1. Chapter Seven: Entropy

2. WHAT IS ENTROPY
Entropy is 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.
Unlike energy, entropy is a nonconserved property, and there is no such thing as a conservation of entropy principle.

3. Derivation of Entropy from First and Second Laws
Consider the following combined system consisting of a heat engine and a piston cylinder system
First Law applied to the combined system
(for example
Carnot cycle)

4. The Carnot Cycle
Idealized thermodynamic cycle consisting of four reversible processes (working fluid can be any substance):
The four steps for a Carnot Heat Engine are:
Reversible isothermal expansion (1-2, TH= constant)
Reversible adiabatic expansion (2-3, Q = 0, THTL)
Reversible isothermal compression (3-4, TL=constant)
Reversible adiabatic compression (4-1, Q=0, TL TH) TL
1-2
2-3
3-4
4-1

5. Carnot Efficiency QR =Q12 = W12 = PdV = mRTHln(V2/V1) (1)
Consider an ideal gas undergoing a Carnot cycle between two temperatures TH and TL.
1 to 2, isothermal expansion, DU12 = 0
QR =Q12 = W12 = PdV = mRTHln(V2/V1) (1)
2 to 3, adiabatic expansion, Q23 = 0
(TL/TH) = (V2/V3)k (2)
3 to 4, isothermal compression, DU34 = 0
QL = Q34 = W34 = - mRTln(V4/V3) (3)
4 to 1, adiabatic compression, Q41 = 0
(TL/TH) = (V1/V4)k (4)
From (2) & (4): (V2/V3) = (V1/V4)  (V2/V1) = (V3/V4)
Since ln(V2/V1) = - ln(V4/V3); substituting for ln(V4/V3) in (1)
 (QL/QR )= (TL/TH)
Hence: th = 1-(QL/QR)= 1-(TL/TH)
Therefore: QL/QR =TL/TH Or

6. Derivation of Entropy from First and Second Laws
Consider the following combined system consisting of a Carnot heat engine and a piston cylinder system
First Law applied to the combined system
For a Carnot Engine (reversible cycle)
Substituting: 6

7. Derivation of Entropy from First and Second Laws
Let the system undergo a cycle while the Carnot cycle undergoes one or more cycles. Integrate over entire cycle :
The net work for one cycle is
The integral must be evaluated at the system boundary
The combined system (cycle) draws heat from a single reservoir while involving work WC. Based on Kelvin–Planck statement (no system can produce a net amount of work while operating in a cycle and exchanging heat with a single thermal energy reservoir) the combined system cannot produce net work output. Therefore, WC
This is the Clausius Inequality which is valid for all thermodynamic cycles, reversible or irreversible.

8. Derivation of Entropy from First and Second Laws
For a reversible cycle
Since the cyclic integral of (dQ/T) is 0, the quantity dQ/T is a state property, it does not depend on the path
We call this new property entropy, S,
For a process where the system goes from state 1 to 2
Specific entropy s= S/M=

9. A Special Case: Internally Reversible Isothermal Heat Transfer Processes
Internally reversible process : is the process that has no irreversibilities that occur within the boundaries of the system during the process. During an internally reversible process, a system proceeds through a series of equilibrium states, and when the process is reversed, the system passes through exactly the same equilibrium states while returning to its initial state.
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 :
where T0 is the constant absolute temperature of the system and Q is the heat transfer for the internally reversible process.

10. EXAMPLE : Entropy Change during an Isothermal Process
A piston-cylinder device contains a liquid–vapor mixture of water at 300 K. During a constant-temperature 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.

11. Solution
The system undergoes an internally reversible, isothermal process, and thus its entropy change can be determined directly to be:
Note that the entropy change of the system is positive, as expected, since heat transfer is to the system.

12. THE INCREASE OF ENTROPY PRINCIPLE
From Clausius inequality:
For complete cycle consisting of irreversible process
from 1 to 2 and internally reversible process from
2 to 1 and:
We conclude from these equations that the entropy change of a closed system during an irreversible process is greater than the integral of evaluated for that process.

13. THE INCREASE OF ENTROPY PRINCIPLE
Remove inequality sign to get the entropy balance equation for a closed system: 13

14. Isolated system (adiabatic closed system)
This is the Increase in Entropy Principle which simply states that
“for an isolated system the entropy always increases or remains the same”.
For isolated system (adiabatic closed system):

15. Remarks about Entropy Entropy is a nonconserved property.
Processes can occur in a certain direction only, not in any direction.
Entropy generation is a measure of the magnitudes of the irreversibilities
The greater the extent of irreversibilities, the greater the entropy generation.

18. Entropy Change of a System
Note that entropy is a property, and the value of a property does not change unless the state of the system changes. Therefore, the entropy change of a
system is zero if the state of the system does not change during the process.

24. Third Law of Thermodynamics
Entropy is the measure of molecular disorder or randomness.
As a system becomes more disordered, the position of the molecules becomes less predictable and the entropy increases.
Entropy is the lowest in solids because molecules are held in place and simply vibrate and highest in gases where the molecules are free to move in any direction.
Third Law of Thermodynamics States that:
“Entropy of a pure crystalline substance at absolute zero temperature (zero Kelvin) is zero since the state of each molecule is known”.

25. Isentropic Process
A process that is both adiabatic and reversible is referred to as isentropic, and for a closed ystem:

26. Assume that the system is at steady state ( )

29. For ideal gas:

31. ENTROPY BALANCE
The second law of thermodynamics states that entropy can be created but it cannot be destroyed.
The entropy change of a system during a process is greater than the net entropy transfer by an amount equal to the entropy generated during the process within the system, and the increase of entropy principle for any system is expressed as:

32. Mechanisms of Entropy Transfer
1. Heat Transfer
2. Mass Flow
Note that there is no entropy change due to work

33. Entropy Generation
Irreversibilities such as friction, mixing, chemical reactions, heat transfer through a finite temperature difference, unrestrained expansion or compression always cause the entropy of a system to increase, and entropy generation is a measure of the entropy created by such effects during a process.
For irreversible process:
For a reversible process, the entropy generation is zero and thus the entropy change of a system is equal to the entropy transfer.
For reversible process:

34. Change in Entropy and Its Rate

35. Open System (Control Volumes)
A control volume involves mass flow across its boundaries, and its entropy change is a combination of entropy transfer accompanying heat and mass transfer and the entropy generation within the system boundaries.

36. Special Cases 1. Steady state process
2. Steady state process, Single-stream
3. Steady state process, Single-stream, Adiabatic

37. Closed Systems
A closed system involves no mass flow across its boundaries, and its entropy change is simply the difference between the initial and final entropies of the system. The entropy change of a closed system is due to the entropy transfer accompanying heat transfer and the entropy generation within the system
boundaries. 37

38. Steps for Solving Entropy Problems
Sketch the process
List data you may need to solve the problem and highlight what is known and what is unknown.
Perform Mass Balance.
Perform Energy Balance.
Perform Entropy Balance. 38

39. EXAMPLE: Entropy Generation in a Mixing Chamber
Water at 20 psia and 50°F enters a mixing chamber at a rate of 300 lbm/min where it is mixed steadily with steam entering at 20 psia and 240°F. The mixture
leaves the chamber at 20 psia and 130°F, and heat is lost to the surrounding air at 70°F at a rate of 180 Btu/min. Neglecting the changes in kinetic and potential
energies, determine the rate of entropy generation during this process.

40. We take the mixing chamber as the system
We take the mixing chamber as the system. This is a control volume since mass crosses the system boundary during the process. We note that there are two inlets and one exit.
Under the stated assumptions and observations, the mass and energy balances for this steady-flow system can be expressed in the rate form as follows:

41. Data =

43. Entropy Balance:
The rate of entropy generation during this process can be determined by applying the rate form of the entropy balance on an extended system that includes the mixing chamber and its immediate surroundings so that the boundary temperature of the extended system is 70°F = 530 R:

44. End of Chapter Seven