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 = u = cvT = 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:

th,rev is the Carnot Efficiency

Comparing With Efficiency

Efficiencies

Work producing devices II =

Work consuming devices II =

Refrigerators II =

General Definition II = xrecovered/xsupplied = 1 – (xdestroyed/xsupplied)

Exergy of a Closed System

The exergy per unit mass () is:

= (u-u0)+P0(v-v0)-T0(s-s0)+V2/2+gz

For a process we can subtract the exergies at the two states

= (u2-u1)+P0(v2-v1)-T0(s2-s1)+(V22-V2

1)/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, :

= (h-h0)-T0(s-s0)+V2/2+gz Exergy change of a fluid stream is:

= (h2-h1)-T0(s2-s1)+(V22-V2

1)/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