ENGR 2213 ThermodynamicsENGR 2213 Thermodynamics

F. C. Lai

School of Aerospace and Mechanical

Engineering

University of Oklahoma

Second Law of ThermodynamicsSecond Law of Thermodynamics

First law of Thermodynamics

Unlike energy, entropy is a non-conserved property.

Second law of ThermodynamicsEnergy

Entropy

Clausius Inequality

Q0

T

The cyclic integral of δQ/T is always

less than or equal to zero.

This inequality is valid for all cycles, reversible or irreversible.

Clausius Inequality

H L

H Lrev

Q Q Q

T T T

H L

H L

Q Q

T T

For reversible heat engines

H LH L

1 1Q Q

T T

= 0

Reservoir TH

Reservoir TL

HE

QH

QL

Clausius Inequality

H L

H Lirrev irrev

Q Q Q

T T T

L irrevH

H L

QQ

T T

H L irrevH L

1 1Q Q

T T

For irreversible heat engines

< 0

From Carnot principleWrev > Wirrev

QL, rev < QL, irrev

QL, irrev = QL, rev+ Qdiff

L rev diffH

H L L

Q QQ

T T T diff

L

Q

T

Example 1Example 1

A heat engine receives 600 kJ of heat from a high-temperature source at 1000 K during a cycle. Itconverts 150 kJ of this heat to work and rejects theremaining 450 kJ to a low-temperature sink at 300 K. Determine if this heat engine violates the2nd law of thermodynamics on the basis of(a) the Clausius inequality.(b) the Carnot principle.

Example 1 (continued)Example 1 (continued)

H L

H L

Q Q Q

T T T

600 450

1000 300

Lth

H

Q 4501 1 0.25

Q 600

(a) Clausius inequality

= - 0.9 kJ/K < 0

(b) Carnot principle

Lrev

H

T 3001 1 0.7

T 1000

ηth < ηrev

Second Law of ThermodynamicsSecond Law of Thermodynamics

int rev

Q0

T

Clausius Inequality

Internally Reversible ProcessesA process is called internally reversible if noirreversibilities occur within the boundaries of thesystem during the process.

2 11 2

int rev A B

Q Q Q0

T T T

1

2A

B

EntropyEntropy

21

B

Q

T

2 11 2

A B

Q Q

T T

int rev

QdS

T

If a quantity whose integral dependsonly on the end states and not theprocess path, then it is a property.

22 1 1

int rev

QS S S

T

EntropyEntropy

21 int rev

0

1Q

T 2 2

1 10int rev int rev

Q QS

T T

0

QS ,

T

Isothermal Processes

Q = T0 ΔS

Increase-in-Entropy PrincipleIncrease-in-Entropy Principle

irrev

Q0

T

2 11 2

irrev A irrev B rev

Q Q Q0

T T T

22 1 1

irrev

QS S

T

Clausius Inequality

1

2A irrev

B rev

= S1 – S2

22 1 1

QS S

T

Increase-in-Entropy PrincipleIncrease-in-Entropy Principle

21

QS

T

The entropy change of a closed system during an irreversible process is greater than the integral ofδQ/T evaluated for that process.

For an adiabatic process, Q = 0

(ΔS)adiabatic ≥ 0 In the absence of heat transfer, entropy change is due to irreversibilitiesonly, and their effect is always to increase the entropy.

Increase-in-Entropy PrincipleIncrease-in-Entropy Principle

(ΔS)adiabatic ≥ 0

A system plus its surroundings constitutes an adiabatic system, assuming both can be enclosed by a sufficiently large boundary across which there is no heat or mass transfer.

(ΔS)total = (ΔS)system + (ΔS)surroundings

≥ 0

system

surroundings

Increase-in-Entropy PrincipleIncrease-in-Entropy Principle

Sgen = (ΔS)total

Causes of Entropy Change► Heat Transfer

Isentropic Process

> 0 irreversible processes= 0 reversible processes< 0 impossible processes

► Irreversibilities

A process involves no heat transfer (adiabatic) and no Irreversibilities within the system (internally reversible).

Remarks about EntropyRemarks about Entropy

1. Process can occur in a certain direction only. A process must proceed in the direction that complies with the increase-in-entropy principle.

2. Entropy is a non-conserved property. There is no such thing as the conservation of entropy principle.

3. The quantity of energy is always preserved during an actual process (the first law), but the quality decreases (the second law). The decrease in quality is always accompanied by an increase in entropy.

What is Entropy?What is Entropy?

Entropt can be viewed as a measure of moleculardisorder, or molecular randomness

From a statistical point of view, entropy is a measureof the uncertainty about the position of molecules atany instant.

The entropy of a pure crystalline substance at absolute zero temperature is zero since there is nouncertainty about the state of the molecules at that instant. - the 3rd Law of Thermodynamics

Second Law of ThermodynamicsSecond Law of Thermodynamics

Heat is, in essence, a form of disorganized energyand some disorganization (entropy) will flow with heat.

Work instead is an organized form of energy, and is free of disorder or randomness and thus free ofentropy.

Example 1Example 1

Saturated water at 100 ºC is contained in a piston-cylinder assembly. The water undergoes aninternally reversible heating process to the corresponding saturated vapor state. Find(a) the work per unit mass for the process.(b) the heat transfer per unit mass for the process.

p

v

1 2

Example 1 (continued)Example 1 (continued)

2g f1

Wp dv p(v v )

m

g f g fQ W

u (u u ) p(v v )m m

g fh h

(a)

(b)

= (101.4)(1.673 – 0.001044) = 170 kJ/kg

= 2257 kJ/kg

2g f1

QT ds T(s s )

m

= (373.15)(7.3549 – 1.3069) = 2257 kJ/kg

Example 2Example 2

Steam at 7 MPa and 450 ºC is throttled through a vavleto 3 MPa. Find the entropy generation through the process.

p2 = 3 MPa

T1 = 450 ºCp1 = 7 MPa

Example 2 (continued)Example 2 (continued)

sgen = Δs

T1 = 450 ºCp1 = 7 MPa

Table A-6h1 = 3287.1 kJ/kgs1 = 6.6327 kJ/kg K

Throttling Process h2 = h1

p2 = 3 MPah2 = 3287.1 kJ/kg

Table A-6 s2 = 6.9919 kJ/kg K

= 6.9919 – 6.6327

= 0.3592 kJ/kg K

Second Law of ThermodynamicsSecond Law of Thermodynamics

Carnot Cycle in T-S Diagram

T

S

1 2

4 3

1 2

4 3

W

T

S

1 4

2 3

1 4

2 3

W