study komp. 3 suct

7/26/2019 Study Komp. 3 Suct

1/11

KOMPRESOR CENTRIFUGAL

3. Kasus Studi - 3.Satu unit kompresor centrifugal multistage untuk LPG plant dengan diagram alir

dan kondisi operasi sebagai berikut :

Jenis gas Propan : Mw= 44,0 lbm !mol, SG = ",#$,

%c= &&&o', Pc = &"( Psia

)ondisi operasi :

*s = *" = ""(&"# +-M P4 = #.,( Psia

Ps = P" = "#,& Psia %4 = &$o- = #/$o'

%s = %" = 4"o- = 4"o'

*a = 4&.#/ +-M Pd = P#= (,/# Psia

Pa = P/= P/1= / Psia %d = %#= .(,#o- = #4(,# o'

%a = %/= %/1= .o- = 4&.o'

*b = "(.44 +-M

Pb = P$= P$1 = 4$,# Psia

%b = %$= %$1= $o- = 4 o'

2itungla3 :

a La5u aliran massa tiap stage 6M, lbm!menit

b 2ead 62 dan 7a8a 62P masing 9 masing stage

c 2ead %otal )ompresor 62, lbfft!lbm

d 7a8a %otal )ompresor 62P, 2P

e 7a8a Penggerak 672P, 2P bila efisiensi transmisi . ;

"

Qs

1

2

2

Qa

Qb

3

!

Qd

3


2/11

P"#$"%"saia# Studi Kasus 3&

1. I'("%%") Sta*" 1 +1st,

a. Lau A%i)a# Massa +'1,

menitlbmTxZx

SGxPxxQm !,

$#,#$

"44

""

""" =

7imana :

*" = ""(&"# +-M, %"= 4"o'

P" = "#,& Psia, SG = ",#$

Meng3itung


3/11


4/11

2P"st= HPx

xHP

x

Hxm

m

87!!7

11!5!32

!!71

2P"st= (#.$ 2P

2. I'("%%") Sta*" 2 +2#d,

a. Lau A%i)a# Massa +'2,

m/1 = menitlbmTxZx

SGxPxxQ!,

CC$#,#$

C"44C

//

//, P/= P/1, %/1 = %/


5/11

%r$= (#,0&&&

4$ ==cT

T

000 lbfft!lbm

M"#*0itu#* pol

:%#

:%#

23

1

23

TT

PP kk

pol

Meng3itung k

?ntuk su3u %/1= .o-, p 6.o- di3itung dengan interpolasi p 60o- =

"#,&4 A%?!lbmo', dan p 6#0o- = "&,.. A%?!lbmo', maka

p 6.o- = "#,&4 D

( )( )

( ) RlbmBTU o!.4,"#

0#0

&4,"#..,"&0. =

k 6.o- = "4,",".4,"#

.4,"#

,".6

.6 =

=FCpFCp

o

o

66474588%#

28!3%# 11111

pol

Jadi

2act= lbmftlbflbmftlbf !:./$$!:,((&,0

&$.=

#


6/11

/. 9a$a K'()"s) 2#d+CP2#d,

2P/nd= HP

x

xHP

x

Hxm

m

act ,,0##0

./$$#$$,

##0

C/ =

2P/nd= .. 2P

3. I'("%%") Sta*" 3 +3)d,

a. Lau a%i)a# 'assa +'3,

m$1 =CC$#,#$

C"44C

$$

$$

TxZx

SGxPxxQ, lbm!min, P$1 = P$= 4$,# Psia


7/11

2pol= lbmftlbfPP

xTxn

nxZx

SG

n

n

av .1:

:1

3!!3

1

3

3

Meng3itung


8/11

p 6$o- = "#,&4 D

( )( )

( ) -lbm!A%?&","&

0#0

&4,"#..,"&0$ o=

k 6$o- = "$&,","&","&

&","&=

64788!23%#

!36!5%# 134111341

pol

Jadi

2act= lbm!ft:lbf&4&$lbm!ft:lbf,(&4,0

4$.=

/. 9a$a K'()"s) 3)d+CP 3)d,

2P 6$rd = 2P,,0@##0

&4&$@*P,

@##0

2@Cm

m

act$ =

2P 6$rd = .#$, 2P

. I'("%%") Sta*" +t0,

a. Lau A%i)a# 'assa +',

m4= m$1 = $/$& lbm!min = lbm!det

maka :

*4 = /,0


9/11

2pol= lbmftlbfP

PxTx

n

nxZx

SG

n

n

av .11

3!!3

1

!

Meng3itung


10/11

6!7,!23!!6+%#

6!52!68%# 1311131

pol

Jadi

2act= lbm!ft:lbf&.($lbm!ft:lbf,(#4,0

#"./=

/. 9a$a K'()"s) t0+CP t0,

2P 4t3= 2P,,0@##0

&.($@*P,

@##0

2@m

m

act4 =

2P 4t3= 0.", 2P

!. "ad K'()"s) Tta% +,

2 = 2act "st D 2act /nd D 2act $rd D 2act 4t3, lbfft!lbm

2 = ""#.# D ./$$ D &4&$ D &.($ = $$"#4 lbfft!lbm

4. 9a$a K'()"s) Tta% +CP,

2P = 2P "st

D 2P /nd

D 2P $rd

D 2P 4t3

, 2P

2P = (#.$ D .. D .#$ D 0.", 2P

2P = $40&. 2P = /#4"# )F

6. 9a$a P"#**")a; +9P,

72P = .,02P,2P

trans

trans

=

72P = 2P,.:0

$40&.

72P = $4(&$ 2P = /#$$ )F

K"si'(u%a# &

Aerdasarkan data operasi dan per3itungan diperole3 3asil sebagai berikut :

o Parameter "st /nd $rd 4t3 %otal

" La5u Hliran Massa "44# $".4 $/$& $/$& $/$&

"0


11/11

/

$

4

6M, lbm!min

2ead 62act,

lbfft!lbm

7a8a )ompresor

62P, 2P

7a8a Penggerak

672P, 2P!kF

""#.#

(#.$

./$$

..

&4&$

.#$

&.($

0."

$$"#4

$40&.

$4(&$

/#$$

""

study komp. 3 suct

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