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Lec 18: Isentropic processes,TdS relations, entropy changes

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For next time: Read: 7-2 to 7-9 Group project subject selection due on

November 3, 2003

Outline: Entropy generation and irreversible processes Entropy as a property Entropy changes for different substances

Important points: Entropy is a property of a system it is not

conserved and is generated by irreversibleprocesses Know how to identify an isentropic processes Know how to use the tables to find values for

entropy

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2

1 revintT

q

Rlb

Btuor

Kkg

kJ

m

s2 - s1 =

Units

are

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Lets look at a simpleirreversible cycle on a p-v

diagram with two processes

P

1

2

.

.A

B

Let A be

irreversible and Bbe reversible

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Irreversible cycle

0)T

QAB

By Clausius Inequality

Evaluate cyclic integral

0T

Q

T

Q

T

Q2

1 B

2

1 Acycle

(non-rev) (rev)

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Irreversible cycle

For the reversible process, B, dS=Q/dT,thus:

0dST

Q

T

Q2

1

2

1 Acycle

Rearranging and integrating dS:

2

1 AT

QS

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Second Law of Thermodynamics

Entropy is a non-conserved property!

2

1 A

12T

QSSS

This can be viewed as a mathematicalstatement of the second law (for aclosed system).

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We can write entropy change as anequality by adding a new term:

gen

2

1 A

12 S

T

QSS

entropychange

entropytransfer

due toheattransfer

entropyproduction

orgeneration

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Entropy generation

Sgen 0 is an actual irreversible process.

Sgen = 0 is a reversible process.

Sgen 0 is an impossible process.

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TEAMPLAY

Consider the equation

You have probably heard, Entropy alwaysincreases.

Could it ever decrease? What are theconditions under which this could happen(if it can)?

gen

2

1 A

12 S

T

QSS

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Entropy transfer and production

What if heat were transferred from thesystem?

The entropy can actually decrease if

gen

2

1 A

ST

Q

and heat is being transferred awayfrom the system so that Q is negative.

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Entropy Production

Sgen quantifies irreversibilities. Thelarger the irreversibilities, the greaterthe value of the entropy production,Sgen .

A reversible process will have no entropyproduction.

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Entropy transfer and production

S2 S1

> 0, Q could be + or ; if,

because Sgen is always positive.

< 0, if Q is negative and

= 0 if Q = 0 and Sgen = 0.

= 0 if Q is negative and

gen

2

1 A

ST

Q

gen

2

1 A

ST

Q

gen

2

1 A

ST

Q

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Isentropic processes

Note that a reversible (Sgen = 0),adiabatic (Q = 0) process is alwaysisentropic (S1 = S2)

But, if the process is merely isentropicwith S1 = S2, it may not be a reversibleadiabatic process.

For example, if Q 0 and gen

2

1 A

ST

Q

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Entropy generation

Consider

What if we draw our system boundariesso large that we encompass all heat

transfer interactions? We wouldthereby isolate the system.

gen

2

1 A

12 S

T

QSS

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Entropy changes of isolated systems

And then

gen

2

1 A

12 ST

QSS

0

gen12 SSS

But Sgen0. So, the entropy of anisolated system always increases. (Thisis the source of the statement, The world

is running down.)

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Entropy

)ss(xss fgf

)T(s)p,T(s f

It is tabulated just like u, v, and h.

Also,

And, for compressed or subcooled liquids,

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The entropy of a pure substance is determined from the tables, just as forany other property

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Ts Diagram for Water

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TEAMPLAY

Use the tables in your book

Find the entropy of water at 50 kPa and500 C. Specify the units.

Find the entropy of water at 100 C anda quality of 50%. Specify the units.

Find the entropy of water at 1 MPa and

120 C. Specify the units.

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Ts diagrams

pdVw

Work was the area under the curve.

Recall that the P-v diagram was veryimportant in first law analysis, and that

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For a Ts diagram

revintT

QdS

TdSQ revint

2

1

revint TdSQ

Rearrange:

Integrate:

If the internally reversible process also isisothermal at some temperature To:

STdSTQ o

2

1

orevint

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On a T-S diagram, the area under the process curve represents theheat transfer for internally reversible processes

d

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Entropy change of a thermalreservoir

For a thermal reservoir, heat transfer occursat constant temperaturethe reservoirdoesnt change temperature as heat isremoved or added:

TQ

S

Since T=constant:

T

QS

Applies ONLY tothermalreservoirs!!!!

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The Tds Equations

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Derivation ofTds equations:

dQ dW = dU

For a simple closedsystem:

dW = PdV

The work is given by:

dQ = dU + PdV

Substituting gives:

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More derivation.

For a reversible process:

TdS = dQ

Make the substitution for Q in the energyequation:

PdV+dU=TdS

Or on a per unit mass basis:

Pdv+du=Tds

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Entropy is a property. The Tds expression

that we just derived expresses entropy interms of other properties. The propertiesare independent of path.We can use theTds equation we just derived to calculatethe entropy change between any twostates:

Tds = du +Pdv

Tds = dh - vdP

Starting with enthalpy, it is possible todevelop a second Tds equation:

Tds Equations

L t l k t th t h

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Lets look at the entropy changefor an incompressible

substance:

dT

T

)T(cds

Tds = cv(T)dT + Pdv

For incompressible substances, v const, sodv = 0.

We also know that cv(T) = c(T), so we canwrite:

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Entropy change of anincompressible substance

dTT

)T(css

2

1

T

T

12

1

212

T

Tlncss

Integrating

If the specific heat does not vary with

temperature:

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TEAMPLAY

Work Problem 7-48

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Entropy change for an ideal gas

dTcdh p And

dpp

RTdTcTds p

Tds = dh - vdp

Remember dh and v for an ideal gas?

v=RT/p

Substituting:

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Change in entropy for an ideal gas

p

dpR

T

dTcds p

Dividing through by T,

Dont forget, cp=cp(T)..a function oftemperature! Integrating yields

1

2

T

T

p12pplnR

TdT)T(css

2

1

Entropy change of an ideal gas

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Entropy change of an ideal gasfor constant specific heats:

approximation

Now, if the temperature range is solimited that c

p constant (and c

v

constant),

1

2pp

T

Tlnc

T

dTc

1

2

1

2p12

p

plnR

T

Tlncss

Entropy change of an ideal gas

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Entropy change of an ideal gasfor constant specific heats:

approximation

Similarly it can be shown from

Tds = du + pdv

that

1

2

1

2v12

vvlnR

TTlncss

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