Lampiran III Spesifikasi Alat

8/19/2019 Lampiran III Spesifikasi Alat

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LAMPIRAN III

SPESIFIKASI PERALATAN

1. ACCUMULATOR-01 (ACC-01)

Fungsi : Tempat menampung kondensat yang berasal dari kondensor-01

Tipe: Silinder horizontal dengan penutup ellipsoidal

Gambar :

input

output

ACC - 01

Kondisi Operasi:

Tekanan = 1atm

Temperatur = !"#o

$%a&u alir = '0('1"1')kg*&am

+ensitas = )1",kg*m'

esiden.e Time = / menit = 0"0' &am

Tekanan +esain = ' atm

Perhitungan e!ain A""u#u$at%r-01

a. Ka&a!ita! A""u#u$at%r' t

olume liuid = %a&u alir*+ensitas 2 holding time

= 3'0('1"1') kg*&am4*3 )1",kg*m'4 2 0"0' &am

= '"50'm'*&am

Faktor keamanan = 106

Kapasitas a..( = 1"1 2 '"50'm'

= !,"))5! m'

olume 7..umulator = Kapasitas a.. 2 esiden.e Time

= !,"))5! 3m'*&am4 2 0"0' &am = 0")1'0 m'

. e!ain U*uran A""u#u$at%r

,,'


2/244

,,!

%$u#e Si$in+er' !

s =

sil inder L D (!

,⋅π

+imana: %silinder = '(+

=

'(

!

' Dπ

%$u#e E$$i&!%i+a$' e

e =,!

4(3 ' Dπ

%$u#e T%ta$ A""u#u$at%r' T

T = s 8 ,(e

ia#eter A""u#u$at%r'

+ =

'

!0,"'

accT V

=

'

#1#)",

)1'0"0

= 0"0#!' m

Sehingga"

! = 0"#!, m'

e = 0"0,)m'

T = 0"#5/ m'

". Pan,ang A""u#u$at%r

% = ' ( +

= ' ( 0"#!' m

= 1"5!!5m

Pan,ang E$$i&!%i+a$' L

h =

D⋅!

1


3/244

,,/

=

m#!'"0!

1⋅

= 0"1#,1 m

Pan,ang T%ta$ A""u#u$at%r' LT

%T = % 8 ,h

= 31"5!!58 3,( 0"1#,144 m

= ,",#51 m

+. Tea$ in+ing A""u#u$at%r

Ketea$an in+ing agian Si$in+er' t!i$in+er

t =

Cc P EjS

r P i +− (#"0(((

+imana:

t = ketebalan dinding bagian head" m

9 = tekanan design = 1 atm = 1!"#5# psi

r = &ari &ari = 0"',! m = 1,")#, inS = working stress yang diizinkan = 5',",,# atm= 1')00 psi

$ = ;aktor korosi yang diizinkan = 0"00'0,,# m = 0"115 in


4/244

,,#

aka didapatkan:

t =

in psi psi

in psi115"0

4#5#"1!,"034/"0)00(1'(3,

4/,!",/43#5#"1!3+

⋅−⋅

= 0"1'/ in = 0"00'!,'1 m

Sehingga"

O+ = >+ 8 ,(tsilinder

= 30"#!' 8 3,( 0"00'!,'144 m

= 0"#//m

IENTIFIKASI

?ama 7lat 7..umulator

Kode 7lat 7$$-01

@umlah 1 Anit

FungsiTempat menampung kondensat yang berasal

dari kondensor-01 3$+-014

ATA ESI/N

TipeSilinder Borizontal dengan penutup ellipsoidal

penutup elipsoidalKapasitas 0")1 m'

Tekanan 1 atm

Temperatur !"# o$

+iameter 0"#/ m

9an&ang ,",) m

Tebal +inding 0"00'! m

Caktu Tinggal / menit

ahan K%n!tru*!i Car%n !tee$

. ACCUMULATOR-0 (ACC-0)

Fungsi : Tempat menampung kondensat yang berasal dari kondensor-0,

Tipe: Silinder horizontal dengan penutup ellipsoidal

Gambar :


5/244

,,)

Kondisi Operasi:

Tekanan = 1atm

Temperatur = ##"0#o$

%a&u alir = '0(/!"!'5kg*&am+ensitas = )1",kg*m'

esiden.e Time = / menit = 0"0' &am

Tekanan +esain = ' atm

Perhitungan e!ain A""u#u$at%r-0

a. Ka&a!ita! A""u#u$at%r' t

olume liuid = %a&u alir*+ensitas 2 holding time

= 3'0(/!"!'5 kg*&am4*3 )1",kg*m'4 2 0"0' &am

= '5"10!m'*&am

Faktor keamanan = 106

Kapasitas a..( = 1"1 2 '5"10!m'

= !'"01/ m'

olume 7..umulator = Kapasitas a.. 2 esiden.e Time

= !'"01/ 3m'*&am4 2 0"0' &am = 0")1# m'

. e!ain U*uran A""u#u$at%r

%$u#e Si$in+er' !

s = sil inder L D (

!

,⋅π

+imana: %silinder = '(+

=

'(!

' Dπ


6/244

,,

%$u#e E$$i&!%i+a$' e

e =

,!

4(3 ' Dπ

%$u#e T%ta$ A""u#u$at%r' T

T = s 8 ,(e

ia#eter A""u#u$at%r'

+ =

'

!0,"'

accT V

=

'

#1#)",

)1#"0

= 0"#!5 m

Sehingga"

s = 0"#!/ m'

e = 0"0,) m'

T = 0"#55 m'

". Pan,ang A""u#u$at%r

% = ' ( +

= ' ( 0"#!5/ m

= 1"5!/m

Pan,ang E$$i&!%i+a$' L

h =

D⋅!

1

=

m#!5/"0!

1 ⋅

= 0"1#,! m

Pan,ang T%ta$ A""u#u$at%r' LT

%T = % 8 ,h

= 31"5!/8 3,( 0"1#,!44 m

= ,",)', m


7/244

,,5

+. Tea$ in+ing A""u#u$at%r

Ketea$an in+ing agian Si$in+er' t!i$in+er

t =

Cc P EjS

r P i +− (#"0(((

+imana:

t = ketebalan dinding bagian head" m

9 = tekanan design = 1 atm = 1!"#5# psi

r = &ari &ari = 0"',! m = 1,")/ in

S = working stress yang diizinkan = 5',",,# atm= 1')00 psi$ = ;aktor korosi yang diizinkan = 0"00'0,,# m = 0"115 in


8/244

,'0

IENTIFIKASI

?ama 7lat 7..umulator

Kode 7lat 7$$-0,

@umlah 1 Anit

FungsiTempat menampung kondensat yang berasal

dari kondensor-0, 3$+-0,4

ATA ESI/N

TipeSilinder Borizontal dengan penutup ellipsoidal

penutup ellipsoidalKapasitas 0")1 m'

Tekanan 1 atm

Temperatur ##"0# o$

+iameter 0"#!5 m

9an&ang ,",) m

Tebal +inding 0"00'! m

Caktu Tinggal / menit

ahan K%n!tru*!i Car%n !tee$

. CILLER-01 (C-01)

Fungsi : engkondensasikan sebagian produk top Flash aporizer

Tipe : Shell and tube Beat


9/244

,'1

Gambar :

Aliran inlet

Aliran outletWater in

ShellTube

HeadRear End

9anas : 9roduk top -01

C = /!(/!1"#/ kg*&am = 1,0(,!'"#' lb*&am

T1 = 1)5"#0 o$ = '//",5 oF

T, = /0"00 o$ = 1,,"00 oF

Fluida +ingin : 7mmonia

E = 11(0)"0 kg*&am = ,!(!,!"/# lb*&am

t1 = -''"#/ o$ = -,"/) oF

t, = /0"00 o$ = 1,,"00 oF

9erhitungan:

1. ean Pana! C-01


10/244

,',

= 1/(105(///"1! k@*&am = 1!(',1('0!")# Dtu*hr

. LMT

Fluida 9anas

3oF4 Komponen Fluida +ingin 3o

F4 Selisih'//",5 Suhu tinggi 1,, ,''"'1,,"00 Suhu rendah -,"# 1/0"/),''",5 Selisih 1/1 ,"),

%T+ 3T4 =4*ln3

-

1,

1,

T T

T T

∆∆∆∆

39ers( /(1!4

= 1"5, oF

. T" +an t"

T. = H 3,',"/, 8 ,',"/,4 = ,'"#!oF

t. = H 31!0 8 #4 = !#"), oF

• 7sumsi"A+ = /0 Dtu * &am ;t,oF 3Table "

Kern4

7 =4(3 T U

Q

D ∆ = 1(/!)"0# ;t,

Karena 7 I ,00 ;t," maka dipilih B< dengan &enis Shell and Tube Beat


11/244

,''

=

,#1("0,0

1/!)"0#

×

= ,5/"!)

dari tabel 5 Kern" didapat nilai yang mendekati ?t perhitungan ,5/

?t =,5/

- Koreksi A+

7 =?t 2 % 2 aLL

=1(/!)"0#;t,

A+ =

A+ =!5 3?ilai A+ sudah mendekati A+ asumsi4

karena nilai Ad perhitungan mendekati dengan nilai Ad asumsi" maka data

untuk shell :

- Shell side

>+ =,5in.h 3Tabel 5( Kern4Da;;le Spa.e 3D = >+*,4 =1!"/in.h

9ass =1

9t =1",/ in triangular pit.h

F$ui+a Pana! 2 Tue Si+e' a$iran t%& F$a!h a&%ri3er

4. F$%5 area +a$a# tue (a6t) = 0"/)# in.h,3Tabel(10" Kern4

7. T%ta$ 8$%5 area (at) = ?t 2 aMt * 1!! 2 n 39ers()(!"

Kern4

=,1!!

0"/)#,5/

x

x

= 1"115 ;t'

9. La,u a$ir' Gt =at

W

F

7(Gt


12/244

,'!

=115"1

,!(!,/

= ,0(###"10 lb* hr( ;t,

:. i$angan Re;n%$+' Ret

9ada T. = !#")1 oF

N = 0"00, lb*;t &am 3aEsM" ,01,4

+ = 0"/#*1, = 0"0) ;t3Tabel(10" Kern4

et = µ

ae D (

=0(00,

,0(###"100)(0 x

= '0!(/'"0,


13/244

,'/

= 10)"!,'

1

)"0

= 51"5/ Dtu * hr( ;t,o

F

F$ui+a ingin 2 She$$ Si+e' a$iran a##%nia

. F$%5 area &a+a !he$$ (a!)

as =

( )

41!!3

Pt

$ xC x #D

39ers()(1" Kern4

= 0"/! ;t,

7. La,u a$ir #a!!a +a$a# !he$$' /!

Gs = sa

W

39ers()(," Kern4

=0"/!

1,0(,!'"#

= ,0/(#"' lb*hr(;t,

9. i$angan Re;n%$+' R e!? @ /! = µ

9ada T. = ,'"#!oF

N = 0"011# .9

= 0"0,0 lb*;t ( &am

.9 = 0"'!55 Dtu*lb(⁰F

k = 0"00/' Dtu*hr(;t⁰F

'1

(

k

c µ

=

'

1

00/'"00,0"00"'!55

x

= 1",,!)

+e = 0"),0 in.h = 0"0#0 ;t 3Fig(," Kern4

es= =

µ

DS

=

0,"0

0#"0 ,0/(#"' ×

= !!1(15/"/,


14/244

,'#

&B = !,0 3Fig(," Kern4

:. K%e8i!ien Per&in+ahan Pana!' h%

Koreksi Piskositas diabaikan karena tidak signi;i.ant" maka diperoleh :ho = &B ( 3k*+4( 3.N*k41*' 39ers( #(," Kern4

= !/"#5 Dtu * &am ;t,oF


15/244

,')

= 0"0) psi

Shell Side

1( Faktor Friksi

e = !!1(15/"/,

; = 0"00051 3Fig(,5" Kern4

,( ?umber o; .ross' 3? 8 14

? 8 1 = 1, % * D 39ers()(!'" Kern4

= 15"#,

+s = >+ * 1,= ') * 1, = ,"!1#) ;t

s = '"5/

9s = s Des

' Ds &s

φ 10

,

10,,"/

413

×+

39ers( 1,(!)"Kern4

= 1"!'/ psi

IENTIFIKASI

?ama 7lat

Kode 7lat

@umlah

Operasi

Fungsi

$hiller

$B 01

1

Kontinyu

enurunkan temperature keluaran dari

reaktor 01 3-014ATA ESI/N

Tipe Shell and Tube Beat +

O+

DCG

9it.h

Duah

,0 ;t

0"/',/ in

1 in

1/

1 in triangular pit.h

>+

Da;;le Spa.e

$learan.e

∆9s

,5 in

1!"/ in

0"'1 in

1"!'/ psi


16/244

,'

9ass

∆9T

!

0"00) psi+irt Fa.tor" d 0"001,

Dahan konstruksi $arbon Steel

4. CONENSER B 01 (C B 01)

Fungsi : engkondensasikan produk top kolom distilasi - 01

Tipe : Shell and Tube Beat


17/244

,'5

Perhitungan e!ain 2

9anas yang ditrans;er pada $+-01

= 10'(1)'(#5,"/1,0 k@*&am

= 5)()51(,,!"1/ Dtu*&am

14 enghitung ∆T 3%T+4

Fluida 9anas 3oF4Temperatur

Fluida +ingin 3oF4Selisih

1!5 Tinggi 1'1 1"01!5 endah ,"! ##(#00 Selisih !5 -!"#

∆T = '#"))/0oF

,4 T. dan t.

T. = TaPg,

F'0,43'5,

o+=

= 1!5oF

t. = taPg

,

F,"!04'113

o+=

= 10#")0oF

7sumsi A+ = ,/0 Dtu*&am ;t,(oF ((3Kern" tabel 4

Sehingga diperoleh :

7T(A

+ ∆=

?F43'#"))/0F4(Dtu*&am(;t,/03

Dtu*&am "1/5)()51(,,! oo,

7 ? 10(#'#"#5'/ ;t,

'4 Karena 7 I ,00 ;t," maka digunakan Shell and Tube


18/244

,!0

!4 @umlah Tube

?tLL%(a

7

=

=*;t;t0"',)1;t,0

;t /10(#'#"#5'

,

,

×

= 1#,/"51

7mbil ?t pada tabel 5 Kern yang mendekati" sesuai dengan ukuran tube yang

telah dipilih" ?t = 1#,#

/4 Koreksi A+

7 = ?t ( % ( aMM

= 1(#,# 2 1# ;t 2 0"',)1 ;t,*;t

= 10(#'#"#5'/ ;t,

A+ =T

Q

∆×7

=F')"1!#/;t /10(#'#"#5'

Dtu*&am "1/5)()51(,,!o, ×

= ,!)"/0 Dtu*&am ;t,(oF

Karena A+ mendekati asumsi" maka dari tabel 5 Kern diperoleh data sebagai

berikut :

?t = 1(#,#

9T = 1 in triangular pit.h

$MM = 39T O+ tube4

= 31 in 0")/ in4

$MM = 0",/ in

>+ Shell = ,1",/ in

Tube 9asses = 1

Tube O+ = 0")/ in

Da;;le Spa.e = 10"# in


19/244

,!1

TUE SIE 2 F$ui+a Pana!' &r%+u* K - 01

#4 FloE area" at

aMt = 0"5/ in, ((( 3Kern" tabel 104

at =n(;t*in1!!

aL( ?

,,

tt

=!;t*in1!!

in0"5/ 1(#,#

,,

,

××

= ,")0! ;t,

)4 %a&u alir massa" Gt

Gt =ta

E

=

,;t,")0!

lb*&am ,()1#(55

= 5))(1//"!, lb*&am ;t,

54 eynold ?umber 3 et4µ = 0"/1/5 .9 = 1",!0 lb*&am ;t

+ = 1"1,0 in = 0"05'' ;t ((( 3Kern" tabel 104

et = µ

+(G t

=

;tlb*&am",!01

;t0"05'';tlb*&am 5))(1//"!, , ×

et = )'(0!5"''

&B = 150 ((( 3Kern" gambar ,!4

104 9ada taPg = 10#")oF

$p = 0"5/#) Dtu*lb(oF

k = 0"10', Dtu*&am(;t(oF


20/244

,!,

9randtl ?umber 39r4 =

k

µ $p(

=

×

F(Dtu*&am(;t0"10',;tlb*&am,!0"1FDtu*lb(0"5/#)

o

o

= 11"/)'

114 hio

hi =

1!"0'

1

$p( &B

w D

k

k µ

µ µ

= 150 2

( )

×

;t0"05''

F(Dtu*&am(;t0"10',11"/)'

o

'1

2 1

= !)!",#Dtu*&am(;t,(oF

hio = hi 2 3>+*O+4

= !)!",#Dtu*&am(;t,(oF 2 30"05'' ;t *0"10!, ;t4

= !,/"!!! Dtu*&am(;t,(oF

SELL SIE 2 F$ui+a ingin' air &en+ingin

TaPg = 1!5oF

Da;;le spa.ing = 10"#' in

$learan.e = 0"'1 in

1,4 %uas area laluan 37s4

7s = Pt

$C #D

×××

1!!

L

=in/#,/"1;t*in1!!

in#'"10in0"'1in,1",/,, ×

××

= 0"'1! ;t,


21/244

,!'

1'4 %a&u alir 3Gs4

Gs =

s A

W

=

,;t0"'1'#

lb*&am #)(#'#"!

= ,1!(1#"5/ lb*&am ;t,

1!4 eynold ?umber 3 es4

iskositas 3 µ

4 = 0"00## .9 = 0"01#1 lb*;t &am

+iameter


22/244

,!!

1:) ho =

1!"0'

1

$p( &B

w D

k

k µ

µ µ

= ,()1'"0! Dtu*&am(;t,(oF

14 $lean oPerall .oe;isien .ondensation" A.

A. =hohio

ho2hio

+

=

F(Dtu*&am(;t4,()1'"0! !!!"!,/3

F(Dtu*&am(;t4,()1'"0!2!,/"!!!3o,

o,

+

= '#)")# Dtu*&am(;t,( oF

154 +irt ;a.tor" d

d =+$

+$

AA

A-A

×

= 0"001' &am ;t, oF*Dtu

PRESSURE ROP

TUE SIE 2 F$ui+a Pana!

,04∆9t

et = )'(0!5"'

Gt = 5))(1//"!,lb*&am ;t,

; = 0"0001) R 3Kern" gambar ,#4

s = "5!

∆9t =t

10

,

t

s(+((10(,,"/

n(%(G(;

Φ

=15!";t0"05''10(,,"/

1;t,04;tlb*&am5))(1//"!,30"0001)

10

,,

××××××

= 0"0)!/ psi


23/244

,!/

,14∆9r

g(,

P,

= 0"00/! R 3Kern" gambar ,)4s = "5!

∆9r =g,

P 2

s

n(!

,

=

0"00/!2"5!

1!

×

= 0"005) psi

,,4∆9T

∆9T = ∆9t 8 ∆9r

= 30"00/!8 0"005)4 psi

= 0"0!, psi

SELL SIE 2 F$ui+a ingin

,'4 es = 1(01!(/'"!/'

; = 0"0005

,!4 @umlah %intasan

?t 8 1D

%(1, =

= ,)1"0#

! ? I She$$

? 1'::0< 8t


24/244

,!#

,/4∆9s

Gs = ,1!(1#"5/ lb*&am ;t,

s = 1,

P! ?t

10

t

,

s

s((+e(10,,"/

143?(+e(G(;

φ

+

=1,;t1"))010(/(,,

,)1"0#;t1"))04;tlb*&am /3,1!(1#"50"0005

10

,,

×××××

= 0"!15) psi

IENTIFIKASI

?ama 7lat

Kode 7lat

@umlah

Operasi

Fungsi

$ondenser

$+ 01

1

Kontinyu

engkondensasikan produk Top K+ - 01ATA ESI/N


O+

DCG

9it.h

9ass∆9T

1#,# buah

,0 ;t

0"/',/ in

1",/ in

1#


!0"0!, psi

>+

Da;;le Spa.e

$learan.e

∆9s

,1",/ in

10"#' in

0"'1 in

0"!15) psi

+irt Fa.tor" d 0"001'Dahan konstruksi $arbon Steel

7. CONENSER B 0 (C B 0)

Fungsi : engkondensasikan produk top kolom distilasi - 0,

Tipe : Shell and Tube Beat


25/244

,!)

F$ui+a Pana! 2 Pr%+u* T%& K B 01E = '0(/!"!'51 kg*&am = #)('!)")0 lb*&am

T1 = ##"0#,!o$ = 1/0"51oF

T, = ##"0#,!o$ = 1/0"51oF

F$ui+a ingin 2 Air &en+ingin

C = '(!!,"0)'5 kg*&am = )(/"!# lb*&am

t1 = ,o$ = ,"!0oF

t, = //o

$ = 1'1o

F

Perhitungan e!ain 2

14 9anas yang ditrans;er pada $+-0,

? 4.99.40:':4 *D=,a#

= !(!,0(1,,"!' Dtu*&am

,4 enghitung ∆T 3%T+4

Fluida 9anas 3oF4Temperatur


1/0"51 Tinggi 1'1 1"01/0"51 endah ,"! #"/10"0000 Selisih !5 -!"#

∆T = '5"''0)oF

'4 T. dan t.

Eater int1

FeedinT1

Eater outt,

FeedoutT,


26/244

,!

T. = TaPg,

F'0,43'5,

o+=

= 1/0"51oF

t. = taPg

,

F,"!04'113 o

+= = 10#")0oF

7sumsi A+ = ,/0 Dtu*&am ;t,(oF ((( 3Kern" tabel 4

Sehingga diperoleh :

7T(A

+ ∆=

?F43'"5')!F4(Dtu*&am(;t,/03

Dtu*&am !'!(!,0(1,,"

oo,

7 ? !/!"0)! ;t,

!4 Karena 7 I ,00 ;t," maka digunakan Shell and Tube


27/244

,!5

= #5 2 1# ;t 2 0"',)1 ;t,*;t

= !/!"0)! ;t,

A+ =T

Q

∆×7

=F'5"''0);t!/!"0)!

Dtu*&am !'!(!,0(1,,"o, ×

= ,!)"/0 Dtu*&am ;t,(oF

Karena A+ mendekati asumsi" maka dari tabel 5 Kern diperoleh data sebagai

berikut:

?t = #59T = 1 in triangular pit.h

$MM = 39T O+ tube4

= 31 in 0")/ in4

$MM = 0",/ in

>+ Shell = ,1",/ in

Tube 9asses = 1

Tube O+ = 0")/ inDa;;le Spa.e = 10"# in

TUE SIE 2 F$ui+a Pana!' &r%+u* K - 01

)4 FloE area" at

aMt = 0"5/ in, ((( 3Kern" tabel 104

at =

n(;t*in1!!

aL( ?

,,

tt

=!;t*in1!!

in0"5/ #5

,,

,

××

= 0"11) ;t,

4 %a&u alir massa" Gt

Gt =ta

E


28/244

,/0

=

,;t0"11)

lb*&am )(/

= #'(5',")/ lb*&am ;t

,

54 eynold ?umber 3 et4

µ = 0"/1/5 .9 = 1",!0 lb*&am ;t

+ = 1"1,0 in = 0"05'' ;t ((( 3Kern" tabel 104

et = µ

+(G t

=;tlb*&am",!01

;t0"05'';tlb*&am #'(5',")/ , ×

et = !())5"!#!#

&B = /0 ((( 3Kern" gambar ,!4

104 9ada taPg = 10#")oF

$p = 0"5/#) Dtu*lb(oF

k = 0"10', Dtu*&am(;t(oF

9randtl ?umber 39r4 =

k

µ $p(

=

×F(Dtu*&am(;t0"10',

;tlb*&am,!0"1FDtu*lb(0"5/#)o

o

= 11"/)'

114 hio

hi =

1!"0'

1

$p( &B

w D

k

k µ

µ µ

= /0 2

( )

×

;t0"05''

F(Dtu*&am(;t0"10',11"/)'

o

'1

2 1

= 1,!"5/!Dtu*&am(;t,

(o

F


29/244

,/1

hio = hi 2 3>+*O+4

= 1,!"5/!Dtu*&am(;t,(oF 2 30"05'' ;t *0"10!, ;t4

= 111"5/5 Dtu*&am(;t,(oF

SELL SIE 2 F$ui+a ingin' air &en+ingin

TaPg = 1/1oF

Da;;le spa.ing = 10"#' in

$learan.e = 0"'1 in

1'4 %uas area laluan 37s4

7s = Pt

$C #D

×××

1!!

L

=in/#,/"1;t*in1!!

in#'"10in0"'1in,1",/,, ×

××

= 0"'1'# ;t,

1!4 %a&u alir 3Gs4

Gs = s A

W

=

,;t0"'1'#

lb*&am #)('!)"!

= ,1!()##"5, lb*&am ;t,

1/4 eynold ?umber 3 es4

iskositas 3 µ

4 = 0"00## .9 = 0"01#1 lb*;t &am

+iameter


30/244


31/244

,/'

d =+$

+$

AA

A-A

×

= 0"00/, &am ;t, o

F*Dtu

PRESSURE ROP

TUE SIE 2 F$ui+a Pana!

,14∆9t

et = !())5"!#!#

Gt = #5(5',")/lb*&am ;t,

; = 0"0001) R 3Kern" gambar ,#4

s = "5!

∆9t =t

10

,

t

s(+((10(,,"/

n(%(G(;

Φ

=15!";t0"05''10(,,"/

1;t,04;tlb*&am#5(5',")/30"0001)

10

,,

××××××

= 0"000' psi

,,4∆9r

g(,

P,

= 0"00/! R 3Kern" gambar ,)4

s = "5!

∆9r = g,

P 2

s

n(!

,

=

0"00/!2"5!

1!

×

= 0"005) psi

,'4∆9T

∆9T = ∆9t 8 ∆9r


32/244

,/!

= 30"000'8 0"005)4 psi

= 0"01 psi

SELL SIE 2 F$ui+a ingin

,!4 es = 1(01!(#1)"051

; = 0"0005

,/4 @umlah %intasan

?t 8 1 D

%(1, =

= ,)1"0#

! ? I She$$

? 1'::0< 8t

,#4∆9s

Gs = ,1!(1#"5/ lb*&am ;t,

s = 1,

P! ?t

10

t

,

s

s((+e(10,,"/

143?(+e(G(;

φ

+

=1,;t1"))010(/(,,

,)1"0#;t1"))04;tlb*&am ,3,1!()##"50"0005

10

,,

××

×××

= 0"!15/ psi

IENTIFIKASI


33/244

,//

?ama 7lat

Kode 7lat

@umlah

Operasi

Fungsi

$ondenser

$+ 0,

1

Kontinyu

engkondensasikan produk Top K+ 0,ATA ESI/N


O+DCG

9it.h

9ass

∆9T

#5 buah

,0 ;t

0"/',/ in

1",/ in1#


!

0"00' psi

>+

Da;;le Spa.e

$learan.e

∆9s

,1",/ in

10"#' in

0"'1 in

0"!15/ psi

+irt Fa.tor" d 0"00/,1!Dahan konstruksi $arbon Steel

9. COOLER-01 (C-01)

Fungsi : eenurunkan produk top dari Kno.k Out +rum-01 3KO+-014

Tipe : Shell and tube Beat


34/244

,/#

Gambar :

Aliran inlet


ShellTube

HeadRear End

Fluida 9anas : 9roduk top -01

C = #(!11")/)5 kg*&am = 1!(1'/"!5 lb*&am

T1 = !1"05# o$ = 10/"5) oF

T, = '0"00

o

$ = #

o

F

Fluida +ingin : 7mmonia

E = !(1'",,# kg*&am = 5(1,!"/' lb*&am

t1 = , o$ = ,"!0 oF

t, = ' o$ = 100"!0 oF

9erhitungan:


35/244

,/)

1. ean Pana! C-01

= 1)'(,/1"11,) k@*&am = 1#!(,1,") Dtu*hr

. LMT

Fluida 9anas

3oF4Komponen Fluida +ingin 3oF4 Selisih

10/"5) Suhu tinggi 100"!0 /"##"00 Suhu rendah ,"!0 '"#015"5) Selisih 1"00 1"5)

%T+ 3T4 = 4*ln3

-

1,

1,

T T

T T

∆∆

∆∆

39ers( /(1!4

= !"/1! oF

. T" +an t"

T. = H 310/"5) 8 #4 = 5/"55oF

t. = H 310"!0 8 ,"!4 = 51"!0 oF

• 7sumsi"A+ = /0 Dtu * &am ;t,oF 3Table "

Kern4

7 =4(3 T U

Q

D ∆ = )!,",!# ;t,



36/244

,/

a x L

A

@umlah tube" ?t =

=,#1"0,0

)!,",!#

×

= 11'"!#

dari tabel 5 Kern" didapat nilai yang mendekati ?t perhitungan 11'

?t =11'

- Koreksi A+

7 =?t 2 % 2 aLL=)!,",/;t,

A+ =

A+ =!5 3?ilai A+ sudah mendekati A+ asumsi4

karena nilai Ad perhitungan mendekati dengan nilai Ad asumsi" maka data

untuk shell :

- Shell side

>+ =,5in.h 3Tabel 5( Kern4

Da;;le Spa.e 3D = >+*,4 =1!"/in.h

9ass =1

9t =1",/ in triangular pit.h

F$ui+a Pana! 2 Tue Si+e' a$iran t%& Kn%"* Out ru#

4. F$%5 area +a$a# tue (a6t) = 0"5/ in.h,3Tabel(10" Kern4

7. T%ta$ 8$%5 area (at) = ?t 2 aMt * 1!! 2 n 39ers( )(!"

Kern4

=,1!!

0"/)#11'

x

x

= 0"1,5! ;t'

F

7(Gt


37/244

,/5


W

=1,5!"0

5(1,/

= )0(/'"#! lb* hr( ;t,


9ada T. = 51"! oF

N = 0"/1# lb*;t &am 3aEsM" ,01,4

+ = 1"1,*1, = 0"05'' ;t3Tabel(10" Kern4

et = µ

ae D (

=0"/1#

)0(/'"#!05''"0 x

= )()'0"//


38/244

,#0

= ,#"11#

1

)"0

= ,!0",', Dtu * hr( ;t,o

F

F$ui+a ingin 2 She$$ Si+e' a$iran air &en+ingin

. F$%5 area &a+a !he$$ (a!)

as =

( )

41!!3

Pt

$ xC x #D

39ers()(1" Kern4

= 0"'1'# ;t,

7. La,u a$ir #a!!a +a$a# !he$$' /!

Gs = sa

W

39ers()(," Kern4

=0"'1'#

1!(1'/"/

= !/(0))"0/ lb*hr(;t,

9. i$angan Re;n%$+' R e!? @ /! = µ

9ada T. = 5/"55oF

N = 0"011' .9

= 0"0,)' lb*;t ( &am

.9 = 0"1'!, Dtu*lb(⁰F

k = 0"0!,5 Dtu*hr(;t⁰F

'1

(

k

c µ

=

'

1

0!,5"00,)'"00"'!55

x

= 0"!!10

+e = 0"510 in.h = 0"0)# ;t 3Fig(," Kern4

es =

µ

DS

=

011'"0

0)#"0 !/(0))"0/ ×

= 1,/(00'"'5


39/244

,#1

&B = '00 3Fig(," Kern4


Koreksi Piskositas diabaikan karena tidak signi;i.ant" maka diperoleh :ho = &B ( 3k*+4( 3.N*k41*' 39ers( #(," Kern4

= )!"!1 Dtu * &am ;t,oF

. irt Fa"t%r' R +

d = Dc

Dc

U U

U U

(

−

=!5 /)"0#'

!5-/)"0#'

x39ers #(1'" Kern4

= 0"00,5

PRESSURE ROP

Tube Side

!( Antuk ?e = )()'0"#

Faktor ;riksi = 0"000, 3Fig ,#" Kern4

s = 1",!!

9t =

= 0"00!' psi

/( , * ,g = 0"00/! 3Fig ,)" Kern4

9r = 3 !n * s 4 3 , * ,g 4

= 0"10! psi

9T = 9t 8 9r

= 0"10/ psi

t e s D x xn Lt &

φ 10

,

10,,"/


40/244

,#,

Shell Side

'( Faktor Friksi

e = 1,/(00!"'5

; = 0"00051 3Fig(,5" Kern4

!( ?umber o; .ross' 3? 8 14

? 8 1 = 1, % * D 39ers()(!'" Kern4

= ,)1"0#

+s = >+ * 1,

= ') * 1, = ,"!1#) ;ts = '"5/

9s = s Des

' Ds &s

φ 10

,

10,,"/

413

×+

39ers( 1,(!)"Kern4

= 0"0/#' psi

IENTIFIKASI

?ama 7latKode 7lat

@umlah

Operasi

Fungsi

$ooler $ - 01

1

Kontinyu

enurunkan temperature dari top produk

Kno.k Out +rum 01 3KO+-014ATA ESI/N


O+

DCG

9it.h

9ass

#5 buah

,0 ;t

0"/',/ in

1",/ in

1/


!

>+

Da;;le Spa.e

$learan.e

∆9s

,1",/ in

10"#' in

0"'1 in

0"0/#' psi


41/244

,#'

∆9T 0"10/ psi+irt Fa.tor" d 0"00/,1!Dahan konstruksi $arbon Steel

:. EPANER ALE-01

Fungsi : Antuk mengalirkan dan menurunkan tekanan produk keluaran dari

eaktor-01 menu&u $ooler-01

Tipe : Turbine

Gambar :

Kondisi Operasi:

Temperatur masuk" T1 = ,/0 $

Tekanan masuk" 91 = /0 atm

Tekanan keluar" 9, = , atm

%a&u alir masuk" C = /!(/!1"#!, kg*&am = 1,0(,!'")! lb*&am

= ,(00!"0# lb*min+ensitas gas" = #"1' kg*mU = 0"'' lb*;tU

Faktor keamanan" ; = 106

k = 1"1'!

enentukan la&u alir Polumetri." :

=W

ρ

= (5)"!5/ mU*&am= /(,')"0, ;tU*min

Sa;ety ;a.tor 106 = 3180"14 2 ;tU*min

= /()#0"),) ;tU*min

aka" = /()#0"),) ;tU*min

Temperatur yang keluar dari e2pander-01:

+imana:


42/244


43/244

,#/

= ,,(/,,(,05"'/ k@*&am 1/(0,!(0!5",1/ k@*&am

= )(!5(1#0"1)0 k@*&am

ange e;isiensi


44/244

,##

Aliran inlet


ShellTube

HeadRear End

/a#ar

2

Fluida 9anas : Saturated steam

C = 1,(550"/)) Kg*&am = ,(#'5",50 lb*&am

T1 = '0 o$ = )1# oF

T, = '0 o$ = )1# oF

Fluida +ingin : 9roduk dari umpan9-01

E = /!(/!1"#/5 kg*&am = 1,0(,!'"#', lb*&am

t1 = !1"#/ o$ = 10#"5# oF

t, = ,/0 o$ = !, oF

9erhitungan:

1. ean Pana! -01

= ,5('1(50,"#5k@*&am = ,)()5('#!"/1 Dtu*&am

. LMT

Fluida 9anas

3o

F4



45/244


46/244


47/244


48/244

,)0

= 0"5)

+e = 0"),0 in.h = 0"0#0 ;t 3Fig(," Kern4

es= = µ

DS

=0!)0"0

0#0"0 51(//0"'#1 ×

= 11#()#1"!5

&B = 10 3Fig(," Kern4


Koreksi Piskositas diabaikan karena tidak signi;i.ant" maka diperoleh :

ho = &B ( 3k*+4( 3.N*k41*' 39ers( #(," Kern4

= #5", Dtu * &am ;t,oF


49/244

,)1

= 0"0015

PRESSURE ROP

Tube Side

!( Antuk ?e = ,'(1,1"!')

Faktor ;riksi = 0"0001' 3Fig ,#" Kern4

s = /"5!0

/( 9t == 0"0001 psi

#( , * ,g = 0(00/! 3Fig ,)" Kern4

9r = 3 !n * s 4 3 , * ,g 4

=

00/!"0!0'"1

!! x

x

= 0"001/ psi

9T = 9t 8 9r

= 0"001/ psi

Shell Side

/( Faktor Friksi

e = 11#()#1"!5

; = 0(001/ 3Fig(,5" Kern4

#( ?umber o; .ross" 3? 8 14

? 8 1 = 1, % * D 39ers()(!'" Kern4

= 31, 2 ,1#4* 11"# = ,!)")!

+s = >+ * 1,

= 1/",/ * 1, = 1"5') ;t

s = 1

t e s D x x

n Lt &

φ 10

,

10,,"/


50/244

,),

9s = s Des

' Ds &s

φ 10

,

10,,"/

413

×+

39ers( 1,(!)"Kern4

= 1"5,# psi

IENTIFIKASI

?ama 7lat Beater Kode 7lat B-01@umlah 1 buahOperasi KontinyuFungsi enaikkan temperatur sebelum masuk ke rea.tor 01

3-014

ATA ESI/N

Tipe Shell and Tube Beat + 1",/ in" 0"/# in 9ass !DCG 1/ 9 0"001 9si9it.h 11*! - in" Triangular pit.h 9 1"5,# 9si 9ass !

>. EATER-0 (-0)

Fung!i 2emanaskan aliran umpan sebelum masuk ke Kolom +istilasi-01Ti&e 2 Shell and tube Beat


51/244

,)'

Aliran inlet


ShellTube

HeadRear End

/a#ar 2

Fluida 9anas : Saturated steam

C = 1(//0"5,5 Kg*&am = '(!15",10 lb*&am

T1 = '0 o$ = )1# oF

T, = '0 o$ = )1# oF

Fluida +ingin : 9roduk dari umpan KO+-01

E = !(1,5"500 kg*&am = 10#(10"1!0 lb*&am

t1 = !1"05# o$ = 10/"5)0 oF

t, = #"10/ o$ = 1#"550 oF

9erhitungan:

1. ean Pana! -0

= 1#(/'(!00"/!k@*&am = 1/()1(,#1"1'0 Dtu*&am

. LMT

Fluida 9anas

3oF4


)1# Suhu tinggi 1) /,5)1# Suhu rendah 10# #10"0'0 Selisih 1 -1"0,


52/244

,)!

%T+ 3T4 =4*ln3

-

1,

1,

T T

T T

∆∆∆∆

39ers( /(1!4

= /#"//#o

F. T" +an t"

T. = H 3)1# 8 )1#4 = /'# oF

t. = H 31) 8 10#4 = 1!#"! oF

• 7sumsi" A+ = /0 Dtu * &am ;t,oF 3Table " Kern4

7 =

4(3 T U

Q

D ∆ = //,"51) ;t,



53/244

,)/

7. T%ta$ 8$%5 area (at) = ?t 2 aMt * 1!! 2 n 39ers()(!" Kern4

=

!1!!

0"/!#,,0

x

x

= 0"1' ;t'


W

=1'"0

10#(10

= /))(!/5",)0 lb* hr( ;t,


9ada T. = 1!#"!1 oF

N = 0"1)5 lb*;t &am

+ = 0"'!*1, = 0"0#5/ ;t 3Tabel(10" Kern4

et = µ

t e D (

=0"1)5

0/))(!/5",)0#5/"0 x

= '#(1!0"!5#


54/244

,)#

Koreksi Piskositas diabaikan karena tidak signi;i.ant" maka didapat :

hi = ,05"#'1 Dtu * hr( ;t,oF

hio = hi

"D

#D

39ers(#(/" Kern4

= ,05"#'1

1

'!"0

= 151"!#' Dtu * hr( ;t,oF

F$ui+a Pana! 2 She$$ Si+e' A$iran Stea#

. F$%5 area &a+a !he$$ (a!)

as =

( )41!!3

Pt

$ xC x #D

39ers()(1" Kern4

= 0"1#1;t,

7. La,u a$ir #a!!a +a$a# !he$$' /!

Gs =

sa

W

39ers()(," Kern4

=1#1/"0

'(!15",10

= ,1(1)1"'/0 lb*hr(;t,

9. i$angan Re;n%$+' R e!? @ /! = µ

9ada T. = )1#oF

$p = 0"!)#,Dtu*lb(oF

k = 0"01) Dtu*lb(oFN = 0"0!)0lb*;t ( &am

'1

(

k

c µ

=

'1

01)"0

0!)0"0!)#,"0

x

= 0"5)


55/244

,))

+e = 0"),0 in.h = 0"0#0 ;t 3Fig(," Kern4

es = µ

DS

=0!)0"0

0#0"0,1(1)1"'/0 ×

= ,)(001"/,,

&B = ,00 3Fig(," Kern4



ho = &B ( 3k*+4( 3.N*k41*' 39ers( #(," Kern4

= ))"#!) Dtu * &am ;t,oF


56/244

,)

PRESSURE ROP

Tube Side

)( Antuk ?e = '#(1!0"!5#

Faktor ;riksi = 0"000// 3Fig ,#" Kern4

s = /"5!0

( 9t =

= 0"0110 psi

5( , * ,g = 0(00/! 3Fig ,)" Kern4

9r = 3 !n * s 4 3 , * ,g 4

=

00/!"001)"1

!! x

x

= 0"001/ psi

9T = 9t 8 9r

= 0"01,/ psi

Shell Side

)( Faktor Friksi

e = ,)(001"/,,

; = 0(00, 3Fig(,5" Kern4

( ?umber o; .ross" 3? 8 14

? 8 1 = 1, % * D 39ers()(!'" Kern4= 31, 2 ,1,4* 11"# = '))")0

+s = >+ * 1,

= 1/",/ * 1, = 1,)0 ;t

s = 1

9s = s Des

' Ds &s

φ 10

,

10,,"/

413

×+

39ers( 1,(!)"Kern4

t e s D x x

n Lt &

φ 10

,

10,,"/


57/244

,)5

= 0"1')! psi

IENTIFIKASI ?ama 7lat Beater Kode 7lat B-0,@umlah 1 buahOperasi KontinyuFungsi enaikkan temperatur sebelum masuk ke Kolom

+istilasi - 01 3K+-014

ATA ESI/N

Tipe Shell and Tube Beat


58/244


59/244

,1

!'

,

W)0"00101!/-W#)0"01!/,,#-

W/0"1)0)!!0-W5)0"1!/0!/-)1"))!)05-e2p=K

= 0"10/'

A =

ρ

ρ−ρ/"0

g

g%0'/"0

= 0"'5)!m*s

4. e!!e$ area #ini#u#' A

7 =(

(

U

Q

= sm

sm

*0"'5)!

*0"#51# '

= 1")!0! m,

7. ia#eter e!!e$ #ini#u#'

+ =

/"07!

π

= 1"!50 m

9. Liui+ %$u#etri" 8$%5rate' L

% =densitas

massaalirla&u

=

'kg*m''!"/!/'

kg*&am !!(1,5"500

= 1!'"')! m'*&am

= 0"0!m'*s

:. Tinggi Liui+' L


60/244

,,

Bolding time" t = / menit

= 0"0'' &am

B% =7

t2

%

= #")m


61/244

,'

t =

$9,"0S


62/244

,!

Fung!i 2engkondensasikan sebagian produk top rea.tor-01

Ti&e 2 Shell and tube Beat


63/244

,/

t, = !0o$ = 10! oF

9erhitungan:

1. ean Pana! PC-01

= ()#'('!/"##! k@*&am = ('0#(1)0"#0 Dtu*&am

. LMT

Fluida 9anas

3oF4


1,, Suhu tinggi 10! 1# Suhu rendah ,"! '"#'# Selisih ,, 1!"!

%T+ 3T4 =4*ln3

-

1,

1,

T T

T T

∆∆∆∆

39ers( /(1!4

= "5!) oF

. T" +an t"

T. = H 31,, 8 #4 = 10! oF

t. = H 310! 8 ,"!4 = 5'", oF

• 7sumsi"A+ = /00 Dtu * &am ;t,oF 3Table " Kern4

7 =4(3 T U

Q

D ∆ = 1(5!"/5/ ;t,



64/244

,#

=

',)1"0,0

1(5!"/5/

×

= ,'1"#

9ada tabel 5 Kern" &umlah tube yang memenuhi adalah ,',

Shell Side Tube Side>+ = ,'",/ in ?umber and %ength = ,', and ,0 ;tD = 11"# in O+" DCG" 9it.h = 1"00-in" 1" 1-in tri

9asses = , 9asses = 1

F$ui+a Pana! 2 Tue Si+e' a$iran ga! &urging

4. F$%5 area +a$a# tue (a6t) = 0"!, in.h, 3Tabel(10" Kern4

7. T%ta$ 8$%5 area (at) = ?t 2 aMt * 1!! 2 n 39ers()(!" Kern4

=,1!!

0"!,,',

x

x

= 0"15!1 ;t'

9. La,u a$ir'Gt =at W

=15!1"0

01)0(5'5"00

= 0(!5)"/#0 lb* hr( ;t,


9ada T. = 5'", oF

N = 0"/1# lb*;t &am

+ = 0"#0#*1, = 0"0/0/ ;t 3Tabel(10" Kern4

et = µ

ae D (

=0"/1#

00(!5)"/#0/0/"0 x

= 1',(#"!1,


65/244

,)

>. Ni$ai hi" pada T. = 5'", oF

$p = 1"0'1 Dtu*lb( oF

k = 0"/5! Dtu*;t( oF( &am

'1

(

k

c µ

=

'1

/5!"0

/',"11"0'1

x

= 1"!)#

hi =

1!"0'*1(

wk

Cp

D

k j!

µ

µ µ

Koreksi Piskositas diabaikan karena tidak signi;i.ant" maka didapat :

hi = !(,1#"0, Dtu * hr( ;t,oF

hio = hi

"D

#D

39ers(#(/" Kern4

= !(,1#"0,

)/"0

#0#"0

= #(!5,")## Dtu * hr( ;t,oF

F$ui+a ingin 2 She$$ Si+e' a$iran air

. F$%5 area &a+a !he$$ (a!)

as =

( )

41!!3

Pt

$ xC x #D

39ers()(1" Kern4

= 0"')/!;t,

7. La,u a$ir #a!!a +a$a# !he$$' /!

Gs = sa

W

39ers()(," Kern4

=')/!"0

01,0(,!'"#0

= ',0('1#"0,0 lb*hr(;t,


66/244

,

9. i$angan Re;n%$+' R e!? @ /! = µ

9ada T. = 10!oF

$p = 0",)'/Dtu*lb(oF

k = 0"/#0) Dtu*lb(oF

N = 0"0,#! lb*;t ( &am

'1

(

k

c µ

=

'1

/#0)"0

0,#!"0#!),"1

x

= 0"5!5'

+e = 0")' in.h = 0"0)# ;t 3Fig(," Kern4

es = µ

DS

=0"0,#!

0)#"00',0('1#"0, ×

= (,)"000

&B = #00 3Fig(," Kern4



ho = &B ( 3k*+4( 3.N*k41*' 39ers( #(," Kern4

= ',,",'1 Dtu * &am ;t,oF

. e!ign Oera$$ C%e88i"ient' U


67/244

,5

= 0"',)1 2 ,0 2 ,', = 1(5)"10 ;t,

A+ =t A

Q

∆⋅ =

"5!)1(5)"10

#0('0#(1)0"

× = !5"''' Dtu*hr(;t,(oF

10. irt Fa"t%r' R +

d = Dc

Dc

U U

U U

(

−

='''"!5'0#"55/

'''"!5-'0#"55/

x39ers #(1'" Kern4

= 0"001,

PRESSURE ROP

Tube Side

1( Antuk ?e = 1',(#"!1,

Faktor ;riksi = 0"000,# 3Fig ,#" Kern4

s = 1",!'/

,( 9t =

= 0"#0!5 psi

'( , * ,g = 0(00/! 3Fig ,)" Kern4

9r = 3 !n * s 4 3 , * ,g 4

=00/!"0,'/"1

#!

x

x

= 0"0#5/ psi

9T = 9t 8 9r

= 0"#)!! psi

Shell Side

1( Faktor Friksi

e = (,)"000

t e s D x x

n Lt &

φ 10

,

10,,"/


68/244

,50

; = 0"00051 3Fig(,5" Kern4

,( ?umber o; .ross' 3? 8 14

? 8 1 = 1, % * D 39ers()(!'" Kern4

= '05"#0

+s = >+ * 1,

= ') * 1, = 1"5')/ ;t

s = '"5/

9s = s Des

' Ds &s

φ 10

,

10,,"/

413

×+

39ers( 1,(!)"Kern4

= '"//1' psi

IENTIFIKASI

?ama 7lat 9artial Kondenser Kode 7lat 9$ 01@umlah 1 buahOperasi Kontinyu

Fungsi engkondensasikansebagianproduk reaktor 013-014

ATA ESI/N

Tipe Shell and Tube Beat + '5 in9an&ang ,0 ;t Da;;le spa.e 15"/ inO+" >+ 0")/ in" 0"#0# in 9ass 1DCG 1/ 9 0"/#! 9si9it.h 11*! - in" Triangular pit.h 9 ,"!/# 9si 9ass #

11. KOLOM ISTILASI B 01 (K B 01)


69/244

,51

Fungsi : emurnikan metanol sebagai produk utama dari komponen lain

Tipe : Sie(e Tra) Tower

Gambar :

A. Menentu*an *%n+i!i %&era!i.

+engan Trial and Error " didapatkan kondisi operasi:

Fee+

9 = 1"! atm = 1(0#!"00) mmBg

T = #"11 o$ = '/5",/# K

Komposisi:

Komponen 9i 3atm4 Wi Ki = 9i * 9 i = Wi ( Ki

$B'OB 1"/1 0"!5) 1"/1 0")#B,O 0"!,# 0"/0' 0"!,# 0",1!

Total 1"000 1"000

T%&

9 = 1"1 atm = '#"00# mmBg

T = !"#)5 o$ = '/)",5 K

Komposisi :


70/244

,5,

Komponen Kmol*&am i Ki = 9i * 9 Wi = i * Ki$B'OB 5/'"'!/ 0"### 1"51/ 0"'!

B,O !)",#, 0"''! 0"/1, 0"#/,

Total 1(!'1"#0) 1(000 1"000

%tt%#

9 = 1"/ atm = 1(1!0"00 mmBg

T = 55 o$ = '),"1/ K

Komposisi :

Komponen kmol*&am Wi Ki = 9i * 9 i = Wi(Ki$B'OB !"')5 0"005 '"!'/ 0"0'0

B,O !5"510 0"551 0"5) 0"5)0 Total !5!",5 1"000 1"000

. e!ain K%$%# e!ti$a!i

Met%+e /ra8i*

1. ata *e!eti#angan Ua&-Cair untu* Air +an Metan%$ &a+a 1 at#

apor %iuid


71/244

,5'

T 1sat =

B

A−ln P−C

T 2

sat =

B

A−ln P−C

P=10( A−( BC +T ))

dimana 9 = mmBg" T = oK

+ata Gra;ik Kesetimbangan $ampuran etanol - 7ir

T1 Saturated = )'",5) $

T, Saturated = 105")11 $

T (K)Pa (##g) P (##g)

@ G(#etan%$) (air)

'!#"!!) 1(0#!"000 ,#5",,5 1 1'!5"!!) 1(1"!,0 '0/", 0"/5 0"5#0'/,"!!) 1(',!"#,/ '!/"',! 0")'! 0"51!

'//"!!) 1(!)'"!/0 '5"#) 0"#,, 0"#,'/"!!) 1(#'/")# !'")10 0"/,, 0"0''#1"!!) 1(1,"!51 !5,")5 0"!'' 0")')'#!"!!) ,(00!"/#/ //,"'!! 0"'/, 0"##!'#)"!!) ,(,1,"5)# #1)()#) 0",0 0"/,')0"!!) ,(!'")!# #5"/05 0",1! 0"!51')'"!!) ,(#,"5'/ )#"0'/ 0"1// 0"'50')#"!!) ,(5!#"#' /'",5 0"100 0",)')5"!!) '(,'0"551 5!)"!00 0"0/1 0"1//',"#1 '(/1"1!' 1(0#!"000 0 0

3 #ntrod*ction to C%emical Engineering T%ermod)namics+Sixt% Ed,+ -,., Smit%

dkk,, Pg,/014


72/244


73/244

,5/

T top = !"#)5 $

T bottom = 55 $

• Temperatur rata-rata

TaPg =

T top+T bottom2 ,

4ottomtop T T +

TaPg = 51"!0 $

+ari gra;ik .( $abe and Thiele didapatkan stage teoritis 3?t4 = ! stages

. E8i!ien!i Tra;

K%#&%nen F H (N!=#) F .H ? Ha

$B'OB 0"!5) 0",!# 0"1,,B,O 0"/0' 0"'01 0"1/1Total 1"000 0",)!

elatiPe Polatility 3Z4 = '")!aka" dengan metode OM$onnel $orrelation 3@( $oulson ol( #4 :


74/244

,5#

+ = '5(1#'")'1kg*&am

% = ( +

= 0"1#/ 2 3'5(1#'")'1kg*&am4

= #(!/!",/ kg*&am

= 1")5' kg*det

= % 8 +

= #(!/!",/ kg*&am 8 '5(1#'")'1kg*&am

= !/(#1"/// kg*&am

= 1,"#), kg*det

+ata Fisik Vapo*r Li5*id .ass 6low rate 3kg*det4

Densit) 3kg*m'4

Vol*metric 6low rate 3m'*det4

1,"#),

5"',

1"'/

1")5'

)5'"#

0"00,S*r&ace tension 3?*m4 0"0'/

. ia#eter *%$%#

Li5*id 7Vapo*r 6low 6actor 86 LV 9

6 LV = L

V

W

W

V

L

ρ

ρ

3$oulson(


75/244

,5)

=

05/"00,"0

0"0'/,"0

= 0"10#•

Ke.epatan 6looding 3u; 4

u; =V

V L ; ρ

ρ ρ −^1

3$oulson(


76/244

,5

= 1"#!0 m,

•

$ross se.tional area dengan 1, 6 doEn.omer area 37.4

7sumsi diambil sebesar 1, 6 berdasarkan rekomendasi pada buku C%emical

Engineering+ (ol, :+ Co*lson+ p, 1:<

7. =1,(01−

An

3$oulson( p(!),4

=1,"01

1"#!0 ,

−m

= 1"#! m,

•

+iameter kolom 3+.4

+. =1!"'

! Ac

3$oulson( p(!),4

=1!"'

4 1"#!3!,m

= 1"/!1

". e!ain plate

• +iameter kolom 3+.4 = 1"/!1 m

•

%uas area kolom 37.4

7. =!

1!"'(, x Dc

3$oulson( p(!)'4

= !

1!"'41"/!13 , x

= 1"#! m,

• +oEn.omer area 37d4

7d = persen doEn.omer 2 7. 3$oulson( p(!)'4

= 0"1, 31"#! m,4

= 0",,! m,

• ?et area 37n4


77/244

,55

7n = 7. 7d

= 1"#! m, 0",,! m,

= 1"#!0 m,

• 7.tiPe area 37a4

7a = 7. , 7d 3$oulson( p(!)'4

= 1"#! m, , 30",,! m,4

= 1"!1) m,

• Bole area 37h4 ditetapkan 10 6 dari 7a berdasarkan rekomendasi pada buku

C%emical Engineering+ (ol, :+ Co*lson+ p, 1:<

7h = 1/ 6 ( 7a

= 0"1/ 31"!1) m,4

= 0",1' m,

• ?ilai weir lengt% 3>E4 ditentukan dari ;igure 11('1" $oulson Pol #

Ordinat =

100 x Ac

Ad

=

100m 1"#!

m 0",,!,

,

×

= 1,

7bsis = Dc

w #

= 0"))

Sehingga :

>E = +. 2 0"))

= 1"/!1m 2 0"))

= 1"1) m

• 9enentuan nilai weir %eig%t 3hE4" %ole diameter 3dh4" dan plate t%ickness" 3nilai

ini sama untuk kolom atas dan kolom baEah4

Weir %eig%t 3hE4 = /0 mm

nilai Eeir height direkomendasikan pada buku C%emical Engineering+ (ol, :+

Co*lson+ p, 1=>

!ole diameter 3dh4 = / mm

ukuran tersebut merupakan pre&erred si?e sesuai rekomendasi pada buku

C%emical Engineering+ (ol, :+ Co*lson+ p, 1=/

Plate t%ickness = / mm 3untuk car4on steel 4

+. Penge"e*an


78/244


79/244


80/244

'0,

Ap

A%

= Aa

A%

= 0"1/

Ap A%

2 100 = 1/

Sehingga didapat nilai Ori;i.e .oe;i.ient 3$o4 = 0"

+ry plate drop 3hd4

hd =

L

V

Co

U%

ρ

ρ

,

/1

∧

3$oulson((


81/244


82/244

'0!

uP = An

maksU(

3$ouldson"p(!)!4

uP =1"#!0

1"'/

= 0", m*s

6 ;looding =

100 x*

*

&

(

3$oulson((p(!)!4

= / 6

Antuk nilai F% = 0"01/ dari ;igure 11(,5 $oulson ed #

+idapat nilai = 0",

Ketentuan bahEa nilai harus lebih ke.il dari 1" telah terpenuhi(

3$oulson((p(!)/4

e. Tria$ &$ate $a;%ut

+igunakan plate type .artridge" dengan /0 mm unper;orted strip mengelilingi

pinggir plate dan /0 mm Eide .alming zones(

+ari ;igure 11(', $oulson ed # pada Dc

#w

= 0"))

+i dapat nilai c$ = 5

Sudut subtended antara pinggir plate dengan unper;orated strip 3c4

c = 10 - c$ 3$oulson((p(!)/4

= 10 5= ,

ean length" unper;orated edge strips 3%m4

%m =

( )

−10

1!(' θ

x%w Dc

3$oulson((p(!)/4

= ,"1'' m


83/244


84/244


85/244

'0)

Tebal dinding ellipsoidal head "

t =

C C P E S

D P

,(0,+

⋅⋅⋅

= !"'#) 2 10-'m

Sehingga :

O+ = >+ 8 ,tsilinder

= 1"#!5 m 8 , 3!"'#) 2 10-'m4

= 1"//0 m

. e!ain *%$%# agian a5ah (Stri&ing !e"ti%n)

a. ata 8i!i* untu* !tri&ing !e"ti%n

F = !(1,5"500 kg*&am

% = #(!/!",/ kg*&am

= !/(#1"/// kg*&am

= 1

=

( )

6

L L −L

3Treyball"


86/244

'0

. ia#eter *%$%#

•

Li5*id 7Vapo*r 6low 6actor 3F%4

F% = L

V

W

W

V

L

ρ

ρ

3$ouldson(


87/244

'05

•

aksimum Polumetri. ;loE rate 3AP maks4

AP maks =

'#00(V

V

ρ

3$oulson( p(!),4

=#00('(*0/!"!

kg*&am !/(#1"///'mkg

= '"1,# m'*s

• ?et area yang dibutuhkan 37n4

7n =

∧

*

maksU V

3$oulson( p(!),4

= sm

sm

*1"!)/

*'"1,# '

= ,"115 m,

•

$ross se.tion area dengan 1, 6 doEn.ormer area 37.4



7. =1,"01−

An

3$oulson( p(!),4

=1,"01

,"115,

−m

= ,"!0 m,

• +iameter kolom 3+.4

+. =1!('

! Ac

3@ ($ouldson( p(!),4

=1!"'

4,"!03! ,m

= 1")/1 m


88/244

'10

". e!ain &$ate

•

+iameter kolom 3+.4 = 1")/1 m

•

%uas area kolom 37.4

7. =!

1!"'(, x Dc

3$oulson( p(!)'4

=!

1!"'(41")/13 , x

= ,"!0 m,

• +oEn.omer area 37d4


= 0"1, 3,"!0 m,4

= 0",5 m,

• ?et area 37n4

7n = 7. 7d

= ,"!0m, 0",5 m,

= ,"115 m,

• 7.tiPe area 37a4

7a = 7. , 7d 3$oulson( p(!)'4

= ,"!0 m, , 30",5 m,4

= 1"'0 m,

• Bole area 37h4 ditetapkan 1/6 dari 7a berdasarkan rekomendasi pada buku


7h = 1/ 6 ( 7a

= 0"1/ 2 1"'0 m,

= 0",)! m,

• ?ilai Eeir length 3>E4 ditentukan dari ;igure 11('1" $oulson ed #

Ordinat =

100 x Ac

Ad

= 1,

7bsis =

Dc

w #

= 0"))


89/244

'11

Sehingga :

>E = +. 2 0"))

= 1"))1m 2 0"))

= 1"'! m

• 9enentuan nilai Eeir height 3hE4 " hole diameter 3dh4" dan plate thi.kness"

3nilai ini sama untuk kolom atas dan kolom baEah4

Ceir height 3hE4 = /0 mm 3$oulson( p(/)14

Bole diameter 3dh4 = / mm 3$oulson( p(/)'4

9late thi.kness = / mm 3$oulson( p(/)'4

+. Penge"e*an

Check weeping

a2imum liuid rate 3%m"ma24

%m"ma2 ='#00

L

3$oulson( p(!)'4

=

#00('

kg*&am /!(/!"),/

= 1/"1#, kg*det

inimum liiud rate 3%m"min4

inimum liuid rate pada )0 6 liuid t*rn down ratio

%m"min = 0") %m" ma2 3$oulson( p(!)'4

= 0") 31/"1#, kg*det4

= 10"#1! kg*det

Ceir liuid .rest 3hoE4

hoE=

',

)/0

#w

Lm

l ρ

3$oulson(


90/244

'1,

=

',

' 1"'!*5#",)!

det* 1/"1#,)/0

m ;g

;g

= ')"551 mm liuid

hoE"min =

',

min")/0

#w

Lm

l ρ

=

',

' 1"'!*5#",)!

det* 10"#1!)/0

m ;g

;g

= ,5"5/1 mm liuid9ada rate minimum

hE 8 hoE = /0 mm 8 ,5"5/1 mm

= )5"5/1 mm

+ari ;igure 11('0 $oulson ed #

K , = '0")

inimum design Papour Pelo.ity 3_h4

`h =

( )[ ]( ) ,1

, !",/50"0V

d% ; ρ

−−

3$oulson(


91/244

'1'

Plate pressure drop

@umlah maksimum Papour yang meleEati holes 3h4

h = A%

maksU("

3$ouldson"p(!)'4

=0",)!

'"1,#

= 11"' m*s

+ari ;igure 11('! $oulson ed #" untuk :

diameter %ole

t%icness Plate

= 1

Ap

A%

= Aa

A%

= 0"1/

Ap

A%

2 100 = 1/

Sehingga didapat nilai Ori;i.e .oe;i.ient 3$o4 = 0"

+ry plate drop 3hd4

hd =

L

V

Co

U%

ρ

ρ

,

/1

∧

3$oulson((


92/244

'1!

ht = hd 8 3hE 8 hoE4 8 hr 3$oulson((p(!)!4

= #"#5 mm 8 )5"5/1 mm 8 1,"#)! mm

= 1)5"',, mm liuid

Ketentuan bahEa nilai ht harus lebih besar dari 100 mm liuid telah terpenuhi(

3$oulson"p(!)!4

Downcomer liuid backup

+oEn.omer pressure loss 3hap4

hap = hE 10 mm 3$oulson((p(!))4

= /0 10

= !0 mm

7rea under apron 37ap4

7ap= hap ( >E 3$oulson((p(!)!4

= !0 2 10-' m 2 1"'! m

= 0"0/! m,

Karena nilai 7ap lebih ke.il dari nilai 7d 30",5 m ,4" maka nilai 7ap yang

digunakan pada perhitungan head loss di doEn.omer 3hd.4

Bead loss in the doEn.omer 3hd.4

hd.=

,

ma2"1##

Aap

Lm

L ρ

3$oulson((


93/244

'1/

tr =maks Lm

%4 Ad L

"

ρ

3$oulson((


94/244

'1#

%m =

( )

−10

1!"' θ

x%w Dc

3$oulson((p(!)/4

= ,"!'! m 7rea o; unper;orated edge strip 37up4

7up = hE ( %m 3$oulson((p(!)/4

= /0 2 10-' m 2 ,"!'! m

= 0"1,, m,

ean length o; .alming zone 3%.z4

%.z =

−,

sin43 C %w Dc θ

3$oulson((p(!)/4

= 1",! m

7rea o; .alming zone 37.z4

7.z = , 3%.z ( hE4 3$oulson((p(!)/4

= , 31",! m 2 /0 (10-' m4

= 0"1, m,

Total area per;orated 37p4

7p = 7a 37up 8 7.z4 3$ouldson((p(!)/4

= 1"'0 m, 30"1,, m,8 0"1, m,4

= 1"/0 m,

+ari ;igure 11('' $oulson ed # di dapat nilai >p*dh = ," untuk nilai 7h*7p =

0"11#

?ilai >p*dh harus berada dalam range ,"/ !"0 3$oulson((p(!#/4

@umlah holes

7rea untuk 1 hole 37oh4

7oh =!

1!"',d%

3$oulson((p(!)/4


95/244


96/244

'1

= '(, 2 10-' m

Tebal dinding silinder "

T =

C C P E S

r P

#(0+

⋅⋅⋅

= !"/ 2 10-' m


t =

C C P E S

D P

,(0,+

⋅⋅⋅

= !"/5 2 10-'

mSehingga :


= 1")/1 m 8 , 3!"/5 2 10-'m4

= 1")#1 m

E. Tinggi *%$%# +e!ti$a!i

B = 3?s 2 tray spa.ing4 8'"0!= 3# 2 0"/4 8 '"0!= #"0!! m

Be = tinggi tutup ellipsoidal

= Y 2 >+

= Y 2 1"/!1

= 0"'/


= Y 2 >+= Y 2 1")/1 m

= 0"!' m

Bt = B 8 Be atas 8 Be baEah

= #"0!! m 8 0"'/ m 8 0"!' m

= #"#) m


97/244

'15

IENTIFIKASI

?ama 7lat Kolom destilasi

7lat Kode K+-01@enis Sie(e Tra) Col*mn@umlah 1 buahOperasi Kontinyu

FungsiAntuk memperoleh produk utama etanol

dengan kemurnian ) 6

ATA ESI/N

T%& %tt%#

Tekanan 1", atm 1"/ atmTemperatur !"# o$ 55 o$

KOLOM

T%& %tt%#

+iameter 1"/!1 m 1")/1 mTray spa.ing 0"/ m 0"/ m

@umlah tray , buah ! bua

hTebal !"'## 2 10-' m !"/ 2 10-' material $arbon steel

PLATE

T%& %tt%#

+oEn.omer area 0",,! m, 0",5 m,

7.tiPe area 1"!1) m, 1"'0 m,

Bole +iameter / mm / mmBole area 0",1' m, 0"1' m,

Tinggi Eeir /0 mm /0 mm9an&ang Eeir 1"1# m 1"'! mTebal pelat / mm / m

9ressure drop per tray 10#"!## mm liuid 1)5"',,mm

liuidTipe aliran .airan Single pass Single pass+esain 6 ;looding / 6 / 6@umlah hole 10(,5 buah 5(',! buah

1. KOLOM ISTILASI B 0 (K B 0)

Fungsi : emurnikan metanol sebagai produk utama dari komponen lain


98/244

',0

Tipe : Sie(e Tra) Tower

Gambar :

A. Menentu*an *%n+i!i %&era!i.

+engan Trial and Error " didapatkan kondisi operasi :

Fee+

9 = 1"1 atm = '#"00# mmBg

T = )!"/' o$ = '!)"#1 K

Komposisi :

Komponen 9i 3atm4 Wi Ki = 9i * 9 i=Wi ( Ki$B'OB 1"!#/ 0"### 1"'', 0")

B,O 0"')' 0"''! 0"''5 0"11'

Total 1"000 1"000

T%&

9 = 1"0!atm = )50"!0/ mmBg

T = ##"0#, o$ = ''5",1, K

Komposisi :

Komponen Kmol*&am i Ki = 9i * 9 Wi = i * Ki$B'OB 5!5"//! 0"55' 1"0,, 0"5),

B,O #")' 0"00) 0",!5 0"0,


99/244

',1

Total 5/#"'') 1(000 1"000

%tt%#

9 = 1"1/ atm = )!"00# mmBgT = 55 o$ = '),"1/ K

Komposisi :

Komponen kmol*&am Wi Ki = 9i * 9 i = Wi(Ki$B'OB '")51 0"00 '"!!# 0"0,)

B,O !)1"!)5 0"55, 0"51 0"5)'Total !)/",)0 1"000 1"000

. e!ain K%$%# e!ti$a!i

Met%+e /ra8i*

7. ata *e!eti#angan Ua&-Cair untu* Air +an Metan%$ &a+a 1 at#

apor %iuid


100/244

',,

P=10( A−( BC +T ))

dimana 9 = mmBg" T = oK

+ata Gra;ik Kesetimbangan $ampuran etanol - 7ir

T1 Saturated = ##"5!5$

T, Saturated = 10,"#5/$

T (K)Pa (##g) P (##g)

@ ;(#etan%$) (air)

'!0(055 '#(000 ,0!(#1 1(000 1(000'!'(055 5'(0)5 ,''(,5' 0(// 0(5#0'!#(055 10/0(,5, ,#/(,5 0(),) 0(51''!5(055 11)'('5 '00(50) 0(#1' 0(#1'/,(055 1'0(15' '!0(!#/ 0(/1, 0(01'//(055 1!//(/10 '!('00 0(!,, 0()'!'/(055 1#1#(,1) !',())0 0('!1 0(#/5'#1(055 1)51(,,1 !#(,/, 0(,# 0(/)!'#!(055 151(!#! /!/(1!' 0(,0' 0(!0'#)(055 ,1)(5, #05(#1 0(1!' 0(')/')0(055 ,!11(#'1 #0(!) 0(050 0(,/5')'(055 ,#/'(#,) )/(/#, 0(0!1 0(1'0')/(!/ ,5,(0! '#(000 0(000 0(000

3 #ntrod*ction to C%emical Engineering T%ermod)namics+Sixt% Ed,+ -,., Smit% dkk,, Pg,/014

/ra8i* ! G


101/244

','

0.000 0.200 0.400 0.600 0.800 1.000 1.200

0.000

0.200

0.400

0.600

0.800

1.000

1.200

"

X

Y

. Men"ari ra!i% re8$u* #ini#u#

+ari gra;ik .( $abe and Thiele didapat nilai :

X D

R m+1

X D

Rm+1=¿

1+ 2m

3D

= 0")#

m 81 = 1"'0#

m = 0"'0#

9. Men"ari ra!i% re8$u*! a"tua$

= 1"/ 2 m

= 1"/ 2 0"'0# = 0"!#0

1+ 2 3D

= 0"#0

T top = ##"0#,$

T bottom = 55 $


102/244

',!

• Temperatur rata-rata

TaPg =

Ttop+Tbottom2 ,

4ottomtop T T +

TaPg = ,"/'1 $

+ari gra;ik .( $abe and Thiele didapatkan stage teoritis 3?t4 = stages

:. E8i!ien!i Tra;

K%#&%nen F H (N!=#) F .H ? Ha

$B'OB 0"### 0",)0 0"10

B,O 0"''! 0"''# 0"11,Total 1"000 0",5,

elatiPe Polatility 3Z4 = !"11aka" dengan metode OM$onnel $orrelation 3@( $oulson ol( #4 :


103/244

',/

= '"501 kg*det

= % 8 +

= 1!(0!,"# kg*&am 8 '0(/!"!'5kg*&am

= !!(/51"1,) kg*&am

= 1,"'# kg*det

+ata Fisik Vapo*r Li5*id .ass 6low rate 3kg*det4

Densit) 3kg*m'4

Vol*metric 6low rate 3m'*det4

1,"'#

#"',

1"5/)

'"501

)!'"#

0"00/S*r&ace tension 3?*m4 0"0'/

. ia#eter *%$%#

a, Li5*id 7Vapo*r 6low 6actor 86 LV 9

6 LV = L

V

W

W

V

L

ρ

ρ

3$oulson(


104/244

',#

u; =V

V L ; ρ

ρ ρ −^1

3$oulson(


105/244

',)

=1,"01

1"50,

−m

= ,"1# m,

h( +iameter kolom 3+.4

+. =1!"'

! Ac

3$oulson( p(!),4

=1!"'

4 ,"1#3!,

m

= 1"##,

". e!ain plate

a( +iameter kolom 3+.4 = 1"##, m

b( %uas area kolom 37.4

7. =!

1!"'(, x Dc

3$oulson( p(!)'4

=!

1!"'41"##,3 , x

= ,"1# m,

.( +oEn.omer area 37d4


= 0"1, 3,"1# m,4

= 0",#0 m,

d( ?et area 37n4

7n = 7. 7d

= ,"1#m, 0",#0 m,

= 1"50 m,

e( 7.tiPe area 37a4

7a = 7. , 7d 3$oulson( p(!)'4

= ,"1#m, , 30",#0m,4

= 1"#!) m,


106/244

',

;( Bole area 37h4 ditetapkan 10 6 dari 7a berdasarkan rekomendasi pada buku


7h = 10 6 ( 7a

= 0"1 31"#!) m,4

= 0"1#/ m,

g( ?ilai weir lengt% 3>E4 ditentukan dari ;igure 11('1" $oulson Pol #

Ordinat =

100 x Ac

Ad

=

100m ,"1#

m 0",#0,

,

×

= 1,

7bsis =

Dc

w #

= 0"))Sehingga :

>E = +. 2 0"))

= 1"##,m 2 0"))

= 1",)5 m

h( 9enentuan nilai weir %eig%t 3hE4" %ole diameter 3dh4" dan plate t%ickness" 3nilai ini

sama untuk kolom atas dan kolom baEah4

i( Weir %eig%t 3hE4 = /0 mmnilai Eeir height direkomendasikan pada buku C%emical Engineering+ (ol, :+

Co*lson+ p, 1=>

ii( !ole diameter 3dh4 = / mm

ukuran tersebut merupakan pre&erred si?e sesuai rekomendasi pada buku

C%emical Engineering+ (ol, :+ Co*lson+ p, 1=/

iii( Plate t%ickness = / mm 3untuk car4on steel 4

+. Penge"e*an

Check weeping


%m"ma2 =#00('

L

=#00('

kg*&am 1!(0!,"#

3$oulson( p(!)'4

= '"501kg*det

inimum liiud rate 3%m"min4


107/244

',5

inimum liuid rate pada )0 6 liuid turn doEn ratio

%m"min = 0") 2 %m" 3$oulson( p(!)'4

= 0") 2 3'"501kg*det4

= ,")'1 kg*det


hoE =

',

)/0

#w

Lm

l ρ

3$oulson(


108/244

''0


`h =

( )[ ]

( ) ,1

, !",/50"0

V

d% ;

ρ

−−

3$oulson(


109/244

''1

Sehingga didapat nilai Ori;i.e .oe;i.ient 3$o4 = 0"!

+ry plate drop 3hd4

hd =

L

V

CoU%

ρ ρ

,

/1

∧

3$oulson((


110/244


111/244

'''

= / 6

Antuk nilai F% = 0"0,5 dari ;igure 11(,5 $oulson ed #

+idapat nilai = 0"1!


3$oulson((p(!)/4


+igunakan plate type .artridge" dengan /0 mm unper;orted strip mengelilingi

pinggir plate dan /0 mm Eide .alming zones(


#w

= 0"))

+i dapat nilai c$ = 5


c = 10 - c$ 3$oulson((p(!)/4

= 10 5

= ,


%m =

( )

−10

1!(' θ

x%w Dc

3$oulson((p(!)/4

= ,"'0/ m

7rea o; unper;orated edge strip 37up4

7up = hE ( %m 3$ouldson((p(!)/4

7up = /0 2 10-'m 2 ,"'0/ m= 0"11/ m,


%.z =

−,

sin43 C %w Dc θ

3$ouldson((p(!)/4


112/244

''!

= 1",1# m


7.z = , 3%.z ( hE4 3$oulson((p(!)/4

= , 31",1# m 2 /0 (10-' m4

= 0"1,, m,


7p = 7a 37up 8 7.z4 3$ouldson((p(!)/4

= 1"#!)m, 30"11/ m,8 0"1,, m,4

= 1"!10 m,

+ari ;igure 11('' $oulson ed # di dapat nilai >p*dh = ,( untuk nilai 7h*7p =

0(11)

?ilai >p*dh harus berada dalam range ,(/ !(0 3$ouldson((p(!#/4

@umlah holes


7oh =!

1!"',d%

3$ouldson((p(!)/4

=!

410/31!"'

,'− x

= 15#(,/ 2 10-)m,

@umlah holes = Ao%

A%

3$ouldson((p(!)/4

@umlah holes =10215#",/

0"1#/)-

= ('5!"!)/

= ('5! holes

8. Ketea$an #ini#u# *%$%# agian ata!.

Ketebalan dinding bagian head" thead

t =

c

j

C P E S

Da P +

− (,(0((,(

39eters Tabel(! Bal /')4


113/244

''/

Ketebalan dinding bagian silinder" tsilinder

t =

c

j

i C P E S

r P +

− (#(0((

39eters Tabel(! Bal /')4Keterangan :

t = tebal dinding"

P = tekanan desain"

@1"0/ atm

D @ diameter"

= 1"##, m

r = &ari-&ari" m= 0"'1 m

S = tekanan ker&a yang diizinkan"

= 5',",,#atm 3untuk material .arbon steel suhu -,0 hingga

#/0oF4

< = e;isiensi pengelasan

= 0"/ 3spot e2amined4

$. = korosi yang diizinkan" m

= '1"/ 2 10-! m


t =

C C P E S

r P

#(0+

⋅⋅⋅

= !",/1 2 10-'m


t =

C C P E S

D P,(0, +⋅⋅ ⋅

= !",/, 2 10-'m

Sehingga :


= 1"##, m 8 , 3!",/, 2 10-'m4

= 1"#)0 m

. e!ain *%$%# agian a5ah (Stri&ing !e"ti%n)


114/244

''#

a. ata 8i!i* untu* !tri&ing !e"ti%n

F = '5(1#'")'1 kg*&am

% = 1!(0!,"# kg*&am

= !!(/51"1,) kg*&am

= 1

=

( )

6

L L −L

3Treyball"


115/244


116/244

''

= sm

sm

*011"

*,"0'5 '

= ,"00' m,

g( $ross se.tion area dengan 1, 6 doEn.ormer area 37.4



7. =1,"01−

An

3$oulson( p(!),4

=1,"01 ,"00'

,

−m

= ,",)) m,

h( +iameter kolom 3+.4

+. =

1!('

! Ac

3@ ($ouldson( p(!),4

=1!"'

4,",))3! ,m

= 1")0' m

". e!ain &$ate

a( +iameter kolom 3+.4 = 1")0' m

b( %uas area kolom 37.4

7. =!

1!"'(, x Dc

3$oulson( p(!)'4

=!

1!"'(41")0'3 , x

= ,",)) m,

.( +oEn.omer area 37d4


117/244

''5


= 0"1, 2 3,",)) m,4

= 0",)' m,

d( ?et area 37n4

7n = 7. 7d

= ,",))m, 0",)' m,

= ,"00' m,

e( 7.tiPe area 37a4

7a = 7. , 7d 3$oulson( p(!)'4

= ,",))m, , 30",)' m,4= 1")'0 m,

;( Bole area 37h4 ditetapkan 106 dari 7a berdasarkan rekomendasi pada buku


7h = 10 6 ( 7a

= 0"1 21")'0 m,

= 0"1)' m,

g( ?ilai Eeir length 3>E4 ditentukan dari ;igure 11('1" $oulson ed #

Ordinat =

100 x Ac

Ad

= 1,

7bsis = Dc

w #

= 0"))

Sehingga :

>E = +. 2 0"))

= 1")0'm 2 0"))

= 1"'11 m

h( 9enentuan nilai Eeir height 3hE4 " hole diameter 3dh4" dan plate thi.kness"

3nilai ini sama untuk kolom atas dan kolom baEah4

Ceir height 3hE4 = /0 mm 3$oulson( p(/)14

Bole diameter 3dh4 = / mm 3$oulson( p(/)'4

9late thi.kness = / mm 3$oulson( p(/)'4


118/244

'!0

+. Penge"e*an

Check weeping


%m"ma2 ='#00

L

3$oulson( p(!)'4

=#00('

kg*&am /'(,0#"!1

= 1!")0 kg*det

inimum liiud rate 3%m"min4inimum liuid rate pada )0 6 liuid t*rn down ratio

%m"min = 0") %m" ma2 3$oulson( p(!)'4

= 0") 31!")0 kg*det4

= 10"'!# kg*det


hoE =

',

)/0

#w

Lm

l ρ 3$oulson(


119/244

'!1

9ada rate minimum

hE 8 hoE = /0 mm 8 ,5")) mm

= )5")) mm

+ari ;igure 11('0 $oulson ed #

K , = '0")


`h =

( )[ ]

( ) ,1

, !",/50"0

V

d% ;

ρ

−−

3$oulson(


120/244

'!,

Ap

A%

= Aa

A%

= 0"1

Ap

A%

2 100 = 10

Sehingga didapat nilai Ori;i.e .oe;i.ient 3$o4 = 0"!

+ry plate drop 3hd4

hd =

L

V

Co

U%

ρ

ρ

,

/1

∧

$oulson((


121/244

'!'

7ap = hap ( >E 3$oulson((p(!)!4

= !0 2 10-' m 21"'11 m

= 0"0/, m,

Karena nilai 7ap lebih ke.il dari nilai 7d 30",)' m ,4" maka nilai 7ap yang

digunakan pada perhitungan head loss di doEn.omer 3hd.4

Bead loss in the doEn.omer 3hd.4

Bd. =

,

ma2"1##

Aap

Lm

L ρ

3$oulson((


122/244

'!!

6 ;looding =

100 x*

*

&

(

3$oulson((p(!)!4

= / Antuk nilai F% = 0"05' dari ;igure 11(,5 $oulson ed #

+idapat nilai = 0"0!


3$oulson((p(!)/4


+igunakan plate type .artridge" dengan /0 mm unper;orted strip mengelilingi pinggir plate dan /0 mm Eide .alming zones(


#w

= 0"))

+i dapat nilai c$ = 5 $


c = 10 - c$ 3$oulson((p(!)/4

= 10 5= ,


%m =

( )

−10

1!"' θ

x%w Dc

3$oulson((p(!)/4

= ,"'#! m

7rea o; unper;orated edge strip 37up4

7up = hE ( %m 3$oulson((p(!)/4

= /0 2 10-'m 2,"'#! m

= 0"11 m,



123/244

'!/

%.z =

−,

sin43 C %w Dc θ

3$oulson((p(!)/4

= 1",! m


7.z = , 3%.z ( hE4 3$oulson((p(!)/4

= , 31",! m 2 /0 (10-'m4

= 0"1,/ m,


7p = 7a 37up 8 7.z4 3$ouldson((p(!)/4

= 1")'0 m, 30"11 m,8 0"1,/ m,4

= 1"!) m,

+ari ;igure 11('' $oulson ed # di dapat nilai >p*dh = ,"/ untuk nilai 7h*7p =

0"11#

?ilai >p*dh harus berada dalam range ,"/ !"0 3$oulson((p(!#/4

@umlah holes


7oh =!

1!"',d%

3$oulson((p(!)/4

=!

410/31!"'

,'− x

= 1"5#'2 10-/m,

@umlah holes = Ao%

A%

3$oulson((p(!)/4

=

/-1021"5#'

0"1)'

= (1#"!0

= (1# holes


124/244

'!#

8. Ketea$an #ini#u# *%$%# agian a5ah.

Ketebalan dinding bagian head" thead

t =

c

j

C P E S

Da P +

− (,(0((,(

39etersTabel(! Bal /')4

Ketebalan dinding bagian silinder" tsilinder

t =

c

j

i C P E S

r P +

− (#(0((

39etersTabel(! Bal /')4

Keterangan :

t = tebal dinding" m

P = tekanan desain" atm

@1", atm

D @ diameter" m

= 1")0' m

r = &ari-&ari" m

= 0"/1 m

S = tekanan ker&a yang diizinkan" psi= 5',",,# atm 3untuk material .arbon steel suhu -,0 hingga

#/0oF4

< = e;isiensi pengelasan

= 0"/ 3spot e2amined4

$. = korosi yang diizinkan" in

= '(, 2 10-' m


t =

C C P E S

r P

#(0+

⋅⋅⋅

= !"!50 2 10-' m



125/244

'!)

t =

C C P E S

D P

,(0,+

⋅⋅⋅

= !"!51 2 10-'m

Sehingga :


= 1")0' m 8 , 3!"!51 2 10-'m4

= 1")1, m

E. Tinggi *%$%# +e!ti$a!i

B = 3?s 2 tray spa.ing4 8'"0!

= 31! 2 0"/4 8 '"0!= 10"'0) m


= Y 2 >+

= Y 2 1"##,

= 0"!1/ m


= Y 2 >+

= Y 2 1")0' m

= 0"!,# m

Bt = B 8 Be atas 8 Be baEah

= 10"'0)m 8 0"!1/ m 8 0"!,# m

= 11"1! m

IENTIFIKASI

?ama 7lat Kolom destilasi7lat Kode K+-0,@enis Sie(e Tra) Col*mn@umlah 1 buahOperasi Kontinyu

FungsiAntuk memperoleh produk utama etanol

dengan kemurnian 55"# 6

ATA ESI/N


126/244

'!

T%& %tt%#

Tekanan 1", atm 1"/ atmTemperatur !"# o$ 55 o$

KOLOM

T%& %tt%#

+iameter 1"##, m 1")0' mTray spa.ing 0"/ m 0"/ m@umlah tray # buah buahTebal !",/1 2 10-' m !"!50 2 10-' material $arbon steel

PLATE

T%& %tt%#

+oEn.omer area 0",#0 m, 0",)' m,

7.tiPe area 1"#!) m, 1")'0 m,

Bole +iameter / mm / mmBole area 0"1#/ m, 0"1)' m,

Tinggi Eeir /0 mm /0 mm9an&ang Eeir 1",)5 1"'11 mTebal pelat / mm / m

9ressure drop per tray 1#")50 mm liuid 1/'"!#/mm

liuidTipe aliran .airan Single pass Single pass+esain 6 ;looding / 6 / 6@umlah hole '5! buah (1# buah

1. KOMPRESOR-01 (K-01)

Fungsi :Antuk mengalirkan udara dan menaikkan tekanan udara yang akan

masuk ke Beater-013B-014

Tipe : $enti;ugal $ompressor

Gambar :


127/244

'!5

1( +ata

%a&u alir massa 3E4 = /!(/!1"#/5kg*&am= 1,0(,!'")5# lb*&am= ,(00!"0#' lb*min

+ensitas 34 = )"//, kg*mU= 0"!), lb*;tU

k =Cp campuran

Cp cam− R =31,016

31,016−8,314 = 1"'##

,( Kondisi Operasi:

Tekanan" 9in = 1/ atm 9out = /1 atm

Temperatur" Tin = '0$ = /!/"#)0 Faktor Keamanan = 106

'( olumetri. FloErate"

=laju alir massa

densitas

=2.004,063lb /min

0,472lb / ft 3

= !(,/0")5# ;tU*min

!( 9oEer Kompressor 39E4

9E =0.0643 kT Q

1

520(k −1) [( P

2

P1 )

(k −1)/k

−1

] 39ers(('0".($abe4

+imana:K = 1"'##V = 069, = /1 atm91 = 1/ atm = !(#)/")# ;t'*min


128/244

'/0

9E =

( )

−

−

−

11/

/1

0"041'##"13/,0

4 )#"#)/(!34/!#34'##"130#!'"0'##"1*1'##"1

= /)1",0, Bp= /)1 Bp

/( asio Kompresi" .

. = 39out*9in4=3/1 atm*1/ atm4= '"!

@umlah stage" ? = 1. perstage = 3.41*?

=(3,4)1 /1

= '"!

#( 9ada Stage 1

9i = 1/ atm $1 = 39o^*9i4aka:9o^ = .1 2 9i

= '"! 2 1/ atm= /1 atm

Temperatur yang keluar dari Kompresor stage 1:

T, = T1 [ P2 P1 ]( ! −1)/ !

39ers( 1,('0 %udEig4

= Tin .3K-14*K

= 3'0$4 2 (3,4)(1,366−1)/ 1,366

= !1"#!) $

)( Kapasitas Kompressor

%a&u alir Polumetrik = !(,/0")5# ;t'*minFaktor keamanan = 106aka:

Kapasitas kompressor = 31006 8 1064 " !(,/0")5# ;t'*min

= !(#)/")# ;t'*min

IENTIFIKASI

?ama alat Kompresor-01


129/244

'/1

Kode alat K-01@umlah 1 buahOperasi Kontinyu

FungsiAntuk menaikkan tekanan gas sebelum masuk

reaktor-01 3 - 014

ATA ESI/N

Tipe $entri;ugal $ompressor Kapasitas" ;t'*min !(#)/")#Temperatur" o$ '0Tekanan asuk" atm 1/Tekanan Keluar" atm /19oEer" hp /)1

@umlah stage 1ahan *%n!tru*!i Car%n !tee$

14. KOMPRESOR-0 (K-0)

Fung!i 2 Antuk mengalirkan udara dan menaikkan tekanan udara yang

akan masuk ke 9ressure SEing 7dsorber-01 39S7-014

Ti&e 2 $entri;ugal $ompressor

/a#ar 2

1. ata 2

%a&u alir massa 3E4 = #!11")/ kg*&am

= 1!1'/"/05 lb*&am

= ,'/"/5, lb*min

+ensitas 3ρ4 = 1"0,) kg*m'


130/244

'/,

= 0"0#!1 lb*;t'

k =

2camCp

camp*ranCp

−

= 1"'#//

. K%n+i!i O&era!i 2

Tekanan" 9in = , atm

9out = ,1 atm

Temperatur" Tin = '00$ = /!# o

Faktor Keamanan= 106

. %$u#etri" F$%5rate'

=densitas

massaalir laj*

=lb*;t'0"0#!1

lb*min,'/"/5,

= '(#)!"#0/ ;t'*min

4. P%5er K%#&re!!%r (P5)

9E =

−

−

−

1413/,0

0#!'"0*413

1

,1

k k

P

P

k

QT k

η

3 pers( ('0" .( $abe4

+imana :K = 1"'#//

η = 0"0

9, = ,1 atm

91 = , atm

= '(#)!"#0/;t'*min


131/244


132/244

'/!

9i = #"!0) atm

$, = 39o^ * 9i4

aka :

9o^ = .1 2 9i

= '",!0! 2 #"!0)atm

= ,1 atm

Temperatur yang keluar dari Kompresor stage 1 :

T' = T,

;' ;

p

P *413

,

'

−

3 pers( 1,('0 %udEig4= Tin .K-1*K?

= 3'/"11,!o$4 2 3'",!0!431"'#//-14 * 31"'#// 2 14

= !1"05#0o$

= /#/"#!


133/244

'//

Temperatur" o$ '0Tekanan asuk" atm ,

Tekanan Keluar" atm ,1

9oEer" hp 1(11#@umlah stage ,ahan *%n!tru*!i Car%n Stee$

17. POMPA (P-01)

Fung!i : engalirkan output bottom KO+-01 menu&u kolom

distilasi 3K+-014

Ti&e : $entri;ugal pump

ahan K%n!tru*!i : $arbon Steel/a#ar :

+ata +esain

Temperatur" T : '0°$

FloErate" ms : !(1,5"500 kg*&am = 10#(10",/lb*&am

+ensitas ;luida" ρ : ),",kg*m' = !")#lb*;t'

iskositas"µ : 0"/0.p = 1",'0lb*;t(hr

Tekanan uap" 9uap : #'/"/'!mmBg = 1,",5psi = 1)#5"#!,lb;*;t,

1. Ka&a!ita! P%#&a' 8

m; = ms

= 10#(10",/ lb*&am

= 1()#"!)1 lb*min

; =

ρ

& m


134/244

'/#

=

'lb*;t!")#

lb*min1()#"!)1

= '#",#, ;t'

*min= 0"#0!! ;t'*se.

= ,)1",#1 gal*min

. Menentu*an U*uran Pi&a

+iameter 9ipa

Antuk aliran turbulent yang mempunyai range Piskositas 0"0, ,0 .p maka

digunakan rumus diameter dalam optimum pipa

+opt = '"5 ; 0"!/ 2 ρ0"1' R(( 9eter hal ,/#

= '"5 2 30"#0!!;t'*se.40"!/ 23!")#) lb*;t'4 0"1'

= /"1/'# in

+ari ta4el >B>< Properties o& steel pipe+ Perr)s c%emical Engineers

!and4ook+ %al >B= 7 >B=0" dimensi pipa yang digunakan adalah :

Antuk Su.tion 9ipe

>9S : / in = 0"!1#) ;t

S? : !0

>+ : /"0!) in = 0"!,0# ;t

O+ : /"/#' in = 0"!#'# ;t

%s : ! m = 1'"1,'! ;t

aJ : 0"1'5 ;t, = ,0"01#0 in,

Antuk +is.harge 9ipe

>9S : ! in = 0"'''' ;t

S? : !0

>+ : '"1) in = 0",#!, ;t

O+ : !"/ in = 0"')/0 ;t

%d : 10 m = ',"0! ;t

aJ : 0"0/1' ;t, = )"'), in,


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'/)

. Perhitungan Pa+a Su"ti%n

a( Su.tion ;ri.tion loss

Su.tion Pelo.ity" s =a

Q &

=

,

'

1'50"0

*se.;t0"#0!!

&t

= !"'!0 ;t*se.

= 1/(#/,"'5 ;t*&am

c g

V

,

,

=

( ) sl4& l4 &t (*(1)!0"',(,

;t*s!"'!0,

= 0",5' ;t( lb;*lb

eynold ?umber" ?e = µ

ρ ((V D

=%r &t l4

&t l4 jam &t &t (*,'0'"1

*)#"!(*'5"#/,(1/(!,0#"0'

= ,#0(5//"01!

aterial yang digunakan untuk konstruksi pipa adalah $ommer.ial Steel

9ipeJ

+ari &ig*re >0>, 6anning &riction &actors &or long straig%t pipes, Peter+ %al

0


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'/

9ada ?e = ,#0(5//"01! an f*+ = 0"000!" dari &ig*re >0>, 6anning &riction

&actors &or long straig%t pipes, Peter+ %al 0


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'/5

K . = 0"! 31",/ 7, * 714

71 = %uas penampang 1 371III 7,4

7, = %uas penampang ," = 0"1'50 ;t,

α = 1 untuk aliran

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Lampiran III Spesifikasi Alat