Chapter Five Applications of Thermodynamic Cycle
19
Advanced Thermodynamics, ME Dept NCHU 頁 1 Chapter Five Applications of Thermodynamic Cycle Updated on 2019/1/20 (5.1). Carnot Cycle (5.1.1). Operating principle of Carnot Cycle The idea of Carnot cycle was proposed by Sadi Carnot at 1824 in a paper. ‘Reflections on the Motive Power of Fire and on Machines Fitted to Develop that Power’ Fig. 5.1.1 Carnot cycle 1 2 3 4 1 Fig. 5.1.2 The Carnot engine operating with ideal gas.
Transcript of Chapter Five Applications of Thermodynamic Cycle
Advanced Thermodynamics, ME Dept NCHU 頁 1
Chapter Five Applications of Thermodynamic Cycle
Updated on 2019/1/20
(5.1). Carnot Cycle
(5.1.1). Operating principle of Carnot Cycle
The idea of Carnot cycle was proposed by Sadi Carnot at 1824 in a paper.
‘Reflections on the Motive Power of Fire and on Machines Fitted to Develop that
Power’
Fig. 5.1.1 Carnot cycle
1 2 3 4 1
Fig. 5.1.2 The Carnot engine operating with ideal gas.
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The working fluid is air. Assume air is an ideal gas with constant heat capacity.
1 2 reversible isothermal compression
1 2 LT T T
1 2 2 1 1 2q u u w
1 2 Lq q
2 1u u
2 2
21 2
11 1
lnL
RT vw Pdv dv RT
v v
2
1
lnL L
vq RT
v
2 3 reversible adiabatic compression
3 2s s
3 3 33 2
2 2 2
ln ln ln ln 0Hv v
L
T v T vs s c R c R
T v T v
3
2
1ln ln ln
1
v H L
L H
v c T T
v R T k T
3 4 reversible isothermal expansion
3 4 HT T T
3 4 4 3 3 4q u u w
3 4 Hq q
4 3u u
4 4
43 4
33 3
lnH
RT vw Pdv dv RT
v v
4
3
lnH H
vq RT
v
4 1 reversible adiabatic expansion
4 1s s
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1 1 11 4
4 4 4
ln ln ln ln 0Lv v
H
T v T vs s c R c R
T v T v
1
4
ln lnv H
L
v c T
v R T
1 2
4 3
ln ln lnv H
L
v c T v
v R T v
2 1
3 4
v v
v v
1 4
2 3
v v
v v
1
2
4
3
ln
ln
L
L L
H HH
vRT
q Tv
vq TRT
v
As a result, the efficiency of Carnot cycle can be expressed as the following.
1 1H L L L
H H H H
w q q q T
q q q T
The thermal efficiency is not related to the stroke. It is determined by the temperature
of reservoirs only.
Carnot cycle
0
0.20.4
0.6
0.8
1
0 2 4 6 8 10
TH/TL
effi
cien
cy
efficiency
Fig. 5.1.3 Efficiency of Carnot cycle
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Example：Find the thermal efficiency of a Carnot cycle with TH = 2000 K and TL =
300 K.
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Assignment 5.1：A Carnot cycle starts with P1=1 bar, T1=300 K, V1=0.1 m3,
V2=0.01 m3. The high temperature is 1000 K. Calculate the properties of each state
and the work output if the working fluid is air with constant heat capacity.
P (bar) T (K) V (L)
1 1 300 100
2 10 300 10
3 676.2 1000 0.493
4 67.62 1000 4.93
qH=RTHln(V4/V3)=66.084 kJ/kg
m=0.1161 kg
QH=76.75 kJ
e=1-300/1000=0.7
W=eQH=53.73 kJ
ΔV=V1-V3=0.099507 m3
Mep=W/ΔV=539.9 kPa
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Carnot cycle is an ideal cycle. It has the highest efficiency that no other cycle
can beat it. However, it has never been realized in real world. Not because the
ideal isentropic process can never be reached, but because there are some intrinsic
features that prohibit the realization of this cycle.
The work output is determined by the stroke of the isothermal process.
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4
3
ln (1 )LH H
H
v TW Q mRT
v T
However, the stroke of the isothermal process can be expressed as
4 4 1 4
3 1 3 1
v v v v
v v v v ， 4
1
1ln ln ln
1
v L L
H H
v c T T
v R T k T
Where 1
3
v
v is the compression ratio of the engine.
4 4
3 1
1ln ln ln ln ln
1
L
H
v v T
v v k T
1 1ln ln (1 ) ln ln ( 1)
1 1
L L L HH L
H H H L
T T T TW mRT mRT
k T T k T T
It is noted that power output is affected by temperature ratio as well as compression
ratio. However, since compression ratio is in the logarithm form, its effect is small
compared with temperature.
Specific work: the work output per unit mass of working fluid.
1ln ln ( 1)
1
L HL
H L
W T Tw RT
m k T T
Non dimensional work:
1ln ln ( 1)
1
L H
L H L
W T Tw
mRT k T T
Mean effective pressure (MEP):
MEP is an index to show how much work can be done in a specific volume of
displacement. The higher the value of MEP, the more work that the engine can
deliver.
11
1 3 11
1 1 1(1 )
W W W Pmep w P
V V m RTV
( 1 LT T )
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1
mepw
P ( 1 )
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Example：Find the work output and the mean effective pressure of a Carnot cycle
with TH = 2000 K, TL = 300 K, P1 = 100 kPa, V1 = 0.1 m3, and compression ratio of
200.
P (bar) T (K) V (L)
1 1 300 100
2 1.743 300 57.38
3 1333.3 2000 0.5
4 765 2000 0.871
1ln ln ( 1)
1
L HL
H L
T Tw RT
k T T
= 271 kJ/kg
L
ww
RT = 3.148
11
mep w P
= 316.4 kPa
4q mepT W P V
4q
mep
TP
V
For comparison
KYMCO 150c.c. moped，Maximum torque 1.15 kg-m/7000 rpm
mep = 944 kPa
Ford Fiesta 1500c.c. automotive, Maximum torque 14.3 kg-m/4400 rpm
mep = 1174 kPa
Hino J08E 7648c.c. Diesel engine, Maximum torque 77 kg-m/1500 rpm
mep = 1240 kPa
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For the same displacement volume, Carnot engine would produce less power than
conventional engine because it has low mean effective pressure.
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(5.1.2). Limitations of Carnot cycle
Maximum work
1ln ln ( 1)
1
L HL
H L
T Tw RT
k T T
1( 1) ln ln
1L
Ww X X
mRT k
H
L
TX
T
For a given value of compression ratio ( ), the specific work is determined by the
temperature ratio( X ) only.
1 1 1ln ln ( 1) 0
1 1
dwX X
dX k k X
1 1ln ln 1
1X
k X
This is the temperature ratio that will produce the highest power output provided that
LT and are given.
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Example：Find the value of TH that will produce maximum output work of a Carnot
cycle with TL = 300 K, and compression ratio of 200.
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Maximum pressure limit
Carnot cycle is a pressure limiting cycle. The maximum pressure increases
sharply with power for a given set of reservoirs.
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1 12 1 2 1 4
max 3 2 1 1 1
3 2 3 2 3
k kk k
k kH H
L L
v v v v T v TP P P P P P
v v v v T v T
1
4
3
H
W
mRTve
v
，1
1
3 1H
kW
kmRT H
L
TP P e
T
P3 is the maximum pressure in the cycle. If Pmax is the limiting factor in the design
of Carnot cycle, the work output is as follows.
14 max
3 0
k
kL
H
v P T
v P T
Where 1 0P P is the reference pressure.
max max1
0 0
(1 )ln[ ( ) ] ( 1)[ln ln ]1
k
L L kH L
H H
T P T P kW mRT mRT X X
T P T P k
H
L
TX
T : temperature ratio that will produce maximum output work while the
maximum pressure is kept at Pmax.
max
0
( 1)[ln ln ]1L
W P kw X X
mRT P k
max
0
1ln (1 ln ) 0
1
dw p kX
dX p k X
max
0
1 11 ln ln
k PX
X k P
，the value of X can be obtained by solving this equation.
max 1
0 3
H H
L L
P v T T
P v T T
max
0
L
H
P T
P T
For 1 kg of working mass, the work and the efficiency for a Carnot cycle under the
maximum output mode are shown in the table.
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max 0/P P X W
100
200
300
400
500
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Example：For a Carnot cycle with Pmax = 100 bar, P1 = 1 bar, and TL = 300 K, find
the thermal efficiency and the specific work output.
Ans：x=2.172，TH = 651.5 K ，η=0.54， 4
3
6.622v
v ，w=5527 kJ/kmole
V1 = 1 liter，n = 4.009×10-5
kmole，W = nw = 0.2216 kJ = 221.6 N-m
Tq × 2π = W，Tq = 35.3 Nm in torque
Mep = W/V = 221.6 kPa
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Example：For a Carnot cycle with Pmax = 300 bar, P1 = 1 bar, V1=0.1m3, and TL =
300 K, find the efficiency, the amount of heat addition, the work output, the mean
effective pressure, and the properties of each state.
1 1
1
PVm
RT = 0.1161 kg
max
0
1 11 ln ln
k px
x k p
= 1.6297
x=2.71，TH = 813 K ，η=0.633
14 max
3 0
k
kL
H
v p T
v p T
=9.156, max
0
L
H
p T
p T = 110.7
1 2 3 4
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T (K) 300 300 813 813
P (bar) 1 9.156 300 32.77
V (m3) 0.1 10.92×10
-3 9.03×10
-4 8.27×10
-3
HQ (kJ) W (kJ) IMEP (kPa)
0.633 59.99 37.97 383.16
4
3
lnH H
vQ mRT
v = 59.99 kJ
HW Q = 37.97 kJ
1 3 1
1
11
W Wmep
V V V
= 383.16 kPa
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Assignment 5.2：For a Carnot cycle with Pmax = 1000 bar, P1 = 1 bar, and TL = 300 K,
find the thermal efficiency and the specific work output.
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With constrain of maximum pressure, the thermal efficiency of Carnot cycle is
no longer the highest.
The mean effective pressure is low compared with other cycle. As a result, the
engine displacement should be very large for Carnot cycle in order to obtain the same
output power.
Limited heat transfer rate
Isothermal process can only be implemented in two ways. The first way is that
the system undergoes an infinitely slow process such that the system and its thermal
reservoir may be at the same temperature. The second way is that the heat transfer
coefficient between the system and its thermal reservoir is infinitely large such that
heat transfer may proceed at regular speed.
If the engine rotates at a regular speed, and the heat transfer coefficient between
the system and its thermal reservoir is NOT infinitely large, there should exist
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temperature deviation between reservoir and system, and the cycle is no longer
isentropic.
In order to enhance heat transfer rate, temperature of reservoir should be higher
than that of system in heating period, and during the cooling period, temperature of
reservoir should be lower that that of system. The resulting system is no more a
reversible system, and the thermal efficiency would be getting lower.
1( )H RH Hq hA T T
1( )L L RLq hA T T
( )1 1
( )
L L RL
H RH H
Q T T
Q T T
The thermal efficiency of the cycle can be shown above in which RHT is the high
thermal reservoir temperature, RLT is the low thermal reservoir temperature,
HT is
the high temperature of the cycle, and LT is the low temperature of the cycle. It is
noted that there is a temperature difference between thermal reservoir and system
such that heat transfer may occur.
However, the efficiency of Carnot cycle depends on the temperature ratio only.
1 L
H
T
T
As a result, the cycle temperature and the thermal reservoir temperature can be
related as the following.
( )
( )
L RL L
RH H H
T T T
T T T
H L H RL L RH H LT T T T T T T T
2
L RHH
L RL
T TT
T T
,
/
2 / 1
H L RL
RH L RL
T T T
T T T
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2 21 1 1 1
2
L L L RL RL L
L RHH RH RH RH
L RL
T T T T T T
T TT T T T
T T
21 ( )
2
RL L L RHH RH
RH RH L RL
hA T T T TW q T
T T T T
/1 2 (1 )
/ 2 / 1
RL RL L L RL
RH RH RH RL L RL
W T T T T T
hAT T T T T T
The value of L
RL
T
T should be confined to the region that efficiency is nonnegative.
1 2 0RL RL L
RH RH RL
T T T
T T T
11 ( 1)
2
L RH
RL RL
T T
T T
It is noted that for a given value of /RH RLT T , the output work varies as /L RLT T
increases from 1.0 to its maximum allowable value, and peaks a certain value of
/RH RLT T . The efficiency decreases all the way as /L RLT T increases, and the value
of HT decreases too.
If we raise the value of /RH RLT T , the allowable value of /L RLT T extends to a
greater range, and the work output increases too.
Performance of Carnot cycle (TRH/TRL=3.0)
0
0.20.4
0.6
0.8
1
1 1.2 1.4 1.6 1.8 2
TL/TRL
perf
orm
ance work
TH/TRH
efficiency
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Fig. 5.1.4 Performance of Carnot cycle with limited heat transfer rate
21 ( )
2
RL L L RHH RH
RH RH L RL
hA T T T TW q T
T T T T
2
2 2 2( ) 1 ( )
2 2 (2 )
L RH RL L RH L RHRH
L RH L RL RH RH L RL L RL
dW hA T T T T T T TT
dT T T T T T T T T T
2
2 1 22 1 (2 2 )
2 (2 )
L RL LL RH RL RH L RH
L RL L RL RH RH
hA T T TT T T T T T
T T T T T T
2
2 12 2
2 (2 )
LRL RH RL RL L RL
L RL L RL
hA TT T T T T T
T T T T
2
18 8 2 4 2 2
(2 )L L L RL RL RL L L RL L RL RH RL RL L RL
L RL
hAT T T T T T T T T T T T T T T T
T T
2
14 4 0
(2 )L L L RL RL RL RL RH
L RL
hAT T T T T T T T
T T
4 4 ( ) 0L L L RL RL RH RLT T T T T T T
1 14 16 16 ( )
8 2L RL RL RL RL RH RL RL RL RHT T T T T T T T T T
11
2
L RH
RL RL
T T
T T
Since 1L
RL
T
T , the only choice is
11
2
L RH
RL RL
T T
T T
11 1
2 1( 1)
2 221 1
RH RH
RL RLH L RL
RH L RL RHRHRH
RLRL
T T
T TT T T
T T T TTT
TT
111 1
111
RH
RLL RL RL
H RH RHRL
RH
T
TT T T
T T TT
T
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21 1 1 1RL L RL RL RH RL
RH RH RH RH RL RH
T T T T T T
T T T T T T
2
12
RH RLH
RH
hA T TW q
T
Work of Carnot cycle
0
0.050.1
0.15
0.2
0.25
1 2 3 4
TL/TRL
wor
k
TRH/TRL=7.0
TRH/TRL=5.0
TRH/TRL=3.0
Fig. 5.1.5: Work of Carnot cycle with limited heat transfer rate
Efficiency of Carnot cycle
0
0.20.4
0.6
0.8
1
0 5 10
Temperature ratio
effi
cien
cy
Carnot cycle
Heat transfer
Fig. 5.1.6: Efficiency of Carnot cycle with finite temperature heat transfer
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Example: A Carnot engine operates between two thermal reservoirs at 1000K and
300 K. Find the efficiency if heat transfer rate is considered.
RHT = 1000 K, RLT = 300 K,
HT = 750 K, LT = 350 K, = 0.452
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Assignment 5.3: A Carnot engine operates between two thermal reservoirs at 3000K
and 300 K. Find the efficiency if heat transfer rate is considered.
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Entropy generation
For ideal Carnot cycle, the process are reversible, no entropy is generated.
However, if limited heat transfer is considered, entropy will be increased.
Entropy generation of Carnot cycle:
( ) 1 ( ) 12H L RH H L RL H L
RH RL RH RL RH RL
q q hA T T hA T T hA T Ts
T T T T T T
21 1
4 4
22 2
RL RH RL RH
RH RL RH RL
hA T T hA T Ts
T T T T
------------------------------------------------------------------------------------------------------
Example: A Carnot engine operates between two thermal reservoirs at 1000K and
300 K. Find the entropy generated if heat transfer rate is considered.
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In summary, the Carnot cycle has the shortcomings of low power, high peak pressure,
and low heat transfer rate. It is no wonder that there is no commercial Carnot engine
even though it has the highest efficiency among all other engines. Besides, the claim
of the highest efficiency seems a specious statement. If the peak pressure is the
limiting constrain, the efficiency of Carnot cycle is no better than that of Otto cycle.
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(5.1.3). Carnot Refrigerator
The Carnot refrigerator is composed of four components, including evaporator,
condenser, compressor, and expander. It’s working principle is just the reverse of
Carnot heat engine.
Fig. 5.1.6: The working principle of Carnot Refrigerator
1 4Lq h h , constant pressure heat addition, evaporation of refrigerant
2 3Hq h h , constant pressure heat removing, condensation of refrigerant
2 1Cw h h , isentropic compression with compressor
3 4Ew h h , isentropic expansion with expander
H H
L L
q T
q T
net C E net H Lw w w q q q
Coefficient of performance : COP
1
1
L L L
HH L H L
L
q q TCOP
Tw q q T T
T
1H
L
T
T , COP
In most applications of refrigerator, H aT T . The lower the value of LT , the more
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difficult to remove heat from thermal reservoir. As a result, the value of COP would
be lower.
2H
L
T
T , 0.5 150L HT T K , 1COP
150LT K , 1COP , Lw q
150LT K , 1COP , Lw q
If L LoadT T , and
H aT T , then
H HH
a H
q qs
T T ， L L
L
Load L
q qs
T T
0H Lnet H L
H L
q qs s s
T T
Carnot refrigeration cycle is an ideal cycle in which net entropy generation is zero.
------------------------------------------------------------------------------------------------------
Example: A Carnot refrigerator uses R134a as the refrigerant. It is known that
TH=26.7℃, and TL=-40℃, find the amount of heat absorption and the COP.
------------------------------------------------------------------------------------------------------
In real applications, L LoadT T , and
H aT T , then
H HH
H a
q qs
T T ， L L
L
L Load
q qs
T T
0H Lnet H L
a Load
q qs s s
T T
------------------------------------------------------------------------------------------------------
Example: Calculate the amount of work to produce one kilogram of ice at -30℃
from water at 25℃ by a Carnot refrigerator using R134a as the refrigerant.
------------------------------------------------------------------------------------------------------
Assignment 5.4: A Carnot engine is used to drive a Carnot refrigerator to make ice.
The engine runs with Pmax = 300 bar, P1 = 1 bar, and TL = 300 K. Find the amount of
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heat to make one kilogram of ice at -30℃ from water at 27℃.
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The reversible work of refrigeration
The minimum work to cool down an object from room temperature to the
desired low temperature in a reversible process is the reversible work of refrigeration.
The net entropy will not increase during this process.
Ta
QH
W
QL
Fig. 5.1.7: The reversible refrigeration process
0Hsys ev sys
a
QS S S S
T
2 1( )H a sys aQ T S T S S
1 2LQ h h
2 1 2 1( )H L aW Q Q h h T S S
1 2
2 1 2 1( )
L
a
Q h hCOP
W h h T S S
------------------------------------------------------------------------------------------------------
Example: Find the reversible work to cool down 1kg of brass block from 25℃ to
-40℃.
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------------------------------------------------------------------------------------------------------
Example: Find the reversible work to make 1kg of ice at -40℃ from water at 25
℃.
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Example: A compression type refrigerator is used to make 1kg of ice at -40℃
from water at 25℃. R134a is used as the refrigerant. The efficiency of compressor is
80%. Find the work of compression and the COP.
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Assignment 5.4: Find the reversible work to liquefy 1kg of nitrogen from 25℃.
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