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