1. Chapter 7
Entropy

2. Objectives Apply the second law of thermodynamics to processes.
Define a new property called entropy to quantify the second-law effects.
Establish the increase of entropy principle.
Calculate the entropy changes that take place during processes for pure substances, incompressible substances, and ideal gases.
Examine a special class of idealized processes, called isentropic processes, and develop the property relations for these processes.
Derive the reversible steady-flow work relations.
Develop the isentropic efficiencies for various steady-flow devices.
Introduce and apply the entropy balance to various systems.

3. Introduction
The second low often leads to expressions that involve inequalities.
One more important inequality in thermodynamics: Clausius Inequality (1888)

4. The Inequality of Clausius
For all cycle reversible or irreversible cycle
The inequality of Clausius is a consequence of the second law of thermodynamics.
Q is the heat transfer to or from the system.
T is the absolute temperature at the boundary.
The symbol is the cyclic integral

5. The Inequality of Clausius
Energy balance to the combined system:
let the system undergo a cycle while the cyclic device undergoes an integral number of cycles.
the combined system is exchanging heat with a single thermal energy reservoir while involving (producing or consuming) work WC during a cycle.
This violating the Kelvin–Planck statement of the second law, which states that no system can produce a net amount of work while operating in a cycle and exchanging heat with a single thermal energy
Reservoir
Then WC cannot be a work output, and thus it cannot be a positive quantity.
Considering that TR is the thermodynamic temperature and thus a positive quantity, we must have
the Clausius inequality.

6. The Inequality of Clausius

7. The Inequality of Clausius

8. The cyclic integral
The cyclic integral indicates that the integral should be performed over the entire cycle and over all parts of the boundary.

9. The cyclic integral

10. Derivation of Clausius Inequality
Irreversible
Reversible
Heat Engine
Refrigeration

11. The cyclic integral of Reversible Heat Engine
Since

12. The cyclic integral of Irreversible Heat Engine
We cannot use this
It is Irreversible
Produced work is larger
For same QH
Since it was zero for reversible when QL rev.<QL irreversible

13. The cyclic integral of Reversible Refrigeration
Since
For reversible Refrigerator

14. The cyclic integral of Irreversible Refrigeration
We cannot use this
It is Irreversible
Consumed work is larger

15. Derivation of Clausius Inequality
Irreversible
Reversible
< 0
0 =
Heat Engine
Refrigeration
The equality in the Clausius inequality holds for totally or just internally reversible cycles and the inequality for the irreversible ones.

16. The Clausius inequality gives the basis for two important ideas
Entropy (S)
Entropy generation (Sg)
These two terms gives quantitative evaluations for systems from second law perspective.

17. Derivation of Entropy (Reversible Process)
For reversible cycle A-B
For reversible cycle C-B
All paths are arbitrary
Subtracting gives
Since paths A and C are arbitrary, it follows that the integral of Q/T has the same value for ANY reversible process between the two sates.

18. Derivation of Entropy (Reversible Process)
Clausius chose the name “Entropy”
Entropy (the unit) S = entropy (kJ/K); s = specific entropy (kJ/kg K)
S2 – S1 depends on the end states only and not on the path,
 it is same for any path
For Reversible Process

19. Entropy is a Property
The entropy change between two specified states is the same whether the process is reversible or irreversible.

20. A Special Case: Internally Reversible Isothermal Heat Transfer Processes
Recall that isothermal heat transfer processes are internally reversible.
Therefore, the entropy change of a system during an internally reversible isothermal heat transfer process can be determined by performing the integration in Eq. 7–5:

21. Derivation of Entropy (Irreversible Process)
Consider 2 cycles AB is reversible and CB is irreversible

22. Derivation of Entropy (Any Process)
This can be written out in a common form as an equality or
2nd law of thermodynamics for a closed system
Entropy Balance Equation for a closed system
Recall: for cycle
In any irreversible process always entropy is generated (Sgen > 0) due to irreversibilities occurring inside the system.

23. Example (7-1) Entropy change during isothermal process.
A friction-less piston-cylinder device contains a liquid-vapor mixture of water at 300 K. During a constant pressure process, 750 kJ of heat is transferred to the water. As a result, part of the liquid in the cylinder vaporizes. Determine the entropy change of the water during this process.
Solution:
This is simple problem.
No irreversibilities occur within the system boundaries during the heat transfer process.
Hence, the process is internally reversible process (Sg = 0).

24. We computed the entropy change for a system using the RHS of the equation.
But we can not get easy form each time.
So, we need to know how to evaluate the LHS which is path independent.

25. Entropy change for different substances (S = S2-S1)
We need to find how to compute the left hand side of the entropy balance for the following substances:
Pure substance like water, R-134, Ammonia etc..
Solids and liquids
Ideal gas

26. 1- S for Pure Substances
The entropy of a pure substance is determined from the tables, just as for any other property
These values were tabulated after conducting a tedious integration.
These values are given relative to an arbitrary reference state.
For water its assigned zero at 0.01 C.
For R-134 it is assigned a zero at -40 C.
Entropy change for a closed system with mass m is given as expected:

27. EXAMPLE 7–3 Entropy Change of a Substance in a Tank
A rigid tank contains 5 kg of refrigerant-134a initially at 20°C and 140 kPa. The refrigerant is now cooled while being stirred until its pressure drops to 100 kPa. Determine the entropy change of the refrigerant during this process.

28. The volume of the tank is constant so V1=V2

29. Important Note:
The negative sign indicates that the entropy of the system is decreasing during this process.
This is not a violation of the second law, however, since it is the entropy generation Sgen that cannot be negative.

30. 7-2: The increase of entropy principle (closed system)
There is some entropy generated during an irreversible process such that
Entropy change
Entropy generation
due to irreversibility
Entropy transfer
with heat
This is the entropy balance for a closed system.

31. The increase of entropy principle (closed system)
The entropy change can be evaluated independently of the process details.
However, the entropy generation depends on the process, and thus it is not a property of the system.
The entropy generation is always a positive quantity or zero and this generation is due to the presence of irreversibilities.
The direction of entropy transfer is the same as the direction of the heat transfer: a positive value means entropy is transferred into the system and a negative value means entropy is transferred out of the system.

32. The increase of entropy principle (closed system)
For an isolated (or simply an adiabatic closed system), the heat transfer is zero, then
This means that the entropy of an adiabatic system during a process always increases or, In the limiting case of a reversible process, remains constant.
In other words, it never decreases.
This is called Increase of entropy principle.
This principle is a quantitative measure of the second law.

33. The increase of entropy principle
Now suppose the system is not adiabatic.
We can make it adiabatic by extending the surrounding until no heat, mass, or work are crossing the boundary of the surrounding.
This way, the system and its surroundings can be viewed again as an isolated system.
The entropy change of an isolated system is the sum of the entropy changes of its components (the system and its surroundings), and is never less than zero.
Now, let us apply the entropy balance for an isolated system:

34. Summary of the increase of entropy principle
Entropy is an extensive property, and thus the total entropy of a system is
equal to the sum of the entropies of the parts of the system.

35. Important Remarks
Processes can occur in a certain direction only , not in any direction. A process must proceed in the direction that complies with the increase of entropy principle. A process that violates this principle is impossible.
Entropy is a non-conserved property. Entropy is conserved during the idealized reversible process only and increases during all actual processes.
The performance of engineering systems is degraded by the presence of irreversibilities, and the entropy generation is a measure of the magnitude of the irreversibilities present during a process.
Entropy generation is used to establish criteria for the performance of engineering devices

36. Example (7-2): entropy generation during heat transfer processes
A heat source at 800 K losses 2000 kJ of heat to a sink at (a) 500 K and (b) 750 K. Determine which heat transfer process is more irreversible.
Solution:
Both cases involve heat transfer via a finite temperature difference and thus are irreversible.
Each reservoir undergoes an internally reversible isothermal process (constant temperature).

37. Take the two reservoirs as your system
Take the two reservoirs as your system. Thus they form an adiabatic system and thus
Now consider each system alone (isothermal with no irreversibility)

38. Let us repeat the same with case b.
Now consider each system alone
Hence the case b involves less irreversibility.
The smaller the temperature difference , the smaller the irreversibility ( if DT =0 no irrevocability due heat transfer)

39. Where does the irreversibility arise from?
Tsource > T sink

40. Where is entropy generated?
Finite Temperature difference cause the entropy to generate

41. What is Entropy? (microscopic point of view)
Entropy can be viewed as a measure of molecular disorder, or molecular
randomness.
As a system becomes more disordered, the positions of the molecules become less predictable and the entropy increases.
The entropy of a substance is lowest in the solid phase and highest in the gas phase.
In the solid phase, the molecules of a substance continually oscillate about their equilibrium positions, but they cannot move relative to each other,
Solid molecules position at any instant can be predicted with good certainty.
In the gas phase, however, the molecules move about at random, collide with each other, and change direction,
It is extremely difficult to predict accurately the microscopic state of a gas phase system at any instant.
Associated with this molecular chaos is a high value of entropy
The entropy of a system is related to the total number of possible microscopic states of that system, called thermodynamic probability p, by the Boltzmann relation, expressed as

42. What is Entropy?
From a microscopic point of view, the entropy of a system increases whenever the molecular randomness or uncertainty (i.e., molecular probability) of a system increases.
entropy is a measure of molecular disorder, and the molecular disorder of an isolated system increases anytime it undergoes a process.

43. What is Entropy?
The molecules of a substance in solid phase continually oscillate, creating an uncertainty about their position.
These oscillations, however, fade as the temperature is decreased, and the molecules supposedly become motionless at absolute zero.
This represents a state of ultimate molecular order (and minimum energy).
The entropy of a pure crystalline substance at absolute zero temperature is zero since there is no uncertainty about the state of the molecules at that instant, This statement is known as the third law of thermodynamics

44. What is Entropy?
The third law of thermodynamics provides an absolute reference point for the determination of entropy.
The entropy determined relative to this point is called absolute entropy, and it is extremely useful in the thermodynamic analysis of chemical reactions.
Notice that the entropy of a substance that is not pure crystalline (such as a solid solution) is not zero at absolute zero temperature.
This is because more than one molecular configuration exists for such substances, which introduces some uncertainty about the microscopic state of the substance.
Disorganized energy does not create
much useful effect, no matter how
large it is.

45. Isentropic Processes
We mentioned earlier that the entropy of a fixed mass can be changed by
(1) heat transfer and
(2) irreversibilities.
Then it follows that the entropy of a fixed mass does not change during a process that is internally reversible and adiabatic
A process during which the entropy remains constant is called an isentropic process.

46. Property diagrams involving entropy
In the second-law analysis, it is very helpful to plot the processes on T-s and h-s diagrams for which one of the coordinates is entropy.
Recall the definition of entropy
Property diagrams serves as great visual aids in the thermodynamic analysis of process.
We have used P-v and T-v diagrams extensively in conjunction with the first law of thermodynamics.

47. Thus
This area has no meaning for irreversible processes!
It can be done only for a reversible process for which you know the relationship between T and s during a process. Let us see some of them.

48. Isothermal internally reversible process.
Temperature
Entropy
Isothermal Process 1 2 Q

49. Adiabatic internally reversible process
In this process Q =0, and therefore the area under the process path must be zero.
This process on a T-s diagram is easily recognized as a vertical-line.
Temperature
Entropy 1 2
Q=0
Isentropic Process

50. T-s Diagram for the Carnot Cycle2
Isothermal expansion 1
Isentropic compression
W=Qin-Qout
Qin 3
Isentropic expansion
Temperature, T
The area inside the box is both the net heat transferred, and the net work
Q-W=0
Isothermal compression 4
Qout
Entropy, S

51. Another important diagram is the h-s Diagram
This diagram is important in the analysis of steady flow devices such as turbines.
In analyzing the steady flow of steam through an adiabatic turbine, for example,
The vertical distance between the inlet and the exit states (h) is a measure of the work output of the turbine,
The horizontal distance (s) is a measure of the irreversibilities associated with the process.

52. Example 7-5: Isentropic Expansion of a Steam in Turbine
Steam enters an adiabatic turbine at 5 MPa and 450°C and leaves at a pressure of 1.4 MPa. Determine the work output of the turbine per unit mass of steam if the process is reversible.

54. Example 7-6: T-ds Diagram of Carnot cycle
Show the Carnot cycle on a T-S diagram and indicate the areas that represent, the heat supplied QH, heat rejected QL, and the net work output Wnet,out on this diagram.
1) On a T-S diagram, the area under the process curve represents the heat transfer for that process
2) the area A12B represents QH
3) the area A43B represents QL,
the difference between these two

55. Property Relationships

56. The T-ds relations:
Apply the differential form of the first law for a closed stationary system for an internally reversible process

57. This equation is known as:
Divide by the mass, you get
This equation is known as:
First Gibbs equation or First T-ds relationship
Divide by T, ..
Although we get this form for internally reversible process, we still can compute Ds for an irreversible process. This is because S is a point function.

58. Second T-ds (Gibbs) relationship
Recall that…
Take the differential for both sides
Rearrange to find du
Substitute in the First Tds relationship
Second Tds relationship, or Gibbs equation

59. Thus We have two equations for ds
Divide by T, ..
Thus We have two equations for ds
To find Ds, we have to integrate these equations. Thus we need a relation between du and T (or dh and T).
Now we can find entropy change (the LHS of the entropy balance) for liquids and solids

60. The T ds relations are valid for both
T-ds Eq. 1 &2 are relations between the properties of a unit mass of a simple compressible system as it undergoes a change of state, Even though they were developed for internally reversible process THEN
The T ds relations are valid for both
reversible and irreversible processes
and for both closed and open systems.

61. 7-8: Entropy Change of Liquids and Solids
Solids and liquids do not change specific volume appreciably with pressure. That means that dv=0, so the first equation is the easiest to use.
Thus For solids and liquids
Recall also that For solids and liquids,

62. Only true for solids and liquids!!
Integrate to give…
What if the process is isentropic?
The only way this expression can equal 0 is if,
Hence, for solids and liquids (incompressible) , isentropic processes are also isothermal.

63. Example(7-7): Effect of Density of a Liquid on Entropy
Liquid methane is commonly used in various
applications. The critical temperature of methane is 191 K (or -82oC), and thus methane must be maintained below 191 K to keep it in liquid phase. The properties of liquid methane at various temperatures and pressures are given next page.
Determine the entropy change of liquid methane as it undergoes a process from 110 K and 1 MPa to 120 K and 5 MPa
using actual data for methane and
approximating liquid methane as an incompressible substance. What is the error involved in the later case?

64. Error % = 12.2

65. Example (7-19): Entropy generated when a hot block is dropped in a lake
A 50-kg block of iron casting at 500 K is dropped in a large lake that is at 285 K. The block reaches thermal equilibrium with lake water.
Assuming an average specific heat of 0.45 kJ/kg.K for the iron, determine:
(a) The entropy change of the iron block,
(b) The entropy change of the water lake,
(c) the entropy generated during this process.

66. Tsurr= 285 K (a) The entropy change of the iron block,
we need also to find Q coming out of the system.
Tsurr= 285 K
(b) The entropy change of the water lake,
T=500K

67. (c) the entropy generated during this process.
Choose the iron block and the lake as the system and treat it is an isolated system.
Tsurr= 285 K
T=500K
Thus Sg = Stot = Ssys + Slake
Sg = Stot =
= 4.32
System boundary

68. 3- The Entropy Change of Ideal Gases, first relation
The entropy change of an ideal gas can be obtained by substituting du = CvdT and P /T= R/ into Tds relations:
integrating

69. Second relation integrating
A second relation for the entropy change of an ideal gas for a process can be obtained by substituting dh = CpdT and /T= R/P into Tds relations:
integrating

70. The integration of the first term on the RHS can be done via two methods:
Assume constant Cp and constant Cv (Approximate Analysis)
Evaluate these integrals exactly and tabulate the data (Exact Analysis)

71. Method 1: Constant specific heats (Approximate Analysis)
First relation
Only true for ideal gases, assuming constant heat capacities
Second relation
Only true for ideal gases, assuming constant heat capacities

72. Ru is the universal gas constant
Sometimes it is more convenient to calculate the change in entropy per mole, instead of per unit mass
kJ/kmol. K
kJ/kmol. K
Ru is the universal gas constant

73. Method 2: Variable specific heats (Exact Analysis)
We use the second relation
We could substitute in the equations for Cv and Cp, and perform the integrations
Cp = a + bT + cT2 + dT3
But this is time consuming.
Someone already did the integrations and tabulated them for us (table A-17)
They assume absolute 0 as the starting point

74. The integral is expressed as:
Where
is tabulated in Table A-17
Therefore

75. Is s = f (T) only? like u for an ideal gas.
Let us see
Temperature dependence
Pressure dependence
From this equation, It can be seen that the entropy of an ideal gas is not a function only of the temperature ( as was the internal energy) but also of the pressure or the specific volume.
The function s° represents only the temperature-dependent part of entropy

76. How about the other relation
We can develop another relation for the entropy changed based on the above relation
but this will require the definition of another function and tabulating it which is not practical.

79. Isentropic Processes The entropy of a fixed mass can be changed by
Heat transfer,
Irreversibilities
It follows that the entropy of a system will not change if we have
Adiabatic process,
Internally reversible process.
Therefore, we define the following:

80. Isentropic Processes of Ideal Gases
Many real processes can be modeled as isentropic
Isentropic processes are the standard against which we should measure efficiency
We need to develop isentropic relationships for ideal gases, just like we developed for solids and liquids

81. Constant specific heats (1st relation)
Recall
For the isentropic case, DS=0. Thus
Recall also from ch 2, the following relations..…
Only applies to ideal gases, with constant specific heats

82. Constant specific heats (2nd relation)
Recall..… or
Only applies to ideal gases, with constant specific heats

83. Since… Third isentropic relationship and HENCE
Which can be simplified to…
Third isentropic relationship

84. 3- Constant specific heats
Full form of Isentropic relations of Ideal Gases
Compact form
Valid for only for
1- Ideal gas
2- Isentropic process
3- Constant specific heats

85. That works if the specific heat constants can be approximated as constant, but what if that’s not a good assumption?
We need to use the exact treatment
This equation is a good way to evaluate property changes, but it can be tedious if you know the volume ratio instead of the pressure ratio

86. s20 is only a function of temperature!!!
Rename the exponential term as Pr , (relative pressure) which is only a function of temperature, and is tabulated on the ideal gas tables ( for air as an ideal gas see table A-17)

87. You can use either of the following 2 equations
This is good if you know the pressure ratio but how about if you know only the volume ratio
In this case, we use the ideal gas law
where
Remember, these relationships only hold for ideal gases and isentropic processes tabulated in A-17

88. Example (7-10):
Isentropic Compression of Air in a Car Engine
Air is compressed in a car engine from 22oC and 95 kPa in a reversible and adiabatic manner. If the compression ratio V1/V2 of this piston-cylinder device is 8, determine the final temperature of the air. <Answer: K>
Sol:

89. Assuming constant specific heat and assuming k we can have alternative solution
The k value at this anticipated average temperature is determined from Table
A–2b to be 1.391

92. Reversible Steady Flow Work

93. For the pump the specific volume is constant so h2-h1 = v (p2-p1)

94. For compressor the specific volume is function of pressure then we use h2-h1 instead of v (p2-p1) which is only used for a pump

95. 7-10: Reversible Steady Flow Work
the energy balance for a steady-flow device
If the changes in kinetic and potential energies are negligible, Then

96. Reversible work relations for steady flow and closed systems.

97. 7–11 MINIMIZING THE COMPRESSOR WORK
When the changes in kinetic and potential energies are negligible, the compressor work is given by
effect of cooling during the compression process ( PVn = Constant)
an isentropic process (involves no cooling), n =k
a polytropic process (involves some cooling), 1<n<k
an isothermal process (involves maximum cooling). n =1
One common way of cooling the gas during compression is to use
cooling jackets around the casing of the compressors.

98. 7–11 MINIMIZING THE COMPRESSOR Work
Win decrease

99. Multistage Compression with Intercooling
cooling a gas as it is compressed is desirable since this reduces the required work input to the compressor.
often it is not possible to have adequate cooling through the casing of the compressor,
effective cooling. One such technique is multistage compression with intercooling,
The gas is compressed in stages and cooled between each stage by passing it through a heat exchanger called an intercooler.
Ideally, the cooling process takes place at constant pressure, and the gas is cooled to the initial temperature T1 at each intercooler.

100. wcomp I,in = wcomp II,in. 4/2=2 4-2=2 8/4=2 8-4=4
The Px value that minimizes the total work is determined by differentiating this expression with respect to Px and setting the resulting expression equal to zero.
to minimize compression work during two-stage compression, the pressure ratio across each stage of the compressor must be the same.
wcomp I,in = wcomp II,in.

102. ENTROPY BALANCE
The property entropy is a measure of molecular disorder or randomness of a system,
The second law of thermodynamics states that entropy can be created but it cannot be destroyed or decreased.

103. Entropy Change of a System

104. Mechanisms of Entropy Transfer, Sin and Sout
Entropy can be transferred to or from system by two mechanisms: heat transfer and mass flow
Entropy transfer by heat transfer:
Recall: Entropy Generated by heat & Irreversabilities
The quantity Q/T represents the entropy transfer accompanied by heat transfer,
and the direction of entropy transfer is the same as the direction of heat
transfer since thermodynamic temperature T is always a positive quantity.

105. Mechanisms of Entropy Transfer, Sin and Sout
When the temperature T is not constant, the entropy transfer during a process 1-2 can be determined by integration (or by summation if appropriate) as
work is entropy-free, and no entropy is transferred by work.
Energy is transferred by both heat and work, whereas entropy is transferred
only by heat. That is,
where Qk is the heat transfer through the boundary at temperature Tk at location k.

106. Entropy Transfer by mass flow
Mass contains entropy as well as energy, and the entropy and energy contents of a system are proportional to the mass.
entropy and energy are carried into or out of a system by streams of matter, and the rates of entropy and energy transport into or out of a system are proportional to the mass flow rate

107. Entropy Generation , Sgen
Irreversibilities such as friction, mixing, chemical reactions, heat transfer through a finite temperature difference, unrestrained expansion, nonquasiequilibrium compression, or expansion always cause the entropy of a system to increase
Entropy generation is a measure of the entropy created by such effects during a process.
For a reversible process (a process that involves no irreversibilities), the entropy generation is zero and thus the entropy change of a system is equal to the entropy transfer.

108. Entropy balance for any system undergoing any process
or, in the rate form
where the rates of entropy transfer by heat transferred at a rate of
And mass flowing at a rate of

109. Mechanisms of entropy transfer for a general system.

110. Entropy Generation – Closed System
The entropy change of a closed system during a process is equal to the sum of the net entropy transferred through the system boundary by heat transfer and the entropy generated within the system boundaries ( No mass flow).
For an adiabatic process (Q = 0) then,
any closed system and its surroundings can be treated as an adiabatic system and the total entropy change of a system is equal to the sum of the entropy changes of its parts, the entropy balance for a closed system
and its surroundings is
where
and

111. Entropy Generation – Closed System
For an adiabatic process (Q = 0) then,
any closed system and its surroundings can be treated as an adiabatic system and the total entropy change of a system is equal to the sum of the entropy changes of its parts, the entropy balance for a closed system
and its surroundings is
where
and

112. Entropy Generation – Open System or Control Volume
In the rate form
The rate of entropy change within the control volume during a process is equal to the sum of the rate of entropy transfer through the control volume boundary by heat transfer, the net rate of entropy transfer into the control volume by mass flow, and the rate of entropy generation within the boundaries of the control volume as a result of irreversibilities.

113. Entropy Generation – Open System or Control Volume
turbines, compressors, nozzles, diffusers, heat exchangers, pipes, and ducts operate steadily,
reversible and adiabatic, then the entropy remains constant, se = si
Reversible adiabatic called Isentropic process

115. Example (7-17): Entropy Generation in a Wall
Steady heat transfer in a wall. The temperatures are shown in the figure. The rate of heat transfer through the wall is 1035 W.
Outside Surface temperature = 5 C
Inside Surface Temperature = 20 C
Outside
T=0 C
Inside home
T= 27 C
Q = 1035 W

116. Example (7-17): Entropy Generation in a Wall-continue
The rate form of the entropy balance for the wall

117. This is a steady state problem.
A) Determine the rate of entropy generation in the wall
This is a steady state problem.
We have to set the CV correctly. For part a of the question, the CV boundary will be as shown in the figure below. CV
Tin= 20 C
=293 K
Tout= 5 C
=278 K

118. B) Determine the rate of total entropy generation for the process
We will extend the CV boundary as follows (system+ immediate surrounding CV
Boundary
Tin= 27 C
=300 K
Tout= 0 C
=273 K
= Sgen, total

119. Special case: Sgen for closed system with constant temperature surroundings Tsurr

120. Example:
water
T1= 100 C
P1 = 5 Mpa
m = 2 kg
s= ?
A piston/cylinder contains 2 kg of water at 5 MPa, 100°C. Heat is added from a reservoir at 700°C to the water until it reaches 700°C.
Find the work, heat transfer, and total entropy production for the system and surroundings.
Solution
This is a constant pressure process.
Hence the work is , W = mP(v2-v1)
To get the heat,
Q-W=m(u2-u1) =>Q=m(h2-h1)

121. To get the entropy change for the system, s=m(s2-s1)
To get the total entropy production for the system and the surrounding, we apply the entropy balance equation for the extended system (system + the immediate surrounding).
So let us begin our solution. State 1 is fixed.
Go to the tables and get the following

122. State 2 is fixed also since the pressure is constant (P2=P1).
Go to the tables and get the following

123. Example 7-18

124. Example 7-18

125. Example 7-18

126. Example 7-20

127. Example 7-20

129. Some Remarks about Entropy
Processes can occur in certain direction only, a direction that complies with the increase of entropy.
Entropy is a non-conserve property. Entropy is conserved during the idealized reversible process only and increasing during all actual processes.
The greater the extent of the irreversibilities, the greater the entropy generation. Therefore, it can be used as a quantitative measure of irreversibilities.