Lecture 22: Carnot Cycle
Download
Report
Transcript Lecture 22: Carnot Cycle
EGR 334 Thermodynamics
Chapter 5: Sections 10-11
Lecture 22:
Carnot Cycle
Quiz Today?
Today’s main concepts:
•
•
•
•
State what processes make up a Carnot Cycle.
Be able to calculate the efficiency of a Carnot Cycle
Be able to give the Classius Inequality
Be able to apply the Classisus Inequality to determine if a cycle is
reversible, irreversible, or impossible as predicted by the 2nd
Law.
Reading Assignment:
Read Chapter 6, Sections 1-5
Homework Assignment:
Problems from Chap 5: 64, 79, 81,86
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
Carnot Cycle
►The Carnot cycle provides a specific example of a reversible cycle
that operates between two thermal reservoirs. Other examples
covered in Chapter 9 are the Ericsson and Stirling cycles.
►In a Carnot cycle, the system executing the cycle undergoes a series
of four internally reversible processes:
two adiabatic processes (Q = 0)
alternated with
two isothermal processes ( T = constant)
p
2
T
3
1
4
2
3
1
v
4
v
Carnot Power Cycles
The p-v diagram and schematic of a gas in a
piston-cylinder assembly executing a Carnot
cycle are shown below:
Carnot Power Cycles
The p-v diagram and schematic of water executing a Carnot cycle
through four interconnected components are shown below:
In each of these cases the
thermal efficiency is given by
max 1
TC
TH
Sec 5.10 : The Carnot Cycle
7
The Carnot cycle:
QH
Gas only cycle
QH
T
2
3
Area = Work
1
QC
4
v
QC
Process 1-2 : Adiabatic Compression.
Process 2 -3 : Isothermal Expansion
receiving QH.
Process 3 – 4 : Adiabatic Expansion.
Process 4 – 1 : Isothermal Compression,
rejecting QC.
Sec 5.10 : The Carnot Cycle
8
Analyzing the Carnot cycle:
Energy Balance: Q U W
U cV T
W
Q cV T
pdV
pdV
First look at the two isothermal processes
Process 2 -3 : Isothermal Expansion receiving QH.
Q 2 3 cV T
Q 23
pdV
pdV
3
pdV
2
3
R TH
2
V
d V R TH
3
dV
2
V
V3
R T H ln
V
2
Process 4 – 1 : Isothermal Compression, rejecting QC.
Q 41 RT C
and
V1
ln
V4
QH
QC
Q 23
Q 41
RT H ln V 3 V 2
RT C ln V1 V 4
T H ln V 3 V 2
T C ln V1 V 4
Sec 5.10 : The Carnot Cycle
9
Analyzing the Carnot cycle:
Energy Balance: Q cV d T
pdV
Then look at the two adiabatic processes (Q = 0)
Process 1-2 : Adiabatic Compression.
The term
Q 1 2 0 cV d T p d V
TH c
TC c
dT
dT
V
V
TC R T T H R T
RT
cV d T p d V
dV
Thus,
V
TH
cV dT
R
TC
T
dV
2
V
1
V2
ln
V1
V2
ln
V1
Process 3 – 4 : Adiabatic Expansion.
TC
TH
c V dT
R
T
4
3
dV
V
V4
ln
V3
V1
V2
V4
V3
V4
V1
ln
V ln V
2
3
or
V1
V4
V2
V3
Sec 5.10 : The Carnot Cycle
10
Analyzing the Carnot cycle:
With
V1
V4
Therefore,
V2
and
V3
QH
QC
QH
QC
T H ln V 3 V 2
T C ln V1 V 4
TH
TC
We have now proven
m ax
1
QC
QH
1
TC
TH
The Carnot Model of a Hurricane
11
Added heat causes
further rising.
As T to dew point, vapor
condenses, releasing hfg
and warming air
Cools & Expands as P
Adiabatic cooling
Warm air rises
Warm moist air
The Carnot Model of a Hurricane
12
Isothermal Compression
Adiabatic
vcore>>v outer
A
Adiabatic
Isothermal Expansion
B
D
C
Sec 5.10 : The Carnot Cycle
13
Example: (5.76) One-half pound of water executes a Carnot power
cycle. During the isothermal expansion, the water is heated at 600°F
from a saturated liquid to a saturated vapor. The vapor then expands
adiabatically to a temperature of 90°F and a quality of 64.3%
(a) Sketch the cycle on a P-v diagram.
(b) Evaluate the heat and work for each process in BTU
(c) Evaluate the thermal efficiency.
p
v
Sec 5.10 : The Carnot Cycle
Example: (5.76) One-half pound of water executes a Carnot power
cycle. During the isothermal expansion, the water is heated at 600°F
from a saturated liquid to a saturated vapor. The vapor then expands
adiabatically to a temperature of 90°F and a quality of 64.3%
(a) Sketch the cycle on a P-v diagram.
(b) Evaluate the heat and work for each process in BTU
(c) Evaluate the thermal efficiency.
14
Sec 5.10 : The Carnot Cycle
15
Example: (5.76) One-half pound of water executes a Carnot power
cycle. During the isothermal expansion, the water is heated at 600°F
from a saturated liquid to a saturated vapor. The vapor then expands
adiabatically to a temperature of 90°F and a quality of 64.3%
(a) Sketch the cycle on a P-v diagram.
(b) Evaluate the heat and work for each process in BTU
(c) Evaluate the thermal efficiency.
Isothermal
state
Adiabatic
Isothermal
Adiabatic
1
2
3
4
T (°F/R)
600
600
90
90
p (psi)
1541
1541
0.6988
0.6988
0
1
0.643
0.643
v (ft3/lb)
0.02363
0.2677
300.74
u (Btu/lb)
609.9
1090.0
687.8
x
Using Table A-2
v 3 0 .0 1 6 1 0 0 .6 4 3( 4 6 7 .7 0 .0 1 6 1 0 ) 3 0 0 .7 4 ft / lb m
3
u 3 5 8 .0 7 0 .6 4 3(1 0 4 0 .2 5 8 .0 7 ) 6 8 7 .8 B tu / lb m
Sec 5.10 : The Carnot Cycle
Example: (5.76)
16
(b) Evaluate the heat and work for each process in BTU
Isothermal Adiabatic Isothermal Adiabatic
1
2
3
4
Process
U
T (°F/R)
600
600
90
90
1-2
240
p (psi)
1541
1541
0.6988
0.6988
2-3
0
1
0.643
3-4
0.02363
0.2677
300.74
4-1
609.9
1090.0
687.8
state
x
v (ft3/lb)
u (Btu/lb)
Process 1-2
Q
274.8 34.81
0
0
U 12 m u 2 u 1 0.5 lb m 1090 609.9 B tu / lb m 240 B tu
W12
pdV
p V 2 V1 p ( m v 2 m v1 ) m p ( v 2 v 1 )
(1541lb f / in ) 0.5 lb m 0.2677 0.02363 ft / lb m
2
3
144 in
1
ft
2
2
1
B tu
778 ft lb f
34.81 B tu
Q1 2 U 1 2 W 1 2 2 4 0 3 4 .8 1 2 7 4 .8 B tu
W
Sec 5.10 : The Carnot Cycle
Example: (5.76)
17
(b) Evaluate the heat and work for each process in BTU
Isothermal Adiabatic Isothermal Adiabatic
1
2
3
4
Process
U
T (°F/R)
600
600
90
90
1-2
240
p (psi)
1541
1541
0.6988
0.6988
2-3
-201
0
1
0.643
3-4
v (ft3/lb)
0.02363
0.2677
300.74
4-1
u (Btu/lb)
609.9
1090.0
687.8
state
x
Q
274.8 34.81
0
0
Process 2-3
Q 23 0
(adiabatic process)
U 23 m u 2 u 1 0.5 lb m 687.8 1090 B tu / lb m 201.1 B tu
W 2 3 Q 2 3 U 2 3 0 ( 2 0 1 .1) 2 0 1 .1 B tu
W
201
Sec 5.10 : The Carnot Cycle
Example: (5.76)
18
(b) Evaluate the heat and work for each process in Btu
Isothermal
Adiabatic Isothermal Adiabatic
1
2
3
4
Process
U
Q
W
T (°F/R)
600
600
90
90
1-2
240
274.85
34.81
p (psi)
1541
1541
0.6988
0.6988
2-3
-201
0
201
0
1
0.643
3-4
-142.6
v (ft3/lb)
0.02363
0.2677
300.74
4-1
0
u (Btu/lb)
609.9
1090.0
687.8
state
x
For Process 3 – 4:
Q 34
for the Carnot cycle:
Q
where QH = Q12 = 274.8 Btu
QC
H
TH
TC
TC
550
QC Q H
274.85 B tu
142.6 B tu
1060
TH
U 34 m ( u 4 u 3 )
W 34
pdV
p 3 (V 4 V 3 ) p 3 m ( v 4 v 3 )
Sec 5.10 : The Carnot Cycle
Example: (5.76)
19
(b) Evaluate the heat and work for each process in Btu
Isothermal
Adiabatic Isothermal Adiabatic
1
2
3
4
Process
U
Q
W
T (°F/R)
600
600
90
90
1-2
240
274.85
34.81
p (psi)
1541
1541
0.6988
0.6988
2-3
-201
0
201
0
1
0.643
3-4
-142.6
v (ft3/lb)
0.02363
0.2677
300.74
4-1
0
u (Btu/lb)
609.9
1090.0
687.8
state
x
continuing for Process 3 – 4:
U 34 Q 34 W 34
m ( u 4 u 3 ) Q 34 m ( p 4 v 4 p 3 v 3 )
recalling h = u + pv
Q 34 m ( u 4 p 4 v 4 ) m ( u 3 p 3 v 3 )
h4 ( u 4 p 4 v 4 )
h4
142.6 B tu
0.5 lb m
Q 34
m
(u 3 p 3v3 )
(687.8 B tu / lb m (0.6988 lb f / in )(300.74 ft / lb m ))
2
3
144 in
1 ft
2
2
1B tu
778 lb f ft
441.5 B tu / lb m
Sec 5.10 : The Carnot Cycle
Example: (5.76)
20
(b) Evaluate the heat and work for each process in Btu
Isothermal
Adiabatic Isothermal Adiabatic
11
22
33
44
Process
U
Q
W
TT (°F/R)
(°F/R)
600
600
600
600
90
90
90
90
1-2
240
274.85
34.81
pp (psi)
(psi)
1541
1541
1541
1541
0.6988
0.6988 0.6988
0.6988
2-3
-201
0
201
00
11
0.643
0.643
0.368
3-4
-142.6
0.02363
0.02363 0.2677
0.2677 300.74
300.74
172.1
4-1
0
state
state
xx
vv (ft
(ft33/lb)
/lb)
uu (Btu/lb)
(Btu/lb)
609.9
609.9
1090.0
1090.0
687.8
687.8
419.5
continuing for Process 3 – 4:
Then using Table A2 at h4 = 441.5 Btu/lbm and T4 = 90 deg.
x4
h4 h4 f
h 4 fg
4 4 1 .5 5 8 .0 7
0 .3 6 8
1 0 4 2 .7
which let the state 4 intensive properties be found:
v 4 0 .0 1 6 1 0 0 .3 6 8( 4 6 7 .7 0 .0 1 6 1 0 ) 1 7 2 .1 ft / lb m
3
u 4 5 8 .0 7 0 .3 6 8(1 0 4 0 .2 5 8 .0 7 ) 4 1 9 .5 B tu / lb m
Sec 5.10 : The Carnot Cycle
21
Example: (5.76) (b) Evaluate the heat and work for each process in BTU
Isothermal Adiabatic Isothermal
Adiabatic
1
2
3
4
Process
U
Q
W
T (°F/R)
600
600
90
90
1-2
240
274.85
34.81
p (psi)
1541
1541
0.6988
0.6988
2-3
-201
0
201
0
1
0.643
0.368
3-4
-134.2
-142.6
-8.32
v (ft3/lb)
0.02363
0.2677
300.74
172.1
4-1
u (Btu/lb)
609.9
1090.0
687.8
419.5
state
x
0
and for Process 4-1:
U 34 m u 4 u 3 0.5 lb m (419.5 687.8) B tu / lb m 134.2 B tu
W 34
p d V p 3 V 4 V3 p 3 m v 4 v3
W 3 4 (0 .6 9 8 8 lb f / in )(0 .5 lb m ) 1 7 2 .1 3 0 0 .7 4 ft / lb m
2
3
1 4 4 in
1
ft
2
2
1
B tu
7 7 8 ft lb f
8.32 B tu
Sec 5.10 : The Carnot Cycle
22
Example: (5.76) (b) Evaluate the heat and work for each process in BTU
Isothermal Adiabatic Isothermal
state
Adiabatic
1
2
3
4
T (°F/R)
600
600
90
90
p (psi)
1541
1541
0.6988
0.6988
0
1
0.643
0.368
v (ft3/lb)
0.02363
0.2677
300.74
172.1
u (Btu/lb)
609.9
1090.0
687.8
419.5
x
Process
1-2
2-3
3-4
4-1
U
240
Q
274.85
W
34.81
-201
-134.2
0
-142.6
201
-8.32
95.2
0
-95.2
Finally, process 4 – 1:
U 41 m u 1 u 4 0.5 lb m (609.9 419.5) B tu / lb m 95.2 B tu
W 4 1 Q 4 1 U 4 1 0 9 5 .2 9 5 .2 B tu
Sec 5.10 : The Carnot Cycle
23
Example: (5.76) (c) Evaluate the thermal efficiency.
Isothermal Adiabatic Isothermal
state
Adiabatic
1
2
3
4
Process
U
Q
W
T (°F/R)
600
600
90
90
1-2
240
274.85
34.81
p (psi)
1541
1541
0.6988
0.6988
2-3
-201
0
201
0
1
0.643
0.368
3-4
-134.2
-142.6
-8.32
v (ft3/lb)
0.02363
0.2677
300.74
172.1
4-1
95.2
0
-95.2
u (Btu/lb)
609.9
1090.0
687.8
419.5
x
Thermal efficiency of the cycle:
m ax 1
W C ycle
Q in
TC
1
TH
550
1060
34.81 201 8.32 95.2
274.85
0 .4 8 1 4 8 %
131.99
274.85
0.480 48%
Clausius Inequality
►The Clausius inequality is developed from the KelvinPlanck statement of the second law and can be expressed
as:
Q
cycle
T b
The nature of the cycle executed is indicated by
the value of cycle:
cycle = 0 no irreversibilities present within the system
cycle > 0 irreversibilities present within the system
cycle < 0 impossible
Sec 5.11 : The Clausius Inequality
We have shown that:
QH
25
QC
Therefore: Q H
TH
QC
TH
TC
and thus
QH
TH
QC
TC
For an ideal/reversible process
0
TC
Now consider a general process
Each part of the cycle is divided into
an infinitesimally small process
dQ
p
H
dQ C
0
TH
TC
Then sum (integrate) the entire process
v
dQ = heat transfer at
Q
boundary
T 0
b
T = absolute T at that
part of the cycle.
Sec 5.11 : The Clausius Inequality
26
For a real process, Qreal > Qreversible
Therefore:
Q
T 0
b
Q
T cycle
b
We can then define σ, where
and σcycle = 0 reversible process
σcycle > 0 irreversible process
σcycle < 0 impossible process
P
We now also have the mathematical
definition of enthalpy.
dQ
dS dQ TdS
T
v
More on this in Chapter 6
Sec 5.11 : The Clausius Inequality
Example: (5.81) A system executes a power cycle while receiving heat
transfer at a temperature of 500 K and discharging 1000 kJ by heat
transfer at a temperature of 300 K. There are no other heat transfers.
Appling the Clausius Inequality, determine σcycle if the thermal
efficiency is (a) 60% (b) 40% (c) 20 %. In each case is the cycle
reversible, or impossible?
TH= 500 K
W=?
Q = 50 kJ
TH= 300 K
27
Sec 5.11 : The Clausius Inequality
28
Example: (5.81) A system executes a power cycle while receiving heat
transfer at a temperature of 1000 K and discharging 1000 kJ by heat
transfer at a temperature of 300 K. There are no other heat transfers.
Appling the Clausius Inequality, determine σcycle if the thermal
efficiency is (a) 60% (b) 40% (c) 20 %. In each case is the cycle
reversible, or impossible?
QC
1
Q C 1 Q H
TH= 500 K
QH
Q
W=?
and cycle
T b
Q = 50 kJ
TH= 300 K
cycle
Therefore:
cycle
QH
QC
QH
TC
TH
1
1
T
T
C
H
1 0 . 6
kJ
1
1000 kJ
0
.
667
300 K
K
500 K
Since σcycle is negative, the cycle is impossible.
Sec 5.11 : The Clausius Inequality
29
Example: (5.81) A system executes a power cycle while receiving heat
transfer at a temperature of 1000 K and discharging 1000 kJ by heat
transfer at a temperature of 300 K. There are no other heat transfers.
Appling the Clausius Inequality, determine σcycle if the thermal
efficiency is (a) 60% (b) 40% (c) 20 %. In each case is the cycle
reversible, or impossible?
TH= 500 K
W=?
Q = 50 kJ
cycle
TH= 300 K
cycle
QH
QC
QH
TC
TH
1
1
T
T
C
H
1 0 . 4
kJ
1
1000 kJ
0
300 K
K
500 K
Since σcycle is zero, the cycle is internally reversible.
Sec 5.11 : The Clausius Inequality
30
Example: (5.81) A system executes a power cycle while receiving heat
transfer at a temperature of 1000 K and discharging 1000 kJ by heat
transfer at a temperature of 300 K. There are no other heat transfers.
Appling the Clausius Inequality, determine σcycle if the thermal
efficiency is (a) 60% (b) 40% (c) 20 %. In each case is the cycle
reversible, or impossible?
TH= 500 K
W=?
Q = 50 kJ
TH= 300 K
cycle
cycle
QH
QC
QH
TC
TH
1
1
T
T
C
H
1 0 . 2
kJ
1
1000 kJ
0 . 667
300 K
K
500 K
Since σcycle is positive, the cycle is irreversible.
31
End of Lecture 21 Slides