Chapter 4: Entropy: an Additional Balance Equation

Important Notation 2

Consider a composite system (A+B) surrounded by a boundary (external wall) that is: Adiabatic Rigid Impermeable Composite system (A+B) is isolated. 3

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. 4

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? 5

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. 6

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? 7

Because the mass and the energy in the initial and final states are equal, they do not serve to characterize the final state as something special. There must be some new property to account for this. Call this property for a moment. 8

If is a property, it should no longer change once equilibrium is achieved. at equilibrium away from equilibrium 9

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.) 10

Entropy balance For a closed system 11

depends on the internal relaxation processes within the system. When analyzing the heat transfer between systems A and B, let us assume it occurs slowly in such a way that the temperature is uniform within each system – a quasistatic process. In this case, is equal to zero in systems A and B. 12

Suppose initially. Then and 13

Note that 14 From experimental observations : The flow of heat is proportional to the T difference; h is a positive constant, and the minus 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:

However we could also use this other proof: 15

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. 16

Further, the isolated composite system (A+B) has no mass and heat flows entering or leaving it:

Entropy balance: integrated form 18

Entropy balance: integrated form If the entropy per unit mass is constant in time at each mass port: 19

Entropy balance: integrated form If the temperature is constant at the point where heat transfer occurs: 20

Entropy balance: integrated form If the simplifications of the two previous slides apply simultaneously: 21

Clausius statement of the 2 nd law It is not possible to build a device that operates in a cycle whose 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: 22

Clausius statement of the 2 nd law It is not possible to build a device that operates in a cycle whose 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: 0 (because the only effect is heat transfer) The entropy balance is: 23

Clausius statement of the 2 nd law It is not possible to build a device that operates in a cycle whose 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: 0 (because the only effect is heat transfer) The entropy balance is: 0 (because S, a state function, returns to its initial value at the end of the cycle) 24

Clausius statement of the 2 nd law It is not possible to build a device that operates in a cycle whose only effect is to transfer heat from a colder body to a hotter body. If is positive, is negative, violating that. 25

Kelvin-Planck statement of the 2 nd law It is not possible to build a device that operates in a cycle whose only effect is the production of work by transferring heat from a single body. At the end of thermodynamic cycle, the system returns to its initial state. The energy balance is: The entropy balance is: 26

Kelvin-Planck statement of the 2 nd law It is not possible to build a device that operates in a cycle whose only effect is the production of work by transferring heat from a single body. is negative, violating that. 27 Since Q >0

Entropy balance and reversibility 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, and the process is said to be reversible. 28

Entropy balance and reversibility Consider a general system open to the flow of mass, heat, and work between two equal time intervals, from 0 to and from to. 29

Entropy balance and reversibility 30

Entropy balance and reversibility Suppose now that all mass, heat and work flows are reversed in the second half of the process, compared to the first half. 31

Entropy balance and reversibility Then we end up with: In general, the rates of entropy generation are positive, their integrals are also positive, and. The system would not have returned to its initial state. The process is irreversible. 32

Entropy balance and reversibility If a process proceeds with infinitesimal gradients within the system, the rates of entropy generation are zero, their integrals are also zero, and. The system would have returned to its initial state. The process is reversible. Examples of reversible processes (negligible viscous dissipation or 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 33

Entropy balance and reversibility Examples 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 34

Entropy balance and reversibility 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. 35

Entropy balance and reversibility 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. 36

Entropy balance and reversibility Few processes are truly reversible but it is sometimes useful to model them to be so. 37

Helmholtz energy (NVT) Consider the energy and entropy balances for a closed, isothermal, constant volume system: Combining these two equations, using that : 38

Helmholtz energy (NVT) We now define a new state function, the Helmholtz energy, as: Then, in an irreversible process: In a reversible process: 39 more work needed to change the state of the system (W s >0) in the irreversible process more work is obtained (W s <0) in a reversible process

Gibbs energy (NPT) Consider the energy and entropy balances for a closed, isothermal, isobaric system: Combining these two equations, using that and : 40

Gibbs energy (NPT) We now define a new state function, the Gibbs energy, as: Then, in an irreversible process : In a reversible process: 41 more work needed to change the state of the system (W s >0) in the irreversible process more work is obtained (W s <0) in a reversible process

Relationship of entropy changes with changes in other state functions Consider a system in which kinetic and potential terms are unimportant, with a single mass flow stream, and the mass and heat flows occur at the common temperature T: 42

Relationship of entropy changes with changes in other state functions Eliminating and multiplying the entropy balance by T: 43

Relationship of entropy changes with changes in other state functions The amounts exchanged over a period of time dt are: Combining these two equations: 44

Relationship of entropy changes with changes in other state functions For a reversible process: For a system with no shaft work (open system): Additionally, if the system is closed: 45

Entropy changes of matter Entropy is very important, but how to evaluate it in practice? 1)Thermodynamic diagrams 2)Property tables, e.g., steam tables 3)Mathematical models, e.g., equations of state 46

Example 1 A 0.5 kg/s steady flow of refrigerant HFC-134a enters a perfectly insulated valve as saturated vapor at 70 o C. 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: Isenthalpic expansion – Joule-Thompson process We now want: 47

f3_3_4 48

Example 1 From the pressure-enthalpy diagram: 49

Entropy changes of matter Consider now the entropy change of one mol of a pure substance: For an ideal gas: 50

Entropy changes of matter 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: 51

Entropy changes of matter The ideal gas equation of state can be used to eliminate the molar volumes (details in the book): For liquids and solids, a good approximation is: 52

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: 53

Example 2 Entropy balance at steady-state (dS/dt=0) The temperature does not change, therefore: 54

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. 55

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 56

Example 4 Energy balance 57

Example 4 Entropy balance Note: this is an isentropic (constant entropy) expansion process. These calculations are very demanding if done manually. 58

Example 5 A well insulated rigid box of volume 6 m 3 is divided in two compartments of equal volume, separated by a rigid, heat-transferring wall. Initially, compartment 1 has air at 100 o C 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 2 nd tank vs the P in the 1 st tank, and the change in total entropy vs. the P in tank 1. Assume air is an ideal gas of constant heat capacity. Amount of air in the system: 59

Example 5 Since U=U 1 +U 2 is constant, therefore T 1 =T 2 =100 o C 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. Total entropy change 60

61 what is the equilibrium state of the system?

Summary of the formulas 62

t4_1_1 63

t4_1_2 64

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. 65

Recommendation Read chapter 4 in the textbook and review all examples. 66