uet bilant sc abur apa

1

BILANT ENERGETIC SI EXERGETIC SCHIMBĂTOR DE CĂLDURĂ ABUR-APĂ

Să se icircntocmească bilanŃul energetic şi exergetic al unui schimbător de căldură orizontal abur-apă icircn contracurent cu suprafaŃa de schimb de căldură S = 1794 m2 Apa circulă prin Ńevi iar aburul condensează icircn spaŃiul dintre Ńevi la ieşirea din aparat de evacuează condensat la saturaŃie Se va admite temperatura mediului ambiant T0 = 293 K NoŃiuni privind funcŃionarea instalaŃiei Schimbătorul de căldură de tip D55-QL-2 este alimentat cu abur de la centrala termică a icircntreprinderii şi are rolul de a prepara apa caldă necesară icircncălzirii incidentelor icircn sezonul rece şi de alimentare a unor consumatori termici icircn tot timpul anului Datele de construcŃie ale schimbătorului sunt - număr tronsoane n1 = 2 - număr Ńevi pe tronson n = 55 - lungimea unui tronson l = 3 m - diametru interior al Ńevii di = 16 mm - diametru exterior al Ńevii de = 21 mm - diametru interior al mantalei Di = 257 mm - grosimea peretelui mantalei 8=δ mm

Fiecare tronson al schimbătorului este de construcŃie semielastică Conturul de bilanŃ - schimbătorul de căldură cu intrările şi ieşirile celor 2 agenŃi termici Mărimi măsurate - debitul de abur G1 = 104 kgs

- temperatura aburului 175

1 =t 0C

- presiunea aburului 6

1 =p bar

- temperatura condensatorului 158

1 =t 0C

- presiunea condensatorului 95

1 =p bar

- debitul de apă G2 = 1855 kgs

- temperatura apei la ieşire 42

2 =t 0C

- presiunea apei la ieşire 4

2 =p bar

- temperatura apei la intrare 69

2 =t 0C

- presiunea apei la intrare 873

2 =p bar

2

Model de calcul 1 Entalpia aburului (

1i )

175

1 =t 0C

6

1 =p bar -gt

1i din tabel apă + abur supraicircncălzit

i170 = kJkg i180 = kJkg i∆ = i180 ndash i170 = kJkg

t∆ = 100Chelliphelliphelliphelliphelliphelliphellip i∆ = kJkg

50Chelliphelliphelliphelliphelliphelliphellip x

1i = kJkg

1i = kJkg

2 Entalpia condensatorului (

1i )

158

1 =t 0C

95

1 =p bar -gt


85i = kJkg

06i = kJkg

i∆ = 06i ndash 85i = 58 kJkg

i∆ = 58helliphelliphelliphelliphelliphelliphellip p∆ = 02 bar

x helliphelliphelliphelliphelliphelliphellip p∆ = 01 bar

95i = 85i + kJkg = kJkg

1i = kJkg

3 Entalpia apei la intrare (

2i )

42

2 =t 0C

4

2 =p bar -gt




20Chelliphelliphelliphelliphelliphelliphellip x kJkg

3

2i = i40 + kJkg = kJkg

4 Entalpia apei la ieşire (

2i )

69

2 =t 0C

873

2 =p bar -gt


Ci 060

= kJkg

Ci 070

= kJkg

i∆ = C

i 070 ndash

Ci 060

= kJkg



2i = C

i 060 + kJkg = 28901 kJkg

2i = kJkg

5 Entropia aburului (

1s )

175

1 =t 0C

6

1 =p bar -gt

1s

S170 = kJkgK s180 = kJkgK s∆ = s180 ndash s170 = kJkgK

t∆ = 100Chelliphelliphelliphelliphelliphelliphellip s∆ = 005 kJkgK


1s = s170 + kJkgK = kJkgK

1s = kJkgK

6 Entropia condensatorului (

1s )

71

1 =t 0C

37

1 =p bar -gt

1s

06s = kJkgK

85s = kJkgK

s∆ =

85

06 ss minus = kJkgK

s∆ = 0013helliphelliphelliphelliphelliphellip p∆ = 02 bar


1s =

85s + kJkgK = kJkgK

4

1s = kJkgK

7 Entropia apei la intrare (

2s )

42

2 =t 0C

4

2 =p bar -gt

2s

S40 = kJkgK S50 = kJkgK s∆ = s50 ndash s40 = kJkgK

t∆ = 100Chelliphelliphelliphelliphelliphelliphellip s∆ = kJkgK


2s = s40 + kJkgK

2s =

8 Entropia apei la ieşire (

2s )

69

2 =t 0C

873

2 =p bar -gt

2s

s60 = kJkgK s70 = kJkgK s∆ = s70 ndash s60 = 0 kJkgK




2s = kJkgK

9 Debitul (fluxul) de căldură cedat (Q1)

)(

1

111 iiGQ minus=

Q1 = kJs 10 Debitul (fluxul) de căldură primit (Q2)

)(

2

222 iiGQ minus=

Q2 = kJs 11 Debitul (fluxul) de căldură pierdut icircn mediul ambiant (Qp) Qp = Q1 ndash Q2

Qp = kJs 12 DiferenŃa medie de temperatură (∆tmed)

5

=

∆∆

∆minus∆=∆

min

max

minmax

lnt

t

tttmed

0C

13 Coeficientul de reŃinere al căldurii (randamentul izolaŃiei) ηr

==1

2

Q

Qrη

ηr = 14Coeficientul global de schimb de căldură (ks)

3

0 10minussdotsdotsdot= StkQ meds ∆ KW

10 3

0

=sdotsdot∆

= minusSt

Qk

med

s

ks = Wm2 0C 15Temperatura medie a apei ( 2t )

meds ttt ∆minus=2 = 0C

16 SecŃiacuteunea de trecere a apei (S2)

4

2

2 =sdot

sdot= idnsπ

m3

17Densitatea apei ( 2ρ )

9152472 =t 0C

4=p bar

040=

Cv m3kg

1

4040

0 ==v

ρ kgm3

050=

Cv m3kg

1

5050

0 ==v

ρ kgm3

6

5040 =minus=∆ ρρρ kgm3

50915247 =+= ρρ kgm3

2 =ρ kgm3

18Viteza de curgere a apei (w2)

22

22 =

sdot=

ρS

Gw ms

w2 = ms 19Vacircscozitatea cinematică a apei ( 2ν )

2 =ν m2s

tm = t2 = 0C

2 =ν m2s

20Numărul Reynolds (Re2)

=sdot

=2

22Re

νidw

Re2 = 21Coeficientul de convecŃie perete-apă (α2)

α2 = Wm2C 22SecŃiunea de curgere a aburului (S1)

44

22

11 =minus

sdot= edn

DS

ππm2

S1 = m2 23Vacircscozitatea dinamică a condensatului la saturaŃie (η1) η1 = Nsm2 24Densitatea vaporilor saturaŃi (ρv)

85vρ = kgm3

06vρ = kgm3

( )20

802

2 04803231430i

mmd

wtt minussdot+=α

7

8506=minus=∆ vv ρρρ kgm3

barp 20=∆ helliphelliphelliphelliphelliphellip kgm3

barp 10=∆ helliphelliphelliphelliphelliphellip x

=vρ kgm3

25Densitatea condensatului la saturaŃie (ρ1)

150ρ = kgm3

160ρ = kgm3


150158 =minus= ρρ kgm3

1 =ρ kgm3

26Numărul Prandtl la saturaŃie (Pr1)

150Pr =

160Pr =

PrPrPr 160150 =minus=∆

Ct 010=∆ helliphelliphelliphelliphelliphelliphelliphellip

Ct 08=∆ helliphelliphelliphelliphelliphelliphelliphellip x

PrPr 150158 =minus=

=1Pr

27Viteza aburului saturat (wv)

1

1 =sdot

=v

vS

Gw

ρms

wv = ms 28Viteza condensatului la saturaŃie (wcd)

11

1 =sdot

=ρS

Gwcd ms

wcd = ms 29Numărul Nunsselt la condensare (Nu1)

80

1

50

1

1

31

1 Pr0260

+

times= cd

v

vve wwd

Nu ρρρ

ρη

8

Nu1 = 30Conductivitatea termică a apei ndash la saturaŃie (λ1)

150λ = Wm0C

160λ = Wm0C

150158 =minus= λλ Wm0C

=1λ Wm0C

31Diametrul echivalent pe partea aburului (1ec

d )

=+minus

==ei

eec

ndD

ndD

P

Sd

22

1

1

141

1=ecd m

32Coeficientul de convecŃie la condensare (α1)

1

111 ==

ecdNu

λα Wm2C

α1 = Wm2C 33Coeficientul global de schimb de căldură al aparatului curat (kso)

e

i

i

e

p

i

so d

d

d

dd

ksdot++=

21

1ln

2

11

αλα

kso = Wm2C 34RezistenŃa termică a depunerilor (Rsd)

11

=minus=sos kk

Rsd Wm2C

Rsd = Wm2C 35EficienŃa procesului de icircncălzire (ηinc)

2

1

2

2 =minusminus

=tt

ttincη

36Randamentul termodinamic (ηtd)

( )( )

0

1

11

2

22

0

1

2 =minusminus

==iiG

iiG

Q

Qtdη

unde

0

1i = entalpia agentului cald la echilibru cu mediu ambiant

p = 59 bar

9

t = 200C rarr =0

1i kJkg

ηtd =

37Exergia specifică a aburului la intrare (

1e )

10

1

1 =minus= sTie kJkg

1 =e kJkg

38Exergia specifică a apei la intrare (

1e )

10

1

1 =minus= sTie kJkg

1 =e kJkg


2e )

20

2

2 =minus= sTie kJkg

40Exergia specifică a apei la ieşire (

2e )

=minus=

20

2

2 sTie kJkg

41Exergia cedată de fluidul primar (∆E1)

( )

1

111 =minus=∆ eeGE kW

42Exergia primită de fluidul secundar (∆E2)

( )

2


43Pierderile de exergie (∆Ep)

21 =∆minus∆=∆ EEEp kW

44Randamentul exergetic (ηex)

1001

2 =sdot∆∆

=E

Eexη

45Pierderea specifică de presiune pentru fluidul rece (

2spp∆ )

=sdot

sdotsdot∆=

∆=∆

Sk

CGp

NTC

pp

s

p

sp2

2

22

2

2 bar

unde NTC = număr de unităŃi de transfer termic de căldură NTC = ksSW

2spp∆ = bar

10

Diagramele SANKEY corespunzătoare bilanŃurilor pentru regimurile analizate Diagrama fluxurilor energetice

Diagrama fluxurilor exergetice

Coeficientul global de schimb de căldură ( )

sk

ks = Wm2C rarr din tabel

sk = 1000divide4000 Wm2C

1

=sdot+

=sdso

sos

Rk

kk

Q2= kJs

Q2rarr

Q1= kJs

Q1rarr100 Qp= kJs

Qprarr

∆E1= kW

∆E1rarr100

∆E2= kW

∆E2rarr

∆Ep= kW

∆Eprarr

11

RezistenŃa termică a depunerilor ( )

sdR

DiferenŃa medie de temperatură ( )

medt

2 =sdot

=∆sk

Qt

s

med

0C

2

1

=∆+== meds tttt0C

Se stabileşte

1p = rarr t1 = C

Se alege presiunea condensatului la ieşire rarr ( )

1i = kJkgK

( )

1i = kJkgK

Coeficientul de reŃinere al căldurii (randamentul izolaŃiei) ηr

1

2 ==Q

Qrη

ηr =

Debitul de abur de icircncălzire

1G

( )

1

1

12 =minus

=ii

QG

rη

1G =

Debitul (fluxul) de căldură cedat (

1Q )

)(

1

1

11iiGQ minus=

1Q = kJs

Debitul (fluxul) de căldură primit (

2Q )

)(

2

2

2

2 iiGQ minus=

2Q = kJs

Debitul (fluxul) de căldură pierdut icircn mediul ambiant (Qp)

2

1

QQQp minus=

pQ = kJs

Se calculează

- Exergia specifică a aburului ( )

1e

( ) ( ) ( )

10

1

1 sTie minus=

1)(i = kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1

1 =minus= sTie kJkgK

12

- Exergia specifică a apei ( )

1e

( ) ( ) ( )

10

1

1 sTie minus=

1 )(i =hellip kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1

1 hellip=minus= sTie kJkgK

- Exergia cedată de abur ( )

1E∆

( )

1

1

1

1 hellip=minus=∆ eeGE kW

- Exergia cedată de apă ( )

2E∆

( )

2

2

2

2 eeGE minus=∆

- Pierderile de exergie ( )

pE∆

2

1

=∆minus∆=∆ EEEpkW

- Randamentul exergetic ( )

exη

1

2 =∆∆

=E

Eexη

BilanŃul energetic - Cantitatea de căldură cedată de abur

1 =Q kJ rarr 100

- Cantitatea de căldură primită de apă

2 =Q kJ rarr hellip

- Cantitatea de căldură pierdută

=pQ kJ rarr hellip

BilanŃul exergetic - Exergie cedată

1 =∆E kW rarr 100

- Exergia primită

2 =∆E kW rarr hellip

- Pierderi de exergie

=∆ pE kW rarr hellip

Ameliorarea funcŃiei aparatului s-a realizat prin - hellip

Efecte - helliphelliphelliphellip

13

Economia anuală de combustibil

( )( )

1

1

11 =sdotsdot

minusminus= n

ic H

iiGGB σ

ηtccan

unde

nσ = timp de utilizare



1Q =hellipkJ

1Q rarr100

2Q =hellip kJ

2Q rarr

pQ =hellip kJ

pQ rarrhellip

1E∆ =hellip kW

1E∆ rarr100

2E∆ =hellip kW

2E∆ rarrhellip

pE∆ =hellip kW

pE∆ rarrhellip

2

Model de calcul 1 Entalpia aburului (

1i )

175

1 =t 0C

6

1 =p bar -gt





1i = kJkg

1i = kJkg

2 Entalpia condensatorului (

1i )

158

1 =t 0C

95

1 =p bar -gt


85i = kJkg

06i = kJkg

i∆ = 06i ndash 85i = 58 kJkg

i∆ = 58helliphelliphelliphelliphelliphelliphellip p∆ = 02 bar


95i = 85i + kJkg = kJkg

1i = kJkg

3 Entalpia apei la intrare (

2i )

42

2 =t 0C

4

2 =p bar -gt




20Chelliphelliphelliphelliphelliphelliphellip x kJkg

3



2i )

69

2 =t 0C

873

2 =p bar -gt


Ci 060

= kJkg

Ci 070

= kJkg

i∆ = C

i 070 ndash

Ci 060

= kJkg



2i = C

i 060 + kJkg = 28901 kJkg

2i = kJkg


1s )

175

1 =t 0C

6

1 =p bar -gt

1s





1s = kJkgK


1s )

71

1 =t 0C

37

1 =p bar -gt

1s

06s = kJkgK

85s = kJkgK

s∆ =

85

06 ss minus = kJkgK



1s =

85s + kJkgK = kJkgK

4

1s = kJkgK


2s )

42

2 =t 0C

4

2 =p bar -gt

2s




2s = s40 + kJkgK

2s =


2s )

69

2 =t 0C

873

2 =p bar -gt

2s





2s = kJkgK


)(

1

111 iiGQ minus=


)(

2

222 iiGQ minus=



5

=

∆∆

∆minus∆=∆

min

max

minmax

lnt

t

tttmed

0C


==1

2

Q

Qrη


3


10 3

0

=sdotsdot∆

= minusSt

Qk

med

s




4

2

2 =sdot

sdot= idnsπ

m3


9152472 =t 0C

4=p bar

040=

Cv m3kg

1

4040

0 ==v

ρ kgm3

050=

Cv m3kg

1

5050

0 ==v

ρ kgm3

6


50915247 =+= ρρ kgm3

2 =ρ kgm3


22

22 =

sdot=

ρS

Gw ms


2 =ν m2s

tm = t2 = 0C

2 =ν m2s


=sdot

=2

22Re

νidw



44

22

11 =minus

sdot= edn

DS

ππm2


85vρ = kgm3

06vρ = kgm3

( )20

802

2 04803231430i

mmd

wtt minussdot+=α

7




=vρ kgm3


150ρ = kgm3

160ρ = kgm3



1 =ρ kgm3


150Pr =

160Pr =




PrPr 150158 =minus=

=1Pr


1

1 =sdot

=v

vS

Gw

ρms


11

1 =sdot

=ρS

Gwcd ms


80

1

50

1

1

31

1 Pr0260

+

times= cd

v

vve wwd

Nu ρρρ

ρη

8


150λ = Wm0C

160λ = Wm0C


=1λ Wm0C


d )

=+minus

==ei

eec

ndD

ndD

P

Sd

22

1

1

141

1=ecd m


1

111 ==

ecdNu

λα Wm2C


e

i

i

e

p

i

so d

d

d

dd

ksdot++=

21

1ln

2

11

αλα


11

=minus=sos kk

Rsd Wm2C


2

1

2

2 =minusminus

=tt

ttincη


( )( )

0

1

11

2

22

0

1

2 =minusminus

==iiG

iiG

Q

Qtdη

unde

0


p = 59 bar

9

t = 200C rarr =0

1i kJkg

ηtd =


1e )

10

1

1 =minus= sTie kJkg

1 =e kJkg


1e )

10

1

1 =minus= sTie kJkg

1 =e kJkg


2e )

20

2

2 =minus= sTie kJkg


2e )

=minus=

20

2

2 sTie kJkg


( )

1



( )

2





1001

2 =sdot∆∆

=E

Eexη


2spp∆ )

=sdot

sdotsdot∆=

∆=∆

Sk

CGp

NTC

pp

s

p

sp2

2

22

2

2 bar


2spp∆ = bar

10




sk



1

=sdot+

=sdso

sos

Rk

kk

Q2= kJs

Q2rarr

Q1= kJs

Q1rarr100 Qp= kJs

Qprarr

∆E1= kW

∆E1rarr100

∆E2= kW

∆E2rarr

∆Ep= kW

∆Eprarr

11


sdR


medt

2 =sdot

=∆sk

Qt

s

med

0C

2

1

=∆+== meds tttt0C

Se stabileşte

1p = rarr t1 = C


1i = kJkgK

( )

1i = kJkgK


1

2 ==Q

Qrη

ηr =


1G

( )

1

1

12 =minus

=ii

QG

rη

1G =


1Q )

)(

1

1

11iiGQ minus=

1Q = kJs


2Q )

)(

2

2

2

2 iiGQ minus=

2Q = kJs


2

1

QQQp minus=

pQ = kJs

Se calculează


1e

( ) ( ) ( )

10

1

1 sTie minus=

1)(i = kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1


12


1e

( ) ( ) ( )

10

1

1 sTie minus=

1 )(i =hellip kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1



1E∆

( )

1

1

1



2E∆

( )

2

2

2

2 eeGE minus=∆


pE∆

2

1



exη

1

2 =∆∆

=E

Eexη


1 =Q kJ rarr 100


2 =Q kJ rarr hellip


=pQ kJ rarr hellip


1 =∆E kW rarr 100

- Exergia primită






13


( )( )

1

1

11 =sdotsdot

minusminus= n

ic H

iiGGB σ

ηtccan

unde




1Q =hellipkJ

1Q rarr100

2Q =hellip kJ

2Q rarr

pQ =hellip kJ

pQ rarrhellip

1E∆ =hellip kW

1E∆ rarr100

2E∆ =hellip kW

2E∆ rarrhellip

pE∆ =hellip kW

pE∆ rarrhellip

3



2i )

69

2 =t 0C

873

2 =p bar -gt


Ci 060

= kJkg

Ci 070

= kJkg

i∆ = C

i 070 ndash

Ci 060

= kJkg



2i = C

i 060 + kJkg = 28901 kJkg

2i = kJkg


1s )

175

1 =t 0C

6

1 =p bar -gt

1s





1s = kJkgK


1s )

71

1 =t 0C

37

1 =p bar -gt

1s

06s = kJkgK

85s = kJkgK

s∆ =

85

06 ss minus = kJkgK



1s =

85s + kJkgK = kJkgK

4

1s = kJkgK


2s )

42

2 =t 0C

4

2 =p bar -gt

2s




2s = s40 + kJkgK

2s =


2s )

69

2 =t 0C

873

2 =p bar -gt

2s





2s = kJkgK


)(

1

111 iiGQ minus=


)(

2

222 iiGQ minus=



5

=

∆∆

∆minus∆=∆

min

max

minmax

lnt

t

tttmed

0C


==1

2

Q

Qrη


3


10 3

0

=sdotsdot∆

= minusSt

Qk

med

s




4

2

2 =sdot

sdot= idnsπ

m3


9152472 =t 0C

4=p bar

040=

Cv m3kg

1

4040

0 ==v

ρ kgm3

050=

Cv m3kg

1

5050

0 ==v

ρ kgm3

6


50915247 =+= ρρ kgm3

2 =ρ kgm3


22

22 =

sdot=

ρS

Gw ms


2 =ν m2s

tm = t2 = 0C

2 =ν m2s


=sdot

=2

22Re

νidw



44

22

11 =minus

sdot= edn

DS

ππm2


85vρ = kgm3

06vρ = kgm3

( )20

802

2 04803231430i

mmd

wtt minussdot+=α

7




=vρ kgm3


150ρ = kgm3

160ρ = kgm3



1 =ρ kgm3


150Pr =

160Pr =




PrPr 150158 =minus=

=1Pr


1

1 =sdot

=v

vS

Gw

ρms


11

1 =sdot

=ρS

Gwcd ms


80

1

50

1

1

31

1 Pr0260

+

times= cd

v

vve wwd

Nu ρρρ

ρη

8


150λ = Wm0C

160λ = Wm0C


=1λ Wm0C


d )

=+minus

==ei

eec

ndD

ndD

P

Sd

22

1

1

141

1=ecd m


1

111 ==

ecdNu

λα Wm2C


e

i

i

e

p

i

so d

d

d

dd

ksdot++=

21

1ln

2

11

αλα


11

=minus=sos kk

Rsd Wm2C


2

1

2

2 =minusminus

=tt

ttincη


( )( )

0

1

11

2

22

0

1

2 =minusminus

==iiG

iiG

Q

Qtdη

unde

0


p = 59 bar

9

t = 200C rarr =0

1i kJkg

ηtd =


1e )

10

1

1 =minus= sTie kJkg

1 =e kJkg


1e )

10

1

1 =minus= sTie kJkg

1 =e kJkg


2e )

20

2

2 =minus= sTie kJkg


2e )

=minus=

20

2

2 sTie kJkg


( )

1



( )

2





1001

2 =sdot∆∆

=E

Eexη


2spp∆ )

=sdot

sdotsdot∆=

∆=∆

Sk

CGp

NTC

pp

s

p

sp2

2

22

2

2 bar


2spp∆ = bar

10




sk



1

=sdot+

=sdso

sos

Rk

kk

Q2= kJs

Q2rarr

Q1= kJs

Q1rarr100 Qp= kJs

Qprarr

∆E1= kW

∆E1rarr100

∆E2= kW

∆E2rarr

∆Ep= kW

∆Eprarr

11


sdR


medt

2 =sdot

=∆sk

Qt

s

med

0C

2

1

=∆+== meds tttt0C

Se stabileşte

1p = rarr t1 = C


1i = kJkgK

( )

1i = kJkgK


1

2 ==Q

Qrη

ηr =


1G

( )

1

1

12 =minus

=ii

QG

rη

1G =


1Q )

)(

1

1

11iiGQ minus=

1Q = kJs


2Q )

)(

2

2

2

2 iiGQ minus=

2Q = kJs


2

1

QQQp minus=

pQ = kJs

Se calculează


1e

( ) ( ) ( )

10

1

1 sTie minus=

1)(i = kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1


12


1e

( ) ( ) ( )

10

1

1 sTie minus=

1 )(i =hellip kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1



1E∆

( )

1

1

1



2E∆

( )

2

2

2

2 eeGE minus=∆


pE∆

2

1



exη

1

2 =∆∆

=E

Eexη


1 =Q kJ rarr 100


2 =Q kJ rarr hellip


=pQ kJ rarr hellip


1 =∆E kW rarr 100

- Exergia primită






13


( )( )

1

1

11 =sdotsdot

minusminus= n

ic H

iiGGB σ

ηtccan

unde




1Q =hellipkJ

1Q rarr100

2Q =hellip kJ

2Q rarr

pQ =hellip kJ

pQ rarrhellip

1E∆ =hellip kW

1E∆ rarr100

2E∆ =hellip kW

2E∆ rarrhellip

pE∆ =hellip kW

pE∆ rarrhellip

4

1s = kJkgK


2s )

42

2 =t 0C

4

2 =p bar -gt

2s




2s = s40 + kJkgK

2s =


2s )

69

2 =t 0C

873

2 =p bar -gt

2s





2s = kJkgK


)(

1

111 iiGQ minus=


)(

2

222 iiGQ minus=



5

=

∆∆

∆minus∆=∆

min

max

minmax

lnt

t

tttmed

0C


==1

2

Q

Qrη


3


10 3

0

=sdotsdot∆

= minusSt

Qk

med

s




4

2

2 =sdot

sdot= idnsπ

m3


9152472 =t 0C

4=p bar

040=

Cv m3kg

1

4040

0 ==v

ρ kgm3

050=

Cv m3kg

1

5050

0 ==v

ρ kgm3

6


50915247 =+= ρρ kgm3

2 =ρ kgm3


22

22 =

sdot=

ρS

Gw ms


2 =ν m2s

tm = t2 = 0C

2 =ν m2s


=sdot

=2

22Re

νidw



44

22

11 =minus

sdot= edn

DS

ππm2


85vρ = kgm3

06vρ = kgm3

( )20

802

2 04803231430i

mmd

wtt minussdot+=α

7




=vρ kgm3


150ρ = kgm3

160ρ = kgm3



1 =ρ kgm3


150Pr =

160Pr =




PrPr 150158 =minus=

=1Pr


1

1 =sdot

=v

vS

Gw

ρms


11

1 =sdot

=ρS

Gwcd ms


80

1

50

1

1

31

1 Pr0260

+

times= cd

v

vve wwd

Nu ρρρ

ρη

8


150λ = Wm0C

160λ = Wm0C


=1λ Wm0C


d )

=+minus

==ei

eec

ndD

ndD

P

Sd

22

1

1

141

1=ecd m


1

111 ==

ecdNu

λα Wm2C


e

i

i

e

p

i

so d

d

d

dd

ksdot++=

21

1ln

2

11

αλα


11

=minus=sos kk

Rsd Wm2C


2

1

2

2 =minusminus

=tt

ttincη


( )( )

0

1

11

2

22

0

1

2 =minusminus

==iiG

iiG

Q

Qtdη

unde

0


p = 59 bar

9

t = 200C rarr =0

1i kJkg

ηtd =


1e )

10

1

1 =minus= sTie kJkg

1 =e kJkg


1e )

10

1

1 =minus= sTie kJkg

1 =e kJkg


2e )

20

2

2 =minus= sTie kJkg


2e )

=minus=

20

2

2 sTie kJkg


( )

1



( )

2





1001

2 =sdot∆∆

=E

Eexη


2spp∆ )

=sdot

sdotsdot∆=

∆=∆

Sk

CGp

NTC

pp

s

p

sp2

2

22

2

2 bar


2spp∆ = bar

10




sk



1

=sdot+

=sdso

sos

Rk

kk

Q2= kJs

Q2rarr

Q1= kJs

Q1rarr100 Qp= kJs

Qprarr

∆E1= kW

∆E1rarr100

∆E2= kW

∆E2rarr

∆Ep= kW

∆Eprarr

11


sdR


medt

2 =sdot

=∆sk

Qt

s

med

0C

2

1

=∆+== meds tttt0C

Se stabileşte

1p = rarr t1 = C


1i = kJkgK

( )

1i = kJkgK


1

2 ==Q

Qrη

ηr =


1G

( )

1

1

12 =minus

=ii

QG

rη

1G =


1Q )

)(

1

1

11iiGQ minus=

1Q = kJs


2Q )

)(

2

2

2

2 iiGQ minus=

2Q = kJs


2

1

QQQp minus=

pQ = kJs

Se calculează


1e

( ) ( ) ( )

10

1

1 sTie minus=

1)(i = kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1


12


1e

( ) ( ) ( )

10

1

1 sTie minus=

1 )(i =hellip kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1



1E∆

( )

1

1

1



2E∆

( )

2

2

2

2 eeGE minus=∆


pE∆

2

1



exη

1

2 =∆∆

=E

Eexη


1 =Q kJ rarr 100


2 =Q kJ rarr hellip


=pQ kJ rarr hellip


1 =∆E kW rarr 100

- Exergia primită






13


( )( )

1

1

11 =sdotsdot

minusminus= n

ic H

iiGGB σ

ηtccan

unde




1Q =hellipkJ

1Q rarr100

2Q =hellip kJ

2Q rarr

pQ =hellip kJ

pQ rarrhellip

1E∆ =hellip kW

1E∆ rarr100

2E∆ =hellip kW

2E∆ rarrhellip

pE∆ =hellip kW

pE∆ rarrhellip

5

=

∆∆

∆minus∆=∆

min

max

minmax

lnt

t

tttmed

0C


==1

2

Q

Qrη


3


10 3

0

=sdotsdot∆

= minusSt

Qk

med

s




4

2

2 =sdot

sdot= idnsπ

m3


9152472 =t 0C

4=p bar

040=

Cv m3kg

1

4040

0 ==v

ρ kgm3

050=

Cv m3kg

1

5050

0 ==v

ρ kgm3

6


50915247 =+= ρρ kgm3

2 =ρ kgm3


22

22 =

sdot=

ρS

Gw ms


2 =ν m2s

tm = t2 = 0C

2 =ν m2s


=sdot

=2

22Re

νidw



44

22

11 =minus

sdot= edn

DS

ππm2


85vρ = kgm3

06vρ = kgm3

( )20

802

2 04803231430i

mmd

wtt minussdot+=α

7




=vρ kgm3


150ρ = kgm3

160ρ = kgm3



1 =ρ kgm3


150Pr =

160Pr =




PrPr 150158 =minus=

=1Pr


1

1 =sdot

=v

vS

Gw

ρms


11

1 =sdot

=ρS

Gwcd ms


80

1

50

1

1

31

1 Pr0260

+

times= cd

v

vve wwd

Nu ρρρ

ρη

8


150λ = Wm0C

160λ = Wm0C


=1λ Wm0C


d )

=+minus

==ei

eec

ndD

ndD

P

Sd

22

1

1

141

1=ecd m


1

111 ==

ecdNu

λα Wm2C


e

i

i

e

p

i

so d

d

d

dd

ksdot++=

21

1ln

2

11

αλα


11

=minus=sos kk

Rsd Wm2C


2

1

2

2 =minusminus

=tt

ttincη


( )( )

0

1

11

2

22

0

1

2 =minusminus

==iiG

iiG

Q

Qtdη

unde

0


p = 59 bar

9

t = 200C rarr =0

1i kJkg

ηtd =


1e )

10

1

1 =minus= sTie kJkg

1 =e kJkg


1e )

10

1

1 =minus= sTie kJkg

1 =e kJkg


2e )

20

2

2 =minus= sTie kJkg


2e )

=minus=

20

2

2 sTie kJkg


( )

1



( )

2





1001

2 =sdot∆∆

=E

Eexη


2spp∆ )

=sdot

sdotsdot∆=

∆=∆

Sk

CGp

NTC

pp

s

p

sp2

2

22

2

2 bar


2spp∆ = bar

10




sk



1

=sdot+

=sdso

sos

Rk

kk

Q2= kJs

Q2rarr

Q1= kJs

Q1rarr100 Qp= kJs

Qprarr

∆E1= kW

∆E1rarr100

∆E2= kW

∆E2rarr

∆Ep= kW

∆Eprarr

11


sdR


medt

2 =sdot

=∆sk

Qt

s

med

0C

2

1

=∆+== meds tttt0C

Se stabileşte

1p = rarr t1 = C


1i = kJkgK

( )

1i = kJkgK


1

2 ==Q

Qrη

ηr =


1G

( )

1

1

12 =minus

=ii

QG

rη

1G =


1Q )

)(

1

1

11iiGQ minus=

1Q = kJs


2Q )

)(

2

2

2

2 iiGQ minus=

2Q = kJs


2

1

QQQp minus=

pQ = kJs

Se calculează


1e

( ) ( ) ( )

10

1

1 sTie minus=

1)(i = kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1


12


1e

( ) ( ) ( )

10

1

1 sTie minus=

1 )(i =hellip kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1



1E∆

( )

1

1

1



2E∆

( )

2

2

2

2 eeGE minus=∆


pE∆

2

1



exη

1

2 =∆∆

=E

Eexη


1 =Q kJ rarr 100


2 =Q kJ rarr hellip


=pQ kJ rarr hellip


1 =∆E kW rarr 100

- Exergia primită






13


( )( )

1

1

11 =sdotsdot

minusminus= n

ic H

iiGGB σ

ηtccan

unde




1Q =hellipkJ

1Q rarr100

2Q =hellip kJ

2Q rarr

pQ =hellip kJ

pQ rarrhellip

1E∆ =hellip kW

1E∆ rarr100

2E∆ =hellip kW

2E∆ rarrhellip

pE∆ =hellip kW

pE∆ rarrhellip

6


50915247 =+= ρρ kgm3

2 =ρ kgm3


22

22 =

sdot=

ρS

Gw ms


2 =ν m2s

tm = t2 = 0C

2 =ν m2s


=sdot

=2

22Re

νidw



44

22

11 =minus

sdot= edn

DS

ππm2


85vρ = kgm3

06vρ = kgm3

( )20

802

2 04803231430i

mmd

wtt minussdot+=α

7




=vρ kgm3


150ρ = kgm3

160ρ = kgm3



1 =ρ kgm3


150Pr =

160Pr =




PrPr 150158 =minus=

=1Pr


1

1 =sdot

=v

vS

Gw

ρms


11

1 =sdot

=ρS

Gwcd ms


80

1

50

1

1

31

1 Pr0260

+

times= cd

v

vve wwd

Nu ρρρ

ρη

8


150λ = Wm0C

160λ = Wm0C


=1λ Wm0C


d )

=+minus

==ei

eec

ndD

ndD

P

Sd

22

1

1

141

1=ecd m


1

111 ==

ecdNu

λα Wm2C


e

i

i

e

p

i

so d

d

d

dd

ksdot++=

21

1ln

2

11

αλα


11

=minus=sos kk

Rsd Wm2C


2

1

2

2 =minusminus

=tt

ttincη


( )( )

0

1

11

2

22

0

1

2 =minusminus

==iiG

iiG

Q

Qtdη

unde

0


p = 59 bar

9

t = 200C rarr =0

1i kJkg

ηtd =


1e )

10

1

1 =minus= sTie kJkg

1 =e kJkg


1e )

10

1

1 =minus= sTie kJkg

1 =e kJkg


2e )

20

2

2 =minus= sTie kJkg


2e )

=minus=

20

2

2 sTie kJkg


( )

1



( )

2





1001

2 =sdot∆∆

=E

Eexη


2spp∆ )

=sdot

sdotsdot∆=

∆=∆

Sk

CGp

NTC

pp

s

p

sp2

2

22

2

2 bar


2spp∆ = bar

10




sk



1

=sdot+

=sdso

sos

Rk

kk

Q2= kJs

Q2rarr

Q1= kJs

Q1rarr100 Qp= kJs

Qprarr

∆E1= kW

∆E1rarr100

∆E2= kW

∆E2rarr

∆Ep= kW

∆Eprarr

11


sdR


medt

2 =sdot

=∆sk

Qt

s

med

0C

2

1

=∆+== meds tttt0C

Se stabileşte

1p = rarr t1 = C


1i = kJkgK

( )

1i = kJkgK


1

2 ==Q

Qrη

ηr =


1G

( )

1

1

12 =minus

=ii

QG

rη

1G =


1Q )

)(

1

1

11iiGQ minus=

1Q = kJs


2Q )

)(

2

2

2

2 iiGQ minus=

2Q = kJs


2

1

QQQp minus=

pQ = kJs

Se calculează


1e

( ) ( ) ( )

10

1

1 sTie minus=

1)(i = kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1


12


1e

( ) ( ) ( )

10

1

1 sTie minus=

1 )(i =hellip kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1



1E∆

( )

1

1

1



2E∆

( )

2

2

2

2 eeGE minus=∆


pE∆

2

1



exη

1

2 =∆∆

=E

Eexη


1 =Q kJ rarr 100


2 =Q kJ rarr hellip


=pQ kJ rarr hellip


1 =∆E kW rarr 100

- Exergia primită






13


( )( )

1

1

11 =sdotsdot

minusminus= n

ic H

iiGGB σ

ηtccan

unde




1Q =hellipkJ

1Q rarr100

2Q =hellip kJ

2Q rarr

pQ =hellip kJ

pQ rarrhellip

1E∆ =hellip kW

1E∆ rarr100

2E∆ =hellip kW

2E∆ rarrhellip

pE∆ =hellip kW

pE∆ rarrhellip

7




=vρ kgm3


150ρ = kgm3

160ρ = kgm3



1 =ρ kgm3


150Pr =

160Pr =




PrPr 150158 =minus=

=1Pr


1

1 =sdot

=v

vS

Gw

ρms


11

1 =sdot

=ρS

Gwcd ms


80

1

50

1

1

31

1 Pr0260

+

times= cd

v

vve wwd

Nu ρρρ

ρη

8


150λ = Wm0C

160λ = Wm0C


=1λ Wm0C


d )

=+minus

==ei

eec

ndD

ndD

P

Sd

22

1

1

141

1=ecd m


1

111 ==

ecdNu

λα Wm2C


e

i

i

e

p

i

so d

d

d

dd

ksdot++=

21

1ln

2

11

αλα


11

=minus=sos kk

Rsd Wm2C


2

1

2

2 =minusminus

=tt

ttincη


( )( )

0

1

11

2

22

0

1

2 =minusminus

==iiG

iiG

Q

Qtdη

unde

0


p = 59 bar

9

t = 200C rarr =0

1i kJkg

ηtd =


1e )

10

1

1 =minus= sTie kJkg

1 =e kJkg


1e )

10

1

1 =minus= sTie kJkg

1 =e kJkg


2e )

20

2

2 =minus= sTie kJkg


2e )

=minus=

20

2

2 sTie kJkg


( )

1



( )

2





1001

2 =sdot∆∆

=E

Eexη


2spp∆ )

=sdot

sdotsdot∆=

∆=∆

Sk

CGp

NTC

pp

s

p

sp2

2

22

2

2 bar


2spp∆ = bar

10




sk



1

=sdot+

=sdso

sos

Rk

kk

Q2= kJs

Q2rarr

Q1= kJs

Q1rarr100 Qp= kJs

Qprarr

∆E1= kW

∆E1rarr100

∆E2= kW

∆E2rarr

∆Ep= kW

∆Eprarr

11


sdR


medt

2 =sdot

=∆sk

Qt

s

med

0C

2

1

=∆+== meds tttt0C

Se stabileşte

1p = rarr t1 = C


1i = kJkgK

( )

1i = kJkgK


1

2 ==Q

Qrη

ηr =


1G

( )

1

1

12 =minus

=ii

QG

rη

1G =


1Q )

)(

1

1

11iiGQ minus=

1Q = kJs


2Q )

)(

2

2

2

2 iiGQ minus=

2Q = kJs


2

1

QQQp minus=

pQ = kJs

Se calculează


1e

( ) ( ) ( )

10

1

1 sTie minus=

1)(i = kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1


12


1e

( ) ( ) ( )

10

1

1 sTie minus=

1 )(i =hellip kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1



1E∆

( )

1

1

1



2E∆

( )

2

2

2

2 eeGE minus=∆


pE∆

2

1



exη

1

2 =∆∆

=E

Eexη


1 =Q kJ rarr 100


2 =Q kJ rarr hellip


=pQ kJ rarr hellip


1 =∆E kW rarr 100

- Exergia primită






13


( )( )

1

1

11 =sdotsdot

minusminus= n

ic H

iiGGB σ

ηtccan

unde




1Q =hellipkJ

1Q rarr100

2Q =hellip kJ

2Q rarr

pQ =hellip kJ

pQ rarrhellip

1E∆ =hellip kW

1E∆ rarr100

2E∆ =hellip kW

2E∆ rarrhellip

pE∆ =hellip kW

pE∆ rarrhellip

8


150λ = Wm0C

160λ = Wm0C


=1λ Wm0C


d )

=+minus

==ei

eec

ndD

ndD

P

Sd

22

1

1

141

1=ecd m


1

111 ==

ecdNu

λα Wm2C


e

i

i

e

p

i

so d

d

d

dd

ksdot++=

21

1ln

2

11

αλα


11

=minus=sos kk

Rsd Wm2C


2

1

2

2 =minusminus

=tt

ttincη


( )( )

0

1

11

2

22

0

1

2 =minusminus

==iiG

iiG

Q

Qtdη

unde

0


p = 59 bar

9

t = 200C rarr =0

1i kJkg

ηtd =


1e )

10

1

1 =minus= sTie kJkg

1 =e kJkg


1e )

10

1

1 =minus= sTie kJkg

1 =e kJkg


2e )

20

2

2 =minus= sTie kJkg


2e )

=minus=

20

2

2 sTie kJkg


( )

1



( )

2





1001

2 =sdot∆∆

=E

Eexη


2spp∆ )

=sdot

sdotsdot∆=

∆=∆

Sk

CGp

NTC

pp

s

p

sp2

2

22

2

2 bar


2spp∆ = bar

10




sk



1

=sdot+

=sdso

sos

Rk

kk

Q2= kJs

Q2rarr

Q1= kJs

Q1rarr100 Qp= kJs

Qprarr

∆E1= kW

∆E1rarr100

∆E2= kW

∆E2rarr

∆Ep= kW

∆Eprarr

11


sdR


medt

2 =sdot

=∆sk

Qt

s

med

0C

2

1

=∆+== meds tttt0C

Se stabileşte

1p = rarr t1 = C


1i = kJkgK

( )

1i = kJkgK


1

2 ==Q

Qrη

ηr =


1G

( )

1

1

12 =minus

=ii

QG

rη

1G =


1Q )

)(

1

1

11iiGQ minus=

1Q = kJs


2Q )

)(

2

2

2

2 iiGQ minus=

2Q = kJs


2

1

QQQp minus=

pQ = kJs

Se calculează


1e

( ) ( ) ( )

10

1

1 sTie minus=

1)(i = kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1


12


1e

( ) ( ) ( )

10

1

1 sTie minus=

1 )(i =hellip kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1



1E∆

( )

1

1

1



2E∆

( )

2

2

2

2 eeGE minus=∆


pE∆

2

1



exη

1

2 =∆∆

=E

Eexη


1 =Q kJ rarr 100


2 =Q kJ rarr hellip


=pQ kJ rarr hellip


1 =∆E kW rarr 100

- Exergia primită






13


( )( )

1

1

11 =sdotsdot

minusminus= n

ic H

iiGGB σ

ηtccan

unde




1Q =hellipkJ

1Q rarr100

2Q =hellip kJ

2Q rarr

pQ =hellip kJ

pQ rarrhellip

1E∆ =hellip kW

1E∆ rarr100

2E∆ =hellip kW

2E∆ rarrhellip

pE∆ =hellip kW

pE∆ rarrhellip

9

t = 200C rarr =0

1i kJkg

ηtd =


1e )

10

1

1 =minus= sTie kJkg

1 =e kJkg


1e )

10

1

1 =minus= sTie kJkg

1 =e kJkg


2e )

20

2

2 =minus= sTie kJkg


2e )

=minus=

20

2

2 sTie kJkg


( )

1



( )

2





1001

2 =sdot∆∆

=E

Eexη


2spp∆ )

=sdot

sdotsdot∆=

∆=∆

Sk

CGp

NTC

pp

s

p

sp2

2

22

2

2 bar


2spp∆ = bar

10




sk



1

=sdot+

=sdso

sos

Rk

kk

Q2= kJs

Q2rarr

Q1= kJs

Q1rarr100 Qp= kJs

Qprarr

∆E1= kW

∆E1rarr100

∆E2= kW

∆E2rarr

∆Ep= kW

∆Eprarr

11


sdR


medt

2 =sdot

=∆sk

Qt

s

med

0C

2

1

=∆+== meds tttt0C

Se stabileşte

1p = rarr t1 = C


1i = kJkgK

( )

1i = kJkgK


1

2 ==Q

Qrη

ηr =


1G

( )

1

1

12 =minus

=ii

QG

rη

1G =


1Q )

)(

1

1

11iiGQ minus=

1Q = kJs


2Q )

)(

2

2

2

2 iiGQ minus=

2Q = kJs


2

1

QQQp minus=

pQ = kJs

Se calculează


1e

( ) ( ) ( )

10

1

1 sTie minus=

1)(i = kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1


12


1e

( ) ( ) ( )

10

1

1 sTie minus=

1 )(i =hellip kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1



1E∆

( )

1

1

1



2E∆

( )

2

2

2

2 eeGE minus=∆


pE∆

2

1



exη

1

2 =∆∆

=E

Eexη


1 =Q kJ rarr 100


2 =Q kJ rarr hellip


=pQ kJ rarr hellip


1 =∆E kW rarr 100

- Exergia primită






13


( )( )

1

1

11 =sdotsdot

minusminus= n

ic H

iiGGB σ

ηtccan

unde




1Q =hellipkJ

1Q rarr100

2Q =hellip kJ

2Q rarr

pQ =hellip kJ

pQ rarrhellip

1E∆ =hellip kW

1E∆ rarr100

2E∆ =hellip kW

2E∆ rarrhellip

pE∆ =hellip kW

pE∆ rarrhellip

10




sk



1

=sdot+

=sdso

sos

Rk

kk

Q2= kJs

Q2rarr

Q1= kJs

Q1rarr100 Qp= kJs

Qprarr

∆E1= kW

∆E1rarr100

∆E2= kW

∆E2rarr

∆Ep= kW

∆Eprarr

11


sdR


medt

2 =sdot

=∆sk

Qt

s

med

0C

2

1

=∆+== meds tttt0C

Se stabileşte

1p = rarr t1 = C


1i = kJkgK

( )

1i = kJkgK


1

2 ==Q

Qrη

ηr =


1G

( )

1

1

12 =minus

=ii

QG

rη

1G =


1Q )

)(

1

1

11iiGQ minus=

1Q = kJs


2Q )

)(

2

2

2

2 iiGQ minus=

2Q = kJs


2

1

QQQp minus=

pQ = kJs

Se calculează


1e

( ) ( ) ( )

10

1

1 sTie minus=

1)(i = kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1


12


1e

( ) ( ) ( )

10

1

1 sTie minus=

1 )(i =hellip kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1



1E∆

( )

1

1

1



2E∆

( )

2

2

2

2 eeGE minus=∆


pE∆

2

1



exη

1

2 =∆∆

=E

Eexη


1 =Q kJ rarr 100


2 =Q kJ rarr hellip


=pQ kJ rarr hellip


1 =∆E kW rarr 100

- Exergia primită






13


( )( )

1

1

11 =sdotsdot

minusminus= n

ic H

iiGGB σ

ηtccan

unde




1Q =hellipkJ

1Q rarr100

2Q =hellip kJ

2Q rarr

pQ =hellip kJ

pQ rarrhellip

1E∆ =hellip kW

1E∆ rarr100

2E∆ =hellip kW

2E∆ rarrhellip

pE∆ =hellip kW

pE∆ rarrhellip

11


sdR


medt

2 =sdot

=∆sk

Qt

s

med

0C

2

1

=∆+== meds tttt0C

Se stabileşte

1p = rarr t1 = C


1i = kJkgK

( )

1i = kJkgK


1

2 ==Q

Qrη

ηr =


1G

( )

1

1

12 =minus

=ii

QG

rη

1G =


1Q )

)(

1

1

11iiGQ minus=

1Q = kJs


2Q )

)(

2

2

2

2 iiGQ minus=

2Q = kJs


2

1

QQQp minus=

pQ = kJs

Se calculează


1e

( ) ( ) ( )

10

1

1 sTie minus=

1)(i = kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1


12


1e

( ) ( ) ( )

10

1

1 sTie minus=

1 )(i =hellip kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1



1E∆

( )

1

1

1



2E∆

( )

2

2

2

2 eeGE minus=∆


pE∆

2

1



exη

1

2 =∆∆

=E

Eexη


1 =Q kJ rarr 100


2 =Q kJ rarr hellip


=pQ kJ rarr hellip


1 =∆E kW rarr 100

- Exergia primită






13


( )( )

1

1

11 =sdotsdot

minusminus= n

ic H

iiGGB σ

ηtccan

unde




1Q =hellipkJ

1Q rarr100

2Q =hellip kJ

2Q rarr

pQ =hellip kJ

pQ rarrhellip

1E∆ =hellip kW

1E∆ rarr100

2E∆ =hellip kW

2E∆ rarrhellip

pE∆ =hellip kW

pE∆ rarrhellip

12


1e

( ) ( ) ( )

10

1

1 sTie minus=

1 )(i =hellip kJkg

1 )(s =hellip kJkgK

( ) ( ) ( )

10

1



1E∆

( )

1

1

1



2E∆

( )

2

2

2

2 eeGE minus=∆


pE∆

2

1



exη

1

2 =∆∆

=E

Eexη


1 =Q kJ rarr 100


2 =Q kJ rarr hellip


=pQ kJ rarr hellip


1 =∆E kW rarr 100

- Exergia primită






13


( )( )

1

1

11 =sdotsdot

minusminus= n

ic H

iiGGB σ

ηtccan

unde




1Q =hellipkJ

1Q rarr100

2Q =hellip kJ

2Q rarr

pQ =hellip kJ

pQ rarrhellip

1E∆ =hellip kW

1E∆ rarr100

2E∆ =hellip kW

2E∆ rarrhellip

pE∆ =hellip kW

pE∆ rarrhellip

13


( )( )

1

1

11 =sdotsdot

minusminus= n

ic H

iiGGB σ

ηtccan

unde




1Q =hellipkJ

1Q rarr100

2Q =hellip kJ

2Q rarr

pQ =hellip kJ

pQ rarrhellip

1E∆ =hellip kW

1E∆ rarr100

2E∆ =hellip kW

2E∆ rarrhellip

pE∆ =hellip kW

pE∆ rarrhellip

uet bilant sc abur apa

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