Module 4
Open Systems
Lecture 4.1
Basic Equations
ANALYSIS OF OPEN SYSTEMSFLOW PROCESS VISUALISED AS A SERIES OF NON-FLOW PROCESSES
System boundaries
Control Surface
Mt
time : t
δmi
δme
δme
Mt+dt
time : t + dt
δmi
ANALYSIS OF OPEN SYSTEMS
Conservation of MASS
ettit mMmM δδ δ +=+ +
tdttei MMmm −=− +δδ{Net flow into C.V.} = {Increase of mass
within CV}
ANALYSIS OF OPEN SYSTEMS
0=−+−+
tm
tm
tMM iettt
δδ
δδ
δδ
or
∑∑ −= eicv mm
dtdM
&&
Rate of increase of mass within CV
Inst. Flow rate of mass leaving the CV
Inst. rate of flow of mass entering the CV
FIRST – LAW FOR C.V.
νδ cW
Qδ
ttM δ+
ttE δ+
2
tttime δ+:
Pe Ve Te ee
Control Surface
System boundaries
Et
Mt
Time : t
Qδ 1νδ cW
Pi Ti
ei Vi
FIRST – LAW FOR C.V.
Total Work done on the System
+= νδ cWWork transfer associated with boundary deformation
FIRST – LAW FOR C.V.
eimmWW eeiic δυδυδ ν Ρ−Ρ+=12
..volsp ..volsp
( )[ ]{ }iii mP δυ−×−
Work transfer across deforming boundaries
{ }eee mP δυ×−
FIRST – LAW FOR C.V.
For system undergoing the process 1-2
first law gives
121221 WEEQ −−=
Where eettiit meEEmeEE δδ δ +=+= +21 ;
Sp. energy of incoming
stream
Sp. energy of outing stream
FIRST – LAW FOR C.V.
or
tmP
tmvP
tW
tmeme
tEE
tQ
iieeecv
iieettt
i
δδυ
δδ
δδ
δδδ
δδδ
−+
−−
+−
= +12
FIRST – LAW FOR C.V.
or( )
tW
ePtm
tEE
Petm
tQ
CVeee
e
tttiii
i
δδ
υδδ
δυ
δδ
δδ
−+
+−
=++ +
)(
12
2
2VgzueWriting ++=
FIRST – LAW FOR C.V.
22
22i
iii
iiiiiiiVgzhVgzPuPe ++=+++=+ υυ
22
22e
eee
eeeeeeeV
gzhV
gzPuPe ++=+++=+ υυ
0→tasitlimtakingand δ
FIRST – LAW FOR C.V.
∑ =+++dtdE
gzVhmQ V
V
Ci
iiiC )(
2
2
&&
∑ −+++VCe
eee WgzV
hm && )2
(2
SPECIAL CASES
}{,
0
timeoffnnotttanConsmmdtdE
eL
CV
→
=
&&
→ SS SF
FIRST – LAW FOR C.V.
∴ For 1 stream entering & 1 stream leaving
mmm ei &&& ==
& S.F.E.E. becomes
)}(2
){(`22
ieie
ieCvCv zzgVV
hhmWQ −+−
+−=+ &&
FIRST – LAW FOR C.V.
Defining energy transfers as per unit mass of flowing stream
mW
wmQ
q cvcv
&
&
&
&==
SFEE becomes :
( ) ( )ieie
io gVV
hhwq Ζ−Ζ+−
+−=+2
22
Lecture 4.2
Basic Equations….contd
FIRST – LAW FOR C.V.Transient Analysis
Uniform State Uniform Flow assumption
The state of mass within c.v. may change with time, but is uniform throughout the cv at every instant
The state of streams entering / leaving the cv is constant with timealthough their may be time varying.m&
FIRST – LAW FOR C.V.Integrating the basic eq. over t, the
process duration
∫ ∫∫ =
+++ dt
dtdE
dtgzV
hmdtQ CVi
iiiCV 2
2
&&
Const.
∫∫∑ −
+++ dtWdtgz
Vhm CVe
eee
&&2
2
Const.
FIRST – LAW FOR C.V.We get
( )+−=
+++∑ 1122
2
2eMeMgz
VhmQ i
iiiCV
∑
++ e
eee gzVhm
2
2
CVW−
Solving few problems
Lecture 4.3Second Law Analysis of
Open systems
SECOND LAW ANALYSIS OF C.V.
Control Surface
System boundaries
Et
Mt
TQδ 1 νδ cW
imδ
st
νδ cWQδ
ttM δ+
ttE δ+
2
emδ
T
ttS δ+
eett smSS δδ += +2iit1 smSS δ+=
SECOND LAW ANALYSIS OF C.V.
Second Law e.q.TQSS δ
≥− 12
TtQ
tSS δδ
δ/12 ≥
−or
TtQ
tmsms
tSS iieettt δδ
δ
δδ
δδ ≥
−+
−⇒ +
Taking limit as 0→tδ
SECOND LAW ANALYSIS OF C.V.
TQmsms
dtdS
iieecv
&&& ≥−+
In order to permit the possibility of additional streams entering / leaving the c.v. & h.t. across different portions of c.v. with different temps.
∑∑∑ ≥−+ TQsmsmdtdS
iieecv &&&
SECOND LAW ANALYSIS OF C.V.
∑∑∑ ≥−+ TQsmsmdtdS
iieecv &&&
Equality sign for reversible process∴ for an irreversible process
σ&&&& +=−+ ∑∑∑ TQsmsmdtdS
iieecv
Where σ is the entropy generationdue to irreversibility
SECOND LAW ANALYSIS OF C.V.
iieecvcv smsmTQ
dtdS ∑∑∑ −+−= &&
&&σThus
Second law statement becomes :
0≥σ&SPECIAL CASES
with one stream entering & one stream leaving
SSSF Processes
SECOND LAW ANALYSIS OF C.V.
0; ===dtdSmmm cv
ie &&&
∴ 2nd law eq. becomes ( ) ∑≥−TQssm cv
ie
&&
for an adiabatic process, 0=cvQ&
0≥−∴ ie ss
SECOND LAW ANALYSIS OF C.V.
Reversible adiabatic process :0=− ie ss
∴ from basic eq. relating properties
dpdhTds υ−=
∫=−∴=e
iie dphhds υ,0it follows
SECOND LAW ANALYSIS OF C.V.
Using first law
wghghq ee
eii
i −Ζ++=Ζ+++22
22 νν
Since 0=q
( )ieiee
izzg
VVdpw −+
−+= ∫ 2
22
υ
SECOND LAW ANALYSIS OF C.V.
Reversible isothermal process
( ) Qssm cv
&& =−Second law eq.
Tie
( ) qmQ
ssT cvie ==−∴
&
&
dpdhTds υ−=Again from basic eq.
SECOND LAW ANALYSIS OF C.V.
( ) ∫−−=−e
iieie dphhssT υ
wgV
hgV
hq ee
eii
i −Ζ++=Ζ+++22
22
( )ieie gVV
dpw Ζ−Ζ+−
+= ∫ 2
22
υ
= q (from above)First law eq.
SECOND LAW ANALYSIS OF C.V.Since above eq. is valid both for rev. isothermal process & rev. adiabatic process, it is valid for any reversible SSSF process since
Any rev. process
≡ series of alternate adiabatic + isothermal
processes
Specific cases : Incompressible flow, processes involving negligible ∆ KE ∆ PE
SECOND LAW ANALYSIS OF C.V.
Specific cases : Incompressible flow, processes involving negligible ∆ KE ∆ PE
∫=e
i
vdPw
Specific cases : Incompressible flow, no work transfer
02
)()(22
=−
+−+− ieieie
VVzzgPPv
Bernoulli’s equation
End of Lecture
Lecture 4.4Second Law Analysis of
Open systems
SECOND – LAW ANALYSIS OF SOME SIMPLE PROCESSES
(i) Steady flow through a turbine
i
Pi
Pe
ee,r
s
h
mmmCont ei &&& ==:.
0, ≈∆∆ PEKEIst Law :
0=CVQ&
SECOND – LAW ANALYSIS OF SOME SIMPLE PROCESSES
i
e
SHW)( ei hhmW −=− &
SHW=
{shaft work output}
SECOND – LAW ANALYSIS OF SOME SIMPLE PROCESSES
)( ie ssm −= &&σIInd Law :
)( ,, reirSH hhmW −= &For rev process
ire ss −== ,0σ&
rei
ri
rSh
Sh
hhhh
WW
,, −−
==η
ηisentropic
)(& ,, reeSHrSH hhmWW −=− &
SECOND – LAW ANALYSIS OF SOME SIMPLE PROCESSES
Now, from basic eq. Tds = dh + vdP it follows that, at constant pressure
( )∫ −≈=−∴=e
rereeavere ssTTdShhTdsdh
,,.,
∫−=
e
reereave Tds
ssT
,,
1where
SECOND – LAW ANALYSIS OF SOME SIMPLE PROCESSES
)(,, ereaveShrSh ssTmWW −=−∴ &&
( ){ } σ&& aveieave TssmT ,, =−=
)1(, η−= rsW
Lost Work = Av. Temp. * Entropy Generation
ave
sr
TW
,
)1( ησ
−=&
Solving a few problems
Show that for an ideal gas passingthrough a diffuser the actualexit pressure Pe is related to
the ideal exit pressure Per by the relation :
)exp(Rm
PP ere &
&σ−=
Where symbols have their usual meanings
SECOND – LAW ANALYSIS OF USUF PROCESS
∑∑∑ ≥−+ TQsmsmdtdS
cviieecv &&&
Integrating over a time interval t, since se, si are constant over t
SECOND – LAW ANALYSIS OF SOME SIMPLE PROCESSES
[ ] ( )∫ ∑∑∑ ≥−+−t cv
iieevc dtTQ
smsmsMsM0..1122
&
( ) dtTQdtTQ cvcv ∑∫∑∫ = )/( &&
Solving a few problems
Filling of an Evacuated Tank
Tank T0
T0 P0
Heat Transfer Q (Environment at T0)
Filling of an Evacuated Tank
Continuitydt
dMmm cv
ii ==− && 0
iicv mtmM ==⇒ &
=The total mass ofair admitted
Filling of an Evacuated Tank
Ist Law ( ) 00 2 −+=+ umhmQ iiicv
( )iicv humQ −= 2{ Final state same as entering air
ii pm υ0
−=
{ }containerofvolVVp :0
−=
iiii phuu υ−==2
op
Filling of an Evacuated Tank
IInd Law
022 T
Qsmsm cvii −−=σ&
02 T
Qsmsm cviii −−=
o
cv
TVp
TQ 0
0
=−=
final state of C.V.= T0, P0 i.e. environ state
∴s2=si
Filling of an Evacuated Tank
Remarks
== VpT00 σ&
Work done by atmosphere during the filing process
How could we do reversible filing ?
Replace valve by reversible expander
End of Lecture
Lecture 4.5Availability
GOUY-STODOLA THEOREM
Gen.
Relationship
between
Entropy
Gen. & lost
work in irrev.
processes
inm&
T1
P0T0 Q0
T2 T3
1Q& 2Q& 3Q&
outm&
dtdV
SHW&
Thermal energy reservoirs
Atmospheric Press & Temp. Reservoir
all modes of Work Transfer
GOUY-STODOLA THEOREMIst Law
∑∑ ∑
+++=
+++ e
eee
CVi
ii
n
ii gzVhmdtdEgzVhmQ
22
22
0
&&&
SHW&+
define : 02
2 iii
i hgzVh =++ : generalised enthalpy or methalpy
∑∑∑ −−++=∴dtdEhmhmQQW CV
eeii
n
iSH00
10 &&&
GOUY-STODOLA THEOREM
IInd Law
∑ ∑∑ ≥−+−==
00
iiee
n
i i
iCV smsmTQ
dtdS
&&&σ
or
σ&&& 01
00 TsmsmTQ
dtdS
TQn
iiiee
i
iCV −
−+−= ∑ ∑ ∑=
GOUY-STODOLA THEOREM
Eliminating Q0 between these two eqs. gives
∑∑ ∑=
+
−+−−=
n
ii
CVeeiiSH Q
TT
dtdEhmhmW
1 1
000 1&&&
( ) σ&&& 0000 TSTdtdTsmTsm CViiee −+−∑ ∑
GOUY-STODOLA THEOREMor ( ) ( )∑∑ −−−= eeeiiiSH sThmsThmW 0
00
0 &&&
( ) i
n
iCVCV Q
TTSTE
dtd ∑
=
−+−−
1 1
00 1
σ&0T−
GOUY-STODOLA THEOREM
( ) ( ) 0max === σ&&&& forWWWSHrevSHSH
and Loss of work due to irreversibility
LAW = I = σ&0Ti.e.
GOUY-STODOLA THEOREM
0;0 ≥≥ Iσ&Since
=> If WSH is +VE {work output}
revSHSHSH WWW ,max, =≤
If WSH is -VE {work input}
revinputinput WW ,≥
GOUY-STODOLA THEOREMFor SSSF process this gives
( ) ∑
−++= ii
iiiSH STgzVhmW 0
2
max 2&&
∑
−++− ei
eee STgzVhm 0
2
2&
∑=
−+
n
iiQT
T1 1
01
define h - T0S = b, availability Is it a thermodynamic property?
AVAILABILITY BALANCE
( ) ( )
−+−= ∑∑∑∑ 22
22
maxe
ei
ieeiiSHVmVmbmbmW &&&&&
flow availability ∆ K.E.
( ) i
n
ieeii Q
TTgzmgzm ∑∑∑
=
−+−+
1 1
01&&
“availability” of heat interaction
∆ P.E.
( )101 TTQi −max,SHW&
LostW&
∑ eebm&∑ iibm&
AVAILABILITY BALANCE
NB While changes in KE & PE are fully converted to work, changes in enthalpy h & h.t. Qi are notfully converted
GOUY-STODOLA THEOREM
K.E. & P.E. are “ordered” forms of energy
Q, U => “disordered” energy stored in the form of random molecular motionWhat are others ordered forms of energy?
MAX WORK IN A CLOSED SYSTEM
0,0,0 === σ&&& ei mm
( )CVCViSH STEdtd
TTQW 0
1
01 −−
−=∑ &&
Integrating over process duration
( ) ( )[ ]20101
01 STESTETT
QWiSH −−−+
−=∑
MAX WORK IN A CLOSED SYSTEMUseful Work = Shaft work –
Work done against atmospheric pressure
( )120 VVPWW ShUS −−=
( )[ 1001
01 STVPETTQi −++
−=∑
( ) ]200 STVPE −+−
Closed system availability
ASTVPE ≡−+ 00
MAX WORK IN A CLOSED SYSTEM
sTPua 00 −+= υ(per unit mass basis)
Max work obtainable from a closed system exchanging heat only with environment
Wmax = a1 – a2
End of Lecture
Lecture 4.6EXERGY
EXERGYMaximum useful work output possible from a given stream in any s.f. process(neglecting ∆KE & ∆PE) ex = b – b0 = (h – h0) – T0 (s-s0)
Components of exergy (thermo-mechanical)
Useful work arising from press. difference w.r.t. P0 & that from temp. difference w.r.t. T0
EXERGYConceptual Visualisation
RH
ISOBARIC COOLINGP1 > P0
T1 > T0
1 I
Isothermal exp.
P1 , T0 P0,T0
∆Pex
exex
∆T
ENVIRONMENT AT T0
EXERGY
( ) ( )∫∫ −−=
−
−=∆
0
11
0
1
001
T
TP
T
T
Tx dsTTdh
TTe
( ) ( )0000 sThsThe iiP
x −−−=∆
( ) ( )ii hhssT −−−= 000
Tx
Pxx eee ∆∆ += {verify}
FLOW EXERGY OF AN IDEAL GAS
−−−=
0000 )(
PPnlR
TTnlCTTTCe ppx
P>P0
P=P0
P<P0
T0T >T0T <T0
ex
T
Tsh
P
=
∂∂
Implies
S
h
e x1
S0S1
S0 – S1
e x=0
e x2=c
onstant
e x1=C
o
1
A
P0
h 1–h
0
T 0(S
0 –S
1)λ λ λ
T0
0tan T=λ
ntFLOW EXERGY FROM h-S CHARTS
FLOW EXERGY FROM h-S CHARTSHence 1 – A ≡ ex1
Further since ex = (h – T0S) – (h0 – T0S0)
0TSh
Se x −
∂∂
=∂∂
0;0 =∂∂
=∂∂
SeT
Sh x∴ For
=> ex = Const for lines II to tangent to const P0 curve.
EXERGY ANALYSIS – SIMPLE PROCESSES
EXPANSION IN TURBINE
2
1
SHW T0
T1
S
2́ 2a bc d
.. . .
. . . .
. . . .
.. . .
. . . .
. . . .
.. . .
. . . .
. . . . .. . . . . . .
.. . . . . . . . .
. . . .
(adiabatic) for simplicity
IWee Shxx ++=21
)( 120 ssTI −=
EXERGY ANALYSIS – SIMPLE PROCESSES
21 xx
SH
eeW−
=ψSecond Law Efficiency
21
1xx ee
I−
−=
21
21
xx eehh
−−
= Grassman Diagr.
SHW
2xe
1xe
I
{assuming mech. losses to be small
EXERGY ANALYSIS – SIMPLE PROCESSES
Compare with isentropic efficiency
21
21
hhhh
S ′−−
=η
( ) )( 12021
21
ssThhhh
−+−−
=ψ
= iSp. Irreversibility{area a b d c a}
EXERGY ANALYSIS – SIMPLE PROCESSES
( ) )( 2221
21
hhhhhh
S ′−+−−
=η
Frictional reheat{area 2 2΄ c d 2}
T0
T1
S
2́ 2a bc d
.. . .
. . . .
. . . .
.. . .
. . . .
. . . .
.. . .
. . . .
. . . . .. . . . . . .
.. . . . . . . . .
. . . .
= r
End of Lecture
Lecture 4.7Exergy Analysis-Flow processes
Exergetic (2nd Law) Efficiency
• 2nd Law implies :
( ) ( ) Iexergyexergyinputoutput
−= ∑∑
Second Law Efficiency
( )( )∑
∑=
input
output
exergy
exergyψ
EXERGY ANALYSIS – SIMPLE PROCESSES
EXPANSION IN TURBINE
2
1
SHW T0
T1
S
2́ 2a bc d
.. . .
. . . .
. . . .
.. . .
. . . .
. . . .
.. . .
. . . .
. . . . .. . . . . . .
.. . . . . . . . .
. . . .
(adiabatic) for simplicity
IWee Shxx ++=21
)( 120 ssTI −=
EXERGY ANALYSIS – SIMPLE PROCESSES
21 xx
SH
eeW−
=ψSecond Law Efficiency
21
1xx ee
I−
−=
21
21
xx eehh
−−
= Grassman Diagr.
SHW
2xe
1xe
I
{assuming mech. losses to be small
EXERGY ANALYSIS – SIMPLE PROCESSES
Compare with isentropic efficiency
21
21
hhhh
S ′−−
=η
( ) )( 12021
21
ssThhhh
−+−−
=ψ
= iSp. Irreversibility{area a b d c a}
EXERGY ANALYSIS – SIMPLE PROCESSES
( ) )( 2221
21
hhhhhh
S ′−+−−
=η
Frictional reheat{area 2 2΄ c d 2}
T0
T1
S
2́ 2a bc d
.. . .
. . . .
. . . .
.. . .
. . . .
. . . .
.. . .
. . . .
. . . . .. . . . . . .
.. . . . . . . . .
. . . . = r
ex2 - ex2’ = r - i (Prove it )
HEAT TRANSFER PROCESSESIsobaric Heat Transfer
2b
2a 1a
1b 1b
2a 1a
2b
2b
2a
T 1a
1b
L
T2b
2a
1a
1b
LPARALLEL FLOW H.E.COUNTER FLOW H.E.
HEAT TRANSFER PROCESSES
T0
T
genS&
Pa
2b2a
1a
1b
Pb
S&(a)
1b
1a2a
2b
T0
T
2b2a
1a
1b
genS& S&(b)
COUNTER FLOW H.E. PARALLEL FLOW H.E.
Isobaric Heat Transfer
2b
2a 1a
1b
HEAT TRANSFER PROCESSES
])()[( 121200 bbbaaa mssmssTTI &&& −+−== σ
)]()[( 12120 bbaa SSSST &&&& −+−=
( ) ( )bxbxbaxaxa eemeemI 1221 −−−= &&
Also from exergy balance
{are these expressions equal to each other}
HEAT TRANSFER PROCESSES
streamhotofdecreaseExergystreamcoldofincreaseExergy
HE =ψ
)()(
12
12
axaxa
bxbxbHE eem
eem−−
=&
&ψ
EXERGY ANALYSIS OF A REFRIG PLANT
Cold Chamber Cooled by brine
V
Environment
(T0)
c
M
cond Evap
I
II
IIIIV
1
2
3 4
5
6
TCQC
Working fluid : NH3 Coolant Brine
EXERGY ANALYSIS OF A REFRIG PLANT (ref: Kotas)
Design ParametersQC = 93.03kW T0 = 200C Tc = -10CComp inlet, T1 = -100C Te = -120CComp outlet, T2 = 1190C Condenser temp = 280CCondensate outlet temp = 250C
EXERGY ANALYSIS OF A REFRIG PLANT
9.083.,.)(==
electmotormechCompressor ηη
T5 = - 50C Brine temps. T6 = - 70C
Cp = 2.85 kj/kg KAssumptionsNegligible heat leaks; adiabatic compressionNegligible Press Drop, ∆KE, ∆PE
Negligible power input to brine pump
End of Lecture
Lecture 4.8Exergy
Analysis…contd
EXERGY ANALYSIS OF A REFRIG PLANT
Cold Chamber Cooled by brine
V
Environment
(T0)
c
M
cond Evap
I
II
IIIIV
1
2
3 4
5
6
TCQC
Working fluid : NH3 Coolant Brine
EXERGY ANALYSIS OF A REFRIG PLANT (ref: Kotas)
Design ParametersQC = 93.03kW T0 = 200C Tc = -10CComp inlet, T1 = -100C Te = -120CComp outlet, T2 = 1190C Condenser temp = 280CCondensate outlet temp = 250C
EXERGY ANALYSIS OF A REFRIG PLANT
9.083.,.)(==
electmotormechCompressor ηη
T5 = - 50C Brine temps. T6 = - 70C
Cp = 2.85 kj/kg KAssumptionsNegligible heat leaks; adiabatic compressionNegligible Press Drop, ∆KE, ∆PE
Negligible power input to brine pump
End of Lecture
LECTURE 4.9
Endo-reversible Engines for maximum power output
ENDOREVERSIBLE ENGINES
( )1TTAUQ HHHH −=&
( )cCCC TTAUQ −= 2&
H
c
TT
TT
−<−= 111
2η
cH QQW &&& −=
021
=+−TQ
TQ cH
&&
E
Tc
TH
W
T2
T1
Qc
QH
Ist law
IInd law
ENDOREVERSIBLE ENGINES
−=
−=∴
1
211TT
QQQ
QW HH
cH
&&
&&&
Combine these eqs. to express in terms of
W&
HcH AUTTM
,, and the ratio τ=1
2
TT
τ1 ==→
TQH 1
2
TQ
c
ENDOREVERSIBLE ENGINES
ccc
c TAU
QT +=&
2
+== ccc
c
c
H
c
H TAU
QQQT
QQT
&
&
&&21
+= ccc
c TAU
Q&
τ1
ENDOREVERSIBLE ENGINES
( )1TTAUQ HHHH −=∴
+−= ccc
cHHHHH T
AUQAUTAU&
τ
τcHH
cc
HHHHHH
TAUAUQAUTAU −−=&
ENDOREVERSIBLE ENGINES
Rearranging
== HHH T
AUT
TAUQ τ&
+
−
+
−
cc
HH
c
cc
HH
c
HHH
AUAU
T
AU
T
11
1.1
τ
τ
hence
( )ττ
τ−
+
−= 1
1cc
HH
H
c
HHH
AUAU
TT
TAUW&
ENDOREVERSIBLE ENGINESFor max output with given
UH AH, Uc Ac, Tc & TH
0=∂∂τW&
−
−+
−
+= 2
11
1τ
τττ
H
c
cc
HH
HHH
TT
AUAUTAU
( )( )( )
+−
+= 2
1.1τH
c
ccHH
HccHH
TT
AUAUTAUAU
ENDOREVERSIBLE ENGINES
H
copt T
T=⇒ 2
.τ
21
1
2
==
H
c
TT
TTτi.e. for max. output
21
1
−=∴
H
copt T
Tη
Optimal allocation of heat transfer areas
→ i.e. relative values of UHAH & UcAc
Suppose UHAH+UcAc=Const. (given) <K>
( )( ) { }ττ
−
−
−= 111AUAU HHHH
H
cH T
TT
KK
W&
Optimal Allocation of Heat Transfer Areas
( ) ( )HHHH
AU20AU
−==∂
∂ KW&
( )2
AUAU2
AU ccHH.HH
+==∴
Kopt
or
ccHH AUAU =→
End of Lecture