ms212 thermodynamics of materials -...

MS212 Thermodynamics of Materials (소재열역학의이해)

Lecture Note: Chapter 3

Byungha Shin (신병하)Dept. of MSE, KAIST

1

2017 Spring Semester

Course InformationSyllabus

Chap 1. Introduction and Definition of Terms (1 lecture)Chap 2. The First Law of Thermodynamics (1.5 lectures)Chap 3. The Second Law of Thermodynamics (3 lectures)Chap 4. The Statistical Interpretation of Entropy (2 lectures)Chap 5. Auxiliary Functions (2.5 lectures)Chap 6. Heat Capacity, Enthalpy, Entropy and the Third

Law of Thermodynamics (3 lectures)Chap 7. Phase Equilibrium in a One-Component System (3 lectures)Chap 8. The Behavior of Gases (3 lectures)Chap 9. The Behavior of Solutions (3 lectures)Chap 10. Gibbs Free Energy Composition and Phase Diagram (3 lectures)

Reversible vs. Irreversible Process

Consider a cycling process (12 then 21)• Reversible:

𝑤𝑤𝑟𝑟𝑟𝑟𝑟𝑟,1→2 = ∫12 𝑃𝑃𝑃𝑃𝑃𝑃 = ∫𝑉𝑉1

𝑉𝑉2 𝑅𝑅𝑅𝑅𝑉𝑉𝑃𝑃𝑃𝑃

𝑤𝑤𝑟𝑟𝑟𝑟𝑟𝑟,2→1 = ∫𝑉𝑉2𝑉𝑉1 𝑅𝑅𝑅𝑅

𝑉𝑉𝑃𝑃𝑃𝑃 = −𝑤𝑤𝑟𝑟𝑟𝑟𝑟𝑟,1→2

𝑤𝑤𝑟𝑟𝑟𝑟𝑟𝑟 = ∮𝛿𝛿𝑤𝑤 = 𝑤𝑤𝑟𝑟𝑟𝑟𝑟𝑟,1→2 + 𝑤𝑤𝑟𝑟𝑟𝑟𝑟𝑟,2→1 = 0∆𝑈𝑈 = ∮𝑃𝑃𝑈𝑈 = 0𝑞𝑞𝑟𝑟𝑟𝑟𝑟𝑟 = ∮𝛿𝛿𝑞𝑞 = ∆𝑈𝑈 + 𝑤𝑤𝑟𝑟𝑟𝑟𝑟𝑟 = 0

• Irreversible:𝑤𝑤𝑖𝑖𝑟𝑟𝑟𝑟,1→2 < 𝑤𝑤𝑟𝑟𝑟𝑟𝑟𝑟,1→2; 𝑤𝑤𝑖𝑖𝑟𝑟𝑟𝑟,2→1 > 𝑤𝑤𝑟𝑟𝑟𝑟𝑟𝑟,2→1

𝑤𝑤𝑖𝑖𝑟𝑟𝑟𝑟 < 0; ∆𝑈𝑈 = ? ; 𝑞𝑞𝑖𝑖𝑟𝑟𝑟𝑟 < 0

P

V

P1

P2

V1 V2

1

2

𝑃𝑃 =𝑅𝑅𝑅𝑅𝑃𝑃

Q: What happens to the surrounding?

Heat Engine𝜃𝜃2

𝜃𝜃1𝑞𝑞1

𝑞𝑞2

𝑤𝑤𝑞𝑞2𝑞𝑞1

𝑤𝑤Cartoon of Watt’s heat engine

system

After the completion of a cycle:

∆𝑈𝑈 = 𝑞𝑞2 − 𝑞𝑞1 − 𝑤𝑤 = 0 𝑤𝑤 = 𝑞𝑞2 − 𝑞𝑞1

𝜂𝜂 =𝑤𝑤𝑞𝑞2

= 1 −𝑞𝑞1𝑞𝑞2

Efficiency,

working substance: water

Carnot EngineWhat is the greatest possible efficiency?

P

V

1

3

2

4

T2

T1

Carnot Cycle1 2: isothermal reversible expansion2 3: adiabatic reversible expansion3 4: isothermal reversible compression4 1: adiabatic reversible compression

𝜂𝜂𝐶𝐶 =𝑤𝑤𝑟𝑟𝑟𝑟𝑟𝑟𝑞𝑞2,𝑟𝑟𝑟𝑟𝑟𝑟

= 1 −𝑞𝑞1,𝑟𝑟𝑟𝑟𝑟𝑟

𝑞𝑞2,𝑟𝑟𝑟𝑟𝑟𝑟

Q: For maximum efficiency, why reversible?

𝑤𝑤𝑟𝑟𝑟𝑟𝑟𝑟

Carnot EngineAny irreversible cycle whose η > ηC ? 𝜂𝜂 =

𝑤𝑤𝑞𝑞2

= 1 −𝑞𝑞1𝑞𝑞2

,

(i) increase 𝑤𝑤𝑞𝑞2 = 𝑞𝑞2,𝑟𝑟𝑟𝑟𝑟𝑟 ; 𝑤𝑤 > 𝑤𝑤𝑟𝑟𝑟𝑟𝑟𝑟 𝑞𝑞1 < 𝑞𝑞1,𝑟𝑟𝑟𝑟𝑟𝑟




𝜃𝜃2

𝜃𝜃1



𝑤𝑤𝑟𝑟𝑟𝑟𝑟𝑟 𝑤𝑤

𝑞𝑞2

𝑞𝑞1

𝜂𝜂𝜂𝜂𝑐𝑐

𝜃𝜃2

𝜃𝜃1𝑞𝑞1,𝑟𝑟𝑟𝑟𝑟𝑟 − 𝑞𝑞1 > 0

𝑞𝑞2,𝑟𝑟𝑟𝑟𝑟𝑟 − 𝑞𝑞2 = 0

𝑤𝑤 − 𝑤𝑤𝑟𝑟𝑟𝑟𝑟𝑟 > 0

Carnot cycle running backward (reversible)

Irreversible cycle whose 𝑤𝑤 > 𝑤𝑤𝑟𝑟𝑟𝑟𝑟𝑟

Heat engine with η = 1 (100% efficiency) which will generate work from a single heat reservoir !• Does not violate the 1st law• But not possible, violates the 2nd law• Perpetual motion machine of the 2nd kind• η ≯ ηC with larger w

Carnot EngineAny irreversible cycle whose η > ηC ? 𝜂𝜂 =

𝑤𝑤𝑞𝑞2

= 1 −𝑞𝑞1𝑞𝑞2

,

(ii) decrease 𝑞𝑞2𝑞𝑞2 < 𝑞𝑞2,𝑟𝑟𝑟𝑟𝑟𝑟 ; 𝑤𝑤 = 𝑤𝑤𝑟𝑟𝑟𝑟𝑟𝑟 𝑞𝑞1 < 𝑞𝑞1,𝑟𝑟𝑟𝑟𝑟𝑟




𝜃𝜃2

𝜃𝜃1



𝑤𝑤𝑟𝑟𝑟𝑟𝑟𝑟= 𝑤𝑤

𝑞𝑞2

𝑞𝑞1

𝜂𝜂𝜂𝜂𝑐𝑐

𝜃𝜃2

𝜃𝜃1𝑞𝑞1,𝑟𝑟𝑟𝑟𝑟𝑟 − 𝑞𝑞1 > 0

𝑞𝑞2,𝑟𝑟𝑟𝑟𝑟𝑟 − 𝑞𝑞2 > 0

𝑤𝑤 − 𝑤𝑤𝑟𝑟𝑟𝑟𝑟𝑟 = 0

Carnot cycle running backward (reversible)

Irreversible cycle whose 𝑞𝑞2 < 𝑞𝑞2,𝑟𝑟𝑟𝑟𝑟𝑟

Spontaneous transfer of heat from the cold to hot reservoir !• Does not violate the 1st law• But not possible, violates the 2nd law• Perpetual motion machine of the 2nd kind• η ≯ ηC with smaller q

Therefore, 𝜼𝜼 ≤ 𝜼𝜼𝑪𝑪 !

Preliminary Formulation of The 2nd Law

The Principle of Thompson:It is impossible, by means of a cyclic process, to take heat from a reservoir and convert it to work without, in the same operation, transferring heat to a cold reservoir.

The Principle of Clausius:It is impossible to transfer hear from a cold to a hot reservoir without, in the same operation, converting a certain amount of work to heat.

2nd Law: “Nothing for something is impossible.”1st Law: “Something from nothing is impossible.”

Thermodynamic Temperature Scale

𝜂𝜂𝑐𝑐

𝜂𝜂𝑐𝑐

𝜃𝜃3

𝜃𝜃2

𝜃𝜃1





𝑤𝑤32= 𝑞𝑞3,𝑟𝑟𝑟𝑟𝑟𝑟− 𝑞𝑞2,𝑟𝑟𝑟𝑟𝑟𝑟

𝑤𝑤21= 𝑞𝑞2,𝑟𝑟𝑟𝑟𝑟𝑟− 𝑞𝑞1,𝑟𝑟𝑟𝑟𝑟𝑟

𝜂𝜂𝐶𝐶 = 1 −𝑞𝑞2,𝑟𝑟𝑟𝑟𝑟𝑟



𝑞𝑞3,𝑟𝑟𝑟𝑟𝑟𝑟= 𝑓𝑓(𝜃𝜃2,𝜃𝜃3)




𝑞𝑞2,𝑟𝑟𝑟𝑟𝑟𝑟=


𝑞𝑞3,𝑟𝑟𝑟𝑟𝑟𝑟�𝑞𝑞3,𝑟𝑟𝑟𝑟𝑟𝑟

𝑞𝑞2,𝑟𝑟𝑟𝑟𝑟𝑟,



𝑓𝑓 𝜃𝜃1,𝜃𝜃2 =𝑓𝑓 𝜃𝜃1,𝜃𝜃3𝑓𝑓 𝜃𝜃2,𝜃𝜃3

: independent of θ3

Thermodynamic Temperature Scale

𝑓𝑓 𝜃𝜃1,𝜃𝜃2 = 𝐹𝐹 𝜃𝜃1 � 𝐹𝐹 𝜃𝜃2 , 𝑓𝑓 𝜃𝜃2,𝜃𝜃3 = 𝐹𝐹 𝜃𝜃2 � 𝐹𝐹 𝜃𝜃3


𝑞𝑞2,𝑟𝑟𝑟𝑟𝑟𝑟= 1 −

𝑅𝑅1𝑅𝑅2


𝑞𝑞2,𝑟𝑟𝑟𝑟𝑟𝑟=


𝑞𝑞3,𝑟𝑟𝑟𝑟𝑟𝑟�𝑞𝑞3,𝑟𝑟𝑟𝑟𝑟𝑟

𝑞𝑞2,𝑟𝑟𝑟𝑟𝑟𝑟, 𝑓𝑓 𝜃𝜃1,𝜃𝜃2 =

𝑓𝑓 𝜃𝜃1,𝜃𝜃3𝑓𝑓 𝜃𝜃2,𝜃𝜃3

: independent of θ3

𝑓𝑓 𝜃𝜃1,𝜃𝜃3 =𝐹𝐹(𝜃𝜃1)𝐹𝐹(𝜃𝜃3)

, 𝑓𝑓 𝜃𝜃2,𝜃𝜃3 =𝐹𝐹(𝜃𝜃2)𝐹𝐹(𝜃𝜃3)

Q: What if ?

• Simplest form of F(θ) is T, F(θ) ≡ T by Kelvin• T : Absolute Temperature or Thermodynamic Temperature

•

Thermodynamic Temperature ScaleIdeal gas temperature scale: 𝑅𝑅 =

𝑃𝑃𝑃𝑃𝑅𝑅 ; 273.16 K at the triple point

P

V

1

3

2

4

T2

T1

q2

w23

q23 = 0

w12

q1 w41

w41

q41 = 0

• Path 1 2: 𝑤𝑤12 = 𝑞𝑞2 = �𝑉𝑉1

𝑉𝑉2𝑃𝑃𝑃𝑃𝑃𝑃 = 𝑅𝑅𝑅𝑅2 ln

𝑃𝑃2𝑃𝑃1

> 0; ∆𝑈𝑈12 = 0

• Path 2 3:

𝑤𝑤23 = −∆𝑈𝑈23 = −�𝑅𝑅2

𝑅𝑅1𝐶𝐶𝑉𝑉𝑃𝑃𝑅𝑅 = 𝐶𝐶𝑉𝑉 𝑅𝑅2 − 𝑅𝑅1 > 0; 𝑞𝑞23 = 0

• Path 2 3: 𝑤𝑤34 = 𝑞𝑞1 = �𝑉𝑉3

𝑉𝑉4𝑃𝑃𝑃𝑃𝑃𝑃 = 𝑅𝑅𝑅𝑅1 ln

𝑃𝑃4𝑃𝑃3

< 0; ∆𝑈𝑈34 = 0

• Path 4 1:𝑤𝑤41 = −∆𝑈𝑈41 = −�

𝑅𝑅1

𝑅𝑅2𝐶𝐶𝑉𝑉𝑃𝑃𝑅𝑅 = 𝐶𝐶𝑉𝑉 𝑅𝑅1 − 𝑅𝑅2 < 0; 𝑞𝑞41 = 0

• Net work: 𝑤𝑤 = 𝑤𝑤12 + 𝑤𝑤23 + 𝑤𝑤34 + 𝑤𝑤41 = 𝑅𝑅𝑅𝑅2 ln𝑃𝑃2𝑃𝑃1

+ 𝑅𝑅𝑅𝑅1 ln𝑃𝑃4𝑃𝑃3

• Efficiency, 𝜼𝜼𝑪𝑪 =𝑤𝑤

𝑞𝑞2(= 𝑤𝑤12)= 1 +

𝑅𝑅𝑅𝑅1 ln 𝑃𝑃4𝑃𝑃3𝑅𝑅𝑅𝑅2 ln𝑃𝑃2𝑃𝑃1

= 𝟏𝟏 −𝑻𝑻𝟏𝟏𝑻𝑻𝟐𝟐

𝑅𝑅2𝑃𝑃1𝛾𝛾−1 = 𝑅𝑅1𝑃𝑃4

𝛾𝛾−1; 𝑅𝑅2𝑃𝑃2𝛾𝛾−1 = 𝑅𝑅2𝑃𝑃3

𝛾𝛾−1 𝑃𝑃2/𝑃𝑃1 = 𝑃𝑃3/𝑃𝑃4

• Thermodynamic T scale = Ideal gas T scale

The Clausius InequalityFor a reversible Carnot engine operating between T2 and T1,


𝑞𝑞2,𝑟𝑟𝑟𝑟𝑟𝑟= 1 −

𝑅𝑅1𝑅𝑅2

;𝑞𝑞1,𝑟𝑟𝑟𝑟𝑟𝑟

𝑞𝑞2,𝑟𝑟𝑟𝑟𝑟𝑟=𝑅𝑅1𝑅𝑅2


𝑅𝑅2−


𝑅𝑅1= 0

(>0) (<0)


𝑅𝑅2+𝑞𝑞1,𝑟𝑟𝑟𝑟𝑟𝑟

𝑅𝑅1= 0

For an irreversible engine operating between T2 and T1,

𝜂𝜂 = 1 −𝑞𝑞1𝑞𝑞2

< 𝜂𝜂𝐶𝐶 = 1 −𝑅𝑅1𝑅𝑅2

𝑞𝑞2𝑅𝑅2

−𝑞𝑞1𝑅𝑅1

< 0

𝑞𝑞2𝑅𝑅2

+𝑞𝑞1𝑅𝑅1

< 0

𝑞𝑞2𝑅𝑅2

+𝑞𝑞1𝑅𝑅1≤ 0 (equality is for reversible process)

The Clausius InequalityFor a general case: an arbitrary closed path can be broken into a series of Carnot cycles (isothermal + adiabatic processes)

P

V

adiabats

isotherms

𝑅𝑅6𝐹𝐹

𝐸𝐸

𝑅𝑅4𝐴𝐴

𝑅𝑅2

𝑅𝑅1 𝑅𝑅5𝑅𝑅3

𝐵𝐵

𝐷𝐷𝐻𝐻

𝐶𝐶 𝐺𝐺 𝑃𝑃

𝑃𝑃

work done by the system = area enclosed by a red loop≈ sum of areas enclosed by Carnot cycles

For ABCD: 𝑞𝑞2𝑅𝑅2

+𝑞𝑞1𝑅𝑅1≤ 0

For EFGH: 𝑞𝑞3𝑅𝑅3

+𝑞𝑞4𝑅𝑅4≤ 0

�𝑖𝑖=1

𝑛𝑛𝑞𝑞𝑖𝑖𝑅𝑅𝑖𝑖≤ 0

lim𝑛𝑛→∞

�𝑖𝑖=1

𝑛𝑛𝑞𝑞𝑖𝑖𝑅𝑅𝑖𝑖

= �𝜹𝜹𝒒𝒒𝑻𝑻≤ 𝟎𝟎

EntropyFor a reversible cyclic process, �

𝛿𝛿𝑞𝑞𝑟𝑟𝑟𝑟𝑟𝑟𝑅𝑅

= 0𝛿𝛿𝑞𝑞𝑟𝑟𝑟𝑟𝑟𝑟𝑅𝑅

is a state function

𝛿𝛿𝑞𝑞𝑟𝑟𝑟𝑟𝑟𝑟𝑅𝑅 ≡ 𝑃𝑃𝑑𝑑, 𝑑𝑑 is entropy

State 1(T1, V1)

State 2(T2, V2)

q12 w12

q21 w21

�𝛿𝛿𝑞𝑞𝑅𝑅

= �1

2 𝛿𝛿𝑞𝑞𝑅𝑅

+ �2

1 𝛿𝛿𝑞𝑞𝑟𝑟𝑟𝑟𝑟𝑟𝑅𝑅

< 0

�1

2 𝛿𝛿𝑞𝑞𝑅𝑅 < −�

2

1 𝛿𝛿𝑞𝑞𝑟𝑟𝑟𝑟𝑟𝑟𝑅𝑅 = �

1

2 𝛿𝛿𝑞𝑞𝑟𝑟𝑟𝑟𝑟𝑟𝑅𝑅

Clausius inequality:

For this inequality to be satisfied,𝛿𝛿𝑞𝑞𝑅𝑅 <

𝛿𝛿𝑞𝑞𝑟𝑟𝑟𝑟𝑟𝑟𝑅𝑅 , or 𝛿𝛿𝑞𝑞 < 𝑅𝑅𝑃𝑃𝑑𝑑

If reversible, 𝛿𝛿𝑞𝑞 = 𝛿𝛿𝑞𝑞𝑟𝑟𝑟𝑟𝑟𝑟 = 𝑅𝑅𝑃𝑃𝑑𝑑

𝛿𝛿𝑞𝑞 ≤ 𝑅𝑅𝑃𝑃𝑑𝑑 𝑃𝑃𝑑𝑑)𝛿𝛿𝑞𝑞=0 ≥ 0 • Any irreversible (= natural = spontaneous) process occurring inside a adiabatic system, entropy always increases.

• At equilibrium, no entropy change𝑃𝑃𝑑𝑑 =𝛿𝛿𝑞𝑞𝑅𝑅 + 𝑃𝑃𝑑𝑑𝑖𝑖𝑟𝑟𝑟𝑟

The 2nd Law of Thermodynamics

Entropy and the criterion for equilibrium: for an isolated systemof constant U and V, the equilibrium is when S is maximum.

• The entropy S, defined by dS = δqrev / T, is a state function.• The entropy of a system in an adiabatic enclosure can never

decrease; it increases during an irreversible process and remains constant during a reversible process.

S

A+B C+D→ X

Isolated system with constant U and VdU = 0, dV = 0 δq = 0𝑃𝑃𝑑𝑑𝑖𝑖𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑟𝑟𝑠𝑠 ≥ 0𝑃𝑃𝑑𝑑𝑖𝑖𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑟𝑟𝑠𝑠 = 0, at equilibrium.

Direction of Process

Isolated system consisting of a hot and cold bodies

hot, T2 cold, T1 (<T2)

δq

𝑃𝑃𝑑𝑑𝑖𝑖𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑟𝑟𝑠𝑠 = 𝑃𝑃𝑑𝑑ℎ𝑠𝑠𝑠𝑠 + 𝑃𝑃𝑑𝑑𝑐𝑐𝑠𝑠𝑠𝑠𝑠𝑠 =− 𝛿𝛿𝑞𝑞𝑅𝑅2

+𝛿𝛿𝑞𝑞𝑅𝑅1

> 0

• S of the system increases for irreversible, natural, spontaneous heat transfer.

• Natural direction of process = increasing S; the 2nd law of thermodynamics often called “time’s arrow”

• Reversible transfer of heat is only possible when T1 = T2

Cyclic Process�𝛿𝛿𝑞𝑞𝑻𝑻≤ 0 in cyclic process (the equality is for a Carnot cycle):

Which T ? T of the working fluid ? T of the reservoirs ?

Master equation: ∆𝑑𝑑12 ≥ �1

2 𝛿𝛿𝑞𝑞𝑅𝑅𝐶𝐶𝐶𝐶

≥ �1

2 𝛿𝛿𝑞𝑞𝑅𝑅𝑅𝑅𝑅𝑅𝐶𝐶

= is for internally (mech. & therm.) reversible processes

= is for externally thermally reversible process

I. A totally reversible process: The piston-cylinder running a Carnot cycle between thermal reservoirs TH and TL.

II. An externally thermally irreversible process: A piston-cylinder heat engine that is operating almost in a Carnot cycle; heat transfer is radiative so there is a true temperature discontinuity at the control surface.

III. An internally thermally irreversible process: Heat conduction down a rod that is held at temperature TH on the top and TL on the bottom end, with a temperature discontinuity in the middle.

IV. An internally mechanically irreversible process: A frictional piston-cylinder device with an ideal gas that is trying to operate in a Carnot cycle. (Handout #1)

I. Totally ReversibleInternally thermally reversible: Tsys is spatially uniform always but may vary with timeInternally mechanically reversible: no friction; quasi-static processExternally thermally reversible: Tsys = T of reservoir that it’s transferring heat to or fromExternally mechanically reversible: no friction at control surface

𝑅𝑅𝐿𝐿

𝑅𝑅𝐻𝐻

𝑄𝑄𝐻𝐻

𝑄𝑄𝐿𝐿

C.S.𝑊𝑊

P

V

𝑄𝑄𝐻𝐻

𝑄𝑄𝐿𝐿

1

2

34

𝑄𝑄 = 0𝑄𝑄 = 0

T

S

𝑄𝑄𝐻𝐻

𝑄𝑄𝐿𝐿

1 2

34

𝑄𝑄 = 0 𝑄𝑄 = 0

Q: Does the working fluid have to be an ideal gas?QH and QL in T-S diagram?

I. Totally Reversible

𝑅𝑅𝐿𝐿

𝑅𝑅𝐻𝐻

𝑄𝑄𝐻𝐻

𝑄𝑄𝐿𝐿

C.S.𝑊𝑊

P

V

𝑄𝑄𝐻𝐻

𝑄𝑄𝐿𝐿

1

2

34

𝑄𝑄 = 0𝑄𝑄 = 0

T

S

𝑄𝑄𝐻𝐻

𝑄𝑄𝐿𝐿

1 2

34

𝑄𝑄 = 0 𝑄𝑄 = 0

�𝛿𝛿𝑞𝑞𝑅𝑅𝑅𝑅𝑅𝑅𝐶𝐶

=𝑄𝑄𝐻𝐻𝑅𝑅𝐻𝐻

−𝑄𝑄𝐿𝐿𝑅𝑅𝐿𝐿

= 0 �𝛿𝛿𝑞𝑞𝑅𝑅𝐶𝐶𝐶𝐶

= 0 (because TCS = TRES; externally thermally reversible)

∆𝑑𝑑12 ≡ �1

2 𝛿𝛿𝑞𝑞𝑅𝑅 𝑟𝑟𝑟𝑟𝑟𝑟

=𝑄𝑄𝐻𝐻𝑅𝑅

, where you can use either TCS or TRES.

𝑃𝑃𝑑𝑑 =𝛿𝛿𝑞𝑞𝑅𝑅

+ 𝛿𝛿𝑑𝑑𝑔𝑔𝑟𝑟𝑛𝑛0

II. Externally Thermally Irreversible• A piston-cylinder heat engine operating between reservoirs at TH and TL.• T of the system (which is always uniform) cycles between TMAX < TH (finite

temperature discontinuity with high-T reservoir) and TMIN = TL (chosen for simplicity).

𝑅𝑅𝐿𝐿

𝑅𝑅𝐻𝐻

𝑄𝑄𝐻𝐻

𝑄𝑄𝐿𝐿

C.S.𝑊𝑊A

𝑅𝑅𝑀𝑀𝑀𝑀𝑀𝑀 ≪ 𝑅𝑅𝐻𝐻

𝑅𝑅𝑀𝑀𝑀𝑀𝑀𝑀 = 𝑅𝑅𝐿𝐿

T

S

𝑅𝑅𝑀𝑀𝑀𝑀𝑀𝑀


𝑅𝑅𝐻𝐻

A

21

3 4

𝑄𝑄𝐻𝐻𝑀𝑀

𝑄𝑄𝐿𝐿

2𝑟𝑟𝑟𝑟𝑠𝑠 1𝑟𝑟𝑟𝑟𝑠𝑠

𝑊𝑊𝑀𝑀𝑄𝑄𝐻𝐻𝑀𝑀

Δ𝑑𝑑12𝑟𝑟𝑟𝑟𝑠𝑠,𝑀𝑀

Irreversible Process A

II. Externally Thermally IrreversibleT

S



𝑅𝑅𝐻𝐻

A

21

3 4


𝑄𝑄𝐿𝐿



Δ𝑑𝑑12𝑟𝑟𝑟𝑟𝑠𝑠,𝑀𝑀 T

S

𝑅𝑅𝐻𝐻

𝑅𝑅𝐿𝐿

B

21

3 4

𝑄𝑄𝐻𝐻𝐵𝐵

𝑄𝑄𝐿𝐿


𝑊𝑊𝐵𝐵𝑄𝑄𝐻𝐻𝐵𝐵

Δ𝑑𝑑12𝑟𝑟𝑟𝑟𝑠𝑠,𝐵𝐵

Reversible Process B:• Same QL as in Process A• Sys. operating b/w TH ↔ TL• Reservoirs at TH and TL

𝜂𝜂𝑀𝑀 < 𝜂𝜂𝐵𝐵: 1 −𝑄𝑄𝐿𝐿𝑄𝑄𝐻𝐻𝑀𝑀

< 1 −𝑄𝑄𝐿𝐿𝑄𝑄𝐻𝐻𝐵𝐵

𝑄𝑄𝐻𝐻𝐵𝐵 > 𝑄𝑄𝐻𝐻𝑀𝑀


=𝑄𝑄𝐻𝐻𝑀𝑀

𝑅𝑅𝐻𝐻−𝑄𝑄𝐿𝐿𝑅𝑅𝐿𝐿

<𝑄𝑄𝐻𝐻𝐵𝐵

𝑅𝑅𝐻𝐻−𝑄𝑄𝐿𝐿𝑅𝑅𝐿𝐿

= 0 (because B is reversible)


< 0

Irreversible Process A:• Sys. operating b/w TMAX ↔ TL• Reservoirs at TH and TL

II. Externally Thermally IrreversibleT

S



𝑅𝑅𝐻𝐻

A

21

3 4


𝑄𝑄𝐿𝐿



Δ𝑑𝑑12𝑟𝑟𝑟𝑟𝑠𝑠,𝑀𝑀

Reversible Process C:• Same QL as in Process A• Sys. operating b/w TMAX ↔ TL• Reservoirs at TMAX and TL

Irreversible Process A:• Sys. operating b/w TMAX ↔ TL• Reservoirs at TH and TL

T

S

𝑅𝑅𝐻𝐻𝐶𝐶 = 𝑅𝑅𝑀𝑀𝑀𝑀𝑀𝑀

𝑅𝑅𝐿𝐿C

21

3 4

𝑄𝑄𝐻𝐻𝐶𝐶

𝑄𝑄𝐿𝐿


𝑊𝑊𝐶𝐶𝑄𝑄𝐻𝐻𝐶𝐶

Δ𝑑𝑑12𝑟𝑟𝑟𝑟𝑠𝑠,𝐶𝐶

States 1 and 2 are identical in A and C; Path 12 identical in A and C; System property including ∆S12 in process 12 identical 𝑄𝑄𝐻𝐻𝑀𝑀 = 𝑄𝑄𝐻𝐻𝐶𝐶

�𝛿𝛿𝑞𝑞𝑅𝑅𝐶𝐶𝐶𝐶

=𝑄𝑄𝐻𝐻𝑀𝑀

𝑅𝑅𝑀𝑀𝑀𝑀𝑀𝑀−𝑄𝑄𝐿𝐿𝑅𝑅𝐿𝐿

=𝑄𝑄𝐻𝐻𝐶𝐶

𝑅𝑅𝑀𝑀𝑀𝑀𝑀𝑀−𝑄𝑄𝐿𝐿𝑅𝑅𝐿𝐿

= 0

∆𝑑𝑑12𝑠𝑠𝑠𝑠𝑠𝑠,𝑀𝑀 =

𝑄𝑄𝐻𝐻𝑅𝑅𝐶𝐶𝐶𝐶

+ 𝑑𝑑𝑔𝑔𝑟𝑟𝑛𝑛,𝑠𝑠𝑠𝑠𝑠𝑠0

internally reversible no entropy generated internally; entropy increases only by the amount of entropy transferred into the system with heat

II. Externally Thermally IrreversibleWhat distinguishes externally thermally irreversible cycle A and totally reversible cycle C ? The entropy change in .

∆𝑑𝑑12𝑟𝑟𝑟𝑟𝑠𝑠,𝑀𝑀 =

−𝑄𝑄𝐻𝐻𝑅𝑅𝐻𝐻

+ 𝑑𝑑𝑔𝑔𝑟𝑟𝑛𝑛,𝑟𝑟𝑟𝑟𝑠𝑠0

∆𝑑𝑑12𝑟𝑟𝑟𝑟𝑠𝑠,𝐶𝐶 =

−𝑄𝑄𝐻𝐻𝑅𝑅𝑀𝑀𝑀𝑀𝑀𝑀

∆𝑑𝑑12𝑟𝑟𝑟𝑟𝑠𝑠,𝐶𝐶 =

𝑄𝑄𝐻𝐻𝑅𝑅𝑀𝑀𝑀𝑀𝑀𝑀

>𝑄𝑄𝐻𝐻𝑅𝑅𝐻𝐻

= ∆𝑑𝑑12𝑟𝑟𝑟𝑟𝑠𝑠,𝑀𝑀

Entropy change of “the universe”

∆𝑑𝑑12𝑠𝑠𝑠𝑠𝑠𝑠,𝑀𝑀 + ∆𝑑𝑑12

𝑟𝑟𝑟𝑟𝑠𝑠,𝑀𝑀 =𝑄𝑄𝐻𝐻𝑅𝑅𝑀𝑀𝑀𝑀𝑀𝑀

+−𝑄𝑄𝐻𝐻𝑅𝑅𝐻𝐻

= 𝑄𝑄𝐻𝐻1

𝑅𝑅𝑀𝑀𝑀𝑀𝑀𝑀−

1𝑅𝑅𝐻𝐻

> 𝟎𝟎

∆𝑑𝑑12𝑠𝑠𝑠𝑠𝑠𝑠,𝐶𝐶 + ∆𝑑𝑑12

𝑟𝑟𝑟𝑟𝑠𝑠,𝐶𝐶 =𝑄𝑄𝐻𝐻𝑅𝑅𝑀𝑀𝑀𝑀𝑀𝑀

+−𝑄𝑄𝐻𝐻𝑅𝑅𝑀𝑀𝑀𝑀𝑀𝑀

= 𝑄𝑄𝐻𝐻1

𝑅𝑅𝑀𝑀𝑀𝑀𝑀𝑀−

1𝑅𝑅𝑀𝑀𝑀𝑀𝑀𝑀

= 𝟎𝟎

positive entropy generation b/c of irreversibility in the universe (specifically between the HT reservoir and the system)

no entropy generation b/c totally reversible

III. Internally Thermally IrreversibleTsys is spatially non-uniform; otherwise reversible

𝑅𝑅𝐿𝐿

𝑅𝑅𝐿𝐿

𝑅𝑅𝐻𝐻

𝑄𝑄

C.S.

𝑄𝑄/2

top half

bottom halfsystem

T

S

𝑅𝑅𝐻𝐻

𝑅𝑅𝐿𝐿 𝑄𝑄2

𝑄𝑄2

𝑸𝑸1𝑠𝑠𝑠𝑠𝑡𝑡2𝑠𝑠𝑠𝑠𝑡𝑡

1𝑏𝑏𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑏𝑏 2𝑏𝑏𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑏𝑏

Δ𝑑𝑑12𝑠𝑠𝑠𝑠𝑡𝑡

2𝑟𝑟𝑟𝑟𝑠𝑠 1𝑟𝑟𝑟𝑟𝑠𝑠Δ𝑑𝑑12𝑟𝑟𝑟𝑟𝑠𝑠

𝑸𝑸

𝑅𝑅𝐻𝐻

Process 12: Q from the HT-reservoir to the upper half of the system while Q/2 transferred from the upper half to the lower half

Process 23: Q (= QL = QH) from the LT-reservoir to the LT reservoir while Q/2 from the upper half to the lower half


=𝑄𝑄𝑅𝑅𝐻𝐻

−𝑄𝑄𝑅𝑅𝐿𝐿

< 0, �𝛿𝛿𝑞𝑞𝑅𝑅𝑠𝑠𝑠𝑠𝑠𝑠

,

undefined b/c Tsysis non-uniform

�𝛿𝛿𝑞𝑞𝑅𝑅𝐶𝐶𝐶𝐶

< 0

b/c TCS = TRES during all heat transfer

Δ𝑑𝑑12𝑏𝑏𝑠𝑠𝑠𝑠

III. Internally Thermally Irreversible∆S12 ?S is an extensive property, S of a system = sum of S of the components

∆𝑑𝑑12𝑠𝑠𝑠𝑠𝑠𝑠 = ∆𝑑𝑑12

𝑠𝑠𝑠𝑠𝑡𝑡 ℎ𝑠𝑠𝑠𝑠𝑎𝑎 + ∆𝑑𝑑12𝑏𝑏𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑏𝑏 ℎ𝑠𝑠𝑠𝑠𝑎𝑎 =

𝑄𝑄𝑅𝑅𝐻𝐻

−𝑄𝑄/2𝑅𝑅𝐻𝐻

+𝑄𝑄/2𝑅𝑅𝐿𝐿

>𝑄𝑄𝑅𝑅𝐻𝐻

�1


= �1



For the entire system,

∆𝑑𝑑12𝑠𝑠𝑠𝑠𝑠𝑠 > �

1


= �1



(can be seen graphically in the left figure)

T

S

𝑅𝑅𝐻𝐻


𝑄𝑄2

𝑸𝑸1𝑠𝑠𝑠𝑠𝑡𝑡2𝑠𝑠𝑠𝑠𝑡𝑡

1𝑏𝑏𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑏𝑏 2𝑏𝑏𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑏𝑏

Δ𝑑𝑑12𝑠𝑠𝑠𝑠𝑡𝑡

2𝑟𝑟𝑟𝑟𝑠𝑠 1𝑟𝑟𝑟𝑟𝑠𝑠Δ𝑑𝑑12𝑟𝑟𝑟𝑟𝑠𝑠

𝑸𝑸

Δ𝑑𝑑12𝑏𝑏𝑠𝑠𝑠𝑠

III. Internally Thermally Irreversible

𝑅𝑅𝐻𝐻

𝑅𝑅𝐿𝐿

𝑄𝑄/2

new C.S.𝑅𝑅𝐿𝐿newsystem

newHTres.

T

S

𝑅𝑅𝐻𝐻


1𝑠𝑠𝑠𝑠𝑠𝑠𝑛𝑛𝑟𝑟𝑛𝑛

Δ𝑑𝑑12𝑛𝑛𝑟𝑟𝑛𝑛 𝑟𝑟𝑟𝑟𝑠𝑠

𝑄𝑄2

2𝑠𝑠𝑠𝑠𝑠𝑠𝑛𝑛𝑟𝑟𝑛𝑛

2𝑟𝑟𝑟𝑟𝑠𝑠𝑛𝑛𝑟𝑟𝑛𝑛 1𝑟𝑟𝑟𝑟𝑠𝑠𝑛𝑛𝑟𝑟𝑛𝑛

Redefining a new system = the bottom half of the old systema new HT reservoir = the old HT reservoir + the top half of the old system

∆𝑑𝑑12𝑛𝑛𝑟𝑟𝑛𝑛 𝑠𝑠𝑠𝑠𝑠𝑠 =

𝑄𝑄/2𝑅𝑅𝐿𝐿

; ∆𝑑𝑑12𝑛𝑛𝑟𝑟𝑛𝑛 𝐻𝐻𝑅𝑅 𝑟𝑟𝑟𝑟𝑠𝑠 =−𝑄𝑄/2𝑅𝑅𝐻𝐻

; ∆𝑑𝑑12𝑛𝑛𝑟𝑟𝑛𝑛 𝐿𝐿𝑅𝑅 𝑟𝑟𝑟𝑟𝑠𝑠 = 0

∆𝑑𝑑12𝑢𝑢𝑛𝑛𝑖𝑖𝑟𝑟𝑟𝑟𝑟𝑟𝑠𝑠𝑟𝑟 = ∆𝑑𝑑12𝑛𝑛𝑟𝑟𝑛𝑛 𝑠𝑠𝑠𝑠𝑠𝑠 + ∆𝑑𝑑12𝑛𝑛𝑟𝑟𝑛𝑛 𝐻𝐻𝑅𝑅 𝑟𝑟𝑟𝑟𝑠𝑠 + ∆𝑑𝑑12𝑛𝑛𝑟𝑟𝑛𝑛 𝐿𝐿𝑅𝑅 𝑟𝑟𝑟𝑟𝑠𝑠 =

𝑄𝑄2

1𝑅𝑅𝐿𝐿−

1𝑅𝑅𝐻𝐻

> 0

positive entropy generation b/c of irreversibility in the universe (specifically in the system)

Carnot Cycle RevisitedIf there are only two heat reservoirs, at θ1 and θ2, then is a Carnot cycle (isothermal expansion adiabatic expansion isothermal contraction adiabatic contraction) the only reversible cycle?

• The ONLY way to transfer heat reversibly is to have infinitesimal T differences between system and surroundings Isothermal heat transfer necessary !

• The ONLY way to change temperatures, and therefore enclose some non-zero area (do non-zero work) on a T-S diagram, is to do it without heat transfer, or else you have not satisfied the conditions of the previous statement Adiabatic temperature changes necessary !

Combined Statement of the 1st and 2nd Laws(completely general)

A new form of differential of U using S and V, natural but inconvenient choices, as independent variables:

𝑃𝑃𝑈𝑈 = 𝛿𝛿𝑞𝑞 − 𝛿𝛿𝑤𝑤

For a closed system (no mass exchange),

𝛿𝛿𝑤𝑤 = 𝑃𝑃𝑃𝑃𝑃𝑃 (no other work than PV)𝛿𝛿𝑞𝑞 = 𝑅𝑅𝑃𝑃𝑑𝑑 (only when reversible)

𝑃𝑃𝑈𝑈 = 𝑅𝑅𝑃𝑃𝑑𝑑 − 𝑃𝑃𝑃𝑃𝑃𝑃 (energetic fundamental equation; only true whenno mass exchange, only PV work, reversible)

𝑃𝑃𝑈𝑈 =𝜕𝜕𝑈𝑈𝜕𝜕𝑑𝑑 𝑉𝑉

𝑃𝑃𝑑𝑑 +𝜕𝜕𝑈𝑈𝜕𝜕𝑃𝑃 𝐶𝐶

𝑃𝑃𝑃𝑃 𝑅𝑅 =𝜕𝜕𝑈𝑈𝜕𝜕𝑑𝑑 𝑉𝑉

;𝑃𝑃 =𝜕𝜕𝑈𝑈𝜕𝜕𝑃𝑃 𝐶𝐶

𝑃𝑃𝑑𝑑 =1𝑅𝑅𝑃𝑃𝑈𝑈 +

𝑃𝑃𝑅𝑅𝑃𝑃𝑃𝑃

(entropic fundamental equation; only true whenno mass exchange, only PV work, reversible)

𝑃𝑃𝑑𝑑 =𝜕𝜕𝑑𝑑𝜕𝜕𝑈𝑈 𝑉𝑉

𝑃𝑃𝑈𝑈 +𝜕𝜕𝑑𝑑𝜕𝜕𝑃𝑃 𝑈𝑈

𝑃𝑃𝑃𝑃1𝑅𝑅

=𝜕𝜕𝑑𝑑𝜕𝜕𝑈𝑈 𝑉𝑉

;𝑃𝑃𝑅𝑅

=𝜕𝜕𝑑𝑑𝜕𝜕𝑃𝑃 𝑈𝑈

Laws of ThermodynamicsThe 1st Law• Conservation of energy• Shows the equivalence of work and heat• For cyclic process, q = w• Suggests one can convert heat into useful work

The 2nd Law• Puts restrictions on the conversion of heat to work• Suggests a directionality to natural processes

The 3rd Law“There exists a lower limit to the temperature that can be attained by matter, called the absolute zero of temperature, and the entropy of all substances is the same (zero) at that temperature.” • At 0 K, Carnot efficiency is 100%. • But, we cannot reach 0K.

�𝑃𝑃𝑈𝑈 = 0

ms212 thermodynamics of materials -...

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