termo2-1-bab7-2013-eksergi

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TopikTopik 11

Analisis Eksergi

Exer Anal sis

1

Dosen: Dr. Ir. I Made Astina, M.Eng.

Copyright 2006-2014, Dr. Ir. I Made Astina, M.Eng.

TujuanTujuan PembahasanPembahasan TopikTopik iniini

Memahami konsep dan definisi eksergi, tingkat keadaanmati (dead state) dan evaluasi eksergi

Memahami pengembangan aliran eksergi pada sistem

tertutup, perpindahan eksergi, dan contoh-contoh

evaluasi dan aplikasinya.

Memahami neraca aliran eksergi pada volume atur

Dapat implementasi aplikasi aliran eksergi pada mesin-

mesin konversi energi

Memahami konsep dan dapat mengevaluasi efisiensi

2

Dapat mengaplikasikan konsep energi dan aspek

ekonomi yang berkaitan dengan proses perancangan

dalam sistem-sistem termal.

Copyright 2006-2014 Dr. Ir. I Made Astina, M.Eng.


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IntroduceIntroduce

Energi dapat dikonservasikan dalam setiap peralatan ataupun

proses. Energi yang dihasilkan dari suatu bahan bakar,elektrisitas, aliran suatu zat, dan lain-lain dapat dihitung

sebagai suatu produk.

What is illustrated

in these figures?

3Copyright 2007-2014, Dr. Ir. I Made Astina, M.Eng.

Availability, Exergy (Kemanfaatan, Eksergi)Availability, Exergy (Kemanfaatan, Eksergi)

Dua sistem pada tingkat keadaan berbeda, akan

dibawa dalam suatu komunikasi untuk kerja

Second Law of Thermodynamics

asar, apa em ang an se aga s s em

diperbolehkan dalam menuju kesetimbangan.

Ketika satu dari dua sistem adalah sebuah sistem ideal

yang disebut sebagai lingkungan dan sistem yang lain

adalah sistem tertutup dari ketertarikan terhadap

lin kun an.

4

Eksergi didefinisikan sebagai kerja teoritik maksimum

yang dapat diperoleh sistem keseluruhan yang

berisikan sistem dan lingkungan karena sistem menuju

kesetimbangan dengan lingkungan.



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Tingkat Keadaan Mati (Tingkat Keadaan Mati (Dead StateDead State))

Suatu keadaan dengan besaran keadaan tetap dari suatu zatdengan suatu kesetimbangan yang dikhayalkan dan dibatasi

dalam kurungan tak tembus aliran massa, dan relatif diam

terhadap lingkungan, kondisi ini disebut sebagai tingkat

keadaan mati.

No interaction between system and environment and no

potential for developing work.

5Copyright 2007-2014 Dr. Ir. I Made Astina, M.Eng.

Dead State

Exergy DiagramExergy Diagram

6

Exergy is a function of the state of the surroundings as well

as the state of the system.


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Perubahan Energi Lingkungan

7

eoeoe VpSTU =

Exergy EvaluationExergy Evaluation

)()()(E ooooo SSTVVpUE +=

Eksergi dari suatu sistem dievaluasi dengan hubungan:

PEKEUE ++= Energi

V Volume

S

8

Uo

Vo

So

po

ToTingkat Keadaan Mati



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Specific ExergySpecific Exergy

Meskipun eksergi adalah sifat ekstensif, sifat ini juga dapat diubahmenjadi sifat intensif. Hubungannya menjadi:

gz2

)()()(e +++= ooooo ssTvvpuu

Divided by mass to find intensive properties!

9

Energy and Exergy ComparisonEnergy and Exergy Comparison

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Take Note!Take Note!

In this course, E and e are used for exergy and specific exergy,

whileEand e denote energy and specific energy, respectively.

.

approximate concept, exergy or energy will be clear in the

context.

Be care with the symbol. Third edition and before using A

and a for exergy

11


Example 1Example 1

It is known that the air inner

piston-cylinder assembly as

shown in the fi ure. Calculate the

exergy and specific exergy, and

take the dead state, To=300 K and

po=1.01325 bar

Assumption?

12



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Exergy Balance DevelopmentExergy Balance Development

First and Second Laws

=2

12 WQEE

1

+=2

112

T

QSS

=2 2

1212 )()(

ooo TWQ

TQSSTEE

13

Exergy Balance for close system becomes


1 1 b

=2

11212 )1()(EE o

b

oo TWQ

T

TVVp

Neraca Eksergi Sistem TertutupNeraca Eksergi Sistem Tertutup

=2

11212 ))(()1(EE oo

b

o TVVpWQT

T

Ed>0 : ada ireversibilitas pada sistem

Exergy ChangeExergy Transfer

Exergy

Destruction

T=E

14

Ed= 0: tidak ada ireversibilitas o


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Exergy Transfer Accompanying HeatExergy Transfer Accompanying Heat

QT

T

b

o

= 1Eq

15


Exergy of WorkExergy of Work

( )12wE VVpW o =

16



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Exergy Rate for Closed SystemExergy Rate for Closed System

dE)()1(E

&&& =dt

dVpWQ

T

T

dt

doj

o

Steady State0

E=

dt

d

17

&&

od T=

E

Example 2Example 2

Takepo=101325

Pa and To=300 K.

Find exergy

destruction in the

wall?

18

Hint: =e

e

b

o

i

i

b

o QT

TQ

T

T&&& )1()1(Ed


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Flow ExergyFlow Exergy

19

)(E 12w VVpW o =

Exergy transfer accompanying work

Exergy Balance Development for Control VolumeExergy Balance Development for Control Volume

Exergy transfer accompanying work

( )eoe vpmW=

Time rate of the exergy transfer accompanying work

( )eoe

tt

vpt

m

t

W

=

limlim

00

W&

On the other hand:

20

eeet vpmt = m0

( ) ( )eoeeoe

t

vpmvpt

m&=

lim

0



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ContdContd

Time rate of energy transfer accompanying mass flow

++= gzVum

2

2

&

me rate o exergy trans er accompany ng mass ow

Time rate of exergy transfer accompanying mass flow and flow work

( ))()()( ooooo ssTvvpuem += &

)(e vppv o+=

21

gz2

)()()(e2

+++= V

ssTvvpuuooooo

gz2

)()(e2

++= V

ssThh ooo

Exergy Rate Balance for Control VolumeExergy Rate Balance for Control Volume

+

=e

dfee

iifi

cvocv

j

j

j

o mmdt

dVpWQ

T

T

dt

dEee1

Ecv &&&&&

0E

==dVd

cvcv Steady State

22

( ) dffcvo mWQT

TEee10

21

&&&& +

=

+

=

e

dfee

iificv

j

j

j

o mmWQT

TEee10 &&&&&

By taking single inlet denoted as 1 and single outlet denoted as 2

where

22

)(2

)()(ee 2121212121 zzgssThh off ++=

&&od T=E

Exergy destruction rate can also be evaluated from the second law


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Example 3Example 3

A rigid tank is divided into two equal parts by a partition.One part is filled with 1.5 kg of compressed liquid water

at 300 kPa and 60oC, while the other side is evacuated.

The partition is removed and water expands into the

entire tank so that the pressure becomes 15 kPa.

Determine the exergy destroyed during the process.

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SolutionSolution

24


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Solution (contd)Solution (contd)

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Example 4Example 4

An insulated cylinder is initially filled with saturated

liquid water at 150 kPa. Initial volume of the cylinder

is 0.02 m3.The water is heated electrically (2200 kJ)

at constant pressure. Determine the minimum work

by which this process can be accomplished and the

exergy destroyed.

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SolutionSolution

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Example 5Example 5

An insulated cylinder is initially filled with saturated

- . .

a reversible manner until the pressure drops to 0.2

MPa. Determine the change in the exergy of the

refrigerant during this process and the reversible

work.

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SolutionSolution

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Second Law (Exergetic) EfficiencySecond Law (Exergetic) Efficiency

Untuk sistem tertutup

yang menerima panas

dengan laju Qs pada Ts

panas Qu pada Tu, serta

sejumlah panas dibuang

ke lingkungan Ql pada Tl

dE&&&&

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dt us

dE111E

=

dt

dVpWQ

T

TQ

T

TQ

T

T

dt

dol

l

ou

u

os

s

o &&&&


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Second Law (contd)Second Law (contd)

Untuk kondisi tunak dan tidak ada interaksi kerja

lus QQQ &&& +=

dl

l

ou

u

os

s

o QT

TQ

T

TQ

T

TE111 &&&& +

+

=

s

u

Q

Q

&

&

=sso

uuo

QTT

QTT

&

&

)/1(

)/1(

=dan

33

=

so

uo

TT

TT

/1

/1

The second-law efficiency is a measure of the

performance of a device relative to its performance

under reversible conditions. It differs from the first-law

Second Law (Contd)Second Law (Contd)

.

Verify!

?revthth

,

=

34

=

==

H

Lin

cv

in

H

o

ocvw

T

TQ

W

QT

T

dt

dVpW

11E

E

in &

&

&

&

&

&

In case: dV/dt= 0 and TL=To


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Definition of Second Law EfficiencyDefinition of Second Law Efficiency

General definition

Work-Producing Device

Work-Consuming device In next slide, the

second law efficiency

35

Refrigerator and Heat Pump

w e er ve rom

the exergy balance for

each component

Turbine EfficiencyTurbine Efficiency

cvdffcvjo mWQ

T

TE)ee(10 21&&&& +

=j

cvdcvff Wm E)ee( 21

&&& +=

cv

mW

= &&

36

21 ee ff


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Pump and Compressor EfficienciesPump and Compressor Efficiencies

cvd Wm &&& =+ E)ee( 12

Wcv

ff

&

&

= 12

ee

37

m

Unmixed Heat ExchangerUnmixed Heat Exchanger

34 mc

cvdfcfhfcfhcv

j

j

j

o mmmmWQT

TE)ee()ee(10 4231&&&&&&& +++

=

&&&

1 2mh

38

cvdffcffh mm eeee 3421 +=

)ee(

)ee(

21

34

ffh

ffc

m

m

=

&

&


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Mixed Fluid Heat ExchangerMixed Fluid Heat Exchanger

2

3

mm Eeeee &&& +=

cvdfchfcfhcv

j

j

j

o mmmmWQT

TEe)()ee(10 321&&&&&&& +++

=

39

cv

)ee()ee(

31

23

ffh

ffc

mm

=

&

&

Valve and Nozzle EfficienciesValve and Nozzle Efficiencies

inlet

exit

E

E

&

&

=

40


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Example 6Example 6

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TermoeconomicsTermoeconomics

42


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Economic BehaviorEconomic Behavior

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Exergy FlowExergy Flow

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Example 7Example 7

Steam is accelerated in an adiabatic nozzle. Thesteam enters at 3.5 MPa, 300oC and exits at 1.6 MPa,

250oC. Determine the exit velocity, the rate of exergy

destruction, and the second-law e iciency.

Steady state

Potential ener chan e are ne lected

Assumption:

45

Velocity of steam at entrance is small -> zero

Analyzed as volume control

SolutionSolution

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47

Example 8Example 8

Refrigerant-134a is expanded adiabatically in an

expansion valve from 1.2 MPa, 20oC to 180 kPa. Dead

state is taken at 100 kPa, 20oC. Determine the work

- , ,

the second-law efficiency.

48


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SolutionSolution

49

SummarySummary

Exergy concept was introduced in this chapter. It

includes development of exergy balance. Exergy

anal sis exam les were also ex lained. Economic

and exergy analysis was also considered to motivate

the further study and application of thermo-

economics.

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termo2-1-bab7-2013-eksergi

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