Perancangan alat asam format

7/26/2019 Perancangan alat asam format

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C.1 Perancangan Flash Drum

Vapour Flow = Stream 20 = 152,490 kg/jam ; P = 3 atm ; T = 303 K

ρvapour=3,85 kg /m3

Per!"tu#ga# $%S%S 200&'5(

)"*u"+ Flow = Stream 21 =&040,41 kg/jam ; P = 3 atm ; T = 303 K

ρliquid=845,7 kg/m3

Per!"tu#ga# $%S%S 200&'5(

Perhitungan gas velocity

u= K √ ρliquid

ρvapour

−1m/ s

u=0,0305√845,7 kg /m

3

3,85 kg/m3 −1 m /s

u=0,4510 m/ s

K -a#g +"gu#aka# .ee.ar 0,0305 t/. +"kare#aka# e..el -a#g +"gu#aka#

ta#pa me#ggu#aka# lua#glua#g ke"l Turto#, 1955, !al24&('

Perhitungan BMrata-rata

kmol kg BM

x x x BM

xBM X xBM X xBM X BM

ratarata

ratarata

COOCH CH H C OCOCH OH CH OH CH COCOratarata

/4,31

(&00,0(32013,0(290,0

(((3333

=++=

++=

−

−

−

Perhitungan Ac

Ac=V .(BM rata−rata)

u .3600 . ρv

¿155 kmol / jam ×31,74 kg /kmol

0,451m/ s x 3600 x 3,85 kg /m3 =0,787 m

3

Perhitungan Diameter

= √ 4 Ac

π

1



= √4 x 0,787m

3

3,14

= 1,0012 m

= 39,3 "#

Perhitungan tinggi total (h total)

Ket"#gg"a# total ter!a+ap +"ameter mem"l"k" pera#+"#ga# 35' "p"l"! 4

.eaga" pera#+"#ga# ket"#gg"a# total ter!a+ap +"ameter'

inh

in xh

D

h D

h

total

total

total

total

4,15

2,394

4

53

==

=

−=

Perhitungan tinggi vapor (hv)

inh

ininh

Dinh

v

v

v

&5,55

3,39213&

2

13&

=

+=

+=

• Perhitugan tinggi umpan (hf )

&5,31

3,392

112

2

112

=

+=

+=

f

f

f

h

ininh

Dinh

• Perhitungan tinggi liqui (h!)

inh

h

hhhh

L

L

f vtotal L

11,0

(&5,31&5,554,15

(

=+−=

+−=

Perhitungan te"al ining

Pe#e#tua# teka#a# +e.a"#6

2



P+e.a"#= 44,0 p."a

Faktor keama#a# u#tuk pera#a#ga# teka#a# +e.a"# a+ala! 207 +ar"

teka#a# opera.", .e!"#gga

P+e.a"#=1,2 8 40,0 = 52,905& p."a

Pe#e#tua# aktor koro."6

)aju koro." 6 0,125 mm/ta!u#

mur paka" 6 10 ta!u#

:aka aktor koro." (6

C C =0,125

mm

tahun ×10 tahun=1,250mm=0,0492∈¿

Pe#e#tua# teal +"#+"#g

<e#". mater"al 6 Carbon Steel Type 25 Grade

P+e.a"# 6 52,905 p."a

"ameter 6 1,0013 m = 39,3 "#

<ar"jar" 6 19,&5 "#

llo!able Stre"" S( 6 &'500

"."e#." Samu#ga# ( 6 0,5

6 0,0492 "#

Teal +"#+"#g t( ¿

P

!"−0,6 P+Cc

¿52,905 ×19,685

0,85 × 6500−0,6 × 52,905+0,0492

¿0,2388∈¿

¿6,606 mm

Perhitungan te"al tutup

<e#". 6 Tori"pheri#al Head

P+e.a"# 6 52,905 p."a

"ameter 6 39,3 "#

llo!able "tre"" S( 6 &'500 p."a

"."e#." .amu#ga# 6 0,5

3



6 0,0492 "#

Teal tutup t( ¿0,885 P#

!"−0,1 P +Cc <awa+,194(

¿0,885 ×52,905 ×39,37

(6500 × 0,85 )−(0,1 ×52,905 )+0,0492

¿0,3832∈¿

¿9,733 mm

C.1. Perancangan #ea$tor C%&#

a. Penentuan 'olume #ea$tor C%&#

• )aju al"r olumetr"k a"ra# al"ra# 13

/jamm1201 kg/m1,250

kg/jam4001

>

m 3

3;?

;?;? ===

/jamm0&,& kg/m91

kg/jam004

>

m

3

3?$;$

?$;$

?$;$

3

3

3 ===

/jamm213,0 kg/m100

&5,3&3

>

m 3

3;$

;$

;$

3

3

3===

O$

O$

O$

O$ 33 ;$?$;$;?0 ++=

231,0&0,&1120 ++=

/jam112&,21m3=

• )aju al"r molar ?( )

;?F

( ) mol/jam 50000kmol/jam 50kg/kmol 2

kg/jam 1400F;? ===

• Ko#.e#tra." awal ?( );?;

3

3

0

;?;?,0 mol/m39,44

/jamm 112&,21

mol/jam 00005

F; ===

4



• Per.amaa# k"#et"ka +e#ga# or+e 1

=a/@T(

0 ek k −

=( )

( )

= ×

−− K 533</mol'K ,314

/95,14

1 em"#59,0k

mol k%

k = 0,55 m"#

k = 35,1 jam1

• Ko#.e#tra." ?

( );?;

( ) x−×= 1;; ;?,0;?

C C$=1400 x (1−0,3)

C C$=4732,287 mol/m3

•

COC k 'r ;? =−)aju reak."

31

;? mol/m 2,432 jam 1,53r ×=− −

jam'mol/m 5120,42r 3

;? =−

• Volume ST@

;?

;?,0;;ST@

r

8FV ×= O

( )81;k

8;

;?,0

;?0;?,0 −××××=

( )81k

8 ;?

0 −××=

( )0,3135,1jam

0,3/jamm 21,112&

1

3

−××= −

3m51,13=

5



•

Volume +e.a"#Faktor keama#a# = 207

alulate+V1,2V ×=

51,132,1V ×=

3m501,1&V =

". Penentuan iameter rea$tor

$ = 1,5

$A4

1V

2=

1,5(A4

1501,1& 2 ×=

311,1501,1& =

1,1

501,1&3 =

019,143 =

3 019,14 =

"# 921,94t910,m411,2 ===

c. Penentuan $etinggian rea$tor

$ = 1,5 B

$ = 1,5 B 2,411 m

$ = 3,&1& m = 11,&3 t = 142,3&2 "#

. Penentuan &e$anan Desain

Pr%aksi=3 atm=3,033 kg /cm2

ρcampuran= [ (0,243× 1,25 )+ (0,731× 791)+ (0,025× 1700 ) ]

6



ρcampuran=621,024kg /m3

Phidrostatik = ρ( g

gc )h=621,024 ×9.8 × 3,616=22007,103

kg

m. s2

¿0,220 atm

Pop%rasi= Pr%aksi+ P hidrostatik

Pop%rasi=3+0,220

Pop%rasi=3,220atm

Faktor keama#a# u#tuk pera#a#ga# teka#a# +e.a"# a+ala! 307 +ar"

teka#a# opera.", .e!"#gga teka#a# +e.a"#6

Pd%sain=1,3 × 3,220atm

Pd%sain=4,186atm=61,536 psi

e. Penentuan Fa$tor orosi

Ca!a# -a#g +"gu#aka# pa+a pera#a#ga# "#" a+ala! High lloy Steel ,laju koro." -a#g +"paka" -a"tu 0,125 mm/ta!u# +a# peramala# pemaka"a#

+e.a"# a+ala! 10 ta!u#' :aka aktor koro." (6

C c=0,125 mm

tahun ×10 tahun=1,250 mm=0,0041 &t =0,0492∈¿

f. Penentuan &e"al Dining #ea$tor (Cylindrical Shell )

<e#". mater"al 6 aro# Steel T-pe 25 Dra+e

Pd%sain=61,536 psia

Pd%sain= P=61,536l'

¿2=61,536 psia

(ari− jari ( )=1

2 #=

1

2×94,921=47,460∈¿

Allo)a'l% !tr%ss%s (! )=13750 psia

7



"&&isi%nsi !am'ungan ( " )=0,85

C c=0,0492∈¿

:aka,

*%'al dinding (t )= P

!"−0,6 P+C c Crow#ell E %ou#g,

25(

*%'al dinding ( t )=61,536 ×43,7480

(13750 × 0,85 )−(0,6 ×61,536)+0,0492

*%'al dinding ( t )=0,231∈¿5,867 mm

g. Penentuan &e"al &utup #ea$tor (Torispherical Head )

<e#". mater"al = aro# Steel T-pe 25 Dra+e

Faktor koro." = 0,0492 "#

Pd%sain=61,536 psia

Pd%sain= P=61,536l'

¿2=61,536 psia

#iam%t%r ( # )=94,921∈¿

Allo)a'l% !tr%ss%s (! )=13750 psia

"&&isi%nsi !am'ungan ( " )=0,85

:aka,

*%'al tutup (t )= P#

2 !"−0,2 P+C c

*%'al tutup (t )=61,536× 87,4961

(2 ×13750 ×0,85 )−(0,2 ×61,536)+0,0492

*%'al tutup (t )=0,230∈¿5,584 mm

8



C. Pompa (P-1*1)

ata per!"tu#ga#6

Temperatur, T = 30

)aju al"r ma..a, m = 150 kmol/! = 1059,2 lm/!

e#."ta., > = 92 kg/m3 = 49,43 l/t3

V"k.o."ta., G = 0,00039& l/t'.

• :e#g!"tu#g e"t

e"t =>m

S:C@(

=.

t0594,0

!

t 214,024

43,49

1059,2 33

==

• :e#g!"tu#g +"ameter +alam opt"mum

H opt"mum =

0,130,45 >8I83,9S:C@(

H opt"mum =

0,134,5 49,4380,059483,9

H opt"mum =1,9 "#

Cer+a.arka# +ata pa+a Tael J'5 Dea#kopl".( +"perole! +ata ukura# p"pa

..ta#+ar +e#ga# +"ameter +alam -a#g me#+ekat" !a."l per!"tu#ga# a+ala! 6

"ameter #om"#al = 2 "#

S!e+ule umer = 0

9



H#."+e +"ameter, H = 1,939 "# = 0,1&1 t = 0,0492 m

?ut."+e +"ameter, ? = 2,35 "#

)ua. pe#ampa#g, J = 0,0205 t2

• :e#g!"tu#g C"la#ga# @e-#ol+

)aju al"r, V =J

I

=.

m0,3

.

t2,9

0,0205

0,0594==

NRE =ρ x v x D

μ

=49,43 x 2,897 x 0,161

0,000396

=58521,84

• :e#g!"tu#g !ea+ lo..

a' $ea+ lo.. kare#a ko#trak."

α 2

V

J1

J21855,0!

2

=

J1 +"a#ggap .a#gat e.ar .e!"#gga J2/J1 = 0

L = 1 u#tuk turule#

( )kg

<214,0

2

0,301855,0!

2

==

' $ea+ lo.. kare#a ek.pa#."

α 2

V

J1

J2181!e8

2

=

$e8 =

( )kg

<391,0

2

0,30181

2

=

' $ea+ lo.. p"pa luru.

"gu#aka# p"pa je#". ommer"al .teel +e#ga# !arga M = 4,&8105

Keka.ara# p"pa =

000934,00492,0

0,00004&

M==

10



e#ga# me#ggu#aka# "g 2'103 gea#kopl"., !al (, +" plot M/

+e#ga# re +"+apat #"la" koe"."e# ge.ek ( pa+a p"pa luru. -a"tu

0,00&

kg

<3,1

2

0,3

0,0492

10 80,00584

2L

V

N)4 F

22

===

+' $ea+ lo.. kera#ga#

Date ale 1 ua!(

K= 0,1

kg<&&5,0

20,3 0,181

2LV k #!

22===

e' $ea+lo.. eloka#

Tee 1 ua!(

kg

<0,391

2

0,3 181

2L

V k #!

22

===

' $ea+lo.. total

Total $ea+lo.. = F O !kera#ga# O !eloka# O ! O!e8

= 3,1 O 0,0&& O 0,391 O 0,214 O 0,391

= 4,kg

<

• :e#g!"tu#g Kerja Pompa

.>

NP −=

(Smith, hal. 279,2005)

kg

<5&29,1

92

4'559'&25101'325

>

PP.

.>

PP

21

21

−=−

=−

=

−=−

• :e#g!"tu#g a-a pompa

11



pompa .e#tr"ugal pa+a 0,1139 mɳ 3/me#"t a+ala! 457 Turto#, !al 3&9(

Hp =-W

s

m

η = -(-5629,17

)×1500,45 = 15,68 kW

C+.Cooler (,-1*)

onisi operasi

Fluia panas

T"# = 199,Q = 391,&4QF

Tout = 45Q = 113QF

)aju al"r = 50 kmol/! = 30&,4 l/!

Fluia ingin

T"# = 30Q = &QF

Tout = 40Q = 104QF

)aju al"r = 1200,5 l/!

12

199,8

o

C



Perhitungan nilai +* ,M*#

−

=

1

2

12

):T

NT

NTl#

(NTNNNT

Sert!, !al' 93, 200(

!"#!D =$279℉ % 27℉&

( '(279℉

27℉ )!"#!D = 107,8 ℉

Perhitungan luas perpinahan panas

ar" tael , !alama# 40 Ker#, 19&5(, +e#ga# je#". lu"+a pa#a. -a"tu ga.

+a# lu"+a +"#g"# a"r #"la" - #=2−50, +a# -a#g +"gu#aka# -a"tu6

)D = 26 *+ /m +2

I = &3,22 k</. = 21514,4& Ctu/!

):T NT8

IJ =

Sert!, !al' 9, 200(

13

45oC40oC

30oC

Flu"+a pa#a.QF(

Flu"+a +"#g"#QF(

NTQF(

T1 391,&4 t1 & 29

T2 113 t2 104 2



¿215714,746 *+

26 *+ +2× 107,8℉

= 76,956 +2

Perhitungan umlah tube

ar" ker#, 19&5( Tael 10, !alama# 43, +"perole! .pe.""ka." tue .eaga"

er"kut 6

? O&t"ide Diameter ( = 0,5 "#

CD = 1&

H 'n"ide Diameter ( = 0,&20 "#

aRt (lo! area per t&be( = 0,3020 "#

2

ao "&rfa#e per line( = 0,19&3 t

/ t =

A

,× ao

! (ft)#equire

o.&u"es

Dshell (num"er of tu"es) in

&ringular %quare

13 30,05&4 30( 2&(

14 2,0024 1052( 1052(

15 2&,135& 122( 12&(

1& 24,5021 13,2510&( 13,2590(

"p"l"! "hell and t&be +e#ga# pa#ja#g 13 t +a# mem"k" 30 t&be +e#ga#

+"ameter "hell .ee.ar "# triang&lar (

Perhitungan $ore$si ilai /

= N+ × / × " Ker#, !al'

150, 19&5(

= 30 × 0,1963 × 13

= 76,557 +2

):T

NJ 8

Ikorek." =

Ker#, !al' 150, 19&5(

= 215714,746 *+/m

76,557 +2

× 107,8 ℉

14



= 26,135 *+-/m +2

Fluia ingin paa tube

Perhitungan luas area

pa..e.8144

tat 8 a t =

Ker#, !al' 150, 19&5(

=$30 x 0,302&

$144 x 2&

= 0,0303 +2

Perhitungan lau alir massa paa tube

ta

al"r lajuDt =

Ker#, !al' 150, 19&5(

= 1200,78 '/m

0,0303 +2

= 38202,278 ' +2 /m

Perhitungan "ilangan #e0nol

T = 99,5Q

= 0,295 l/t' jam Ker#, "g' 15, 19&5(

t = 0,&2 "# = 0,051 t Ker#, tael 10, 19&5(

G

Dtt 8 re =

Ker#, !al 150, 19&5(

=0,0517 + ×38202,278 ' +

2 /m

0,2975 '+. /m

= 6631,64

Perhitungan nilai transfer panas paa tube ()

j$ = 45 Ker#, "g' 2, 19&5(

k = 0,3&3 Ctu/!'t'F Ker#' Tael 5, 19&5(

15



1/3

$"k

pG

H

k j!

=

Ker#, !al' 150, 19&5(

= 45( 0,363

0,62 )(1088,06 x 0,2976

0,0135 )13

¿295,8 *+/m +2 ℉

=

?

H!! ""o

Ker#, !al' 150, 19&5(

= 295,8 (0,62

0,75 ) = 244,541 *+/m +

2 ℉

Fluia panas paa shell

Perhitungan flow area

Baffle "pa#e = "#

)it#h = 1 "#? = 0,5 "#

R = PT U ?

= 0,25 "#

PT8144

.pae ale8;8.!ellHa . =

Ker#, !al' 150, 19&5(

2

. 0,09t

18144

80,25"# 8a ==

Perhitungan lau alir massa paa shell

0s=lajualir

as

=3086,42 '/m

0,0889 +2

=27777,7 ' +2 /m

16



Perhitungan "ilangan #e0nol paa shell

= 0,0532 l/t'jam

)it#h = 1 "# = 0,033 t

? = 0,5 "# = 0,0&25 t

4 ×( P* 2−(3,14 ×

$#2

4 ))¿

¿ #s=¿

=

$4 × (0,08332+ -(3,14 ×

0,06252+

4 ))( 3,14 × 0,06253

2+ )

= 1,264 +

N =Ds × :s

;

=1,264 + × 27777,7 ' +2

/m0,0532 '+./m

= 660117,64

Perhitungan nilai transfer panas paa shell ()

ar" gamar 2 Ker#, 19&5( +"+apatka# #"la" j$, .ee.ar 500

= H(k

Ds )(<p μ

k )13

ϕ

= 500

(0,0135

1,264

)( 0,25 × 0,532

0,0135

)

13

× 1

= 60,781 *+/m +2 ℉

Perhitungan clean overall coefficient (/c)

o"o

o"o

!!

!8!

+=

Ker#, !al' 150, 19&5(

17



¿(244,541 × 60,781 )(244,541 60,781 )

¿48,681 *+/m +2℉

Perhitungan fa$tor $e$otoran (# )

+

+

+ 8

@

−=

Ker#, !al' 150, 19&5(

=48,681 - 26,135

48,681 x 26,135

= 0,0177 /m +2 ℉ *+

Perhitungan pressure drop paa tube

= 0,00009 Ker#, "g' 2&, 19&5(

. = 1 Ker#, tale &, 19&5(

81.t 881085,22

pa..e.8)8Dt8 NP

10

2

t =

Ker#, !al' 151, 19&5(

=$0,00009 × 38202,278

2 × 13 × 2&

(5,22 × 1010 ) × 0,0516 × 1 × 1

= 0,0052 ps>?

2

2@= 0,001

Ker#, "g' 2, 19&5(

2g

V8

.

pa..e.84P

2

t2 =∆

Ker#, !al 151, 19&5(

18



=4 × 2

1× 0,001

= 0,08 ps>

A++/' = A+1 A+2

A++/' = 0,0052 ps> 0,08 ps>

A++/' = 0,0813 ps>

+P pada tu'%<10 psi makam%m%nuhi s1arat Ker#, !al 153, 19&5(

Perhitungan pressure drop paa shell

= 0,00025 Ker#, "g' 29, 19&5(

. = 1,29 Ker#, tale &, 19&5(

.paeCale

) 8121# =+

Ker#, !al 151, 19&5(

=12 ×13 +

0,67 += 234

As'' =$ × :s

2 × BD s'' × $(1&&

(5,22×1010 )× Ds × s × 1

=$0,000285 × 27777,7

2 × 0,67 × 234&

(5,22 × 1010 ) × 1,267 × 1,29 × 1

= 0,009 ps>

+P pada sh%ll<10 psi maka m%m%nuhi s1arat Ker#, !al 153, 19&5(

19



eater (,-1*1)

onisi operasi

Fluia panas

T"# = 100Q = 212QF

Tout = 100Q = 212QF

)aju al"r = 915,2& l/!

Fluia ingin

T"# = 33,&Q = 92,4QF

Tout = 0Q = 1&QF

)aju al"r = 1052,1 l/!

20100oC



Perhitungan nilai +* ,M*#

−

=

1

2

12):T

NT

NTl#

(NTNNNT

Sert!, !al' 93, 200(

!"#!D =$120℉ % 36℉&

('(120℉36℉ )

!"#!D = 69,6 ℉

Perhitungan luas perpinahan panas

ar" tael , !alama# 40 Ker#, 19&5(, +e#ga# je#". lu"+a pa#a. -a"tu ga.

+a# lu"+a +"#g"# a"r #"la" )D =200-700 2 +a# -a#g +"gu#aka# -a"tu6

)D = 200 *+ /m +2

I = 11,5& k</. = 5533,1 Ctu/!

):T NT8

IJ =

Sert!, !al' 9, 200(

¿585383,71 *+

200 *+ +2× 69,9℉

= 42,052 +2

Perhitungan umlah tube

21

100oC

80oC

33,6oC

Flu"+a pa#a.

QF(

Flu"+a +"#g"#

QF(

NT

QF(

T1 212 t1 92 120

T2 212 t2 1& 3&



ar" ker#, 19&5( Tael 10, !alama# 43, +"perole! .pe.""ka." tue .eaga"

er"kut 6

? O&t"ide Diameter ( = 1 "#

CD = 14

H 'n"ide Diameter ( = 0,340 "#

aRt (lo! area per t&be( = 0,54&0 "#2

ao "&rfa#e per line( = 0,2&1 t

/ t =

A

,× ao

! (ft)#equire

o.&u"es

Dshell (num"er of tu"es) in

&ringular %quare

10 1&,0&2 1&( 1&(

11 14,&025 1032( 1032(

12 13,35& 1252( 1245(

13 12,35&0 13,25&&( 13,255&(

"p"l"! "hell and t&be +e#ga# pa#ja#g 10 t +a# mem"k" 1& t&be +e#ga#

+"ameter "hell .ee.ar "# triang&lar (

Perhitungan $ore$si ilai /

= N+ × / × " Ker#, !al'

150, 19&5(

= 16 × 0,2618 × 10

= 41,88 +2

):T

NJ 8

I

korek." =Ker#, !al' 150, 19&5(

=585383,71 *+/m

41,88 +2 × 69,6 ℉

= 200,784 *+/m +2

Fluia ingin paa tube

Perhitungan luas area

22



pa..e.8144

tat 8 a t =

Ker#, !al' 150, 19&5(

=$16 x 0,546&

$144 x 2&

= 0,0303 +2

Perhitungan lau alir massa paa tube

ta

al"r lajuDt =

Ker#, !al' 150, 19&5(

= 10582,18 '/m

0,0303 +2

= 181318,68 ' +2 /m

Perhitungan "ilangan #e0nol

= 0,144 l/t' jam Ker#, "g' 15, 19&5(

t = 0,340 "# = 0,0&95 t Ker#, tael 10, 19&5(

G

Dtt 8 re =

Ker#, !al 150, 19&5(

=0,0695 + × 181318,68 '- +

2 /m

0,1447 '-+. /m= 87098,992

Perhitungan nilai transfer panas paa tube ()

j$ = 250 Ker#, "g' 2, 19&5(

k = 0,124 Ctu/!'t'F Ker#' Tael 5, 19&5(

1/3

$"k

pGHk j!

=

Ker#, !al' 150, 19&5(

= 250( 0,124

0,069 )(30,753 x 0,1447

0,124 )13

¿1471,193 *+/m +2 ℉

=

?

H!! ""o

Ker#, !al' 150, 19&5(

23



= 1471,193(0,069

0,083 ) = 14723,7 *+/m +

2 ℉

Fluia panas paa shell

Perhitungan flow area

Baffle "pa#e = "#

)it#h = 1,25 "#

? = 1 "#

R = PT U ?

= 0,25 "#

PT8144

.pae ale8;8.!ellHa . =

Ker#, !al' 150, 19&5(

2

. 0,09t18144

80,25"# 8a ==

Perhitungan lau alir massa paa shell

0s=lajualir

as

=915,265 '/m

0,0889 +2

=10296,73 ' +2 /m

Perhitungan "ilangan #e0nol paa shell

= 0,050 l/t'jam

)it#h = 1,25 "# = 0,1042 t

? = 1 "# = 0,03 t

4 ×( P* 2−(3,14 ×

$#2

4 ))¿¿

#s=¿

24



=

$4 ×

(0,1042

2+ -(3,14 ×

0,0832+

4 ))( 3,14 × 0,083

2+ ) = 0,99 +

N =Ds × :s

;

=0,99 + × 10296,73 ' +

2 /m

0,0508 '+./m=200750,712

Perhitungan nilai transfer panas paa shell ()

ar" gamar 2 Ker#, 19&5( +"+apatka# #"la" j$, .ee.ar 520

= H(k

Ds )(<p μ

k )13

ϕ

= 520(0,0145

0,99 )( 639,58 × 0,508

0,0145 )13

× 1

= 99,618 *+/m +2 ℉

Perhitungan clean overall coefficient (/c)

o"o

o"o

!!

!8!

+=

Ker#, !al' 150, 19&5(

¿(14723,7 × 99,618 )(14723,7 99,618 )

¿98,95 *+/m +2℉

Perhitungan fa$tor $e$otoran (# )

+

+

+ 8

@

−=

Ker#, !al' 150, 19&5(

=98,95 - 200,78

98,95 x 200,78

= 0,0051 /m +2 ℉ *+

Perhitungan pressure drop paa tube

25



= 0,00015 Ker#, "g' 2&, 19&5(

. = 0,9 Ker#, tale &, 19&5(

81.t 881085,22

pa..e.8)8Dt8 NP

10

2

t =

Ker#, !al' 151, 19&5(

=$0,00015 × 181318,7

2 × 10 × 2&

(5,22 × 1010 ) × 0,0695× 0,79 × 1

= 0,0344 ps>?

2

2@= 0,42

Ker#, "g' 2, 19&5(

2g

V8

.

pa..e.84P

2

t2 =∆

Ker#, !al 151, 19&5(

=4 × 2

0,79× 0,42

= 4,253 ps>

A++/' = A+1 A+2

A++/' = 0,034 ps> 4,253 ps>

A++/' = 4,2876 ps>

+P pada tu'%<10 psi makam%m%nuhi s1arat Ker#, !al 153, 19&5(

Perhitungan pressure drop paa shell

= 0,0012 Ker#, "g' 29, 19&5(

. = 1 Ker#, tale &, 19&5(

.paeCale

) 8121# =+

Ker#, !al 151, 19&5(

=12 ×10 +

0,67 += 180

26



As'' =$ × :s

2 × BD s'' × $(1&&

(5,22×1010 )× Ds × s × 1

=$0,0012 × 10296,73

2 × 0,67 ×180&

(5,22 × 1010 ) × 0,99 × 1 × 1

= 0,03 ps>

+P pada sh%ll<10 psi maka m%m%nuhi s1arat Ker#, !al 153, 19&5(

27

Perancangan alat asam format

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