Lecture 24: Introduction to Entropy

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EGR 334 Thermodynamics
Chapter 6: Sections 1-5
Lecture 24:
Introduction to Entropy
Quiz Today?
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
Introducing Entropy Change and the Entropy Balance
• The entropy change and entropy balance concepts are
developed using the Clausius inequality expressed as:
 Q 


s


  T b cycle
where
scycle = 0 no irreversibilities present within the system
scycle > 0 irreversibilities present within the system
scycle < 0 impossible
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.
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.
Defining Entropy Change
• Recalling (from Sec. 1.3.3) 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
 2 Q 
S 2  S1   

 1 T int
rev
where the subscript “int rev” signals that the integral
is carried out for any internally reversible process
linking states 1 and 2.
Defining Entropy Change
 Q 
S2  S1   

 1 T int
rev
2
• 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.
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

in US Customary

• Units for specific entropy, s, are:
in
SI

in US Customary

[kJ/K]
[Btu/oR]
[kJ/kg-K]
[Btu/lbm-oR]
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
x
s  sf
sg  s f
or
s  s f  x( sg  s f )
For slightly compressed liquids, the entropy may be
approximated as:
s (T , p)  s f (T )
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 6.7871 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
x  (v  v f ) /(vg  v f )  (0.20  .0015) /(0.4821  0.0015)  0.413
and then
s  s f  x( sg  s f )  0.4753  0.413(5.5190  0.4753)  2.5583kJ / kg  K
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) = 0.0356 Btu/lbm-oR
10
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 = 47.91 BTU/lb, T1 = -40 °F
saturated vapor, p2 = 40 psi
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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
1
545°F
2
v
Table A-4E
@ T1 = 800 °F, p1 = 1000 psi
s1 = 1.5665 BTU/lb·°R
Table A-5E
@ T2 = 100 °F, p2 = 1000 psi
s2 = 0.12901 BTU/lb·°R
s2 - s1 = 0.12901 - 1.5665 = -1.43749 BTU/lb·°R
13
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 = 47.91 BTU/lb, T1 = -40 °F
saturated vapor, p2 = 40 psi
T
Table A-10E,
@ T1 = -40 °F
hf=0, hg=95.82 BTU/lb
47 .91  0
x
 0.5
95 .82  0
2
-40°F
1
v
Table A-11E, sat. vapor, p2 = 40 psi
s2  0.2197 Btu / lbm  R
sf=0, sg=0.2283 BTU/lb·°R
s1  s f  x  sg  s f

 0   0.5  0.2283  0 
 0.1142 Btu / lbm  R
s2 - s1 = 0.2197 - 0.11415 = 0.1056 BTU/lb·°R
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.
T-s diagram:
Vertical axis: Temperature
Horizontal axis: Entropy
Shows other constant
property lines:
Constant Pressure
Constant Quality
Constant Enthalpy
Constant Specific Volume
h-s diagram:
Vertical axis: Enthalpy
Horizontal axis: Entropy
Shows other constant
property lines:
Constant Temperature
Constant Pressure
Constant Quality
Example 2: Using T-s diagram:
a) Given: H20 at T=900 oF
and h = 1400 Btu/lbm find s:
s ≈ 1.5 Btu/lbm-R
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) =
0.0356 Btu/lbm-oR
b) Given: H20 at s=1.5 Btu/lbm-R
and x=85%
find h:
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) =
0.0356 Btu/lbm-oR
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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=0.1992 kJ/kg-K
Sec 6.3 : The TdS Equations
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How is S related to U?
Consider a pure, simple, compressible system undergoing a internally
reversible process.
Energy Balance:
KE  PE  U  Q  W
U  Q  W
then differentiate
where,
therefore,
dU  Qint rev  W int rev
 W int rev  pdV
 Q int rev  TdS
dU  TdS  pdV
TdS  dU  pdV
1st TdS equation
Sec 6.3 : The TdS Equations
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How is S related to H?
Recall that:
H  U  pV
dH  dU  d  pV   dU  pdV  Vdp
dU  pdV  dH  Vdp
and
dU  TdS  pdV
from 1st Tds equation
TdS  dU  pdV
therefore,
TdS  dH  Vdp
2nd T dS equation
can also write these on a per mass basis
du  Tds  pdv
dh  Tds  vdp
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.
Finding Entropy: Method 4 for incompressible fluid
When a substance is compressible (like many liquids),
du  Tds  pdv
du pdv
ds 

T
T
Recall that for incompressible fluids:
du  c dT
dv  0
and
where c is the specific heat capacity
then
cdT
ds 
T
s2  s1  
c dT
T
if the specific heat is treated as constant
T2
dT
s2  s1  c 
 c ln 1
T
T
Δs for incompressible
liquids
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:
pv  RT
du  cv dT
k  c p / cv
dh  c p dT
c p  cv  R
Combining ideal gas relations with the Tds equations
du pdv
ds 

T
T
dh vdp
ds 

T
T
dT
dv
ds  cv
R
T
v
dT
dp
ds  c p
R
T
p
give
ds = f(T,v)
ds = f(T,p)
Sec 6.4 Idea Gases
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So for Ideal Gas, change in entropy can be evaluated by integrating
these relations
cv T  dT R
ds 
 dv
T
v
and
cP T  dT R
ds 
 dp
T
P
 v2 
cV T dT
s  
 R ln 
T
 v1 
cP T  dT
 p2 
s  
 R ln 

T
p
 1
Sec 6.4 Idea Gases
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Options to evaluate:
 v2 
cV T dT
s  
 R ln 
T
 v1 
and
cP T  dT
 p2 
s  
 R ln 

T
p
 1
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)
cP    T  T 2  T 3  T 4
or constant heat capacities from Tables A20, A22, or A23.
cv 
R
k 1
or
cP 
kR
k 1
then use
or
p 
v 
T
T
s  c p ln 2  R ln  2 
s  cv ln 2  R ln  2 
T1
T1
 p1 
 v1 
ii) Use Tabulated values( Table A22):
s  sT2   sT1 
The s° indicates that it is based on a reference temp
 p2 
s  s T2   s T1   R ln 

p
 1
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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
cP T  dT
 p2 
s  
 R ln 

T
 p1 
 T2
s  cP ln 
 T1
where:
Tave = 230 F = 690 R
R = 0.06855 Btu/lb-R

 p2 
  R ln 

p

 1
cp = 0.242 Btu/lb-R
 880 
1
o
s  0.242 ln 

0.06855
ln

0.1843
Btu
/
R

 
 500 
2
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
 P2
s  sT2   sT1   R ln
 P1



from Table A-22E
@ T1 = 500 °R, s° = 0.58233 BTU/lb·°R
@ T2 = 880 °R, s° = 0.71886 BTU/lb·°R
 1 atm 
s  0.71886  0.58233  0.06855 ln 

2
atm


BTU
s  0.1841
lb   R
27
end of lecture 24 slides

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