EXERGYEXERGYBasic DefinitionsBasic Definitions

ExergyExergy: : is property used to determine the useful work is property used to determine the useful work potential of a given amount of energy at some specified state.potential of a given amount of energy at some specified state.

It does not represent the amount of work that a work-producing It does not represent the amount of work that a work-producing device will actually deliver upon installation. Rather, it represents device will actually deliver upon installation. Rather, it represents the upper limit on the amount of work a device can deliver the upper limit on the amount of work a device can deliver without violating any thermodynamic laws.without violating any thermodynamic laws.A system delivers the maximum possible work as it undergoes a A system delivers the maximum possible work as it undergoes a reversible process from the specified initial state to the state of reversible process from the specified initial state to the state of its environment (dead state)its environment (dead state)

Dead StateDead State: a system is said to be in dead state when it is in : a system is said to be in dead state when it is in thermodynamic equilibrium with its environment. Also it has no thermodynamic equilibrium with its environment. Also it has no potential or kinetic energy. And it is chemically inert (no reaction potential or kinetic energy. And it is chemically inert (no reaction with the environment)with the environment)

Exergy of potential energyExergy of potential energy: (work potential) of a : (work potential) of a system is equal to the potential energy itself system is equal to the potential energy itself regardless of the temperature and pressure of the regardless of the temperature and pressure of the environment.environment.

Exergy of kinetic energyExergy of kinetic energy: (work potential) of a : (work potential) of a system is equal to the kinetic energy itself system is equal to the kinetic energy itself regardless of the temperature and pressure of the regardless of the temperature and pressure of the environment. (not necessarily true)environment. (not necessarily true)

REVERSIBLE WORK AND IRREVERSIBILITY:REVERSIBLE WORK AND IRREVERSIBILITY:

At difference to exergy, actual processes do not At difference to exergy, actual processes do not occur from an initial point to a final point equal to occur from an initial point to a final point equal to the dead state. On the other hand isentropic the dead state. On the other hand isentropic efficiencies are limited to adibatic processes.efficiencies are limited to adibatic processes.

Surrounding Work:Surrounding Work: is the work done by or against the is the work done by or against the surroundings during a process. It has significance only for a surroundings during a process. It has significance only for a process where boundary work occurs (closed system).process where boundary work occurs (closed system).

WWsurrsurr = P = P00(V(V22 – V – V11))

Then the useful work would be:Then the useful work would be:

WWuu = W – W = W – Wsurrsurr = W – P = W – P00(V(V22 – V – V11))

How is WHow is Wsurrsurr for a rigid tank? for a rigid tank?

Reversible WorkReversible Work: maximum amount of useful work : maximum amount of useful work that can be produced (or the minimum work that that can be produced (or the minimum work that needs to be supplied) as a system undergoes a needs to be supplied) as a system undergoes a process between the specified initial and final process between the specified initial and final states.states.

Whats is the difference between exergy and Whats is the difference between exergy and reversible work?reversible work?

Any difference between reversible and useful work is Any difference between reversible and useful work is due to irreversibilities.due to irreversibilities.

I = WI = Wrev,outrev,out – W – Wu,outu,out I = W I = Wu,inu,in – W – Wrev,inrev,in

SECOND LAW EFFICIENCYSECOND LAW EFFICIENCY

rev

thII

60.050.0

30.0, AII

In cases like this the first-law efficiency alone is not a realistic measure of performance of engineering devices.

The second-law efficiency is defined as the ratio of the thermal efficiency to the maximum possible thermal efficiency under the same conditions

Then, in the example:

43.070.0

30.0, BII

%50600

30011,

K

K

T

T

AH

LArev

%701000

30011,

K

K

T

T

BH

LBrev

In other words:

v

uII W

W

Re

Work producing devices

u

vII W

WRe

vII COP

COP

Re

Work consuming devices

Refrigerators and Heat Pumps

In general

SuppliedExergy

DestroyedExergy

SuppliedExergy

eredExergyII 1

covRe

00000 SSTVVPUUWuseful

Closed SystemTotal useful work delivered in a reversible process to the dead state:

The total exergy for a closed process would be given by:

mgzV

mSSTVVPUUX 2

2

00000

The total exergy per unit mass:

gzV

ssTvvPuu 2

2

00000

(kJ)

(kJ/kg)

EXERGY CHANGE FOR A CLOSED SYSTEM

)(2

)( 12

21

22

1201201212 zzmgVV

mSSTVVPUUmX

1201201212 )( SSTVVPEEmX

)(2

)( 12

21

22

1201201212 zzgVV

ssTvvPuu

1201201212 )( ssTvvPee

Per unit mass:

Exergy of a flow stream

flowflowingnoflowing xxx _

vPPvPPvx flow )( 00

vPPgzV

ssTvvPuu )(2 0

2

00000

gzV

ssThh 2

2

000

Exergy change of a flow stream

)(2

) 12

21

22

1201212 zzgVV

ssThh

Isolated system

System (closed or

open)

Q

W

Decrease or Exergy Principle (Exergy Destruction)

00 XST gen

Exergy destructiongendestroyed STX 0

Exergy decreases

Exergy Balance

systemtheof

exergytotal

theinChange

destroyed

exergy

Total

leaving

exergy

Total

entering

exergy

Total

Exergy transfer by heat, work and mass

By Heat: QT

TX heat

01

By Work:

By mass:

W

WW

Xsur

work

No boundary work

If boundary work

mX mass

XXXX destroyedworkheat Closed system (no mass flowing)

XSTVVPWQT

Tgenk

k

0120

0 )(1

dt

dXST

dt

dVPWQ

T

Tgenk

k

0001

mgzV

mSSTVVPUUX 2

2

00000

Sol.

00000 ssmTvvmPuumX

From table A6 & A4, for water:

u=2594.7 kJ/kg u0 = 104.83 kJ/kh

@ 180°C v=0.2472 m3/kg @ 25°C v0 = 0.001003 m3/kg

800 kPa s=6.7155 kJ/kg.K 100 kPa s0 = 0.3672 kJ/kg.K

From table A13 (R-134a superheated vapor):

u=386.99 kJ/kg u0 = 252.615 kJ/kh

@ 180°C v=0.044554 m3/kg @ 25°C v0 = 0.23803 m3/kg

800 kPa s=1.3327 kJ/kg.K 100 kPa s0 = 1.10605 kJ/kg.K

Kkg

kJkgK

kg

mkgkPa

kg

kJkgX w

.3672.01392.21.298

001003.0001127.01.10083.10492.76113

kJX w 7.622

Kkg

kJkgK

kg

mkgkPa

kg

kJkgX R

.10605.13327.11.298

23803.0044554.01.100615.25299.38613

kJX R 5.47

XXXX destroyedoutin

0destroyedX

Sol.

For a reversible process, therefore Wrev=X2-X1

1201201212 ssTvvPuu

1212 TTCvuu

1

2

1

212 lnln

v

vR

T

TCvss

but

and

Cv = 0.164 Btu/lbm.R R = 0.0621 Btu/lbm.R

3

3

12

.4039.5125.17.14

12

5.1ln

.0621.0

535

985ln

.164.0535

535985.

164.0

ftpsi

Btu

lbm

ftpsia

Rlbm

Btu

Rlbm

BtuR

RRlbm

Btu

lbm

Btu77.6012

Solution:

120°C

100

P (kPa)

v (m3/kg)

20 °C

1

180 21’

First the process is at constant volume until the pressure is enough to move the piston (1-1’), then the process is at constant pressure (1’-2)

States 1 and 2 are in the region of superheated vapor, the summarized data from table is:

P1=140kPa v1=0.1652 m3/kg P2=180kPa v2=0.17563 m3/kg

u1=246.01 kJ/kg u2=331.96 kJ/kg

T1=20°C s1=1.0532 kJ/kgK T2=120°C s2=1.3118 kJ/kgK

Also at (1’) v1’ = v1 and P1’=P2

a) The work done is the boundary work, from (1) to (1’) is zero since it is a constant volume process; from (1’) to (2) is a constant pressure process. Then the boundary work is given by:

Wb = P2.m.(v2-v1) = (180)(1.4)(0.17563 – 0.1652) = 2.63 kJ

b) Doing a energy balance we obtain:

Q-W = m(u1’ – u1) + m(u2 – u1’) = m(u2 – u1)

Then Q = m(u2 – u1) + W = (1.4)(331.96 – 246.01) + 2.63 kJ

Q = 122.96 kJ

Since there is no kinetic nor potential energy involved the exergy change can be expressed by:

The useful work at the exit is given by the boundary work minus the work against the environment:

Wu = Wb – m.P0(v2-v1) = 2.63kJ – (1.4)(100)(0.17563–0.1652)

Wu = 1.17 kJ

From the total exergy change the only amount of useful work is 1.17kJ everything else is the exergy destroyed, therefore:

d) The second law efficiency is given by:

12012012 ssmTvvmPuumX

)0532.13118.1)298)(4.1(1652.017563.0)100)(4.1(01.24696.3314.1 X

kJX 90.13

kJkJkJWXX udestroyed 73.1217.190.13

084.09.13

17.1

X

WuII

XXXXX destroyedmassworkheat

Exergy Balance Open System

Notice that now we are including the exergy entering and leaving with mass, then:

(kJ)

In rate form:

Fortunately we usually have to deal with steady flow devices, then our equation becomes:

(kW)

1212001 XXXmmVVPWQ

T

Tdestroyed

outink

k

dt

dXXmm

dt

dVPWQ

T

Tdestroyed

outink

k

001

01 0

destroyed

outink

k

XmmWQT

T

01 120

destroyedkk

XmWQT

T

01 0

outinrevk

k

mmWQT

T

0 destroyedrev XW

01 120

destroyedk

k

xwqT

T (kJ/kg)

Per unit mass:

(kW)

Previous equations can be used to determine the reversible work by making the exergy destruction term equal to zero since (i.e. no irreversibilities implies no exergy destruction)

for a single stream this last equation becomes:

Then:

or: 01 0

kkinout

rev QT

TmmW

Single stream (one inlet - one outlet):

01 012

kk

rev QT

TmW

Example:

Solution:

We need to determine the exergy destroyed during this process. In this case the easiest way is by:

This equation leads us to find the entropy generated which is given by doing an entropy balance:

From table at 200 psia:

State (1) sat. liquid: h1=355.46 Btu/lbm s1=0.54379Btu/lbm.R

State (2) sat. vapor: h2=1198.8 Btu/lbm s2=1.5460 Btu/lbm.R

At environment conditions (P0=14.7 psia, To=80°F comp. liq.):

h0=48.07 Btu/lbm s0=0.09328 Btu/lbm.R

gendestroyed sTx 0

T

qsssss

T

qgengen )( 12

We need to determine q in the previous equation, we obtain this by doing an energy balance:

q – w = (h2 – h1) since there is no work we have:q = 1198.8 – 355.46 = 843.34 Btu/lbm

Now we can determine the entropy generated:

960R is the absolute gas temperature (500°F)

And the exergy destroyed will be:

The exergy 9or work potential) of the steam is given by:

Therefore the temperature of the gases does not affect the exergy of the steam. However it does affect sgen and therefore xdestroyed too.

Rlbm

Btu

R

RlbmBtusgen .

124.0960

./34.843)54379.05460.1(

lbm

Btu

Rlbm

BtuRsTx gendestroyed 96.66

.124.0)540(0

12012 ssThh

a) We find the actual work by doing an ener4gy balance:

Kinetic and potential energy changes are assumed to be zero.

Consider specific heats for the enthalpy change. Solving for work we have:

Cp=1.134 kJ/kg.K (from table)

Q = -30kW (lost)

PEKEhhmWQ a

12

12 TTCpmQW a

Substitue in the previous equation to obtain:

We have an expression for the reversible work from the exergy balance for a single stream:

The ideal situation for a turbine occurs when there are no heat losses, therefore:

The entropy change is obtained from:

Then the rev. work is:

kWKkgK

kJ

s

kgkWWa 67.4321023903134.14.330

01 012

kk

rev QT

TmW

][ 1201212 ssThhmmW rev

Kkg

kJ

Kkg

kJ

Kkg

kJ

P

PR

T

TCpss

.11.0

1200

500ln

.287.0

1023

903ln

.134.1ln

1

2ln

1

212

kJKkg

kJKK

Kkg

kJ

s

kgW rev 574]

.11.0[298]1023903[

.134.14.3

The exergy destroyed will be given by:

Finally, the second law efficiency would be:

kJWWX arevdestroyed 45.14167.43212.574

754.012.574

67.432

rev

a

II

W

W