chapter two the thermodynamic laws - 國立中興 …...one kg of air at 1 bar and 300k is...

Advanced Thermodynamics, ME dept. NCHU 頁 1

Chapter Two

The Thermodynamic Laws

Update on 2017/9/19

(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 BT T , and

B CT T , then A CT 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 20th century, long after the first three laws were already

widely in use and named as such, hence the zero numbering.


(2.2)、The first law of thermodynamics

The first law of thermodynamics is an expression of the universal law of

conservation of energy, and identifies heat transfer as a form of energy transfer.

The increase in the internal energy of a thermodynamic system is equal to the amount

of heat transfer added to the system minus the work done by the system on the

surroundings.

Q U W

History

The first explicit statement of the first law of thermodynamics was given by Rudolf

Clausius, who was a German physicist and mathematician, in 1850. The original

statement of the first law was "There is a state function E, called ‘energy’, whose

differential equals the work exchanged with the surroundings during an adiabatic

process."

(2.2.1). Isolated system

If a system exchanges neither mass nor energy with its environment, its energy

is invariant with time.

.U Const

If the system is composed of several parts with different temperature, the

conservation of internal energy is

.A A B BU 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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Air Vacuum

3 bars

25℃


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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.

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Assignment 2.1

An insulated rigid chamber is divided into two equal parts with a diaphragm. The

left part is filled with air at 25℃ and 3 bars, and the right part is filled with air at

50℃ and 1 bar. The diaphragm ruptures and the air mixes together. Find the final

temperature of the air in the chamber.

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(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

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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

1kg 400 L

3 bars 2.5 bars

25℃ 300℃

Air Air

3 bars 1 bar

25℃ 50℃


the exponent of 1.6. Assuming that air is an ideal gas with constant heat capacity,

determine the work and heat transfer during these processes.

P

V

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Example

A rigid chamber with a volume of 10 liters is divided into two parts of the same

volume with a diaphragm. The diaphragm does not conduct heat, but can move

without friction. The left part is filled with CO2 and the right part is filled with N2.

Both sides are at 300 K and 100 kPa. The right side is heated gradually until N2

expands to a volume of 7 liters while the left side is insulated during this process.

Assuming both N2 and CO2 are ideal gases with constant specific heats, find the heat

transfer during this process.

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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.

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CO2 N2

100 kPa 100 kPa

300K 300K

CO2 N2

3L 7L


(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

dmm m

dt

2 2

( ) ( )2 2

i ei i i e e e

V dE VQ m h gz m h gz W

dt

(2.2.3.1). Steady state steady flow system (SSSF)

0dm

dt

i em m m

2 2

( ) ( )2 2

i ei i i e e e

V VQ m h gz m h gz W

2 2

2 2

i ei i e e

V Vq h gz h gz w

i eq h h w

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Example

The pressure of saturated liquid R134a at 10 bars declines to 1 bar as passing through

a throttle valve. If the variation in kinetic energy is neglected, find the quality of

R134a.

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Assignment 2.3

Air at 25 ℃ and 200 kPa flows through an insulated throttle valve and drops to 100

kPa. Calculate the final temperature.

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(2.2.3.2). Uniform state steady flow system (USUF)

2 1 i em m m m

,i i e em m dt m m dt

2 2

2 1( ) ( )2 2

i ei i i e e e

V VQ m h gz E E m h gz W

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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.

2 1 im m m

2 2 1 1i im h m u m u

2 1 2 2 1 1( ) im m h m u m u

2 1 2 12 1

2 1 2 1

( ) p i v v

PV PV PV PVc T c T c T

RT RT RT RT

2 1 2 12 1

2 1 2 1

( ) i

P P P PkT T T

T T T T

2

50 5( )1.4 298 50 5

298T

2T = 401 K

22

2

PVm

RT =4.343 kg

11

1

PVm

RT =0.585 kg

2 1m m m = 3.758 kg

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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℃.


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Assignment 2.4

10 kg of water at 25 ℃ is contained in a vessel with the volume of 100L. A valve is

installed on top of the vessel. This valve is preset at the pressure of 500 kPa. This

vessel is heated on a stove. When the pressure of steam reaches 500 kPa, the valve

opens to keep the pressure inside vessel constant. Calculate the amount of heat

transfer when 5 kg of water remains inside this vessel.

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(2.2.3.3). Uniform state uniform flow (USUF)

Flow out of a system adiabatically through a small hole:

e

dmm

dt

( )

0e e e e

d mu du dmm h m u m h

dt dt dt


0v

dT dm dmmc u h

dt dt dt

v

dT dmmc RT

dt dt

1 1

v

dT R dm

T dt m c dt

1

2 2 2

1 1 1

v

Rk

cT m m

T m m

2 2

1 1

k

P m

P m

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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.

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Example

A rigid chamber with a volume of 10 liters is divided into two parts of the same

volume with a diaphragm. The left part is filled with air at 25℃ and 3 bars, and the

right part is in vacuum. A small hole is drilled on the diaphragm to let air leak into

the right part until pressures on both sides are equal. Find the final temperature of

air in the right part.

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Assignment 2.5

A bottle containing 100 liters air at 1 bar and 25℃ is filled adiabatically with

compressed air at 25 ℃ until a pressure of 50 bars is reached. The air in bottle is

then cooled to 25℃ without leakage. The valve is open accidently until pressure

reaches 10 bars before it is closed again. If the bottle should be filled with air at 50

bars and 25℃, what the final pressure should be if compressed air at 25℃is used

Air Vacuum

25℃

3 bars


again to refill the bottle?

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The hot water system

( )T T t ， ( )m m t

i e

dmm m

dt

v i i e e

dE dT dmmc u m h m h Q S W

dt dt dt

im

iT

aT

S

em

eT

(1). Mixing process

0Q ， 0S

0dm

dt ， i em m m

i ih cT ， e eh cT

( )v i i e e i e

dTmc m h m h mc T T

dt

eT T


Let iT T

0d m

dt m

Let1

m

m

m , where

m is the time constant of mixing process.

10

m

d

dt

1m

t

c e

At 0t , 0T T ,

0 0iT T

0m

t

e

As t , 0 , iT T

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Example

A tank contains 1000 kg of cold water at 30℃. Hot water at 80℃ is pouring into

this tank at the rate of 1 kg/sec. At the same time, mixed water is flowing out of the

tank at the same rate. Find the temperature of water inside tank 10 minutes after

this mixing process.

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Assignment 2.6

A tank contains 200L of hot water at 60℃. Hot water is consumed at the rate of 6

LPM while the same rate of cold water at 25℃ is supplemented into this tank to keep

the volume of water constant. An electric heater is actuated automatically when the

water temperature drops to 45℃. If heat loss is not considered, find out when the

heater will be turned on.

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(2). Accumulation process

0Q ， 0S

i e

dmm m m

dt


i em m m

0m m m t

dm mdt ，dm

dtm

( )i i i

dTmc m c T T m c

dt

0

id m

dt m m t

0

i i

i

d m dm m dm

m m t m m m

0

0 0

i im mm mm m

m m

im

m

, the ratio of inlet flow to net flow

i em m , 0m , 0 , is decreasing.

i em m , 0m , 0 ， is also decreasing.

0

0

m

m

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Example



tank at the rate of 0.5 kg/sec. Find the temperature of water inside tank 10 minutes

after this mixing process.

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Assignment 2.7



tank at the rate of 1.5 kg/sec. Find the temperature of water inside tank 10 minutes

after this mixing process.


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(3). Heating process

v i i e e

dE dT dmmc u m h m h S

dt dt dt

( )i

dT m ST T

dt m mc

m

d S

dt mc

1

1m

t Sc e

mc

Let f

S

mc , the equilibrium temperature of heating process

0t ， 0 ，則 1 0 fc

0

m

t

f

f

e

As t , f , the system has reached the final equilibrium temperature.

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Example

A tank contains 1000 kg of warm water at 45℃. Water is consumed at the rate of 1

kg/sec while cold water at 25℃ is supplemented into the system at the same rate.

An electric heater is installed inside the tank with the capacity of 250 kW. The

heater has to be turned off as the water temperature has reached 50℃. Find out

when the heater will be turned off.

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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.

adbQ 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 A(P1, V1) can be reached from state B (P2,

V2) with an adiabatic process. There exists another state C(P3, V2) with the same

volume as sate B, but the values of pressure are different. Suppose that state C can

also be reached from state A with an adiabatic process. It is noted that the three

processes of AB, BC, and CA compose a cycle. Since AB and CA are adiabatic,

there is no heat transfer along these two processes. Heat transfer must occur during

the process BC. However, process BC does not output work since the volume

keeps the same during the process. As a result, the cycle absorb heat during the

process BC and then converts heat totally to work during the process AB and CA.

P A (P1, V1)

C (P3, V2)

B (P2, V2)

V


If the state of C exists, then the Kelvin Plank statement of the second law of

thermodynamics is violated because the system absorbs heat with single reservoir

and converts this heat to work. That is, if axiom II of Caratheodory can be violated,

then the Kelvin Plank statement of the second law of thermodynamics can also be

violated.

In order not to violate the second law of thermodynamics, the state of C must not

exist. As a result, in the immediate neighborhood of state A, state C can not be

reached from A by an adiabatic process.

(2.3.3). Entropy Rate Balance of Isolated system

Principle of increase of entropy

0net sys envS S S

Reversible process: rev

QdS

T

Irreversible process: Q

dST

For all processes: Q

dST

, TdS Q

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

0TdS Q ， 0S

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 2A A B B A A B BU m u m u U m u m u

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


1 12

A vA A B vB B

A vA B vB

m c T m c TT

m c m c

2 2 2 2

1 1 1 1

ln ln ln( ) ( )A vA B vBm c m c

A vA B vB

A B A B

T T T TS m c m c

T T T T

In the case that A vA B vBm c m c

22

A BT TT

2

2 2 ( )1

4

A B

A B A B

T T T T

T T T T

0S

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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 vAm c 10 × 0.447 = 4.47 kJ

B vBm c 100 × 4.186 = 418.6 kJ

Tav = 27.9 ℃

2 2

1 1

ln lnA vA B vB

A B

T TS m c m c

T T = -2.8792 + 4.05393 = 1.1747 kJ/K

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Assignment 2.8

An iron block of 10 kg at 300 ℃ is cooled in an open air at 25 ℃. Calculate the

entropy change.

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(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.

QS

T

，

QS

T

j

j j

QdS

dt T

0

1 1( )net sys evS S S QT T

0

1 1( ) 0j

jnet

dSQ

dt T T

(2.3.4.1). Adiabatic process

In an adiabatic process, we have 0jQ . As a result, the entropy increase of

the system is 0net

dS

dt

.

0Q U W

U W

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Example

An insulated chamber with a volume of 100 liters is filled with air at 100 kPa and

25℃. A peddle rotates inside the chamber, doing work on the air until its

temperature has been raised to 50℃. Suppose air is an ideal gas with constant

specific heat, find the amount of work being done and the net entropy change of the

system. The atmospheric temperature is 25 ℃.


PVm

RT = 0.1169 kg

2 1 0Q U U W

1 2 1 2( )vW U U mc T T = 2.095 kJ

2 2

1 1

ln lnv

T VS mc mRT

T V = 6.757×10-3 kJ/K

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Example

A chamber with a volume of 100 liters is filled with air at 100 kPa and 25℃. Heat

is added to the chamber to raise the temperature of air to 50℃. Find the amount of

heat transfer and the net entropy change of the system. The atmospheric temperature

is 25 ℃.

Discussion: Why is that the entropy generation by work is higher than that by heat

transfer?

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Assignment 2.9: An insulated chamber with a volume of 0.1 m3 is filled with air at

100 kPa and 298 K. A peddle rotates inside the chamber, doing work on the air.

If the amount of work being done is 1 kJ, calculate the net entropy change of the

system.

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Assignment 2.10: There are two blocks of iron. One is 10 kg weight at 300 ℃.

The other is 20 kg weight at 100℃. The atmospheric temperature is 25℃ and the

heat capacity of iron is 450 J/kg-K. (20%)

(1). If these two iron blocks are contacted to each other until an equilibrium

temperature is reached, find out the entropy generated in this process?

TH


(2). If the high temperature block is used as the heat source, and the low temperature

block is used as the heat sink, how much work can be delivered theoretically?

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Assignment 2.11: A rigid chamber with a volume of 20 liters is divided into two

parts with a diaphragm. The left part is 2 liters, and is filled with CO2 at 300K and

50 bars. The right part is 18 liters and is kept at vacuum. If the diaphragm ruptures

suddenly and the whole chamber is filled with CO2, find the net entropy change of

this process and the final pressure. Assume that CO2 is an ideal gas with constant

heat capacity, and the value of constant volume heat capacity is vc = 0.653 kJ/kg-K.

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Assignment 2.12 : There are two bottles of CO2, one is at 100 bars, 300K, and the

other one is at 10 bars, 300K. They both have the same volume of 100L. These two

bottles are connected to each other with a valve. Now open the valve and let CO2

flow from the high pressure bottle to the low pressure bottle until an equilibrium

pressure is reached. The whole process may be assumed to be adiabatic. Assume

that CO2 is an ideal gas with constant heat capacity. (20%)

(1). Find the final temperature in each bottle.

(2). Find the net entropy change of this process.

capacity is pc = 0.842 kJ/kg-K. (20%)

10 bars 100 bars

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(2.3.4.2). Polytropic process

.nPV const Work in a polytropic process

1 1 2 2

1( )

1revW PdV PV PV

n

Heat transfer in a polytropic process:

11 1 11 1 2 2

2

1( ) [1 ( ) ]

1 1 1 1

nk n k n Pv vq Pv P v

k n k n v


1 1 1 1

2 2 2 2

( 1)ln ln [ ( 1) ]ln ln1

p p

v v v n k vs c n Rn c n Rn R

v v v k v

1 2v v ，n k ， 0s ，heat absorption

2 1v v ，n k ， 0s ，heat rejection

1 2v v ，n k ， 0s ，heat rejection

2 1v v ，n k ， 0s ，heat absorption

11 1 1

0 0 2

1( ) [1 ( ) ]

1 1

n

en

q k n Pv vs

T T k n v

11 1 1 1

0 2 2

1( ) [1 ( ) ] ln

1 1 1

n

net

k n Pv v n k vs R

T k n v k v

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

℃ to a volume of 0.5 m3.in a polytropic process with n=1.3.

P1 = 100 kPa，T1 = 298 K，v1 = 0.8553 m3/kg

P2 = P1 × (v1 / v2 )1.3 = 200.94 kPa

T2 = T1 × (v1 / v2 )0.3 = 350.1 K

Δssys =-0.03843 kJ/kg-K

q= -12.455 kJ/kg

Δsen =0.04179 kJ/kg-K

Δsnet =0.00336 kJ/kg-K

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Example： CO2 in a cylinder is at 300K and 50 bars with a volume of 2 liters. It

expands to a volume of 20 liters and the expansion process is a polytropic process

with n=1.1. Assume that CO2 is an ideal gas with constant heat capacity. Calculate

the work of compression and the associated heat transfer.

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Assignment 2.13 : Calculate the net entropy change and the work to compress 1 kg

of air at 100 kPa and 25℃ adiabatically to a volume of 0.3 m3.in a polytropic

process with n=1.5.

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(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

QdSm s m s

dt T


Rate of entropy change = Rate of entropy transfer +Rate of exchange + Rate of

entropy production

0

1 10j

jnet j

dSQ

dt T T

steady state， 0dS

dt

0j

i i e e

j j

Qm s m s

T

for single input and single output system

0j

i e

j

Qms ms

T

j

e i

j

qs s

T

outlet entropy = inlet entropy + entropy transfer + entropy generated

In an adiabatic process, we have 0jQ . As a result, the entropy increase of

the system is 0net

dS

dt

.

e i is s s

The entropy at the outlet of a steady system is always greater than that ath the

inlet.

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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?

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Assignment 2.14 : 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 .nPV const , the entropy generation is as the following.


Work in a polytropic process 1

1

1 1 2 21( )

1 1

nnn

rev

n

n nW VdP dP P PV PV

n nP

Heat transfer in a polytropic process:

i eq h h w

2 1 2 2 1 1 1 1 2 2( ) ( ) ( )1 1

p

k nq c T T w Pv Pv Pv Pv

k n

1

1 21 1 2 2

1

1( ) [1 ( ) ]

1 1 1 1

n

nk n k n RT P

q Pv P vk 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 Ps c R R R

T P k n P k n P

2 1P P ，n k ， 0s ， 0q heat absorption

2 1P P ，n k ， 0s ， 0q ，heat rejection

2 1P P ，n k ， 0s ， 0q ，heat rejection

2 1P P ，n k ， 0s ， 0q ，heat absorption

1

1 2

0 0 1

1( ) [1 ( ) ]

1 1

n

nen

q k n RT Ps

T T k n P

1

1 2 2

0 1 1

1( ) [1 ( ) ] ln

1 1 ( 1)

n

nnet

k n RT P n k Ps 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 m3.in a polytropic process with n=1.3.

P1 = 100 kPa，T1 = 298 K，v1 = 0.8553 m3/kg

P2 = P1 × (v1 / v2 )1.3 = 200.94 kPa

T2 = T1 × (v1 / v2 )0.3 = 350.1 K

Δssys =-0.03843 kJ/kg-K

q= -12.455 kJ/kg

Δsen =0.04179 kJ/kg-K

Δsnet =0.00336 kJ/kg-K

------------------------------------------------------------------------------------------------------

Assignment 2.15 : CO2 is compressed from 1 bar, 300K to the pressure of 100 bars.

The process may be expressed as a polytropic process with n=1.2. Find the work

required to execute the compression process and the associated heat transfer. Assume

that CO2 is an ideal gas with constant heat capacity, and the value of constant volume


heat capacity is pc = 0.842 kJ/kg-K. (20%)

100 bars

1 bar 300K Q

------------------------------------------------------------------------------------------------------

(2.3.6). Efficiency of real process

Compressor:

2 1

2 1

s sc

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

22 1

1

11 1

k

k

c

PT 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

at

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

22 1

1

1 1

k

k

t

PT 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.

------------------------------------------------------------------------------------------------------

Assignment 2.16 : Air is compressed adiabatically in a two-stage compression

process. The inlet condition is 100 kPa and 25℃, and the final pressure is 10 MPa.

The efficiency of the first stage compressor is 80%, and the efficiency of the second

stage compressor is 90%. An intercooler is placed between these two compressors

to cool down the compressed air to the temperature of 100℃. Assume that air is an

ideal gas with constant heat capacity, and the environmental temperature is 25℃.

P2=? 100℃

10MPa

100 kPa

25℃ η=80% η=90%

(1). Find the intermediate pressure between these two compressors such that the total

compression work is minimal.

(2). Calculate the power required for the compression process if the inlet air flow is

100 L/min.

------------------------------------------------------------------------------------------------------

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

2 222 1 1

1

1 11

2 2

k

k

n p

PV 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 is

90% and the outlet pressure is 100 kPa.

------------------------------------------------------------------------------------------------------

Assignment 2.17 : In a jet engine, air is compressed from 25 ℃ and 100 kPa to the

Intercooler


pressure of 1 MPa, and then heated to 1200 K in the combustor. If the compressor

efficiency is 85%, the turbine efficiency is 90%, and the nozzle efficiency is 95%,

calculate the thrust of engine assuming that air is an ideal gas with constant heat

capacity.

2 3

1 4 5

------------------------------------------------------------------------------------------------------

Fluid nozzle:

2 21 1

2 2i i e eh V h V

2 21 1

2 2i i i i e e e eu Pv V u Pv V

For a perfect nozzle with isentropic process, the entropy change is zero.

ln 0e

i

Ts c

T ,

es iT T

2 21 1

2 2i i i i es e e esu Pv V u Pv V

2 21 1( )

2 2es i e iV P P v V

2 21 1/

2 2e sV V

2 2 21 1 1[( ) ]

2 2 2i i i i e e e es e e e i e iu Pv V u Pv V u Pv P P v V

2 2 2 21 1 1 1[( ) ](1 )

2 2 2 2e i i i i e e es es i e iu u Pv V Pv V V P P v V

21[( ) ](1 )

2e i i e iu 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

23 1( ) [( ) ](1 )

2e i e i i e iu 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. The area ratio of nozzle is 10, and the inlet velocity is 10 m/sec.

------------------------------------------------------------------------------------------------------

Assignment 2.18 : 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 did the distinguished physicists say about the second law?

[A law] is more impressive the greater the simplicity of its premises, the more

different are the kinds of things it relates, and the more extended its range of

applicability. Therefore, the deep impression which classical thermodynamics

made on me. It is the only physical theory of universal content, which I am

convinced, that within the framework of applicability of its basic concepts will

never be overthrown.

Albert Einstein, quoted in M.J. Klein, Thermodynamics in Einstein's Universe,

in Science, 157 (1967), p. 509.

The law that entropy always increases -- the second law of thermodynamics --

holds I think, the supreme position among the laws of Nature. If someone

points out to you that your pet theory of the universe is in disagreement with

Maxwell's equations - then so much worse for Maxwell equations. If it is

found to be contradicted by observation - well these experimentalists do

bungle things sometimes. But if your theory is found to be against the second

law of Thermodynamics, I can give you no hope; there is nothing for it but to

collapse in deepest humiliation.

Sir Arthur Stanley Eddington, in The Nature of the Physical World.

Maxmillan, New York, 1948, p. 74.


(2.4.2). Statistical insight of the second law

Can you fix a cup of Latte by mixing one half cup of coffee and one half cup of milk?

Can we stir a cup of Latte into one half of coffee and one half of milk?

2

1

1!1!2

2!C =1

4

2

2!2! 12

4! 3C

6

3

3!3! 12

6! 10C

8

4

4!4! 12

8! 35C

100 30

50

50!50!2 9.9 10

100!C

1000 301

500

500!500!2 1.9 10

1000!C

10000 3010

5000

5000!5000!2 1.1 10

10000!C

If a cup of Latte contains 6×1024 molecules, what is the probability that we can

separate it into coffee and milk by keep stirring?

Stirling’s formula:


ln ! lnN N N N

N =6×1024 , ln N 57.05, ln !N 3.36×1026 263.36 10!N e

263.36 10

/ 2

( / 2)!( / 2)!2

!

N

N

N NC e

N

The validity of the second law is based on the statistics. The probability of a system

which is composed of millions of particles to violate the second law is zero.


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

(2.4.3.1). The principle of vortex tube

Fig. 2.4.3.1 A real vortex tube

The vortex tube is a device that produces hot and cold air streams

simultaneously at its two ends from a source of compressed air. Unlike the

traditional compression type cycle that requires several components, a vortex tube is

very simple with no moving parts.

Vortex tube was invented by French engineer Georges Ranques at 1928.

However, it was until 1946 that people started to show interest in vortex tube after a

research paper had been published by Rudolph Hilsch.

Fig. 2.4.3.2 Structure of vortex tube

A cold orifice is placed at the centre of the left end with a suitable sized hole.

Compressed air is introduced into the tube through a tangential inlet nozzle which is

located near the left end. At the right end, a conical valve is inserted to confine the

exiting air to outer regions and restrict it to the central portion of the tube. The

tangential flow imparts a vortex motion to the inlet air, and creates a cold stream in

the left end and a warm stream in the right end.

There is no theory to give a satisfactory explanation of the vortex tube


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 im，the outlet cold air is

cm , and the outlet warm air

is hm . The conservation of mass results in the relationship amnong the flow rates

as folliwing.

c h im m m

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.

( )c h i c c h hm m h m h m h

Assume that air is an ideal gas with constant heat capacity, then we have a

relationship among the temperatures of inlet and outlet flows.

i i c c h hmT m T m T

Define the cold fraction as /c ix m m , the temperature relationship can be

expressed as

(1 )i c hT xT x T

In which iT is the inlet temperature,

cT is the outlet cold temperature, and hT

is the outlet warm temperature.


For any given cold fraction value, the cold temperature and the warm temperature

are in linear relationship.

For example,

0.3x , 0.3 0.7c h iT T T

0.5x , 2c h iT T T

0.6x , 0.6 0.4c h iT T T

The worst case is that no cooling effect occurs in which the cold temperature and

the warm temperature are equal to the inlet temperature.

1 c hT T T

Since 1cT T and

1hT T , all possible solutions would be lying on the first quarter

of the point of 1 c hT T T , as shown in the figure below. However, not all points

lying on the segment OA are possible solution, because there is another restriction

set by the seond law.

Tc

O

T1

x=0.5

x=0.3 x=0.6 x=0.7

A

T1 Th

The entropy change of the process is

( ) ( )c c h h i i c c i h h iS m s m s m s 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

ln( / ) ln( / )c i p c i o is s c T T R P P

ln( / ) ln( / )h i p h i o is s c T T R P P

where iP is the upstream pressure, and

oP is the down stream pressure pressures.

The total entropy change rate can be expressed as

( ) ( )c c h h i i c c i h h iS m s m s m s m s s m s s

And the specific entropy change is as the following.

ln( / ) ln( / ) (1 ) ln( / ) ln( / )p c i o i p h i o i

i

Ss x c T T R P P x c T T R P P

m

The entropy change can then be expressed as

1

1ln( / ) ( / ) ln( / )k

x x kp c i h i o is 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.

The lowest temperature that can be reached is no generation of entropy.

0s

1

1ln( / ) ( / ) ln( / )k

x x kc i h i o iT T T T P P

1

1( / ) ( / ) =( / )k

x x kc i h i o iT T T T P P

------------------------------------------------------------------------------------------------------

Example: A vortex tube runs on the cold fraction of 0.5. The inlet temperature is

300K. Find the cold temperature and the hot temperature for the pressure ratio of 2.0,

3.0, 4.0, and 5.0.

0.5x 4

7( / )( / )=( / )c i h i o iT T T T P P


2c h iT T T 4

7( / )(2 / )=( / )c i c i o iT T T T P P

c

i

T

T

4

7(2 )=( / )o iP P 4

2 72 ( / ) 0o iP P 4

47

72 4 4( / )

1 1 ( / )2

o i

o i

P PP P

Pressure ratio 2.0 3.0 4.0 5.0

0.4281 0.3172 0.2603 0.2245

Tc 128 95 78 67

Th 472 505 522 533

Tc

O

T1

x=0.5 PR=2.0

A

T1 Th

(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 ka cT T T T P P

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

following.

32

1 1

(1 ) 1TT

x xT T

3 2

1 1

1 11

1 1 1

T T xx

T x T x x

, where 2

1

T

T

1

11( ) =( / )

1

k

x x ka c

xP P

x

The equation may be solved with the Newton Raphson method. 1

11( ) =( / )

1

k

x x ka c

xP P

x

1

11( ) ( ) -( / ) 0

1

k

x x ka c

xf P P

x

1 1 11 1 1 1( ) (1 ) ( ) ( )

1 1 1 1 1

x x x x x xdf x x x x xx x x

d x x x x x

1

1

1

1( ) -( / )

11 1/

( )1 1

k

x x ka c

new old

x x

xP P

f xx xdf d

xx x

------------------------------------------------------------------------------------------------------

Assignment 2.19 : A vortex tube runs on the pressure of 2.8. The inlet temperature

is 300K. Find the cold temperature and the hot temperature for the cold fraction of

0.3, 0.5, 0.6, and 0.7. 1

11( ) =( / )

1

k

x x ka c

xP P

x

=0.7451

x 0.3 0.5 0.6 0.7

Tc

Th

------------------------------------------------------------------------------------------------------


Cold air fraction

00.2

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. 2.4.3.3 Temperature ratio of vortex tube

For any given value of x, can be obtained if the pressure ratio is known. The

cooled air produced in an isentropic process is

2 1 2 1 1( ) (1 )L p pQ m c T T m c T x

1 1

(1 )L

p

Qx

m c 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. 2.4.3.4 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 is the theoretical temperature ratio. It is noted that the

entropy change should be positive according to the second law of thermodynamics,

------------------------------------------------------------------------------------------------------

Example: A vortex tube operates with the pressure of 5 bars. The cold fraction is

0.5, and the efficiency is 0.7. Calculate the outlet temperature at the cold side.

------------------------------------------------------------------------------------------------------

(2.4.3.4). COP of vortex tube

The work that is required in the second law is from the compressor that is used

to raise the pressure of the inlet air so that the vortex tube may perform its function

The COP of the vortex tube can be expressed as the ratio of the amount of

chilled air produced to the work that must be done, i.e.

2 1 2( )

a

m h hCOP

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

11 1

0

11

k

k

a p

c

PW m c T

P

For the best case, the process is isentropic, and the COP becomes

For real vortex tube, the COP is

2

2 1 2 1

11

111 1

00

(1 )( )

111

cp

kk

kk

p

c

Tx

m c T T TCOP

PPm c T

PP

For ideal vortex tube, the COP is

1

1

1(1 ) ( )

11

1 ( )1

x x

c

x x

xx

xCOPx

x

------------------------------------------------------------------------------------------------------

Assignment 2.20 :


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.

-----------------------------------------------------------------------------------------------------

Exergy analysis of vortex tube

02 1 0 2 1 0 2 1( ) (1 ) ( )rev H

H

Tw h h T s s q T s s

T (2.4.3.5).

Real performance of vortex tube

------------------------------------------------------------------------------------------------------

Assignment 2.21 : Use the performance data of a commercial vortex tube to calculate

the efficiency as well as the entropy generation.

Pressure

Supply

Cold Fraction %

BAR 20 30 40 50 60 70 80

1.4 34 33 31 28 24 20 16

8 14 20 28 26 46 59

2.8 48 46 42 39 34 28 20

11 18 28 38 50 62 80

4.1 57 55 51 46 40 33 25

14 22 33 44 57 73 92

5.5 63 62 56 51 45 36 28

14 24 35 47 63 80 100

6.9 68 65 61 55 48 39 30

14 25 37 50 66 84 106

8.4 72 69 64 58 50 41 31

14 26 38 52 68 86 108

Temperature drop of cold air,

(°C) in blue

Temperature rise of hot air,

(°C) in red

-----------------------------------------------------------------------------------------------------

(2.4.3.6). Applications of vortex tube

Spot cooling in machining


Spot cooling in welding

Temperature controlled high chair

Intake manifold heating of diesel engine during cold start


(2.4.4). Maxwell’s demon – what happens inside a chamber in which a tiny creature

lives?

Maxwell's Demon is an imaginary creature that the physicist James Clerk

Maxwell created to demonstrate the limitation of the second law of thermodynamics.

Suppose that you have a box filled with a gas at some temperature. This means that

the average speed of the molecules is a certain amount depending on the temperature.

The Maxwell-Boltzmann velocity distribution is 3

22

22( ) 4 exp( )

2

kT mvf v dv v dv

m kT

Mean speed: 0

8( )

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

The lessons we learn from the thermodynamics laws are as the following.

The first law: You can’t get something from nothing.

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

However, it is very tempting to invent a device that may run forever without any

input energy. For past decades of years, many applications have been send to the

bureau of patents claiming that they have the feature of perpetual motion. None of

them has been approved. As a result, the National Institute of Standards and

Technology ( NIST) has announced officially that all applications regarding

perpetual motion machine will be refused.

Perpetual motion machines can be classified as three categories.

The zeroth kind of perpetual motion machine

The first kind of perpetual motion machine

The second kind of perpetual motion machine

Some of the examples shown in the following content are taken from the web site

http://www.lhup.edu/~dsimanek/museum/

(2.5.1). The zeroth kind

The zeroth kind perpetual motion machine is a fault in the interpretation of

mechanics. It was presumed as a perpetual motion machine because no solid

analysis of mechanics was performed.


Water fall that never ends

Magic wheel may rotate forever


Floating ball will rotate forever

A wheel in the form of a perfect sphere or cylinder rotates about a frictionless

horizontal shaft. The left side is in a chamber filled with water, perfect (frictionless

and leakage proof) seals around the rotating wheel prevent the liquid from escaping.

The left side of the wheel therefore experiences an upward buoyant force due to the

liquid it displaces. So that side will rise, and the wheel rotates clockwise.

(2.5.2). The first kind

A perpetual motion machine of the first kind produces strictly more energy than it

uses, thus violating the law of conservation of energy. Over-unity devices, that is,

devices with a thermodynamic efficiency greater than 1.0 (unity, or 100%), are

perpetual motion machines of this kind.

(2.5.2.1). Magnetic generator

People’s dream to generate power without ant cost


磁能永動機稱為第三類機械永動機，它是唯一不違反能量守恆第一二定律。違

反第一定律指永動機可產生自己所需的能量，例如用一台發電機一開始借有電

池起動後，再去帶動更多台發電機來產生電能後一直運轉下去。違反第二定律

是指永動機利用該機械熱庫可重複利用﹐進而到達永續運轉目的﹐事實上機械

所做的功產生熱能會有一半自耗另一半穫得利用所以能量會越用越少.

唯有磁能發電機是屬於自然界所產生的動能，它不需任何外界給予動力本身就

是一個自發體，我們只要懂得利用機械貫性原理就有辦法製造出電能，因此它

既不違反目前科學界所建立的物理原則。所以請不要再給永動機冠上不可能的

名詞。

20071211(yahoo.com)磁能永動機…▲唯一不違反能量守恆第一

二定律★

(2.5.2.2). Water fuel vehicle

自由時報電子報 2008 年 6 月 17 日星期二

水燃料車一公升跑八十公里


〔記者陳英傑／綜合外電報導〕日本 Genepax，推出一套水能量系統（Water

Energy System，簡稱 WES），將水分解成氫和氧產生能量，讓汽車只吃水就可

以發動，雨水、河水甚至茶都可使用，目前共有 120W 及 300W 的原型機，一

公升的水就可以時速八十公里連續行駛一小時，未來量產後，預計價錢將可壓

低到每套系統新台幣十四萬元左右，不過目前上市時間未定。

這款水燃料車與其他廠商一樣，都是將水分解成氫與氧來產生能量，但由

於採用薄膜電導組件，擁有較長的使用壽命。Genepax 執行長平澤表示，只要

有瓶裝水，車子就可連續行駛。Genepax 在日本大阪實地用水來發動並行駛車

子測試，雨水或河水都可使用。

目前這套系統高達兩百萬日圓（約新台幣五十六萬元），不過 Genepax 認

為，未來大量生產後可降至五十萬日圓（約新台幣十四萬元），除有 120W 及

300W 的原型機，還會有 1kW 的系統，將打進家庭用電市場。

------------------------------------------------------------------------------------------------------

Assignment 2.22

Give a comment on the operation of water engine and explain why it doesn’t work.

-----------------------------------------------------------------------------------------------------

(2.5.2.3). Super electrolysis

You may produce more energy than that you input.

Stanley MEYER is the most famous inventor in the 'Super-efficient Electrolysis' field,

and many took inspiration from his work. His original Electrolysis Concept was able

to produce many times as much hydrogen gas as permits the Faraday's Law of

electrolysis, and the Law of Conservation of Energy, then he brought it up to a still

un-understood high level he called 'Thermal Explosive Energy'. We can say that he

was at least 50 years in advance of his time ...


The evidence, then, is that his concept of electrolysis is tapping in another source of

energy, not 'recognized' yet in the classical physics, that must be in another

dimension, outside our three dimensions, and it must be a non thermal form of

energy, because Stan Cell is not producing heat while electrolyzing. There is still not

a complete consensus of the Experts in Zero Point Energy.

(2.5.3). The second kind

A perpetual motion machine of the second kind is a machine which spontaneously

converts thermal energy into mechanical work. This need not violate the law of

conservation of energy, since the thermal energy may be equivalent to the work done;

however, it does violate the second law of thermodynamics.

Note that such a machine is different from real heat engines, which always involve a

transfer of heat from a hotter reservoir to a colder one, the latter being warmed up in

the process. The signature of a perpetual motion machine of the second kind is that

there is only one single heat reservoir involved, which is being spontaneously cooled

without involving a transfer of heat to a cooler reservoir. This conversion of heat into

useful work, without any side effect, is impossible by the second law of

thermodynamics.


(2.5.3.1). Dipping bird

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Assignment 2.23

Explain the principle of work for the dipping bird, and have some comments on

whether it violates the second law or not.

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(2.5.3.2). Brownian ratchet

The simple machine, consisting of a tiny paddle wheel and a ratchet(棘輪), appears

to be able to extract useful work from random fluctuations (heat) in a system at

thermal equilibrium in violation of the second law of thermodynamics.

The device consists of a ratchet that rotates freely in one direction but is prevented

from rotating in the opposite direction. The ratchet is connected by an axle to a

paddle wheel that is immersed in a fluid of molecules at temperature T1. The

molecules undergo random Brownian motion. The device is small enough that the

http://en.wikipedia.org/wiki/Paddle_wheel

http://en.wikipedia.org/wiki/Ratchet_(device)

http://en.wikipedia.org/wiki/Thermal_fluctuations

http://en.wikipedia.org/wiki/Thermal_equilibrium

http://en.wikipedia.org/wiki/Second_law_of_thermodynamics

http://en.wikipedia.org/wiki/Ratchet_(device)

http://en.wikipedia.org/wiki/Paddle_wheel

http://en.wikipedia.org/wiki/Brownian_motion


impulse from a single molecular collision can turn the paddle. Although such

collisions would tend to turn the rod in either direction with equal probability, the

pawl allows the ratchet to rotate in one direction only. The net effect of many such

random collisions should be sufficient for the ratchet to rotate continuously in that

direction. The ratchet's motion then can be used to do work on other systems.

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(2.5.3.3). A sustainable power system

A gas turbine engine is working with a heat pump. Air is compressed with the

compressor, and then heated with a heat pump instead of a burner. Give a comment

on the feasibility of this system.

Q4

Wp

Q5

2 3

Wc Wt

1 4

1

22 1 1

1

1( ) 1

k

k

c p p

c

PW c T T c T

P

1

43 4 3

3

( ) 1

k

k

t p p t

PW c T T c T

P

3 1

1 11 1T p t p

c

W c T c T XX

1

2

1

k

kPX

P

5 3 2( )pQ c T T

http://en.wikipedia.org/wiki/Work_(physics)


4 4

5 5 3

aTQ T

Q T T

5 4 3 2

3

( )(1 )ap p

TW Q Q c T T

T

3 1 3 2

3

1 11 1 ( )(1 )a

net T p p t p p

c

TW W W c T c T X c T T

X T

1t

1c

2 1T T X

3 3 1

1 1 1 3

2

3 3 3 3 31 1 1

1 1 1 3 1 3 1 3

11 1 ( )(1 )

1 1 11 1 2 0

net

p

W T T TX X

c T T X T T

T T T T TT T TX X X X X

T T X T T T X T T X T

(2.5.3.4). CO2 engine

Liquid carbon dioxide is stored in a bottle. It absorbs heat from environment and

vaporizes. High pressure CO2 vapor flows into cylinder as the ball valve is activated

and pushes the piton downwards.

It seems that it may convert heat from atmosphere to work. Does it 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.

0lim( ) 0TT

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

dTs T p c

T

0

( , )m b

m b

T T T

m bp p p

m bT T

dT L dT L dTs T p c c c

T T T T T

Entropies at 25℃ and 1 atm ( 0 /mS R )

CH4 22.39 H2 15.705 N2 23.03

CO 23.76 H2O(l) 8.41 O2 24.66

CO2 25.70 H2O(g) 22.70 NH3 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

http://en.wikipedia.org/wiki/Walther_Nernst

http://en.wikipedia.org/wiki/Zero

http://en.wikipedia.org/wiki/Ground_state

http://en.wikipedia.org/wiki/Degeneracy


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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Example:

Methane is burned with oxygen to form carbon dioxide and water vapor. If all the

reactants and products are at 25℃ and 1 atm, find the entropy generation for 1 kg of

methane.

4 2 2 2 2 22( 3.76 ) 2 7.52CH O N CO H O N

Mixing combustion cooling

Initial state:

4CH : s=22.39×8.314= 186.15 kJ/kmole-K

2O : s=24.66×8.314= 205.02 kJ/kmole-K

2N : s=23.03×8.314= 191.47kJ/kmole-K

186.15×1/10.52+205.02×2/10.52+191.47×7.52/10.52=193.54 kJ/kmole-K

Mixing process:

0

0 0

0

( , ) ( , ) ln

T

p

T

P dTs T P s T P R c

P T

4CH : s=186.15-8.314×ln(1/10.52)= 205.72 kJ/kmole-K

2O : s=205.02-8.314×ln(2/10.52)= 218.82 kJ/kmole-K

2N : s=191.47-8.314×ln(7.52/10.52)= 198.07 kJ/kmole-K

Mixture: s= (205.72×1+218.82×2+198.07×7.52)/10.52= 202.74 kJ/kmole-K

s =202.74-193.54= 9.20 kJ/kmole-K

O2

CH4

CH4+2O2+7.52N2 CO2+2H2O+7.52N2 CO2+2H2O+7.52N2 N2


Combustion process:

Constant pressure adiabatic flame temperature: 2324 K

2CO : 318.35-8.314×ln(1/10.52)= 337.92 kJ/kmole-K

2H O : 272.41-8.314×ln(2/10.52)= 286.21 kJ/kmole-K

2N : 257.42-8.314×ln(7.52/10.52)= 260.21 kJ/kmole-K

Products: s= 272.54 kJ/kmole-K

s =272.54- 202.74= 69.80 kJ/kmole-K

Cooling process:

Final temperature: 298 K

2CO : 213.69-8.314×ln(1/10.52)= 233.26 kJ/kmole-K

2H O : 188.72-8.314×ln(2/10.52)= 202.52 kJ/kmole-K

2N : 191.50-8.314×ln(7.52/10.52)= 194.29 kJ/kmole-K

Cool products: s= 199.46 kJ/kmole-K

s =199.46-272.54= -73.08 kJ/kmole-K

2 2 22 7.52CO H O NQ h h h =802310 kJ

q 76265 kJ/kmole

0

EV

qs

T =255.92 kJ/kmole-K

Net EV syss s s =182.84 kJ/kmole-K=120.22 kJ/kg-K

QI QE

QH

W

QL

T0


Hq =76265 kJ

Net EV syss s s =0

0

LEV

qs

T =73.08 kJ/kmole-K

Lq =21778 kJ/kmole

H Lw q q = 54487 kJ/kmole=35825 kJ/kg CH4

In summary:

Mixing process: s =9.20 kJ/kmole-K

Combustion process: s =69.80 kJ/kmole-K

Cooling process: Net EV syss s s =120.22 kJ/kg-K

Maximum work: w = 54487 kJ/kmole=35825 kJ/kg CH4

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chapter two the thermodynamic laws - 國立中興 …...one kg of air at 1 bar and 300k is...

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