termodinamika chapter 09

8/21/2019 termodinamika Chapter 09

1/68

Chapter 9

Gas Power Cycles

Study Guide in PowerPoint

to accompany

Thermodynamics: An Engineering Approach, 5th edition

by unus A! "engel and #ichael A! $oles


2/68

2

Our study of gas power cycles will involve the study of those heat engines in which

the working fluid remains in the gaseous state throughout the cycle. We often study

the ideal cycle in which internal irreversibilities and complexities (the actual intake of

air and fuel, the actual combustion process, and the exhaust of products of

combustion among others) are removed.

We will be concerned with how the maor parameters of the cycle affect the

performance of heat engines. !he performance is often measured in terms of the

cycle efficiency.

η thnet

in

W

Q=


3/68

"

Carnot Cycle

!he #arnot cycle was introduced in #hapter $ as the most efficient heat engine that

can operate between two fixed temperatures T H and T L. !he #arnot cycle is

described by the following four processes.

#arnot #ycle%rocess &escription

'2 sothermal heat

addition

2" sentropic expansion

"* sothermal heat

reection*' sentropic

compression


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*

+ote the processes on both the P-v and T-s diagrams. !he areas under the process

curves on the P-v diagram represent the work done for closed systems. !he net

cycle work done is the area enclosed by the cycle on the P-v diagram. !he areas

under the process curves on the T-s diagram represent the heat transfer for the

processes. !he net heat added to the cycle is the area that is enclosed by the cycle

on the T-s diagram. or a cycle we know W net - Qnet therefore, the areas enclosed on

the P-v and T-s diagrams are e/ual.

η th Carnot L

H

T

T , = −1

We often use the #arnot efficiency as a means to think about ways to improve thecycle efficiency of other cycles. One of the observations about the efficiency of both

ideal and actual cycles comes from the #arnot efficiency0 !hermal efficiency

increases with an increase in the average temperature at which heat is supplied to

the system or with a decrease in the average temperature at which heat is reected

from the system.


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$

Air-Standard Assumptions

n our study of gas power cycles, we assume that the working fluid is air, and the air

undergoes a thermodynamic cycle even though the working fluid in the actual power

system does not undergo a cycle.

!o simplify the analysis, we approximate the cycles with the following assumptions0

1!he air continuously circulates in a closed loop and always behaves as an ideal gas.

1 ll the processes that make up the cycle are internally reversible.

1!he combustion process is replaced by a heataddition process from an external

source.

1 heat reection process that restores the working fluid to its initial state replaces the

exhaust process.

1!he coldairstandard assumptions apply when the working fluid is air and has

constant specific heat evaluated at room temperature (2$o# or 33o).


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4

Terminology for Reciprocating Devices

!he following is some terminology we need to understand for reciprocating engines5

typically pistoncylinder devices. 6et7s look at the following figures for the definitions

of top dead center (!), bottom dead center (8), stroke, bore, intake valve,

exhaust valve, clearance volume, displacement volume, compression ratio, and mean

effective pressure.


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3

!he compression ratio r of an engine is the ratio of the maximum volume to the

minimum volume formed in the cylinder.

r V

V

V

V

BDC

TDC

= =max

min

!he mean effective pressure (MEP ) is a fictitious pressure that, if it operated on the

piston during the entire power stroke, would produce the same amount of net work as

that produced during the actual cycle.

MEP W

V V

w

v v

net net =−

=−max min max min


8/68

9

Otto Cycle T!e "deal Cycle for Spar#-"gnition Engines

#onsider the automotive sparkignition power cycle.

%rocesses

ntake stroke

#ompression stroke

%ower (expansion) stroke

:xhaust stroke

Often the ignition and combustion process begins before the completion of the

compression stroke. !he number of crank angle degrees before the piston reaches

! on the number one piston at which the spark occurs is called the engine timing.

What are the compression ratio and timing of your engine in your car, truck, or

motorcycle;


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<

T!e air-standard Otto cycle is the ideal cycle that approximates the sparkignition

combustion engine.

%rocess &escription

'2 sentropic compression2" #onstant volume heat addition

"* sentropic expansion

*' #onstant volume heat reection

!he P-v and T-s diagrams are


10/68

'=


11/68

''

!hermal :fficiency of the Otto cycle0

η thnet

in

net

in

in out

in

out

in

W

Q

Q

Q

Q Q

Q

Q

Q= = =

−= −1

+ow to find Qin and Qout.

pply first law closed system to process 2", V - constant.

!hus, for constant specific heats,

Q U

Q Q mC T T

net

net in v

,

, ( )

23 23

23 3 2

=

= = −

∆


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'2

pply first law closed system to process *', V - constant.

!hus, for constant specific heats,

Q U

Q Q mC T T

Q mC T T mC T T

net

net out v

out v v

,

, ( )

( ) ( )

41 41

41 1 4

1 4 4 1

=

= − = −

= − − = −

∆

!he thermal efficiency becomes

η th Ottoout

in

v

v

Q

Q

mC T T

mC T T

,

( )

( )

= −

= − −−

1

1 4 1

3 2


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'"

η th OttoT T

T T

T T T

T T T

,

( )

( )

( / )

( / )

= − −

−

= − −

−

1

1

1

1

4 1

3 2

1 4 1

2 3 2

>ecall processes '2 and "* are isentropic, so

?ince V " - V 2 and V * - V ', we see that

T

T

T

T

or T

T

T

T

2

1

3

4

4

1

3

2

=

=


14/68

'*

!he Otto cycle efficiency becomes

η th OttoT

T , = −1

1

2

s this the same as the #arnot cycle efficiency;

?ince process '2 is isentropic,

where the compression ratio is r - V '@V 2 and

η th Otto k r , = − −1

11


15/68

'$

We see that increasing the compression ratio increases the thermal efficiency.

Aowever, there is a limit on r depending upon the fuel. uels under high temperature

resulting from high compression ratios will prematurely ignite, causing knock.


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'4

E$ample %-&

n Otto cycle having a compression ratio of


17/68

'3

!he first law closed system for process 2" was shown to reduce to (your homework

solutions must be complete that is, develop your e/uations from the application of

the first law for each process as we did in obtaining the Otto cycle efficiency e/uation)

Q mC T T in v= −( )3 2

6et qin - Qin @ m and m - V '@v '

v RT

P

kJ

kg K K

kPa

m kPa

kJ

m

kg

11

1

3

3

0 287 290

95

0875

=

= ⋅

=

. ( )

.


18/68

'9

Q

mQ

v

V

kJ

m

kg

m

kJ

kg

inin

in= =

= ⋅

=

−

1

1

3

3 37 5

0875

38 10

1727

.

.

.

!hen,

T T

C

K

kJ

kg

kJ

kg K

K

in

v3 2

6984

1727

0718

31037

= +

= +

⋅

=

.

.

.


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'<

Esing the combined gas law (F" - F2)

P P T

T MPa3 2

3

2

9 15= = .

%rocess "* is isentropic therefore,

1 1 1.4 1

34 3 3

4

1 1(3103.7)

9

1288.8

k k V

T T T K V r

K

− − − = = =

=


20/68

2=

%rocess *' is constant volume. ?o the first law for the closed system gives, on a

mass basis,Q mC T T

Q

mC T T

kJ

kg K K

kJ

kg

out v

out out

v

= −

= = −

=⋅

−

=

( )

( )

. ( . )

.

4 1

4 1

0 718 1288 8 290

7171

!he first law applied to the cycle gives (>ecall ∆ucycle - =)

w

kJ

kg

kJ

kg

net net in out = = −

= −

=

( . )

.

1727 717 4

1009 6


21/68

2'

!he thermal efficiency is

η th Ottonet

in

w

kJ

kg

kJ

kg or

,

.

. .

= =

=

1009 6

1727

0 585 58 5%

!he mean effective pressure is

max min max min

1 2 1 2 1 1

3

3

(1 / ) (1 1/ )

1009.6

12981

0.875 (1 )9

net net

net net net

W w MEP

V V v v

w w w

v v v v v v r

kJ

m kPakg kPa

m kJ

kg

= =− −

= = =− − −

= =

−


22/68

22

!he back work ratio is (can you show that this is true;)

2 112 2 1

exp 34 3 4 3 4

( ) ( )

( ) ( )

0.225 22.5%

!om" v

v

w C T T u T T BWR

w u C T T T T

or

−∆ −= = = =

−∆ − −

=Air-Standard Diesel Cycle

!he airstandard &iesel cycle is the ideal cycle that approximates the &iesel

combustion engine

%rocess &escription'2 sentropic compression

2" #onstant pressure heat addition

"* sentropic expansion

*' #onstant volume heat reection

!he P-v and T-s diagrams are


23/68

2"


24/68

2*

T!ermal efficiency of t!e Diesel cycle

η th Die#e$ net

in

out

in

W

Q

Q

Q, = = −1

+ow to find Qin and Qout.

pply the first law closed system to process 2", P - constant.

!hus, for constant specific heats

Q U P V V Q Q mC T T mR T T

Q mC T T

net

net in v

in "

,

,

( )( ) ( )

( )

23 23 2 3 2

23 3 2 3 2

3 2

= + −= = − + −

= −

∆


25/68

2$

pply the first law closed system to process *', V - constant (ust as we did for the

Otto cycle)

!hus, for constant specific heats

Q U

Q Q mC T T Q mC T T mC T T

net

net out v

out v v

,

, ( )( ) ( )

41 41

41 1 4

1 4 4 1

=

= − = −= − − = −

∆


η th Die#e$ out

in

v

"

Q

Q

mC T T

mC T T

,

( )

( )

= −

= − −

−

1

1 4 1

3 2


26/68

24

η th Die#e$ v

"

C T T

C T T

k

T T T

T T T

,

( )

( )

( / )

( / )

= − −

−

= − −

−

1

1

1 1

1

4 1

3 2

1 4 1

2 3 2

What is T "@T 2 ;

PV

T

PV

T P P

T

T

V

V r !

3 3

3

2 2

2

3 2

3

2

3

2

= =

= =

where

where r c is called the cutoff ratio, defined as V " @V 2, and is a measure of the duration

of the heat addition at constant pressure. ?ince the fuel is inected directly into the

cylinder, the cutoff ratio can be related to the number of degrees that the crank

rotated during the fuel inection into the cylinder.


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23

What is T *@T ' ;

PV

T

PV

T V V

T

T

P

P

4 4

4

1 1

1

4 1

4

1

4

1

= =

=

where


PV PV PV PV k k k k 1 1 2 2 4 4 3 3= = and

?ince V * - V ' and P " - P 2, we divide the second e/uation by the first e/uation andobtain

!herefore,


28/68

29

η th Die#e$

!

k

!

k

!

k

!

k

T T T

T T T

k

T

T

r

r

r

r

k r

,

( / )

( / )

( )

( )

= − −

−

= − −

−

= − −

−−

1 1 1

1

1

1 1

1

1 1 1

1

1 4 1

2 3 2

1

2

1

What happens as r c goes to '; ?ketch the P-v diagram for the &iesel cycle and showr c approaching ' in the limit.

%

v

When r G ' for a fixed r 8ut


29/68

2<

η η th Die#e$ th Otto, ,< r r Die#e$ Otto> η η th Die#e$ th Otto, ,>When r

c G ' for a fixed r , . 8ut,

since , .

'rayton Cycle

!he 8rayton cycle is the airstandard ideal cycle approximation for the gasturbine

engine. !his cycle differs from the Otto and &iesel cycles in that the processes

making the cycle occur in open systems or control volumes. !herefore, an open

system, steadyflow analysis is used to determine the heat transfer and work for thecycle.

We assume the working fluid is air and the specific heats are constant and will

consider the coldairstandard cycle.


30/68

"=

T!e closed cycle gas-turine engine


31/68

"'

%rocess &escription

'2 sentropic compression (in a

compressor)

2" #onstant pressure heat addition

"* sentropic expansion (in a turbine)

*' #onstant pressure heat reection

!he T-s and P-v diagrams are


32/68

"2

Thermal e%%iciency o% the $rayton cycle

η th Bra%tonnet

in

out

in

W

Q

Q

Q, = = −1

+ow to find Qin and Q

out.

pply the conservation of energy to process 2" for P - constant (no work), steady

flow, and neglect changes in kinetic and potential energies.

E E

m h Q m h

in out

in

=

+ =2 2 3 3!he conservation of mass gives

m m

m m m

in out == =2 3

or constant specific heats, the heat added per unit mass flow is

( )

( )

( )

Q m h h

Q mC T T

Q

m C T T

in

in "

in

in

"

= −

= −

= = −

3 2

3 2

3 2

f f f f ( 7


33/68

""

!he conservation of energy for process *' yields for constant specific heats (let7s

take a minute for you to get the following result)

( )

( )

( )

Q m h h

Q mC T T

Q

mC T T

out

out "

out out

"

= −

= −

= = −

4 1

4 1

4 1


η th Bra%tonout

in

out

in

"

"

Q

Q

C T T

C T T

,

( )

( )

= − = −

= − −

−

1 1

1 4 1

3 2

η th Bra%ton T T T T

T T T

T T T

, ( )( )

( / )

( / )

= − −−

= − −

−

1

1 1

1

4 1

3 2

1 4 1

2 3 2

> ll ' 2 d " * i t i


34/68

"*


?ince P " - P 2 and P * - P ', we see that

3 32 4

1 4 1 2

or T T T T

T T T T = =

!he 8rayton cycle efficiency becomes

η th Bra%tonT

T , = −1 1

2

s this the same as the #arnot cycle efficiency;

?ince process '2 is isentropic,

h th ti i P @P d


35/68

"$

where the pressure ratio is r p - P 2@P ' and

η th Bra%ton "

k k r

, ( )/= − −1

11

E$tra Assignment

:valuate the 8rayton cycle efficiency by determining the net work directly from the

turbine work and the compressor work. #ompare your result with the above

expression. +ote that this approach does not re/uire the closed cycle assumption.

E l % )


36/68

"4

E$ample %-)

!he ideal airstandard 8rayton cycle operates with air entering the compressor at


37/68

"3

or constant specific heats, the compressor work per unit mass flow is

( )

( )

( )

W m h h

W mC T T

wW

mC T T

!om"

!om" "

!om"

!om"

"

= −

= −

= = −

2 1

2 1

2 1

?ince the compressor is isentropic

w C T T= ( )


38/68

"9

w C T T

kJ

kg K K

kJ

kg

!om" "= −

=⋅

−

=

( )

. ( . )

.

2 1

1005 492 5 295

19815

!he conservation of energy for the turbine, process "*, yields for constant specific

heats (let7s take a minute for you to get the following result)

( )

( )

( )

W m h h

W mC T T

w W

mC T T

tur&

tur& "

tur&tur&

"

= −

= −

= = −

3 4

3 4

3 4

?ince process "* is isentropic

?ince P - P and P - P we see that


39/68

"<

?ince P " - P 2 and P * - P ', we see that

( 1) /

4

3

( 1) /(1.4 1)/1.4

4 3

1

1 11100 659.1

6

k k

"

k k

"

T

T r

T T K K r

−

−−

=

= = =

w C T T kJ

kg K K

kJ kg

tur& "= − = ⋅ −

=

( ) . ( . )

.

3 4 1 005 1100 659 1

4425

We have already shown the heat supplied to the cycle per unit mass flow in process

2" is

( ) . ( . )

.

m m m

m h Q m h

Q

mh h

C T T kJ

kg K K

kJ

kg

in

inin

"

2 3

2 2 3 3

3 2

3 2 1005 1100 492 5

609 6

= =

+ =

= = −

= − =⋅

−

=

!he net work done by the cycle is


40/68

*=


w w w

kJ

kg kJ

kg

net tur& !om"= −

= −

=

( . . )

.

4425 19815

244 3

!he cycle efficiency becomes

η th Bra%ton net

in

w

kJ

kg

kJ

kg

or

,

.

.

.

=

= =244 3

609 6

0 40 40%

!he back work ratio is defined as


41/68

*'


BWR w

w

w

w

kJ

kg

kJ

kg

in

out

!om"

tur&

= =

= =19815

4425

0448.

.

.

+ote that T * - 4$


42/68

*2

6etCs take a closer look at the effect of the pressure ratio on the net work done.

w w w

C T T C T T

C T T T C T T T

C T r

C T r

net tur& !om"

" "

" "

"

"

k k " "

k k

= −

= − − −

= − − −

= − − −−−

( ) ( )

( / ) ( / )

( ) ( )( ) /

( )/

3 4 2 1

3 4 3 1 2 1

3 1 1

1

1 1

11

1

+ote that the net work is *ero when


43/68

*"

+ote that the net work is *ero when/( 1)

3

1

1

k k

" "

T r an' r

T

−

= =

or fixed T " and T ', the pressure ratio that makes the work a maximum is obtained

from0'w

'r

net

"

= 0

!his is easier to do if we let X - r p(k ')@k

w C T (

C T ( net " "= − − −3 11 1 1( ) ( )

'w

'( C T ( C T net " "= − − − − =

−3

2

10 1 1 0 0[ ( ) ] [ ]

?olving for X

!hen the r that makes the work a maximum for the constant property case and fixed


44/68

**

!hen, the r p that makes the work a maximum for the constant property case and fixed

T " and T ' is

or the ideal 8rayton cycle, show that the following results are true.

1When r p = r p, max work, T * - T 21When r p < r p, max work, T * G T 21When r p > r p, max work, T * H T 2

!he following is a plot of net work per unit mass and the efficiency for the aboveexample as a function of the pressure ratio.

0 2 4 6 8 10 12 14 16 18 20 22

120

140

160

180

200

220

240

260

280

0.15

0.20

0.25

0.30

0.35

0.40

0.45

0.50

0.55

0.60

Pratio

w n e t

k J / k g

t h , B r a y t o n

T1 = 22C

P1 = 95 kPa

T3 = 1100 K

t = c = 100%

rp,a!

Regenerative 'rayton Cycle


45/68

*$

Regenerative 'rayton Cycle

or the 8rayton cycle, the turbine exhaust temperature is greater than the

compressor exit temperature. !herefore, a heat exchanger can be placed between

the hot gases leaving the turbine and the cooler gases leaving the compressor. !his

heat exchanger is called a regenerator or recuperator. !he sketch of the regenerative8rayton cycle is shown below.


46/68

*4

We define the regenerator effectiveness

εregen as the ratio of the heat transferred to

the compressor gases in the regenerator to the maximum possible heat transfer to

the compressor gases.

h h

h h h h

h h

h h

regen a!t

regen

regen

regen a!t

regen

,

, max

,

, max

= −

= − = −

= = −−

5 2

5 2 4 2

5 2

4 2

ε

or ideal gases using the coldairstandard assumption with constant specific heats


47/68

*3

or ideal gases using the cold air standard assumption with constant specific heats,

the regenerator effectiveness becomes

5 2

4 2

regen

T T

T T ε

−≅

−

Esing the closed cycle analysis and treating the heat addition and heat reection as

steadyflow processes, the regenerative cycle thermal efficiency is

η th regenout

in

h hh h

, = −

= − −−

1

1 6 1

3 5

+otice that the heat transfer occurring within the regenerator is not included in the

efficiency calculation because this energy is not heat transferred across the cycle

boundary.

ssuming an ideal regenerator εregen - ' and constant specific heats, the thermalefficiency becomes (take the time to show this on your own)


48/68

*9

When does the efficiency of the airstandard 8rayton cycle e/ual the efficiency of the

airstandard regenerative 8rayton cycle; f we set ηth,8rayton - ηth,regen then

>ecall that this is the pressure ratio that maximiIes the net work for the simple

8rayton cycle and makes T * - T 2. What happens if the regenerative 8rayton cycle

operates at a pressure ratio larger than this value;

or fixed T " and T ', pressure ratios greater than this value cause T * to be less than


49/68

*<

" ', p g *T 2, and the regenerator is not effective.

What happens to the net work when a regenerator is added;

What happens to the heat supplied when a regenerator is added;

!he following shows a plot of the regenerative 8rayton cycle efficiency as a function

of the pressure ratio and minimum to maximum temperature ratio, T '@T ".

E$ample %-+ Regenerative 'rayton Cycle


50/68

$=

E$ample % + Regenerative 'rayton Cycle

ir enters the compressor of a regenerative gasturbine engine at '== k%a and "== D

and is compressed to 9== k%a. !he regenerator has an effectiveness of 4$ percent,

and the air enters the turbine at '2== D. or a compressor efficiency of 3$ percent

and a turbine efficiency of 94 percent, determine(a) !he heat transfer in the regenerator.

(b) !he back work ratio.

(c) !he cycle thermal efficiency.

#ompare the results for the above cycle with the ones listed below that have the

same common data as re/uired. !he actual cycles are those for which the turbineand compressor isentropic efficiencies are less than one.

(a) !he actual cycle with no regeneration, ε - =.(b) !he actual cycle with ideal regeneration, ε - '.=.(c) !he ideal cycle with regeneration, ε - =.4$.(d) !he ideal cycle with no regeneration, ε - =.(e) !he ideal cycle with ideal regeneration, ε - '.=.

We assume air is an ideal gas with constant specific heats, that is, we use the cold

airstandard assumption.

!he cycle schematic is the same as above and the T-s diagram showing the effects of


51/68

$'

!he cycle schematic is the same as above and the T s diagram showing the effects of

compressor and turbine efficiencies is below.

s

T

'== k%a

9== k%a

T-s Diagram for ,as Tur(ine it! Regeneration

&

)s)a

.

+

/s

/a

0

Summary of Results


52/68

$2

Summary of Results

!"#$e "pe 'a$ 'a$ 'a$ dea$ dea$ dea$

εreen

0.00 0.65 1.00 0.00 0.65 1.00

η#omp

0.75 0.75 0.75 1.00 1.00 1.00

η'r*

0.86 0.86 0.86 1.00 1.00 1.00

in +/+ 578.3 504.4 464.6 659.9 582.2 540.2

w#omp

+/+ 326.2 326.2 326.2 244.6 244.6 244.6

w'r*

+/+ 464.6 464.6 464.6 540.2 540.2 540.2

w#omp

/w'r*

0.70 0.70 0.70 0.453 0.453 0.453

ηh

24.0% 27.5% 29.8% 44.8% 50.8% 54.7%

Compressor analysis


53/68

$"

p y

!he isentropic temperature at compressor exit is( 1) /

2 2

1 1

( 1) /

(1.4 1)/1.422 1

1

800300 ( ) 543.4

100

k k

#

k k

#

T P

T P

P kPaT T K K

P kPa

−

−

−

=

= = =

!o find the actual temperature at compressor exit, T 2a, we apply the compressor

efficiency

η

η

!om"

i#en !om"

a!t !om"

#

a

#

a

a

!om"

#

w

w

h h

h h

T T

T T

T T T T

K K

K

= = −

− ≅

−−

= + −

= + −

=

,

,

( )

.(543. )

.

2 1

2 1

2 1

2 1

2 1 2 1

1

300 1

0 754 300

624 6

?ince the compressor is adiabatic and has steadyflow


54/68

$*

p y

2 1 2 1( )

1.005 (624.6 300) 326.2

!om" a " aw h h C T T

kJ kJ K

kg K kg

= − = −

= − =

⋅Turine analysis

!he conservation of energy for the turbine, process "*, yields for constant specific

heats (let7s take a minute for you to get the following result)

( ) ( )

( )

W m h hW mC T T

w W

mC T T

tur& a

tur& " a

tur&tur&

" a

= −= −

= = −

3 4

3 4

3 4

?ince P " - P 2 and P * - P ', we can find the isentropic temperature at the turbine exit.


55/68

$$

" 2 * '

( 1) /

4 4

3 3

( 1) /

(1.4 1)/1.444 3

3

1001200 ( ) 662.5800

k k

#

k k

#

T P

T P

P kPaT T K K P kPa

−

−

−

=

= = =

!o find the actual temperature at turbine exit, T *a, we apply the turbine efficiency.

η

η

tur&

a!t tur&

i#en tur&

a

#

a

#

a tur& #

a

w

w

h h

h h

T T

T T

T T T T

K K

K T

= =

−

− ≅

−

−= − −= − −= >

,

,

( )

. ( . )

.

3 4

3 4

3 4

3 4

4 3 3 4

2

1200 0 86 1200 662 5

737 7

!he turbine work becomes


56/68

$4

w h h C T T

kJ

kg K K

kJ

kg

tur& a " a= − = −

=⋅

−

=

3 4 3 4

1 005 1200 737 7

464 6

( )

. ( . )

.


BWR w

w

w

wkJ

kg

kJ

kg

in

out

!om"

tur&

= =

= =3262

464 6

0 70

.

.

.

Regenerator analysis


57/68

$3

!o find T $, we apply the regenerator effectiveness.

ε

ε

regena

a a

a regen a a

T T

T T T T T T

K K

K

≅ −

−= + −

= + −=

5 2

4 2

5 2 4 2

624 6 0 65 737 7 624 6

6981

( )

. . ( . . )

.

!o find the heat transferred from the turbine exhaust gas to the compressor exit gas,


58/68

$9

( )

. ( . . )

.

m h Q m h

m m m

Q

mh h

C T T

kJ kg K

K

kJ

kg

a a regen

a

regen

regen

a

" a

2 2 5 5

2 5

5 2

5 2

1005 6981 624 6

739

+ =

= =

= = −

= −

= ⋅ −

=

apply the steadyflow conservation of energy to the compressor gas side of the

regenerator.

Esing qregen, we can determine the turbine exhaust gas temperature at the regenerator


59/68

$<

4 4 6 6

4 6

4 6 4 6

6 4

( )

73.9

737.7

1.005

664.2

a a regen

a

regen

regen a " a

regen

a

"

m h Q m h

m m m

Q h h C T T

m

kJ

kg T T K

kJ C kg K

K

= +

= =

= = − = −

= − = −

⋅=

&& &

& & &

&

&

Es g qregen, e ca de e e e u b e e aus gas e pe a u e a e ege e a o

exit.

1eat supplied to cycle


60/68

4=

pply the steadyflow conservation of energy to the heat exchanger for process $".

We obtain a result similar to that for the simple 8rayton cycle.

h h C T T

kJ

kg K K

kJ

kg

in "

= − = −

=⋅

−

=

3 5 3 5

1 005 1200 6981

504 4

( )

. ( . )

.

Cycle t!ermal efficiency


(464.6 326.2) 138.4

net tur& !om"w w w

kJ kJ

kg kg

= −

= − =

!he cycle efficiency becomes


61/68

4'

η th Bra%tonnet

in

w

kJ

kg

kJ

kg

or

,

.

.

. .

=

= =1384

504 4

0 274 27 4%

Jou are encouraged to complete the calculations for the other values found in the

summary table.

Ot!er 2ays to "mprove 'rayton Cycle Performance


62/68

42

ntercooling and reheating are two important ways to improve the performance of the

8rayton cycle with regeneration.

!he !s diagram for this cycle is shown below. ?ketch the %v diagram.


63/68

4"

v

%

"ntercooling


64/68

4*

"ntercooling

When using multistage compression, cooling the working fluid between the stages will

reduce the amount of compressor work re/uired. !he compressor work is reduced

because cooling the working fluid reduces the average specific volume of the fluid

and thus reduces the amount of work on the fluid to achieve the given pressure rise.

!o determine the intermediate pressure at which intercooling should take place to

minimiIe the compressor work, we follow the approach shown in #hapter 3.

or the adiabatic, steadyflow compression process, the work input to the compressorper unit mass is

4 32 4

1 31 2

- -!om"w v 'P v 'P v 'P v 'P + +∫ ∫ ∫ ∫ =

or the isentropic compression process

k k


65/68

4$

[ ]

2 2 1 1 4 4 3 3

2 1 4 3

1 2 1 3 4 3

( 1) /( 1) /

2 41 3

1 3

- ( ) ( )1 1

( ) ( )

1 1

( / 1) ( / 1)1

1 11

!om"

k k k k

k k w P v Pv P v P v

k k

k kR R T T T T

k k k

R T T T T T T k

k P P R T T

k P P

−−

− + −

= − + −

= − + −

= − + −

+otice that the fraction kR/(k-' ) = C p.

#an you obtain this relation another way; Aint0 apply the first law to processes '*.

or twostage compression, let7s assume that intercooling takes place at constant

d th b l d t th i l t t t f th


66/68

44

pressure and the gases can be cooled to the inlet temperature for the compressor,

such that P " - P 2 and T " - T '.

!he total work supplied to the compressor becomes

!o find the unknown pressure P 2 that gives the minimum work input for fixed

compressor inlet conditions T ', P ', and exit pressure P *, we set

'w P

'P

!om" ( )2

20=

!his yields


67/68

43

P P P 2 1 4=

or, the pressure ratios across the two compressors are e/ual.

P

P

P

P

P

P 2

1

4

2

4

3= =

ntercooling is almost always used with regeneration. &uring intercooling the

compressor final exit temperature is reduced therefore, more heat must be supplied

in the heat addition process to achieve the maximum temperature of the cycle.

>egeneration can make up part of the re/uired heat transfer.

!o supply only compressed air, using intercooling re/uires less work input. !he next

time you go to a home supply store where air compressors are sold, check the larger

air compressors to see if intercooling is used. or the larger air compressors, the

compressors are made of two pistoncylinder chambers. !he intercooling heat

exchanger is often a pipe with a attached fins that connects the large pistoncylinderchamber with the smaller pistoncylinder chamber. ?ometimes the fly wheel used to

drive the compressor has fan type blades a spokes to increase air flow across the

compressor and heat exchanger pipe to improve the intercooling effect.

E$tra Assignment


68/68

Obtain the expression for the compressor total work by applying conservation of

energy directly to the low and highpressure compressors.

Re!eating

When using multistage expansion through two or more turbines, reheating

between stages will increase the net work done (it also increases the

re/uired heat input). !he regenerative 8rayton cycle with reheating was shown

above.

!he optimum intermediate pressure for reheating is the one that maximiIes the

turbine work. ollowing the development given above for intercooling and assuming

reheating to the highpressure turbine inlet temperature in a constant pressure

steadyflow process, we can show the optimum reheat pressure to be

P P P 7 6 9=or the pressure ratios across the two turbines are e/ual.

P

P

P

P

P

P

6

7

7

9

8

9

= =

termodinamika chapter 09

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