l26 - regeneration.pptx

7/24/2019 L26 - Regeneration.pptx

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Department of Mechanical Engineering

ME 322 – Mechanical Engineering

Thermodynamics

Lecture 26

Use of Regeneration in Vapor Power Cycles



What is Regeneration?

• Goal of regeneration

– Re!ce the f!el inp!t re"!irements #$in%

– &ncrease the temperat!re of the feewater entering the 'oiler

#increases a(erage )h in the cycle

• Res!lt of regeneration

– &ncrease thermal efficiency

• Energy so!rce for regeneration

– *igh press!re steam from the t!r'ines

• Regeneration e"!ipment

– +eewater heater #+W*%

– )his is a heat e,changer that !tili-es the high press!re steam

e,tracte from the t!r'ine to

heat the 'oiler feewater

.



Regeneration – /pen +W*

0

&ncrease temperat!reinto the 'oiler !e to

regenerati(e heating



1eeping )rac2 of Mass +low 3plits

4

Define a mass flow fraction5

Determination of the flow

fractions re"!ires applicationof the conser(ation of mass

thro!gho!t the cycle an the

conser(ation of energy

aro!n the feewaterheater#s%6

Note: &f a mass flow rate is 2nown or can 'e calc!late5 then

the flow fraction approach is not necessary7

11 y =

2 y

3 y

4 y

5 y6

y7 y

1

mass flow rate at any state

mass flow rate entering the HPT

n

n

m n y

m= =

&

&



Regeneration – Close +W*

8

)here are two types of close feewater heaters

Close +W* withDrain P!mpe

+orwar

Close +W* withDrain Cascae

9ac2war



Regeneration – Close +W*

:

E,ample – Close +W* with Drain Cascae 9ac2war

1 1 y =

2 y3 y

4 y

5 y6 y

7 y

8 y



Regeneration – M!ltiple +W*

;



Regeneration E,ample

<

Given= > Ran2ine cycle is operating with one open

feewater heater6 3team enters the high press!re t!r'ine at

8@@ psia5 A@@B+6 )he steam e,pans thro!gh the high

press!re t!r'ine to @@ psia where some of the steam is

e,tracte an i(erte to an open feewater heater6 )he

remaining steam e,pans thro!gh the low press!re t!r'ineto the conenser press!re of psia6 3at!rate li"!i e,its

the feewater heater an the conenser6

Find:

#a% the 'oiler heat transfer per l'm of steam entering thehigh press!re t!r'ine

#'% the thermal efficiency of the cycle

#c% the heat rate of the cycle



Regeneration Cycle

A

1

1

1500 psia900 F

P T

=

= °

2 100 psia P =

3 1 psia P =

4

4

1 psia0

P x

=

=

5100 psia P =

6

6

100 psia0

P x

=

=

7 1 P P =



1nown Properties

@

)he ne,t step is to '!il the property ta'le

1

1

1500 psia900 F

P T

== °

2100 psia P =

3 1 psia P =

4

4

1 psia0

P x

=

=

5100 psia P =

6

6

100 psia

0

P x

=

=

7 1 P P =



Un2nown Properties

1

1

1500 psia900 F

P T

=

= °

2100 psia P =

3 1 psia P =

4

4

1 psia

0

P

x=

=

5100 psia P =

6

6

100 psia

0

P x

=

=

7 1 P P =



>rray )a'le

.

)he res!lting property ta'le 666

ow5 we can procee with the thermoynamics7

1

1

1500 psia900 F

P T

=

= °

2100 psia P =

3 1 psia P =

4

4

1 psia

0

P

x=

=

5100 psia P =

6

6

100 psia

0

P x

=

=

7 1 P P =



9oiler Moeling

0

)he heat transfer rate at the 'oiler

can 'e fo!n 'y applying the +irst

aw5

o flow rate information is gi(en6 *owe(er5 we can fin the

heat transferre per l'm of steam entering the *P)5

1

1

1500 psia900 F

P T

=

= °

2100 psia P =

3 1 psia P =

4

4

1 psia

0

P

x=

=

5100 psia P =

6

6

100 psia

0

P x

=

=

7 1 P P =( )1 1 7inQ m h h= −& &

1 7

1

inin

Qq h hm

= = −

&

&



)!r'ine Moeling

4

)he thermal efficiency of the cycle

is gi(en 'y5

)he t!r'ine power eli(ere is5

)he flow fractions nee to

'e etermine7

1

1

1500 psia900 F

P T

=

= °

2100 psia P =

3 1 psia P =

4

4

1 psia

0

P

x=

=

5100 psia P =

6

6

100 psia

0

P x

=

=

7 1 P P =

t pnet th

in in

W W W

Q Qη

−= =

& &&

& &

1 1 2 2 3 3t W m h m h m h= − −& & & &

2 3

1 2 3

1 1 1

t m mW h h h

m m m

= − −

& & &

& & &

1 2 2 3 3

1

t t

W w h y h y h

m= = − −

&

&

1 1

1

t p

in

W m W m

Q m

−=

& && &

& &



P!mp Moeling

8

)here are two p!mps in the cycle6

)herefore5

)hen 666

)his is an important step

in the analysis6 >ll

specific energy transfers

nee to 'e 'ase on thesame flo rate6 )he

common (al!e is chosen

to 'e the inlet to the high

press!re t!r'ine #*P)%6

1

1

1500 psia900 F

P T

=

= °

2100 psia P =

3 1 psia P =

4

4

1 psia

0

P

x=

=

5100 psia P =

6

6

100 psia

0

P x

=

=

7 1 P P =1 2 p p pW W W = +& & &

( ) ( )4 5 4 6 7 6 pW m h h m h h= − + −& & &

( ) ( )4 6

5 4 7 6

1 1 1

p m mW

h h h hm m m

= − + −

& & &

& & &

( ) ( )4 5 4 6 7 6

1

p

p

W w y h h y h h

m= = − + −

&

&

1 1

1

t p t p

th

in in

W m W m w w

Q m qη

− −= =

& && &

& &



Mass Conser(ation

:

)he flow fractions m!st 'e fo!n6

)he easy flow fractions are 666

Conser(ation of mass applie to the +W* gi(es5

1

1

1500 psia900 F

P T

=

= °

2100 psia P =

3 1 psia P =

4

4

1 psia

0

P

x=

=

5100 psia P =

6

6

100 psia

0

P x

=

=

7 1 P P =

2 5 6m m m+ =& & &

2 5 6

1 1 1

m m m

m m m+ =

& & &

& & &

2 5 6 y y y+ =

1 6 71 y y y= = =

3 4 5 y y y= =



Closing the 3ystem

;

Where is the missing e"!ation?

Mass is conser(e in the +W*5 '!t

so is energy6 )herefore5 we nee

to apply the +irst aw to the +W*5

)he e"!ations can 'e sol(e7 )he

res!lt is a new property ta'le with a

col!mn for the mass flow fractions6

1

1

1500 psia900 F

P T

=

= °

2100 psia P =

3 1 psia P =

4

4

1 psia

0

P

x=

=

5100 psia P =

6

6

100 psia

0

P x

=

=

7 1 P P =

2 2 5 5 6 6m h m h m h+ =& & &

2 5 6

2 5 6

1 1 1

m m mh h h

m m m+ =

& & &

& & &

2 2 5 5 6 6 y h y h y h+ =



>!gmente >rray

<

)he !pate property ta'le 666

+rom pre(io!s analysis5

1

1

1500 psia900 F

P T

=

= °

2100 psia P =

3 1 psia P =

4

4

1 psia

0

P

x=

=

5100 psia P =

6

6

100 psia

0

P x

=

=

7 1 P P =

t p

thin

w w

qη

−

=

1 2 2 3 3t w h y h y h= − −

( ) ( )4 5 4 6 7 6 pw y h h y h h= − + −



Cycle Performance Parameters

A

)he heat rate of the cycle is5

EE! 3ol!tion #1ey Varia'les%=

1

1

1500 psia900 F

P T

=

= °

2100 psia P =

3 1 psia P =

4

4

1 psia0

P x

=

=

5100 psia P =

6

6

100 psia

0

P x

=

=

7 1 P P =

1

1 1

H!

in in in

net t p t p

Q Q m q

W W m W m w w= = =

− −

& & &

& & && &

l26 - regeneration.pptx

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