Post on 22-Feb-2018
7/24/2019 L26 - Regeneration.pptx
http://slidepdf.com/reader/full/l26-regenerationpptx 1/19
Department of Mechanical Engineering
ME 322 – Mechanical Engineering
Thermodynamics
Lecture 26
Use of Regeneration in Vapor Power Cycles
7/24/2019 L26 - Regeneration.pptx
http://slidepdf.com/reader/full/l26-regenerationpptx 2/19
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
.
7/24/2019 L26 - Regeneration.pptx
http://slidepdf.com/reader/full/l26-regenerationpptx 3/19
Regeneration – /pen +W*
0
&ncrease temperat!reinto the 'oiler !e to
regenerati(e heating
7/24/2019 L26 - Regeneration.pptx
http://slidepdf.com/reader/full/l26-regenerationpptx 4/19
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= =
&
&
7/24/2019 L26 - Regeneration.pptx
http://slidepdf.com/reader/full/l26-regenerationpptx 5/19
Regeneration – Close +W*
8
)here are two types of close feewater heaters
Close +W* withDrain P!mpe
+orwar
Close +W* withDrain Cascae
9ac2war
7/24/2019 L26 - Regeneration.pptx
http://slidepdf.com/reader/full/l26-regenerationpptx 6/19
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
7/24/2019 L26 - Regeneration.pptx
http://slidepdf.com/reader/full/l26-regenerationpptx 7/19
Regeneration – M!ltiple +W*
;
7/24/2019 L26 - Regeneration.pptx
http://slidepdf.com/reader/full/l26-regenerationpptx 8/19
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
7/24/2019 L26 - Regeneration.pptx
http://slidepdf.com/reader/full/l26-regenerationpptx 9/19
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 =
7/24/2019 L26 - Regeneration.pptx
http://slidepdf.com/reader/full/l26-regenerationpptx 10/19
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 =
7/24/2019 L26 - Regeneration.pptx
http://slidepdf.com/reader/full/l26-regenerationpptx 11/19
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 =
7/24/2019 L26 - Regeneration.pptx
http://slidepdf.com/reader/full/l26-regenerationpptx 12/19
>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 =
7/24/2019 L26 - Regeneration.pptx
http://slidepdf.com/reader/full/l26-regenerationpptx 13/19
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
= = −
&
&
7/24/2019 L26 - Regeneration.pptx
http://slidepdf.com/reader/full/l26-regenerationpptx 14/19
)!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
−=
& && &
& &
7/24/2019 L26 - Regeneration.pptx
http://slidepdf.com/reader/full/l26-regenerationpptx 15/19
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
− −= =
& && &
& &
7/24/2019 L26 - Regeneration.pptx
http://slidepdf.com/reader/full/l26-regenerationpptx 16/19
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= =
7/24/2019 L26 - Regeneration.pptx
http://slidepdf.com/reader/full/l26-regenerationpptx 17/19
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+ =
7/24/2019 L26 - Regeneration.pptx
http://slidepdf.com/reader/full/l26-regenerationpptx 18/19
>!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= − + −
7/24/2019 L26 - Regeneration.pptx
http://slidepdf.com/reader/full/l26-regenerationpptx 19/19
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= = =
− −
& & &
& & && &