chapter 8 exergy - 서강대학교 청년광장 -...

Fundamentals of Thermodynamics

Chapter 8Exergy

Thermal Engineering Lab. 2

• Exergy ↔ Availability, available energy

• Anergy ↔ Unavailable energy

• Irreversible energy, reversible work, and irreversibility

• Exergy analysis : Pure Thermodynamics

• EGM(Entropy Generation Minimization)‑ Thermodynamics + Heat transfer + Fluid mechanics

Chapter 8. Exergy


8.1 Exergy, reversible work, and irreversibility

Chapter 8. Exergy

• Efficiency

① ; 1st law

② ; 2nd law

③ exergy ; 2nd law + environment (T0 , P0)

* Possible work we can extract from a given physical setup when it is allowed to interact with the ambient and the process end state is at T0 , P0.

WQ

h =&

&

( ) at s

s

WW

h =


Chapter 8. Exergy

• Exergy(Available energy) of heat‑ 일정 온도 T에서 Q만큼의 열을 받아 얻을 수 있는 최대 work


Chapter 8. Exergy

0revW Q Q= -

0(1 )TQT

= -

QEº 0revW Q T S= - D

0

0

Q Q ST T

= = D

revI W Wº -Irreversibility :


Chapter 8. Exergy

- T 가 변하는 경우

0revW Q Q= -

0 0Q T S= D

Q Tds= ò


Chapter 8. Exergy

. .

. ., , . .

. .,

c vi e

c vj i tot i e tot e c v ac

jc vi i e e gen ac

j

dm m mdt

dE Q m h m h Wdt

QdS m s m s Sdt T

= -

= + - -

= + - +

å å

å å å

å å å

& &

& && &

&&& &


Chapter 8. Exergy

oQ& 소거

0 , ,1

n

j i tot i e tot ej i e

dE Q Q W m h m hdt =

= + - + -å å å& & & & &

0

10

nj

i i e e genj i ej

QdS Q m s m s Sdt T T=

= + + - +å å å&&

&& &


Chapter 8. Exergy

( ) ( ) ( )

( )

, ,1 1

, ,1

,1

1

1

n nj

o o o i i o e e o gen j i tot i e tot ej jj

no o

j i tot i o e tot e o o genj j

no o

j i totj j

QdE dST T T m s T m s T S Q W m h m hdt dt T

d E T S T Q W m h T s m h T s T Sdt T

d E T S TW Q m hdt T

= =

=

=

= - - + - + - + -

æ ö-= - - + - - - -ç ÷ç ÷

è ø

æ ö-= - + - +ç ÷ç ÷

è ø

å å å å å å

å å å

å

&& & && & & &

& && & &

&& & ( ) ( )

{

,

0

i o e tot e o o gen

rev o gen

rev o gen

I

rev

T s m h T s T S

W T S

W W T S

I W W

- - - -

Þ =

= -

= -

å å

&

&&

&&

&& &

& & &


Chapter 8. Exergy

• General equation expressing

( ) ( )

( ) ( )

0 0. ., , 0 , 0 0

1

0 0. ., , 0 , 0

1

0 0.

( ) 1

( ) 1

( ) 1

n

c v j i tot i i e tot e e genj i ej

n

rev c v j i tot i i e tot e ej i ej

rev c vj

d E T S TW Q m h T s m h T s T Sdt T

d E T S TW Q m h T s m h T sdt T

d E T S TW Qdt T

=

=

æ ö-= - + - + - - - -ç ÷ç ÷

è øæ ö-

= - + - + - - -ç ÷ç ÷è øæ ö-

= - + -ç ÷ç ÷è ø

å å å

å å å

& && & &

&& & &

&& ( ) ( )., , 0 , 00

0

00 0 0

0

. ., , ,0

n

j i tot i i e tot e ej i e

rev gen

n

j i i e ej i ej

n

c v j i tot i e tot ej i e

m h T s m h T s

I W W T S

TdST Q m T s m T sdt T

dEW Q m h m hdt

=

=

=

+ - - -

= - =

= - - +

æ ö= - + + -ç ÷

è ø

å å å

å å å

å å å

& &

&& & &

& & &

&& & &Q

W&


Chapter 8. Exergy

• SSSF process

Assumption : Single inlet & single exit à ,i om m m= =& & &

0, 0 , 0 0

0, 0 , 0

(1 ) ( ) ( )

(1 ) ( ) ( )

j tot i i tot e e genj

rev j tot i i tot e ej

Tw q h T s h T s T sT

Tw q h T s h T sT

= - + - - - -

= - + - - -

å

å

( ) ( ) gene

eetotei

iitotijvcj

STsThmsThmQTTW &&&&&

00,0,.,.01 ----+÷÷ø

öççè

æ-= ååå


Chapter 8. Exergy

• SSSF process, Single inlet & Single exit

0, 0 , 0 0

0, 0 , 0

1

0, 0 , 0

0

00 0

0

, ,

(1 ) ( ) ( )

(1 ) ( ) ( )

(1 ) ( ) ( )

( )

(

j tot i i tot e e genj

n

rev j tot i i tot e ej j

n

rev j tot i i tot e ej j

nrev

gen e i jj j

tot i tot e

Tw q h T s h T s T sT

Tw q h T s h T sT

Tw q h T s h T sT

Ti w w T s T s s qT

w h h q

=

=

=

= - + - - - -

= - + - - -

= - + - - -

= - = = - -

= - +

å

å

å

å

Q0

)n

jj=å


Chapter 8. Exergy

• Reversible work à Maximum or minimum work0³-= acrev WWI &&&


Ex. 8.1 A feedwater heater has 5 kg/s water at 5 MPa and 40℃ flowing through it, being heated from two sources, as shown in Fig. 8.6. One source adds 900 kW from a 100℃ reservoir, and the other source transfers heat from a 200℃ reservoir such that the water exit condition is 5 MPa, 180℃. Find the reversible work and the irreversibility.

Chapter 8. Exergy


Ex. 8.2 Consider an air compressor that receives ambient air at 100 kPa and 25℃. It compresses the air to a pressure of 1 MPa, where it exits at a temperature of 540 K. Since the air and compressor housing are hotter than the ambient surroundings, 50 kJ per kilogram air flowing through the compressor are lost. Find the reversible work and the irreversibility in the process.

Chapter 8. Exergy


Chapter 8. Exergy

• Control mass process

[ ]

00 . .

01 2, 1 2 2 1 0 2 1

01 2 1 2, 1 2 0 2 1 1 2

1 ( )

1 ( )

( )

rev j c vj

rev jj

rev jj

T dW Q E T ST dt

TW Q E E T S ST

TI W W T S S QT

æ ö= - - -ç ÷ç ÷

è øæ ö

= - - - - -ç ÷ç ÷è ø

= - = - -

å

å

å

&&


Ex. 8.3 An insulated rigid tank is divided into two parts, A and B, by a diaphragm. Each part has a volume of 1 m3. Initially, part A contains water at room temperature, 20℃, with a quality of 50 %, while part B is evacuated. The diaphragm then ruptures and the water fills the total volume. Determine the reversible work for this change of state and the irreversibility of the process.

Chapter 8. Exergy


Chapter 8. Exergy

• The transient process‑ Uniform ; E=me, V=mv, S=ms

‑ Integration : 1→2

( ) ( )

00 . . . .,

0

, 0 ,

( ) 1n

rev c v c v jj j

i tot i i i tot e o ei i

TdW E T S Qdt T

m h T s m h T s

=

æ ö= - - + -ç ÷ç ÷

è ø

+ - - -

å

å å

&&

& &

01 2, 1 2 , 0 , 0

2 2 1 1 0 2 2 1 1

1 2 1 2, 1 2 0 2 2 1 1 1 2

(1 ) ( ) ( )

[ ( )]1[( ) ]

rev j i tot i i e tot e ej

rev i i e e jj

TW Q m h T s m h T sT

m e m e T m s m s

I W W T m s m s m s m s QT

= - + - - -

- - - -

= - = - + - -

å å å

å å å


Ex. 8.4 A 1 m3 rigid tank, Fig. 8.8, contains ammonia at 200 kPa and ambient temperature 20℃. The tank is connected with a valve to a line flowing saturated liquid ammonia at -10℃. The valve is opened, and the tank is charged quickly until the flow stops and the valve is closed. As the process happens very quickly, there is no heat transfer. Determine the final mass in the tank and the irreversibility in the process.

Chapter 8. Exergy


8.2 Exergy and second-law efficiency

Chapter 8. Exergy

( ) ( )

00 . . . .,

0

, 0 ,

( ) 1n

rev c v c v jj j


TdW E T S Qdt T

m h T s m h T s

=

æ ö= - - + -ç ÷ç ÷

è ø

+ - - -

å

å å

&&

& &

01 q jj j

T QT

æ öF = -ç ÷ç ÷

è øå &&

• Exergy of heat


Chapter 8. Exergy

• Flow exergy

( )

( ) ( )( ) ( )

2

0 0 0 0 0

0 ,0 0 0

, 0 , 0

2

tot tot

i e tot i i tot e e

Vh gz T s h gz T s

h T s h T s

h T s h T s

y

y y

æ ö= + + - - + -ç ÷è ø

= - - -

- = - - -


Chapter 8. Exergy

• Nonflow exergy

( )2

0 0 0 0 0 0 0 002

Vu gz P v T s u gz P v T sfé ùæ ö

é ù= + + + - - + + + -ê úç ÷ ë ûè øë û

e 0e

0 . .

0 0 0 0 . .

0 0 0 0

( )

[ ( ) ( )]( ) ( )

surr o

availrev rev surr o c v

availrev o c v

oavail

revavail

rev

W PVdW W W E T S PVdt

W E E T S S P V VE E T S S P V V

W

W

=

= - = - - +

= - - - - + -

F = - - - + -

= -F

= -F

& &

& & &

& &

[ ] [ ]2 1 2 0 2 0 2 1 0 1 0 1e P v T s e P v T sf f- = + - - + -


Chapter 8. Exergy

• Exergy balance equation

( ) ( )

00 0 . . . .,

0

, 0 ,

. .

. .

( ) 1n

availrev c v c v j

j j


availrev q i i e e C V

i e

availC V q rev i i e e

i e

TdW E T S PV Qdt T

m h T s m h T s

W m m

W m m

y y

y y

=

æ ö= - - + + -ç ÷ç ÷

è ø

+ - - -

= F + - -F

F = F - + -

å

å å

å å

å å

&&

& &

& & && &

&& & & &


Chapter 8. Exergy

• Exergy efficiency and isentropic efficiency

isentropica

ss

ww

h =

,aexergy

i e

why y

=-

1 lossexergy

input

EE

h = -&

&


Ex. 8.5 An insulated steam turbine (Fig. 8.10), receives 30 kg of steam per second at 3 MPa, 350℃. At the point in the turbine where the pressure is 0.5 MPa, steam is bled off for processing equipment at the rate of 5 kg/s. The temperature of this steam is 200℃. The balance of the steam leaves the turbine at 15 kPa, 90 % quality. Determine the exergy per kilogram of the steam entering and at both points at which steam leaves the turbine, the isentropic efficiency and the second-law efficiency for this process.

Chapter 8. Exergy


Chapter 8. Exergy

• Heat exchanger

1 2 1

3 3 4

. .

( )( )exer

wanted source c vexer

source source

mm

GenerallyI

y yhy y

h

-=

-

F F -= =F F

&

&

&& &

& &


Ex. 8.6 In a bolier, heat is transferred from the products of combustion to the steam. The temperature of the products of combustion decreases from 1100℃to 550℃, while the pressure remains constant at 0.1 MPa. The average constant-pressure specific heat of the products of combustion is 1.09 kJ/kg K. The water enters at 0.8 MPa, 150℃, and leaves at 0.8 MPa, 250℃. Determine the second-law efficiency for this process and the irreversibility per kilogram of water evaporated.

Chapter 8. Exergy


8.3 Exergy balance equation

Chapter 8. Exergy

1 0 0 0 0( ) ( ) ( )m m e e P m v v T s sfF = = - + - - -

2

2ve u gz= + +

0. .1 c v

Td Qdt TF æ ö= -ç ÷

è øå &

. . 0c vdVW Pdt

æ ö- -ç ÷è ø

i ei em my y+ -å å& &

DE- & 0( )D genE T S= &&

heat

work

flow

exergy destruction


Ex. 8.7 Let us look at the flows and fluxes of exergy for the feedwater heater in Example 8.1. The feedwater heater has a single flow, two heat transfers, and no work involved. When we do the balance of terms in Eq. 8.38 and evaluate the flow exergies from Eq. 8.22, we need the reference properties (take saturated liquid instead of 100 kPa at 25℃);

Chapter 8. Exergy


Ex. 8.8 Assume a 500 W heating element in a stove with an element surface temperature of 1000 K. On top of the element is a ceramic top with a top surface temperature of 500 K, both shown in Fig. 8.15. Let us disregard any heat transfer downward, and follow the flux of exergy, and find the exergy destruction in the process.

Chapter 8. Exergy


8.4 Engineering applications

Chapter 8. Exergy

HHEIHE QηW =

HH

HEIIHHEIIHE QTTηηW ÷÷ø

öççè

æ-=F= 01

HH

HEIIHHEIIHE QTTηηW ÷÷ø

öççè

æ-=F= 01

HPHHHP

HHPII WQ

TT

Wη /1 0

÷÷ø

öççè

æ-=

F=


Chapter 8. Exergy

• 공기 압축기의 Exergy analysis

1h

Q&

2h

inW&


Chapter 8. Exergy

• 내연기관의 Exergy analysis

BDC TDC BDC

Expansion work

Heat transfer

combustion

availability (exergy)

irreversibility


Chapter 8. Exergy

• Adiabatic SSSF process


Chapter 8. Exergy

• SSSF process with heat and work transfer


Chapter 8. Exergy

• Heat transfer process


Chapter 8. Exergy

• Isothermal compression process


Chapter 8. Exergy

• Steam power plant


Chapter 8. Exergy

• Combustion process


Chapter 8. Exergy

• Gas turbine process

chapter 8 exergy - 서강대학교 청년광장 -...

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