Chapter 4: Entropy: an Additional Balance Equation

Chapter 4:Entropy: an Additional Balance

Equation

2

Important Notation

3

f4_1_1

Consider a composite system (A+B) surrounded bya boundary (external wall) that is:

• Adiabatic• Rigid• Impermeable

Composite system (A+B) is isolated.

4

f4_1_1

Suppose that the internal wall between systems A and B initially is:

• Adiabatic• Rigid• Impermeable

Systems A and B are initially isolated. The thermodynamic properties are generally different in each of them.

5

f4_1_1

Suppose now that the internal wall between systems A and B becomes:

• Diathermal (allows heat transfer)• Rigid• Impermeable

If we wait long enough for equilibration, what happens to the temperatures of systems A and B?

6

f4_1_1

The mass and the energy in the initial and final states are identical.

Nonetheless, of all states possible, system (A+B) evolves to a stable equilibrium state in which the temperatures in systems A and B are equal.

7

f4_1_1

The mass and the energy in the initial and final states are identical.

Nonetheless, of all states possible, system (A+B) evolves to a stable equilibrium state in which the temperatures in systems A and B are equal.

What makes this state so special?

8

f4_1_1

Because the mass and the energy in the initial and final states are equal, they do not serve to characterize the final state assomething special.

There must be some new property to account for this.Call this property for a moment.

𝑑𝜃𝑑𝑡 = ൬𝑅𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓𝜃 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 ൰= ൭

𝑅𝑎𝑡𝑒 𝑎𝑡 𝑤ℎ𝑖𝑐ℎ 𝜃𝑖𝑠 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑤𝑖𝑡ℎ𝑖𝑛 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 ൱

9

f4_1_1𝑑𝜃𝑑𝑡 = ൬𝑅𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓𝜃 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 ൰= ൭

𝑅𝑎𝑡𝑒 𝑎𝑡 𝑤ℎ𝑖𝑐ℎ 𝜃𝑖𝑠 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝑤𝑖𝑡ℎ𝑖𝑛 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 ൱

0gen

d

dt

If is a property, it should no longer change once equilibrium is achieved.

0gen

d

dt

at equilibrium

away from equilibrium

10

Entropy definition• Entropy ( ) is a state function. • If there are flows of both heat by conduction and work across the system boundaries, the conductive heat flow, but not the work flow, causes a rate of entropy change T is the absolute thermodynamic temperature of the system at the point of heat flow. •If there are mass flows across the system boundary, the total

entropy of the system will also change due to this convected flow. Each element of mass entering or leaving the system carries with it its entropy (as well as internal energy, enthalpy, etc.)

S

Q T

11

Entropy balance

1

ˆK

k k genk

dS QM S S

dt T

gen

dS QS

dt T

For a closed system

12

genS depends on the internal relaxation processes within the system.

When analyzing the heat transfer between systems A and B, let usassume it occurs slowly in such a way that the temperature isuniform within each system – a quasistatic process. In this case, is equal to zero in systems A and B.genS

13

f4_1_1A A

A

dS Q

dt T

B B

B

dS Q

dt T

A B A B

A B

dS dS Q QdS

dt dt dt T T

Suppose initially . Then and B AT T 0AQ B AQ Q

1 10A

A B

dSQ

dt T T

14

Note that

From experimental observations :ThQ

The flow of heat is proportional to the T difference; h is a positive constant, and theminus sign means that heat flows opposite to the temperature difference

That means that heat is always <0 when flows from high T to low T

Therefore as shown in our previous slide:

1 10A

A B

dSQ

dt T T

15

However we could also use this other proof:

A

BA

A

AA

T

TTh

T

Q

dt

dS

B

BA

B

BB

T

TTh

T

Q

dt

dS

0

2

BA

BA

B

BA

A

BAgen

TT

TTh

T

TTh

T

TThS

16

f4_1_1A A

A

dS Q

dt T

B A

B

dS Q

dt T

Note:

1) The entropy of system A increases;2) The entropy of system B decreases;3) The entropy of the isolated composite system (A+B) increases.

17

f4_1_1

1

1 1ˆ 0K

k k gen Ak A B

dS QM S S Q

dt T T T

Further, the isolated composite system (A+B) hasno mass and heat flows entering or leaving it:

0 0

0genS

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Entropy balance: integrated form

1

ˆK

k k genk

QdS M S dt dt S dt

T

2 2 2

1 1 1

2 11

ˆt t tK

k k genk t t t

QS S M S dt dt S dt

T

2 2

1 1

2 11

ˆt tK

k k genk t t

QS S M S dt dt S

T

19


2 2

1 1

2

1

2 11

1

ˆ

ˆ

t tK

k k genk t t

tK

k k genk t

QS S S M dt dt S

T

QS M dt S

T

If the entropy per unit mass is constant in time at each mass port:

20


2 2

1 1

1t t

t t

Q Qdt Qdt

T T T

If the temperature is constant at the point where heat transfer occurs:

21


2 11

ˆK

k k genk

QS S S M S

T

If the simplifications of the two previous slides apply simultaneously:

22

Clausius statement of the 2nd law

It is not possible to build a device that operates in a cyclewhose only effect is to transfer heat from a colder body to a hotter body.

At the end of thermodynamic cycle, the system returns to its initial state.

The energy balance is:

1 20f iU U Q Q W

23





1 20f iU U Q Q W

0 (because the only effect is heat transfer)

The entropy balance is:

1 2

1 2

0f i gen

Q QS S S

T T

24





1 20f iU U Q Q W

0 (because the only effect is heat transfer)


1 2

1 2

0f i gen

Q QS S S

T T

0 (because S, a state function,returns to its initial value atthe end of the cycle)

25



1 21

1 2 2 1

1 1gen

Q QS Q

T T T T

2 12 1

1 10T T

T T

If is positive, is negative, violating that . 1Q genS 0genS

26

Kelvin-Planck statement of the 2nd law

It is not possible to build a device that operates in a cyclewhose only effect is the production of work by transferringheat from a single body.



0f iU U Q W W Q


0f i gen

QS S S

T

27

Kelvin-Planck statement of the 2nd law

It is not possible to build a device that operates in a cyclewhose only effect is the production of work by transferringheat from a single body.

is negative, violating that . 0genS genS

0T

QSgen

Since Q >0

28

Entropy balance and reversibility

0genS

The rate of entropy generation is proportional to the temperature and velocity gradients in the system.

If, in a given process, these gradients are very small, the rate of entropy generation is essentially zero, andthe process is said to be reversible.

29


1 2

1

2 01 10

t tK K

k kk k t

M M M dt M dt

Consider a general system open to the flow of mass, heat, and work between two equal time intervals, from 0 to and from to .1t 1t 2t

1

2

1

2 010

1

ˆ

ˆ

t K

k k sk

t K

k k skt

dVU U M H P W Q dt

dt

dVM H P W Q dt

dt

30

Entropy balance and reversibility1 2

1

1 2

1

2 01 10

0

ˆ ˆ

t tK K

k k k kk kt

t t

gen gen

t

Q QS S M S dt M S dt

T T

S dt S dt

31

Entropy balance and reversibilitySuppose now that all mass, heat and work flows are reversed in the second half of the process, compared to the first half.

1 2

10

t t

k k

t

M dt M dt 1 2

10

t t

t

Qdt Qdt

1 2

10

ˆ ˆt t

k k k k

t

M H dt M H dt

1 2

10

ˆ ˆt t

k k k k

t

M S dt M S dt

1 2

10

t t

t

Q Qdt dt

T T

1 2

10

t t

s s

t

W dt W dt

1 2

10

t t

t

dV dVP dt P dtdt dt

32

Entropy balance and reversibilityThen we end up with:

2 0M M

2 0U U1 2

1

2 0

0

t t

gen gen

t

S S S dt S dt

In general, the rates of entropy generation are positive, their integrals are also positive, and . The systemwould not have returned to its initial state. The process isirreversible.

2 0S S

33

Entropy balance and reversibilityIf a process proceeds with infinitesimal gradients within the system, the rates of entropy generation are zero, their integrals are also zero, and . The systemwould have returned to its initial state. The process isreversible.

2 0S S

Examples of reversible processes (negligible viscous dissipationor internal heat flow)

• Uniform and slow expansion or compression of a fluid• Processes with slow changes such that gradients do not

appear in the system

34

Entropy balance and reversibilityExamples of irreversible processes

• Abrupt expansion or compression of a fluid• Heat conduction process in which temperature

gradient exists• Processes in which friction is important• Mixing of fluids of different temperatures, pressures,

or compositions

35


If the surroundings are extracting work from the system, the maximum amount of work is obtained, for a given change of state, if the process is carried out reversibly.

36


If the surroundings are doing work on the system, the minimum amount of work is needed, for a given change of state, if the process is carried out reversibly.

37


Few processes are truly reversible but it is sometimes useful to model them to be so.

38

Helmholtz energy (NVT)Consider the energy and entropy balances for aclosed, isothermal, constant volume system:

2 1 sU U Q W

2 1 2 1gen gen

QS S S Q T S S TS

T

Combining these two equations, using that :2 1T T

2 2 2 1 1 1s genW U T S U T S TS

39

Helmholtz energy (NVT)

We now define a new state function, the Helmholtz energy, as:

A U TS

Then, in an irreversible process:

2 1s genW A A TS

In a reversible process:

2 1revsW A A

• more work needed to change the state of the system (Ws >0) in the irreversible process• more work is obtained (Ws <0) in a reversible process

40

Gibbs energy (NPT)Consider the energy and entropy balances for aclosed, isothermal, isobaric system:

2 1 2 2 1 1sU U Q W PV PV

2 1 2 1gen gen

QS S S Q T S S S

T

2 2 2 2 2 1 1 1 1 1s genW U T S PV U T S PV TS

Combining these two equations, using that and :2 1T T 2 1P P

41

Gibbs energy (NPT)

We now define a new state function, the Gibbs energy, as:

G U TS PV

Then, in an irreversible process :

2 1s genW G G TS

In a reversible process:

2 1revsW G G

• more work needed to change the state of the system (Ws >0) in the irreversible process• more work is obtained (Ws <0) in a reversible process

42

Relationship of entropy changes with changes in other state functions

Consider a system in which kinetic and potentialterms are unimportant, with a single mass flow stream, and themass and heat flows occur at the common temperature T:

dMM

dt

ˆs

dU dVMH Q P W

dt dt

ˆgen

dS QMS S

dt T

43


ˆs

dU dM dVH Q P W

dt dt dt

ˆgen

dS dMT TS Q TSdt dt

Eliminating and multiplying the entropy balance by T: M

44


ˆsdU HdM Q PdV W

ˆ ˆgen genTdS TSdM Q TS Q TdS TS TSdM

The amounts exchanged over a period of time dt are:

Combining these two equations:

ˆˆ

ˆ

gen s

gen s

dU TdS TS PdV W H TS dM

TdS TS PdV W GdM

45


For a reversible process:

ˆrev

sdU TdS PdV W GdM

For a system with no shaft work (open system):

ˆdU TdS PdV GdM

Additionally, if the system is closed:

dU TdS PdV dU TdS PdV

dMSTTdSQrev ˆ

46

Entropy changes of matter

Entropy is very important, but how to evaluate it in practice?

1) Thermodynamic diagrams2) Property tables, e.g., steam tables3) Mathematical models, e.g., equations of state

47

Example 1

A 0.5 kg/s steady flow of refrigerant HFC-134a enters a perfectly insulated valve as saturated vapor at 70oC. In the stream exiting the valve, the pressure is equal to 0.1 MPa. What is the change in the specific entropy of refrigerant HFC-134a across the valve? Neglect changes in kinetic and potential energies.

Draw a sketch, it often helps

In Chapter 3, we found that:

1 2ˆ ˆH H

Isenthalpic expansion – Joule-Thompson process

We now want:

2 1ˆ ˆS S

48f3_3_4

f3_3_4

49

Example 1

From the pressure-enthalpy diagram:

2 1ˆ ˆ 1.90 1.70 0.20

. .

J JS S

kg K kg K

1 1.70 .S J kg K

2 1.90 .S J kg K

50

Entropy changes of matterConsider now the entropy change of one mol of a pure substance:

1 PdU TdS PdV dS dU dV

T T

For an ideal gas:

P P VC C T dU C dT

PV RT

1 VCP RdS dU dV dT dV

T T T V

51


2 2 2

1 11

22 1

1

lnVT T

V V

T V T

C CR VS S dT dV dT R

T V T V

The molar entropy change of an ideal gas between two states, 1 and 2, is:

If the heat capacity at constant volume is independent of temperature:

222 1

11

ln lnV

T VS S C R

T V

52


The ideal gas equation of state can be used to eliminatethe molar volumes (details in the book):

2 22 1

1 1

ln lnP

T PS S C R

T P

For liquids and solids, a good approximation is:

2

1

2 1

T

P

T

CS S dT

T

53

Example 2

A 2 mol/s steady flow of nitrogen enters a perfectly insulated valve at 250 K and 100 bar. In the stream exiting the valve, the pressure is equal to 10 bar. What is the rate of entropy generation as nitrogen flows across the valve? Neglect changes in kinetic and potential energies. Assume nitrogen to be an ideal gas with heat capacity at constant pressure equal to 29.5 J/(mol.K).

In Chapter 3, we found that:

1 2 2 1ˆ ˆ ˆ ˆ 0H H H H

2

1

1 2 2 1ˆ ˆ 0 250

T

P

T

H H C dT T T K

54

Example 2

Entropy balance at steady-state (dS/dt=0)

22 1

1

10ln 8.314 ln 19.14

. 100 .

P J JS S R

P mol K mol K

1 2 2 10 gen gen

dSN S S S S N S S

dt

The temperature does not change, therefore:

2 1 2 19.14 38.28. .gen

mol J JS N S S

s mol K K s

55

Example 3

A 2 mol/s steady flow of nitrogen enters a perfectly insulated valve at 250 K and 100 bar. In the stream exiting the valve, the pressure is equal to 10 bar. What is the rate of entropy generation as nitrogen flows across the valve? Neglect changes in kinetic and potential energies. Model nitrogen’s behavior using the Soave-Redlich-Kwong equation of state and temperature-dependent heat capacity in the ideal gas state.

These calculations are very demanding if done manually. You can use an equation of state package to carry them out.

56

Example 4

A 2 mol/s steady flow of nitrogen enters a perfectly insulated reversible turbine at 250 K and 100 bar. In the stream exiting the turbine, the pressure is equal to 70 bar. Neglect changes in kinetic and potential energies. Model nitrogen’s behavior using the Soave-Redlich-Kwong equation of state and temperature-dependent heat capacity in the ideal gas state. What is the temperature of nitrogen as it exits the turbine? What is the shaft power?

Mass balance

Energy balance

Entropy balance

1 2 1 20dN

N N N N Ndt

1 21 2 0 (steady state)s

dUN H N H W

dt

1 2 0 (steady state)dS

N S Sdt

57

Example 4

Energy balance

2 1sW N H H

58

Example 4

Entropy balance

1 2 0 (steady state)dS

N S Sdt

Note: this is an isentropic (constant entropy) expansion process.

These calculations are very demanding if done manually.

59

Example 5

A well insulated rigid box of volume 6 m3 is divided in two compartments of equal volume, separated by a rigid, heat-transferring wall. Initially, compartment 1 has air at 100oC and 2 bar and compartment 2 is evacuated. The valve between the two compartments is opened and air slowly flows into compartment 2. Assume that, at all times, the temperatures in the two compartments are equal. Plot the pressure in the 2nd tank vs the P in the 1st tank, and the change in total entropy vs. the P in tank 1. Assume air is an ideal gas of constant heat capacity.

1 2N N N

1 2U U U

1 2S S S

Amount of air in the system:

3

1 25

2 3193.4

.8.314 10 373.15

.

i PV bar mN mol

bar mRTK

mol K

60

Example 5

1 10.95 iN N

1 2 1

1 21 2 1 2 1

1 21 1 2 1

f i i

i

i i

S S S S S S

N S N S N N S

N S S N S S

Since U=U1+U2 is constant , therefore T1=T2 =100oC at all times

When the amount in compartment 1 is, for example, , the amount in compartment 2 is because of mass conservation. With these values, the volume, and the temperature, it is possible to evaluate the pressure in each compartment.

2 10.05 iN N

Total entropy change

11 1

1

lnii

PS S R

P

22 1

1

lnii

PS S R

P

61

f4_5_4

what is the equilibrium state of the system?

62

Summary of the formulas

63t4_1_1

t4_1_1

64t4_1_2

t4_1_2

65

Summary

There is a favored direction in natural phenomena and there are limits on the conversion of heat into work. Altogether, these observations lead to the second law of thermodynamics.

Unlike energy, entropy can be generated within a system. The entropy generation term is either zero or positive, but never negative.

Processes with zero entropy generation are called reversible processes. Processes with positive entropy generation are called irreversible processes. In practice, most processes are irreversible.

Entropy is an extensive state property (or state function). The Helmholtz and Gibbs energies are also extensive state properties.

Entropy cannot be measured directly but its value can be inferred from other properties. Entropy values necessary for practical calculations are usually available as tables, diagrams, or analytical (mathematical) formulas.

66

Recommendation

Read chapter 4 in the textbook and review all examples.

Documents

Chapter 4: Entropy: an Additional Balance Equation