

reversibleengine

operating

betw

eenthe

same

two

reservoirs.

2All

enginesthat

operate

onthe

Carnot

cyclebetw

eentw

ogiven

constant-temperature

reservoirs

havethe

same

efficiency.

T114

3An

absolutetem

perature

scalem

aybe

defined

which

isindep

endentof

them

easuringsubstances.

©D

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Zahuru

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odynam

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/56

The

Second

Law

ofT

herm

odynam

ics

Pro

of:ηrev>ηirr

T032

Ifηirr>ηrev⇒

|WIE|>

|WRE|for

sameQ

H.

Hen

ce,co

mposite

systempro

duces

net

work

outp

ut

while

exchan

gin

gheat

with

asin

gle

reservoir⇒

violatio

nofK

-P

statemen

t.

©D

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d.

Zahuru

lH

aq

(BU

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)Second

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odynam

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/56

The

Second

Law

ofT

herm

odynam

ics

Pro

of:ηrev

=sa

me

for

sam

eTH&

TL

The

proof

ofthis

proposition

issim

ilarto

thepro

ofjust

outlined,w

hich

assumes

thatthere

isone

Carnot

cyclethat

ism

oreeffi

cientthan

another

Carnot

cycleop

eratingbetw

eenthe

same

temperature

reservoirs.Let

the

Carnot

cyclew

iththe

highereffi

ciencyreplace

theirreversible

cycleof

the

previousargum

ent,and

letthe

Carnot

cyclew

iththe

lower

efficiency

operate

asthe

refrigerator.

©D

r.M

d.

Zahuru

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aq

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odynam

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/56

The

Second

Law

ofT

herm

odynam

ics

Therm

odynam

icTem

perature

Scale

Therm

aleffi

ciencyof

areversible

heatengine

ata

givenset

ofreservoirs

is

independent

ofconstruction,

designand

working

fluid

ofthe

engine.

T057

•ηth

=1−

QL

QH=

1−ψ(T

L ,TH)

•Q

1Q

2=ψ(T

1 ,T2 ),

Q2

Q3=ψ(T

2 ,T3 )

•Q

1Q

3=ψ(T

1 ,T3 )

=Q

1Q

2 ·Q

2Q

3

•ψ(T

1 ,T3 )

=ψ(T

1 ,T2 ).ψ

(T2 ,T

3 )︸

︷︷︸

Notafunctio

nofT

2

⇒ψ(T

1 ,T2 )

=f(T

1)

f(T

2) ,ψ

(T2 ,T

3 )=

f(T

2)

f(T

3)

⇒ψ(T

1 ,T3 )

=f(T

1)

f(T

3)=

f(T

1)

f(T

2) ·

f(T

2)

f(T

3)

⇒Q

H

QL=ψ(T

H,T

L)=

f(TH)

f(TL)

Kelvin

proposed

that,f(T

)=

T

QH

QL=

TH

TL⇛

ηrev.engine=

1−

TL

TH

©D

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aq

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odynam

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/56

The

Second

Law

ofT

herm

odynam

ics

Mora

nEx.

5.1

⊲An

inventorclaim

sto

havedevelop

eda

pow

ercycle

capableof

deliveringa

network

outputof

410kJ

foran

energyinput

byheat

transferof

1000kJ.

The

systemundergoing

thecycle

receivesthe

heattransfer

fromhot

gasesat

atem

perature

of500

Kand

dischargesenergy

byheat

transferto

theatm

osphereat

300K

.Evaluate

thisclaim

.

T143

•Claim

edeffi

ciency:

⇒η=

WQin=

410

1000=

0.41

=41

%.

•M

aximum

possible

thermal

efficiency:

⇒ηmax=

1−

TL

TH=

1−

300

500=

0.40

=40

%.

The

Carnot

corollariesprovide

abasis

forevaluating

theclaim

:Since

thetherm

aleffi

ciencyof

theactual

cycleexceeds

them

aximum

theoreticalvalue,

theclaim

cannotbe

valid.

©D

r.M

d.

Zahuru

lH

aq

(BU

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)Second

Law

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odynam

ics

ME6101

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/56

Entro

py

Consequences

ofSecond

Law

ofT

hermodynam

ics

•If

asystem

istaken

througha

cycleand

produces

work,

itm

ustbe

exchangingheat

with

atleast

two

reservoirsat

2diff

erent

temperatures.

•If

asystem

istaken

througha

cyclew

hileexchanging

heatw

itha

singlereservoir,

work

must

be

zeroor

negative.

•H

eatcan

neverbe

convertedcontinuously

andcom

pletelyinto

work,

butwork

canalw

aysbe

convertedcontinuously

andcom

pletelyinto

heat.•

Work

isa

more

valuableform

ofenergy

thanheat.

•For

acycle

andsingle

reservoir,W

net6

0.

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

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odynam

ics

ME6101

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/56

Entro

py

Clausius

Inequity

T141

Clausius

Inequality∮δQT6

0

⇒δW

net ≡

(δW

rev+δW

sys )

=δQ

R−dU

⇒δQ

R

δQ

=TR

T

⇒δW

net=

TRδQT−dU

⇒W

net=

TR

∮δQT

⇒∮δQT6

0asW

net6

0

∮δQT

{=

0rev

ersible

process

<0

irreversib

leprocess

©D

r.M

d.

Zahuru

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aq

(BU

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)Second

Law

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herm

odynam

ics

ME6101

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/56

Entro

py

Exam

ple

:⊲

Clausius

Inequality:Steam

Pow

erPlant:

T037•

At

0.7M

PaTsat=

TH=

164.95

oC

•At

15kP

aTsat=

TL=

53.97

oC

•Q

H=

Q12=

h2−h

1=

2.066

MJ/kg

•Q

L=

Q34=

h4−h

3=

−1.898

MJ/kg

⇒∮

δQT=

QH

TH+

QL

TL=

−1.086

kJ/kg⊳

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

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/56

Entro

py

Entropy

(S):

AT

hermodynam

icProp

erty

T038

For

reversiblepro

cess:∮δQT

=0

⇒∮δqT=

∫21

(

δqT

)

A+∫

12

(

δqT

)

B=

01○

⇒∮δqT=

∫21

(

δqT

)

C+∫

12

(

δqT

)

B=

02○

•1○

−2○

:⇛∫

21

(

δqT

)

A=

∫21

(

δqT

)

C=

···

Since

∫δq/T

issam

efor

allreversible

processes/paths

betw

eenstate

1&

2,this

quantityis

independent

ofpath

andis

afunction

ofend

statesonly.

This

property

iscalled

Entro

py,

S.

ds≡

(

δqT

)

rev ⇒

∆s=

s2−s1=

∫21

(

δqT

)

rev

δqrev=

Tds

©D

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d.

Zahuru

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Law

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odynam

ics

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/56

Entro

py

T187

•Entropy

isa

property,

hencechange

in

entropybetw

eentw

oend

statesis

same

for

allpro

cesses,both

reversibleand

irreversible.

•If

noirreversibilities

occur

within

thesystem

boundaries

ofthe

systemduring

the

process,

thesystem

isinternally

reversible.

For

aninternally

reversiblepro

cess,the

changein

theentropy

isdue

solely

forheat

transfer.So,

heattransfer

acrossa

boundary

associated

with

itthe

transferof

entropyas

well.

∫21

(

δqT

)

rev

≡E

ntro

py

transfer

(or

flux)

©D

r.M

d.

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lH

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(BU

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)Second

Law

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odynam

ics

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/56

Entro

py

Exam

ple

:⊲

Steam

generationin

Boiler.

T152

•s2−s1=

∫21

(

δqT

)

rev=

∫21δq

Tsat

⇒s2−s1=

q12

Tsat=

h2−h1

Tsat

=hfg

Tsat

⇒hfg=

sfgTsat

•q

23=

∫32δq=

∫32Tds

•q

12=

Tsat (s

2−s1 )

=area

(1−

2−

b−

a)

•q

23=

∫32Tds=

area

(2−

3−

c−

b)

•qnet=

q12+q

23=

area

(1−

2−

3−

c−

a)

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

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odynam

ics

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/56

Entro

py

Mora

nEx.

6.1

:⊲

Internallyreversible

heatingin

apiston-cylinder

system.

T169

⇒w

=∫gfPdv=

P(v

g−vf )

=101

.325(1

.673−

0.001044

)=

170kJ/kg⊳

⇒q=

∫gfTds=

T(s

g−sf )

=Tsfg=

373.15·6

.0486=

2256.8

kJ/kg⊳

Also

notethat,

hfg=

2257kJ/kg⊳

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

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32

/56

Entro

py

Carnot

Cycle

T153

1-2:

Isentropiccom

pression.

2-3:

Isothermal

heataddition

andexpansion.

3-4:

Isentropicexpansion.

4-1:

Isothermal

heatrejection

andcom

pression.©

Dr.

Md.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

33

/56

Entro

py

T156

T154

δqrev=

Tds

•q23=

∫32Tds=

TH

∫32Tds=

TH(s

3−s2 )

=TH∆s

•q41=

∫14Tds=

TL

∫14Tds=

TL (s

1−s4 )

=−TC∆s

•wnet=

qnet=

q23+q41=

(TH−TC)∆

s

•qin=

q23

ηCarnot ≡

wnet

qin

=(TH−TC)∆s

TH∆s

=1−

TC

TH

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

34

/56

Entro

py

Entropy

Change

forIrreversible

CM

Pro

cess

T041

•For

reversiblepro

cess:∮δqT=

0

⇒∮δqT=

∫21

(

δqT

)

A+∫

12

(

δqT

)

B=

01○

•For

irreversiblepro

cess:∮δqT<

0

⇒∮δqT=

∫21

(

δqT

)

C+∫

12

(

δqT

)

B<

02○

•1○

−2○

:⇛∫

21

(

δqT

)

A>

∫21

(

δqT

)

C

•Since

pathA

isreversible,

andsince

entropyis

aprop

erty

∫21

(

δqT

)

A

=

∫21dsA=

∫21dsC⇒

∫21dsC>

∫21

(

δqT

)

C

•Since

pathC

isarbitrary,

forirreversible

process

ds>

δqT⇒

s2−s1>

∫21

(

δqT

)

irrev

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

35

/56

Entro

py

T170•

For

irreversiblepro

cess:

w12 6=

∫21Pdv

:q12 6=

∫21Tds

So,

thearea

underneaththe

pathdo

esnot

representwork

andheat

on

theP−v

andT

−s

diagrams,

respectively.

•In

irreversiblepro

cesses,the

exactstates

throughw

hicha

system

undergoes

arenot

defined.

So,

irreversiblepro

cessesare

shown

as

dashedlines

andreversible

processes

assolid

lines.

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

36

/56

Entro

py

Entropy

Generation

dS≧

δQT⇒δσ≡

dS−δQT≧

0

σ,

Entropy

produced

(generated)by

internalirreversibilities.

T173

σ:

>0

irreversiblepro

cess

=0

internallyreversible

process

<0

impossible

process

•For

CM

system:dSCM

=δQT+δσ

•CM

system,w

ithheat

transferoccurring

atseveral

boundaries,

ifTiis

thetem

perature

atpoint

where

δQ

itakes

place,then

dSCM

dt

=∑

δQ

i

Ti+δσ

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

37

/56

Entro

py

•Change

inentropy

ofany

CM

systemis

dueto

only2

physicaleff

ects:

1H

eattransfer

to/fromthe

systemas

measured

byentropy

transfer/flux,

δQ/T

.2

Presence

ofirreversibilities

within

thesystem

&its

contributionis

measured

byentropy

production,

σ>

0.

•O

nly

way

todecrease

the

entropy

ofa

closedsystem

isto

transfer

of

heat

fromit.

Inth

iscase,

heat

transfer

contrib

ution

must

be

more

-ve

that

the

+ve

contrib

ution

ofan

yin

ternal

irreversibility.

•Reversib

lepro

cess:ds=δq/T

&ad

iabatic

process:

δq=

0

δq/T

=0⇒

s=

consta

nt:

forreversib

lead

iabatic

process.

•All

isentrop

icpro

cessesare

not

necessarily

reversible

&ad

iabatic.

Entropy

canrem

aincon

stant

durin

ga

process

ifheat

removal

balan

ces

the

contrib

ution

due

toirreversib

ility.

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

38

/56

Entro

py

Mora

nEx.

6.2

:⊲

Irreversiblepro

cessof

water.

T172•

du+dke+dpe=δq−δw

⇒du=

−δw

⇒Wm

=−∫gfdu=

−(u

g−uf )

=−

2087.56

kJ/kg⊳

Note

that,the

work

inputby

stirringis

greaterin

magnitude

thanthe

work

doneby

thewater

asit

expands(170

kJ/kg).

•δ(σ/m)=

ds−

δqT=

ds−

0=

ds

⇒σm=

sg−sf=

6.048

kJ/kg.K⊳

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

39

/56

Entro

pyP

rincipleof

Increaseof

Entropy

ds≧

δqT

Contro

lMass,

tempera

ture

=T

Surro

undings,

tempera

ture

=To

δq

δw

Adiabatic

oriso

lated

system

T171

•For

CM

:dsCM≧

δqT.

•For

surrou

ndin

gs,reversib

leheat

transfer:

dssurr=

−δq

To

.

⇒dsnet=

dssys+dssurr≧δq[

1T−

1To

]

•IfT>

To⇒δq<

0⇛

dsnet≧

0

•IfT<

To⇒δq>

0⇛

dsnet≧

0

dsnet≧

0

Entropy

chan

gefor

anisolated

systemcan

not

be

negative.

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

40

/56

Entro

py

T188

Entro

pych

ange

ofan

isolated

systemis

the

sum

ofth

een

tropy

chan

ges

ofits

com

ponen

ts,an

dis

never

lessth

anzero

.

T189

Asystem

and

itssu

rroundin

gs

forman

isolated

system.

dsiso

lated>

0

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

41

/56

Entro

py

Exam

ple

:⊲

Supp

osethat

1kg

ofsaturated

water

vapour

at100

oCis

toa

saturatedliquid

at100

oCin

aconstant-pressure

process

byheat

transferto

thesurrounding

air,w

hichis

at25

oC.W

hatis

thenet

increasein

entropyof

thewater

plussurroundings?

∆snet=∆ssys+∆ssurr

•∆ssys=

−sfg=

−6.048

kJ/kg.K

•∆ssurr=

qTo=

hfg

To=

2257

298=

7.574

kJ/kg.K

⇒∆snet=

1.533

kJ/kg.K⊳

So,

increasein

netentropy.

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

42

/56

Second-L

aw

ofT

herm

odynam

ics

for

CV

Syste

ms

Second

Law

Analysis

forCV

System

att

att+∆t

T320

CM

(timet)

:region

A+

CV

SCM

,t=

SA+SCV

,t

CM

(t+∆t)

:CV

+region

B

SCM

,t+∆t=

SB+SCV

,t+∆t

•SCM

,t+∆t −

SCM

,t

∆t

=SCV

,t+∆t −

SCV

,t

∆t

+SB−SA

∆t

•dSCM

dt

=dSCV

dt

+m

BsB−m

AsA

•dSCM

dt

=∑

Qi

Ti+σ

dSCV

dt

=∑

Qi

Ti+∑

i (ms)

i−∑

e (ms)

e+σCV

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

43

/56

Second-L

aw

ofT

herm

odynam

ics

for

CV

Syste

ms

T042

T175

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

44

/56

Second-L

aw

ofT

herm

odynam

ics

for

CV

Syste

ms

dSCV

dt

=∑

jQ

j

Tj+∑

(ms)

i−∑

(ms)

e+σCV

•For

CM

systems:

mi=

0,m

e=

0⇒

dSCM

dt

=∑

jQ

j

Tj+σ

•For

steady-state

steady-fl

ow(S

SSF)

process:

dScv/dt=

0.

•For

1-inlet

&1-ou

tletSSSF

process:

mi=

me .

⇒(s

e−si )=

∑jqj

Tj+σCV

m

•For

adiab

atic1-in

let&

1-outlet

SSSF

process:

⇒(s

e−si )=

σCV

m⇒

se≧

si

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

45

/56

Second-L

aw

ofT

herm

odynam

ics

for

CV

Syste

ms

T757

(a)Energ

ybalan

ce(b

)Entro

pybalan

ce.

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

46

/56

Second-L

aw

ofT

herm

odynam

ics

for

CV

Syste

ms

Mora

nEx.

6-6

⊲D

etermine

therate

atw

hichentropy

ispro

ducedw

ithinthe

turbineper

kgof

steamflow

ing,in

kJ/kgK.

T1082

•SSSF:

dSCV

dt

=0,

mi=

me=

m

•∑

Qi

Ti=

QTb :

Tb

=350

K.

•zi=

ze

•States

1○&

2○:

defined.

•dECV

dt

=Q

−W

cv+m

i

(

hi+

V2i

2+gzi

)

−m

e

(

he+

V2e

2+gze

)

⇒Q

=√

•dSCV

dt

=∑

Qi

Ti+(m

s)i−(m

s)e+σCV

⇒σCV=

√

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

47

/56

Second-L

aw

ofT

herm

odynam

ics

for

CV

Syste

msP

rincipleof

Increaseof

Entropy

∑

mi s

i

∑

mese

surru

nding systemT

0

δQ

cv

T

T321

•dSCV

dt

=∑

Qi

Ti+∑

i (ms)

i−∑

e (ms)

e+σCV

⇒dSCV

dt

=∑

Qi

Ti+∑

i (ms)

i−∑

e (ms)

e+σCV

•dSsurr

dt

=∑

Qi

Ti+∑

i (ms)

i−∑

e (ms)

e+σsurr

⇒dSsurr

dt

=−

QCV

To−∑

i (ms)

e+∑

e (ms)

e+✟✟✟✯

0σsurr

•dSnet

dt

=dSCV

dt

+dSsurr

dt

=∑

Qi

Ti−

QCV

To+σtot

•σtot>

0,so

dSnet

dt

=dSCV

dt

+dSsurr

dt>

0

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

48

/56

Second-L

aw

ofT

herm

odynam

ics

for

CV

Syste

ms

Observations:

Principle

ofIncrease

ofEntropy

•T

he

increase-in

-entropy

princip

lesare

direction

alstatem

ents.

A

decrease

inen

tropyis

not

possib

lefor

closed,ad

iabatic

systems

orfor

composite

systems

which

interact

amon

gth

emselves.

•T

he

entropy

function

isa

non

-conserved

property,

and

the

increase-in

-entropy

princip

lesare

non

-conservation

laws.

Irreversible

effects

createen

tropy.T

he

greaterth

em

agnitu

de

ofth

e

irreversibilities,

the

greaterth

een

tropych

ange.

•T

he

second

lawstates

that

astate

ofeq

uilibriu

mis

attained

when

the

entropy

function

reaches

the

maxim

um

possib

levalu

e,con

sistent

with

the

constrain

tson

the

systems.

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

49

/56

Entro

py

ofa

Pure

Substa

nce

Entropy

ofa

Pure

Substance

T1080

T-s

and

h-s

diag

rams

forsteam

.

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

50

/56

Entro

py

ofa

Pure

Substa

nce

Entropy

ofIdeal

Gas

•First

Law

:δq−δw

=du

•Reversib

lepro

cess:

δw

=Pdv

:δq=

Tds

•Id

ealgas:

Pv=

RT

:du=

cvdT

:dh=

cPdT

⇒du=

−Pdv+Tds⇒

Tds=

du+Pdv

:1st

Tds

Equation

.

⇒ds=

cV

dTT+

PTdv=

cV

dTT+R

dvv

:For

ideal

gas.

⇒s(T

2 ,v2 )

−s(T

1 ,v1 )

=∫T

2

T1cV

dTT+R

ln(

v2v1

)

⇒s2−s1=

cV

ln(

T2

T1

)

+R

ln(

v2v1

)

:Id

ealgas

with

cV

=con

stant.

⇒h=

u+Pv⇒

dh=

du+Pdv+vdP

=δqrev+vdP=

Tds+vdP

⇒dh=

Tds+vdP

⇒Tds=

dh−vdP

:2ndTds

Equation

.

⇒s(T

2 ,P2 )

−s(T

1 ,P1 )

=∫T

2

T1cPdTT−R

ln(

P2

P1

)

⇒s2−s1=

cP

ln(

T2

T1

)

−R

ln(

P2

P1

)

:Id

ealgas

with

cP

=con

stant.

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

51

/56

Entro

py

ofa

Pure

Substa

nce

IsentropicPro

cess:s=

consta

nt⇒∆s=

0

•s2−s1=

cP

ln(

T2

T1

)

+R

ln(

v2v1

)

⇒ln

(

T2

T1

)

=−

RcV

ln(

v2v1

)

=−(k

−1)ln

(

v2v1

)

=ln

(

v1v2

)

(k−

1)

⇒T

2T

1=

(

v1v2

)

(k−

1)

(idea

lgas,s

1=

s2 ,co

nsta

ntk)

•s2−s1=

cP

ln(

T2

T1

)

−R

ln(

P2

P1

)

⇒ln

(

T2

T1

)

=RcP

ln(

P2

P1

)

=(k−

1)

kln

(

P2

P1

)

=ln

(

P2

P1

)

(k−

1)

k

⇒T

2T

1=

(

P2

P1

)

(k−

1)

k(id

ealgas,s

1=

s2 ,co

nsta

ntk)

⇒Pvk=

consta

nt

(idea

lgas,s

1=

s2 ,co

nsta

ntk)

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

52

/56

Entro

py

ofa

Pure

Substa

nce

T1081

Polytro

pic

processes

on

P-v

and

T-s

diag

rams

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

53

/56

Entro

py

ofa

Pure

Substa

nce

CengelEx.

7.2

:⊲

Air

iscom

pressedin

acar

enginefrom

22oC

and95

kPa

ina

reversibleand

adiabaticm

anner.If

thecom

pressionratio,

rc=

V1 /V

2of

thisengine

is8,

determine

thefinal

temperature

ofthe

air.

T190

T2

T1

=

(

v1

v2

)

(k−

1)

⇒T

2=

T1

(

v1

v2

)

(k−

1)

=295

(8)1.4−

1=

677.7

K⊳

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

54

/56

Entro

py

ofa

Pure

Substa

nce

Exam

ple

:⊲

Determ

inethe

changein

specifi

centropy,

inK

J/kg-K,of

airas

anideal

gasundergoing

apro

cessfrom

300K

,1

barto

400K

,5

bar.Because

ofthe

relativelysm

alltem

perature

range,we

assume

aconstant

valueof

cP

=1.008

KJ/kg-K

.

∆s

=cP

lnT

2

T1

−R

lnP

2

P1

=

(

1.008

kJ

kg.K

)

ln

(

400K

300K

)

−

(

8.314

28.97

kJ

kg.K

)

ln

(

5bar

1bar

)

=−

0.1719

kJ/kg.K⊳

•N

otethat,

forisentropic

compression,

T2s=

T1 (P

2 /P

1 )(k−

1)/k=

475K

.H

ence,entropy

changeis

(-)ve

because

ofco

olingof

airfrom

475K

to400

K.

•Com

ment

onthe

resultsis

final

stateis

5bar

and500

K.

©D

r.M

d.

Zahuru

lH

aq

(BU

ET

)Second

Law

ofT

herm

odynam

ics

ME6101

(2021)

55

/56

Entro

py

ofa

Pure

Substa

nce

Exam

ple

:⊲

Air

iscontained

inone

halfof

aninsulated

tank.T

heother

sideis

completely

evacuated.T

hem

embrane

ispunctured

andair

quicklyfills

theentire

volume.

Calculate

thesp

ecific

entropychange

ofthe

isolatedsystem

.

T022•

du=δq−δw

•s2−s1=

cvln

(

T2

T1

)

+R

ln(

v2

v1

)

⇒w

=0,q

=0⇒

du=

0⇒

cVdT

=0⇒

T2=

T1 .

⇒s2−s1=

cvln

(

T2

T1

)

+R

ln(

v2

v1

)

=0+

287ln(2)=

198.93

kJ/kg.K⊳

•N

otethat:

s2−s1=

198.93

kJ/kg.K>

(

δqT

)

︸︷︷

︸0

⇒∆s>

0

©D

r.M

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