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

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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.

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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

(2.2.3.1). Steady state steady flow system

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.

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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?

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(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.

(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.

adb

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

(2.3.4.1). Adiabatic process

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

steady state

，

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

research paper had been published by Rudolph Hilsch.

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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捯潬敤楲⁰牯摵捥搠楮c 獥湴牯灩挠灲潣敳猠楳

㈱ 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

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瑯慩獥⁴桥⁰牥獳畲攠潦⁴桥湬整楲漠 瑨慴⁴桥⁶潲瑥砠瑵扥 礠灥牦潲y瑳畮捴楯渠

周攠䍏倠潦⁴桥⁶潲瑥砠瑵扥慮攠數灲 敳獥搠慳⁴桥慴楯映瑨攠慭潵湴映

捨楬汥搠慩爠灲潤畣敤⁴漠瑨攠睯牫 ⁴桡琠獴攠摯湥Ⱐ椮攮s

㈱ 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.

Radius of cylinder: R

Radius of ball: r

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.

People’s dream to generate power without ant cost

20071211(yahoo.com)磁能永動機…▲唯一不違反能量守恆第一

二定律★

http://tw.myblog.yahoo.com/chinlingg/article

﹖

mid=7&prev=10&next=5

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

.

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

［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。

(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.

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