Transcript Lecture 7.5 Chemical Exergy

Lecture 7.5
Chemical Exergy
CONCEPT OF CHEMICAL
EXERGY
Exergy : Max. work that can be
obtained from any form of
energy( e.g. in a flowing
stream)
Environment can interact with systems
in three ways:
Thermal interaction
Mechanical Interaction
Chemical interaction
CONCEPT OF CHEMICAL
EXERGY
Environmental State
& Dead State
Mech & Thermal
eqlbm. With
environment
{Thermomechanical exergy}
M+T+Chemical
eqlbm.
{Total exergy}
Ex : CO2 flowing in a pipe at To, P0.
CONCEPT OF CHEMICAL
EXERGY
Chemical Exergy :
Max amount of work obtainable when
the substance under consideration
is brought from environmental state
to Dead State by processes involving
heat transfer & exchange of matter
only with the environment.
or
work obtainable due to diff. in
the chemical potential
Reference Substances for exch.
(i) Gaseous
components of
atmosphere
3 kinds of
(ii)Solid substances
ref.
from the Earth’s crust
substances
(iii)Ionic-Nonionic
substances from
oceans
Reference Substances for exch.
for ref. substances
exch  Diff in chem. Potentials of the
substance in stream & that in
the environment
e.g. for gaseous phase, say CO2, at
To, Po we would visualise a
ch
system for extracting
x
e
Chemical Exergy
TER
To, Po
1
2
Wsh
Environment
S.P.M.
Chemical Exergy
exch  Work output in isothermal exp.
 g1  g2  1  2
 RT n
f1
f2
Making ideal gas assumption
Chemical Exergy
e
ch
x
 Po 
e
 RT n o    RT n yco
2
 Pco 
2 



mol. fraction of
CO2 in
environment
Ideal gas Mixture
ex  ?
ch
e  RT0 {ln( yi / y )} yi
ch
e
i
i
Total Exergy
Flow Process
2
V
ch
ex  (h  T0 s)  (h0  T0 s0 )   gz  ex
2
Non-flow Process
V2
ch
  (u  P0v  T0 s)  (u0  P0v0  T0 s0 )   gz  ex
2
Chemical Exergy of fuels
For substances not present in
environment
consider
an
idealised
reaction
of
the
substance with other substances
(usually ref. substances for
which exergies are known) so
that the final products also are
reference substances
Chemical Exergy of fuels
ex. Hydrocarbon fuel.
Wsh
T0
Ca Hb
P0
O2
(T0, P0)
CO2
H2O (liq)
Q
Chemical Exergy of fuels
b
b

Ca H b   a  O2  aCO2 
H 2O
4
2

Assuming no irreversibility
e
ch
x , fuel
b  ch
b ch

ch
  a  exo2  Wshmax  a exco2  exH 2O
4
2

Chemical Exergy of fuels
As already proved
ch
xi
& e


exchfu el
Wsh,max  Gr
  RT n y
e
i
 y e a b / 4 
o2


 Gr  RT n e a
e b 2 
y
 co2 y H 2o 
Standard Environment for Calculating the
Chemical Exergy of Hydrocarbon Fuels
Temperature
T0
Pressure,P0
298.15 K
101.325 kPa
Composition
Substance
Mole
(All in Gas Fraction
Phase)
xi
N2
0.7567
02
0.2035
H20
0.0303
C02
0.0003
Other
0.0092
Example
Calculate the chemical exergy in methane gas at
250C and 101.325 kPa.
Solution : The reaction equation is
CH4  202  C02  2H 2 0 
ex
Ch

 
  gC 02  2 g H 2 0   gCH 4  2 g 02
 R T0


y 
ln
y y 
2
e
02
e
2
e
C 02
H2 0
SOLUTION
y e 02  0.2035
e
y C 02  0.0003
e
y H 2 0  0.0303
At 250C and 101.325 kPa
g C 0  394,630 kJ / kg mol
g H 0  228,730 kJ / kg mol
2
2
g CH 4  50,844 kJ / kg mol
g 02  0
SOLUTION
ex
Ch
  394,630  2 228,730
  50,844  0

0.2035
 8.3143  298.15 ln
2
0.00030.0303
2
 801,250  29,550 kJ kg mol
 830,800 kJ kg mol
End of Lecture