1. PRESENTED BY:
NISHANT R. SHAH
GUIDE:
PROF. ARVIND PRASAD

2. INTRODUCTION
• What do you mean by exergy?
• Maximum work that can be obtained from any form of
energy or in simple words
• Exergy is a available energy
Environment can interact with system in three ways:
Thermal interaction
Mechanical interaction
Chemical interaction

3. Environmental State & Dead State
Mech & Thermal
eqlbm. With
environment {Thermo-
mechanical exergy}
M+T+Chemical
eqlbm.
{Total exergy}

4. 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 & mass
transfer only with the environment
Or
Work obtainable due to the difference in
chemical potential

5. Chemical Exergy
Determination of Chemical Exergy for Gases:
Turbine
T,P 1
Environment
2 T0,P0
Wsh

6. Chemical Exergy
Making ideal gas assumption
Exch
= RT ln Po = -RT ln (yCO2)e
PCO2
mole fraction of CO2 in
environment
Similarly for ideal gas mixture
Exch
= RT0 ∑ ln(yi/ye)ye

7. Total Exergy
Flow Process
ex = (h-Ts) – (h0-T0s0) + V2
+ gz + Exch
2
Non-Flow Process
ex = (U-U0) + P0(V-V0) –T0(S-S0) + V2
+ gz + Exch
2

8. ex. Hydrocarbon fuel
O2
Ca Hb
Q
Wsh
H2O (liq)
CO2
(T0, P0)
T0
P0

9. Chemical Exergy of fuels
CaHb + a + b O2 aCO2 + b H2O
4 2
Assuming no irreversibility,
Exch
fue l + a + b Exch
O2 Wshmax + a(Exch
CO2) + b(Exch
H2O)
4 2
As, (Wshmax = -∆Gr & Exch
= -RT lnyi )
Exch
fuel = -∆Gr + RT ln yO2^a+b/4
yCO2^a + yH2O^b/2

10. Standard Environment for Calculating the
Chemical Exergy of Hydrocarbon Fuels
0.7567
0.2035
0.0303
0.0003
0.0092
N2
02
H20
C02
Other
101.325 kPa298.15 K
Mole
Fraction
xi
Substance
(All in Gas
Phase)
Pressure,P0Temperature
T0
Composition

11. Example
Calculate the chemical exergy in methane gas at
25degC and 101.325kPa
Solution: The reaction equation is
CH4 + 2O2 CO2 + 2H2O
Exch
= - (gCO2 + 2gH2O) – (gCH4 + 2gO2 )
+ RT0 ln (yO2)^2
(yCO2)(yH2O)^2

12. SOLUTION
yO2 = 0.2035
yCO2 = 0.0003
yH2O = 0.0303
At 25degC and 101.325 kPa
gCO2 = -394,630 kJ/kgmol
gH2O = -228,730 kJ/kgmol
gCH4 = -50,844 kJ/kgmol

13. SOLUTION
Exch
= - (-394,630) + 2(-228,730) -
(-50,844 + 0)
+ 8.314*298.15 ln (0.22035)^2
(0.0003)(0.0303)^2
= 801,250 + 29,550 kJ/kgmol
= 830,800 kJ/kgmol

14. Application
Exergy analysis is performed in the field of industrial
ecology to use energy more efficiently
Design engineers have recognised that a higher exergy
efficiency involves building a more expensive plant & a
balance between capital investment & operating efficiency
must be determined in the context of economic
competition
Exergy methods are computed before setting up any plant
to select the best possible & efficient design

15. THANK YOU