### Entropy and Closed Systems

```EGR 334 Thermodynamics
Chapter 6: Sections 6-8
Lecture 25:
Entropy and closed system
analysis
Quiz Today?
Today’s main concepts:
• Heat transfer of an internally reversible process
can be represented as an area on a T-s diagram.
• Learn how to evaluate the entropy balance for a
closed system
Homework Assignment:
Problems from Chap 6: 36, 38, 59, 66
3
Recall from last time:
Energy Balance:
 E sys  Q sys  W sys
dE sys
Energy Rate Balance:
d E sys
dt
W
Q
dt
 Q sys  W sys
Entropy Balance:
 S sys  
Q
T
Q
  g en
T
Entropy Rate Balance:
d S sys
dt
 
Q
T

d S sys
dt
gen
 g en
Entropy Balance for Closed Systems
 Q 
S   

 T b
change of
entropy
entropy
transfer


entropy
production
where the subscript b
indicates the integral is
evaluated at the system
boundary.
• Unlike mass and energy balances, the entropy balance
doesn’t represent a conserved quantity.
:
= 0 (no irreversibilities present within the system)
> 0 (irreversibilities present within the system)
< 0 (impossible)
Change of Entropy of the System
 Q 
S   

 T b


As seen in the previous lecture, ∆S = S2-S1, represents a
difference of state properties and may be evaluated by
-- looking up s values on substance property tables.
-- applying Tds relationships for ideal gas
• If two or more properties of the end states of a process are
known, then the Change of Entropy per unit mass, ∆s, is
completely defined and discernible.
Entropy Transfer
• Consider the Entropy Transfer term of the Entropy
Balance. On a differential basis it can expressed
• This expression indicates that when a closed system
undergoing an internally reversible process receives
energy by heat transfer, the system experiences an
increase in entropy. Conversely, when energy is removed
by heat transfer, the entropy of the system decreases.
From these considerations, we say that entropy transfer
accompanies heat transfer. The direction of the entropy
transfer is the same as the heat transfer.
Entropy and Heat Transfer
►In an internally reversible, adiabatic process (no heat
transfer), entropy remains constant. Such a constantentropy process is called an isentropic process.
► Rearranging the differential expression gives
which can be integrated from state 1 to state 2,
2
Q int 
rev
 T dS
1
Entropy and Heat Transfer
Consider how this integral would represented
on a T-S diagram:
2
Q in t 
rev
 T d S   A rea
u n d er T -S
1
An energy transfer by heat to a
closed system during an
internally reversible process is
represented by an area on a
temperature-entropy diagram:
6.6&7 : Entropy and Closed Systems
9
For the Carnot cycle:
What does the area represent?
Carnot Work:
W 
 PdV
What does the area represent?
Carnot Heat Transfer:
Q 
In the Carnot Cycle, only the constant temperature
processes contribute to Heat (and Entropy) transfer.
 TdS
6.6&7 : Entropy and Closed Systems
10
For the Carnot cycle:
3
Q 23 
 TdS
2
Q 
dS  

T

 in t
Q  TdS
0
1
Q 41 
rev.
 TdS
4
0
Q  Q1 2  Q 2 3  Q 3 4  Q 4 1
(note that Q41 is negative)
Entropy Balance for Closed Systems
That  has a value of zero when there are no internal
irreversibilities and is positive when irreversibilities are
present within the system leads to the interpretation that
 accounts for entropy produced (or generated) within
the system by action of irreversibilities.
 Q 
S   

 T  int

rev.
Expressed in words, the entropy balance is
change in the amount
of entropy contained
within the system
during some
time interval
net amount of
entropy transferred in
across the system boundary +
accompanying heat transfer
during some time interval
amount of
entropy produced
within the system
during some
time interval
A general approach to analyze a Closed System
with entropy balance.
• Step 1: Identify properties at each state
including T, p, v, u, x, and s.
• Step 2: Apply 1st law and attempt to evaluate
∆U, Q, and W for each process.
• Step 3: Write the 2nd Law (Entropy balance) and
attempt to evaluate ∆S and ΣQ/T for each
process. Usually σ will be determined from
σ = ∆S - ΣQ/T
• Step 4: Explain the significance of σ.
Example: One kg of water vapor contained within a pistoncylinder assembly, initially at 5 bar, 400oC, undergoes an
adiabatic expansion to a state where pressure is 1 bar and the
temperature is (a) 200oC, (b) 100oC. Using the entropy
balance, determine the nature of the process in each case.
Boundary
Q  and entropy balance reduces to give
0
S 2  S1 

1
2
 Q

 T

 
b
→
Find property values:
using m = 1 kg and Table A-4
m(s2 – s1) = 
get s1 = 7.7938 kJ/kg∙K.
Example 1 continued:
(a) At p = 1 bar and T = 200 deg C.
Table A-4 gives, s2 = 7.8343 kJ/kg∙K.
then
  m(s2 – s1)
 = (1 kg)(7.8343 – 7.7938) kJ/kg∙K = 0.0405 kJ/K
(Since s is positive, irreversibilities are present
within the system during expansion )
(b) At p = 1 bar and T = 100 deg C.
Table A-4 gives, s2 = 7.3614 kJ/kg∙K.
then
  m(s2 – s1)
 = (1 kg)(7.3614 – 7.7938) kJ/kg∙K = –0.4324 kJ/K
(Since s is negative, expansion (b) is impossible.
Example 1 continued:
Just a little more analysis of part b)
The result of part b was that  is negative.
<0 = <0 + ≥0
► Since  cannot be negative and
► For expansion in (part b) S is negative, then
► By inspection the integral must be negative and so heat
transfer from the system must occur in expansion (b).
Entropy Rate Balance for Closed Systems
Some problems are presented in the form of a closed
system entropy rate balance given by
where
dS

the time rate of change of the entropy of
the system

the time rate of entropy transfer through the
portion of the boundary whose temperature is Tj
dt
Q j
Tj
  time rate of entropy production due to
irreversibilities within the system
Example 2:
An inventor claims that the device shown generates electricity at a
rate of 100 kJ/s while receiving a heat transfer of energy at a rate of
250 kJ/s at a temperature of 500 K, receiving a second heat transfer at
a rate of 350 kJ/s at 700 K, and discharging energy by heat transfer at a
rate of 500 kJ/s at a temperature of 1000 K. Each heat transfer is
positive in the direction of the accompanying arrow. For operation at
Q 1  250 kJ/s
T1 = 500 K
Q 2  350 kJ/s
+
–
T2 = 700 K
T3 = 1000 K
Q 3  500 kJ/s
Example 2 continued:
Applying an energy rate balance
dE
dt
Solving
0
 0  Q 1  Q 2  Q 3  W e
W e  250 kJ/s  350 kJ/s  500 kJ/s  100 kJ/s
The claim is in accord with the first law of thermodynamics.
Applying an entropy rate balance
Solving
 250 kJ/s
   
 500 K

350 kJ/s
700 K
   0 . 5  0 . 5  0 . 5 
kJ/s
K

dS
0
0
dt
Q 1
T1

Q 2
T2

Q 3
 
T3
500 kJ/s 

1000 K 
  0.5
kJ/s
K
∙ negative, the claim is not in accord with the 2nd Law of
Since σ is
Thermodynamics and is therefore denied.
19
Example 3 (6.14):
One kilogram of water contained in a piston-cylinder assembly, initially
at 160°C, 150 kPa undergoes an isothermal compression process to
saturated liquid. For the process, W= –471.5 kJ. Determine for the
process.
(a)
(b)
(c)
(d)
Sketch the process on a T-s diagram.
The heat transfer, in kJ
The change in entropy in kJ/K
The entropy generated in the process.
State 1: T1= 160 oC
p1 = 150 kPa = 1.50 bar
super heated vapor
State 2: T2= T1 =160 oC
sat. liquid.
T
2
1
s
20
Example 3 (6.14): One kilogram of water contained in a piston-cylinder
assembly, initially at 160°C,150 kPa undergoes an isothermal expansion
from compression process to saturated liquid. For the process,
W= –471.5 kJ. b) find the heat transfer
State
1
2
Look up the rest of the state properties.
Then apply the 1st Law.
U  Q  W
Q  m u 2  u 1   W
T (°C)
160
160
p (kPa)
1.50
150
617.8
SH
0
u (kJ/kg
2595.2
674.86
s (kJ/kg K)
7.4665
1.9427
x
Q   1kg   674.86  2595.2  kJ / kg  (  471.5 kJ )   2391.84 kJ / K
(c) To find the change in entropy
 S  m ( s 2  s1 )
 S   1kg   1 .9 4 2 7  7 .4 6 6 5  kJ / kg  K   5 .5 2 3 8
kJ
kg K
21
Example 3 (6.14): One kilogram of water contained in a piston-cylinder
assembly, initially at 160°C,150 kPa undergoes an isothermal
compression process to saturated liquid. For the process, W= –471.5 kJ.
Determine for the process.
(d) To find the entropy generation
  S 

where
Q
T
 S  m  s 2  s1    5.5238 kJ / kg  K
Q
T

 2391.84 kJ
(160  273) K
  S  
Q
State
1
2
T (°C)
160
160
p (kPa)
150
617.8
x
SH
0
u (kJ/kg
2595.2
674.86
s (kJ/kg K)
7.4665
1.9427
  5.5239 kJ / K
  5.5238  (  5.5239)   0.0001  0 kJ / K
T
What does this mean?
6.11 : Isentropic Processes
States may be given as having the same entropy (two-phase,
saturated vapor, superheated vapor)
Any process where the entropy does not change is called isentropic.
22
6.11 : Isentropic Processes
23
Consider isentropic process for an ideal gas
Starting with

s 
and
Next with
c V T dT
T
cV 
s 
R

k 1
 T1

T
 2
s 

cP T  dT
T
and
R
cP 
k 1
v 
 R ln  2 
T
 v1 
dT
 T2
ln 
k  1  T1
1
thus
v 
 R ln  2 
 v1 
s 
k 1

kR
k 1
 p2 
 R ln 

T
p
 1 
dT
 T2 
 p2 
ln 
  ln 

k  1  T1 
p
 1 

 v2 
   ln 


 v 

 1 
  v2 


 v 
  1 
kR
 p2 
 R ln 

p
 1 
k
1  k 1 
thus
and
An isentropic process is a  p 2    v1 
type of polytropic process  p   v 
 1   2 
 T2   p 2 



T
p
 1   1 
( k  1)
k
k
p 1 v1  p 2 v 2
k
k
24
Example (6.27): Air in a piston-cylinder assembly and modeled as an
ideal gas undergoes two internally reversible processes in series from
State 1 where T1 = 290 K, p1 = 1 bar.
Process 1 – 2 : Compression to p2 = 5 bar during which pV1.19 = constant
Process 2 – 3 : Isentropic expansion to p3 = 1 bar.
(a) Sketch the two processes on T-s coordinates
(b) Determine the temperature at State 2 in K
Air
(c) Determine the net work in kJ
5 bar
2
TT
1
3
SS
290 K
1 bar
25
Example (6.27): Air in a piston-cylinder assembly and modeled as an
ideal gas undergoes two internally reversible processes in series from
state 1 where T1 = 290 K, p1 = 1 bar.
Process 1 – 2 : Compression to p2 = 5 bar during which pV1.19 = constant
Process 2 – 3 : Isentropic expansion to p3 = 1 bar.
(a) Sketch the two processes on T-s coordinates
(b) Determine the temperature at state 2 in K
(c) Determine the net work in kJ
p1V1
p 2V 2
Process 1-2:

N
N
ideal gas: T1
T2
polytropic: p1V1  p 2V 2
V2
V1
 p 
 1 
 p2 
1/ N
V2
 p1 


 p2 
 p2 
T 2  T1 

 p1 
1/ N
 n 1

p1T 2

p1T 2
V1
p 2T2
1 .1 9  1 
1 .1 9
p 2T2
n
5
  290 K   
1
 375 K
26
Example (6.27):
Process 1 – 2 : Compression to p2 = 5 bar during which pV1.19 = constant
Process 2 – 3 : Isentropic expansion to p3 = 1 bar.
Find the state properties
State 1: p1 = 1 bar
T1 = 290 K
State 2: p2 = 5 bar
T2 = 375 K
State
1
2
T (K)
290
375
1
5
u (kJ/kg)
206.91
268.075
s °(kJ/kg K)
1.66802
1.92657
p (bar)
State 3: p3 = 1 bar
s3 = s2
From Table A-22: at T = 290 K: u = 206.91
so = 1.66802
From Table A-22: at T = 375 K: u = 268.075
so = 1.92657
3
1
27
Example (6.27):
Process 2 – 3 : Isentropic expansion to p3 = 1 bar.
State
State
11
22
33
TT (K)
(K)
290
290
375
375
236.83
11
55
11
uu (kJ/kg)
(kJ/kg)
206.91
206.91
268.075
268.075
168.86
ss °(kJ/kg
°(kJ/kg K)
K)
1.66802
1.66802
1.92657
1.92657
1.46466
pp (bar)
(bar)
At State 3:
 p3 
s 3  s 2  s   T3   s   T 2   R ln 

p
 2 
 p 
0  s   T3   s   T 2   R ln  3 
 p2 
 p3 
s   T3   s   T 2   R ln 

p
 2 
 1b a r 
s   T3   (1 .9 2 6 5 7 kJ / kg  K )  (0 .2 8 7 0 kJ / kg  K ) ln 
  1 .4 6 4 6 6 kJ / kg  K
 5b a r 
From Table A-22: for so=1.46466 find T and u
T3 = 236.82 K
and u3 = 168.86 kJ/kg
28
Example (6.27): (c) Determine the net work in kJ
Process
ΔU
1–2
+61.2
Q
W
State
1
2
3
-67.2
-128.4
T (K)
290
375
636.82
1
5
1
u (kJ/kg)
206.91
268.075
168.86
s °(kJ/kg K)
1.66802
1.92657
1.46466
2–3
p (bar)
Process 1 – 2:
 U 12  Q12  W 12
where
 U 12  m ( u 2  u 1 )
 U 12
m
 268.08  206.91
W 12 
 pdV
W 12
 0.287 kJ
m
 61.17 kJ / kg
Q12
m

 U 12
m



p 2V 2  p1V1
1 n

m R  T 2  T1 
1 n
/ kg  K  (375  290) K
1  1.19 
  128.39 kJ / kg
W 12
m
 61.17  (  128.39)   67.22 kJ / kg
29
Example (6.27):
Process
ΔU
Q
W
1–2
+61.2
-67.2
-128.4
2–3
-99.2
0
+99.2
Process 2-3:
State
1
2
3
T (K)
290
375
236.82
1
5
1
u (kJ/kg)
206.91
268.075
168.86
s °(kJ/kg K)
1.66802
1.92657
1.46468
P (bar)
 U 23  Q 23  W 23
3
where
Q 23   T ds  0
 U 23  m ( u 3  u 2 )
 U 23
2
 168.86  268.08   99.22 kJ / kg
m
W 23

m
Q 23

 U 23
m
 0  (  99.22)  99.22 kg / kg
m
Net Work over both processes:
W net
m

W12
m

W 23
m
  128 . 39
kJ
kg
 99 . 22
kJ
kg
  29 . 17
kJ
kg
6.8 : Directionality of Processes
30
Second Law statement: “It is impossible for a system to operate such
that entropy is destroyed.”
This can be seen in the entropy balance, but first look at the
energy balance.
E
E
isolated system
system
 E
0
en viro n m en t
0
Energy is neither created nor destroyed.
S
Isolated System
S
system
 Q 
 

 T  int
 S

rev.
en viro n m en t
 
31
End of Slides for Lecture 25
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