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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
r.M
d.
Zahuru
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
1/
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
r.M
d.
Zahuru
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
3/
56
Heat
Engin
es
T055
©D
r.M
d.
Zahuru
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
4/
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
r.M
d.
Zahuru
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
5/
56
Heat
Engin
es
Heat
Engine:
Classifi
cations
T058
©D
r.M
d.
Zahuru
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
6/
56
Heat
Engin
es
(Heat)
Engine
T106
T105
T107
©D
r.M
d.
Zahuru
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
7/
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)
8/
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
(2021)
9/
56
Heat
Engin
es
T144
©D
r.M
d.
Zahuru
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
10
/56
Heat
Engin
es
T111
T138
Coeffi
cientof
Perform
ance,COPR=
Desire
dO
utp
ut
Require
dIn
put=
QL
Wnet,in
©D
r.M
d.
Zahuru
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
11
/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
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
12
/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
r.M
d.
Zahuru
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
13
/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
r.M
d.
Zahuru
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
14
/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
r.M
d.
Zahuru
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
15
/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
r.M
d.
Zahuru
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
16
/56
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.
Zahuru
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
17
/56
The
Second
Law
ofT
herm
odynam
ics
T113
Vio
lation
ofK
evlin-P
lanck
(KP)
statemen
t⇒
violatio
nofClau
sius
(C)
statemen
t.
©D
r.M
d.
Zahuru
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
18
/56
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
r.M
d.
Zahuru
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
19
/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
r.M
d.
Zahuru
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
20
/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
r.M
d.
Zahuru
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
21
/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
r.M
d.
Zahuru
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
22
/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
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
23
/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
r.M
d.
Zahuru
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
24
/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
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
25
/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
herm
odynam
ics
ME6101
(2021)
26
/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
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
27
/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
ME6101
(2021)
28
/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
r.M
d.
Zahuru
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
29
/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.
Zahuru
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
30
/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
herm
odynam
ics
ME6101
(2021)
31
/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
ME6101
(2021)
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
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/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
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aq
(BU
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ics
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/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
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aq
(BU
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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
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aq
(BU
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/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
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Zahuru
lH
aq
(BU
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/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
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aq
(BU
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/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
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aq
(BU
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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
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aq
(BU
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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
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(BU
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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
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lH
aq
(BU
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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
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aq
(BU
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Second-L
aw
ofT
herm
odynam
ics
for
CV
Syste
ms
T042
T175
©D
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aq
(BU
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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
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d.
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aq
(BU
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Second-L
aw
ofT
herm
odynam
ics
for
CV
Syste
ms
T757
(a)Energ
ybalan
ce(b
)Entro
pybalan
ce.
©D
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aq
(BU
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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
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d.
Zahuru
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aq
(BU
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ofT
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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
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aq
(BU
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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
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d.
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aq
(BU
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Entro
py
ofa
Pure
Substa
nce
Entropy
ofa
Pure
Substance
T1080
T-s
and
h-s
diag
rams
forsteam
.
©D
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aq
(BU
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herm
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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
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aq
(BU
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/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
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aq
(BU
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Entro
py
ofa
Pure
Substa
nce
T1081
Polytro
pic
processes
on
P-v
and
T-s
diag
rams
©D
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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
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Zahuru
lH
aq
(BU
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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
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aq
(BU
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/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
d.
Zahuru
lH
aq
(BU
ET
)Second
Law
ofT
herm
odynam
ics
ME6101
(2021)
56
/56
