lecture notes on thermodynamics 2008

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1

Lecture Notes on Thermodynamics 2008Chapter 7 Entropy

Prof. Man Y. Kim, Autumn 2008, ⓒ[email protected], Aerospace Engineering, Chonbuk National University, Korea


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Entropy (1/3)

• The Inequality of Clausius

0 QT

The inequality of Clausius is a corollary or a consequence of the 2nd law of thermodynamics. It is valid for all possible cycles, including both reversible and irreversible ones The entropy is defined from this formulation, i.e.,

2

2 11rev rev

Q QdS S ST T

and


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Entropy (2/3)

Consider first a reversible (Carnot) heat engine cycle : 0H LQ Q Q From the definition of absolute temperature scale ( )

0H L H H H L

H L L L H L

Q Q Q Q T Q QT T T Q T T T

H L H LQ Q T T

If , and H LT T 0H LQ Q Q 0H L

H L

Q Q QT T T

Finally, we conclude that for all reversible heat engines,

0Q 0QT

andNow consider an irreversible cycle heat

engine :, , , ,irr rev H L irr H L rev L irr L revW W Q Q Q Q Q Q

Consequently, for the irreversible cycle engine,,

, 0 0L irrHH L irr

H L

QQ QQ Q QT T T and

If we make the engine become more and more irreversible, but keep , , and fixed,

H H LQ T T0 0QQ

T and

Finally, we conclude that for all irreversible heat engine cycles, 0Q 0Q

T

andSimilarly, the same procedure can be applied for both reversible and irreversible

refrigeration cycles.

• Proof of the Inequality of Clausius


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Entropy (3/3)

Reversible process along path A-B2 1

1 20

A B

Q Q QT T T

Reversible process along path C-B

2 1

1 20

C B

Q Q QT T T

Subtracting the second equation from the first, we have

2 2

1 1A C

Q Q QT T T

This property is called entropy

is independent of the path → point function → property

rev

QdST

and

1rev

Qdsm T

2

2 11 rev

QS ST

• Entropy – A Property of a System


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Principle of the Increase of Entropy (1/2)

0Q

T

From the Clausius Inequality

or

2 1

1 2 intrev0Q Q

T THere, you can find that

Entropy generation

• Increase of Entropy Principle

• Entropy Generation

2 1 2 2

1 2 1 21 2 1 1intrev

0 0Q Q Q Q QS S S S dST T T T T

2

2 11

genQS ST

S

genQdS ST

where, :entropy generation due to irreversibility occurring inside the system ( because of friction, unrestricted expansion, internal energy transfer over a finite temp. difference, etc.)Reversible process : and

0genS

Q TdS W PdV Irreversible process : irr gen genQ T dS S TdS T S 1st law : irr irrQ dU W

Thermodynamic property relation :TdS dU PdV irr genW PdV T S

2 2

2 1 1 21 1

genQS S dS ST

Thus we have an expression for the change of entropy for an irreversible process as an equality, whereas in the last slide we had an inequality.

Lost Work → Exergy (Chapter 8)


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Principle of the Increase of Entropy (2/2)

Discussion 1 : There are 2 ways in which the entropy of a system can be increased by (1) transferring heat to the system (2) having an irreversible process Note : There is only one way in which entropy can be decreased by transferring heat from the systemDiscussion 2 : For an adiabatic system, the increase of entropy is always associated with the irreversibilityDiscussion 3 : The presence of irreversibility will cause the work to be smaller than the reversible work

• Discussions on Entropy Generation

0gen gen irr genQdS S S W PdV T ST


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Entropy Change of a Pure Substance

• Isentropic Process

• see Examples 7–3 (p.326) and 7–4 (p.327)

0s or 2 1s s


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Isentropic RelationsConsider the case of an ideal gas undergoing an isentropic process,

02 2 2 2 2 20

1 1 1 0 1 1 12 1 ln ln ln n0 l p

RC

pp

T P T R P T PC RT P T C P

sT

sP

However,0 0 0

0 0 0

1p v p

p p v

C C CR k kC C k C

, where : specific heat ratio

Finally we can obtain

1 12 2 2 1 2 1

1 1 1 2 1 2

k k kkT P T v P v

T P T v P v

, and : Isentropic Relation

Note : constant is a special case of a polytropic process in which the polytropic exponent n is equal to the specific heat ratio k

kPv


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T–s Diagram of the Carnot CycleConsider the Carnot cycle, i.e.,

③ → ④ : reversible isothermal heat rejection process0

rev

QdST

4 4

3 43

33 44

3

1 0 0rev L L

QQS S QT T T

Q

Efficiency

Comments on efficiency :

④ → ① : reversible adiabatic processArea 3-4-a-b-3 : heat transferred from the working fluid to the low-temperature reservoir.

0rev

QdST

→ isentropic process

Area 1-2-3-4-1 : net work of the cycle1 2 3 4 11 2 1

out

in

netth

H

areaWQ area b

WaQ

, , 0 100%H th L th thLT T T

2 2

1 21

11 22

1

1 0 0rev H H

Q QS S QT T T

Q

① → ② : reversible isothermal heat addition process

② → ③ : reversible adiabatic processArea 1-2-b-a-1 : heat transferred to the working fluid during the process

→ isentropic process

• see Example 7–6


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What is Entropy ?

Figure 7–20 Figure 7–21

Figure 7–22

Figure 7–23

Figure 7–24 Figure 7–25

Figure 7–26

Figure 7–27


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Thermodynamic Property Relations• Gibbs Equations (T–ds Relations)

TdS dU PdV TdS dH VdP For the simple compressible substance with no motion or gravitational effects, the 1st law becomes Q dU W

For a reversible process of a simple compressible substance, Q TdS W PdV and

TdS dU PdV Since enthalpy is defined asH U PV

Tds dh vdP

dh vdPdsT T

and

For a unit mass,

and

Tds du Pdv

du PdvdsT T

dH dU VdP dU VdPPdV TdS dU TdS dH VdP


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Entropy Change during Irreversible ProcessReversible cycle : reversible process along path A-B

2 1

1 20

A B

Q Q QT T T

Irreversible cycle : irreversible path C and reversible path B 2 1

1 20

C B

Q Q QT T T

Subtracting the second equation from the first, we have

2 2 2 2 2 2 2

1 1 1 1 1 1 1C A C

A C C A

Q Q Q QdS dS dST T T T

As path C was arbitrary, the general result is (both reversible and irreversible cases)

2

2 11

Q QdS S ST T

and

This is one of the most important equations of thermodynamics ! rev irr

Q Q QdS dS dST T T

and

Therefore, we can find that the entropy change for an irreversible process is larger than the change in a reversible process for the same and T.Q


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Entropy Change for a Solid(Liquid) and Ideal Gas

Tds du Pdv du specific volume is very small, and dh du CdT 2

2 11

lndu C Tds dT s s CT T T

• For a Solid or Liquid

We know that , and 0vP RTds du Pdv du C dTT v

• For an Ideal Gas

22

0 2 1 01 1

lnv vdT Rdv dT vds C s s C RT v T v

and

Similarly, , and 0pv RTds dh vdP dh C dTT P

2

20 2 1 0

1 1lnp p

dT dP dT Pds C R s s C RT P T P

and

If we assume that the specific heat is constant,2 2 2 2

2 1 0 2 1 01 1 1 1

ln ln ln lnv pT v T Ps s C R s s C RT v T P

and


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Reversible Polytropic Process for an Ideal Gas

If n is a constant,

ln ln ln 0lnd P n d P nd Vd V

1 1 2 2n n nPV PV PVconstant

• Polytropic Process

• Work done during a reversible polytropic process2

1 21

nW PdV PV and constant

2 2

1 21 1

2 12 2 2 21 1

ndVW PdVV

mR T TPV PVn n

constant

Isobaric process (P=constant) : n=0 Isothermal process (T=constant) : n=1 Isentropic process (s=constant) : n=k Isochoric process (v=constant) : n=∞

for any value of n except n=1•The reversible isothermal process

2 2

2 11 1 2 2 1 2 1 1 1 1

1 1 1 2ln lndV V PPV PV P V W PdV PV PV

V V Pconstant

constant

2 11 2

1 2ln lnV PW mRT mRT

V P or

1 12 1 2 2 1

1 2 1 1 2

nn nnP V T P V

P V T P V

and : Polytropic Relation


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Heat Transfer and Entropy Generation


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Examples (1/3)

• Turbine : Example 7–14


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Examples (2/3)

• Compressor : Example 7–14


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Examples (3/3)

• Nozzle : Example 7–16


19Homework #7Solve the Examples 7–1 ~ 7–23

Saemangum @ Jellabukdo

lecture notes on thermodynamics 2008

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