- 131 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' EXERGY' - philip-bennett

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

### EXERGY

Basic Definitions

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 potential energy: (work potential) of a system is equal to the potential energy itself regardless of the temperature and pressure of 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.

Surrounding Work: is the work done by or against the surroundings during a process. It has significance only for a process where boundary work occurs (closed system).

Wsurr = P0(V2 – V1)

Then the useful work would be:

Wu = W – Wsurr = W – P0(V2 – V1)

How is Wsurr for a rigid tank?

Reversible Work: maximum amount of useful work that can be produced (or the minimum work that needs to be supplied) as a system undergoes a process between the specified initial and final states.

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

SECOND LAW EFFICIENCY

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:

Closed System

Total useful work delivered in a reversible process to the dead state:

The total exergy for a closed process would be given by:

(kJ)

The total exergy per unit mass:

(kJ/kg)

EXERGY CHANGE FOR A CLOSED SYSTEM

Per unit mass:

Exergy of a flow stream

Exergy change of a flow stream

Sol.

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

From table A13 (R-134a superheated vapor):

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

P (kPa)

120°C

1’

180

2

20 °C

1

100

v (m3/kg)

Solution:

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)

States 1 and 2 are in the region of superheated vapor, the summarized data from table is:

- P1=140kPa v1=0.1652 m3/kg P2=180kPa v2=0.17563 m3/kg
- u1=246.01 kJ/kg u2=331.96 kJ/kg
- T1=20°C s1=1.0532 kJ/kgK T2=120°C s2=1.3118 kJ/kgK
- Also at (1’) v1’ = v1 and P1’=P2
- The work done is the boundary work, from (1) to (1’) is zero since it is a constant volume process; from (1’) to (2) is a constant pressure process. Then the boundary work is given by:
- Wb = P2.m.(v2-v1) = (180)(1.4)(0.17563 – 0.1652) = 2.63 kJ
- Doing a energy balance we obtain:
- Q-W = m(u1’ – u1) + m(u2 – u1’) = m(u2 – u1)
- Then Q = m(u2 – u1) + W = (1.4)(331.96 – 246.01) + 2.63 kJ
- Q = 122.96 kJ

Since there is no kinetic nor potential energy involved the exergy change can be expressed by:

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:

Exergy Balance Open System

Notice that now we are including the exergy entering and leaving with mass, then:

(kJ)

In rate form:

(kW)

Fortunately we usually have to deal with steady flow devices, then our equation becomes:

for a single stream this last equation becomes:

(kW)

Per unit mass:

(kJ/kg)

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)

Then:

Solution:

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

We need to determine q in the previous equation, we obtain this by doing an energy balance:

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 find the actual work by doing an ener4gy balance:

- Kinetic and potential energy changes are assumed to be zero.
- Consider specific heats for the enthalpy change. Solving for work we have:
- Cp=1.134 kJ/kg.K (from table)
- Q = -30kW (lost)

Substitue in the previous equation to obtain:

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:

The exergy destroyed will be given by:

Finally, the second law efficiency would be:

Download Presentation

Connecting to Server..