MECH 330: APPLIED THERMODYNAMICS II LECTURE 04
2. AVAILABILITY ANALYSIS / EXERGY
Exergy is the potential for use (i.e., to do work). Unlike Energy, it
is not conserved but is destroyed by irreversabilities (Energy can
be converted to new forms (e.g., thermal energy) due to
irreversibilities (e.g., friction)).
For example, consider a fuel-air mixture in an isolated system.
After combustion, what remains are warm air and combustion
products. Energy is conserved, but the potential for use (i.e., to do
work) is greater for the unburned mixture than for the end-products
after combustion. Thus, due to the irreversible process of
combustion, the system has lost exergy.
Exergy analysis involves a consideration of
- a system of interest
- an “exergy reference environment”
The immediate surroundings are considered part of the system. The
surroundings which are far enough away to be unaffected by
processes (e.g., Q transfer) involving the system and its immediate
surroundings is taken to be the environment.
The boundary of this combination of the system and the
environment (i.e., the combined system) is located such that the
only energy transfers across the boundary are work (and not heat).
Another term for Exergy is Availability. The text uses a bold E for
exergy and an italicized E for energy. To avoid confusion, we will
write out the term “Exergy” instead of using a symbol, or use the
impromptu symbol “Ex”.
Specific Exergy will be denoted as e, as in the text.
38
MECH 330: APPLIED THERMODYNAMICS II LECTURE 04
Dead State
When the system is in equilibrium with the environment, at
temperature To and pressure Po, and is at rest with respect to the
environment, it has no potential to do work. This condition is
referred to as the dead state.
Unless otherwise specified, take the environment conditions to be:
To = 20oC (293 K)
Po = 1 atm (or 1 bar if that is what happens to be conveniently
tabulated when ∆P is small.)
Exergy Equations
The exergy of a given state:
Exergy = ( E − U o ) + Po (V − Vo ) − To ( S − So )
where,
E = U + KE + PE of the system
V = volume of the system
S = entropy of the system
Uo, Vo and So are the same properties at the dead state
On a unit mass basis, specific exergy is:
v2
e = ((u + + gz ) − uo ) + Po (υ − υo ) − To ( s − so )
2
or
v2
e = (u − uo ) + Po (υ − υo ) − To ( s − so ) + + gz
2
39
MECH 330: APPLIED THERMODYNAMICS II LECTURE 04
The change in exergy between two states of a closed system:
v22 − v12
e2 − e1 = (u2 − u1 ) + Po (υ 2 − υ1 ) − To ( s2 − s1 ) + + g ( z2 − z1 )
2
The work obtainable from the combined system as it passes to the
dead state:
Wc = ( E − U o ) + Po (V − Vo ) − To ( S − S0 ) − Toσ c
“c” denotes combined system entropy production
The Toσ c term reflects irreversibilities in the process to the dead
state.
Toσ c = 0 if no irreversibilities present
Toσ c > 0 if irreversibilities present
Note: Wc is at a maximum when Toσ c = 0.
40
MECH 330: APPLIED THERMODYNAMICS II LECTURE 04
Example
Cylinder of an IC engine contains 2450 cm3 of gaseous combustion
products at: 7 bar and 867oC. What is the specific exergy, e?
Let To = 27oC and Po = 1.013 bar.
2450 cm3
of air at
7 bars and
867oC
Solution
Assumptions:
- Model combustion products as air assumed to be an ideal gas.
- PE and KE effects are ignored (Thus, E = U)
e = (u − uo ) + Po (υ − υo ) − To ( s − so )
From Table A-22: u = 880.35 kJ/kg
uo = 214.07 kJ/kg
∴ (u − uo ) = 666.28 kJ / kg
Based on ideal gas model:
R P
s − so = s o (T ) − s o (To ) − ln
M Po
Table A-22
8.314 7
s − so = (3.11883 − 1.70203)kJ / kg − ln
28.97 1.013
s − so = 0.8621 kJ / kg Table A-1
41
MECH 330: APPLIED THERMODYNAMICS II LECTURE 04
∴ To ( s − so ) = (300 K )(0.8621 kJ / kg ⋅ K )
To ( s − so ) = 258.62 kJ / kg
Recall that we are assuming that this is an ideal gas. Thus, using
the ideal gas equation:
R T R To
V= and Vo =
M P M Po
R PoT
∴ Po (V − Vo ) = − To
M P
8.314 (1.013)(1140)
Po (V − Vo ) = − 300
28.97 7
Po (V − Vo ) = −38.75 kJ / kg
Thus,
e = 666.28 + (−38.75) − 258.62
e = 368.91 kJ / kg
Note that, as expected, e ≥ 0 so this is, in terms of sign, realistic
since system was well above the dead state.
42
