Exergy is a quantity that is destroyed, not generated, in a process by irrevers-ibility. Therefore, an equivalent way to write the general balance equation (Equation 3.44) is
∑
Ω − Ω − Ω =∑
dΩin out des dsys
t (3.74)
The subscript “des” signifies a destroyed quantity (like exergy). Notice that this term is opposite in sign to a quantity that is generated (like entropy).
Making the substitution Ω = X into Equation 3.74 results in the exergy balance.
∑
X in−∑
X out−X des=dXdtsys (3.75)The variables X or x will be used in this book to specify total or specific exergy, respectively. Subscripts will be used, particularly in the case of spe-cific exergy, x, so it is not confused with quality.
Exergy is carried in and out of a system by energy (heat and work), and by mass flow. Rewriting Equation 3.75 to reflect this results in
∑ (
X in−X out)
Q+∑ (
X in−X out)
W+∑ (
m x in f,in−m x out f,out)
−X des=dXdtsys (3.76)To apply Equation 3.76 to a system, the exergy terms need to be quantified.
The equations that follow in the remainder of this section are presented without derivation. The interested reader is encouraged to consult an undergraduate-level textbook in mechanical engineering thermodynamics for the details of these derivations, such as the works by Moran (2014), Cengel (2011), and Klein (2012).
Exergy transport rate due to heat transfer, the first term on the left-hand side of Equation 3.76, can be written as
∑ (
−)
=∑
− − −
X X Q T
T Q T
T
Q b b
in out in 1 0 1
in
out 0
out
(3.77)
In Equation 3.77, Tb is the system boundary temperature where the heat is being transferred and T0 is the dead-state temperature, both expressed as absolute temperatures.
The exergy transferred to and from the system by work, the second term on the left-hand side of Equation 3.76, is given by
∑ (
X in−X out)
W=∑ (
W in−W out)
+ dP0 dVt (3.78) In Equation 3.78, the term P0(
dV t represents the rate of work that is done d)
on or by the surroundings due to a change in volume of the system over time. If the time derivative of the system volume is negative, then the system is being compressed by the surroundings. This is a free source of exergy that can be capitalized on. If the system is expanding, then the volume derivative is positive. This represents work that is done on the environment and is not useful. Therefore, it ends up reducing the exergy output in the form of work.
In a steady-flow scenario or a case where the system volume does not change over time, the volume derivative vanishes.
The third term on the left-hand side of Equation 3.76 is the net exergy transported by mass as it crosses the system boundary. In this equation, xf, is known as flow exergy and it is defined as
( )
( )= − − − + +
f 0 0 0 2
x h h T s s V2 gz (3.79) The subscripts “0” in Equation 3.79 represent the properties at the dead state, which is defined by P0 and T0.
The fourth term on the left-hand side of Equation 3.76 is the exergy destruction rate. Exergy destruction is related to entropy production. Exergy destruction is also called irreversibility. The exergy destruction rate term is given by
=
X des T S0 gen (3.80)
The magnitude of the exergy destruction in a process is of great interest to engineers. Equation 3.80 shows how the abstract concept of entropy genera-tion is used in a very practical sense to design and analyze systems.
If the flow of exergy into and out of a system is unsteady, then there will be storage of exergy in the system. This is represented by the derivative on the right-hand side of Equation 3.76. The exergy of a system is given by
( ) ( ) ( )
= − − − + − + +
sys 0 0 0 0 0 2
X U U T S S P V V V2
gz (3.81) Using Equation 3.81, the rate of exergy storage inside the system can be written as
As with previous equations, the subscript “0” refers to the dead-state properties.
Substituting Equations 3.77 through 3.82 into Equation 3.76 would make a very complex equation! Therefore, in exergy analysis, it is easier to start with Equation 3.76 and substitute Equations 3.77 through 3.82 as needed.
Fortunately, for most exergy analyses, the resulting exergy balance equation is quite manageable.
In the development of the exergy balance, notice that there are terms which are representative of the first and second laws of thermodynamics. The equa-tions above contain enthalpy and internal energy (first-law properties) and entropy (a second-law property). This happens because the equations for the exergy balance were derived using a combination of the first and second laws. Therefore, the exergy balance is sometimes known as the combined law.
EXAMPLE 3.10
Steam at 1.6 MPa and 350°C enters a steam turbine at a flow rate of 16 kg/s.
The steam leaves the turbine as a saturated vapor at 30°C. The turbine deliv-ers 9 MW of power. The turbine is insulated, but not perfectly. The average temperature of the insulation on the outer surface of the turbine is 70°C. The environment surrounding the turbine can be considered to be the dead state at 25°C, 100 kPa. Figure E3.10 shows this turbine. Determine the exergy destruc-tion rate in the turbine.
FIGURe e3.10
SOLUTION: This is the same turbine that was considered in Examples 3.5 and 3.8. In Example 3.5, the turbine was analyzed from a first-law point of view.
Example 3.8 considered the turbine from a second-law perspective. In this example, the performance of the turbine will be considered relative to exergy.
There are several ways to determine the exergy destruction. Perhaps the sim-plest is to consider Equation 3.80. The entropy generation rate has previously been determined in Example 3.8. Therefore,
= =( + )
= ←
25 273.15 K 23.39kW
K 6974 kW
des 0 gen
X T S
Notice that the temperature used in this calculation must be on the absolute scale. Although this provides the required answer, it really does not reveal
Turbine
sat. vapor 2
1
Q˙ =?
m˙ =16 kg/s
W˙ =9 MW T2=30˚C T1=350˚C
P1=1.6 MPa T0=25˚C P0=100 kPa
much about the turbine. More can be learned about the turbine by considering application of Equation 3.76 to the turbine.
This formulation is particularly informative, because each term has signifi-cance to how the turbine is performing as shown below:
des= ( f1− f2) − −
Net exergy flow input out,
Exergy loss due to heat out, Exergy output
X m x x X Q X W
This equation is helpful because it shows the magnitude of each of the exergy components of the turbine: the net incoming exergy, the exergy loss due to heat transfer, the exergy output (power), and the exergy destruction.
Neglecting kinetic and potential energy differences, the net exergy transfer rate due to mass flow can be written as
Using the property values determined in Examples 3.5 and 3.8, the net exergy transfer rate into the turbine is
s 3146.0 2555.6 kJ
kg 298.15 K 7.0713 8.4520 kJ kg K
The rate that exergy is lost from the turbine due to heat transfer is given by
( ) ( )
The heat transfer rate from the turbine was determined in Example 3.5. The exergy transfer rate from the turbine is the power delivered
out, = = 9000 kW X W W
Therefore, the exergy destruction rate in the turbine is
des= ( f1− f2)− out, − out, =(16, 033 59 9, 000 kW 6, 974 kW− − ) = ←
X m x x X Q X W
This is the same result that was found using the simpler equation, X des=T S0 gen . However, this result is more meaningful because the relationship between the net exergy input and the exergy destruction is clearly seen. Because of the steam flow entering and leaving, the turbine is provided with 16,033 kW of flow exergy. This exergy rate is converted to power (9000 kW), an exergy transfer due to heat (59 kW), and the remainder is destroyed (6974 kW).
As demonstrated in Example 3.10, the exergy destruction rate is more phys-ically meaningful than the entropy generation rate. For example, the entropy generation rate of the turbine in Example 3.10 is 23.39 kW/K. But what does this really mean? Unfortunately, the only thing this indicates is that there are irreversibilities in the turbine (the entropy generation rate is positive).
However, the number, by itself, does not tell us how bad the irreversibility is.
On the other hand, the full exergy analysis of the turbine clearly reveals how much exergy is input to the turbine, and how it is distributed. The exergy analysis gives a much clearer and meaningful picture to how the turbine is operating.