Transcript 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
Reading Assignment:
Read Chapter 6, Sections 9-10
Homework Assignment:
Problems from Chap 6: 36, 38, 59, 66
Recall from last time: Energy Balance:
E sys
Q sys
W sys
Energy Rate Balance:
dE sys dt
Q sys
W sys
Entropy Balance :
S sys
Q T
gen
Entropy Rate Balance:
dS sys dt
Q T
gen Q T Q dE sys dt W dS sys dt
gen
3
Entropy Balance for Closed Systems
change of entropy
Q T b
entropy transfer entropy production
where
the subscript indicates the integral is evaluated at the system boundary.
b
:
• Unlike mass and energy balances, the entropy balance doesn’t represent a conserved quantity.
= 0 > 0 < 0
(no irreversibilities present within the system) (irreversibilities present within the system) (impossible)
Change of Entropy of the System
Q T
b
As seen in the previous lecture, ∆S = S 2 -S 1 , 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 constant entropy process is called an
isentropic
process.
► Rearranging the differential expression gives which can be integrated from state 1 to state 2 ,
Q
int
rev
2 1
TdS
Entropy and Heat Transfer
Consider how this integral would represented on a T-S diagram:
Q
int
rev
2 1
TdS
Area under T-S
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 For the Carnot cycle: What does the area represent?
Carnot Work:
W
PdV
What does the area represent?
Carnot Heat Transfer:
Q
TdS
9 In the Carnot Cycle, only the constant temperature processes contribute to Heat (and Entropy) transfer.
6.6&7 : Entropy and Closed Systems For the Carnot cycle:
Q
23 3 2
TdS Q
Q dS
Q
T TdS Q
12 0
Q
23 int rev.
Q
34 0
Q
41
Q
41 (note that Q 41 is negative) 1 4
TdS
10
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.
S
Q
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 1 st 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 piston cylinder assembly, initially at 5 bar, 400 o C , undergoes an adiabatic expansion to a state where pressure is 1 bar and the temperature is
(a)
200 o C ,
(b)
100 o C . Using the entropy balance, determine the nature of the process in each case.
Since the expansion occurs adiabatically, Q and entropy balance reduces to give
0
S
2
S
1
1 2
Q T
b
→ m(s 2 – s 1 ) =
Boundary
Find property values: using m = 1 kg
and
Table A-4
get
s
1 = 7.7938 kJ/kg∙K
.
Example 1 continued:
(a) At p = 1 bar and T = 200 deg C.
Table A-4
then gives,
s
2 = 7.8343 kJ/kg∙K
.
m(s 2
– s 1 )
= (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
then gives,
s
2 = 7.3614 kJ/kg∙K
m(s 2
– s 1 )
= (1 kg)(7.3614 – 7.7938) kJ/kg∙K .
= –0.4324 kJ/K
(
Since s is negative, expansion (b) is impossible. It cannot occur adiabatically.)
Example 1 continued: Just a little more analysis of part b) The result of part b was that
is negative . The process cannot occur
adiabatically
.
► ► ► Since
< 0 = < 0 + ≥ 0
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 dt
the time rate of change of the entropy of the system
T j j
the time rate of entropy transfer through the portion of the boundary whose temperature is
T j
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 rate of
350 kJ/s
at
700 K
, and
discharging
a second heat transfer at 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 steady state, evaluate this claim.
250 kJ/s
T
1
350 kJ/s
T
2
500 kJ/s
T
3
Example 2 continued:
Applying an energy rate balance at steady state
dE
0
0
dt
1 2
Q
3
e
Solving
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 at steady state
dS
0
dt
0
Q
1
T
1
T
2 2
T
3 3 Solving 250 kJ/s 500 K 350 kJ/s 700 K 500 kJ/s 1000 K 0 .
5 0 .
5 0 .
5 kJ/s K
0.5
kJ/s K
Since σ is negative, the claim is
not
in accord with the 2 nd Thermodynamics and is therefore denied.
Law of
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) Sketch the process on a T-s diagram.
(b) The heat transfer, in kJ (c) (d) The change in entropy in kJ/K The entropy generated in the process.
19 State 1: T 1 = 160 o C p 1 = 150 kPa = 1.50 bar super heated vapor 2 1 State 2: T 2 = T 1 =160 o C sat. liquid.
s 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 Look up the rest of the state properties.
Then apply the 1 st Law. 617.8
Q
Q W
m
u
2
u
1
W Q
(c) To find the change in entropy ( 2
s
1 )
1
kg
1.9427
7.4665
2595.2
kJ
) 2391.84
674.86
1.9427
5.5238
kJ kgK
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.
State 1 2
(d) To find the entropy generation where 2
Q T
s
1 5.5238
T (°C) p (kPa) x u (kJ/kg 160 150 SH 2595.2
s (kJ/kg K) 7.4665
160 617.8
0 674.86
1.9427
Q T
2391.84
kJ
5.5239
K
Q
T
What does this mean? Process had no irreversibilities 21
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 Consider isentropic process for an ideal gas Starting with and
s
c V c V
dT T
R
R
ln
v
2
v
1
k
1
s
k R
1
dT T
R
ln
v
2
v
1 thus
k
1 1 ln
T
2
T
1
T T
2 1
v
2
v
1 1 ln
k
v
1
v
1 2 An isentropic process is a type of polytropic process
p
2
p
1 and Next with and
c P T c P
R
ln
kR k
1
p
2
p
1
k kR
1
dT T
R
ln
p
2
p
1
k k
1 ln
T
2
T
1 ln
p
2
p
1
v v
2 1
k
thus
T
2
T
1
p
2
p
1 (
k
1)
k p v
1 1
k
p v
2 2
k
23
Example (6.27): State 1 where T 1 Air in a piston-cylinder assembly and modeled as an ideal gas undergoes two internally reversible processes in series from = 290 K , p 1 = 1 bar . Process 1 – 2 : Compression to p 2 = 5 bar during which pV 1.19
= constant Process 2 – 3 : Isentropic expansion to p 3
(a)
= 1 bar .
Sketch the two processes on T-s coordinates
(b) (c) Determine the temperature at State 2 in K Determine the net work in kJ Air 24 2 5 bar 3 1 290 K 1 bar S S
25 Example (6.27): state 1 where T 1 Air in a piston-cylinder assembly and modeled as an ideal gas undergoes two internally reversible processes in series from = 290 K, p 1 = 1 bar. Process 1 – 2 : Compression to p 2 = 5 bar during which pV 1.19
= constant Process 2 – 3 : Isentropic expansion to p 3 = 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 Process 1-2: polytropic:
p V
1 1
N V
2
p
1
p V
2 2
N
1/
N
ideal gas:
p V
1 1
V
2
T
1
p T
1 2
p V
2 2
T
2
V
1
p
2
p
1
p
2 1/
N
p T
2 2
V
1
p T
2 2
T
2
T
1
p
2
p
1
n
1
n
290
K
5 375
K
Example (6.27): Process 1 – 2 : Compression to p 2 = 5 bar during which pV Process 2 – 3 : Isentropic expansion to p 3 = 1 bar.
1.19
= constant Find the state properties State 1: p 1 = 1 bar T 1 = 290 K State 2: p 2 = 5 bar T 2 = 375 K 206.91
1.66802
268.075
1.92657
State 3: p 3 = 1 bar s 3 = s 2 From Table A-22: at T = 290 K: u = 206.91
s o = 1.66802
From Table A-22: at T = 375 K: u = 268.075
s o = 1.92657
26
27 Example (6.27): Process 2 – 3 : Isentropic expansion to p 3 = 1 bar.
1 1 5 5 236.83
1 1 168.86
At State 3: 1.46466
s
3 0
s
2
s
3 3 3 2 (1.92657
R
ln
R
2 ln
p
3
p
2
R
ln
p p
2 3
p
3
p
2 ) (0.2870
/ ) ln 1
bar
1.46466
/ 5
bar
From Table A-22: for s o =1.46466 find T and u T 3 = 236.82 K and u 3 = 168.86 kJ/kg
28 Example (6.27): (c) Determine the net work in kJ +61.2 -67.2
-128.4
State
T (K) p (bar) u (kJ/kg)
1
290 1 206.91
Process 1 – 2:
U
12
Q
12
W
12 s °(kJ/kg K) 1.66802
where
U
12 ( 2
u
1 )
U
12
m
61.17
W
12
W
12
m
pdV
0.287
2
375 5 268.075
1.92657
3
636.82
1 168.86
1.46466
p V
2 2 1
n p V
1 1 2 1
n
T
1
K
128.39
/
Q
12
m
U
12
m
W
12
m
67.22
/
29 Example (6.27): -99.2
0 +99.2
State
T (K) P (bar) u (kJ/kg)
1
290 1 206.91
2
375 5 268.075
Process 2-3:
U
23
Q
23
W
23 s °(kJ/kg K) 1.66802 1.92657
where
U
23
U
23
m
( 3
u
2 ) 99.22
W
23
Q
23
U m m
Net Work over both processes:
m
23
Q
23 99.22
3 2
Tds
0 /
W net m
W
12
m
W
23
m
128 .
39
kJ kg
99 .
22
kJ kg
29 .
17
kJ kg
3
236.82
1 168.86
1.46468
6.8 : Directionality of Processes 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
0
isolated system
E
system
E
environment 0 Energy is neither created nor destroyed.
S
Isolated System
Q T
int rev.
30
S
system
S
environment
End of Slides for Lecture 25
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