Introduction to the
“Exergy” Concept
2.83/2.813
T. G. Gutowski
2
Readings
1.
Ch.2 Thermodynamics Revisited (p 7-13), and
2.
Ch 6 Exergy, a Convenient Concept (p
59-62), Jakob de Swaan Arons, 2004
3.
Ch 6 Thermodynamic Analysis of Resources
Used in Mfg Processes, Gutowski & Sekulic,
in TDR 2011
4.
Appendix, Tables of Standard Chemical
Exergy, Szargut, Morris & Steward, 1988
3
Exergy Accounting
• Exergy Units: Joules J, rate in Watts, W
• Exergy symbols: B, Ex, X, and b, ex, x
• Also called “Availability” by Keenan, 1941
and “Available Energy”, ΩR, by
Gyftopoulos and Beretta, 1991
4
Definition of Exergy
“Exergy is the amount of work obtainable
when some matter is brought to a state of
thermodynamic equilibrium with the
common components of the natural
surroundings by means of reversible
processes, involving interaction only with
the above mentioned components of
nature” [Szargut et al 1988].
5
Exergy
System State
Maximum work obtainable
between System and Reference
States.
Reference State
6
Exergy
System State
The minimum work needed
to raise System from the reference
state to the System State
Reference State
7
Aggregate Exergy
Accounting
Exin
Exout
Exlost
Exin ! Exout = Exlost
8
Thermodynamics Overview
•
•
•
•
•
Systems
Heat Interactions
Exergy of heat interaction
Entropy and Enthapy
Physical and Chemical Exergy
9
Open System
Work interaction
W
massin
Mass
massout
Q
Heat interaction
10
Closed System
X
W
massin
Mass
X
massout
Q
11
Isolated System
X
X
W
massin
Mass
X
massout
X
Q
12
The 1st Law
#U = Qin " Wout
dU = !Qin " !Wout
dQ
"
dW
=
0
!
!
13
Heat Interaction Q
TH
Q
Wout = Q ! Qo
Qo
To
Wout
Qo
!=
= 1"
Q
Q
14
Carnot’s “Reversible” Heat
Engine
!reversible = f (TH ,TL )
Qout TL
=
Qin TH
TL
" = 1!
TH
15
Maximum Work Output
max
out
W
TL
= Qin (1 ! )
TH
• Qin is at TH, let TL be Tref or To
• This gives the “available energy” of a heat
interaction at TH in reference To.
• Work and Heat are no longer equivalent!
16
Exergy “Ex” of Heat
Interaction
To
E = Q(1 ! )
T
Q
X
• Exergy, “Ex” is the available energy w.r.t.
a reference environment, T0, and P0…
• Ex(work) = W;
Ex (heat) = Q(1-T0/T)
17
Availability
“The First Law states that in every cyclic
process either work is converted into heat
or heat is converted into work. In this
sense it makes no distinction between
work and heat except to indicate a means
of measuring each in terms of equivalent
units. Once this technique of
measurement is established, work and
heat become entirely equivalent for all
applications of the First Law.”
Keenan, 1941
18
Availability
• “The Second Law, on the other hand,
marks the distinction between these two
quantities by stating that heat from a
single source whose temperature is
uniform cannot be completely converted
into work in any cyclic process, whereas
work from a single source can always be
completely converted into heat.”
Keenan, 1941
19
Availability
“The efforts of the engineer are very largely
directed toward the control of processes
so as to produce the maximum amount of
work, or so as to consume the minimum
amount of it. The success of these efforts
can be measured by comparing the
amount of work produced in a given
process with the maximum possible
amount of work that could be produced in
the course of a change of state identical
with that which occurs in the process.” 20
Keenan, 1941
Energy, E and Exergy, B = Ex
E2, B2
E1, B1
B1- B2 = E1- E2
B1- B2 > E1- E2
Properties for two different states
of the system shown by the boxes.
This change may come about due to
spontaneous changes or due to heat
or work interaction, or mass transfer.
reversible process
irreversible process
Ref: Gyftopoulos and Beretta
21
Define Entropy
1
"#( E1 ! Eo ) ! (B1 ! Bo ) $%
S1 = So +
CR
• They show CR = TR = T0
• Entropy is a Property
• Entropy is a measure of something lost
Ref: Gyftopoulos and Beretta
22
Entropy Difference
1
"#( E2 ! E1 ) ! (B2 ! B1 ) $%
S2 ! S1 =
CR
• ∆S = 0, reversible process
• ∆S > 0, irreversible process
Ref: Gyftopoulos and Beretta
23
Example, Heat Interaction
Q, T
T
T0
E2 = E1 +Q
B2 = B1 + Q(1-T0/T)
ΔS = (1/T0)(Q – Q + Q(T0/T)) = Q/T
ΔS = Q/T
24
Example, Work Interaction
W
E2 = E1 +W
B2 = B 1 + W
ΔS = (W - W) = 0
25
Homeworks 1 & 2
TH
Qin
Qout
1. Calculate the entropy change
for a reversible heat engine,and
2. Calculate the entropy loss for a
reversible heat engine.
Use the results given in this
Presentation.
TL
26
Answers for 1 & 2
Consider the process in two stages; 1) you transfer heat in, and
2) You transfer heat and work out. Use the result from Carnot to
Show that the change in entropy is zero. This leads to the result that
The exergy lost is also zero.
1
QL QH
!S = S2 " S1 = #$( E2 " E1 ) " (B2 " B1 ) %& =
"
To
TL TH
Blost = QH " QL + To !S " W
27
Properties or State Variables
•
•
•
•
•
•
•
•
T = temperature
intensive variables
P = pressure
V = volume
U = internal energy
E = energy
extensive
B = exergy
and
H = enthalpy (H = U + PV)
intensive
variables
S = entropy
28
State Variables
2
dU
=
0
!
d
!
=
!
"
!
2
1
#
dQ
=
0
!T
d
(
U
+
PV
)
=
0
!
1
29
Enthalpy H=U+PV
2
1
Here the Work done is
The First Law can be written as
W = P(V2 – V1)
Q = (U+PV)2 – (U + PV)1
The quantity in parenthesis is Enthalpy
H = U + PV
The First Law can be written as
Qin = ΔH
Constant Pressure Equilibrium Process
30
For Flow System Energy
1
1
2
U + P! + mV + mgz = H + mV 2 + mgz
2
2
Control Volume
F
d
Wboundary = Fd = Pν
31
Open System with H, S
H, S
W
po, To
H, S
Q, To
p, T
Consider the Work to bring the system from the
reference environment at standard conditions,
To, po to the state at T, p
See Ch 6 de Swaan Arons
32
From EQ 1 & 2(Ch 6), de Swaan Arons
H! in ! H! out ! Q! out + W!in = 0
!
Q
S!in ! S!out ! out + S! generated = 0
To
! = !H! " T !S! + T S!
W
o
o generated
Steady State Work to bring system from Po, To to P, T
33
Minimum Work = Exergy
Wmin
W! rev
=
= "H ! To "S
!
m
= ( H p ,T ! H po ,To ) ! To ( S p ,T ! S po ,To )
B = ( H ! To S ) ! ( H ! To S ) o
Wmin
W! rev
=
= Bout ! Bin
m!
34
Lost Work & Lost Exergy
Recall:
! = B! out ! B! in + To S!generated
W
Let:
! =W
! min + Wlost then
W
! lost = B
! lost = To S!generated
W
35
Exergy also is
p, T
W
W! max
b!
= wmax
!
m
po, To
Q
B = ( H ! To S ) ! ( H ! To S ) o
the maximum amount of work that can be
obtained from a system in reference to the
environment at standard conditions, To, Po
Standard ref. values T0 = 298.2! K , Po = 101.3kPa
36
Open flow system
37
First Law for a Flow System
'
$
ui2
' dU $
! i %% hi + + gzi ""
%
" = )m
2
& dt # cv in &
#
'
$
ui2
! j %% h j + + gz j ""
!)m
2
out
&
#
+ ) Q! in ! ) Q! out + ) W! sh ,in ! ) W! sh ,out
'
$ !
(u 2
! %% (H +
m
+ g(z "" + Q in ! W!out = 0
2
&
#
EQ 1
38
one stream steady state
Second Law for a Flow System
' dS $
! i Si ! ( m
! jSj
% " = (m
& dt # cv in
out
*Q! k
*Q!1 !
+()
!()
+ Sgenerated
T
T
in
out
Q! surr !
! !S +
m
+ Sgenerated = 0
T
one stream steady state
EQ 2
39
From EQ 1 & 2
'
$ !
(u 2
! %% (H +
m
+ g(z "" + Q ! W!out = 0
2
&
#
! !S +
m
!
W
rev
Q!
T0
+ S! generated = 0
2
(
%
!u
! && !H +
! To !S
=m
+ g!z ## " m
2
'
$
40
Physical and Chemical
Exergy
• B = Bph + Bch
• Bph(T=To, p=po , µ= µ* ≠ µο) =0
– this is the “restricted dead state”
• Bch(µ* = µο) = 0
• when B = Bph + Bch = 0
– this is the “dead state”
41
Thank you Jan Szargut
42
Chemical Reaction, at To, po
n1R1 + n2 R2 → n3Π3
R1
Π3
R2
Q
43
Chemical Reactions
stoichiometric mass balance
va Ra + vb Rb + .... # v j " j + vk " k + ...
exergy " balance"
va bRa + vbbRb + .... ! v j b" j ! vk b" k = Blost
where exergy b is given in kJ/mole
44
Example: Burning Carbon
C
BC
+
O2  CO2
+ BO2 -
410.3 kJ
+ 3.97 kJ
mol
BCO2
=
ΔB
– 19.9 kJ = 394.4kJ
mol
mol
The maximum work you can get out of one mol of carbon is
394.4 kJ
= 32.9 MJ
mol of carbon
kg
These exergy values come from Szargutʼs Appendix Tables
45
Burning Octane
2C8H18(l) + 25 O2(g)  16 CO2(g) + 18H2O(g)
2(5413.1) + 25(3.97) - 16(19.87) - 18(9.5) = ΔB
ΔB = 10,436.53 kJ/2 mols of octane
10,436.53
= 45.8 MJ
(2[(8 x 12) + 18]= 228g)
kg
Note:
ΔB = -ΔGºf = ΔH - ToΔS ≈ LHV
46
Note that Δg°≈ Δh°
(lower heating value) for fuels
Ref
Gyftopolous
& Beretta
47
48
Example: Oxidation of Aluminum
3
2Al +
O2
!
Al2O3
2
kJ
3
kJ
kJ
2 " 888.4
+
" 3.97
# 200.4
mol
2
mol
mol
Blost
= Blost
kJ
= (1776.8 + 6.0 # 200.4 ) = 1582.4
mol(Al2O3 )
See Appendix of Szargut for exegy values49
Materials can do work?
System State (TH)
QH
Wrev
Insert reversible
heat engine between
high and low
temperatures
QL
Reference State (TL)
50
Chemical Properties referenced to
the “environment”
Crust
Oceans
Atmosphere
T0 = 298.2 K, P0 = 101.3 kPA
51
Exergy Reference System
pure metal, element
chemical
reactions
oxides, sulfides…
crustal component
earthʼs crust (ground state)
extraction
52
Exergy Reference System
Aluminum (c=1)
888.4 kJ/mol
Al2O3 (c=1)
200.4 kJ/mol
Al2SiO5 (c=1)
Al2SiO5 (c = 2 x 10-3)
15.4kJ/mol
0 kJ/mol (ground)
53
Example; making pure iron from
the crust
Fe (c = 1)
376.4 kJ/mol
reduction
Fe2O3 (c=1)
16.5 kJ/mol
Fe2O3 (c = 1.3 x 10-3)
extraction
0 kJ/mol (ground)
54
55
Extraction from the crust
Extracting Fe 2O 3 from c = 1.3x10 !3 (crust) to c = 1
1
B = To R ln
1.3x10 !3
J
1
kJ
B = 298.2 K " 8.314
" ln
= 16.5
!3
mol K
1.3 " 10
mol
o
Note: R = k Navo (Boltzmannʼs constant X Avogadroʼs number)
56
Reduction of Fe2O3 (Hematite)
2Fe2O3 + 3C 4Fe + 3CO2
2 x 16.5 + 3 x 410.3 – 4 x 376.4 – 3 x 19.9 =
Blost = - 301.4 kJ
this is an endothermic reaction
i.e. minimum energy required to reduce 2 mole of hematite
57
Iron Ore Reduction
Recall C + O2  CO2 produces 394.8 kJ/mol C
301.4
= 0.76
We need
394.8
mols of carbon
2Fe2O3 + 3.76C + 0.76O2  4Fe + 3.76 CO2
58
Iron Ore Reduction
but the efficiency of the use of carbon
is only 30.3%
therefore the actual reaction is
2Fe2O3 + 12.42C + 9.42O2  4Fe + 12.42CO2
33kJ + 5095.9 + 37.7 - 1505.6 - 247.2kJ =
3,413.8 kJ for 4 mol of Fe
this is 15.2 MJ/kg (Fe)
59
Iron Ore Reduction
12.42 ! 394.8 4.903MJ
MJ
Fuel (C) Intensity =
=
= 22
4 ! 55.85 g
.2234kg
kg
12.42 ! 44
kg CO2
CO2 Intensity =
= 2.5
4 ! 55.85
kg Fe
Exergy value of pure Fe is 376.4 kJ/mole = 6.7 MJ/kg
60
Summary for Iron Ore
fuel used = 22 MJ/kg ≈ 15.2(Blost) + 6.7(BFe)
“Credit” for producing
pure iron from the crust
Lost exergy from making iron
from Fe2O3
See Smil Table A.12, iron from ore 20 - 25 MJ/kg
61
Exergy Balance, Open System
B!W ,in
B!Q,in
B! in
B!W ,out
B! loss
B!Q,out
B! out
B! in + B!W ,in + B!Q,in = B! out + B!W ,out + B!Q,out + B! loss
Includes: materials flows, heat and work interactions
62