Transcript Exergy Thermodynamics Professor Lee Carkner Lecture 15

Exergy
Thermodynamics
Professor Lee Carkner
Lecture 15
PAL # 14 Reversibility
Air compressed with constant specific heats
R = 0.287 (Table A-1), k = 1.4 (Table A-2)

(T2/T1) = (P2/P1)(k-1)/k
T2 = T1(P2/P1)(k-1)/k = (290)(800/100)(0.4/1.4) = 525.3 K

w = Du = cvDT = 0.727(525.3-290) =
PAL # 14 Reversibility
Air compressed with non-constant specific
heats
Need to use reduced pressure table (A-17)
For T1 = 290, Pr1 = 1.2311 and u1 = 206.91

Pr2 = (P2/P1)Pr1 = (800/100)(1.2311) = 9.849
For table A-17 this corresponds to T2 =
522.4 K and u2 = 376.16
w = u2-u1 = (376.16-206.91) =
Exergy

Exergy (x) is a measure of the work potential of
an energy source

Defined as:

The dead state is defined as the state in
thermodynamic equilibrium with the environment
Exergy is the upper limit for the work an actual
device could produce
Exergy Systems

 e.g. the amount of work you can generate from a geothermal
well depends on where you dump the waste heat
 Kinetic energy

 Potential Energy

 Both KE and PE can be completely converted to work
 n.b. V and z are relative to the environment
Kinds of Work
 Surroundings Work

 Wsurr = P0(V2-V1)
 Useful work

 Wa = W – P0(V2-V1)
 Reversible work

 If the final state is the dead state the reversible work equals the
exergy
 Irreversibility

 I = Wrev - Wu
Second Law Efficiency
Our standard thermal efficiency has 100%
as an upper limit

We instead want to compare the work
output to the true maximum; that given by
a reversible engine
The second law efficiency is:
hth,rev is the Carnot Efficiency
Comparing With Efficiency
Efficiencies
Work producing devices
hII =
Work consuming devices
hII =
Refrigerators
hII =
General Definition
hII = xrecovered/xsupplied = 1 – (xdestroyed/xsupplied)
Exergy of a Closed System

The exergy per unit mass (f) is:
f = (u-u0)+P0(v-v0)-T0(s-s0)+V2/2+gz
For a process we can subtract the exergies at
the two states
Df = (u2-u1)+P0(v2-v1)-T0(s2-s1)+(V22-V21)/2+g(z2-z1)
Flow Exergy

The flow energy is Pv and we can find its exergy
by subtracting the work needed to displace the
fluid against the atmosphere

By including this in our previous relationship we
find the flow or stream exergy, y:
y = (h-h0)-T0(s-s0)+V2/2+gz
Exergy change of a fluid stream is:
Dy = (h2-h1)-T0(s2-s1)+(V22-V21)/2+g(z2-z1)
Exergy Transfer: Heat

The most work that a given amount of
heat can generate is through a Carnot
cycle, so we can use the reversible
efficiency to find the exergy:
Where T0 is the temperature of the
environment
Transferring Exergy
Exergy Transfer: Work

One exception is overcoming atmospheric
pressure for moving boundary work
Xwork = W – Wsurr = W – P0(V2-V1)

e.g. shaft work, electrical work, etc.
Exergy Transfer: Mass
Mass flow carries exergy into or out of a
system just as it does energy

May have to integrate if fluid properties
are variable
Xmass =
Xheat =
Next Time
Next class Tuesday, April 18
Exam #2 Wednesday, April 19
Read: 8.6-8.8
Homework: Ch 8, P: 38, 42, 64, 75