Chap IV Exergy Analysis(火用 - 國立中興大學 · 2020. 2. 3. · Chap IV Exergy...

Advanced Thermodynamics NCHU ME dept 頁 1

Chap IV Exergy Analysis(火用)

Update on 2019/11/1

Contents:

(4.1). Exergy, Reversible Work and Availability

(4.1.1). Dead State

(4.1.2). Reversible Work for a Non-reacting closed System

(4.1.3). Reversible Work for a Non-reacting open System

(4.2). Global Exergy Reservoirs, Flux, and Anthropogenic Destruction

(4.3). Reversible work of process

(4.3.1). Closed system compression process

(4.3.1.1). Adiabatic process

(4.3.1.2). Compression with heat transfer

(4.3.1.3). Isothermal compression

(4.3.2). Open system compression process

(4.3.2.1). Adiabatic axial type compressor

(4.3.2.2). Adiabatic axial type compressor with heat transfer

(4.3.3). Adiabatic expansion process

(4.3.4). Unconstrained expansion process

(4.3.5). Separation process

(4.3.6). Mixing process

(4.3.7). Heat transfer process

(4.4). Exergy analysis of refrigeration cycles

(4.4.1). Reversible work of refrigeration

(4.4.2). How much work is required to produce ice?

(4.4.3). Basic refrigeration cycle

(4.4.4). Multi stage expansion system

(4.4.5). Multi stage cooling system

(4.4.6). Methane liquefaction


(4.5). Conversion efficiency of exhaust energy recovery

(4.5.1). How much work can be obtained from the exhaust of an engine?

(4.5.2). Single stage engine

(4.5.3). Single stage with finite heat transfer rate

(4.5.4). Multi stage engines

(4.5.5). Thermal energy recovery with ORC

(4.5.6). Thermal energy recovery with TEM

(4.6). Compressed air energy

(4.6.1). Compressed air as energy storage

(4.6.2). How much energy is required to produce a bottle of compressed air?

(4.6.3). Energy in a bottle of high pressure air

(4.6.4). How to obtain the energy stored in bottle?


(4.6.4.2). Air preheating

(4.6.4.3). Bottle heating

(4.7). Chemical Exergy and Bio Energy

(4.8). Hydrogen energy

(4.8.1). How much power can be delivered from a continuous flow of air and

hydrogen mixture at T0 and P0?

(4.8.2). Constant pressure combustion

(4.9). Fuel Cell


(4.1). Exergy, Reversible Work and Availability

Exergy analysis is used to find out the energy utilization efficiency of an energy

conversion system. It is known that any energy conversion system should obey the

second law of thermodynamics. However, the energy conversion system is usually

composed of several processes. It is necessary to know the conversion efficiency of

each process such that efforts of improvement could be focused on the least efficient

process.

According to the second law of thermodynamics, it is not possible to totally

convert thermal energy into work. The energy contained inside any system could

be only partially converted into work theoretically. However, due to the irreversible

effect inside the system, the actual work is even less than expected.

Exergy is the maximum available work that can be extracted from a system

when the system undergoes a process from its initial state to the state in equilibrium

with its environment. Equivalent names of exergy are availability, available energy,

exergic energy, essergy, utilizable energy, available work, maximum work,

reversible work, and ideal work.

Exergy is not a thermodynamic property. Its value depends on the states of the

system as well as the state of its environment.

(4.1). Dead State

Exergy is a measure of the potential of a system to do work as it undergoes a

process from its original state to a final state that it achieves equilibrium with it

surroundings. The surroundings are often also called a reference environment. After

the system and surroundings reach equilibrium, the system won't change or be

changed. This is known as the system dead state, and it has an exergy of zero.

When a system is in equilibrium with its environment, it is not possible to do any

work by itself.

All kinds of difference between a system and its environment can be used to do

work. A difference in temperature would cause heat transfer which could in turn be


converted to work. A difference in pressure would promote work directly, and a

difference in concentration would induce mass transfer which could also be

converted to work.

T Q W , the work of coal power plant

u W , the work of wind power plant

P W , the work of hydro power plant

C m W , the work of osmotic power plant

W , the work of dipping bird

As difference in potential vanishes, changes would cease, and no more work can be

derived.

A dead state is the state that all the properties of a system are identical to those

of the environment such that no driving force exists between the system and its

environment, and no work can be derived.

0T T ，0P P ，

0C C ， 0u ，0

The reference states of the concentrations of some gases are defined to be the

same as those of atmospheric concentrations. Air consists of N2=75.67%, O2=20.35%,

H2O(g)=3.12%, CO2=0.03% and other gases=0.83%.

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Assignment 4.1:

It is estimated that 31000 TW of solar energy is absorbed by atmosphere. Since the

atmosphere contains huge amount of thermal energy, is it possible to make use of

this energy and convert it to work? Please place your own comment.

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November 24, 2009 6:05 AM PST

Norway opens world's first osmotic power plant

by Reuters

Norway opened on Tuesday the world's first osmotic power plant, which produces

emissions-free electricity by mixing fresh water and sea water through a special

membrane.

State-owned utility Statkraft's prototype plant, which for now will produce a

tiny 2 kilowatts to 4 kilowatts of power or enough to run a coffee machine, will

enable Statkraft to test and develop the technology needed to drive down

production costs.

The plant is driven by osmosis that naturally draws fresh water across a membrane

and toward the seawater side. This creates higher pressure on the sea water side,

driving a turbine and producing electricity. "While salt might not save the world

alone, we believe osmotic power will be an interesting part of the renewable energy

mix of the future," Statkraft Chief Executive Baard Mikkelsen told reporters.

Statkraft, Europe's largest producer of renewable energy with experience in

hydropower that provides nearly all of Norway's electricity, aims to begin building

commercial osmotic power plants by 2015.

The main issue is to improve the efficiency of the membrane from around 1 watt

per square meter now to some 5 watts, which Statkraft says would make osmotic

power costs comparable to those from other renewable sources. The prototype, on

the Oslo fjord and about 40 miles south of the Norwegian capital, has about 2,000

square meters of membrane. Future full-scale plants producing 25 megawatts of

electricity, enough to provide power for 30,000 European households, would be as

large as a football stadium and require some 5 million square meters of membrane,

Statkraft said. Once new membrane "architecture" is solved, Statkraft believes the

global production capacity for osmotic energy could amount to 1,600 to 1,700

terawatt hours annually, or about half of the European Union's total electricity

demand. Europe's osmotic power potential is seen at 180 terawatts, or about 5

percent of total consumption, which could help the bloc reach renewable energy

goals set to curb emissions of heat-trapping gases and limit global warming.

Osmotic power, which can be located anywhere where clean fresh water runs into

the sea, is seen as more reliable than more variable wind or solar energy.

http://www.statkraft.com/


Here is the company's illustration of how the plant works.

(Credit: Statkraft)

(4.1.2). Reversible Work

Reversible work is the maximum work that can be extracted from a system if

the system moved from the current state to the dead state.

1

A B C

2

It is noted that the amount of work that a system may deliver depends on the

process that the system undergoes from the first state to the second state. However,

there exists a maximum value of work that no other process may produce more work

than this value.


The evaluation of exergy is based on the principle of increase of entropy. If a

system proceeds from its original state to a final state in equilibrium with its

surroundings, no matter what that process is, the net entropy changes of the system as

well as its environment would never be negative.

0net sys envS S S

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

0S

In a closed system, the increase of entropy can be attributed to the external

irreversibility and the internal irreversibility.

0

1 1( )net sys evS S S QT T

In an open system, the net increase can be attributed to the external irreversibility and

the internal irreversibility.

0

1 10j

jnet j

dSQ

dt T T


(4.1.3 ). Reversible Work for a Non-reacting closed System

State 2

QH

State 1

Wa

If a system undergoes a process from state 1 to state 2, and receives an amount of

heat QH from a high temperature reservoir at TH during the process, the work output

and the net entropy change are

Hrev

H

QS

T ，

2 1H

net rev sys

H

QS S S S S

T

2 1( )a HW Q U U

Since the temperature of the reservoir (TH) must be higher than that of the

system (T1), there is a finite difference of temperature (△T = TH－T1)between the

reservoir and the system. Entropy is produced during the heat transfer process from

the reservoir to the system. In order to reduce the entropy increase due to the finite

temperature difference heat transfer, a new approach should be conducted.

A Carnot engine should be operated between the reservoir and the system in

order to avoid the finite temperature difference heat transfer. However, as the same

amount of heat (QH) is transferred from the reservoir to the Carnot engine, part of it

will be converted to work (WE= QH-QL), and only the rest of heat (QL) will be

transferred to the system. As a result, the amount of heat transferred to the system

TH


is less than QH, and the system will not undergo the process from state 1 to state 2

because the heat transfer amount is not adequate.

In order to get sufficient amount of heat to complete the process, A Carnot

refrigerator has to be operated between the system and the environment to supply the

shortage of heat (QH－QL). If the heat absorbed from the environment by the

Carnot refrigerator is Q0, the work input to the refrigerator should be WR= QH－QL－

Q0.

QH

WE= QH-Q State 2

QL

State 1

Wa

QH－QL

WR= QH－QL－Q0

Q0

Since in the new approach, entropy increase is reduced to zero, a entropy

balance could be conducted as follows.

02 1

0

0Hnet

H

Q QS S S

T T

TH

T0


The heat transfer from the environment to the Carnot refrigerator (Q0) should be

0 0 0 2 1( )H

H

QQ T T S S

T

The work delivered by the Carnot engine is

E H LW Q Q

The work delivered to the Carnot refrigerator is

0R H LW Q Q Q

The total work of the process is the sum of three terms.

2 1 0( ) ( )rev a E R H H L H LW W W W Q U U Q Q Q Q Q

2 1 0 2 1 0 0 2 1( ) ( ) ( )HH H

H

QQ U U Q Q U U T T S S

T

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

H

TU U T S S Q

T

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

H

TU T S U T S Q

T

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

H

Tw u u T s s P v v q

T

0rev a carnot netI W W W T S

Second law efficiency: the ratio of actual work to reversible work.

2a

nd

rev

W

W for positive work

2rev

nd

a

W

W for negative work

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Example: An iron block of 10 kg at 300 ℃ is cooled down naturally in an

atmosphere of 25 ℃. Calculate the reversible work and the irreversibility of the

process. The heat capacity of iron is 425 J/kg-K.


0( )HQ mc T T

0

0 0

( )Hev

Q mc T TS

T T

0lnsys

H

TS mc

T

0 0

0

( )ln H

net sys ev

H

T mc T TS S S mc

T T

00 0 0 0

0 0

ln ( ) 1 lnH Hnet H

H

T T TI T S mcT mc T T mcT

T T T

rev aI W W

0

0 0

1 lnH Hrev

T TW I mcT

T T

What can we do to obtain the reversible work?

HQ

W

LQ

dQ mcdT

0 01 1T T

W dQ mc dTT T


0

0 0 00 0 0

0

1 ( ) ln 1 ln

H

T

HH

H HT

T T T TW mc dT mc T T mcT mcT

T T T T

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

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, calculate the reversible work and the irreversibility

of the process assuming the atmospheric temperature is 25 ℃.

T (K) P (bar) V (L)

1 298 3 1

2 225.8 1.137 2

3 298 1.5 2

QL=U3-U2

P

V

We lose 0.2079 kJ of work in this process.

How do we get it back.? We need two processes.

3 bars Vacuum

25℃

1.5 bars

25℃

1.5 bars

25℃


The first one is to replace the diaphragm with a piston and let the piston move

rightwards to the limit. We may get 0.1815 kJ of work during this expansion process.

The second step is to use a Carnot engine to transfer heat from environment to the

system until the temperature of the system reaches 298K. We may get 0.0265 kJ of

work in this process.

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

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. Assuming the atmospheric temperature is 25 ℃,

calculate the reversible work and the irreversibility of the process.

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

00 0(1 )rev H

H

TW T S U Q T S

T =350.06 kJ

rev aI W W =350.06 kJ,

2a

nd

rev

W

W = 0

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

Assignment 4.2(育陽)

Two iron blocks are of the same weight and at different temperature. One is at 300

℃ and the other one is at 30℃. A Carnot engine uses these two blocks as thermal

reservoir and converts heat to work. How much work can be converted at

most?(18.11 kJ/kg)


300 ℃ 30℃

W=?

143.7℃ 143.7℃

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

Assignment 4.3: (士傑)

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. Calculate the reversible work

and the irreversibility of the process assuming the atmospheric temperature is 25 ℃.

(29.9 kJ)

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

If the system undergoes a process to the dead state, the available work is

00 1 0 0 1( ) ( ) (1 )rev H

H

Tw u u T s s q

T

Since the environment is at the pressure of P0，expansion of the system would

expel the air and do work to its environment, and the available work would be

reduced if the work to the environment is counted.

00 0 0 1 0 0 1 0 0( ) ( ) ( ) ( ) (1 )avail rev H

H

Tw w P v v u u T s s P v v q

T

This is the maximum work that is available if the system undergoes a process to the

dead state. We define the availability of the system as


00 0 0 0 0( ) ( ) (1 )H

H

Tu u T s s P v v q

T

And the maximum work that is available if the system undergoes a process from state

1 to state 2 could be obtained as

0 02 1 0 2 1 0 2 1 1 2( ) ( ) ( ) (1 ) (1 )avail H H

H H

T Tw u u T s s P v v q q

T T

If 1 2 , there is a positive work, which means that the process would do work out.

If 1 2 , the process requires work in.

(4.1.3). Reversible Work for a Non-reacting open System

2 2

2 1 02 2 1 1 0 2 1( ) ( ) ( ) (1 )

2 2rev H

H

V V Tw h gz h gz T s s q

T

If kinetic energy and gravitational potential energy can be neglected, the reversible

work is

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

H

Tw h h T s s q

T

If the heat transfer is towards the environment, TH=T0, then the reversible work is

2 1 0 2 1( )revw h h T s s

If the system undergoes a process to the dead state, the available work is

00 0 0( ) ( ) (1 )rev H

H

Tw h h T s s q

T 0

0 0 0 0[ ] [ ] (1 )H

H

Th T s h T s q

T

The exergy of the system is defined as 0h T s such that the available work if

the system undergoes a process to the dead state can be obtained as

00 (1 )rev H

H

Tw q

T

is the potential of the system to deliver work.

Thus the reversible work of a system going from state 1 to state 2 is


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

H H

T Tw h h T s s q q

T T

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

Example: Air at 300 kPa and 25 ℃ flows through a throttle valve and drops to the

pressure of 100 kPa at a rate of 1 kg/sec. Calculate the reversible work of the

process.

22 1 0 2 1 0

1

( ) lnrev

Pw h h T s s RT

P =93.96 kJ/sec

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

Assignment 4.4: (家軒)

Air at 300 kPa and 25 ℃ flows through a throttle valve and drops to the pressure of

100 kPa at a rate of 1 kg/sec. How to obtain the reversible work of the process?

298K 217.7K

A turbine is installed in the pipe. Air drives the turbine to do work.

1

22 1

1

( )k

kP

T TP

=217.7K

1 2( )T pw c T T =80.66 kW

Since the temperature is very low, heat transfer may occur from environment to the

298K


flow. A series of Carnot engines are installed between environment and flow. Work

can be done during the heat transfer process.

3 2( )L pq c T T =80.66 kW

3

2

lnsys p

Ts c

T =0.3154 kW/K

0

Hev

qs

T =-0.3154 kW/K

0H evq T s =93.99 kW

C H Lw q q =13.33 kW

T Cw w w =93.99 kW

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

Example: A stream of hot air flow at 300 ℃ is cooled down to 25 ℃at a rate of 1

kg/sec. Calculate the reversible work of the process.

iT

eT

( )H p i eQ mc T T

ln esys p

i

Ts mc

T

0

Lev

Qs

T


0

ln 0e Lnet sys ev p

i

T Qs s s mc

T T

0 0ln lne iL p p

i e

T TQ mc T mc T

T T

0( ) ln iH L p i e p

e

TW Q Q mc T T mc T

T

We need a series of Carnot engine to convert waste thermal energy to work.

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

Example: Hot air at 300 kPa and 300 ℃ flows through a turbine at a rate of 1 kg/sec,

and drops to a pressure of 100 kPa. If the efficiency of the turbine is 85%, calculate

the reversible work of the process and the second law efficiency. (89.1%)

1

22 1 1 2 1

1

( ) 1

k

k

s s p s p

Pw h h c T T c T

P

1 2( )a T s pw w c T T

2 22 1 0 2 1 1 2 0

1 1

( ) ( ) ( ln ln )rev p p

T Pw h h T s s c T T T c R

T P

2a

nd

rev

w

w

2a a rev rev

s nd

s rev s s

w w w w

w w w w

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


Example: Air is cooled by water in a heat exchanger. Air flows into the heat

exchanger at 100 ℃ with the flow rate is 1 kg/sec. Water flows into the heat

exchanger at 30℃ with the flow rate 5 kg/sec. If air is to be cooled to 40℃,

calculate the entropy generation inside the heat exchanger, and the reversible work of

this process. The atmospheric temperature is 30℃.

( ) ( )a p ai ae w w wi wem c T T m c T T

1*1.0045*(100-40)=5*4.186*(Tw-30)

Tw=32.88℃

ln aea a p

ai

TS m c

T =-0.1762 kJ/K

ln wew w w

wi

TS m c

T =0.1980 kJ/K

net w aS S S =0.0218 kJ/k

0 0( ) ( )rev a p ai ae a w w wi we wW m c T T T s m c T T T s

( ) ( ) 0a p ai ae w w wi wem c T T m c T T

0 0 0rev a w w netW m T s m T s T S =6.605 kJ

Note: The reversible of heat exchanger is the lost work of heat transfer.

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

Example: Air is cooled by water in a heat exchanger. Air flows into the heat

exchanger at 100 ℃ and is cooled to 40℃ with the flow rate of 10 kg/sec. Water


flows into the heat exchanger at 20℃ with the flow rate of 5 kg/sec. How much

work can be obtained if Carnot engines are placed between the air flow and the water

flow.

Air

100℃ QH

W

QL

Water

20℃

( )H a p ai aeQ m c T T

( )L w w we wiQ m c T T

H LW Q Q

( ) ( )a p ai ae w w wi wem c T T m c T T W

ln ln 0ae wea p w w

ai wi

T TS m c m c

T T

a p

w w

m c

m c =10*1.0045/(5*4.186)=0.48

1ae we

ai wi

T T

T T

aiwe wi

ae

TT T

T

=318.7K=45.7℃

( )H a p ai aeQ m c T T =10*1.0045*(100-40)=602.7 kW

( )L w w we wiQ m c T T =5*4.186*(45.7-20)=537.9 kW

H LW Q Q =64.8kW


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(4.2). Global Exergy Reservoirs, Flux, and Anthropogenic Destruction

The cascading flow of exergy from primary reservoirs into secondary reservoirs and

on to its eventual natural or anthropogenic destruction can be summarized as a

system of exergy reservoirs and flows.

The major exergy flux is supplied by direct sun light which is about 162000

tera-watts. The amount of exergy from sun light in a year is about 5110 ZJ. It is

noted that the global energy consumption at the year of 2005 is 462 QBtu, which is

equivalent to 0.487 ZJ. That is, the amount of exergy from sun light in a year can

be used to supply 10485 years of energy consumption at the level of 2005.

However, only a very small portion of the exergy flux is utilized by human and

consumed in anthropogenic activities. Most of the exergy flux has been destructed

and dissipated in the nature.

Huge amount of exergy is stored under ground as the form of crustal thermal

energy which is equivalent of 2935 times of the amount of exergy from sun light in a

year. An even greater amount of exergy is stored inside the nucleus of deuterium

which is the isotope of hydrogen.


(4.3). Reversible work of process

(4.3.1). Closed system compression process

In a closed system, the reversible work is as the following.

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

H


T


0Hq

2 1 0 2 1 0 2 1( ) ( ) ( )revw u u T s s P v v

If the process is isentropic

2 22 1

1 1

ln( ) ln( ) 0v

T Vs s c R

T V

1

12 1

2

k

s

VT T

V

2 1 0 2 1 1 2 0 2 1( ) ( ) ( )rev s v sw u u P v v c T T P v v

However, if the process is not isentropic

2 22 1

1 1

ln( ) ln( ) 0v

T Vs s c R

T V

1

12 1

2

k

VT T

V

2 1 0 2 1 1 2 0 2 1( ) ( ) ( )a vw u u P v v c T T P v v

rev aw w

We need to do more work for the compression process if entropy increases in this

process. This would result in a higher temperature after compress.

1

11 0 2 1

2

2

1 2 0 2 1

1 ( )

( ) ( )

k

v

revnd

a v

vc T P v v

vw

w c T T P v v

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


Example：Calculate the reversible work of an insulated piston to compress 1 kg of

air at 100 kPa and 25℃ to a volume of 0.2 m3.

P1 = 100 kPa，T1 = 298 K，v1 = 0.8553 m3/kg，v2 = 0.2 m

3/kg

2T 2P

2 1s s revw

aw I

532.9 764.8 0 -168.54 -168.54 0

550 789.3 0.02265 -174.06 -180.891 6.75

600 861.1 0.08508 -191.33 -216.69 25.36

Note: The extra energy to compress air is used to overcome the friction force, and

will dissipate as heat to air. As a result, the final temperature is getting higher.

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

(4.3.1.2). Compression with heat transfer

Piston compression process with heat transfer to or from the cylinder wall:

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

w


T

2 1 0 2 1( )w u u P v v q

02 1 0 2 1

2

2 1

( ) (1 )rev w

nd

a

Tu u T s s q

w T

w u u q

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

Example：Calculate the reversible work 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. The atmospheric

temperature is 25℃, and the atmospheric pressure is 100 kPa.

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

02 1 0 2 1 0 2 1( ) ( ) ( ) (1 )avail H

H


T

= -298×0.03843-0.7175×(350.1-298) – 100×(0.5-0.8553) = -13.304 kJ

1 1 2 2 0 2 1

1( ) ( )

1aW PV PV P V V

n

=-14.27 kJ

rev aI W W =0.966 kJ

2rev

nd

a

W

W = 0.932

Note: Heat transfer occurs between compressed air and cylinder wall. The wall

temperature is not necessary identical to that of atmospheric temperature. However,

finally the heat transfer will go to the atmosphere at all.

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

Assignment: Calculate the reversible work to compress 1 m3 of air at 100 kPa and 25

℃ to a pressure of 1 MPa in a polytropic process with n=1.5. The atmospheric

temperature is 25℃, and the atmospheric pressure is 100 kPa。

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

(4.3.1.3). Isothermal compression

During the compression process, the temperature is kept at T0.

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

w


T

20 0 2 1

1

ln ( )v

T R P v vv

2 1 0 2 1 0 2 1( ) ( )w u u P v v q P v v q

20 0 2 1

12

0 2 1

ln ( )

( )

revnd

a

vRT P v v

w v

w q P v v


In an actual isothermal compression process, the heat that should be removed

out of the system is higher than that of a reversible isothermal compression process.

The extra heat transfer is converted from the friction work.

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


to a volume of 0.5 m3 isothermally.

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

P2 = P1 × (v1 / v2 ) = 171.06 kPa

wa =2

1

ln( )P

RTP

=-45.91 kJ/kg

Δssys =- 2

1

ln( )P

RP

=-0.1541 kJ/kg-K

q= -45.91 kJ/kg

Wrev= T0Δssys =-45.91 kJ/kg

The actual work is more than 45.90 kJ/kg to air at 100 kPa and 25℃ to a volume of

0.5 m3 isothermally. Once the actual work is measured, the second law efficiency

can then be obtained.

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

(4.3.2). Open system compression process

2 2

2 1 02 2 1 1 0 2 1( ) ( ) ( ) (1 )

2 2rev H

H

V V Tw h gz h gz T s s q

T

(4.3.2.1). Adiabatic axial type compressor

2 1 0 2 1( )revw h h T s s

2 1 2 1aw h h q h h

For isentropic compression

2 1s s , rev aw w , 2 1

2 1

1s sc

a

w h h

w h h

,

2 1revnd

a

w

w


For compressor with efficiency less than 1.0

2 1s s , rev aw w ,

2 1nd , 1c

2 1

2 1

s sc

a

w T T

w T T

=isentropic work/adiabatic work

1

22 1

1

11 1

k

k

c

PT T

P

1

2 2 2 22 1

1 1 1 1

1ln( ) ln( ) ln 1 1 ln( )

k

k

p p

c

T P P Ps s c R c R

T P P P

1 1

1 2 2 22 1 0 2 1 0 0

1 1 1

1( ) 1 ln 1 1 ln( )

k k

k k

rev p p

c c

T P P Pw h h T s s c c T RT

P P P

1

22 1 1

1

11

k

k

a p

c

Pw h h c T

P

1

2 20 0

1 1

1ln 1 1 ln( )

k

k

rev a p

c

P PI w w c T RT

P P

1

2 20 0

1 1

2 1

21

1

1ln 1 1 ln( )

1

11

k

k

p

c

revnd k

a k

p

c

P Pc T RT

P Pw

wP

c TP

For a compressor with given efficiency, its second law efficiency can be

calculated if the pressure ratio and the environmental temperature are known.

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

Example：Calculate the reversible work of an insulated compressor to compress 1


kg of air at 100 kPa and 25℃ to a pressure of 10 bars if the compressor efficiency is

0.85. The atmospheric temperature is 25℃.

P1 = 100 kPa，T1 = 298 K，v1 = 0.8553 m3/kg，v2 = 0.5 m

3/kg

1

22 1

1

11 1

k

k

c

PT T

P

=624.3 K

wa = -327.76 kJ/kg

Δssys =0.08202 kJ/kg-K

I = 24.44 kJ/kg

Wrev= -303.32 kJ/kg

2nd 0.9254

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

Assignment: Calculate the reversible work to compress 1 m3 of air at 100 kPa and 25

℃ to a pressure of 1 MPa in a polytropic process with n=1.5. The atmospheric

temperature is 25℃, and the atmospheric pressure is 100 kPa。

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

(4.3.2.2). Adiabatic axial type compressor with heat transfer

02 1 0 2 1( ) (1 )rev

w

Tw h h T s s q

T

2 1aw h h q

00 2 1( )rev a

w

TI w w T s s q

T

02 1 0 2 1

2

2 1

( ) (1 )rev w

nd

a

Th h T s s q

w T

w h h q

= reversible work/actual work

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



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

wa = 1 1 2 2( )1

nPV PV

n

=-64.74 kJ/kg

Δssys =-0.03843 kJ/kg-K

q= -12.455 kJ/kg

Wrev= -63.738 kJ/kg

I = 1.0013 kJ/kg

2nd 0.9853

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

(4.3.3). Adiabatic expansion process

Example: gas turbine

2 1 0 2 1( )revw h h T s s

For expansion with efficiency less than 1.0, 2 1s s ,

rev aw w , 2 1nd

2 1

2 1

ac

s s

w T T

w T T

1

22 1

1

1 1

k

k

T

PT T

P

1

2 2 2 22 1

1 1 1 1

ln( ) ln( ) ln 1 1 ln( )

k

k

p p T

T P P Ps s c R c R

T P P P

0 2 1( )rev aI w w T s s

1 22

1 2 0 2 1( )nd

h h

h h T s s


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

Example：Calculate the reversible work of an insulated turbine to expand 1 kg of air

at 1000 kPa and 800℃ to 100 kPa if the turbine efficiency is 0.85.

P1 = 1000 kPa，T1 = 1073 K，P2 = 100 kPa

T2 = 633.3 K ，wa = 441.6 kJ/kg


I = 39.098 kJ/kg

Wrev= 480.78 kJ/kg

2nd 0.9185

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

Assignment: In a gas turbine engine, air is compressed from 25 ℃ and 100 kPa to

the pressure of 1 MPa, and then heated to 1200 K in the combustor. The inlet air

flow rate is 10 m3/sec. If the compressor efficiency is 85%, the turbine efficiency is

90%, calculate the power of engine and the reversible work.

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

(4.3.4). Unconstrained expansion process

Example: throttle valve, expansion valve, flash chamber

2 2

2 2

i ei i e e

V Vq h gz h gz w

0q ， 0w

i eh h

2 1 0 2 1 0 2 1( ) ( )revw h h T s s T s s

For ideal gas with constant heat capacity,

2 1T T

2 2 22 1

1 1 1

ln( ) ln( ) ln( )p

T P Ps s c R R

T P P


20

1

ln( )rev

Pw RT

P

0aw

20

1

ln( )P

I RTP

Since there is no work output for unconstrained expansion process, the reversible

work is the exergy lost due to the unconstrained expansion.

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

Example：Calculate the reversible work of air flowing through a throttle valve with

the upstream pressure of 100 kPa and the downstream pressure of 50 kPa.

20

1

ln( )rev

Pw I RT

P =59.282 kJ/kg

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

Example：Calculate the reversible work of methane at 8 MPa and 200 K expanding

to the pressure of 101 kPa.

P1 = 8 MPa，T1 = 200 K，h1=88.54，s1=7.2069

P2 = 101.3 kPa，h2 =-286.5 +x*510.33

x = 0.7349，s2 = 4.9336 + 0.7349*4.5706 = 8.2925


Wrev= 323.51 kJ/kg

I = 323.51 kJ/kg

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

(4.3.5). Separation process

Example: flash chamber, separation of vapor and liquid.

2 2

2 2

i ei i e e

V Vq h gz h gz w

0q ， 0w


(1 )i e g f f fgh h xh x h h xh

It is noted that ih is the enthalpy of condensed fluid after condenser and

fc is the

heat capacity of fluid.

( )i f f i e fgh h c T T xh

( )f i e

fg

c T Tx

h

2 (1 )g fs xs x s

2 1 0 2 1 0 0( ) ( (1 ) ) ( )rev g f i fg f iw h h T s s T xs x s s T xs s s

ln ef i f

i

Ts s c

T

2 1 ( ) ln ( ) ln ( 1 ln )f fg fe e i i

i e f i e f f

fg i e i e e

c s cT T T Ts s T T c T T c c

h T T T T T

0 0 ( 1 ln )i irev f

e e

T Tw T s c T

T T

According to the Clapeyron Equation, the saturation pressure and saturation

temperature during phase transition is related as the following.

ln ( 1)fg fg fgi i

e e i i e

h h hP T

P RT RT RT T


1 lni i i

e fg e

T RT P

T h P

ln 1 ln 1 lni i i i

f fg e fg e

s RT P RT P

c h P h P

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

Example：Calculate the reversible work of compressed water at 10 MPa and 180℃

undergoing an expansion process to 100 kPa in a flash chamber.

P1 = 10000 kPa，T1 = 180℃，h1=767.84，s1=2.1275

P2 = 100 kPa，h2 =417.46 +x*2258

x = 0.1445，s2 = 1.3026*0.8555 + 7.3594*0.1445 = 2.1778


Wrev= 14.99 kJ/kg

I = 14.99 kJ/kg

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

Assignment 4.8: A refrigeration system works with R134a as the refrigerant. The

cooling load is 10 tons at the temperature of -40℃and the compressor efficiency is

90%. Find out the COP of the system, the refrigerant rate, and the entropy increase in

each process. It is known that T0=25℃.

(1). A single stage system is used with the following states:

P1=0.5164 bar, T1=-40℃, P2=9.0 bar, T3=30℃

(2). A two-stage system is used with the following states:

P1=0.5164 bar, T1=-40℃, P2=2.8 bar, P4=9.0 bar, T5=30℃

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

(4.3.6). Mixing process

Example: feed water heater, mixing of different components

i i e e

i e

m h m h


e e i i

e i

s m s m s

0( )rev e e i i g

e i

w T m s m s xs

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

Example：Methane mixes with stoichiometric air at 100 kPa 1nd 25℃, calculate the

reversible work of the mixing process.

P1 = 10000 kPa，T1 = 180℃，h1=767.84，s1=2.1275

P2 = 100 kPa，h2 =417.46 +x*2258

x = 0.1445，s2 = 1.3026*0.8555 + 7.3594*0.1445 = 2.1778


Wrev= 14.99 kJ/kg

I = 14.99 kJ/kg

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

(4.3.7). Heat transfer process

Example: heat exchanger, condenser, regenerator

A Ai B Bi A Ae B Bem h m h m h m h

( ) ( )A A Ai Ae B B Be Bim c T T m c T T

( ) ( )A Ae Ai B Be Bis m s s m s s

2 1 0 2 1 0( ) [ ( ) ( )]rev A Ae Ai B Be Biw h h T s s T m s s m s s

Assume that Ai BiT T , A is used to heat up B.

If A A B Bm c m c , then

Ai Ae Be BiT T T T . The maximum achievable level of the

outlet temperature of B is Be AiT T . However,


( )B BAe Ai Be Bi Ai Bi Be

A A

m cT T T T T T T

m c

Define the effectiveness of the heat exchanger as

Be Bi

Ai Bi

T T

T T

, ( )Be Bi Ai BiT T T T

( )Ae Ai Bi Be Ai Ai BiT T T T T T T

ln( ) ln( ) [ln(1 ) ln(1 )]Ae Be Bi AiA A B B A A

Ai Bi Ai Bi

T T T Ts m c m c m c

T T T T

0 , 0s

1 , [ln( ) ln( )] 0Bi AiA A

Ai Bi

T Ts m c

T T

0.5 , [ln(1 )(1 ) ln 4] [ln(2 ) ln 4]Bi Ai Bi AiA A A A

Ai Bi Ai Bi

T T T Ts m c m c

T T T T

2 ln ( ) / 2Ai BiA A

Bi Ai

T Tm c

T T

If A A B Bm c m c , then

Ai Ae Be BiT T T T , and Be AiT T . The maximum achievable

level of the outlet temperature of B is Be AiT T . However,

( ) (1 )B B B B B BAe Ai Be Bi Ai Bi Bi

A A A A A A

m c m c m cT T T T T T T

m c m c m c

Define the effectiveness of the heat exchanger as

Be Bi

Ai Bi

T T

T T

, ( )Be Bi Ai BiT T T T

( ) (1 )B B B B B BAe Ai Ai Bi Ai Bi

A A A A A A

m c m c m cT T T T T T

m c m c m c

ln( ) ln( ) [ln(1 ) ln(1 )]Ae Be B B B B Bi B B AiA A B B A A

Ai Bi A A A A Ai A A Bi

T T m c m c T m c Ts m c m c m c

T T m c m c T m c T


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

Example：Hot water at 100 ℃ is used to heat up oil at 0 ℃. It is known that

1

2

B B

A A

m c

m c . TA1 = 100℃，TB1 = 0℃，TA2=100-100×ε ℃，TB2=100×ε ℃

ε 0 0.2 0.4 0.5 0.6 0.8 1.0

TA2 100 90 80 75 70 60 50

TB2 0 20 40 50 60 80 100

s 0 0.008175 0.01326 0.01471 0.01549 0.01506 0.01213

It is noticed that even ε＞1, it is still that s ＞0.

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

Example：Sea water is used to condense the steam flowing out of a steam turbine.

It is known that the steam pressure is 10 kPa with the quality of 95%. The inlet

water is at 25℃ and the outlet water is at 35℃, calculate the reversible work.

hA1 = 104.87，hA2 = 146.66，hB1=2664.99，hB2=191.81

mA (hA2 -hA1)= mB (hB1 -hB2)

mA =59.181 mB

sA1 = 0.3673，sA2 = 0.5052，sB1=7.7752，sB2=0.6492

s 59.181*0.1379-7.126=1.0351 kJ/kg-steam-K

Wrev= 308.45 kJ/kg steam

However, the warm sea water would cool to 25℃ after flowing back to sea. As a

result, exergy would lose again for the cooling process. The total loss of reversible

work combing both the condensation and the cooling would be

Wrev= 308.45 kJ/kg steam

2 1 0 2 1( )rev B B B Bw h h T s s =349.63 kJ/kg steam


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

Example：Hot air from the outlet of gas turbine at 700 K is used to heat the outlet of

compressor at 500 K. Calculate the reversible work if the regenerator effectiveness

is 0.8.

TA2 TA1

TB1 TB2

TA1 = 700 K，TB1 = 500 K，TA2=700-200×0.8 = 540 K，TB2=660 K

2 2 2 2

1 1 1 1

ln( ) ln( ) ln( )A B A Bp p p

A B A B

T T T Ts c c c

T T T T =0.0182 kJ/kg- K

Wrev= 5.424 kJ/kg

The effect of effectiveness on reversible work:

2 2 2 2

1 1 1 1

ln( ) ln( ) ln( )A B A Bp p p

A B A B

T T T Ts c c c

T T T T

1 2

1 1

A A

A B

T T

T T

2 1 1 1 1 1( ) (1 )A A A B A BT T T T T T

2 1 1 2 1 1( ) (1 )B B A A B AT T T T T T

2 2 1 1

1 1 1 1

[1 ][1 ]A B B A

A B A B

T T T TX

T T T T

1 1

1 1

4 2 (1 2 )( ) 0B A

A B

dX T T

d T T

1 1

1 1

(1 2 )( 2) 0B A

A B

T T

T T

21 1

1 1

(1 2 )( ) 0B A

A B

T T

T T


1

2 ，

1 1A BT T

The regenerator will lose the maximum amount of exergy when 0.5 .

ε 0 0.2 0.4 0.5 0.6 0.8 1.0

TA2 700 660 620 600 580 540 500

TB2 500 540 580 600 620 660 700

s 0 0.01812 0.02706 1.0286 0.02706 0.01812 0

It is interesting to note that the maximum increase of entropy occurs at 0.5 .

The case of 1 and 0 are of the same result in terms of entropy increase.

However, these two cases are two extremes for energy recovery. One indicates

perfect energy exchange, while the other one means no exchange at all. The second

law analysis is not consistent with the first law analysis because the exergy analysis

is focusing on the conversion of thermal energy to work only. The exchange of

thermal energy is not considered by the exergy analysis unless the exchange will

deteriorate the potential of work conversion.

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


(4.4). Exergy analysis of refrigeration cycles

(4.4.1). Reversible work of refrigeration

The minimum work to cool down an object from normal temperature to a

specified low temperature theoretically is called the reversal work. Net entropy does

not increase during the process of doing reversal work.

Ta

QH

W

QL

The net entropy change is as the following.

0Hnet sys ev sys

a

QS S S S

T

2 1sysS S S is the change of entropy of the system. Since the temperature of the

system drops, the entropy is decreased.

H

a

Q

T is the change of entropy of the environment. Since heat is transferred from the

system to the environment, the entropy is increased.

If the process is reversal, no net entropy increase occurs.

0Hnet sys ev sys

a

QS S S S

T

The amount of heat transfer to the environment would be

2 1( )H a sys aQ T S T S S

Tload


However, the amount of heat transfer from the system is

1 2( )LQ m h h

The work that needs to be conducted would be

2 1 2 1( ) ( )H L aW Q Q m h h T S S

As a result, the coefficient of performance of the system would be

1 2

2 1 2 1( )

L

a

Q h hCOP

W h h T s s

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

Example：

Find the reversal work to cool 1 kg of copper from 25℃ to -40℃.

( ) ln LL a a

a

TW mc T T mcT

T

( ) ln LL a a a

a

Tw c T T cT q T s

T

c=0.385 kJ/kg-K

HT 25℃=298K

LT -40℃=233K

ln L

a

TS mc

T =-0.0947 kJ/K

H aQ mT S =28.230 kJ

( )L a H LQ mc T T =25.025 kJ

W=28.230-25.025=3.20 kJ

LQCOP

W 7.82

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

Example： Find the work to cool 1 kg of copper from 25℃ to -40℃ with a

Carnot refrigerator.


25℃

QH

W

QL

L

H L

TCOP

T T

3.58

LQ =0.385×(-40-25)=25.025 kJ

LQW

COP =6.98 kJ

H LQ Q W =32.006 kJ

Hev

a

QS

T =0.1074 kJ/k

2

1

lnsys

TS mc

T =-0.0947 kJ/K

net env sysS S S =0.0127 kJ/K>0

Comments:

It is noted that the COP of a reversal work if higher than that of a Carnot cycle

operating on the same temperatures.

During the cooling process, the low temperature side of the system remains at -40℃.

Its temperature does not adjust to the load temperature.

-40℃

Copper block


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

Example： Find the reversal work to cool 1 kg of water from 25℃ to -30℃.

2 31 2 2 3

1 2 2

( ) ln ( 1) ( ) lnaw w a i i a

T T TW mc T T mc T m h mc T T mcT

T T T

2 31 2 2 3

2 1 2

( ) ( ) ( 1) ln lnaw i w a i a

T T Tw c T T c T T h c T c T

T T T

1 2 2 3 3 4 1 2 2 3( ) ( )L w ls iq h h h h h h c T T h c T T

Where T1 =25℃，T2 = 0℃，T3 = -30℃，cw=4.186 kJ/kg-K，ci=2.11 kJ/kg-K

lsh 333.4 kJ/kg

W=4.186×(0-25)+2.11×(-30-0)+333.4×(298/273-1)-4.186×298×ln(273/298)-2.11×29

8×ln(263/273) = 45.08 kJ/kg

Lq =501.35 kJ/kg

L

a

qCOP

w =11.12

It is noted that to make 1 kg of ice from water at 25℃ to -30℃ needs 45.08 kJ of

work theoretically.

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

Example： Find the reversal work to liquefy 1 kg of methane from 27℃ at 100 kPa

to -161.3℃.

Boiling Temp. = 111.7 K (1.01bar)， lsh 510.33 kJ/kg

1h =627.58 kJ/kg, 2h =223.83 kJ/kg,

3h =-286.50 kJ/kg,

1s =11.6286 kJ/kg, 2s =9.5042 kJ/kg-K,

3s =4.9336 kJ/kg-K,

25℃ water 0℃ water 0℃ ice -30℃ ice


1 3Lq h h =914.08 kJ/kg

1 3s s s =6.6950 kJ/kg-K

H L aq w q T s =2008.5

H Lw q q =1094.42 kJ/kg

COP=914.08/1094.42=0.835

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

Example： Find the reversal work to liquefy 1 kg of nitrogen from 27℃ at 100

kPa.

Boiling Temp. = 77.3 K (1.01bar)，lsh 201.82 kJ/kg

1h =311.16 kJ/kg, 2h =76.69 kJ/kg,

3h =-122.15 kJ/kg,

1s =6.8457 kJ/kg, 2s =5.4033 kJ/kg-K,

3s =2.8326 kJ/kg-K,

1 3Lq h h =433.31 kJ/kg

1 3s s s =4.0131 kJ/kg-K

H L aq w q T s =1203.93

H Lw q q =770.62 kJ/kg

COP=433.31/770.62=0.562

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

Example： Find the reversal work to liquefy 1 kg of hydrogen from 27℃ at 100

kPa.

Boiling Temp. = 20.2 K (1.01bar)，lsh 201.82 kJ/kg

1h =4199.4 kJ/kg, 2h =188.9 kJ/kg,

3h =-256.7 kJ/kg,

1s =64.778 kJ/kg, 2s =29.994 kJ/kg-K,

3s =7.956 kJ/kg-K,

1 3Lq h h =4456.1 kJ/kg

1 3s s s =56.822 kJ/kg-K

H L aq w q T s =17046.6


H Lw q q =12590 kJ/kg=3.5 kWh/kg

It is estimated that the energy requirement to produce liquid hydrogen ranges from

6~10 kWh/kg.

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


(4.4.2). How much work is required to produce ice in a real process?

A basic refrigeration system is composed of four major components,

compressor, condenser, expansion valve, and evaporator.

1→2, isentropic compression process, 2 1Cw h h

2→3, constant pressure heat rejection process, 2 3Hq h h

3→4, constant enthalpy process, 3 4h h

4→1, constant pressure process heat addition process, 1 4Lq h h

Coefficient of performance (COP), 1 4

2 1

Lq h hCOP

w h h

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

Example： Find the power to produce 10 tons/h of ice at -30℃from water at 25℃

using a compression type refrigerator with R134a. The ambient temperature is 25℃.

T(℃) P(kPa) h(kJ/kg) s(kJ/kg-K) x

1 -40.7 50 373.01 1.7631

2s 45.7 800 430.75 1.7631

2 60 800 445.19 1.8076

3 31.2 800 243.61 1.1496

4 -40.7 50 243.61 1.2063 0.4246


2 1s sw h h =57.74 kJ/kg

2 1s

a

c

ww h h

=72.18 kJ/kg

1 4Lq h h =129.40 kJ/kg

2 3Hq h h =201.58 kJ/kg

L

a

qCOP

w =1.793

This is the work for 1 kg of R134a. We need to know how much R134a is required to

make 1 kg of ice.

q=4.186*25+333.4+2.04*30=499.25 kJ/kg

10 tons/h=10000/3600=2.778 kg/sec

=499.25/129.40=3.858 kg/kg

To make 1 kg of ice, we need to remove 499.25 kJ/kg of heat. For 1 kg of R134a, it

may absorb 129.40 kJ/kg of heat as it evaporates. As a result, we need 3.858 kg of

R134a to make 1 kg of ice.

-30℃


aI aw w =3.858*72.18=278.48 kJ/kg

2.778*278.48=773.63 kW

The work required to make 1 kg of ice from water at 25℃ to -30℃is 278.48 kJ.

The power to produce ice at 10 ton/h is 773.63 kW.

It is noted that to make 1 kg of ice from water at 25℃ to -30℃ needs 45.08 kJ of

work theoretically.

(4.4.3). How to improve the performance of this system?

Deviation from theoretical performance of refrigeration system:

Entropy increase:

1→2: 2 1s s s =0.0445 kJ/kg-K

2→3: 3 2syss s s =-0.658 kJ/kg-K

2 3

0 0

Hev

q h hs

T T

=0.6764 kJ/kg-K

s -0.658+0.6764=0.0184 kJ/kg-K

3→4: 4 3s s s =0.0567 kJ/kg-K

4→1: 1 4syss s s =0.5568 kJ/kg-K

The entropy change for water is

273 333.4 243

4.186ln 2.04ln298 273 273

ws =-1.8255 kJ/kg-K

s 0.5568-1.8255/3.858=0.0836 kJ/kg-K

Process s ( kJ/kg-K) Cause of entropy increase

1→2 0.0445 Compressor efficiency

2→3 0.0184 Heat transfer with finite temperature difference

3→4 0.0567 Free expansion


Total 0.2032


0 0 0

1( ) (1 )H L L

total w w w

q q w qs s s s

T T T

Since the entropy reduction of water (ws ) and the heat removal of water (

Lq ) are

fixed values no matter what kind of refrigeration system is used, the total entropy

increase depends on coefficient of performance ( ) only.

System loss:

(1). Components efficiency: 2 1s s

A more efficient compressor could be used to improve the system performance.

(2). Finite temperature difference heat transfer:

Over heat during the heat rejection process. 3 aT T

a HT T , 2 3Hq h h , 2 3H

H

a a

q h hs

T T

T3 should be as close to Ta as possible. The pressure after compression could be

reduced.

(3). Free expansion. 4 3s s

Entropy increases during the constant enthalpy process in the expansion valve. A

two stage system could be used to reduce the entropy increase of this process.

(4). Finite temperature difference heat transfer: Load LT T

Over cool during the heat addition process.


1 4Lq h h , 1 4LL

Load Load

q h hs

T T

T1 should be as close to Tload as possible. The pressure after expansion could be

raised. Separation cooling could reduce the temperature difference of heat transfer.

(4.4.4). Multistage Expansion System

The two stage expansion system is used to reduce the entropy generation in the

flash chamber.

3 4 (1 )f gh h xh x h

3 f

g f

h hx

h h

3

4 3 3 3(1 )f

f g f fg

g f

h hs s s xs x s s s s s

h h

3 3 4( )fh h c T T

4

fg

fg

hs

T

43

3

lnf

Ts s c

T

4 3 4 4 3 3 3

3 4 3 4 4 4

( )ln ln 1 1 ln

fg

fg

hT c T T T T T Ts c c c c

T h T T T T T

3 3

4 4

1 lns T T

c T T

Entropy increase depends on temperature ratio before and after expansion.


3

4

T

T

s

c

1.01 0.00005

1.1 0.0047

2 0.3069

3 0.9014

Two stages of compression is used to raise the pressure of refrigerant. An flash

chamber is used to separate the liquid and vapor in the upper stage. The vapor after

expansion has no contribution of heat absorbing in evaporator. Only liquid

refrigerant may absorb heat by evaporation. The vapor part just flows through

evaporator and gets compressed again. The advantage of the multistage system is that

less vapor is compressed and cooled in vain.

Work of the lower stage compressor: 2 1( )(1 )CLw h h x

Work of the upper stage compressor: 4 3CHw h h

4 5Hq h h

5 6h h

6 7 8(1 )h xh x h


8 9h h

Heat addition in the evaporator: 1 9( )(1 )Lq h h x

Mixing process: 3 7 2(1 )h xh x h

2 1 4 3( )(1 )Cw h h x h h

1 9

2 1 4 3

( )(1 )

( )(1 )

L

C

q h h xCOP

w h h x h h

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

Example：

Find the work to cool 1 kg of water from 25℃ to -30℃ using a two stage type

refrigerator. The upper stage pressure is 800 kPa. The lower stage pressure is 200

kPa. The refrigerant is R134a and a compressor efficiency is 80%.


1 -40.7 50 373.01 1.7631

2s -0.5 200 400.48 1.7631

2 7.5 200 407.35 1.7879

3 2.6 200 403.12 1.7727

4s 48.7 800 433.77 1.7726

4 56.3 800 441.44 1.7961

5 31.24 800 243.61 1.1496

6 -10.23 200 243.61 1.1671 0.2780

7 -10.23 200 392.14 1.7321

8 -10.23 200 186.42 0.9495

9 -40.7 50 186.42 0.9602 0.1703


P fh

gh fs

gs

165 180.19 389.20 0.9258 1.7354

200 186.42 392.14 0.9495 1.7321

201.7 186.72 392.28 0.9507 1.7319

6 (1 ) f gh x h xh

243.61=(1-x)*186.42+x*392.14

xU=0.2780

1 9(1 )( )L Uq x h h =134.72 kJ/kg

3 7 2(1 )U Uh x h x h =403.12

3T 2.6℃

4 5Hq h h =200.15 kJ/kg

2 1 4 3( )(1 )C Uw h h x h h =63.11 kJ/kg

L

C

qCOP

w =2.135

=499.25/134.72=3.706 kg/kg

We need 3.706 kg of R134a to make 1 kg of ice.

aI aw w =3.706*63.11=233.88kJ/kg

2.778*233.88=649.73 kW


Entropy increase:

Compression: comps 0.0713 kJ/kg-K

1→2: 2 1( )(1 )s s s x =0.0478 kJ/kg-K

3→4: 4 3s s s =0.0235 kJ/kg-K

Expansion: exps 0.0252 kJ/kg-K

5→6: 6 5s s s =0.0175 kJ/kg-K


8→9: 9 8(1 )( )s x s s =0.0077 kJ/kg-K

Evaporation: evas 0.0871 kJ/kg-K

4→1: 1 9(1 )( )syss x s s =0.5797 kJ/kg-K

The entropy change for water is

273 333.4 243

4.186ln 2.04ln298 273 273

ws =-1.8255 kJ/kg-K

s 0.5797-1.8255/3.706=0.0871 kJ/kg-K

Mixing: exps 0.0004 kJ/kg-K

2,7→3: 3 2 7(1 )s s x s xs =0.0004 kJ/kg-K

Condensing: conds 0.0251 kJ/kg-K

4→5: 5 4s s s =-0.6465 kJ/kg-K

2 3

0 0

Hev

q h hs

T T

=0.6716 kJ/kg-K

s -0.6465+0.6716=0.0251 kJ/kg-K

Entropy change in each process

Process Original Two stage

Compression 0.0445 0.0713

Condensing 0.0184 0.0251

Expansion 0.0567 0.0252

Evaporation 0.0836 0.0871

Mixing 0.0004

Total 0.2032 0.2109

s /kg ice 0.7839 0.7816


Three stage Cooling System

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

Example：

Find the work to cool 1 kg of water from 25℃ to -30℃ using a three stage type

refrigerator. The upper stage pressure is 800 kPa. The middle stage pressure is 300

kPa, and the lower stage pressure is 150 kPa. The refrigerant is R134a and a

compressor efficiency is 80%.

7

6

wc1

8 9 Ux 5

15 4

10 (1-Ux ) wc2

11 12 (1-Ux )

Mx 3

2

13 (1-Ux )(1-

Mx ) wc3

1

14

Lq



1 -40.7 50 373.01 1.7631

2s -0.5 200 400.48 1.7631

2 7.5 200 407.35 1.8293

3 200 405.49 1.7812

4s 400 420.74 1.7812

4 400 424.55

5 400 421.07 1.7823

6s 800 436.91 1.7823

6 800 440.87

7 31.2 800 243.61 1.1496

8 400 243.61 0.1654

9 400 403.52

10 400 211.92

11 200 211.92 0.1226

12 200 392.14

13 200 186.42

14 50 186.42

P fh gh

fs gs

165 180.19 389.20 0.9258 1.7354

200 186.42 392.14 0.9495 1.7321

201.7 186.72 392.28 0.9507 1.7319


P fh

gh fs

gs

350.9 206.75 401.32 1.0243 1.7239

400 211.92 403.52 1.0426 1.7223

415.8 213.58 404.23 1.0485 1.7218

P fh

gh fs

gs

39.6 143.18 370.32 0.7740 1.7695

50 148.12 373.01 0.7954 1.7631

51.8 148.98 373.48 0.7991 1.7620

8 10

9 10

U

h hx

h h

=0.1645

11 13

12 13

M

h hx

h h

=0.1226

1 14(1 )(1 )( )L U Mq x x h h =136.64 kJ/kg

3 2 1(1 )(1 )( )c U Mw x x h h =25.15kJ/kg

2 4 3(1 )( )c Uw x h h =15.91 kJ/kg

1 6 5cw h h =19.80

1 2 3c c c cw w w w =60.86 kJ/kg

L

C

qCOP

w =2.245

=499.25/136.64=3.654 kg/kg


aI aw w =3.654*60.86=222.37kJ/kg

2.778*222.37=617.74 kW



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

Comparison of single stage, two stage, and three stage system:

cw (kJ/kg)

Lq (kJ/kg) COP Rm (kg/s) CW (kW) Saving

single 75.24 129.40 1.738 10.718 806.4

two stage 63.11 134.72 2.135 10.295 649.7 19.4%

three stage 60.86 136.64 2.245 10.150 617.7 23.4%

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

What if more and more stages are installed?

For single stage of expansion, we have

0( ) ( )f f fgh T h T xh

0 0( ) ( ) ( )f f f fgh T h T c T T xh

0( )f

fg

c T Tx

h

0 0

lnfg fgh hP

P RT RT

What is the ultimate case that if infinity multi stages can be installed?

( ) ( ) ( ) ( )f f gmh T m dm h T dT h T dT dm

( ) ( ) ( )f

f f f f

dhh T dT h T dT h T c dT

dT

( ) ( ) ( )( ( ) ) ( ) ( )f f f f f fm dm h T dT m dm h T c dT mh T h T dm mc dT

( ) ( ) ( ) ( )f f f f gmh T mh T h T dm mc dT h T dT dm

( ) ( )f f g fgmc dT h T dm h T dT dm h dm

f

fg

cdmdT

m h


If fc and

fgh are constant, we have

0

0

ln ( )f

fg

cmT T

m h

0 0

1fmm

xm m

0ln(1 ) ( )f

fg

cx T T

h

0( )

1

f

fg

cT T

hx e

If fgh is a function of temperature, we have

0 0 0 0( )fg fg fgh h T T h T T

0 0

0 0 0 0

( )f f fg

fg fg

c c d h T TdmdT

m h T T h T T

0 0

ln lnf fg

fg

c hm

m h

0

1fc

fg

fg

hx e

h

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

Example：

Saturated liquid of R134a at 30℃ expands to -30℃. Find the fraction of saturated

liquid.

(1). Single stage of expansion.

241.79=161.12+218.68x

x=0.369

1-x=0.631

(2). Two stages of expansion. The temperature in the upper stage is 0℃.


241.79=200.0+198.36x

x=0.2107, 1-x=0.7893

200.0=161.12+218.68x

x=0.1778, 1-x=0.8222

0.7893*0.8222=0.6490

(3). Infinity number of stages. Assume fc =1.345 kJ/kg-K,

fgh =198.0 kJ/kg.

0ln(1 ) ( )f

fg

cx T T

h =-0.4076

1-x=0.6653

(4). Infinity number of stages. Assume fc =1.345 kJ/kg-K, =0.7576 kJ/kg-K.

fgh =216.68 kJ/kg, 0fgh =173.29 kJ/kg,

0

ln(1 ) lnf fg

fg

c hx

h

=-0.3973

1-x=0.6721

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

Work to compress vapor to the designated pressure is as the following.

f

fg

cdm m dT

h

0( )

0

f

fg

cT T

hm m e

1

2 1

k

k

p

Pw c T

P

0

1( )

201

f

fg

k cT Tk h f

p

fg

cPdW wdm c T m e dT

P h


0

0 0

0

1 1exp

fg

fg

h

RTfg

h

RT

h eP P P

R T Te

2 0P P

0 00

0

1

1 1 111 ( )2

fgfg fg fgfgfg

p p p

fg

kh kk h h hh kh kRTk c T c T c T TRT kRT kh

RT

P ee e e e e

Pe

00

1 1( )

0 e 1

fg f

p fg

h cT T

c T T hf

p

fg

cdW m c e TdT

h

(4.4.5). Multistage Cooling System

This system is composed of one stage of compression and two stages of expansion.

Two evaporators at different temperature may exit.

qH

5

4

wc2

6 8 3

7

2

qL1 9

wc1

10 1

qL2


4

5 3 2

9 6 7 8

10 1

1 7 6Lq h h

7 8 9(1 )h xh x h

7 9

8 9

h hx

h h

8 7

8 9

1h h

xh h

8 72 1 10 1 9

8 9

(1 )( ) ( )L

h hq x h h h h

h h

8 7 8 1 1 91 2 7 6 1 9 7 8 6

8 9 8 9 8 9

( )L L L

h h h h h hq q q h h h h h h h

h h h h h h

The ratio of upper heat transfer to the total heat transfer is defined as .

1 7 6

8 1 1 97 8 6

8 9 8 9

L

L

q h h

h h h hqh h h

h h h h

8 1 1 97 8 6 7 6

8 9 8 9

h h h hh h h h h

h h h h

8 1 1 97 6 8 6

8 9 8 9

1h h h h

h h h hh h h h


1 96 8 6

8 9

78 1

8 9

1

h hh h h

h hh

h h

h h

Work of the lower stage compressor: 2 1( )(1 )CLw h h x

Work of the upper stage compressor: 4 3CHw h h

4 5Hq h h

5 6h h

6 7 8(1 )h xh x h

Mixing process: 3 8 2(1 )h xh x h

2 1 4 3( )(1 )Cw h h x h h

7 6 1 9

2 1 4 3

(1 )( )

( )(1 )

L

C

q h h x h hCOP

w h h x h h

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

For example, 5P =800 kPa,

6 7 8 9P P P P =200 kPa, 10P =50 kPa

6h =243.62 kJ/kg

8h =392.14 kJ/kg

9 10h h =185.89 kJ/kg

1h =376.64 kJ/kg

7243.62 392.14h

7

243.62 119.05

1 0.0752h

70.0752 119.05Lq h

7h Lq 7h

Lq

0.1 257.46 138.41 0.5 314.99 142.74

0.2 271.51 139.49 0.6 329.94 143.86


0.3 285.78 140.54 0.7 345.12 145.99

0.4 300.27 141.63 0.8 360.55 146.16

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

Example： Find the work to cool 1 kg of water from 25℃ to -30℃ using a two

stage type refrigerator. The upper stage pressure is 800 kPa. The middle stage

pressure is 200 kPa, and the lower pressure is 50 kPa. The refrigerant is R134a and

the compressor efficiency is 80%.

(1). Water is cooled from 25℃ to 0℃ and totally becomes ice in the upper

evaporator, and then cooled further to -30℃ in the lower evaporator.

1 2 2 3( ) ( )L w ls iq c T T h c T T =4.186*25+333.4+2.04*30=499.25 kJ/kg

1 1 2( )L w lsq c T T h =438.05 kJ/kg

1L

L

q

q 0.877

1 96 8 6

8 9

78 1

8 9

1

h hh h h

h hh

h h

h h

7

243.62 119.05

1 0.0752h

=372.09 kJ/kg

7 9

8 9

h hx

h h

=0.9033

1 7 8Lq h h =128.64 kJ/kg

2 1 10(1 )( )Lq x h h =18.04 kJ/kg

1 2L L Lq q q =146.68 kJ/kg

1 2 1(1 )( )cw x h h =3.32 k J/kg

2 4 3cw h h =36.33 kJ/kg


1 2c c cw w w =39.65 kJ/kg

L

c

qCOP

w =146.68/39.65=3.699


1 -40.7 50 373.01 1.7631

2s -0.5 200 400.48 1.7631

2 7.5 200 407.35 1.8293

3 -8.5 200 393.61 1.7375

4s 37.9 800 422.68 1.7375

4 45 800 429.94

5 31.24 800 243.61

6 -10.23 200 243.61 0.2780

7 -10.3 200 372.25 0.9033

8 -10.3 200 392.14

9 -10.3 200 186.42

10 -40.7 50 186.42

=499.25/146.68=3.404 kg/kg


aI aw w =3.404*39.65=134.96kJ/kg

2.778*134.96=374.91 kW


(2). Water is cooled from 25℃ to 0℃ and one half becomes ice at 0℃ in the upper

evaporator, and then cooled further to -30℃ in the lower evaporator.

1 2 2 3( ) ( )L w ls iq c T T h c T T =4.186*25+333.4+2.04*30=499.25 kJ/kg

1 1 2( ) / 2L w lsq c T T h =271.35 kJ/kg


1L

L

q

q 0.544

7h 320.80 kJ/kg

7 9

8 9

h hx

h h

=0.6532

1 7 6Lq h h =77.19 kJ/kg

2 1 10(1 )( )Lq x h h =64.71 kJ/kg

1 2L L Lq q q =141.90 kJ/kg

1 2 1(1 )( )cw x h h =11.91 kJ/kg

2 4 3cw h h =37.13 kJ/kg

1 2c c cw w w =49.04 kJ/kg

L

c

qCOP

w =141.90/49.04=2.894


1 -40.7 50 373.01 1.7631

2s -0.5 200 400.48 1.7631

2 7.5 200 407.35 1.8293

3 200 397.41 1.7518

4s 800 427.12 1.7518

4 49.4 800 434.54

5 31.24 800 243.61

6 -10.23 200 243.61 0.2780

7 -10.3 200 320.81 0.6532

8 -10.3 200 392.14

9 -10.3 200 186.42

10 -40.7 50 186.42


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

Assignment： Find the work to cool 1 kg of water from 25℃ to -40℃ using a two

stage expansion refrigerator. The upper stage pressure is 800 kPa. The middle stage

pressure is 300 kPa, and the lower pressure is 30 kPa. The refrigerant is R134a and

the compressor efficiency is 80%.

Water is cooled from 25℃ to 0℃ and totally becomes ice in the upper evaporator,

and then cooled further to -40℃ in the lower evaporator.(士傑)

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

(4.4.6). Compressor efficiency

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

Example： Effect of compressor efficiency.

Find the work to cool 1 kg of water from 25℃ to -30℃ using a compression type

refrigerator with R134a. The evaporator temperature is -35℃. The condenser

pressure is 1000 kPa. The compressor efficiency is 100%.


1 -40.7 50 373.01 1.7631

2 -0.5 200 400.48 1.7631

3 -8.5 200 392.95 1.7377

4 38.0 800 422.74 1.7377

5 31.24 800 243.61

6 -10.3 200 243.61 0.2780

7 -10.3 200 372.25 0.9033

8 -10.3 200 392.14

9 -10.3 200 186.42


10 -40.7 50 186.42

1 7 8Lq h h =128.64 kJ/kg

2 1 10(1 )( )Lq x h h =18.04 kJ/kg

1 2L L Lq q q =146.68 kJ/kg

1 2 1(1 )( )cw x h h =2.66 k J/kg

2 4 3cw h h =29.79 kJ/kg

1 2c c cw w w =32.45kJ/kg

L

c

qCOP

w =146.68/32.45=4.521

=499.25/146.68=3.404 kg/kg


aI aw w =3.404*39.65=110.46kJ/kg

2.778*110.46=306.86 kW


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

In summary

COP Work (kJ/kg) s ( kJ/kg-K)

original 1.793 278.48

Two stage expansion 2.135 233.88 -44.6 0.0567

Two stage cooling 3.699 134.96 -98.92 0.0836

Perfect compressor 4.521 110.46 -24.5 0.0445

Theoretical limit 11.12 45.08

Process s ( kJ/kg-K) Cause of entropy increase


1→2 0.0445 Compressor efficiency


3→4 0.0567 Free expansion


Total 0.2032


(4.5). Conversion efficiency of exhaust energy recovery

Exhaust energy recovery is an important issue as energy efficiency becomes a

common concern for saving our environment. However, the extent that the exhaust

can be recovered is not as much as we expect.

There are three ways to recover exhaust energy. The first way is to convert the

exhaust energy to another form of thermal energy. Air pre heater in the power plant

and the regenerator in the gas turbine cycle are examples of thermal energy

utilization. The efficiency of this kind of energy recovery depends on the

effectiveness of heat exchanger. It might be quite high. Usually at least 50% of

energy can be recovered.

The second kind of energy recovery is to convert thermal energy to work, either

mechanical work or electric work. Examples of converting tools include low

temperature organic Rankine cycle, Stirling engine, and thermal electric module

(TEM). Thermodynamic cycles are used for this conversion. The efficiency of this

kind of energy recovery is limited by the Second Law of thermodynamics.

The third way of energy recovery is to convert thermal energy to chemical

energy which is stored in reformulated fuel. Steam reformulation of methane to

hydrogen is one example for waste heat recovery.

CH4+2H2O→4H2+CO2

(4.5.1). How much work can be obtained from the exhaust of an engine?

The maximum work that can be obtained is the reversible work of the exhaust.

The reversible work for a non-reacting open system is

2 2

2 1 02 2 1 1 0 2 1( ) ( ) ( ) (1 )

2 2rev H

H

V V TW H gz H gz T S S Q

T (1.1)

If kinetic energy and gravitational potential energy can be neglected, the reversible

work is


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

H

Tw h h T s s q

T (1.2)

If the heat transfer is towards the environment, TH=T0, then the reversible work is

2 1 0 2 1( )revw h h T s s

Since the exhaust is rejected to the environment, we have 2 0h h ,

2 0s s . Assume

that exhaust is an ideal gas with constant specific heat, the specific available work is

0 0 00 0 0 0 0

0

( ) ( ) ( ) ln 1 ln irev i i p i p p i

i i i

T T T Tw h h T s s c T T T c c T

T T T T

(1.3)

The specific energy that can be released if the exhaust is cooled to the environmental

temperature is

0 0( ) ( )i p ih h h c T T (1.4)

As a result, the maximum efficiency that can be reached for the conversion of

exhaust energy is

00 0

max

0

( ) ln1 ln

( ) 1

p prev

p

Tc T T T c

w Th c T T

(1.5)

Where 0

i

T

T is the ratio of environmental temperature to the temperature of

exhaust stream just issuing out engine exhaust valve.

Maximum efficiency of exhaust energy recovery

0

0.2

0.4

0.6

0 200 400 600 800 1000

exhaust temperature (C)

effi

cien

cy

eff

Fig.4.4.1 The maximum conversion efficiency


Fig. 4.4.1 shows the maximum conversion efficiency for the temperature range

of exhaust that is often encountered in normal operations of internal combustion

engines. For example, if the exhaust temperature is 400℃, then the maximum

efficiency of energy conversion would be 35.3%. It is noted that if a Carnot engine

runs between two heat reservoirs of 25℃ and 400℃, the efficiency of engine would

be 55.7%.

1

T

2

S

The cooling process may be represented as the constant pressure curve in the TS

diagram. The maximum work that can be obtained is the shaded area.

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

Example: The exhaust flow of an internal combustion engine is at 600℃, and the

flow rate is 0.088 kg/sec ( 3000 c.c. engine running at 3000 rpm). Calculate the

reversible work of the exhaust flow.

00 0 0 0 0 0

0 0

( ) ( ) ( ) ln 1 lnrev p p p

T T Tw h h T s s c T T T c c T

T T T

= 253.69 kJ/kg

rev evW mw = 22.26 kW

0

0

ln

1

1

rev

T

w T

Th

T

=0.4408


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

Assignment 4.8: A gas turbine engine is shown below. Inlet air is at 100 kPa and

300K. The pressure is raised to 1 MPa after compression, and temperature becomes

1500K after combustion. Assume that air is an ideal gas with constant heat capacity.

The compressor and the turbine can be assumed to be isentropic. The waste energy in

the exhaust flow is recovered by infinity number of Carnot engines. The atmospheric

temperature is 300K. (30%)

(1). Find out the temperature in the exit of turbine.

(2). How much work can be obtained by waste energy recovery?

(3). Find out the total efficiency if waste energy recovery is included.

QH

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

Assignment 4.9: The same amount of heat is used to heat up air at 100 kPa and

300K in a constant pressure tube. The thermal energy is then recovered by infinity

number of Carnot engines. How much work can be obtained by thermal energy

recovery?

QH

300K 300K

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


(4.5.2). Single stage engine

The maximum efficiency expressed in Equ. (1.1) can only be realized if we

have infinity number of Carnot engines arranged in order along the length of exhaust

pipe and the operating temperatures of these engines decrease continuously such that

all the heat contained in the exhaust can be utilized. Besides, the length of the pipe

should be long enough that the hot exhaust stream may cool down to the

environmental temperature as it leaves the exit of pipe. However, if only finite

number of Carnot engines are available, the exit temperature would not be as low as

T0. There also exists a finite temperature difference between exhaust gas and the

surface of engine hot side because the gas temperature decreases along the length of

pipe and the engine hot side remains at a constant temperature.

If only one set of Carnot engine is installed to extract the exhaust energy, the hot

side temperature of the engine would be lower or at most equal to the temperature of

exhaust stream such that heat transfer may occur from exhaust to engine. As a

result, the exhaust left out of the engine would be

( )e i eQ mc T T (2.1)

Where iT is the gas temperature prior to the surface of Carnot engine hot side, and

eT is the gas temperature just behind the Carnot engine.

The engine efficiency is

1 c

e

T

T (2.2)

Where eT is the hot side temperature, and

cT is the cold side temperature of Carnot

engine. If the heat sink of Carnot engine is exposed to the atmosphere, the heat sink

temperature may be equal to the atmospheric temperature. It is noted that we

assume the hot side temperature of Carnot engine is the same as the gas temperature

just behind the Carnot engine.

The work output is the product of the amount of heat transfer at the Carnot engine

hot side and the engine efficiency.


( )(1 )ce i e

e

TW Q mc T T

T (2.3)

In the above equation, the exhaust mass flow ( m ), the heat capacity (c ), the exhaust

temperature (iT ), and the engine cold side temperature (

cT ) are all known for a

specific operating condition. Only the engine hot side temperature (eT ) is not

known. It can be varied. If eT is high, the engine efficiency is high. However,

the amount of heat transfer is low. On the other side, if eT is low, the amount of

heat transfer is high. But the associated engine efficiency is low. There exists a

value of eT that would maximize the output work.

2( 1) 0i c

e e

dW TTmc

dT T (2.4)

It can be solved that the value of eT is

e i cT TT (2.5)

This is the engine hot side temperature that would maximize the output work, and the

associated output work is

( )(1 ) (1 )(1 )ci i c i

i c

TW mc T TT mcT

TT (2.6)

Where 0

i

T

T is the ratio of environmental temperature to the temperature of

exhaust stream just issuing out engine exhaust valve.

As a result, the maximum conversion efficiency for single stage engine is

2

max

(1 ) 1

1 1

revw

h

(2.7)

For example, if the exhaust temperature is 400℃ and the environmental temperature

is 25℃, then the maximum efficiency of energy conversion with single stage of

Carnot engine would be 20.1%. We can see that the maximum conversion efficiency

for single stage engine is much lower than that for a series of infinite engines.


1

T

2

S

The maximum work that can be obtained with single Carnot engine can be

represented as the red rectangular area.

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


flow rate is 0.088 kg/sec ( 3000 c.c. running at 3000 rpm). Calculate the reversible

work of the exhaust flow if single Carnot engine is used for energy conversion.

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

(4.5.3). Single stage with finite heat transfer rate

If the heat transfer coefficient between the exhaust gas and the plate is of finite

magnitude, the temperature of plate would be even lower such that heat transfer rate

and the associated work output are reduced.

( )w

dTmc hp T T

dx (2.1)

( ) 0w

dT hpT T

dx mc

1

x

wT T c e

1 hp

mc


0x , iT T

1

x

w

i w

T Tc e

T T

x L , eT T

L

e w

i w

T Te

T T

( )L

e w i wT T T T e

( ) ( )(1 )L

e i e i wQ mc T T mc T T e

( ) ( )(1 )(1 )L

ce e i e i w

w

TW Q Q mc T T mc T T e

T

w i cT TT

2

(1 )(1 )(1 ) (1 ) 1L L

i c c ci i

i ii c

TT T TW mcT e mcT e

T TTT

2

0

1(1 )

( ) 1

L

i

We

mc T T

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

Assignment 4.10: The exhaust flow of an internal combustion engine is at 600℃,

and the flow rate is 0.088 kg/sec ( 3000 c.c. running at 3000 rpm). The diameter of

exhaust pipe is 10 cm, and the length is 2 meters. Calculate the reversible work of the

exhaust flow if single Carnot engine is used for energy conversion and the heat

transfer effect is considered.

Ti

Tw

Te


0.8 0.30.0332Re Prhd

Nuk

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

(4.5.4). Multi stage engines

If more than one set of Carnot engines are installed in the exhaust pipe and these

engines are arranged in series one by one, more energy could be extracted because

these engines could be operated at different temperatures. Engines located at

downstream of exhaust flow operate at lower temperature than thoese located at

upstream. As a result, energy that could not be extracted by upstream engines could

be trapped by downstream engines. The net conversion efficiency is thus higher.

For the case that two engines are installed, the total heat transfer is

( ) ( )e i m m eQ mc T T mc T T (3.1)

Where mT is the hot side temperature of the first engine, and

eT is the hot side

temperature of the second engine. It is noted that we assume as the hot gas leaves

the first engine, it is in equilibrium with the engine such that its temperature is

identical to the engine hot side temperature. For the same reason, as the hot gas

leaves the second engine, its temperature is the same as eT .

The total work output is

( )(1 ) ( )(1 )c ci m m e

m e

T TW mc T T mc T T

T T

There are two variables, mT and

eT , that could be varied to change the work output.

The maximum work could be obtained as the following.

2( 1) (1 ) 0i c c

m m e

dW TT Tmc mc

dT T T

2( 1) 0m c

e e

dW T Tmc

dT T

The values of mT and

eT that could maximize the output work output are


e m cT T T

2

i c c

m e

TT T

T T , 2

m i m cT T T T , 2/3 1/3

m i cT T T

1/ 3 2 / 3

e i cT T T

The output work and the conversion efficiency are thus

2 / 3 1/ 3 2 / 3 1/ 3 1/ 3 2 / 3

2 / 3 1/ 3 1/ 3 2 / 3( )(1 ) ( )(1 )c c

i i c i c i c

i c i c

T TW mc T T T mc T T T T

T T T T

1 2 1 2 1

3 3 3 3 3

(1 )(1 ) ( )(1 )c c c c ci

i i i i i

T T T T TW mcT

T T T T T

1 2 1 2 1

3 3 3 3 3

max

(1 )(1 ) ( )(1 )

1

1

T

2

S

The maximum work that can be obtained with two Carnot engines can be

represented as the two red rectangular.

For example, if the exhaust temperature is 400℃, then the maximum efficiency of

energy conversion with two stages of Carnot engine would be 25.6%.

For the case that three engines are installed, the total heat transfer and the total


work are

1 1 2 2( ) ( ) ( )e i m m m m eQ mc T T mc T T mc T T

1 1 2 2

1 2

( )(1 ) ( )(1 ) ( )(1 )c c ci m m m m e

m m e

T T TW mc T T mc T T mc T T

T T T

The maximization process is quite the same as that of two engine.

2

1 1 2

( 1) (1 ) 0i c c

m m m

W TT Tmc mc

T T T

1

2

2 2

( 1) (1 ) 0m c c

m m e

W T T Tmc mc

T T T

2

2( 1) 0m c

e e

W T Tmc

T T

2

1 2

i c c

m m

TT T

T T

1

21 2m i mT TT

1

2

2

m c c

m e

T T T

T T

1

22 1m m eT T T

1 1 1

2 4 22 1 2m m e i m eT T T TT T

1 2

3 32m i eT T T

1

22e m cT T T

1 1 1 1

2 6 3 22e m c i e cT T T T T T

1 3

4 4e i cT T T

1 2 2 2

3 3 4 42m i e i cT T T T T


1 3 1

2 4 41 2m i m i cT TT T T

1 3 1 2 2 2 3 1

4 4 4 4 4 4 4 4(1 )(1 ) ( )(1 ) ( )(1 )iW mcT

1 3 1 2 2 2 3 1

4 4 4 4 4 4 4 4

max

(1 )(1 ) ( )(1 ) ( )(1 )

1

For example, if the exhaust temperature is 400℃, then the maximum efficiency of

energy conversion with three stages of Carnot engine would be 28.2%.

1

T

2

S

The maximum work that can be obtained with three Carnot engines can be

represented as the three red rectangular boxes.

The optimization process could be extrapolated to the case that N stages of engines

are installed in series. The conversion efficiency is

1 1

1 1 1

1max

( )(1 )

1

i i N iN

N N N

i


Energy recovery efficiency

0

0.2

0.4

0.6

0 5 10 15 20 25

stages of Carnot engine

max

imum

eff

icie

ncy

300 C

400 C

500 C

600 C

700 C

800 C

Fig.4.4.2 The conversion efficiency for multi stage engines

Fig. 4.4.2 shows the maximum conversion efficiency for multi stage engines

installed in series operated at several exhaust temperatures. It is noted that

increasing the stage number of the Carnot engines would enhance the exhaust energy

utilization efficiency. However, there is a limit of efficiency that can be reached.

As a matter of fact, the increase in efficiency gets lower and lower as number of

stages exceeds five.

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


flow rate is 0.088 kg/sec (3000 c.c. running at 3000 rpm). Calculate the reversible

work of the exhaust flow if two Carnot engines in series are used for energy

conversion.

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

Assignment 4.11: The exhaust flow of an internal combustion engine is at 600℃,

and the flow rate is 0.088 kg/sec (3000 c.c. running at 3000 rpm). Calculate the

reversible work of the exhaust flow and the working temperature of the Carnot

engines if three Carnot engines in series are used for energy conversion.

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


(4.5.5). Thermal energy recovery with ORC

iT eT

3 4

2

6 5

1

iT =400℃=673K, 0T =27℃=300K,

im =1kg/sec, pc =1.05 kJ/kg-K

0( )i i p iQ m c T T =391.7 kJ/sec

A three stage accumulator is design, with the stage temperature as

367 K(94℃), 449 K(176℃), 550K(277℃)

89℃ 240℃

The maximum coefficient is

0

i

T

T =0.4458

Accumulator Tc

Heat exchanger

Radiator

Regenerator

94℃

176℃

277℃


If three stage of conversion is installed, the efficiency is

1 3 1 2 2 2 3 1

4 4 4 4 4 4 4 4

max

(1 )(1 ) ( )(1 ) ( )(1 )

1

=0.280

iT =400℃=673K, 0T =27℃=300K,

( )H i p i eQ mc T T =321.3 kJ/sec

An organic Rankine cycle is adopted to convert thermal energy to work. Ammonia is

the working fluid.

Expander efficiency: 80%

Pump efficiency: 60%

Regenerator effectiveness: 70%

T(℃) P (kPa) h (kJ/kg) s (kJ/kg-K) x

1 49.4 2000 418.16

2 50 5000 427.0

3 88.9 5000 708.7 0.0135

4 240 5000 1957.9 5.5600

5s 149 2000 1762.7 5.5600

5 164.4 2000 1801.7

6 84.3 2000 1585.9

2000 kPa, fh =418.16 kJ/kg,

gh =1053.28 kJ/kg, fv =0.00176 m

3/kg

fs =1.5019 kJ/kg-K, gs =4.7683 kJ/kg

/p f pw v P =0.00176*(5000-2000)/0.6=8.8 kJ/kg

2 1 ph h w =427.0 kJ/kg

5 6

5 2

T T

T T

=0.7


6 5 5 2( )T T T T =84.3℃

6h =1585.9 kJ/kg

Energy balance in regenerator:

3 2 5 6h h h h

3 f fgh h xh

5000 kPa, fh =631.90 kJ/kg,

gh =1441.31kJ/kg, fv =0.00206 m

3/kg

3 2 5 6h h h h =642.8 kJ/kg

3 f

fg

h hx

h

=(642.8-631.9)/809.4=0.0135

4 5exw h h =156.2 kJ/kg

4 3Hq h h =1249.2 kJ/kg

exORG

H

w

q =0.125

H A HQ m q

HA

H

Qm

q =0.2572 kg/sec

A exW m w =40.17 kW

i

W

Q =0.103


(4.5.6). Thermal energy recovery with TEM

iT

Hot plate

LED

Heating side heat transfer:

( )(1 ) ( )e

e

L

e e e ie e e ie eQ m c T T e T T

2( ) 1

2

e ce P e P

P

T TQ T I R I

2( ) 1( )

2

e cP e P e ie e

P

T TT I R I T T

Thermal accumulator

TEM

Cooler (cold plate)


21

2

e cP e e e P e ie

P P

T TT I T R I T

Cooling side heat transfer:

( )(1 )c

c

L

c c c ic cQ m c T T e

2( ) 1

2

e ce P c P

P

T TQ T I R I

Characteristics of thermal electric module

( )P e c PV T T R I

Characteristics of LED

C CV V R I

Operating point


( )P e c P C CT T R I V R I

( ) ( )P e c C P CT T V R R I

( )P e c C

P C

T T VI

R R

2

( )P e c C

P C

T T VW IV

R R


(4.6). Compressed air energy

(4.6.1). Compressed air as energy storage

Air is compressed with off peak power. Air is heated up during the

compression process. However, extra heat is removed from the air with inter coolers

following compression, and is dissipated into the atmosphere as waste. Upon

removal from storage, the air must be re-heated prior to expansion in the turbine to

power a generator. The heat discarded in the intercoolers degrades efficiency, but

thus far it is the only system which has been implemented commercially. The

McIntosh CAES plant requires 0.69kWh of electricity and 1.17kWh of gas for each

1.0kWh of electrical output.

(1.0-0.69)/1.17=0.265


(4.5.2). Compressed air as power of vehicle.

【聯合報╱記者陳智華／台北報導 2008/09/04

中央大學機械系研發綠能一號氣動機車，機車不吃油，也不會吐廢氣，只要有

空氣就會跑，可節能省碳，不過，目前只能跑 1.2 公里，時速最高 30 公里，

還不知何時可量產上路。

中央大學機械系教授黃衍任與博士生沈毓達，兩年前開始研發不需用油的氣動

機車，利用高壓氣體讓機車行動，可解決空氣汙染問題，也節省能源，研究成

果 8 月刊登在 Applied Energy 期刊。

黃衍任表示，目前所知的動力裝置，大多使用汽油或電能來提供動能，讓機器

運轉，以機車為例，使用汽油經過燃燒、爆炸，產生能量，經汽缸活塞，轉換

為機械動力，再連動鏈條來帶動機車行進。

他指出，氣動機車是直接改良動力裝置，利用高壓氣體來帶動氣動馬達，以輸

出有效動力，耗能較少，除灌充高壓氣體要使用電力，此外，也不會排放廢氣。

沈毓達說，他們研發的空氣概念車是以高壓氣瓶、氣動馬達，取代油箱、引擎，

在處理控制器的介面操作下，讓機車行進，目前概念車實驗的高壓氣瓶，容量

有 10 公升，氣瓶有 15 公斤重，壓力為 100 個大氣，可跑約 1.2 公里的距離，

時速最快可達每小時 30 公里。


早期的壓縮空氣摩托車

Jem Stansfield 的壓縮空氣摩托車


The first commercial car to be powered by compressed air could be about to hit the

production lines, as Indian automaker Tata Motors prepares to build ex-Formula One

engineer Guy Nègre's design. The City Cat runs on nothing but compressed air --

which can be refueled at "air stations," and overnight using a built-in compressor --

with a not too shabby top speed of 68MPH and a range of 125 miles.

印度的 Tata Motors 設計出空氣動力汽車 City Cat，已經準備交付量產了！

City Cat 僅僅用壓縮空氣作為啟動能源，可達 68 MPH 每小時 68 英里的速

度，加滿一次可跑 125 英里。

https://www.engadget.com/search/?q=India


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

PSA Peugeot Citroën Hybrid Air concept exhibited at the 2013 Geneva Motor Show.

With Hybrid Air technology, Groupe PSA combines the environmental

advantages of compressed air and the performance of a petrol engine

without using electricity.

2.9 l / 100 km

Fuel consumption observed in certification testing for a standard body

type (Peugeot 208 or Citroën C3) with no special adaptation

https://en.wikipedia.org/wiki/Geneva_Motor_Show


The AIRPod is a compressed-air vehicle in development by Motor Development

International . The AIRPod is planned to be produced in three different

configurations that will vary the number of seats and amount of cargo storage while

keeping the same basic chassis. It is designed as an zero-emission urban vehicle.

Prototypes have been tested by KLM/AirFrance for use as emission-free vehicles in

airports.

MDI has been promising production of the AirPod each year since 2000. As of

October 2018 not a single production car has been created. Zero Pollution Motors is

now promising production by mid 2019.

https://en.wikipedia.org/wiki/Compressed-air_vehicle

https://en.wikipedia.org/wiki/Motor_Development_International

https://en.wikipedia.org/wiki/Motor_Development_International

https://en.wikipedia.org/wiki/KLM/AirFrance


(4.5.3). How much energy is required to produce a bottle of compressed air?

Assume that the compression process is carried out in two stages with a perfect

intercooler placed in between. The compressors are running with 100% efficiency.

Pm T0

PH T0 PH

T0, P0

The work of the first stage compression

1

1 2 0 0 1k

kc p s pw c T T c T

The pressure ratio of the first stage compression

0.5

0 0

m HP P

P P

If a perfect intercooler is installed between two compressors, the inlet temperature of

the second stage compressor would be the same as that of the first stage, and total

work of compression would be

1

1 2 02 1k

kc c c pw w w c T

For air, pc =1.0045 kJ/kg-K.

0T =298K. 0P =1 bar.

HP (bar) 10 50 100 300

cw (kJ/kg) 350 903 1232 1907

T0

P0

V


Assume that the bottle is filled with air at 0T and

0P initially. The final pressure

would be cP at

0T . The volume of the bottle is V .

The initial mass is 00

0

PVm

RT

The final mass is 0

CC

P Vm

RT

During the filling process, the temperature change is

2 0 0 0C C im u m u m m h

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

2 2vu c T , 0 0vu c T

2 0 0 0C C im T m T m m kT

A perfect intercooler is also installed after the second compressor so that the

compressed air is cooled to the atmospheric temperature.

0iT T after cooling

0 0 0 0 0 0 02 0 0 01 1C i

i

C C C C C

m m kT m T m m P PT kT T kT T

m m m P P

2 0 0 0

0

1 (1 )C C C

T P P Pk k k

T P P P

It is the temperature in the bottle after the filling process is finished.

2 22 0 0

0

CC C

m RT TP P P P k P

V T

2

0 0 0

1 1 1C CP P Pk k k

P P P

It is the pressure in the bottle after the filling process is finished. It is required that

2 HP P in order to fill air into the bottle to reach the target pressure.


CP (bar) 10 50 100 300

2P (bar) 13.6 69.6 139.6 419.6

2T (K) 405 415 416 417

The energy to fill the bottle would be

1

0 0 0( ) 2 ( ) 1k

kC C c p CW m m w c T m m

1

0

0 0

11

kC CkPV P P

W k kk P P

11

0

0 00 0

1 1

00

0 0 0

111

2 ( ) 1 2( 1) 1

kC C k

C C

k k

C C Ck kp

kPV P Pk k P P

k kk P PP PW

W P V PV Pc T

RT RT P

2. The compressed air is cooled down to 350K after the second stage of compression.

Energy storage efficiency without heating

Lq

Compressor Storage Tank Turbine

Compression work:

1

1

2 1 2 1 1k

p p kc p s

c c

c c Tw c T T T T


2

1

P

P

Cooling load

2 3L p cq c T T w

Expansion work

4 3 3 4 3 1

11t p t p s t p k

k

w c T T c T T c T

Efficiency

3 1 1

11 11

1 11 1

1 1

t p t ck k

k kt t c

kk kpc kk k

c

c T

w

c Tw


Energy storage efficiency with heating Hq

Lq

Compressor Storage Tank Turbine

Heating load

3H p Hq c T T

Expansion work

1

11t t p H k

k

w c T

Efficiency

1 11

1 1

3

1 11 1 1 1

1

k kp k k

t p H tk k

c ck kt c

H p H

c Tc T

w w

q c T T

1

H

T

T


(4.5.2). Energy in a bottle of high pressure air

Now we have a bottle of high pressure air at PH and T0. What is the maximum

work that can be delivered from this bottle of air? Assume that the environmental

temperature is T0 and the environmental pressure is P0.

The air contained in the bottle can be released out and then expands through a

turbine, and produces work. The final state of the air should be identical to that of

the environment. As a result, the maximum work would be the availability of the

system.

00 0 0( ) (1 )ev H

H

Tw u u T s s q

T

If the process is adiabatic, 0Hq , the reversible work is

00 0

0

ln lnev

v Pw T R RT

v P

0 0

lnevw P

RT P

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

Example: Compressed air is stored in a bottle of 100 L at 100 bars and 25℃,

calculate the maximum work that can be delivered from this bottle of air.

PVm

RT = 11.692 kg

0

0

lnrev

PW mRT

P = 4605 kJ

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


If heat transfer occurs during this process from a heat source other than the

environment, the reversible work is

00

0

ln (1 )ev H

H

P Tw RT q

P T

The heat transfer from other heat source is used to heat up the air flowing out of

the bottle. If the temperature of air is raised to TH as it flows into the working

device, the reversible work is

00 0

0

ln ( )(1 )ev p H

H

P Tw RT c T T

P T

2

0 0 00 0 0

0 0

( )(1 ) ( 1)(1 )1

H Hp H p

H H H

T T T k T Tc T T c T RT

T T T k T T

2

00

0 0

ln1

Hrev

H

P k T TW mRT

P k T T

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


calculate the maximum work if the temperature of air is raised to 200℃ as it flows

out of the bottle.

TH


2

00

0 0

ln1

Hrev

H

P k T TW mRT

P k T T

=5365 kJ

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

(4.5.3). How to obtain the energy stored in bottle?

Assume that a perfect turbine is used to convert the energy of compressed air.

The turbine can operate at any pressure ratio, and the efficiency is 1.0. The work is

1

01

k

k

p i

i

Pdw c T dm

P

where Ti and Pi are the temperature and the pressure of air at the entrance of turbine.

Initially, the bottle contains air at PH and T0, and the volume of bottle is VH.

As the pressure of bottle gets down to P0, the air ceases to flow out of bottle, and no

more work can be done.


iT

If all the process is adiabatic, the variations of pressure and mass inside the

bottle are


1

0

k

k

H

T P

T P

,

1

0

k

k

i

H

PT T

P

1

k

H H

m P

m P

0

H HH

P Vm

RT

11

1( )

k

H

H H

P Pdm m d

k P P

As a result, the work of turbine is

11 1

00

1 1

0 0 0 0 0

0 0 0

1

0

1 11

1 11

11

kk k

kk ki i

p H i

H i H H

k k

k kp H H i H i

H i i

k

H k

P P Pdw c T m dP

P P k P P

c T P V P P P PV P Pd d

k RT P P P k P P

PVX dX

k

0

iPX

P

00

11 1 1

0 0

1

0

0 0

11 1

11

HH

k

H Hk k

PPPP

kH H H

PV PVW dw X dX kX X

k k

PV P Pk k

k P P

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


calculate the turbine work as it flows out of the bottle adiabatically.


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

(4.5.3.2). Air preheating

iT

0T

If a heater is used in the upstream of turbine to heat up air to the temperature of

T0, the work is

1 11111

0 0 00 0

0 0

1 11 1 2 2 1

0 0

11 1 ( )

11 1

kkk kk

kp p H

i H H

k k k kk kH H H Hk k k k

H H

P P P P PdW c T dm c T X m d

P k P P P P

P V P P V PX X dX X X dX

k P k P

0

11

2 1

0

2 11

0

0 0

1 2

1 2 2

H

kkH H k k

PHP

k

k kkH H H H

H

P V P kW X kX

k P k

P V P k k P Pk k

k P k k P P

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



calculate the turbine work as air is heated to 25 ℃ just prior to turbine.

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

(4.5.2.3). Bottle heating

0T

If the bottle is heated such that its temperature remains at T0 during the whole

process, the variation of mass inside the bottle is

H H

m P

m P

H

H

dPdm m

P

The work is

1 1

0 0 00 0

1

0

1 1

11

k k

k k

p p H

i i H

k

H k

P P PdW c T dm c T m dX

P P P

kPVX dX

k

0

11

1

0 0

0 0

11 1

H

kH H H Hk

P

P

kPV kPV P PW kX X k k

k k P P


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


calculate the work that can be delivered from this bottle of air by ways of adiabatic

process, air preheating, and bottle heating.

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


(4.7). Chemical Exergy

Chemical exergy is defined as the maximum work that can be obtained when

the system is reacted with air in the environment. Chemical exergy depends on the

temperature and pressure of a system as well as on the composition. If the

temperature, pressure or composition of a system differs from the environment's state,

then the overall system will have exergy.

Fuel is burned with air to form products in a constant volume chamber. In

general, the environment is defined as the composition of air at 25°C and 1 atm of

pressure. Air consists of N2=75.67%, O2=20.35%, H2O(g)=3.12%, CO2=0.03% and

other gases=0.83%.

Temperature as well as pressure in the chamber will be raised up due to the heat

release of combustion. Work can be done as the temperature of system is higher

than the environment temperature and the system pressure is also higher that the

atmospheric pressure. Besides, the products of combustion contain high levels of

CO2 and H2O, and the O2 concentration is much less than that in the environment.

The difference in concentrations can also be used to deliver work. The total work

that is obtained by the temperature difference, pressure difference and the

concentration difference is called the chemical exergy.

For example, methane is burned with stoichiometric amount of air in a constant

pressure process. The reaction is as following.

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

(1) (2) (3) (4)

Fuel Products Exhaust

Air

Physical work Mixing work


State 1: methane and air at 298K and 101.3 kPa.

State 2: high temperature products of combustion.

State 3: combustion products at 298K and 101.3 kPa.

State 4: combustion products mix with environment air.

1→2

Methane and air is burned in an adiabatic chamber in which pressure remains the

same. The products of combustion are at the adiabatic flame temperature. If the

air is at the standard condition (298K, 101.3 kPa), the adiabatic flame temperature

can be determined with the conservation of enthalpy.

1 2R PH H H H

4 2 2 4,2 7.52R CH O N f CHH h h h h

2 2 2 2 2 2 2 2, ,2 7.52 2( ) 7.52P CO H O N f CO CO f H O H O NH h h h h h h h h

In a constant pressure process, the enthalpy of the system remains the same.

4 2 2 2 2 2, , ,2( ) 7.52f CH f CO CO f H O H O Nh h h h h h

2 2 2 4 2 2

0

, , , , , ,2 7.52 2

abfT

p CO p H O p N f CH f CO f H O

T

c c c dT h h h

Where 4,f CHh =-74870 kJ/kmole,

2,f COh =-393520 kJ/kmole, and 2,f H Oh =-241820

kJ/kmole.

4 2 2 4 2 2 2 2

0 0 0

1 2 7.52 2( ln ) 7.52( ln )CH O N CH O O N NS s s s s s R x s R x

2 2 2 2 2 2 2 2 2

2 2 2

0

0 0 0

2 O

, , ,

2 7.52 ln 2( ln ) 7.52( ln )

2 7.52

abf

CO H O N CO CO H O H N N

T

p CO p H O p N

T

S s s s s R x s R x s R x

dTc c c

T


2 2 2

2 2 2 4 2 2

2 2

2 2 2

0

2 7.52

O0 0 0 0 0 0

2 1 2 7.52

, , ,

( )2 7.52 2 7.52 ln

( )

2 7.52

abf

CO H N P

CO H O N CH O N

O N R

T

p CO p H O p N

T

x x xS S s s s s s s R

x x

dTc c c

T

Where the entropy of each species at the standard condition can be found in the table

below.

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

The reversible work in this process is

2 1 0 2 1( ) ( )revW H H T S S

The actual work is zero. So the combustion process is a loss of reversible work.

2→3

The high temperature products flow through a channel in which some conversion

devices are used to convert thermal energy to mechanical energy until the

temperature of products gets in equilibrium with the environment.

0

2 2 2 2 2 23 2 , , ,2 7.52 2 7.52

abf

T

CO H O N p CO p H O p N

T

H H h h h c c c dT


0

2 2 23 2 , , ,2 7.52

abf

T

p CO p H O p N

T

dTS S c c c

T

The reversible work that the burned gas may produce as it cools down to the

environment temperature is

3 2 0 3 2( )revW H H T S S

3→4

The products of combustion mix with environmental air such that the molar

fraction of each component comes to the equivalent value that exits in the

atmospheric air. During this process, the temperature remains the same.

3 4H H

2 2 2

2 2 2

2 7.52

O 0

4 3 2 7.52

O

( )ln

( )

CO H N

CO H N P

x x xS S R

x x x

The reversible work of the mixing process is

4 3 0 4 3( )revW H H T S S

The reversible work of the whole process is the sum of the reversible work of each

individual process.

2 1 0 2 1 3 2 0 3 2 4 3 0 4 3

1 3 0 4 1

( ) ( ) ( ) ( )

( )

revW H H T S S H H T S S H H T S S

H H T S S

2 2 2 4 2 2

0

1 3 , , , , , ,2 7.52 2

abfT

p CO p H O p N f CH f CO f H O f

T

H H c c c dT h h h h

This is the heating value of fuel.

2 2

2 2 4 2

2

2

O 00 0 0 0

4 1 2

0

( )2 2 ln

( )

CO H

CO H O CH O

O

x xS S s s s s R

x

It is noted that entropy will increase in the combustion process. 4 1S S .

As a result, the exergy of a fuel is greater than its heating value.


4 4 2 2 2 2 2 2

2 2

2

2 2

4 2 2 2

2

0 0 0 0

, 0 , 0 , 0 , 0

2

O 0

0 2

0

2

O 00 0 0 0

0 2

0

( ) 2( ) 2( )

( )ln

( )

( )2 2 ln

( )

rev f CH CH f CO CO f H O H O f O O

CO H

O

CO H

CH O CO H O

O

W h T s h T s h T s h T s

x xRT

x

x xg g g g RT

x

where

4 4 4

0 0

, 0CH f CH CHg h T s

2 2 2

0 0

, 0CO f CO COg h T s

2 2 2

0 0

, 0O f O Og h T s

2 2 2

0 0

, 0H O f H O H Og h T s

In general, for the fuel of a b cC H O , the reaction of the fuel with stoichiometric air is

2 2 2 2 2( )( 3.76 ) ( )3.764 2 2 4 2

a b c

b c b b cC H O a O N aCO H O a N

The reversible work of this fuel is

2 2 2 2 2 2

2

2 2

2

2 2 2

2 2

0 0 0 0

, 0 , 0 , 0 , 0

4 2

0

2

0

4 20 0 0 0

0

2

0

( ) ( ) ( )( )2 4 2

ln

( ) ln4 2 2

rev f F F f CO CO f H O H O f O O

b ca

O

b

a

CO H O

b ca

O

F O CO H O b

a

CO H O

b b cW h T s a h T s h T s a h T s

yRT

y y

yb c bg a g ag g RT

y y

Where 0 0

, 0 0X f Xg h T s

Air consists of N2=75.67%, O2=20.35%, H2O(g)=3.12%, CO2=0.03% and other

gases=0.83%.

2Oy =0.2035

2COy =0.0003

2H Oy =0.0312


Fuel a b c

2

2 2

4 2

0

2

0

ln

b ca

O

b

a

CO H O

yRT

y y

H2 0 2 0 6618

CH4 1 4 0 29390

CH3OH 1 4 1 31362

C 1 0 0 16153

C3H8 3 8 0 74932

C2H5OH 2 6 1 54133

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

Assignment : Find the reversible work of hydrogen at 298K and 101.3 kPa .

2 2 2 2 2

1( 3.76 ) 1.88

2H O N H O N

0a , 2b , 0c

2

2 2 2

2

1

20 0 0

0

0

1ln

2

O

rev H O H O

H O

yW g g g RT

y

2 2 2

0 0 0

, 0H f H Hg h T s =0-298××15.705×8.314=

2 2 2

0 0 0

, 0O f O Og h T s =0-298××24.66×8.314=

2 2 2

0 0 0

, 0H O f H O H Og h T s =-241820-298×22.70×8.314=

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

Assignment : Find the reversible work of methanol at 298K and 101.3 kPa .

3 2 2 2 2 2

3( 3.76 ) 2 5.64

2CH OH O N CO H O N

1a , 4b , 1c

2

3 2 2 2

2 2

3

20 0 0 0

0 2

0

32 ln

2

O

rev CH OH O CO H O

CO H O

yW g g g g RT

y y

3 3 3

0 0 0

, 0CH OH f CH OH CH OHg h T s =


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

It is noted that 0revW LHV T S . If 0S , then

revW LHV . We can see

that all the fuels listed in the above table have a higher chemical exergy than the

heating value. That is, the entropy change of reaction is positive.

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

Example: Find the reversible work of the process of hydrogen generation with

methanol reformation at 298K and 101.3 kPa .

3 ( ) 2 ( ) 2 23l lCH OH H O CO H

QH1 QH2

(1) (2) (3) (4)

CH3OH(l) CH3OH(g) CO2+3H2

H2O(l) H2O(g)

25℃ 250℃ 250℃ 25℃

TH TH


QL

01 4 0 4 1( ) (1 )rev H

H

TW H H T S S Q

T

3 ( ) 2 ( )1 , ,l lf CH OH f H OH h h

3 ( ) 2 ( ) 3 2

0

2 , , , ,

R

g g

T

f CH OH f H O p CH OH p H O

T

H h h c c dT

2 2 2 2

0

3 , , , ,3 3RT

f CO f H p CO p H

T

H h h c c dT

2 24 , ,3f CO f HH h h

3 ( ) 2 ( )

0 0

1 l lCH OH H OS s s

3 ( ) 2 ( ) 3 2

0

0 0

2 , ,

R

g g

T

CH OH H O p CH OH p H O

T

dTS s s c c

T

2 2 2 2

0

0 0

3 , ,3 3RT

CO H p CO p H

T

dTS s s c c

T

2 2

0 0

4 3CO HS s s

3 2 3 2

0

1 2 1 , , , ,

RT

H fg CH OH fg H O p CH OH p H O

T

Q H H h h c c dT

2 3 2HQ H H

2 2

0

1 2 3 1 4 1 , ,3RT

H H H p CO p H

T

Q Q Q H H H H c c dT

2 2 3 ( ) 2 ( )4 1 , , , ,3

l lf CO f H f CH OH f H OH H H h h h h

2 2 3 ( ) 2 ( )

0 0 0 0

4 1 3l lCO H CH OH H OS S S s s s s


2 2

0

2 2

0

00 , ,

0 00 , ,

3 (1 )

(1 ) 3

R

R

T

rev p CO p H

HT

T

p CO p H

H H T

TW H T S H c c dT

T

T TH T S c c dT

T T

CH3OH(l) CH3OH(g) H2O(l) H2O(g) CO2 H2

fh -239220 -201300 -285830 -241826 -393522 0

0s 126.809 239.709 69.950 188.834 213.795 130.678

The reforming process is undertaken at the temperature of 250℃. The temperature

of thermal reservoir should be higher than or at least equal to the reactor temperature.

It is assumed that the thermal reservoir is operated at the temperature of 400℃.

T0=25℃=298K， TR=250℃=523K，TH=400℃=673K

2

0

, 0( ) ( )RT

p CO R

T

c dT h T h T =9362 kJ/kmole

2

0

, 0( ) ( )RT

p H R

T

c dT h T h T = 6554 kJ/kmole

2 2

0

, ,3RT

p CO p H

T

c c dT = 34640 kJ

2 2 3 ( ) 2 ( ), , , ,3

l lf CO f H f CH OH f H OH h h h h 131528 kJ

2 2 3 ( ) 2 ( )

0 0 0 03l lCO H CH OH H OS s s s s = 409.07 kJ/K

2 2

0

00 , ,3 (1 )

RT

rev p CO p H

HT

TW H T S H c c dT

T

=-131528+121903+92590=-9625+92590= 82965 kJ

It is noted that the first part is a negative value, meaning that the exergy of methanol

solution is lower than the mixture of CO2 and hydrogen. In other words, the reformed


gas may do more work than the original solution. However, the increase in exergy

is accomplished with the expense of heat transfer from a thermal reservoir at higher

temperature. The heat transfer may do more work compared with the increase in

exergy. As a result, the reversible work of the whole process is a positive value,

indicating that work could be done in this process. The reality is that no work was

done.

2 2

0

, ,3RT

H p CO p H

T

Q H c c dT =166168 kJ

H

H

Q

0.7915

Remarks:

A positive value of reversible work implies that work can be done during the

reforming process.

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


(4.7). Hydrogen energy

(4.6.1). How much power can be delivered from a continuous flow of air and

hydrogen mixture at T0 and P0?

If the kinetic energy and potential energy variations can be neglected, the

reversible work for streams of air and fuel at an appropriate ratio is

0( )rev p p a a f f p p a a f fW m h m h m h T m s m s m s

The purest form of the reaction of hydrogen and oxygen is

2 2 2

1

2H O H O

For one kmole of hydrogen, the heat of reaction at T0 and P0 is

2Hh =0, 2Oh =0,

2 ( )H O gh =-241820 kJ/kmole

2Hs =130.57, 2Os =205.03,

2H Os =188.72 kJ/kmole-K,

2 2 20.5H O H OLHV h h h =-241820 kJ/kmole

2 2 20 ( )( 0.5 )rev H O g H OW LHV T s s s =241820 – 298×44.365 = 228599 kJ/kmole

This is the maximum energy that can be obtained from the reaction of pure

hydrogen and pure oxygen.

(4.6.1). Constant pressure combustion

If hydrogen and oxygen is mixed together to form combustible mixture, and

then is ignited at constant pressure environment. There will two steps to complete

this process. The initial state is1 1( , )S H .

The first step is the mixing of hydrogen and oxygen. Enthalpy remains

constant and entropy increases. After mixing process, the state is2 2( , )S H , but

2 1S S , 2 1H H .

The second step is adiabatic constant pressure combustion. Enthalpy remains

constant and entropy increases. After combustion process, the state is 3 3( , )S H , but

3 2S S , 3 2H H .


AS a result, the reversible work after the combustion process becomes

0 0 3 0 0 1( ) ( )revW LHV T S S LHV T S S

The lost work is caused by mixing and combustion process.

Now we have a bottle of high pressure air at PH and T0. What is the maximum

work that can be delivered from this bottle of air? Assume that the environmental

temperature is T0 and the environmental pressure is P0.

The air contained in the bottle can be released out and then expands through a

turbine, and produces work. The final state of the air should be identical to that of

the environment. As a result, the maximum work would be

00 1 0 0 1 0

0

( ) ( ) (1 ) ln Hrev H

H

T Pw u u T s s q RT

T P

2 2 2 2 2

1( 3.76 ) 1.88

2H O N H O N

For one kmole of hydrogen, the heat of reaction is

2Hh =0, 2Oh =0,

2Nh =0, 2 ( )H O gh =-241820 kJ/kmole

2Hs =130.57, 2Os =205.03,

2Ns =191.50, 2H Os =188.72 kJ/kmole-K,

2 2 2 2 21.88 0.5 1.88H O N H O NLHV h h h h h =241820 kJ/kmole

2 2 2 2 20 ( )(1.88 0.5 1.88 )rev N H O g H O NW LHV T s s s s s =241820 – 298×43.365 =

Chap IV Exergy Analysis(火用 - 國立中興大學 · 2020. 2. 3. · Chap IV Exergy...

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Transcript of Chap IV Exergy Analysis(火用 - 國立中興大學 · 2020. 2. 3. · Chap IV Exergy...