# Chapter Two The Thermodynamic Laws

Mechanics

Oct 27, 2013 (4 years and 6 months ago)

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Chapter Two
The Thermodynamic Laws

(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
A
B
T T
, and
B
C
T T
, then
A
C
T T

.

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 20
th
century, long after the first three laws were already
widely in use and named as such, hence the zero numbering.

Ralph H. Fowler

(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."

Rudolf Clausius (1822 – 1888)

(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
.
A A B B
U m u m u 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.
------------------------------------------------------------------------------------------------------
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.
------------------------------------------------------------------------------------------------------
Assignment 2.1 : An insulated rigid chamber is divided into two equal parts with a
diaphragm. T The left part is filled with air at 25℃ and 3 bars, and the right part is
filled with air at 25℃ and 1 bars. The diaphragm ruptures and the air mixes
together. Find the final temperature of the air in the chamber.
------------------------------------------------------------------------------------------------------

(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, 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.
------------------------------------------------------------------------------------------------------
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 CO
2
and the right part is filled
with N
2
. Both sides are at 300 K and 100 kPa. The right side is heated gradually
until N
2
expands to a volume of 7 liters while the left side is insulated during this
process. Assuming both N
2
and CO
2
are ideal gases with constant specific heats,
find the heat transfer during this process.
------------------------------------------------------------------------------------------------------
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.
------------------------------------------------------------------------------------------------------

(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.

i e
dm
m m
dt
   
 

2 2
( ) ( )
2 2
i e
i i i e e e
V dE V
Q m h gz m h gz W
dt
         

 

0
dm
dt

i e
m m m 
  

2 2
( ) ( )
2 2
i e
i i i e e e
V V
Q m h gz m h gz W        

 

2 2
2 2
i e
i i e e
V V
q h gz h gz w      

i e
q h h w  

------------------------------------------------------------------------------------------------------
Example：The pressure of saturated liquid R134a at 10 bars declines to 1 bas 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.
------------------------------------------------------------------------------------------------------

(2.2.3.2). Uniform state steady flow system

2 1
i e
m m m m  

,
i i e e
m mdt m m dt 
 
 

2 2
2 1
( ) ( )
2 2
i e
i i i e e e
V V
Q m h gz E E m h gz W        

------------------------------------------------------------------------------------------------------
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.
------------------------------------------------------------------------------------------------------
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℃.
------------------------------------------------------------------------------------------------------

(2.2.3.3). Uniform state uniform flow

Flow out of a system adiabatically through a small hole:

e
dm
m
dt
 

( )
0
e e e e
d mu du dm
m h m u m h
dt dt dt
    
 

0
v
dT dm dm
mc u h
dt dt dt
  

v
dT dm
mc RT
dt dt

1 1
v
dT R dm
T dt m c dt

1
2 2 2
1 1 1
v
R
k
c
T m m
T m m

   
 
   
   

2 2
1 1
k
P m
P m
 

 
 

------------------------------------------------------------------------------------------------------
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 left part.
------------------------------------------------------------------------------------------------------
Assignment 2.4 : 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 again to refill the bottle?
------------------------------------------------------------------------------------------------------

(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
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.

(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
1
L
H
H
W Q
Q Q
  

Refrigerator is a machine used to transfer heat from a cold body to a hot body. The
performance of a refrigerator is defined as
L L
H
H
Q Q
W Q Q
 

(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,
0Q

, 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 W 

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 B(P
1
, V
1
) can be reached from state A (P
2
,
V
2
) with an adiabatic process. There exists another state C(P
3
, V
2
) 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.
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.

(2.3.3). Entropy Rate Balance of Isolated system
Principle of increase of entropy
0
net sys env
S S S     

Reversible process:
rev
Q
dS
T

Irreversible process:
Q
dS
T

For all processes:
Q
dS
T

,
TdS Q

In an isolated system, the entropy always increases all the time.
0
TdS Q

 

0
S 

dS
dt

Thermal equilibrium
Two bodies at different temperatures reach thermal equilibrium by contacting each
other for a long period.
1 1 1 2 2 2
A
A B B A A B B
U m u m u U m u m u    

1 1 2 2A vA A B vB B A vA B vB
m c T m c T m c T m c T  

1 1
2
A
vA A B vB B
A vA B vB
m c T m c T
T
m c m c

2 2 2 2
1 1 1 1
ln ln ln( ) ( )
A vA B vB
m c m c
A vA B vB
A B A B
T T T T
S m c m c
T T T T
   

In the case that
A
vA B vB
m c m c

2
2
A
B
T T
T

2
2 2
( )
1
4
A B
A B A B
T T T T
T T T T

 

0S 

------------------------------------------------------------------------------------------------------
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.
A vA
m c 
10 × 0.447 = 4.47 kJ
B vB
m c 
100 × 4.186 = 418.6 kJ
T
av
= 27.9

2 2
1 1
ln ln
A vA B vB
A
B
T T
S m c m c
T T
  
= -2.8792 + 4.05393 = 1.1747 kJ/K
------------------------------------------------------------------------------------------------------
Assignment 2.5 : 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
irreversibility and the internal irreversibility.
Q
S
T

 

Q
S
T

  

j
j
j
Q
dS
dt T

 

0
1 1
( )
net sys ev
S S S Q
T T

        

0
1 1
( ) 0
j
j
net
dS
Q
dt T T

 
    
 
 

In an adiabatic process, we have
0
j
Q

. As a result, the entropy increase of
the system is
0
net
dS
dt

 
 
 
 

.
0Q U W   

U W  

------------------------------------------------------------------------------------------------------
Example

An insulated chamber with a volume of 0.1 m
3
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.

P
1
= 100 kPa

T
1
= 298 K

V
1
= 0.1 m
3

m = 0.1169 kg
ΔU = mc
v
ΔT = -W = 1 kJ
ΔT = 11.92 K

T
2
= 309.92 K

P
2
= 104 kPa
Δ
s
sys
= 0.003289 kJ/kg-K
Δ
s
en
=0.04179 kJ/kg-K
------------------------------------------------------------------------------------------------------

(2.3.4.2). Polytropic process
.
n
P
V const

Work in a polytropic process
1 1 2 2
1
( )
1
rev
W PdV PV PV
n
  

Heat transfer in a polytropic process:
1
1 1 1
1 1 2 2
2
1
( ) [1 ( ) ]
1 1 1 1
n
k n k n Pv v
q Pv Pv
k n k n v

 
   
   

1 1 1 1
2 2 2 2
( 1)ln ln [ ( 1) ]ln ln
1
p p
v v v n k v
s c n Rn c n Rn R
v v v k v

       

1 2
v v

n k

0
s
 

heat absorption
2 1
v v

n k

0
s
 

heat rejection
1 2
v v

n k

0
s
 

heat rejection
2 1
v v

n k

0
s
 

heat absorption
1
1 1 1
0 0 2
1
( ) [1 ( ) ]
1 1
n
en
q k n Pv v
s
T T k n v

 
    
 

1
1 1 1 1
0 2 2
1
( ) [1 ( ) ] ln
1 1 1
n
net
k n Pv v n k v
s R
T k n v k v

 
    
  

------------------------------------------------------------------------------------------------------
Example

Calculate the net entropy change to compress 1 kg of air at 100 kPa and 25

to a volume of 0.5 m
3
.in a polytropic process with n=1.3.
P
1
= 100 kPa

T
1
= 298 K

v
1
= 0.8553 m
3
/kg
P
2
= P
1
×
(v
1
/ v
2
)
1.3
= 200.94 kPa
T
2
= T
1
×
(v
1
/ v
2
)
0.3
= 350.1 K
Δ
s
sys
=-0.03843 kJ/kg-K
q= -12.455 kJ/kg
Δ
s
en
=0.04179 kJ/kg-K
Δ
s
net
=0.00336 kJ/kg-K
------------------------------------------------------------------------------------------------------
Assignment 2.6 : 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 m
3
.in a polytropic process
with n=1.5.
------------------------------------------------------------------------------------------------------

(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.

j
i i e e
j
j
Q
dS
ms m s
dt T

   
  

  

Rate of entropy change = Rate of entropy transfer +Rate of exchange + Rate of
entropy production
0
1 1
0
j
j
net
j
dS
Q
dt T T

 
 
   
 
 
 
 
 

0
dS
dt

0
j
i i e e
j
j
Q
ms m s
T
   
  

  

for single input and single output system
0
j
i e
j
Q
ms ms
T
   

  

j
e i
j
q
s s
T

  

outlet entropy = inlet entropy + entropy transfer + entropy generated

In an adiabatic process, we have
0
j
Q

. As a result, the entropy increase of
the system is
0
net
dS
dt

 
 
 
 

.
e i i
s
s s

  

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?
------------------------------------------------------------------------------------------------------
Assignment 2.7 : 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
.
n
PV const

, the entropy generation is as the following.

Work in a polytropic process
1
1
1 1 2 2
1
( )
1 1
n
n
n
rev
n
n n
W VdP dP P PV PV
n n
P

       
 
 

Heat transfer in a polytropic process:
i e
q h h w  

2 1 2 2 1 1 1 1 2 2
( ) ( ) ( )
1 1
p
k n
q c T T w Pv Pv Pv Pv
k n
      
 

1
1 2
1 1 2 2
1
1
( ) [1 ( ) ]
1 1 1 1
n
n
k n k n RT P
q Pv Pv
k n k n P

 
   
   

2 2 2 2
1 1 1 1
1
( ) ln ln [ 1]ln ln
1 ( 1)
sys p
T P k n P n k P
s c R R R
T P k n P k n P

     
 

2 1
P P

n k

0
s
 

0q 
heat absorption
2 1
P
P

n k

0
s
 

0q 

heat rejection
2 1
P P

n k

0
s
 

0q 

heat rejection
2 1
P P

n k

0
s
 

0q 

heat absorption

1
1 2
0 0 1
1
( ) [1 ( ) ]
1 1
n
n
en
q k n RT P
s
T T k n P

 
    
 

1
1 2 2
0 1 1
1
( ) [1 ( ) ] ln
1 1 ( 1)
n
n
net
k n RT P n k P
s R
T k n P k n P

 
    
  

------------------------------------------------------------------------------------------------------
Example

Calculate the net entropy change to compress 1 kg of air at 100 kPa and 25

to a volume of 0.5 m
3
.in a polytropic process with n=1.3.
P
1
= 100 kPa

T
1
= 298 K

v
1
= 0.8553 m
3
/kg
P
2
= P
1
×
(v
1
/ v
2
)
1.3
= 200.94 kPa
T
2
= T
1
×
(v
1
/ v
2
)
0.3
= 350.1 K
Δ
s
sys
=-0.03843 kJ/kg-K
q= -12.455 kJ/kg
Δ
s
en
=0.04179 kJ/kg-K
Δ
s
net
=0.00336 kJ/kg-K
------------------------------------------------------------------------------------------------------

(2.3.6). Efficiency of real process

Compressor:
2 1
2 1
s s
c
a
W h h
W h h

 

If air is assumed to be an ideal gas with constant heat capacity, the outlet temperature
of a compressor would be
1
2
2 1
1
1
1 1
k
k
c
P
T T
P

 
 
 
 
 
  
 
 
 
 
 
 
 
 

------------------------------------------------------------------------------------------------------
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:
1 2
1 2
a
t
s
s
W h h
W h h

 

If air is assumed to be an ideal gas with constant heat capacity, the outlet temperature
of a turbine would be
1
2
2 1
1
1 1
k
k
t
P
T T
P

 
 
 
 
 
  
 
 
 
 
 
 
 
 

------------------------------------------------------------------------------------------------------
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:
2
2
2
2
n
s
V
V
 

If air is assumed to be an ideal gas with constant heat capacity, the outlet velocity
of a nozzle would be
1
2 2
2
2 1 1
1
1 1
1
2 2
k
k
n p
P
V c T V
P

 
 
 
 
 
  
 
 
 
 
 
 
 
 

------------------------------------------------------------------------------------------------------
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 2.8 : 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.
------------------------------------------------------------------------------------------------------

Fluid nozzle:
2 2
1 1
2 2
i i e e
h V h V  

2 2
1 1
2 2
i i i i e e e e
u Pv V u Pv V    

For a perfect nozzle with isentropic process, the entropy change is zero.
ln 0
e
i
T
s c
T
  
,
es i
T T

2 2
1 1
2 2
i i i i es e e es
u Pv V u Pv V    

2 2
1 1
( )
2 2
es i e i
V P P v V  

2 2
1 1
/
2 2
e s
V V

2 2 2
1 1 1
[( ) ]
2 2 2
i i i i e e e es e e e i e i
u Pv V u Pv V u Pv P P v V

         

2 2 2 2
1 1 1 1
[( ) ](1 )
2 2 2 2
e i i i i e e es es i e i
u u Pv V Pv V V P P v V

         

2
1
[( ) ](1 )
2
e i i e i
u u P P v V

    

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
2
3 1
( ) [( ) ](1 )
2
e i e i i e i
u u c T T P P v V
r

       

------------------------------------------------------------------------------------------------------
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 2.9 : 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.
-----------------------------------------------------------------------------------------------------

(2.4). Can the second law be violated?

(2.4.1). What 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.

(2.4.2). Statistical insight of the second law

Can we stir a cup of Latte into one half of coffee and one half of milk?
2
1
1!1!
2
2!
C  
=1
4
2
2!2!1
2
4!3
C   

6
3
3!3!1
2
6!10
C   

8
4
4!4!1
2
8!35
C   

100 30
50
50!50!
2 9.9 10
100!
C

   

If a cup of Latte contains 6
×
10
24
molecules, what is the probability that we can
separate it into coffee and milk by keep stirring?

(2.4.3 ). Vortex tube – does it follow the second law?

(2.4.3.1). The principle of 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

Fig. 2.3.8.1 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
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
1
m

the outlet cold air is
2
m

, and the outlet warm air
is
3
m

. The conservation of mass results in the relationship amnong the flow rates
as folliwing.

  
m m m
2 3 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 m h mh mh
2 3 1 2 2 3 3
  

Assume that air is an ideal gas with constant heat capacity, then we have a
relationship among the temperatures of inlet and outlet flows.

1 1 2 2 3 3
mT m T mT 
  

Define the cold fraction as
2 1
/
x
m m

 
, the temperature relationship can be
expressed as

1 2 3
(1 )T xT x T  

In which
1
T
is the inlet temperature,
2
T
is the outlet cold temperature, and
3
T

is the outletwarm temperature. The entropy change of the process is

2 2 3 3 1 1 2 2 1 3 3 1
( ) ( )S m s m s ms m s s m s s       
    

If air is assumed to be ideal gas with constant specific heat, then the entropy change
can be expressed as

2 1 2 1 2 1
ln(/) ln(/)
p
s s c T T R P P  

3 1 3 1 3 1
ln(/) ln(/)
p
s s c T T R P P  

where
1
P
is the upstream pressure, and
2
P
and
3
P
are the down stream pressure
pressures. The total entropy change rate can be expressed as

2 2 3 3 1 1 2 2 1 3 3 1
( ) ( )S m s m s ms m s s m s s       

    

And the specific entropy change is as the following.

2 1 2 1 3 1 3 1 3 1
1
ln(/) ln(/) (1 ) ln(/) ln(/)
p p
S
s x c T T R P P x s s c T T R P P
m

   
        
   

However, since
2 3
a
P P P 
, the entropy change can then be expressed as

1
1
2 1 3 1
ln(/) (/) ln(/)
k
x x
k
p a c
s c T T T T P P

 
  
 
 

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.

1
1
2 1 3 1
(/) (/) =(/)
k
x x
k
a c
T T T T P P

Coupling the realtionship above with the energy balance would result in the
following.

3
2
1 1
(1 ) 1
T
T
x x
T T
  

3
2
1 1
1 1
1
1 1 1
T
T
x
x
T x T x x

 
   
 
  
 
, where
2
1
T
T

1
1
1
( ) =(/)
1
k
x x
k
a c
x
P P
x

The equation may be solved with the Newton Raphson method.
1
1
1
( ) =(/)
1
k
x x
k
a c
x
P P
x

1
1
1
( ) ( ) -(/) 0
1
k
x x
k
a c
x
f P P
x

 

 

1 1 1
1 1 1 1
( ) (1 ) ( ) ( )
1 1 1 1 1
x x x x x x
df x x x x x
x x x
d x x x x x
   

  

    
    

    

    

1
1
1
1
( ) -(/)
1
1 1
/
( )
1 1
k
x x
k
a c
new old
x x
x
P P
f
x
x x
df d
x
x x

 
 

 

  
 
 


 
 

Cold air fraction
0
0.2
0.4
0.6
0.8
0 0.2 0.4 0.6 0.8 1
x
T2/T1
Pr=2
Pr=5
Pr=10

Fig. 2.4.1 Temperature ratio of vortex tube

For any given value of x,

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㈱ 2 1 1
⠩ (1 )
L p p
Q m c T T mc T x

   
 

1 1
(1 )
L
p
Q
x
mc T

  

is a diemnsionless index to show the amount of cooled air
produced.

0
0.1
0.2
0.3
0.4
0.5
0.6
0 0.2 0.4 0.6 0.8 1
T2/T1
q

Fig. 2.4.2 Temperature ratio and cooling load of vortex tube

Efficiency of a vortex tube is defined as
1
1
a a
s
s

 

, where
a

is the actual
temperature ration, and
s

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⠲⸴⸳⸴⤮⁃佐⁯映癯牴數⁴畢攠

    周攠睯牫⁴桡琠楳⁲敱畩牥搠楮⁴桥⁳散T 湤⁬慷⁩猠晲潭⁴桥⁣潭p牥獳潲⁴桡琠楳⁵獥搠

㈱ 2

a
m h h
COP
W

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
1
1
1 1
0
1
1
k
k
a p
c
P
W m c T
P

 
 
 
 
 
 
 
 
 

For the best case, the process is isentropic, and the COP becomes
1
2 1 2
1
1
1
1 1
0
1
(1 ) ( )
( )
1
1
1 ( )
1
1
1
x
x
c
p
k
x x
k
p
c
x
x
m c T T
x
COP
x
P
x
m c T
P

 

 

   

 
   

 
 
 
 
 
 

------------------------------------------------------------------------------------------------------
Assignment 2.11 : 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.
-----------------------------------------------------------------------------------------------------

(2.4.4). Perpetual Motion Machines – human’s dream to break the second law.

It is customary to classify perpetual motion machines as follows:
1.

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.

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
(see also entropy). Note that such a machine is different from real heat engines (such
as car 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.

The first law: You can not get something from nothing.

The second law: You can get even less from something.

The following examples are taken from the web site
http://www.lhup.edu/~dsimanek/museum/

Buoyancy motor #1

A J-shaped tube A, is open at both ends but tapers at the lower end. A well-greased
cotton rope C passes over the wheel B and through the small opening of the tube with
little or no friction, and also without leakage. The tube is then filled with water. The
rope above the line WX balances over the pulley, and so does that below the line YZ.
The rope in the tube between these lines is lifted by the water, while the rope on the
other side of the pulley between these lines is pulled downward by gravity.

Comments: There is no buoyancy acting on the rope because the pressure on every
section of the rope is balanced by each other.
Buoyancy motor #2
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 leakproof) 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.

Comments: The pressure force exerting on the surface of sphere passes through the
center of sphere such that no moment is produced.

Buoyancy motor #3
The main drum is filled with a liquid. In it are round chambers filled with air (or a
vacuum) and connected by rods to the weights outside. The rods slide in frictionless
leakproof seals, of course. When in position 1, the buoyancy of the lower sphere is
enough to lift the weight to its highest position. If the drum is now pushed so it
moves counter clockwise, the weight stays at this large radial distance at least until it
has rotated 90°.

During the next quarter turn the weight has a large lever arm. At the end of this
quarter turn, position 3, the air chamber rises to the top of the drum, and the weight is
now is at its smallest radial distance, (and smallest lever arm) where it stays for the
next quarter turn. During the last quarter turn the air chamber's buoyancy causes the
weight to rise until it is at its largest radius.

Comments: The floating ball is not able to rotate the cylinder.
Length of bar: 2(R-r)
Weight of ball: m
W

Weight of water expelled by the ball: m
L

Since the bar will be lifted upwards in the vertical position 1, the weight of ball
should be less than that the water diplaced by the ball inside cylinder.
L W
m m

The loss the potential energy when the cylinder rotates from position to position 3 is
2 ( ) 2 ( 2( ) )
d L W
E m R r g m R R r r g       

The loss of potential energy is converted to kinetic energy or work output if the
cylinder rotates at a constant speed.
The energy required to rotate the cylinder from position 3 to position 1 is
2 ( ) 2 ( )
u L W
E m R r g m R r g    

The net energy output is
2 ( ) 2 (3 ) 2 ( ) 2 ( )
d u L W L W
E E E m R r g m R r g m R r g m R r g            

2 2 6 2 2 2 2 2
L L W W L L W W
g m R m r m R m r m R m r m R m r        

4 4 ( )( ) 0
L L W W W L
g m R m r m R m r g R r m m        

Since the net output energy is negative, it is not possible that the cylinder could run
by itself.

Buoyancy motor #4
The sealed container has two vertical tubes. The right one contains a liquid (blue)
such as water, and a very light ball (red), much lighter than the liquid. As usual we'll
let you use a liquid with zero viscosity.
Two "gates" G1 and G2 are made like iris diaphragms that can open and close
quickly. They are, of course, watertight when closed.
Now we all know that when a light object, like a cork, is underwater, then released, it
pops to the surface and can even pop above the surface. We take advantage of that
fact. Our machine, with its viscosity-free liquid, should allow even greater speed at
the top. The machine is started with the ball at the bottom. As it rises, a high-tech
sensor quickly opens gate G1 to let it through, closing the gate immediately, and
then
opening gate G2 in time for the ball to pass through.
Since one of the gates is closed at all times the water levels are maintained. The ball
pops above the surface with some momentum, and the curved top of the apparatus
deflects it to the other tube, where it falls, gaining speed and momentum in the fall,
enough so that it goes under the liquid surface there and is bumped over into the right
tube, where, of course, it begins to rise. This should go on forever, gaining speed
each cycle.

Comments: In the limiting case that the viscous force is not considered, the ball will
rise and fall cyclicly with the same initial velocity.
Initial velocity at G1: v
1

Rising speed at G2: v
2

Falling speed at G1: v
3

Falling speed at G2: v
4

Weight of ball: m
b

Weight of water expelled by the ball: m
L

L b
m m

Assume that initially the ball stands still in water. The kinetic energy of the ball is
obtained by conversion of potential energy of the wall expelled by the ball.
2
2
1
2
b L b
mV m hg m hg 

The ball rises upwards at G1 position. When the ball falls back again at G1 position,
it retains its velocity if no friction occurs.
3 2
V V

When tha ball falls to position G2, it is accelerated to the velocity v
3
.
2 2
4 3
1 1
2 2
b b b L
mV mV m hg m gh  

Since the ball falls back to its original height, the kinetic energy actually is gained
from the loss of potential energy of the water expelled by the ball. The level of
water becomes lower as the ball leaves the water.
However, when the ball dives into water again and returns to its original place at G1,
water has been expelled by the ball and the level of water gets back to original height.
The kinetic energy of the ball has been totally converted to potential energy of water.
The velocity of the ball becomes zero and the whole process resumes again.

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(2.4.5). 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
2
2
2
( ) 4 exp( )
2
kT mv
f
v dv v dv
m kT

 
 
 
 

Mean speed:
0
8
( )
RT
v vf v dv

 

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
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
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?

(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.

0
lim( ) 0
T
T
s

 

This law provides an absolute reference point for the determination of entropy. The
entropy determined relative to this point is the absolute entropy.

0
(,)
T
p
dT
s T p c
T

0
(,)
m b
m b
T T
T
m b
p p p
m b
T T
dT L dT L dT
s T p c c c
T T T T T
    
  

Entropies at 25

and 1 atm (
0
/
m
S R
)
CH
4
22.39 H
2
15.705 N
2
23.03
CO 23.76 H
2
O(l) 8.41 O
2
24.66
CO
2
25.70 H
2
O(g) 22.70 NH
3
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
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.