1. EGR 334 Thermodynamics Chapter 6: Sections 1-5
Lecture 24:
Introduction to Entropy
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2. Today’s main concepts:
Explain key concepts about entropy
Learn how to evaluate entropy using property tables
Learn how to evaluate changes of entropy over reversible processes
Introduction of the temperature-entropy (T-s) diagram
Reading Assignment:
Read Chapter 6, Sections 6-8
Homework Assignment:
Problems from Chap 6: 1,11,21,28

3. Introducing Entropy Change and the Entropy Balance
The entropy change and entropy balance concepts are developed using the Clausius inequality expressed as:
where
scycle = 0 no irreversibilities present within the system
scycle > 0 irreversibilities present within the system
scycle < 0 impossible

4. Defining Entropy Change
Consider two cycles, each composed of two internally reversible processes, process A plus process C and process B plus process C, as shown in the figure.
Applying Classius Equation to these cycles gives,
where scycle is zero because the cycles are composed of internally reversible processes.

5. Defining Entropy Change
Subtracting these equations:
Since A and B are arbitrary internally reversible processes linking states 1 and 2, it follows that the value of the integral is independent of the particular internally reversible process and depends on the end states only.

6. Defining Entropy Change
Recalling (from Sec ) that a quantity is a property if, and only if, its change in value between two states is independent of the process linking the two states, we conclude that the integral represents the change in some property of the system.
We call this property entropy and represent it by S. The change in entropy is
where the subscript “int rev” signals that the integral is carried out for any internally reversible process linking states 1 and 2.

7. Defining Entropy Change
The equation allows the change in entropy between two states to be determined by thinking of an internally reversible process between the two states. But since entropy is a property, that value of entropy change applies to any process between the states – internally reversible or not.

8. Entropy Facts Entropy is an extensive property.
Like any other extensive property, the change in entropy can be positive, negative, or zero:
Units of entropy, S, are: in SI  [kJ/K] in US Customary  [Btu/oR]
Units for specific entropy, s, are: in SI  [kJ/kg-K] in US Customary  [Btu/lbm-oR]

9. Finding Entropy: Method 1
For problem solving, specific entropy values are provided in Tables A-2 through A-18. Values for specific entropy are obtained from these tables using the same procedures as for specific volume, internal energy, and enthalpy.
For two phase liquid/vapor mixtures, quality, x, may be used as or
For slightly compressed liquids, the entropy may be approximated as:

10. Example 1: find s using property tables
a) Given: H20 at T=520 oC and p= 8 MPa find s:
From Table A-2…substance is found to be superheated vapor.
From Table A-4…..s can be found ad kJ/kg-K
b) Given: Ammonia at p =2.5 bar and v = 0.20 m3/kg find s:
From Table A14…substance is found to be liquid/vapor mixture.
the quality can be found as
and then
c) Given: R22 at T = 20 oF and p = 80 psi find s:
From Table A-7E or A-8E…substance is found to be compressed liquid
then s can be found using s(T,p)=sf(T) = Btu/lbm-oR

11. Example (6.4): Using the appropriate tables determine the change in specific enthalpy between the specified states, in BTU/lb·°R.
(a) Water p1 = 1000 psi, T1 = 800 °F
p2 = 1000 psi, T2 = 1000 °F
(b) Refrigerant 134a h1 = BTU/lb, T1 = -40 °F
saturated vapor, p2 = 40 psi

12. Example (6.4): Using the appropriate tables determine the change in specific enthalpy between the specified states, in BTU/lb·°R.
(a) Water p1 = 1000 psi, T1 = 800 °F
p2 = 1000 psi, T2 = 100 °F T
1000 psi
Table A-4E
@ T1 = 800 °F, p1 = 1000 psi
s1 = BTU/lb·°R 1
545°F 2
Table A-5E
@ T2 = 100 °F, p2 = 1000 psi
s2 = BTU/lb·°R v
s2 - s1 = = BTU/lb·°R

13. (b) Refrigerant 134a h1 = 47.91 BTU/lb, T1 = -40 °F
Example (6.4): Using the appropriate tables determine the change in specific enthalpy between the specified states, in BTU/lb·°R.
(b) Refrigerant 134a h1 = BTU/lb, T1 = -40 °F
saturated vapor, p2 = 40 psi T
Table A-10E,
@ T1 = -40 °F
hf=0, hg=95.82 BTU/lb 2
-40°F 1
sf=0, sg= BTU/lb·°R v
Table A-11E, sat. vapor, p2 = 40 psi
s2 - s1 = = BTU/lb·°R

14. Finding Entropy: Method 2
For problem solving, states often are shown on property diagrams having specific entropy as a coordinate: the Temperature-Entropy (T-s) diagram and the Enthalpy-Entropy (h-s) diagram
the h-s diagram is also know as the Mollier diagram
These diagrams
are in your appendix as Figures 7 and 8
and
Figures 7E and 8E.

15. T-s diagram: Vertical axis: Temperature Horizontal axis: Entropy
Shows other constant property lines:
Constant Pressure
Constant Quality
Constant Enthalpy
Constant Specific Volume

16. h-s diagram:
Vertical axis: Enthalpy
Horizontal axis: Entropy
Shows other constant property lines:
Constant Temperature
Constant Pressure
Constant Quality

17. Example 2: Using T-s diagram:
b) Given: H20 at s=1.5 Btu/lbm-R and x=85% find h:
a) Given: H20 at T=900 oF
and h = 1400 Btu/lbm find s:
s ≈ 1.5 Btu/lbm-R
h ≈ 1.5 Btu/lbm
From Table A-7E or A-8E…substance is found to be compressed liquid
then s can be found using s(T,p)=sf(T) = Btu/lbm-oR
From Table A-7E or A-8E…substance is found to be compressed liquid
then s can be found using s(T,p)=sf(T) = Btu/lbm-oR

18. Finding Entropy: Method 3…IT
IT can also be used for working with entropy. Once again it’s recommended that you don’t compare individual values from IT with the values you might find on the Appendix tables from you book, but changes in entropy between two states will be consistent and can be compared.
For H20, find the difference in entropy between
State 1: T = 500 deg C
and p = 20 bar
State 2: T = 200 deg C
and p = 10 bar
Δs= kJ/kg-K

19. Sec 6.3 : The TdS Equations
How is S related to U?
Consider a pure, simple, compressible system undergoing a internally reversible process.
Energy Balance:
then differentiate
where,
therefore,
1st TdS equation

20. How is S related to H? Recall that: and therefore,
Sec 6.3 : The TdS Equations
How is S related to H?
Recall that:
and
from 1st Tds equation
therefore,
2nd T dS equation
can also write these on a per mass basis
Relationships between p, T, v, u, h, and s
Even though this derivation is based on a reversible process, these equations hold for any process… even an irreversible process.

21. Finding Entropy: Method 4 for incompressible fluid
When a substance is compressible (like many liquids),
Recall that for incompressible fluids:
and
where c is the specific heat capacity
then
if the specific heat is treated as constant
Δs for incompressible liquids

22. Finding Entropy: Method 5 for Ideal Gas
Recall that if a substance can be treated as an ideal gas, the following relationships may be applied to the defining the properties:
Combining ideal gas relations with the Tds equations
give
ds = f(T,v)
ds = f(T,p)

23. So for Ideal Gas, change in entropy can be evaluated by integrating
Sec 6.4 Idea Gases
So for Ideal Gas, change in entropy can be evaluated by integrating
these relations
and

24. or constant heat capacities from Tables A20, A22, or A23.
Sec 6.4 Idea Gases
Options to evaluate:
and
there are several options to evaluate the heat capacity. Remember that cV & cP are functions of temperature.
i) Find values for cp and or cv (Using Table A21)
or constant heat capacities from Tables A20, A22, or A23.
then use or or
ii) Use Tabulated values( Table A22):
The s° indicates that it is based on a reference temp

25. Example (6.4): Using the appropriate tables determine the change in specific enthalpy between the specified states, in BTU/lb·°R.
Air as an ideal gas with T1 = 40°F = 500° R , p1 = 2 atm
and T2 = 420°F = 880° R, p2 = 1 atm
(a) using a constant cv at Tave
where: Tave = 230 F = 690 R
R = Btu/lb-R cp = Btu/lb-R

26. Example (6.4): Using the appropriate tables determine the change in specific enthalpy between the specified states, in BTU/lb·°R.
Air as an ideal gas with T1 = 40°F = 500° R , p1 = 2 atm
and T2 = 420°F = 880° R, p2 = 1 atm
(c) Air as an ideal gas using so
from Table A-22E
@ T1 = 500 °R, s° = BTU/lb·°R
@ T2 = 880 °R, s° = BTU/lb·°R

27. end of lecture 24 slides