Download Chapter Two The Thermodynamic Laws

Chapter Two
The Thermodynamic Laws
Update on 2014/9/17
(2.1)、The zeroth law of thermodynamics
The zeroth law of thermodynamics is to define the relationship between two
systems in thermal equilibrium. A system in thermal equilibrium is a system whose
properties are invariant with time.
When two bodies are in thermal equilibrium with a third body, they are in
thermal equilibrium with one another.
Thus, thermal equilibrium is a relation between thermodynamic systems.
Mathematically, the zeroth law expresses that this relation is an equivalence relation.
(A) = (C) = (B)
(B) Fig. 2.1.1 The systems in equilibrium
When two bodies are in thermal equilibrium, they are of the same temperature.
If TA  TB , and TB  TC , then TA  TC .
History
The term zeroth law was coined by Ralph H. Fowler(1889 – 1944), who was a
British physicist and astronomer. In many ways, the law is more fundamental than
any of the others. However, the need to state it explicitly as a law was perceived
until the first third of the 20th century, long after the first three laws were already
widely in use and named as such, hence the zero numbering.
Advanced Thermodynamics, ME dept. NCHU 頁 1 (2.2)、The first law of thermodynamics
The first law of thermodynamics is an expression of the universal law of
conservation of energy, and identifies heat transfer as a form of energy transfer.
The increase in the internal energy of a thermodynamic system is equal to the amount
of heat transfer added to the system minus the work done by the system on the
surroundings.
Q  U  W
History
The first explicit statement of the first law of thermodynamics was given by Rudolf
Clausius, who was a German physicist and mathematician, in 1850. The original
statement of the first law was "There is a state function E, called ‘energy’, whose
differential equals the work exchanged with the surroundings during an adiabatic
process."
(2.2.1). Isolated system
If a system exchanges neither mass nor energy with its environment, its energy
is invariant with time.
U  Const.
If the system is composed of several parts with different temperature, the
conservation of internal energy is
U  m Au A  mBu B  Const.
-----------------------------------------------------------------------------------------------------Example
An insulated rigid chamber is divided into two equal parts with a diaphragm. The
diaphragm neither moves nor conducts heat. The left part is filled with air at 3 bars
and 25℃, and the right part is kept at vacuum. If the diaphragm ruptures and the
whole chamber is filled with air, find the final temperature of the air in the chamber.
Air
3 bars
25℃
Vacuum
-----------------------------------------------------------------------------------------------------Advanced Thermodynamics, ME dept. NCHU 頁 2 -----------------------------------------------------------------------------------------------------Example
An insulated rigid chamber is divided into two parts with a diaphragm. The
diaphragm is heat conductive and can move without friction. The left part is filled
with air of 1 kg at 3 bars and 25℃, and the right part is filled with air at 300℃ and
2.5 bars with a volume of 400 liters. Now let the diaphragm move freely until the
pressure and the temperature on both sides are equal. Find the final temperature of
the air in the chamber.
1kg
3 bars
25℃
400 L
2.5 bars
300℃
-----------------------------------------------------------------------------------------------------Assignment 2.1
An insulated rigid chamber is divided into two equal parts with a diaphragm. The
left part is filled with air at 25℃ and 3 bars, and the right part is filled with air at
50℃ and 1 bar. The diaphragm ruptures and the air mixes together. Find the final
temperature of the air in the chamber.
Air
3 bars
25℃
Air
1 bar
50℃
-----------------------------------------------------------------------------------------------------(2.2.2). Closed system
If a system exchanges energy with its environment, the increase in the internal
energy of the system equals to the amount of heat transfer added to the system minus
the work done by the system on the surroundings.
Q  U  W
-----------------------------------------------------------------------------------------------------Example
One kg of air at 50 bars and 500K expands isothermally to the pressure of 10 bars,
Advanced Thermodynamics, ME dept. NCHU 頁 3 and then continues to expand to the pressure of 5 bars in a polytropic process with
the exponent of 1.6. Assuming that air is an ideal gas with constant heat capacity,
determine the work and heat transfer during these processes.
P
V
-----------------------------------------------------------------------------------------------------Example
A rigid chamber with a volume of 10 liters is divided into two parts of the same
volume with a diaphragm. The diaphragm does not conduct heat, but can move
without friction. The left part is filled with CO2 and the right part is filled with N2.
Both sides are at 300 K and 100 kPa. The right side is heated gradually until N2
expands to a volume of 7 liters while the left side is insulated during this process.
Assuming both N2 and CO2 are ideal gases with constant specific heats, find the heat
transfer during this process.
CO2
100 kPa
300K
CO2
3L
N2
100 kPa
300K
N2
7L
-----------------------------------------------------------------------------------------------------Assignment 2.2
One kg of air at 1 bar and 300K is compressed isothermally to the pressure of 10 bars,
and then expands adiabatically to the pressure of 1 bar. Assuming that air is an
ideal gas with constant heat capacity, determine the work and heat transfer during
these processes.
Advanced Thermodynamics, ME dept. NCHU 頁 4 -----------------------------------------------------------------------------------------------------(2.2.3). Open system
An open system allows both mass and energy flow through it. As a result, the
mass and energy may vary with time.
dm
  m i   m e
dt
Vi 2
dE
Ve2

Q   m i ( hi 
 gzi ) 
  m e ( he 
 gze )  W
2
2
dt
(2.2.3.1). Steady state steady flow system
dm
0
dt
m i  m e  m
V2
V2
Q   m i ( hi  i  gzi )   m e ( he  e  gze )  W
2
2
2
2
V
V
q  hi  i  gzi  he  e  gze  w
2
2
q  hi  he  w
-----------------------------------------------------------------------------------------------------Example
The pressure of saturated liquid R134a at 10 bars declines to 1 bar as passing through
a throttle valve. If the variation in kinetic energy is neglected, find the quality of
R134a.
-----------------------------------------------------------------------------------------------------Assignment 2.3
Air at 25 ℃ and 200 kPa flows through an insulated throttle valve and drops to 100
kPa. Calculate the final temperature.
-----------------------------------------------------------------------------------------------------Advanced Thermodynamics, ME dept. NCHU 頁 5 (2.2.3.2). Uniform state steady flow system
m2  m1  mi  me
mi   m i dt , me   m e dt
Q  mi ( hi 
Vi 2
V2
 gzi )  E2  E1  me ( he  e  gze )  W
2
2
-----------------------------------------------------------------------------------------------------Example
A bottle containing 100 liters air at 5 bars and 25℃ is filled with compressed air at
25 ℃ until a pressure of 50 bars is reached. Suppose the process is adiabatic and air
is an ideal gas with constant specific heat, find the amount of air injected into the
bottle.
m2  m1  mi
mi hi  m2u2  m1u1
( m2  m1 ) hi  m2u2  m1u1
PV PV
PV
PV
( 2  1 )c pTi  2 cvT2  1 cvT1
RT2 RT1
RT2
RT1
P P
P
P
( 2  1 ) kTi  2 T2  1 T1
T2 T1
T2
T1
50
5
( 
)1.4  298  50  5
T2 298
T2 = 401 K
PV
m2  2 =4.343 kg
RT2
PV
m1  1 =0.585 kg
RT1
m  m2  m1 = 3.758 kg
-----------------------------------------------------------------------------------------------------Example
One kg of water at 25℃ is contained in an insulated bottle. A vacuum pump is used
to suck the vapor out of the bottle. Determine how much vapor will be sucked out
when the bottle temperature reached 0℃.
Advanced Thermodynamics, ME dept. NCHU 頁 6 -----------------------------------------------------------------------------------------------------Assignment 2.4
10 kg of water at 25 ℃ is contained in a vessel with the volume of 100L. A valve is
installed on top of the vessel. This valve is preset at the pressure of 500 kPa. This
vessel is heated with a stove. When the pressure of steam reaches 500 kPa, the
valve will open to keep the pressure inside vessel constant. Calculate the amount of
heat transfer when 5 kg of water remains inside this vessel.
-----------------------------------------------------------------------------------------------------(2.2.3.3). Uniform state uniform flow
Flow out of a system adiabatically through a small hole:
dm
  m e
dt
d ( mu )
du
dm
 m e he  m
u
 m e he  0
dt
dt
dt
Advanced Thermodynamics, ME dept. NCHU 頁 7 dT
dm
dm
u
h
0
dt
dt
dt
dT
dm
mcv
 RT
dt
dt
1 dT 1 R dm

T dt m cv dt
mcv
R
cv
 m2 
T2  m2 

 

T1  m1 
 m1 
k 1
k
P2  m2 
 
P1  m1 
-----------------------------------------------------------------------------------------------------Example
Air at 5 bars and 25℃ is contained in a bottle of 100 liters. Open the valve and let
air flow out until pressure reaches 1 bar. Calculate the amount of air remains in the
bottle.
-----------------------------------------------------------------------------------------------------Example
A rigid chamber with a volume of 10 liters is divided into two parts of the same
volume with a diaphragm. The left part is filled with air at 25℃ and 3 bars, and the
right part is in vacuum. A small hole is drilled on the diaphragm to let air leak into
the right part until pressures on both sides are equal. Find the final temperature of
air in the right part.
Air
25℃
3 bars
Vacuum
-----------------------------------------------------------------------------------------------------Assignment 2.5
A bottle containing 100 liters air at 1 bar and 25℃ is filled adiabatically with
compressed air at 25 ℃ until a pressure of 50 bars is reached. The air in bottle is
then cooled to 25℃ without leakage. The valve is open accidently until pressure
reaches 10 bars before it is closed again. If the bottle should be filled with air at 50
bars and 25℃, what the final pressure should be if compressed air at 25℃is used
Advanced Thermodynamics, ME dept. NCHU 頁 8 again to refill the bottle?
------------------------------------------------------------------------------------------------------
The hot water system
T  T (t ) ， m  m(t )
dm
 m i  m e
dt
dE
dT
dm
 mcv
u
 m i hi  m e he  Q  S  W
dt
dt
dt
m i
Ti
Ta
S
m e
Te
(1). Mixing process
Q  0 ， S  0
dm
 0 ， m i  m e  m
dt
hi  cTi ， he  cTe
mcv
dT
 (Ti  Te )
 m i hi  m e he  mc
dt
Te  T
Advanced Thermodynamics, ME dept. NCHU 頁 9 Let   T  Ti
d m
  0
dt m
Let
1
m
m
, where  m is the time constant of mixing process.
m

d 1
  0
dt  m
  c1e

t
m
At t  0 , T  T0 ,   T0  Ti   0
   0e

t
m
As t   ,   0 , T  Ti
-----------------------------------------------------------------------------------------------------Example
A tank contains 1000 kg of cold water at 30℃. Hot water at 80℃ is pouring into
this tank at the rate of 1 kg/sec. At the same time, mixed water is flowing out of the
tank at the same rate. Find the temperature of water inside tank 10 minutes after
this mixing process.
-----------------------------------------------------------------------------------------------------Assignment 2.6
A tank contains 200L of hot water at 60℃. Hot water is consumed at the rate of 6
LPM while the same rate of cold water at 25℃ is supplemented into this tank to keep
the volume of water constant. An electric heater is actuated automatically when the
water temperature drops to 45℃. If heat loss is not considered, find out when the
heater will be turned on.
-----------------------------------------------------------------------------------------------------(2). Accumulation process
Q  0 ， S  0
dm
 m i  m e  m
dt
Advanced Thermodynamics, ME dept. NCHU 頁 10 m i  m e  m
m  m0  m  t
 ， dt 
dm  mdt
mc
dm
m
dT
 m i c(Ti  T )   m i c
dt
d
 m i

dt m0  m  t
d
i

m i
dm
m dm
 i
m0  m  t m
m m
 m
 
 0  m0 


m i
m
m 
 0 
m
m i
m
m i
, the ratio of inlet flow to net flow
m
m i  m e , m  0 ,   0 ,  is decreasing.
m i  m e , m  0 ,   0 ，  is also decreasing.
  m0 

 0  m 

-----------------------------------------------------------------------------------------------------Example
A tank contains 1000 kg of cold water at 30℃. Hot water at 80℃ is pouring into
this tank at the rate of 1 kg/sec. At the same time, mixed water is flowing out of the
tank at the rate of 0.5 kg/sec. Find the temperature of water inside tank 10 minutes
after this mixing process.
-----------------------------------------------------------------------------------------------------Assignment 2.7
A tank contains 1000 kg of cold water at 30℃. Hot water at 80℃ is pouring into
this tank at the rate of 1 kg/sec. At the same time, mixed water is flowing out of the
tank at the rate of 1.5 kg/sec. Find the temperature of water inside tank 10 minutes
after this mixing process.
Advanced Thermodynamics, ME dept. NCHU 頁 11 -----------------------------------------------------------------------------------------------------(3). Heating process
dE
dT
dm
 mcv
u
 m i hi  m e he  S
dt
dt
dt
dT m
S
 (Ti  T ) 
dt m
mc
d 
S


dt  m mc
  c1e

1
m
Let  f 
t

S

mc
S
, the equilibrium temperature of heating process

mc
t  0 ，    0 ，則 c1   0   f
t

  f

e
0   f
m
As t   ,    f , the system has reached the final equilibrium temperature.
-----------------------------------------------------------------------------------------------------Example
A tank contains 1000 kg of warm water at 45℃. Water is consumed at the rate of 1
kg/sec while cold water at 25℃ is supplemented into the system at the same rate.
An electric heater is installed inside the tank with the capacity of 25 kW. The
heater has to be turned off as the water temperature has reached 45℃. Find out
when the heater will be turned off.
------------------------------------------------------------------------------------------------------
Advanced Thermodynamics, ME dept. NCHU 頁 12 (2.3)、The second law of thermodynamics
(2.3.1). History
The first theory on the conversion of heat into mechanical work is due to Nicolas
Léonard Sadi Carnot in 1824.
Rudolf Clausius was the first to formulate the second law in 1850.
Established in the 19th century, the Kelvin-Planck statement of the Second Law says,
"It is impossible for any device that operates on a cycle to receive heat from a single
reservoir and produce a net amount of work." This was shown to be equivalent to the
statement of Clausius.
(2.3.2). Statements of the second law
(2.3.2.1). Thermal reservoir
Thermal reservoir, characterized by its temperature, is a reservoir of infinite heat
capacity. Thermal reservoir can play the roles of either heat sink or heat source.
No matter how much heat is delivered, temperature of the reservoir will never
change.
There are many statements of the second law which use different terms, but are all
equivalent.
(2.3.2.2). Kelvin-Plank Statement
It is impossible for any system to operate in a thermodynamic cycle and deliver a net
amount of work to its surroundings while receiving energy by heat transfer from a
single thermal reservoir.
An equivalent statement by Lord Kelvin is:
"A transformation whose only final result is to convert heat, extracted from a source
at constant temperature, into work, is impossible."
This statement implies an inequality of conversion between heat and work. Work
can be totally converted to heat. However, heat can only be partially converted to
work.
Advanced Thermodynamics, ME dept. NCHU 頁 13 (2.3.2.3). Clausius Statement
It is impossible for any system to operate in such a way that the sole result would be
an energy transfer by heat from a cooler to a hotter body.
Another statement by Clausius is:
"Heat cannot of itself pass from a colder to a hotter body."
This statement implies an inequality of the heat transfer between a hot body and a
cold body. Heat transfer from a hot body to a cold body can spontaneously occur.
However, heat transfer in the reversed direction can not happen without the
intervention of work.
The most common enunciation of second law of thermodynamics is essentially due
to Rudolf Clausius: The entropy of an isolated system not in equilibrium will tend to
increase over time, approaching a maximum value at equilibrium.
The second law holds in a statistical sense. That is, the second law will hold on
average, with a statistical variation on the order of 1/√N where N is the number of
particles in the system. For everyday (macroscopic) situations, the probability that
the second law will be violated is practically nil. However, for systems with a small
number of particles, thermodynamic parameters, including the entropy, may show
significant statistical deviations from that predicted by the second law. Classical
thermodynamic theory does not deal with these statistical variations.
Consequences of the Kelvin-Plank Statement are as the following.
For all the cycles working on the same reservoirs, the reversible cycle is of the
highest thermal efficiency.
For all the reversible cycles working on the same reservoirs, they are of the same
thermal efficiency.
Heat engine is a machine used to convert heat into work. The thermal efficiency of a
heat engine is defined as
W
Q

1 L
QH
QH
Refrigerator is a machine used to transfer heat from a cold body to a hot body. The
performance of a refrigerator is defined as
Advanced Thermodynamics, ME dept. NCHU 頁 14 
QL
QL

W QH  QH
(2.3.2.4). Caratheodory’s Two Axioms
Axiom I: The work is the same in all adiabatic processes that take a system from a
given initial state to a given final state.
Q  U  W
In an adiabatic process, the heat transfer is zero, Q  0 , which would produce the
result that W  U .
The Caratheodory’s first axiom is equivalent to the conventional statement of the
First Law of Thermodynamics.
The heat transfer interaction is defined as the difference between the actual work
transfer and the adiabatic work transfer associated with the given end states.
Q  W  Wadb
Axiom II: In the immediate neighborhood of every state of a system, there are other
states that can not be reached from the first by an adiabatic process.
In the PV diagram, assume that state A(P1, V1) can be reached from state B (P2,
V2) with an adiabatic process. There exists another state C(P3, V2) with the same
volume as sate B, but the values of pressure are different. Suppose that state C can
also be reached from state A with an adiabatic process. It is noted that the three
processes of AB, BC, and CA compose a cycle. Since AB and CA are adiabatic,
there is no heat transfer along these two processes. Heat transfer must occur during
the process BC. However, process BC does not output work since the volume
keeps the same during the process. As a result, the cycle absorb heat during the
process BC and then converts heat totally to work during the process AB and CA.
P
A (P1, V1)
C (P3, V2)
B (P2, V2)
V
Advanced Thermodynamics, ME dept. NCHU 頁 15 If the state of C exists, then the Kelvin Plank statement of the second law of
thermodynamics is violated because the system absorbs heat with single reservoir
and converts this heat to work. That is, if axiom II of Caratheodory can be violated,
then the Kelvin Plank statement of the second law of thermodynamics can also be
violated.
In order not to violate the second law of thermodynamics, the state of C must not
exist. As a result, in the immediate neighborhood of state A, state C can not be
reached from A by an adiabatic process.
(2.3.3). Entropy Rate Balance of Isolated system
Principle of increase of entropy
S net  S sys  Senv  0
Reversible process: dS 
Q
Irreversible process: dS 
For all processes: dS 
T
Q
Q
T
rev
T
, TdS   Q
In an isolated system, the entropy always increases all the time.
TdS   Q  0 ， S  0
dS
 
dt
Thermal equilibrium
Two bodies at different temperatures reach thermal equilibrium by contacting each
other for a long period.
U1  m Au A1  mBu B1  U 2  m Au A 2  mBu B 2
m AcvATA1  mB cvBTB1  m AcvAT2  mB cvBT2
Advanced Thermodynamics, ME dept. NCHU 頁 16 T2 
m AcvATA1  mB cvBTB1
m AcvA  mB cvB
S  mAcvA ln
T2
T
T
T
 mB cvB ln 2  ln( 2 ) mAcvA ( 2 ) mBcvB
TA1
TB1
TA1
TB1
In the case that m AcvA  mB cvB
T2 
TA  TB
2
T2 T2 (TA  TB ) 2

1
TA TB
4TATB
S  0
-----------------------------------------------------------------------------------------------------Example
An iron block of 10 kg at 300 ℃ is immersed into a basin of water at 25 ℃. The
volume of water is 100 liters. Calculate the final temperature as well as the entropy
change.
mAcvA  10 × 0.447 = 4.47 kJ
mB cvB  100 × 4.186 = 418.6 kJ
Tav = 27.9 ℃
T
T
S  mAcvA ln 2  mB cvB ln 2 = -2.8792 + 4.05393 = 1.1747 kJ/K
TA1
TB1
-----------------------------------------------------------------------------------------------------Assignment 2.8
An iron block of 10 kg at 300 ℃ is cooled in an open air at 25 ℃. Calculate the
entropy change.
-----------------------------------------------------------------------------------------------------(2.3.4). Entropy Rate Balance of Closed system
In a closed system, the increase of entropy can be attributed to the external
Advanced Thermodynamics, ME dept. NCHU 頁 17 irreversibility and the internal irreversibility.
S  
Q
T
， S  
Q
T

Q j
dS


dt
j Tj
S net  S sys  Sev  Q  (
1 1
 ) 
T T0
 dS 
 1 1

   Q j  (  )   0
T T0
 dt  net
j
(2.3.4.1). Adiabatic process
In an adiabatic process, we have Q j  0 . As a result, the entropy increase of
 dS 
the system is      0 .
 dt  net
Q  U  W  0
U  W
-----------------------------------------------------------------------------------------------------Example
An insulated chamber with a volume of 100 liters is filled with air at 100 kPa and
25℃. A peddle rotates inside the chamber, doing work on the air until its
temperature has been raised to 50℃. Suppose air is an ideal gas with constant
specific heat, find the amount of work being done and the net entropy change of the
system. The atmospheric temperature is 25 ℃.
Advanced Thermodynamics, ME dept. NCHU 頁 18 PV = 0.1169 kg
RT
Q  U 2  U1  W  0
W  U1  U 2  mcv (T1  T2 ) = 2.095 kJ
T
V
-3
S  mcv ln 2  mRT ln 2 = 6.757×10 kJ/K
T1
V1
m
-----------------------------------------------------------------------------------------------------Example
A chamber with a volume of 100 liters is filled with air at 100 kPa and 25℃. Heat
is added to the chamber to raise the temperature of air to 50℃. Find the amount of
heat transfer and the net entropy change of the system. The atmospheric temperature
is 25 ℃.
TH
Discussion: Why is that the entropy generation by work is higher than that by heat
transfer?
-----------------------------------------------------------------------------------------------------Assignment 2.9: An insulated chamber with a volume of 0.1 m3 is filled with air at
100 kPa and 298 K. A peddle rotates inside the chamber, doing work on the air.
If the amount of work being done is 1 kJ, calculate the net entropy change of the
system.
------------------------------------------------------------------------------------------------------
Advanced Thermodynamics, ME dept. NCHU 頁 19 (2.3.4.2). Polytropic process
PV n  const.
Work in a polytropic process
1
( PV
Wrev   PdV 
1 1  PV
2 2)
n 1
Heat transfer in a polytropic process:
k n 1
k  n Pv
v
1 1
q
( Pv
[1  ( 1 ) n1 ]
1 1  P2 v2 ) 
k 1 n 1
k 1 n 1
v2
v
v
v
n  k v1
s  c p ( n  1) ln 1  Rn ln 1  [c p ( n  1)  Rn]ln 1  R
ln
v2
v2
v2
k  1 v2
v1  v2 ， n  k ， s  0 ，heat absorption
v2  v1 ， n  k ， s  0 ，heat rejection
v1  v2 ， n  k ， s  0 ，heat rejection
v2  v1 ， n  k ， s  0 ，heat absorption
q
1 k  n Pv
v
1 1
( s )en 

[1  ( 1 ) n1 ]
T0
T0 k  1 n  1
v2
1 k  n Pv
v
n  k v1
1 1
( s ) net  
[1  ( 1 ) n1 ]  R
ln
T0 k  1 n  1
v2
k  1 v2
-----------------------------------------------------------------------------------------------------Example：Calculate the net entropy change to compress 1 kg of air at 100 kPa and 25
℃ to a volume of 0.5 m3.in a polytropic process with n=1.3.
P1 = 100 kPa，T1 = 298 K，v1 = 0.8553 m3/kg
P2 = P1 × (v1 / v2 )1.3 = 200.94 kPa
T2 = T1 × (v1 / v2 )0.3 = 350.1 K
Δssys =-0.03843 kJ/kg-K
q= -12.455 kJ/kg
Δsen =0.04179 kJ/kg-K
Δsnet =0.00336 kJ/kg-K
-----------------------------------------------------------------------------------------------------Assignment 2.10 : Calculate the net entropy change and the work to compress 1 kg
of air at 100 kPa and 25℃ adiabatically to a volume of 0.3 m3.in a polytropic
process with n=1.5.
------------------------------------------------------------------------------------------------------
Advanced Thermodynamics, ME dept. NCHU 頁 20 (2.3.5). Entropy Rate Balance of Open system
In an open system, the change of entropy is balanced among the exchange
process, the transfer process, and the production process. The net increase can be
attributed to the external irreversibility and the internal irreversibility.
Q j
dS

  m i si  m e se 
dt
j Tj
Rate of entropy change = Rate of entropy transfer +Rate of exchange + Rate of
entropy production


 dS 
 1 1

   Q j      0
 dt  net
j
 T j T0 
dS
steady state，
0
dt
Q j
j T   m i si  m e se   0
j
for single input and single output system
Q j
 i  ms
 e    0
 ms
Tj
se 
qj
Tj
 si  
outlet entropy = inlet entropy + entropy transfer + entropy generated
In an adiabatic process, we have Q j  0 . As a result, the entropy increase of
 dS 
the system is      0 .
 dt  net
se  si    si
The entropy at the outlet of a steady system is always greater than that ath the
inlet.
-----------------------------------------------------------------------------------------------------Example ：Air flows through an device. It is known that the pressure and the
temperature at one end is 100 kPa and 298 K, and at the other end is 200 kPa and 380
K. Determine which kind of device it is. Is it a compressor, or a turbine?
Advanced Thermodynamics, ME dept. NCHU 頁 21 -----------------------------------------------------------------------------------------------------Assignment 2.11 : Air at 25 ℃ and 200 kPa flows through an insulated throttle valve
and drops to 100 kPa. Calculate the entropy generated.
-----------------------------------------------------------------------------------------------------For polytropic process with PV n  const. , the entropy generation is as the following.
Work in a polytropic process
Wrev    VdP   

1
n
1
n
n 1
n
n
( PV
dP  
Pn 
1 1  PV
2 2)
n 1
n 1
P
Heat transfer in a polytropic process:
q  hi  he  w
k
n
q  c p (T2  T1 )  w 
( P2v2  Pv
( Pv
1 1) 
1 1  P2 v2 )
k 1
n 1
k n 1
k  n RT1
P2 nn1
q
( Pv
[1  ( ) ]
1 1  P2 v2 ) 
k 1 n 1
k 1 n 1
P1
T
P
k n 1
P
nk
P
( s ) sys  c p ln 2  R ln 2  R[
 1]ln 2  R
ln 2
T1
P1
k 1 n
P1
( k  1) n P1
P2  P1 ， n  k ， s  0 ， q  0 heat absorption
P2  P1 ， n  k ， s  0 ， q  0 ，heat rejection
P2  P1 ， n  k ， s  0 ， q  0 ，heat rejection
P2  P1 ， n  k ， s  0 ， q  0 ，heat absorption
q
1 k  n RT1
P2 nn1
( s )en 

[1  ( ) ]
T0
T0 k  1 n  1
P1
1 k  n RT1
P n 1
nk
P
[1  ( 2 ) n ]  R
ln 2
T0 k  1 n  1
P1
( k  1) n P1
-----------------------------------------------------------------------------------------------------Example：Calculate the net entropy change to compress 1 kg of air at 100 kPa and 25
℃ to a volume of 0.5 m3.in a polytropic process with n=1.3.
P1 = 100 kPa，T1 = 298 K，v1 = 0.8553 m3/kg
P2 = P1 × (v1 / v2 )1.3 = 200.94 kPa
T2 = T1 × (v1 / v2 )0.3 = 350.1 K
Δssys =-0.03843 kJ/kg-K
q= -12.455 kJ/kg
Δsen =0.04179 kJ/kg-K
( s ) net  
Advanced Thermodynamics, ME dept. NCHU 頁 22 Δsnet =0.00336 kJ/kg-K
------------------------------------------------------------------------------------------------------
(2.3.6). Efficiency of real process
Compressor:
W
h h
c  s  2 s 1
Wa h2  h1
If air is assumed to be an ideal gas with constant heat capacity, the outlet temperature
of a compressor would be
k 1



1  P2  k


 1 
T2  T1 1 




 c  P1 


 

-----------------------------------------------------------------------------------------------------Example: Find the power of an air compressor that raise the pressure of air from 1
bar and 300 K to 10 bars with an efficiency of 85%.
-----------------------------------------------------------------------------------------------------Turbine:
W
h h
t  a  1 2
Ws h1  h2 s
If air is assumed to be an ideal gas with constant heat capacity, the outlet temperature
of a turbine would be
k 1



k


P


T2  T1 1  t 1   2   
  P1  


 

-----------------------------------------------------------------------------------------------------Example: Hot air at 10 bars and 1000 K flows through a turbine. Find the exit
pressure if the work delivered by the turbine is 230 kJ/kg.
-----------------------------------------------------------------------------------------------------Gas nozzle:
V22
n  2
V2 s
Advanced Thermodynamics, ME dept. NCHU 頁 23 If air is assumed to be an ideal gas with constant heat capacity, the outlet velocity
of a nozzle would be
k 1




k


1 2
P
1


V2   n c pT1 1   2    V12 
  P1   2
2






-----------------------------------------------------------------------------------------------------Example: Air flows through a nozzle at 10 m/sec and 500 K. If the velocity of air is
supposed to reach 600 m/sec, find the pressure required if the nozzle efficiency if
90%.
-----------------------------------------------------------------------------------------------------Assignment 2.12 : In a jet engine, air is compressed from 25 ℃ and 100 kPa to the
pressure of 1 MPa, and then heated to 1200 K in the combustor. If the compressor
efficiency is 85%, the turbine efficiency is 90%, and the nozzle efficiency is 95%,
calculate the thrust of engine assuming that air is an ideal gas with constant heat
capacity.
2
1
3
4
5
-----------------------------------------------------------------------------------------------------Fluid nozzle:
1
1
hi  Vi 2  he  Ve2
2
2
1 2
1
ui  Pv
Vi  ue  Pe ve  Ve2
i i 
2
2
For a perfect nozzle with isentropic process, the entropy change is zero.
T
s  c ln e  0 , Tes  Ti
Ti
1 2
1
ui  Pv
Vi  ues  Pe ve  Ves2
i i 
2
2
1 2
1
Ves  ( Pi  Pe )v  Vi 2
2
2
1
1
  Ve2 / Vs2
2
2
Advanced Thermodynamics, ME dept. NCHU 頁 24 1 2
1
1
ui  Pv
Vi  ue  Pe ve  Ves2  ue  Pe ve  [( Pi  Pe )v  Vi 2 ]
i i 
2
2
2
1 2
1 2
1 2
1 2
ue  ui  Pv
Vi  Pe ve  Ves  Ves  [( Pi  Pe )v  Vi ](1   )
i i 
2
2
2
2
1
ue  ui  [( Pi  Pe )v  Vi 2 ](1   )
2
This is the internal energy change if fluid flows through a nozzle with given
efficiency. If fluid spray at the outlet of nozzle is composed of fine droplets, the
internal energy change would be
3
1
ue  ui  c(Te  Ti ) 
 [( Pi  Pe )v  Vi 2 ](1   )
r
2
-----------------------------------------------------------------------------------------------------Example: A water nozzle is operated at 10 bars and issues water jet composed of fine
droplets with averaged diameter of 0.01 mm. The inlet water temperature is 25℃,
and the efficiency of nozzle is 85%. Find the power required and compute the
distribution of energy among the kinetic energy, the potential energy, and the thermal
energy.
-----------------------------------------------------------------------------------------------------Assignment 2.13 : A pump raises the pressure of water from 1 bar to 10 bars. If the
pump efficiency is 65%, calculate the temperature rise as water flows through the
pump.
-----------------------------------------------------------------------------------------------------
Advanced Thermodynamics, ME dept. NCHU 頁 25 (2.4). Can the second law be violated?
(2.4.1). What did the distinguished physicists say about the second law?

[A law] is more impressive the greater the simplicity of its premises, the more
different are the kinds of things it relates, and the more extended its range of
applicability. Therefore, the deep impression which classical thermodynamics
made on me. It is the only physical theory of universal content, which I am
convinced, that within the framework of applicability of its basic concepts will
never be overthrown.
Albert Einstein, quoted in M.J. Klein, Thermodynamics in Einstein's Universe,
in Science, 157 (1967), p. 509.

The law that entropy always increases -- the second law of thermodynamics -holds I think, the supreme position among the laws of Nature. If someone
points out to you that your pet theory of the universe is in disagreement with
Maxwell's equations - then so much worse for Maxwell equations. If it is
found to be contradicted by observation - well these experimentalists do
bungle things sometimes. But if your theory is found to be against the second
law of Thermodynamics, I can give you no hope; there is nothing for it but to
collapse in deepest humiliation.
Sir Arthur Stanley Eddington, in The Nature of the Physical World.
Maxmillan, New York, 1948, p. 74.
Advanced Thermodynamics, ME dept. NCHU 頁 26 (2.4.2). Statistical insight of the second law
Can you fix a cup of Latte by mixing one half cup of coffee and one half cup of milk?
Can we stir a cup of Latte into one half of coffee and one half of milk?
1!1!
 2 =1
2!
2!2!
1
C24 
2
4!
3
3!3!
1
C36 
2
6!
10
4!4!
1
C48 
2
8!
35
50!50!
100

 2  9.9  10 30
C50
100!
500!500!
1000
C500

 2  1.9  10301
1000!
5000!5000!
10000
C5000

 2  1.1  10 3010
10000!
C12 
If a cup of Latte contains 6×1024 molecules, what is the probability that we can
separate it into coffee and milk by keep stirring?
Stirling’s formula:
Advanced Thermodynamics, ME dept. NCHU 頁 27 ln N !  N ln N  N
N =6×1024 , ln N  57.05, ln N !  3.36×1026
26
N !  e3.3610
26
( N / 2)!( N / 2)!
C NN / 2 
 2  e 3.3610
N!
The validity of the second law is based on the statistics. The probability of a system
which is composed of millions of particles to violate the second law is zero.
Advanced Thermodynamics, ME dept. NCHU 頁 28 (2.4.3 ). Vortex tube – does it follow the second law?
(2.4.3.1). The principle of vortex tube
Fig. 2.4.3.1 A real vortex tube
The vortex tube is a device that produces hot and cold air streams
simultaneously at its two ends from a source of compressed air. Unlike the
traditional compression type cycle that requires several components, a vortex tube is
very simple with no moving parts.
Vortex tube was invented by French engineer Georges Ranques at 1928.
However, it was until 1946 that people started to show interest in vortex tube after a
research paper had been published by Rudolph Hilsch.
Fig. 2.4.3.2 Structure of vortex tube
A cold orifice is placed at the centre of the left end with a suitable sized hole.
Compressed air is introduced into the tube through a tangential inlet nozzle which is
located near the left end. At the right end, a conical valve is inserted to confine the
exiting air to outer regions and restrict it to the central portion of the tube. The
tangential flow imparts a vortex motion to the inlet air, and creates a cold stream in
the left end and a warm stream in the right end.
There is no theory to give a satisfactory explanation of the vortex tube
Advanced Thermodynamics, ME dept. NCHU 頁 29 phenomenon at the present time.
Does the vortex tube violate the second law of thermodynamics?
According to the Clausius Statement of the second law of thermodynamics, it is
impossible to transfer heat from a cooler body to a hotter body without doing work.
Does the separation of warm and cold air streams violate the second law?
Where is the work?
(2.4.3.2). Analysis of vortex tube
The whole process can be analyzed as the following:
The flow rate of inlet air is m 1，the outlet cold air is m 2 , and the outlet warm air
is m 3 . The conservation of mass results in the relationship amnong the flow rates
as folliwing.
m 2  m 3  m1
Since the process is adiabatic, and no shaft work is carried out, the energy
balance would resultthe following relationship among the enthalpies of the inlet and
outlet flows.
( m 2  m 3 ) h1  m 2h2  m 3h3
Assume that air is an ideal gas with constant heat capacity, then we have a
relationship among the temperatures of inlet and outlet flows.
m 1T1  m 2T2  m 3T3
Define the cold fraction as x  m 2 / m 1 , the temperature relationship can be
expressed as
T1  xT2  (1  x)T3
In which T1 is the inlet temperature, T2 is the outlet cold temperature, and T3
is the outletwarm temperature. The entropy change of the process is
S  m 2 s2  m 3 s3  m 1s1  m 2 ( s2  s1 )  m 3 ( s3  s1 )
Advanced Thermodynamics, ME dept. NCHU 頁 30 If air is assumed to be ideal gas with constant specific heat, then the entropy change
can be expressed as
s2  s1  c p ln(T2 / T1 )  R ln( P2 / P1 )
s3  s1  c p ln(T3 / T1 )  R ln( P3 / P1 )
where P1 is the upstream pressure, and P2 and P3 are the down stream pressure
pressures. The total entropy change rate can be expressed as
S  m 2 s2  m 3 s3  m 1s1  m 2 ( s2  s1 )  m 3 ( s3  s1 )
And the specific entropy change is as the following.
s 
S
 x c p ln(T2 / T1 )  R ln( P2 / P1 )   (1  x)  s3  s1  c p ln(T3 / T1 )  R ln( P3 / P1 ) 
m 1
However, since P2  P3  Pa , the entropy change can then be expressed as
k 1


s  c p  ln(T2 / T1 ) x (T3 / T1 )1 x  ln( Pa / Pc ) k 


So long as the temperature of cold air is lower that that obtained in the equation
above, the net change of entropy is positive. There is no violation of the second
law.
(2.4.3.3). Efficiency of vortex tube
Efficiency of a vortex tube can be defined as the amount of cooled air produced
to that if the process is isentropic for a given value of pressure ratio and cold fraction.
If the process is isentropic, the entropy does not change, the cold temperature and
warm temperature are relared with the pressure ratio as following.
(T2 / T1 ) x (T3 / T1 )1 x =( Pa / Pc )
k 1
k
Coupling the realtionship above with the energy balance would result in the
following.
x
T
T2
 (1  x) 3  1
T1
T1
Advanced Thermodynamics, ME dept. NCHU 頁 31 T3
T
T2 
1 
1
x
 , where   2


1  x  
T1
T1 1  x 
T1  1  x 1  x
k 1
1  x 1 x
) =( Pa / Pc ) k
1 x
 x(
The equation may be solved with the Newton Raphson method.
k 1
1  x 1 x
) =( Pa / Pc ) k
 (
1 x
k 1
1  x 1 x
f ( )   x (
) -( Pa / Pc ) k  0
1 x
df
1  x 1 x
1  x  x  x
1  x  x 1  x

)  (1  x) x (
)
) 
 x x 1 (
 x x 1 (
 
d
1 x
1 x
1 x
1 x
 1 x

k 1
1  x 1 x
 x(
) -( Pa / Pc ) k
f
new
old
1

x

  
1  x  x 1  x
df / d

) 
 
x x 1 (
1 x
1

x


x
T2/T1
Cold air fraction
0.8
0.6
0.4
0.2
0
Pr=2
Pr=5
Pr=10
0
0.2
0.4
0.6
0.8
1
x
Fig. 2.4.3.3 Temperature ratio of vortex tube
For any given value of x,  can be obtained if the pressure ratio is known. The
cooled air produced in an isentropic process is
QL  m 2 c p (T1  T2 )  m 1c pT1 x(1   )

QL
 x(1   ) is a diemnsionless index to show the amount of cooled air
m 1c pT1
produced.
Advanced Thermodynamics, ME dept. NCHU 頁 32 0.6
0.5
0.4
T2/T1
0.3
q
0.2
0.1
0
0
0.2
0.4
0.6
0.8
1
Fig. 2.4.3.4 Temperature ratio and cooling load of vortex tube
Efficiency of a vortex tube is defined as  
 a 1  a

, where  a is the actual
 s 1  s
temperature ration, and  s is the theoretical temperature ratio. It is noted that the
entropy change should be positive according to the second law of thermodynamics,
-----------------------------------------------------------------------------------------------------Example: A vortex tube operates with the pressure of 5 bars. The cold fraction is
0.5, and the efficiency is 0.7. Calculate the outlet temperature at the cold side.
-----------------------------------------------------------------------------------------------------(2.4.3.4). COP of vortex tube
The work that is required in the second law is from the compressor that is used
to raise the pressure of the inlet air so that the vortex tube may perform its function
The COP of the vortex tube can be expressed as the ratio of the amount of
chilled air produced to the work that must be done, i.e.
COP 
m 2 (h1  h2 )
Wa
However, the work that has been done is carried out by the compressor to
generate the high pressure air in the upstream. The work of compressor can be
calculated as
k 1


k


P

1
Wa  m 1 c pT1    1
 P0 

c


1
For the best case, the process is isentropic, and the COP becomes
Advanced Thermodynamics, ME dept. NCHU 頁 33 For real vortex tube, the COP is
COP 
m 2 c p (T1  T2 )
k 1


k


1
P

1
m 1 c pT1    1
 P

c
 0 

xc (1 

 P1 
 
 P0 
T2
)
T1
k 1
k
1
For ideal vortex tube, the COP is
COP 
1  x 1 x
)
1 x
1  x 1 x
1  x (
)
1 x
c x(1   ) x (
-----------------------------------------------------------------------------------------------------Assignment 2.14 :
A vortex tube operates with the pressure of 5 bars. The cold fraction is 0.5, and the
efficiency is 0.7. The compressed air is obtained with a compressor with 75%
efficiency. Calculate the COP of the system.
----------------------------------------------------------------------------------------------------(2.4.3.5). Real performance of vortex tube
Pressure
Supply
BAR
1.4
20
30
34
33
8
14
2.8
48
46
11
18
4.1
57
55
14
22
5.5
63
62
14
24
6.9
68
65
14
25
8.4
72
69
14
26
Temperature drop of cold air,
(°C) in blue
Advanced Thermodynamics, ME dept. NCHU Cold Fraction %
40
31
20
42
28
51
33
56
35
61
37
64
38
50
28
28
39
38
46
44
51
47
55
50
58
52
60
70
80
24
20
16
26
46
59
34
28
20
50
62
80
40
33
25
57
73
92
45
36
28
63
80
100
48
39
30
66
84
106
50
41
31
68
86
108
Temperature rise of hot air,
(°C) in red
頁 34 -----------------------------------------------------------------------------------------------------Assignment 2.15 :
Use the performance data of a commercial vortex tube to calculate the efficiency as
well as the entropy generation
----------------------------------------------------------------------------------------------------(2.4.3.6). Applications of vortex tube
Spot cooling in machining
Spot cooling in welding
Temperature controlled high chair
Intake manifold heating of diesel engine during cold start
Advanced Thermodynamics, ME dept. NCHU 頁 35 (2.4.4). Maxwell’s demon – what happens inside a chamber in which a tiny creature
lives?
Maxwell's Demon is an imaginary creature that the physicist James Clerk
Maxwell created to demonstrate the limitation of the second law of thermodynamics.
Suppose that you have a box filled with a gas at some temperature. This means that
the average speed of the molecules is a certain amount depending on the temperature.
The Maxwell-Boltzmann velocity distribution is

3
2
mv 2
 2 kT  2
f (v) dv  4 
) dv
 v exp( 
2kT
 m 

8 RT
Mean speed: v   vf (v ) dv 
0

Figure 2.4.3 The velocity distribution of ideal gas
One half of the molecules will be flying with speed faster than the average value
and the other one half will be flying slower than the average value. For example, air
at 300 K and 1 bar has the average speed of 468 m/sec.
Suppose that a partition is placed across the middle of the box separating the
two sides into left and right. Both sides of the box are now filled with the gas at the
same temperature. Maxwell imagined a molecule sized trap door in the partition with
his tiny demon poised at the door who is observing the molecules. When a faster than
average molecule approaches the door he makes certain that it ends up on the left
Advanced Thermodynamics, ME dept. NCHU 頁 36 side (by opening the tiny door if it's coming from the right) and when a slower than
average molecule approaches the door he makes sure that it ends up on the right side.
So after these operations he ends up with a box in which all the faster than average
gas molecules are in the left side and all the slower than average ones are in the right
side. So the box is hot on the left and cold on the right. Then one can use this
separation of temperature to run a heat engine by allowing the heat to flow from the
hot side to the cold side.
Fig. 2.4.4 The Maxwell demon and the partition that it operates.
Another possible action of the demon is that he can observe the molecules and
only open the door if a molecule is approaching the trap door from the right. This
would result in all the molecules ending up on the left side. Again this setup can be
used to run an engine. This time one could place a piston in the partition and allow
the gas to flow into the piston chamber thereby pushing a rod and producing useful
mechanical work.
The demon is trying to create more useful energy from the system than there
was originally. Equivalently he was decreasing the randomness of the system (by
ordering the molecules according to a certain rule) which is decreasing the entropy.
No such violation of the second law of thermodynamics has ever been found.
The demon is not a real creature, it owns the capability that we do not have. It is
of course that what it can do is not an evidence of the collapse of the second law of
thermodynamics because the second law is founded on the observations of real life.
However, it is interesting to know what should be equipped to the demon should it
could do its job, and what the cost would be for the demon to do it job.
The demon should be able to identify the particles flying towards it and to
measure the velocity of those particles such that it can make decision to open the gate
or to close the gate in time. As a result, the demon needs a light source, and uses the
light to detect the location as well as the velocity of particles. The wavelength of the
Advanced Thermodynamics, ME dept. NCHU 頁 37 light should be less than the particle size and the light intensity, which is equivalent
to the number of photons emitted from light source per second, should be comparable
to the frequency of collision of particles on wall to ensure that every particle flying
towards the gate could be detected.
Energy from a high temperature reservoir should be supplied to the light source
such that light at the required frequency and intensity could be emitted continuously.
Entropy will be generated when energy is supplied to the system. The net entropy
change would be positive if the entropy generated is considered. As a result, the
second law is not violated even a miniature demon is operating the gate.
So, does the Maxwell’s demon violate the second law?
Advanced Thermodynamics, ME dept. NCHU 頁 38 (2.5). Perpetual Motion Machines – human’s dream to break the second law.
The lessons we learn from the thermodynamics laws are as the following.
The first law: You can’t get something from nothing.
The second law: You can get even less from something.
However, it is very tempting to invent a device that may run forever without any
input energy. For past decades of years, many applications have been send to the
bureau of patents claiming that they have the feature of perpetual motion. None of
them has been approved. As a result, the National Institute of Standards and
Technology ( NIST) has announced officially that all applications regarding
perpetual motion machine will be refused.
Perpetual motion machines can be classified as three categories.
The zeroth kind of perpetual motion machine
The first kind of perpetual motion machine
The second kind of perpetual motion machine
Some of the examples shown in the following content are taken from the web site
http://www.lhup.edu/~dsimanek/museum/
(2.5.1). The zeroth kind
The zeroth kind perpetual motion machine is a fault in the interpretation of
mechanics. It was presumed as a perpetual motion machine because no solid
analysis of mechanics was performed.
Advanced Thermodynamics, ME dept. NCHU 頁 39 Water fall that never ends
Magic wheel may rotate forever
Advanced Thermodynamics, ME dept. NCHU 頁 40 Floating ball will rotate forever
A wheel in the form of a perfect sphere or cylinder rotates about a frictionless
horizontal shaft. The left side is in a chamber filled with water, perfect (frictionless
and leakage proof) seals around the rotating wheel prevent the liquid from escaping.
The left side of the wheel therefore experiences an upward buoyant force due to the
liquid it displaces. So that side will rise, and the wheel rotates clockwise.
(2.5.2). The first kind
A perpetual motion machine of the first kind produces strictly more energy than it
uses, thus violating the law of conservation of energy. Over-unity devices, that is,
devices with a thermodynamic efficiency greater than 1.0 (unity, or 100%), are
perpetual motion machines of this kind.
(2.5.2.1). Magnetic generator
People’s dream to generate power without ant cost
Advanced Thermodynamics, ME dept. NCHU 頁 41 磁能永動機稱為第三類機械永動機，它是唯一不違反能量守恆第一二定律。違
反第一定律指永動機可產生自己所需的能量，例如用一台發電機一開始借有電
池起動後，再去帶動更多台發電機來產生電能後一直運轉下去。違反第二定律
是指永動機利用該機械熱庫可重複利用﹐進而到達永續運轉目的﹐事實上機械
所做的功產生熱能會有一半自耗另一半穫得利用所以能量會越用越少.
唯有磁能發電機是屬於自然界所產生的動能，它不需任何外界給予動力本身就
是一個自發體，我們只要懂得利用機械貫性原理就有辦法製造出電能，因此它
既不違反目前科學界所建立的物理原則。所以請不要再給永動機冠上不可能的
名詞。
20071211(yahoo.com)磁能永動機…▲唯一不違反能量守恆第一
二定律★
(2.5.2.2). Water fuel vehicle
自由時報電子報 2008 年 6 月 17 日 星期二
水燃料車 一公升跑八十公里
Advanced Thermodynamics, ME dept. NCHU 頁 42 ［記者陳英傑／綜合外電報導］日本 Genepax，推出一套水能量系統（Water
Energy System，簡稱 WES），將水分解成氫和氧產生能量，讓汽車只吃水就可
以發動，雨水、河水甚至茶都可使用，目前共有 120W 及 300W 的原型機，一
公升的水就可以時速八十公里連續行駛一小時，未來量產後，預計價錢將可壓
低到每套系統新台幣十四萬元左右，不過目前上市時間未定。
這款水燃料車與其他廠商一樣，都是將水分解成氫與氧來產生能量，但由
於採用薄膜電導組件，擁有較長的使用壽命。Genepax 執行長平澤表示，只要
有瓶裝水，車子就可連續行駛。Genepax 在日本大阪實地用水來發動並行駛車
子測試，雨水或河水都可使用。
目前這套系統高達兩百萬日圓（約新台幣五十六萬元），不過 Genepax 認
為，未來大量生產後可降至五十萬日圓（約新台幣十四萬元），除有 120W 及
300W 的原型機，還會有 1kW 的系統，將打進家庭用電市場。
-----------------------------------------------------------------------------------------------------Assignment 2.16
Give a comment on the operation of water engine and explain why it doesn’t work.
-----------------------------------------------------------------------------------------------------
(2.5.2.3). Super electrolysis
You may produce more energy than that you input.
Stanley MEYER is the most famous inventor in the 'Super-efficient Electrolysis' field,
and many took inspiration from his work. His original Electrolysis Concept was able
to produce many times as much hydrogen gas as permits the Faraday's Law of
electrolysis, and the Law of Conservation of Energy, then he brought it up to a still
un-understood high level he called 'Thermal Explosive Energy'. We can say that he
was at least 50 years in advance of his time ...
Advanced Thermodynamics, ME dept. NCHU 頁 43 The evidence, then, is that his concept of electrolysis is tapping in another source of
energy, not 'recognized' yet in the classical physics, that must be in another
dimension, outside our three dimensions, and it must be a non thermal form of
energy, because Stan Cell is not producing heat while electrolyzing. There is still not
a complete consensus of the Experts in Zero Point Energy.
(2.5.3). The second kind
A perpetual motion machine of the second kind is a machine which spontaneously
converts thermal energy into mechanical work. This need not violate the law of
conservation of energy, since the thermal energy may be equivalent to the work done;
however it does violate the more subtle second law of thermodynamics.
Note that such a machine is different from real heat engines, which always involve a
transfer of heat from a hotter reservoir to a colder one, the latter being warmed up in
the process. The signature of a perpetual motion machine of the second kind is that
there is only one single heat reservoir involved, which is being spontaneously cooled
without involving a transfer of heat to a cooler reservoir. This conversion of heat into
useful work, without any side effect, is impossible by the second law of
thermodynamics.
e.g. Brownian ratchet, dipping bird
Advanced Thermodynamics, ME dept. NCHU 頁 44 -----------------------------------------------------------------------------------------------------Assignment 2.17
Explain the principle of work for the dipping bird, and have some comments on
whether it violates the second law or not.
----------------------------------------------------------------------------------------------------Assignment 2.18
A gas turbine engine is working with a heat pump. Air is compressed with the
compressor, and then heated with a heat pump instead of a burner. Give a comment
on the feasibility of this system.
-----------------------------------------------------------------------------------------------------
Advanced Thermodynamics, ME dept. NCHU 頁 45 (2.5)、The third law of thermodynamics
History
The third law was developed by Walther Nernst, during the years 1906-1912, and is
thus sometimes referred to as Nernst's theorem or Nernst's postulate. The third law
of thermodynamics states that the entropy of a system at zero is a well-defined
constant. This is because a system at zero temperature exists in its ground state, so
that its entropy is determined only by the degeneracy of the ground state; or, it states
that "it is impossible by any procedure, no matter how idealised, to reduce any
system to the absolute zero of temperature in a finite number of operations". the third
law relates to energy.
Nernst Simon Statement
The entropy change associated with any isothermal reversible process of a condensed
system approaches zero as the temperature approaches absolute zero.
lim(s )T  0
T 0
This law provides an absolute reference point for the determination of entropy. The
entropy determined relative to this point is the absolute entropy.
T
s (T , p )   c p
0
Tm
dT
T
T
T
dT Lm b dT Lb
dT
s (T , p )   c p

  cp
   cp
T
Tm Tm T
Tb Tb T
0
Entropies at 25℃ and 1 atm ( S m0 / R )
CH4
CO
CO2
22.39
23.76
25.70
H2
H2O(l)
H2O(g)
15.705
8.41
22.70
N2
O2
NH3
23.03
24.66
23.13
One application of the third law is with respect to the magnetic moments of a
material. Paramagnetic materials (moments random) will order as T approaches 0 K.
They may order in a ferromagnetic sense, with all moments parallel to each other, or
Advanced Thermodynamics, ME dept. NCHU 頁 46 they may order in an antiferromagnetic sense, with all moments antiparallel to each
other.
Yet another application of the third law is the fact that at 0 K no solid solutions
should exist. Phases in equilibrium at 0 K should either be pure elements or
atomically ordered phases.
-----------------------------------------------------------------------------------------------------Example:
Methane is burned with oxygen to form carbon dioxide and water vapor. If all the
reactants and products are at 25℃ and 1 atm, find the entropy generation for 1 kg of
methane.
CH 4  2(O2  3.76 N 2 )  CO2  2 H 2O  7.52 N 2
Mixing
combustion
cooling
CH4
CH4+2O2+7.52N2
N2
CO2+2H2O+7.52N2
CO2+2H2O+7.52N2
O2
Initial state:
CH 4 : s=22.39×8.314= 186.15 kJ/kmole-K
O2 : s=24.66×8.314= 205.02 kJ/kmole-K
N 2 : s=23.03×8.314= 191.47kJ/kmole-K
186.15×1/10.52+205.02×2/10.52+191.47×7.52/10.52=193.54 kJ/kmole-K
Mixing process:
T
P
dT
s (T , P )  s (T0 , P0 )  R ln   c p
P0 T0 T
CH 4 : s=186.15-8.314×ln(1/10.52)= 205.72 kJ/kmole-K
O2 : s=205.02-8.314×ln(2/10.52)= 218.82 kJ/kmole-K
N 2 : s=191.47-8.314×ln(7.52/10.52)= 198.07 kJ/kmole-K
Mixture: s= (205.72×1+218.82×2+198.07×7.52)/10.52= 202.74 kJ/kmole-K
s =202.74-193.54= 9.20 kJ/kmole-K
Advanced Thermodynamics, ME dept. NCHU 頁 47 Combustion process:
Constant pressure adiabatic flame temperature: 2324 K
CO2 : 318.35-8.314×ln(1/10.52)= 337.92 kJ/kmole-K
H 2O : 272.41-8.314×ln(2/10.52)= 286.21 kJ/kmole-K
N 2 : 257.42-8.314×ln(7.52/10.52)= 260.21 kJ/kmole-K
Products: s= 272.54 kJ/kmole-K
s =272.54- 202.74= 69.80 kJ/kmole-K
Cooling process:
Final temperature: 298 K
CO2 : 213.69-8.314×ln(1/10.52)= 233.26 kJ/kmole-K
H 2O : 188.72-8.314×ln(2/10.52)= 202.52 kJ/kmole-K
N 2 : 191.50-8.314×ln(7.52/10.52)= 194.29 kJ/kmole-K
Cool products: s= 199.46 kJ/kmole-K
s =199.46-272.54= -73.08 kJ/kmole-K
Q  hCO2  2hH 2O  7.52hN 2 =802310 kJ
q  76265 kJ/kmole
q
sEV  =255.92 kJ/kmole-K
T0
s Net  sEV  ssys =182.84 kJ/kmole-K=120.22 kJ/kg-K
QI
QE
QH
W
QL
T0
Advanced Thermodynamics, ME dept. NCHU 頁 48 qH =76265 kJ
s Net  sEV  ssys =0
qL
=73.08 kJ/kmole-K
T0
qL =21778 kJ/kmole
w  qH  qL = 54487 kJ/kmole=35825 kJ/kg CH4
sEV 
In summary:
Mixing process: s =9.20 kJ/kmole-K
Combustion process: s =69.80 kJ/kmole-K
Cooling process: s Net  sEV  ssys =120.22 kJ/kg-K
Maximum work: w  = 54487 kJ/kmole=35825 kJ/kg CH4
------------------------------------------------------------------------------------------------------
Advanced Thermodynamics, ME dept. NCHU 頁 49