an introduction to thermodynamics and heat engines...
TRANSCRIPT
An Introduction to
Thermodynamics and Heat engines
HUANG, Yrjö Jun
http://en.wikipedia.org/wiki/Heat_engines
4
1
Cyclic process
A , or a , is a
process for which the initial and end
states are same. The change in the value
of any property for a cyclic process is
zero. A two-process and a four-process
cycles
cyclic process cycle
are in the figure. Hence, the
change in an intensive property for a
cycle is given by,
0
y
dy =∫�
2
Initial: 开始的
Cyclic process cont.
Two primary classes of thermodynamic cycles are and
. Power cycles are cycles which convert some heat
input into a mechanical work output, while heat pump cycles transfer
power cycles
heat pump cycles
heat from low to high temperatures using mechanical work input.
Cycles composed entirely of quasistatic processes can operate as
power or heat pump cycles by controlling the process direction.
On a pressure volume diagram or temperature entropy diagram, the
clockwise and counterclockwise directions indicate power and heat
pump cycles, respectively.
3
Primary: 首要的，主要的 entire 完整的 -static: 静止的
Clockwise: 顺时针 counterclockwise: 逆时针
Heat engine
Because the net variation in state properties during a thermodynamic
cycle is zero, it forms a closed loop on a - diagram. A - diagram's
Y axis shows pressure ( ) and X axis shows volume ( ). The
p v p v
p v area
enclosed by the loop is the work ( ) done by the process:
This work is equal to the balance
of heat ( ) transferred into the system:
If the cyclic process moves clockwise
= =
= = −
∫ ∫� �o
o in out
w
w dw pdv
q
w q q q
around the loop, then will be positive,
and it represents a heat engine. If it moves
counterclockwise, then will be negative,
and it represents a heat pump.
w
w
4
Loop：环线，绳套 enclosed: 封闭 positive: 正方向的 negative: 负的
Represent: 描绘，表示
Heat engine cont.
The engineers frequently measure the performace of devices by
the ratio of the desired result to the required or costly input. The
performace of a heat engine is called , which
is def
thermal efficiency
0
ined as
1
In the - diagram, equals the area
in 5-6-7- - -5, and equals the
area in 7-8-5- - -7. Hence, it is easier
to get the the thermal efficiency from the
-
in out outt
in in in
in
out
w q q q
q q q
T s q
f e q
e f
T s
η −= = = −
diagram more than - diagram.p v
0 = −in out
w q q
5
device: 装置
Heat pump
Thermodynamic heat pump cycles are the models for heat pumps
and refrigerators. The difference is that heat pumps are intended
to keep a place warm while refrigerators are designed to cool it.
It is a reverse process of power cycle, namly,
= −o out inw q q
= −o out in
w q q
6
Refrigerator: 冰箱
Heat pump cont.
Like the thermal efficiency for heat engines,
( ) is defined as the desired result divided by the costly or required
input. For refrigeration processes, the desired result
coefficient of performace
COP
0
is the heat transfer
to the cycle from a low temprature region and the requied input is the net
input, and, Therefore, for any refrigeration process,
COP =
The desired result for a he
in inR
out in
q q
w q q=
−at pump is the heat transfer from the cycle to
a high temprature region and the requied input is the net input. Hence, for
any heat pump, COP ( ) is given as,
COP (EER)=
HP
HP
q
Energy efficiency ratio, EER
0
out out
out in
q
w q q=
−
7
Refrigeration：冷冻
Thermal-energy reservoirs
A thermal reservoir, a short-form of thermal energy reservoir,
is a thermodynamic system with a heat capacity that is large
enough that when it is in thermal contact with another system of
interest or its environment, its temperature remains effectively
constant. The temperature of the reservoir does not change，
irrespective of whether heat is added or extracted. As it can act as
a source and sink of heat it is often also referred to a heat
reservoir.
8
Reservoir: 蓄水池，库 irrespective: 不受影响的 extract: 抽取 sink：汇
Statements of the Second Law
Rudolf Clausius
William Thomson,
1st Baron Kelvin
In 1850, Clausius gave the first statement of the second laws of thermodynamics.
Thomson gives an alternative statement of the second law.
No process is possible in which the sole result is the
absorption of heat from a reservoir and its complete
conversion into work.
No process is possible whose sole result is the transfer
of heat from a body of lower temperature to a body of
higher temperature.
0 0
COP = , COP =in outR HP
q q
w w< ∞ < ∞
1 1out
t
in
q
qη = − <
9
Sole:唯一的 alternative: 等价的
The Carnot cycle is a theoretical thermodynamic cycle proposed byNicolas Léonard Sadi Carnot in 1823 and expanded by Benoit Paul ÉmileClapeyron in the 1830s and 40s. It is the most efficient cycle forconverting a given amount of thermal energy into work, or conversely,creating a temperature difference by doing a given amount of work.
Carnot cycle
2
1 Carnot
Thermal efficiency, 1 = 1in out outt
in in
q q q T
q q Tη
−= = − −
outq
inq
10
Theoretical: 理论的 converse: 逆向的
Carnot cycle cont.
The Carnot cycle acting as a heat engine contain of the following steps:
1. at the "hot" temperature (A to B) . The gas
expansion is propelled by absorption of heat energy andinq
Isothermal expansion
1
of entropy
/ from the high temperature reservoir.
2. (B to C). The gas neither gain nor
lose heat. The gas does work on the surroundings, and losing an equiva
inds q T=
Isentropic (adiabatic) expansion
out
2
lent
amount of internal energy.
3. at the "cold" temperature, (C to D). The
surroundings do work on the gas, causing an amount of heat energy
and of entropy / to out
q
ds q T=
Isothermal compression
flow out of the gas to the low temperature
reservoir.
4. (D to A). The surroundings do work on the
gas, increasing its internal energy and compressing it, causing the
temperatur
Isentropic compression
1e to rise to . T11
Propel: 推动
Reversed Carnot Cycle
inq
outq
A schematic of reversed Carnot Cycle , operating as a heat pump or
a refreigerator, as below:
2
0 1 2 Carnot Cyc.
1
0 1 2 Carnot Cyc.
COP =
COP =
in inR
out in
out outHP
out in
q q T
w q q T T
q q T
w q q T T
= = − −
= = − −
12
schematic 图示
If a cycle includes two isothermal processes, it may have the same heat
efficiency as the Carnot cycle. The cycles with the same efficiency as
the Carnot cycle is called . Thegenerality Carnot Cycle ideal Stirling
cycle is an example, which is represented in the figure and consists of
four processes which combine to form a closed cycle: two isothermal
and two isochoric processes. The processes are shown on both a -
diagram and a - diagram.
p v
T s
Generality Carnot Cycle
13
Generality Carnot Cycle
The Ericsson cycle is another example, named after inventor John Ericsson, and it consists of four thermodynamic processes as:• Process 1-2: Isobaric heat addition. • Process 2-3: Isothermal expansion.• Process 3-4: Isobaric heat removal. • Process 4-1: Isothermal compression.
14
1st Carnot’s Principle
The efficiency of a reversible cyclewill always be greater than theefficiency of an irreversible cycleoperating between the same twothermal reservoirs.
Consider the two heat engines (HE's) shown here. Both HE's receive the same amount of heat, Q
H, from the hot reservoir.
Let's assume that the 1st Carnot Principle is false and see if that violates the 2nd Law?
The 2nd Law is violated. then we will know for sure that the 1st Carnot Principle is true.
http://www.learnthermo.com
15
If ,
namely / /
then
irr rev
irr H rev H
irr rev
W Q W Q
W W
η η>>
>
violate：违背
2nd Carnot’s Principle
Let's assume that the HE2is more
efficient than HE1. Just as:
Because it is reversible, we can
reverse HE1. It becomes a
reversible heat pump, HP1.
Consider the system within the
dashed line that includes the HE,
the HP and the hot reservoir.
This system has an efficiency of
100% and that is a violation of
the Kelvin Statement of the 2nd
Law. Therefore, the 2nd Carnot
Principle is true.
2 1 1 2c cW W Q Q> >
No engine can be more efficient than areversible engine operating between thesame temperature limits, and all reversibleengines operating between the sametemperature limits have the sameefficiency.
16
efficient:高效的 operate: 运作 efficiency: 效率
dashed line: 虚线
Entropy in Cycle
0
0
in a Carnoit cycle,
1
in a reverse Carnoit cycle,
COP =
1
OP
1
C =
η − = −
=− −
=
−= = − =
= =−
= =− −−
in out out
t
in in
in in
R
out in
out o
L L
H H
L L
H L H L
H Hut
HP
out in H L H L
q q q
q q
q q
w q q
q q
w
q T
q T
q T
q q T T
q T
q q T Tq q
Thermal efficiency
COP
⇒ =H H
L L
q T
q T
Considering the direction of the heat flow, 0 or 0 in
a Carnot cycle. The difinition of entropy, = , Hence, 0.
In other words, the entropy doesn't change in a Carnot cycle.
H L
H L
q q q
T T T
dqds s ds
T
− = =
∆ = =
∑
∑
17
Entropy in Cycle (2)
For any reversible cycle as shown
in the figure, a group isentropic
curves can divided the cycle into a
series of elements, such as - - - -
and - - - - in the figure. Each
element is a generalit
a b f g a
b c e f b
2
1
1 2
1 2
y Carnot cycle
and the increase in entropy is zero in
every element. For the element
- - - - ,
1-
and 0
t
a b f g a
T
T
dq dq
T T
η =
+ =
• Process a-b: any process of expansion.
• Process b-f: Isothermal expansion.
• Process f-g: any process of compression.
• Process g-a: Isothermal compression.
18
T1
T2
dq1dq
2
Clausius Equality
1 2
1 21 2 2 1
1 2 2 1
For all elements,
0
or be written as,
0
Hence,
0
− − − −
− − − −
+ =
+ =
= = ∆ =
∑ ∑
∫ ∫
∫ ∫� �
a f
a f
dq dq
T T
dq dq
T T
dqds s
T
Conclusion: The increase of entropy in a reversible cycle is zero.
0 is given by Clausius firstly and it is named as
or
=∫�dq
T Clausius integration
Clausius equality.
19
T1
T2
dq1dq
2
integration: 积分
1 21 2
1 2
1 2
1 21 2 2 1 1 2 2 1
If considerding the heat flows and are vectors, 0.
For all elements, 0 or 0
Hence, 0
This inequality is called
− − − − − − − −
+ <
+ < + <
<
∑ ∑ ∫ ∫
∫
a f a f
irr
dq dqdq dq
T T
dq dq dq dq
T T T T
dq
T
Clausius inequality.
Clausius Inequality
1 2 1 2
1 1
1 2
1 2
If one of the cyclic element (e,g, - - - - )
is irreversible, Carnot's first priciple tell us
, we can obtain
or
η η
η η
<− −
= < =
<
irr rev
irr rev
b f g a b
dq dq T T
dq T
dq dq
T T
20
T1
T2
dq1dq
2
inequality: 不等式 vector: 矢量
Clausius Inequality
2 1 1 2
1 2 2 1
0
or
If one cycle includes two parts: 1- -2 is irrever-
siable and 2- -1 is reversible. The whole cycle
is irreversible,
− − − −
− − − −
+ < =
< −
∫ ∫
∫ ∫
∫�B A
irrA B
irr
dq
T
dq
T
A
B
dq dq
T T
dq
T
2 1 1 2
2
2 11
1 2 2 1
Because 2- -1 is a reversible process
Entropy is a state function and both 1 and 2 are quasi-stable states,
-
− − − −
− − − −
= −
= = = −
∫ ∫
∫ ∫ ∫
B B
B B
dq dqB
T T
dq dq dqs s
T T T21
Clausius Inequality
1 2
1 2
1 2
1 2
2 1
2 1
2 1
2 1
2 1
2 1
--
-
If 1- -2 is a reversable process, -
− −
− −− −
− −
− −
− −
< −
⇒ < ≤ = − =
∫
∫ ∫
∫
∫
∫
irrA
irrA
A
revA
B
B
dq
T
dq dq
T T
dqs s
s sTdq
s sT
dqA s s
T
This is the
mathematical
statement for the
2nd law.
1 2
2 1The differential of - is
Entropy is a state function, 0 and therefore 0
− −
≤ ≤
= ≤
∫
∫ ∫� �
A
dq
T
dqs s ds
T
dqds
T
22
Principle of increase of entropy
isolated sys.
Conclustion in last page: For an irreversible process,
For an isolated system, no mass and energy transfer between the system and
surroundings, 0. Hence,
dS 0
dqds
T
dq
≤
=≥
The entropy of an isolated system always increases due to internal
irreversibility. In the limiting case of an internally reversible process, the
entropy will remain a constant. An isolated process in which dS isolated sys.< 0
is physically impossible.
23
Principle of increase of entropy (2)
substance
hot res. cold res substance
substance
Here is an isolated system with a hot reservoir,
a cold reservoir and subtance, thus,
.
For a reversible cycle, 0. The
hot reservoir releases heat flow
= ∆ + ∆ + ∆∆ =
dS S S S
S
hot res.
and the
entropy decreases as,
∆ = −
H
H
H
Q
QS
T
cold res.
hot res. cold res substance
hot
By the same way, . As proofed, , ( are scalars), thus
. 0
For an irreversible, , leads to , thus
L L H
L L H
rev
H L H L H Lirr rev
H H H L
Q Q QS Q
T T T
dS S S S
Q Q T T Q Q
Q T T T
dS S
η η
∆ = =
= ∆ + ∆ + ∆ =
− −= < = <
= ∆ res. cold res substance. >0irrS S+ ∆ + ∆24
Principle of increase of entropy (3)
dQA B
Heat flow
Two objects with temperature and . If = , and considering the
heat flow from A to B, the change of entropy in each of them,
and
For the whole system, 0.
If
= − =
= + =
A B A B
A B
A B
A B
T T T T
dQ
dQ dQdS dS
T T
dS dS dS
> , the heat transfer is an irriversible process, and ,hence
0
>
= + = − + >
A B
B A
irr A B
A B
dQ dQT T
T T
dQ dQdS dS dS
T T
Isolated
system
25
Principle of increase of entropy (4)
Friction is associated with bodies in motion that are in contact with each other.
The bodies could be:
1. Two solids( blocks or planes)
2. A solid and a fluid (a car moving through air)
3. Two fluids at different velocities (wind blowing across the surface of a lake)
Friction irreversibly dissipates work into internal energy. The inscrease of
entropy object which accepts the internal energy: =
W
dWdS
T> 0.
Conclusion: In an isolated system, the entropy of the whole system always
increases, if any irreversible process (e.g. irreversible heat transfer,
mechanical energy dissipation) happens in the system.
26
dissipate 耗散
Questions
The following figure shows a heat engine and a heat pump. The heat engine
works between two thermal reservoirs of TE=1200K, and T0=300K. Total heat
added to heat engine cycle QE=100kJ and the thermal efficiency is 0.6. The work
output W from the heat engine is transferred into the heat pump cycle and raise
the surrounding temperature to 400K. If COPR of the heat pump is 2.0, find
(a) QP in kW.
(b) If both engines are Carnot engines, what is QP and COPR?
(c) If QP > QH, if this violates the second law? Why?
HE
η=0.6
QE=100kJ
TE=1200K
T0=300K
WHP
COP=2.0
QP=?
TP=400K
27
28
Entropy Equation
Control
volume
(CV)
in in inS m s=ɺ ɺ
out out outS m s=ɺ ɺ
0
CVq
QS
T=
ɺɺ
gSɺ
0
For adiabatic process, 0,
0
a) If the process is reversiable, 0
b) For steady problem, namely ,
0
c) For steady a
CV in g q out
q
CV in g out
CV
g out in
in out
g out in
S S S S S
S
S S S S
S
S S S
m m
s s s
= + + − ≥
=
= + − ≥
=
= −
== − ≥
ɺ ɺ ɺ ɺ ɺ
ɺ
ɺ ɺ ɺ ɺ
ɺ
ɺ ɺ ɺ
ɺ ɺ
nd reversiable process:
0g out ins s s= − =
, ,
in in inh c Z
out
out
out
h
c
Z
CVWɺ
2 2
0 0
0
Energy equation: ( ) ( - ) 02
Entropy equation: ( ) 0
in outCV CV in out in out
CVin g out CV in out g
c cQ W m h h g Z Z
Qms S ms q T s s T s
T
−− + − + + =
+ + ≤ ⇒ + − + ≤
ɺ ɺ ɺ
ɺɺɺ ɺ
29
0
2 2
0
0
2
0
0
2
( ) ( - )2
( ) ( - )
0
( ) 0
( ) 0
2( )
CV in ou
in outCV C
CVin out
V in out in out
in outCV CV in out in ou
t g
CV in out
g
t g
Q mT
c cQ W m h h g Z Z
c
Qm s s T S
Q mT s s T Sc
Q W m h h g Z Z
s s ST
=⇒
+ − + ≤
−− + − + +
−− + − + +
+ − + ≤
+
⇒
−
− +
ɺ ɺ ɺ
ɺ
ɺ ɺɺ
ɺ ɺ ɺ ɺ
ɺ
ɺ
ɺ
ɺ
0 0
0
2 2
2 2
0
( ) ( - )2
( ) ( - )
0
( )
( )2
When the left side equals the right side, the process is rev
in outCV in out in out
CV in outin ou
in out
t
g
in outut gin o
c cW m h h g Z Z
W u uh h
mT s s T S
T sg Z Z Tm
s s
− +
−≤ − + +
−≤
− −−
−
+ + −
≥
ɺ ɺ ɺɺ
ɺ
ɺ
ersiable.
Energy-Entropy Balance
30
Energy-Entropy Balance
CV
1
1
air 1bar
1 /
300
c m s
T K
==
（ ）
2
2
1 /
250
c m s
T K
==0 200 1barT K= ，
1 1
2 1
From the table, we know that,
300.2kJ/kg, 1.70kJ/kg K
250.1kJ/kg, 1.52kJ/kg K
h s
h s
= = ⋅= = ⋅
2 2
:
( ) ( - ) 02
and simplited as: ( ) 0, Subsitute the values:
(300.2 250.1) 0 50.1 /
Energy equation
in outCV CV in out in out
CV in out
CV CV
c cQ W m h h g Z Z
q h h
q q kJ kg
−− + − + + =
+ − =+ − = ⇒ = −
ɺ ɺ ɺ
0 0
0 0
0 0
: ( ) ,where 0.
ubstitute the values: -50.1 / 200 (1.70-1.52) 0
Hence, this process is possible.
If T =300K, ( ) 50.1 300(1.70 1.52) 3.9
If T =250K, (
Entropy CV in out g g
CV in out
CV i
q T s s T s s
S KJ Kg K
q T s s
q T s
+ − + =
+ <
+ − = − + − =+
0 0
) 50.1 250(1.70 1.52) 5.1
If T =277K, ( ) 50.1 277(1.70 1.52) 0.
n out
CV in out
s
q T s s
− = − + − = −+ − = − + − =
Thanks for your attention!
31