EXERGY. Basic Definitions Exergy : is property used to determine the useful work potential of a given amount of energy at some specified state.
Exergy: is property used to determine the useful work potential of a given amount of energy at some specified state.
It does not represent the amount of work that a work-producing device will actually deliver upon installation. Rather, it represents the upper limit on the amount of work a device can deliver without violating any thermodynamic laws.
A system delivers the maximum possible work as it undergoes a reversible process from the specified initial state to the state of its environment (dead state)
Dead State: a system is said to be in dead state when it is in thermodynamic equilibrium with its environment. Also it has no potential or kinetic energy. And it is chemically inert (no reaction with the environment)
Exergy of kinetic energy: (work potential) of a system is equal to the kinetic energy itself regardless of the temperature and pressure of the environment. (not necessarily true)
REVERSIBLE WORK AND IRREVERSIBILITY:
At difference to exergy, actual processes do not occur from an initial point to a final point equal to the dead state. On the other hand isentropic efficiencies are limited to adibatic processes.
Wsurr = P0(V2 – V1)
Then the useful work would be:
Wu = W – Wsurr = W – P0(V2 – V1)
How is Wsurr for a rigid tank?
Whats is the difference between exergy and reversible work?
Any difference between reversible and useful work is due to irreversibilities.
I = Wrev,out – Wu,out I = Wu,in – Wrev,in
In cases like this the first-law efficiency alone is not a realistic measure of performance of engineering devices.
The second-law efficiency is defined as the ratio of the thermal efficiency to the maximum possible thermal efficiency under the same conditions
Then, in the example:
Work producing devices
Work consuming devices
Refrigerators and Heat Pumps
Total useful work delivered in a reversible process to the dead state:
The total exergy for a closed process would be given by:
The total exergy per unit mass:
Per unit mass:
Exergy change of a flow stream
Decrease or Exergy Principle (Exergy Destruction)
If boundary work
No boundary work
From table A6 & A4, for water:
u=2594.7 kJ/kg u0 = 104.83 kJ/kh
@ 180°C v=0.2472 m3/kg @ 25°C v0 = 0.001003 m3/kg
800 kPa s=6.7155 kJ/kg.K 100 kPa s0 = 0.3672 kJ/kg.K
u=386.99 kJ/kg u0 = 252.615 kJ/kh
@ 180°C v=0.044554 m3/kg @ 25°C v0 = 0.23803 m3/kg
800 kPa s=1.3327 kJ/kg.K 100 kPa s0 = 1.10605 kJ/kg.K
For a reversible process, therefore Wrev=X2-X1
Cv = 0.164 Btu/lbm.R R = 0.0621 Btu/lbm.R
First the process is at constant volume until the pressure is enough to move the piston (1-1’), then the process is at constant pressure (1’-2)
The useful work at the exit is given by the boundary work minus the work against the environment:
Wu = Wb – m.P0(v2-v1) = 2.63kJ – (1.4)(100)(0.17563–0.1652)
Wu = 1.17 kJ
From the total exergy change the only amount of useful work is 1.17kJ everything else is the exergy destroyed, therefore:
d) The second law efficiency is given by:
Notice that now we are including the exergy entering and leaving with mass, then:
In rate form:
Fortunately we usually have to deal with steady flow devices, then our equation becomes:
Per unit mass:
Previous equations can be used to determine the reversible work by making the exergy destruction term equal to zero since (i.e. no irreversibilities implies no exergy destruction)
Single stream (one inlet - one outlet):
We need to determine the exergy destroyed during this process. In this case the easiest way is by:
This equation leads us to find the entropy generated which is given by doing an entropy balance:
From table at 200 psia:
State (1) sat. liquid: h1=355.46 Btu/lbm s1=0.54379Btu/lbm.R
State (2) sat. vapor: h2=1198.8 Btu/lbm s2=1.5460 Btu/lbm.R
At environment conditions (P0=14.7 psia, To=80°F comp. liq.):
h0=48.07 Btu/lbm s0=0.09328 Btu/lbm.R
q – w = (h2 – h1) since there is no work we have:
q = 1198.8 – 355.46 = 843.34 Btu/lbm
Now we can determine the entropy generated:
960R is the absolute gas temperature (500°F)
And the exergy destroyed will be:
The exergy 9or work potential) of the steam is given by:
Therefore the temperature of the gases does not affect the exergy of thesteam. However it does affect sgen and therefore xdestroyed too.
We have an expression for the reversible work from the exergy balance for a single stream:
The ideal situation for a turbine occurs when there are no heat losses, therefore:
The entropy change is obtained from:
Then the rev. work is:
Finally, the second law efficiency would be: