OUTLINE OF CHAPTER 4

Entropy Recall Observation 5 (Chapter 1): “All spontaneous processes in an isolated constant-volume (rigid) system result in the evolution to equilibrium (time invariant and uniform state)” Let us try to express this through a balance equation for an isolated and constant-volume

or

But in a time invariant system 0ddtθ=

Thus, at equilibrium 0genθ = Assume we can identify a variable that has a positive

generation away from equilibrium and 0d

dtθ= at equilibrium…

Since 0ddtθ> then θ is increasing and should be

MAXIMUM at equilibrium. Thus, we define:

Closed System

0genS ≥ by definition, which also means 0genS = at equilibrium.

Before using entropy, we need to establish that:

a) S is a variable of State. b) 0genS ≥

Equation is a BLACK BOX equation and cannot tell anything about genS , which is coming from internal relaxation processes within the system. Special Case in which we can gain insight on genS

A and B can interact transferring heat only. A+B is isolated

Assumption: Heat transfer takes place, but the systems remain

uniform Assume now that the heat transfer takes place very slowly so that no internal generation of entropy takes place (bunch of successive uniform states, or “equilibrium” states. Then

But for the combined system (Q=0, S= SA+SB), we have

Then

Here we assumed that the parts are uniform, but we realize the whole system is not!!! Note:

- genS is positive

- genS is proportional to the second power of the non-uniformity

Integral form of entropy balance

SECOND LAW (SECOND PRINCIPLE)

Clausius Statement of the Second Law It is not possible to construct a device that operates in a cycle and whose sole effect is to transfer heat from a cold source to a hot sink.

Consider one cycle

Thus Q1=-Q2 and the entropy balance says

or

BUT Sgen >0, which is a contradiction. The machine works well in reverse. Kelvin Planck Statement of the Second Law It is not possible to construct a device that operates in a cycle and whose sole effect is to get heat from a source and produce work.

Consider one cycle

Therefore

which means that Q<0, which is a contradiction. In other words: HEAT CANNOT BE CONVERTED IN WORK IN A CYCLIC PROCESS. THE INVERSE (Work into heat) is possible.

THUS, HEAT IS CONSIDERED A LESS “USEFUL” FORM OF ENERGY. Mechanical work converted into heat (friction) is said to be “degraded” Summary:

1st Law establishes constraints on processes (Q2=-Q1, or W=-Q above). 2nd Law establishes feasibility!!!

REVERSIBILITY Observation: genS is proportional to the SQUARE of gradients (T, v, etc). Thus in processes with small gradients, we can achieve 0genS ≈ . We associate this with slow processes The designation of “reversible” comes from the following observation:

Consider a system changing from a state at t=0 to another state at t1 and then to another state at t2=2t1. Then

Assume everything is reversed between t1 and t2.

and we get

In general , but if we keep it zero, the system is returning to its original condition. THAT IS WHY WE CALL IT REVERSIBLE. (For this we need infinitesimally small steps so gradients do not show up, obviously an idealization). When 0genS > , the process is IRREVERSIBLE In addition, if work is extracted, the maximum is obtained if it is done reversibly.

Consider a closed (constant volume) system. Then

Then, eliminating Q

HELMOLTZ FREE ENERGY

Therefore

Consider a closed (constant pressure) system. Then

GIBBS FREE ENERGY

We can consider TSgen as mechanical energy that is transformed into heat. Indeed a reversible process in a closed system gives.

Then

So that

Heat is smaller than in the reversible case and work is larger. The additional work is converted to heat!!!! THESE ARE CALLED “FREE” ENERGIES, BECAUSE THEY REPRESENT THE MAXIMUM WORK THAT CAN BE EXTRACTED FROM A SYSTEM

ADDITION OF MASS

Assume one entering stream at the same temperature.

Substituting

Now:

Then in the interval dt, we have

From the second we get

Substituting in the first

Thus, for a reversible process

and therefore

Thus, when there is no shaft work

If in addition, the system is closed

HEAT ENGINES

At steady state, or at the end of one cycle if the machine operates cyclically

From the second equation.

Substituting

Go back to

If we set Q2 =0, we get 1

1gen

QS

T−

=

But Q1 >0 , hence the contradiction because Sgen is positive. This reinforces the idea that although work and heat are equivalent, heat can only be converted to work under certain conditions.

CARNOT ENGINE Can a cyclic process be built? YES

1) Isothermal (reversible) expansion from Va to Vb adding Q1 2) Adiabatic (reversible) expansion from Vb to Vc reducing pressure. 3) Isothermal (reversible) volume reduction from Vc to Vd by removing Q2 4) Adiabatic (reversible) compression from Vd to Va using a compressor.

Thus

With

Eliminating Q2

which is what we got earlier. In addition

Qc d is the same but using T2. the work is the area inside the rectangle in the TS diagram.

TURBINES

At steady state

SPECIAL CASES Temperatures of the system = to the temperature of source/sink

Adiabatic case

MAXIMUM WORK OBTAINED OR MINIMUM NEEDED WORK AS A

FUNCTION OF P AND V

Consider the machine

Steady state equations

Thus

IDEAL GASES

Polytropic process

We prove this next, but we first need to derive the formula for IDEAL GAS change of entropy

For one mol of gas

Then

Or in terms of temperature and pressure

Or in terms of pressure and volume

We now derive that for an ideal gas /P Vc cγ = . We start with

2 22 1

1 1

ln ln 0VT Vs s c RT V

⎛ ⎞ ⎛ ⎞− = + =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

(Isentropic)

2 2

1 1

11

2 2

1 1

11

2 2

1 1

Note that / 1 //( 1)

10 ln ln1

0 ln ln

0 ln

Take exponential of both sides

exp(0) 1

P V P V V

V

c c R c c R cor c RThen

T vRT v

T vT v

T vT v

T

γ

γ

γγ

γ

−

−

= + ==> = = +

= −

⎡ ⎤⎛ ⎞ ⎛ ⎞= +⎢ ⎥⎜ ⎟ ⎜ ⎟− ⎝ ⎠ ⎝ ⎠⎣ ⎦

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥= ⋅⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

= =

11

2 2

1 1

vT v

γ −⎛ ⎞ ⎛ ⎞⋅⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

1

2 1

1 2p

s constc const

T vT v

γ −

==

⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

But for an idela gas 2 2 2

1 1 1

T PVT PV

= . Substituting 1

2 2 1

1 1 2

PV vPV v

γ −⎛ ⎞

= ⎜ ⎟⎝ ⎠

this yields 2 1

1 2p

s constc const

P vP v

γ

==

⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ or 1 1 2 2Pv P vγ γ=

My equation editor does not allow me to us an underline in V. So I switched to small v

We now rewrite

In a more familiar way . We first substitute

1

12 2 1 1

2

vP v Pvv

γ −⎛ ⎞

= ⎜ ⎟⎝ ⎠

(from above)

( )

1

12 2 1 1 1 1

2

11 1

sW vP v Pv PvN v

γγ γ

γ γ

−⎛ ⎞⎛ ⎞⎛ ⎞⎜ ⎟= − = −⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠⎝ ⎠

and also write 1 1 1Pv RT=

as well as 1/1/ 1/

1 1 21/ 1/

2 2 1

//

v Const P Pv Const P P

γγ γ

γ γ

⎛ ⎞= = ⎜ ⎟

⎝ ⎠ to get:

1

21

1

11

sW PRTN P

γγγ

γ

−⎛ ⎞⎛ ⎞⎛ ⎞ ⎜ ⎟= −⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠⎜ ⎟⎝ ⎠

Efficiencies of Turbines and Compressors Recall, for a turbine First Law (steady-state, neglecting KE and PE effects and heat losses) yields

Expansion (P2 < P1)

1 2ˆ ˆsW H H

N= −

An entropy balance yields 2 1

ˆ ˆ 0genSS S

N− = ≥

Actual turbine, irreversibilities are present, S2 > S1

The maximum theoretical amount of turbine work output is obtained for an isentropic expansion

1 2ˆ ˆs

sW H HN

= −

P2 T P1

T

s

1

2

2s

P1

P2

Ws

Then the isentropic turbine efficiency is defined by

( )1 2

1 2

ˆ ˆ/ˆ ˆ/

st

ss rev

W N H HH HW N

η −= =

−

Note, ηt < 1

Compressor

The isentropic compressor efficiency is then defined by

( )1 2

1 2

ˆ ˆ/ˆ ˆ/

s sc

s rev

W N H HH HW N

η −= =

−

Note, ηc < 1

T

s

1

2 2s

P2

P1

ENTROPY CHANGES OF MATTER

SOLIDS AND LIQUIDS

Therefore