5 Exergy
5.1 Introduction
Evaluation
of heat
Quantity
Quality
Exergy and Anergy
Full convertible energy: mechanical
Partial convertible energy: heat
Unconvertible energy: environmental
Exergy: useful work potential; available energy;
The maximum useful work a system can delivered
from a specified state to the state of its environment
in theory.
Anergy: unavailable energy, unconvertible energy
Conditions for definition of exergy
work  f (initial state, path, final state)
（1）based on environment;
exergy of environmental energy is zero
（2）reversible process
（3）there is no other heat resource in the process.
E  Ex  An
e  ex  an
5.2 Calculation of exergy
Type
Chemical
Physical
unbalance
Chemical potential
Temperature and pressure
Kinetic
Velocity
Position
Position
Distribution
Concentration
Source
Electricity
Unbalance
voltage
Hydraulic
Water level
Pneumatic
Wind pressure
Wave
Difference between
interior and surface
5.2.1 Work resource
Electricity, mechanical energy, pneumatic energy,
hydraulic energy, can be converted to work
entirely.
Exergy of work resource = its total energy
1 2
Exk  mc
2
Exp  mg Z



5.2.2 Heat exergy
Potential work of heat
T
 ExQ
W0
T0
T0
 (1  ) Q
T
ExQ  Q  T0 
Q
T
Q  T0  S
AnQ  Q  ExQ  T0  S
T0
For constant temperature heat resource
ExQ
T0
 Q(1  )  Q  T0 S
T
AnQ
T0

Q  T0 S
T
T
T0
T0
For finite heat resource
T
E XQ
2
T0
T0
  (1  ) Q   (1  )mcP dT
T
T
T1
 mcP [(T2  T1 )  T0
 Q(1 
T0
)
T
T2  T1 T2
ln ]
T2  T1 T1
T2  T1
T 
T2
ln
T1
Influential factors:
Heat quantity
Heat resource temperature
T 
ExQ , S 
T ,  ExQ , S 
Environmental temperature T0
T  T0

ExQ  0
If the system absorbs heat, it absorbs exergy;
If the system discharges heat, it discharges exergy;
Example
1kg air with temperature of 200℃ was
cooled to 40℃. Please calculate the heat
exergy.
The
specific
cp=1.004kJ/(kg.K).
the
temperature is T0=25℃。
heat
of
air
is
environmental
Solution ：
 q  c p dT
exq
T0
  (1  )c p dT
473
T
313
313
  1.004  [(313  473)  298 ln
]
473
473
 37 kJ/kg
313
5.2.3 Low temperature heat exergy
Potential work of heat at a temperature below
the environmental temperature
E xQ
T0
T
 (1 
)Q0
T0
'
Q0  Wmax  Q
W0
E x Q
Ex Q'  
T
AnQ'
'
'
T0
(
 1)Q '
T
T0
(  1) Q '  T0 S  Q '
T
T0

 Q '  T0  S
T
For constant temperature
ExQ'
AnQ'
T0
'
 (  1)Q
T
T0 '
 Q
T
T
For finite heat resource
T
E XQ
2
T0
T0
  (  1) Q   (  1)mcP dT
T
T
T1
 mcP [T0
 Q(
T2  T1 T2
ln  (T2  T1 )]
T2  T1 T1
T0
 1)
T
T2  T1
T 
T2
ln
T1
Influential factors:
Heat quantity
Heat resource temperature
T , 
ExQ , S 
T 
ExQ , S 
T0
when T  ,
2
ExQ  Q
Environmental temperature T0
T  T0

ExQ  0
If the system absorbs heat, it discharges exergy;
If the system discharges heat, it absorbs exergy;
5.2.4 Inner energy exergy
For closed system
From initial state ( p, v, s, u ) to final state ( p0 , v0 , s0 , u0 )
For A→B→O
q  u0  u  w
q  qAB  qBO  0  T0 (s0  s)
w  (u  u0 )  T0 (s  s0 )
The dead work to resist environment
p0 (v0  v)
The maximum work
ex  (u  u0 )  T0 (s  s0 )  p0 (v  v0 )
For m kg substance
Ex  (U  U 0 )  T0 (S  S0 )  p0 (V  V0 )
The maximum available work
(w12 )max  ex1  ex 2  (u1  u2 )  T0 (s1  s2 )  p0 (v1  v2 )
Exergy in p  v diagram
Example
Please calculate the inner energy exergy
of air in the state of 1MPa and 50℃. The
environmental pressure is p0=0.1MPa，the
temperature is T0=25℃ and the specific
heat is cv=0.716kJ/(kg·K)。
ex  (u  u0 )  T0 (s  s0 )  p0 (v  v0 )
T
v
 cv (T  T0 )  T0 (cv ln  R ln )  p0 (v  v0 )
T0
v0
RT 0.287  323
3
v

 0.927 m /kg
p
1000
RT0 0.287  298
3
v0 

 0.855 m /kg
p0
100
ex  0.716 （50 — 25）
323
—298 （0.716ln
298
0.0927
0.287 ln
)
0.855
100  (0.0927 — 0.855)
 1145kJ/kg
5.2.5 Enthalpy exergy
For steady flow system
From initial state ( p, v, s, h) to final state ( p0 , v0 , s0 , h0 )
Exergy in p  v diagram
For A→B→O
q  h0  h  wt
q  qAB  qBO  0  T0 (s0  s)
ex  wt  (h  h0 )  T0 (s  s0 )
For m kg substance
Ex  ( H  H 0 )  T0 (S  S0 )
The maximum available work
(wt12 )max  ex1  ex 2  (h1  h2 )  T0 (s1  s2 )
Exergy in T  s diagram
under isotonic，h  q
 h  h0  area enclosed by1abs1
Example
Please compare the enthalpy exergy
value of saturated steam of 0.5MPa
with 5MPa. The environmental state
is p0=0.1MPa and T0=20℃。
Solution :
Look up properties in the table of water or steam
h0=84 kJ/kg
s0=0.2963 kJ/(kg·K)
h1=2747.5 kJ/kg
s1=6.8192 kJ/(kg·K)
h2=2794.2 kJ/kg
s2=5.9735 kJ/(kg·K)
ex1 =(h1－h0)－T0(s1－s0)
=(2747.5－84)－293×(6.8192－0.2963)
=752.3kJ/kg
ex2=(h2－h0)－T0(s2－s0)
=(2791.2－84)－293×(5.9735－0.2963)
=1046.8kJ/kg
5.3 Exergy loss
Exergy loss caused by temperature difference
TA
Q
H.E
T0
ExQA
T0
 Q(1  )
TA
TA
Q
TB
Q
H.E
T0
ExQB
T0
 Q(1  )
TB
Exergy loss
El  ExQA  ExQB
1
1
 QT0 (  )
TB TA
Entropy production
1 1
S g1  Q(  )
TB TA
El1  T0  S g1
T
S g 2
H.E
T0
'
T0
1
1
 Q0 (  )
T0 T0
El 2  T0  S g 2
Exergy loss caused by temperature difference
S g 3
Wl

 S7  S6
T0
If discharging temperature is T0 '
Wl

T0 '
S g 3
El 3  T0  S g 3
The whole exergy loss
El 
 E   (T
li
i
0
i
 S gi )  T0  S g
5.4 exergy equation
ds  ds f  dsg
ds f 
q
T
T0
  q  T0 ds  T0 dsg
T
and
 q  dh   wt
T0
(1  ) q  d (h  T0 s)   wt  T0 dsg
T
exq  (ex1  ex 2 )  wt  el
For many streams
 (E
)  ( ( Ex1 )i   ( Ex 2 )i )  Wt  El
xQ i
Heat exergy
Technical work
Enthalpy exergy
Exergy loss
5.5 Exergy efficiency
1
2
ex
exergy utilized
 ex 
exergy supplied
（E x）
a

（E x）
th
whole exergy out of system

whole exergy into system
( E x ) out

( E x )in
El
 1
( E x )in
exQ
W

E xQ
W
t 
Q
T0 ExQ
c  1  
T
Q
exQ
t

c
Example
The high temperature is TH=1800K and the low
temperature is the environmental temperature that is
T0=290K. A heat engine absorbs heat at T1=900K and
discharges heat at T2=320K. The engine efficiency is
70% of that of corresponding carnot cycle。If each
kilogram substance absorbs heat 100kJ ， please
calculate: (1)the practical work of heat engine; (2)the
heat exergy at given temperature; (3)the entropy
production and exergy loss of each process; (4)the
entropy increase of isolated system and the whole
exergy loss.
TH = 1800 K
qH
T1 = 900 K
q1
W
R
q2
T2 = 320 K
q2
T0 = 290 K
TH
T1
T2
1
2
3
4
T0
s g 1 s g 2 s g 3
s
Solution:
(1) The carnot cycle works between T1 and T2
T2
320
c  1   1 
 0.644
T1
900
t  0.7c  0.7  0.644  0.451
Practical work of engine
w  t  q1  0.451100  45.1 kJ/kg
Discharged heat
q2  q1  w  100  45.1  54.9 kJ/kg
(2) Heat exergy
Heat exergy at 1800K
exq1
T0
290
 q1 (1  )  100(1 
)  83.9 kJ/kg
TH
1800
Heat exergy at 900K
exq 2
T0
290
 q1 (1  )  100(1 
)  67.8 kJ/kg
T1
900
Heat exergy at 320K
T0
290
exq 3  q2 (1  )  54.9  (1 
)  5.2 kJ/kg
T2
320
(3) Entropy production and exergy loss
Entropy production caused by temperature
difference in absorbing heat process
1 1
1
1
sg1  q1 (  )  100  (

)  0.0556 kJ/kg
T1 TH
900 1800
Exergy loss
el1  T0 sg1  290  0.0556  16.1 kJ/kg
or
el1  exq1  exq 2  83.9  67.86  16.1 kJ/kg
Friction work loss
wl  wc  w  64.4  45.1  19.3 kJ/kg
Entropy production
wl 19.3
sg 2  
 0.0603 kJ/(kg  K)
T2 320
Exergy loss
el 2  T0 sg 2  290  0.06  17.5 kJ/kg
0r
el 2  exq2  exq3  w  67.8  5.2  45.1  17.5 kJ/kg
Entropy production caused by temperature
difference in discharging heat process
1 1
1
1
sg 3  q2 (  )  54.9  (  )  0.177 kJ/(kg  K)
T0 T2
290 320
Exergy loss
el 3  T0 sg 3  290  0.1776  5.2 kJ/kg
0r
el 2  exq3  exq 4  5.2 kJ/kg
(4)Entropy increase of isolated system
siso  sg1  sg 2  sg 3
 0.0556  0.0603  0.0177
 0.1338 kJ/(kg  K)
0r
siso
q1
q2

0
TH
T0
100 54.9


1800 290
 0.1338 kJ/(kg  K)
The whole exergy loss
el  el1  el 2  el 3
 16.1  17.5  5.2
 38.8 kJ/kg
0r
el  T0 siso
 290  0.1338
 38.8kJ/kg