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APPENDIX A

NERACA MASSA

Rencana Kapasitas Produksi

Kapasitas Produksi Ethil Benzen = 70000 ton / tahun

Jumlah hari kerja = 1 tahun =Waktu kerja tiap hari = 1 hari =

Laju Produksi = 70000 tonx

1 tahun

= 111!1111 k"x

1 hari

B$ etil%enzen = 10&!1&7

J'() Laju Produksi tiap jam = **+*!+*+*+*+*+* k" / jam

1. REAKTOR (R 11!"

,un"si - mereaksikan etilena dan %enzena

$ $#

$1

$+

Keteran"an

- 'liran Etilena masuk reaktor

- 'liran masuk reaktor .&& %enzena

- 'liran Etil%enzena keluar reaktor

- 'liran Benzena keluar reaktor

Spesi#ikasi A$iran Masuk Reaktor (R 11!"

a! 2pesi3ikasi EtilenaKo%posisi Et&i$ena 'M ru%us Titik didi&

Etilena *!04&1* .# 510+!&

Etana +0!071 .& 5**!&

Propilena #!0*#7 .+& 5#7!74

Kar%ondioksida ##!01014 .6 57*!44

%! 2pesi3ikasi Benzena

Ko%posisi 'enena 'M ru%us Titik didi&

Benzena 7*!107 .&& *0!0

'ir 1*!014 6 100

8oluena !1#**1 .7* 110!&2ul3ur .oump! +!07 5 ###!&

$1

$

$+

$#

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Men)&itun) Neraca Massa Reaktor

a. *a+u a$ir ke$uar reaktor

(iharapkan reaksi dalam reaktor men"hasilkan -

&49 Etil%enzena: 09 (ietil%enzena dan 149 8rietil%enzena 9Wt

(en"an kon;ersi reaksi 0 9

Basis 1410*!+#*##1&*1* k" produk reaktor

Komponen B$ 9

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Ko%ponenReaksi I Reaksi II Reaksi

.# 51 !#1 5 !411 5+

.&& 51 !#1 51 !411 51

.&4.4 1 !#1

(EB 1 !411

8EB 1

Komponen B$

$asuk

k"mol/jam k"/jam k"mol/jam

.# *!04&1* 1!++1&+*11& 44!&4#&0#&* 1!++1&+*

.&& 7*!107 1#+!4&+7+4 111#!+4*+&+1##1 1#!+4&+7+

.&4.4 10&!17& 0 0 !#1177++(EB 1+#!++0* 0 0 !410&1*74

8EB 1&!*& 0 0 1+!

8otal 1&7*7!04+*#00

Jika jumlah massa keluar Etil%enzena (EB 8EB reaktor 1&7*7!04+*#1 K" /

ana men"hasilkan Etil%enzena *0!#*70 K" /

(an produk Etil%enzena an" diharapkan **+*!+*+*+*+* K" /

Jadi di%utuhkan jumlah massa keluar reaktor 1410*!+#*##17 K" /

Komponen B$

$asuk

k"mol/jam k"/jam k"mol/jam

.# *!04&1* 1!++1&+*11& 44!&4#&0#&* 1!++1&+*

.&& 7*!107 1#+!4&+7+4 111#!+4*+&+1##1 1#!+4&+7+

.&4.4 10&!17& 0 0 !#1177++

(EB 1+#!++0* 0 0 !410&1*74

8EB 1&!*& 0 0 1+!

8otal 1&7*7!04+*#00

41

/1

41

/-

4

+= iiii nn++11&+&+

+++= in OHCout

OHC nn

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. *a+u a$ir %assa a&an %asuk reaktor a$ki$asi

Ko%ponen B$

$1

5t 6 K)7+a% k"mol/jam

Eti$ena *!04&1* 9 44!&4#&0#&* 1!++1&+*1

Etana +0!071 0!709 #0!&7#1#**0& 1!+444+#1

Propilena #!0*#7 0!1&09 !0+*&7+# 0!1#77&0Kar%ondioksida ##!01014 0!109 &!770#*011 0!14#0++1

'enena 7*!107

'ir 1*!014

8oluena !1#**1

2ul3ur Coump! +!07

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Neraca Massa Reaktor (R11!"

Masuk

Ko%ponen k)7+a% Ko%ponen

M1 (Eti$ena" M %e

Etilena 44!&44 .&4.4

Etana #0!&7#1 (EBPropilena !0+*7 8EB

Kar%ondioksida &!770 8oluena

2ul3ur Coump!

Jumlah 4!1*7+

M- ('enena" M3 %e

Benzena 111#!+4*# .#

'ir 110!+7++ .&&

8oluena &0!+071 'ir

2ul3ur Coump! 1+!&4## Kar%ondioksida

etanaPropilena

Jumlah 11+7*!&+

TOTA* 18!-8.99!:

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-. Ko$o% Disti$asi E' (D 1-!"

$emisahkan etil %enzena dari pen"otor

$4

$+

$&

Keteran"an

- 'liran produk etil%enzen / keluar distilasi

- 'liran Etil%enzena keluar reaktor / masuk striper

- 'liran keluar distilasi dietil%enzena

a! 'liran %ahan masuk kolom destilasi ( 5 10

Komponen B$

$+ $+

k"/jam k"mol/jam

8oluena !1#**1 110!& &0!+0707+ 0!&4##4+0

Etil%enzena 10&!17& 1+&!1 *0!#*70+ !#1177++

(EB 1+#!++0* 1*+!4 +01!&&&**++ !410&1*74

8EB 1&!*& 17!& &&!4&&4+ 1+!

sul3ur Coump! +!07 ###!& 1+!&4##+1*+1# 0!#47&&

868'L 1+0!0#&&+

Dntuk menentukan K dan LK ditentukan dari %e%an distilasi terendah

(iperoleh %e%an destilasi terendah = ( = !11+70017

d8

2ehin""a ditentukan Etil %enzena se%a"ai Li"ht ke dan (EB se%a"ai ea; ke

'sumsi- Etil%enzena li"ht ke an" keluar se%a"ai %ottom se%esar

(ietil%enzena hea; ke an" keluar se%a"ai destilat se%esar

Ko%ponen;eed2 ; (M" Desti$at (M:"

k%o$7+a% k%o$7+a%

8oluena 0!&4#4 0!0040 0!&4#4 0!0

Etil%enzena LK !#1 0!711 1!4&&+ 0!

(EB K !410& 0!17+1 0!41 0!0

8EB 1+!+ 0!107#

sul3ur Coump! 0!#4* 0!00++

-.33:9

. Menentukan su&u 'u$e Point dan De? Point pada ko$o% desti$asi D1-!

Dntuk mendapatkan komponen an" CoCok untuk proses destilasi maka dilakukan

;un)si,

$4

$+

$&

8 didih o.

@;

@

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Persamaan 'ntoine

(ata 'ntoine

Ko%ponen A ' C8oluena #!4### 17+*!1+0 0!+#

Etil%enzena #!#04# 1&4!0&0 5+!&

(EB #!1+0 147&!&10 57!1#

8EB #!70+& 1++!44*0 5+7!#&

2ul3ur #!40& 1+7!+4#0 57!4+

lo"10

P = ' B / 8 .

P = ;apor pressure %ar

8 = temperature K

B(, +=

B(, F!BF!(F!, +=

11 +=

i

ii

K

zx

101 +=

i

iii

K

Kzy

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1. ;eed ko$o% desti$asi

$enentukan suhu %u%le point pada 3eed kolom destilasi

P = 1 %ar = 7&0 mm"

8 = 1##!*00 = #17!700 K

Ko%ponen Ki iBKi

8oluena 0!0040 !#4+4 !#4+4 0!01+

Etil%enzena 0!711 1!771 1!771 0!0*+(EB 0!17+1 0!+7#0 0!+7#0 0!0

8EB 0!107# 0!1#* 0!1#* 0!01+#

2ul3ur 0!00++ 0!+#71 0!+#71 0!0011

Jumlah 1.!!! !.>1: 1.!!!

G = 0!0000

$enentukan suhu de< point pada 3eed kolom destilasi

8 = 1&7!110* = ##0!&0*+44 K

Ko%ponen X; i Ki i7Ki8oluena 0!0040 +!*0& +!*0& 0!001+

Etil%enzena 0!711 !1&1 !1&1 0!+7

(EB 0!17+1 0!70& 0!70& 0!#41

8EB 0!107# 0!441 0!441 0!#10

2ul3ur 0!00++ 0!+# 0!+# 0!0041

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8 = 1&!44* = #+4!&74* K

Ko%ponen Ki iBKi

.*10 0!0#& 1!4#+ 1!4#+ 0!0#*1 0!0

(EB 0!47 0!&+& 0!&+& 0!+&& 0!4

8EB 0!+71# 0!17 0!17 0!0*+ 0!+

2ul3ur 0!011+ 0!4&* 0!4&* 0!4000 0!0

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Neraca Massa Ko$o% Desti$asi D1-!

Masuk

Ko%ponen k)7+a% Ko%ponen

M dari R 11! M: %

8oluena &0!+071 8oluena

Etil%enzena *0!#&4 Etil%enzena(EB +01!&&7 (EB

8EB &&!4+

2ul3ur .ompound 1+!&4##

Tota$ 1:19-.!>> Tota$

M= %

Etil%enzena

(EB

8EB

2ul3ur .ompound

Tota$

>

c. $en"hitun" minimum sta"e kolom destilasi

= +!4#& de.::9= Tota$

M9 %

(EB

8EB

2ul3ur

Tota$.::9=

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++0 hari# jam

1000 k"x

1 tahun

1 ton ++0 hari

hari

# jam

= *+!#*1*& k"mol / jam

= *# k"mol / jam

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&1

&1

&*

&*

4 = !#1177++7

EB = !410&1*74&1

EB = 1+!&*

A1

A

A+

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II k%o$7+a%

+! 1!++ 1!+++! 1#+!4& 1#!+&

0 !#1

0 !410&

+! 0 1+!+

Keluar

k"/jam

1 44!&4#&0#7

4& 111!#+4*+&+1##

*0!#*70+&1 +01!&&&**++

&* &&!4&&4+

1&7*7!04+*#00

am

am

am

am

Keluar

k"/jam

1 44!&4#&0#7

4& 111!#+4*+&+1##

*0!#*70+

&1 +01!&&&**++

&* &&!4&&4+

1&7*7!04+*#00

/

ni

in ni

out

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$

5t 6 K)7+a% k"mol/jam

1&

*

#*

*!#9 111#!+4*+&+1##1 1#+!4&+7+4

0!79 110!+7+++70* &!1&&77&*

0!4+9 &0!+0707+ 0!&4##4+04

0!19 1+!&4##+1*+1# 0!#47&&+7

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e$uar

k)7+a%

u+u D 1-!

*0!#*70+

+01!&&&**++&&!4&&4+

&0!+071

1+!&4##

141*!+0#7#+4#

u+u stora)e

44!&4#&0#7

111!#+4*+&+1##

110!+7++

&!770

#0!&7#1!0+*7

1*#4!470477*

18!-8.99!:

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i

0!0040

0!711

&1 0!17+1

&* 0!107#

7 0!00++

7 1!000000

= 1!09

= 1!09

'otto% (M="

k%o$7+a%

70*

0# 0!# 0!0#&

## !*44 0!47

1+!+ 0!+71#

0!#4* 0!011+

8.=!!: 1.!!!

D@

'

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110!#

0 1+&!1

0 1*+!4

0 17!&

0 ###!&

8 didih o.

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Xi Fi

0!0040 0!01+

0!711 0!0*+0!17+1 0!0

0!107# 0!01+#

0!00++ 0!0011

!.>>: !.>99

Xi Fi0!001+ 0!0040

0!+7 0!711

0!#41 0!17+1

0!#10 0!107#

0!0041 0!00++

!.>>> !.>>:

i Fi

71 0!01+*

04 0!*44

# 0!0007

00 1!0000

i Fi

71 0!01+

04 0!+#

# 0!0007

00 1!00*0

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i Fi

#& 0!0#*1

7 0!+&&

1# 0!0*+

11+ 0!00&4

00 0!40&4

i Fi

4 0!0#&

47 0!47

0 0!+71#

* 0!011+07 1!0000

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e$uar

k)7+a%

enu+u ; 1-3

&0!+071

7!+0!1&7

>91-.83=!

nu+u D 1!

*!0#+

1!#4+0

&&!4+

1+!&4##

:=>.:=3!

1:19-.!>>

$C.a%e:1+

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&

F3am

0#+ 0!01*+

#4+0 0!44714+ 0!#1

4## 0!004

4 1!0000

= 1!09= 1!09

'otto% (M9"

k%o$7+a%

41+ 1+&!1

++ 0! 0!*7 1*+!4

44# 0!1+& 0!177 17!&

0!#4* 0!4#01 ###!&

!.899 1

D@

'

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i Fi

&0 0!041

1 0!4+

# 0!+744

00 1!0000

i Fi

7 0!++&*

7 0!0*0

01 0!4*11

00 0!

i Fi

14 0!*7

7# 0!177

00 0!4#01

1!0000

e$uar

k)7+a%

nu+u stora)e

*!0##

&1!4+&7

#+!4*&1

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:!.-8-

nu+u stora)e

!1#4

!&&4

1+!&4##

==.-13:=>.::9=

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APPENDIX '

NERACA PANAS

1. Neraca Panas Reaktor (R 11!"

$ $#

$1

$+

Keteran"an

- 'liran Etilena masuk reaktor = +0 =

- 'liran masuk reaktor .&& %enzena = +4 =

- 'liran Etil%enzena keluar reaktor = 00 =

- 'liran Benzena keluar reaktor = 10 =

(iketahui -

KomponenPanas 2ensi%el

a % C d

Etilena #0!74 1!1#7 50!000&*1 1!7&&E5007

etana #!+7 1!+ 50!0004*1& 7!*E500*

propilena 4!4* 1!771 50!001017 0!000000#&

kar%ondioksida +&!11 0!#++ 0!000**7 7!#E500*

Benzena 57!7+ 7!704# 50!01* 0!00001*7

'ir 1*! #!71 50!01++** 0!00001+1#

toluena 1!*0*+ *!1 50!01417 0!00001&+

sul3ur Comp 14! 0!&* 0 0.&4.4 7!044 5+! 0!0++0** 50!00001477

(EB 5!&7&E5007 5!&11+7 0!0+11* 51!+&4&E5004

8EB 0 5!*+ 0!0+###& 50!000017*1

a. A$iran Masuk Reaktor

= *!14 K = 4 .

KomponenPanas sensi%el 'liran dari $1

J/mol 81 kmol/jam 1 kCal/jam

Etilena +4* *&* +0 17 #04#7#+ 14+** 1#1

$1

o.

$

o.

$+

o.

$#

o.

M(ikutip dari immel%lauN .p= a%8C8d8+

8re3

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etana #+&!04# +0 1!17+ 1&!*&7

propilena 4+7!4* +0 0!1++ #!*+&4

kar%ondioksida +!*4& +0 0!1+*& 7!#7#

Benzena 1+#&!74#0

'ir 4*4!*0

toluena 147!**0

sul3ur Comp !##0

Jumlah 1*0!4#4 144#7!7

. A$iran Ke$uar Reaktor

= *!14 K = 4 .

KomponenPanas sensi%el 'liran dari $#

J/mol kmol/jam

.# 414!+0& 10 17!#04 10*0+&!#*+

etana +107!&0&& 10 1!17+ 0#7!7+71

Propilena +*#&&!##& 10 0!1++ 1777!1&1+

Kar%ondioksida 1+0!##* 10 0!1+*& #&0!*7#7

.&& 4&&+!71* 10 1!*&& 71+7!+717

'ir &740!10 10 4!407+ 17+4!07

8oluena +0#747!4&4

2ul3ur Coump! 17*!4#+0

.&4.4 #+70!#&*4

(EB 41#4#!*&0

8EB 444#70!1+4Jumlah +7!*+7 10+71!1&&

e. Men)&itun) panas reaksi dari %asin)%asin) ko%ponen

1! $en"hitun" panas reaksi )

Ko%ponen k%o$7+a%k

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+! $en"hitun" panas reaksi )))

Ko%ponen k%o$7+a%k

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DE' 1*1!1 0!4&

TE' 17!& 1!4&7*

To$uena 0!4**+

su$#ur co%p 0!+*7

$4

$+

$&

(ata an" diperoleh dari perhitun"an temperatur %u%le point dan de< point

8emperatur pada ,EE(-

Bu%le point = 1##!*00 = #17!700 K

(e< point = 1&7!110* = ##0!&0* K

8emperatur pada (E28)L'8-

Bu%le point = 1+4!1+ = #0*!7+ K

(e< point = 1+4!## = #0*!47# K

8emperatur pada B6886$-

Bu%le point = 1&!44* = #+4!&74* K

(e< point = 1*!&+ = #71!77+ K

(irenCanakan-

2uhu %ahan masuk = 1*0 = #4+!14 K

2uhu liRuid keluar re%oiler = 14 = #&*!14 K

2uhu air pendin"in masuk = * = +01!14 K

2uhu air pendin"in keluar = 0 = +&+!14 K

a. Komposisi masin"5masin" komponen pada tiap aliran %ahan dalam kolom destilasi

Komponen8 %oil

,eed: , (estilat: (

kmol/jam kmol/jam kmo

8oluena 110!& 0!4**+ 0!0040 0!4*# 0!00& 0!0

1+&!1 *+!#1 0!711 *+!#1 0!07 0!0

(EB 1*+!4 0!4& 0!17+1 0!0& 0!00# 0!0

8EB 17!& 1!4&7* 0!107# 0!0000 1!4

2ul3ur ###!& 0!+*7 0!00++ 0!0000 0!+Jumlah 117!0#04 1 *#!071 1 ++!0

o.

o.

o.

o.

o.

o.

o.

o.

o.

o

.

x,

x(

.*

10

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. $enentukan re3luks minimum

(en"an men""unakan persamaan 11:751 dan 11:750 Seankoplis: 1+

(imana 3eed masuk dalam keadaan liRuid pada titik didihna: R = 1 Kusnarjo: 01

8rial T = +&*+!74*#*

Dmpan masuk kolom destilasi u$e

Komposisi Ki H

8oluena 0!0040 !#4+4 &!44# 0!0000

0!711 1!771 +!#1#+ 50!0007

(EB 0!17+1 0!+7#0 1!0000 0!00008EB 0!107# 0!1#* 0!+++* 0!0000

sul3ur 0!00++ 0!+#71 0!*0 0!0000

Jumlah 1 50!0007

(estilat keluar kolom destilasi de?

Komposisi Ki H

toluena 0!00& 1! 7!04*4 0!0000

0!07 1!00 +!4#& 50!0010

(EB 0!00# 0!70 1!0000 0!0000

Jumlah 1 50!0010

= 50!0010 0!000*0

= 51!0010

(irenCanakan re3luks rasio = 1!4 x

= 0!001

c. $en"hitun" panas pada laju alir %a"ian atas dan laju alir %a"ian %a

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omponenkmol/jam 8* * kCal/jam

8oluena 0!000 0!0140 *!14 #71!77 11+&!711

0!0000 0!0000 *!14 #71!77 0!0000

0!&074 41!1#* *!14 #71!77 +7&!7*

0!+*07 +!0+4 *!14 #71!77 #+441&!&

2ul3ur 0!011& 0!74& *!14 #71!77 #7!0#41Jumlah 1!0000 *#!140& 107**7!#1

d. $en"hitun" panas 3eed masuk: destilat keluar dan %ottom keluar

1! $en"hitun" panas pada 3eed masuk kolom destilasi

KomponenPanas liRuid masuk kolom destilasi: ,

kmol/jam 8 kCal/jam

8oluena 0!0040 0!4**+ *!14 #17!7 007!&++#

0!711 *+!#1 *!14 #17!7 4141!&&

(EB 0!17+1 0!4& *!14 #17!7 1#4*!040&8EB 0!107# 1!4&7* *!14 #17!7 101&*!+14&

2ul3ur 0!00++ 0!+*7 *!14 #17!7 11*!+77*

Jumlah 1!0000 117!0#04 7701**4!0#+7

! $en"hitun" panas liRuid keluar se%a"ai destilat

KomponenPanas liRuid keluar se%a"ai destilat: (

kmol/jam 810

8oluena 0!00& 0!4*# *!14 #0*!7 &11!1#+&

0!07 *+!#1 *!14 #0*!7 #4**&*!4#(EB 0!00# 0!0& *!14 #0*!7 1++#!4+4#

Jumlah 1!0000 *#!071 #&*00!0+1

+! $en"hitun" panas liRuid keluar se%a"ai %ottom

KomponenPanas liRuid keluar se%a"ai %ottom: B

kmol/jam 811 11 kCal/jam

8oluena 0!000 0!004 *!14 #+4!&* ++!+7*#

(EB 0!&074 0!0470 *!14 #+4!&* 17+0+!177

8EB 0!+*07 1!4&7* *!14 #+4!&* 1&7!04*#

2ul3ur 0!011& 0!+*7 *!14 #+4!&* 1#*!&4#

Jumlah 1!0000 ++!01+# +017*7!*7#

e. $en"hitun" IeraCa Panas 6;erall

Kesetim%an"an panas o;erall pada kolom destilasi-

=

7701**4!0#+7 = #&*00!0+1 +017*7!*7#

= 5400&!&4&

1! $en"hitun" panas air keluar kondensor

Kesetim%an"an panas pada kondensor-

B 8re3

.*

10

.10

1#

.1

1*

x, 8

re3

.*

10

x( 8

re3

10kCal/jam

.*

10

xB 8

re3

Q

10

11 .

Q

.

Q

.

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#*#!17 = 7114!&&&+ #&*00!0+1

= *7*!+#7* kCal/jam

! $en"hitun" panas steam pada re%oiler

= 5400&!&4&

= 5400&!&4& *7*!+#7*

= 17&71!+* kCal/jam

#. $en"hitun" ke%utuhan air pendin"in

= m x .p x 8

*7*!+#7* = m x #!1*1& kJ/k" K x +&+!14 5 +01!14

m = **!40*1 k"/jam

). $en"hitun" neraCa panas pada re%oiler

Kesetim%an"an panas pada re%oiler-=

107#70!4## = 107**7!#1+ +017*7!*7#

= +0#!7##1

(imana-

= 19 ? * 11 @

= 19 ? 107**7!#1+ +017*7!*7# @

= 1+7*&&!74*7*4 kCal/jam

2ehin""a-

= +0#!7##1 = +0#!7##1 1+7*&&!74

= +0071!#70 kCal/jam

&. $en"hitun" ke%utuhan steam

Kondisi steam masuk- 8 = 00

P = 1444 kPa

= 1+*!& kJ/k"

= #&+!+##* kCal/k"

$aka steam an" di%utuhkan =

= +0071!#70

#&+!+##*

= &0&!*01 k"/jam

Neraca Panas Ko$o% Desti$asi

'liran Panas $asuk 'liran Panas Keluar

Komponen Ener"i kCal/jam Komponen Ener"i kCal/jam

7701**4!0#+7 #&*00!0+1

.

.

Q

.

Q

Q

.

7 Q

* 11 Loss

Q

Loss

Q

Loss

Loss

Q Loss

o.

10

QP

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(e< point = 1!#7 = #&&!07 K

(irenCanakan-

2uhu %ahan masuk = 1*0 = #4+!14 K

2uhu liRuid keluar re%oiler = 14 = #&*!14 K

2uhu air pendin"in masuk = * = +01!14 K



Komponen8 %oil

,eed: , (estilat: (


Etil%enzena 1+&!1 0!# 0!0#& 0!# 0!041 0!0

(EB 1*+!4 !*44 0!47 !0&7 0!4+ 0!

8EB 17!& 1+!+ 0!+71# 1+!*#& 0!+744 0!12ul3ur .omp ###!& 0!#4* 0!011+ 0!0000 0!0000 0!#

Jumlah +7!&004 1 +&!*1 1 0!7




8rial T = 404!4++&

Dmpan masuk kolom destilasi u$e

Komposisi Ki H

Etil%enzena 0!0#& 1!400 *!7*+& 50!0001

(EB 0!47 0!&+& !*0** 50!0007

8EB 0!+71# 0!0 1!0000 50!0001

2ul3ur .omp 0!011+ 0!4&& !4&47 0!0000Jumlah 1 50!000

(estilat keluar kolom destilasi de?

Komposisi Ki H

Etil%enzena 0!041 #!14 !71 0!0000

(EB 0!4+ 1!4#0 1!0000 50!000

8EB 0!+744 0!&0* 0!#0+1 50!0001

Jumlah 1 50!000+

= 50!000+ 0!000+&*

= 1 000+

o.

o.

o.o.

o.

x,

x(

x,

x(

Qm 1

Q

=

i

i,i xR1

=

i

i,ix

R1

=+ (i

m

x1Q

=+ (im

x1Q

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= 0!000


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=

+0+&47!4+* = #+7*04#!4&+ 1+1!41#4

= 1+&0*+!44+0



1+ =

#0*0*!4**# = 1#!+7* #+7*04#!4&+

= 47*1+! kCal/jam


= 1+&0*+!44+0

= 1+&0*+!44+0 47*1+!

= 1***&4+!00 kCal/jam


= m x .p x 8

47*1+! = m x #!1*1& kJ/k" K x +&+!14 5 +01!14

m = 0+4!*4+1 k"/jam


Kesetim%an"an panas pada re%oiler-=

#+!#0# = 1+1!41# 5&4#1!7*04

= 5#47*71!&&&

(imana-

= 19 ? 17 14 @

= 19 ? 1+1!41# 5&4#1!7*04 @

= 540!&&00#+ kCal/jam

2ehin""a-

= 5#47*71!&&&

= 5#47*71!&&& 540!7

= 5#474+7+!& kCal/jam



P = 1444 kPa

= 1+*!& kJ/k"

= #&+!+##* kCal/k"


11 Q

1&

17 .

Q

.

Q

.

1 1&

.

.

.

Q

.

Q

Q

.

1# Q

17 14 Loss

Q

Loss

Q

Loss

Loss

Q Loss

o.

QP

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+0+!14 K

+0*!14 K

#7+!14 K

#&+!14 K

'liran dari $

8 kmol/jam kCal/jam

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+4 1*!&&1 #1+*&*!700+

+4 4!407+ 770#!1#*+

+4 0!4**+ 1#*!#+

+4 0!+*7 **!&

1+4!#### #+*0!4+4&

'liran dari $+

kmol/jam

00 0!4**+ #*41!11*4+774

00 0!+*7 1*0!&1#04&*+#

00 *+!#1 *&4&&!*##+#+1

00 0!4& #*1#7!7&0*1

00 1!4&7* 1&&*4##!+*470&117!0#04 1*0+!70#

8+

+kCal/jam

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!0+7

11* kCal/jam

+1+ kJ/jam

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Bottom: B

/jam

4 0!000

00 0!0000

470 0!&074

&7* 0!+*07

7 0!011&1+# 1

xB

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K

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K

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APPENDIX C

SPESI;IKASI A*AT

1. Tan)ki Peni%panan 'enena (; 11"

Direncanakan

%ahan konstruksi = .2 2' #0 "rade $ tpe +1&

, allo

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Benzena 1007#!4&*7 0!*+* **!

'ir !++4147+7 0!007 1000

8oluena 4#!7&+&&4+ 0!004+ *70

2ul3ur .oump! 1!*****+ 0!001 1*10

10#0!*+*7+4*17 1

kon;ersi

1k" = !0#& l%

1m = +!*0* 3t

= +!+7 in

13t = 1!0001114 in

= +4!+1#&&&

= &!* %%l

=

= **+!&*7&&&

44!1&&*0+1&&

sehin""a dapat dihitun" ;olume tan"ki

;olume liRuid = mliRuid

= +*+#0!1*0+*#

**+!&*7&&&

= #+&!#&1*700&

= 147*7!4440*01 4+!#&00#47+#

= 741!40&7##& %%l

o;er desi"n = 09 peter and timmerhaus:11 hal +7

Jo$u%e tan)ki = 100/*0 x Uolume liRuid

= 1!4 x 741!40&7##&

= +#0!+*+#*074 %%l

= 4#0*!077++774*&

= 10*#!###**411+

dari appendix E %ro

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=

= 0!4

maka

te%al tutup = 0!7## in

spesi#ikasi stora)e enena

Iama = 2tora"e Benzena

,un"si = untuk menmpan %ahan %aku %enzena

8pe = silinder te"ak lurus den"an tutup atas ConiCal dan tutup %a

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X Campuran = 0!&0& l%/3t!h

Y Campuran = 44!1&&*

Ko%posisi Eti$ena k)7+a% ?i

Benzena 1007#!4&*7 0!*+* **!

'ir !++4147+7 0!007 1000

8oluena 4#!7&+&&4+ 0!004+ *70

2ul3ur .oump! 1!*****+ 0!001 1*10

10#0!*+*7+4*17 1

=

= **+!&*7&&&

= 44!1&&*

=

= 0!&0&

Men)&itun) Rate 4o$u%etrik

Rate #eed %asuk = 141*!+0#7#+4# k"/jam =

= Qate %ahan masuk =

Y %ahan masuk

= x

= 1!74014+& x 1!&*##*++0477

= !#7774*#* in

= 0!#4# ft

2tandarisasi (i = !#& in sCh #0 Seankoplis: 'pp '!4 hal

2ehin""a diperoleh -

6( = !*74 in

= 0!+4*+++++ 3t

)( = !#& in

= 0!0474 3t

= &!71 mm

l%/3t+

(k)7%"

ca%puran

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Panjan" pipa total L =

= 4 \

= 1&!#+#* 3t

= 1417!17 in

E. Menentukan #riksion *oss

1! ,riksi pada pipa lurus

,3 = #3 ]L

(

= # x

= +!+7# l%3!3t/l%m

! Kontraksi =KC = 0!44 x

= 0!44 x

= 0!44

hC = KC

= 0!00 l%3!3t/l%m

+! Ekspansi= 1 5

= 1 5

= 1

hex =

= 1

= 1!*&+ l%3!3t/l%m

K3 = 0!74 8a%el !1051 [Seankoplis

h3 = 1K3

= 1 Z

Pipa lurus el%o< 0o Sate ;al;e "lo%e

;

"C

Kex

Kex

#! El%o< 0o: 1 %uah

;

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3. Coo$er

;un)si , $enurunkan 8emperatur dari 8 = 1#4 5 1+4

8pe - 2hell and 8u%e

Bahan -

Direncanakan

5 3aktor kekotoran "a%un"an minimum Qd =

5 penurunan tekanan aliran maksimum ]p =

5 ]p maksimum steam =

5 di"unakan pipa ukuran -

6( =

BWS =

L =

P8 =

susunan se"iti"a trian"ular

Kondisi operasi ,

5 massa %ahan masuk W = 4+&!4

= 11*+7!7#07

5 = 10

5 = 1#4

5 ke%utuhan steam m = *00!*4

= 17&4!445 panas an" di%a

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1! $en"hitun" IQe

=

n x Pt x 1##

= * 0!+

1 1!4

= 0!017*

= $

= 11*+7!7#07

0!017*

= &&4*7!1#0+11#*

X = 0!100 CP

= Ss x de

X x :#

= &&4*7!1#0+114

0!100

= 1+7477!04

! J = 740

+! $en"hitun" har"a koe3isien 3ilm

.P = 0!4 Btu/l%! ,

k = 0!1

= 0!1#01ho/s = +&7!*#&&71#*

t< = +0&!0717004 ,

X< = 0!01 CP

s = 1!&77+77

ho = #&&!1&&

D. Mencari ta&anan panas pipa ersi&

= ho x hio

ho hio

= #&&!1&& x 1400

#&&!1&& 1400

= +4*! Btu/jam!3t! ,

E. Mencari dirt #actor (#aktor kekotoran" pipa terpakai

Qd = 5

x

= +4*! 5 &0#!&*

Ba"ian Shell asam sul3at

as

)(2x . x B

Ss

as

l%/jam!3t

Ires

Btu/jam!3t! ,/3t

k .p!X/k1/+

Btu/jam!3t ,

D.

D.

D(

D.

D(

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1#00

=

. Menentukan di%ensi stora)e

( = !4 Bro

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= (

#+0 x t

=

= 0!4

maka

te%al tutup = 0!7## in

spesi#ikasi stora)e enena

Iama = 2tora"e Etil Benzena

,un"si = untuk menimpan produk Etil %enzena

8pe = silinder te"ak lurus den"an tutup atas ConiCal dan tutup %a

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Direncanakan


, allo

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= &+!4+0&1#7

+*!471

sehin""a dapat dihitun" ;olume tan"ki

;olume liRuid = mliRuid

= +70*!&4+&47+#

&+!4+0&1#7

= 4*7!7&&*&11+14

= 1*&7+4!704*&7* 47!147*+*7&*

= +++07!11#&*14*1# %%l

o;er desi"n = 09 peter and timmerhouse:11 hal +7

Jo$u%e tan)ki = 100/*0 x Uolume liRuid= 1!4 x +++07!11#&*14*1#

= #1&++!*++417&* %%l

= &&0!70*47##

= ++#1!&407++#11

dari appendix E %ro

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jumlah = 1 %uah

8. Stora)e &ea4 (&asi$ reaksi sa%pin)" (; 1="

;un)si , $enampun" komponen hea; hasil reaksi sampin"

Direncanakan


, allo

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&&!+1#+ 1

kon;ersi

1k" = !0#& l%

1m = +!*0* 3t

= +!+7 in

13t = 1!0001114 in

= +4!+1#&&&

= &!* %%l

=

= 4&*!*4&*110#

+4!40

sehin""a dapat dihitun" ;olume tan"ki;olume liRuid = mliRuid

= 4+!7&0471714&

4&*!*4&*110#

= +!111*#

= 1+*1!01+#&* 11!1+&4&11

= #&!+47#4**1 %%l

o;er desi"n = 09 peter and timmerhouse:11 hal +7

Jo$u%e tan)ki = 100/*0 x Uolume liRuid

= 1!4 x #&!+47#4**1

= +07!#7#*+41 %%l

= #*!**0*0+4

= 17&!4014#+711

dari appendix E %ro

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0!*4

1& = 0!0&4

jamx

1# hari

hari

?i7

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0!0011141&&&

0!000007

0!00000&0

0!000000&&+

0!0011

3t

1#00 %%l

0!14*414+017

1 tan"ki

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x

in

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datar

0!00440+1*

!0#177&7*E5004

0!00000+071

0!000001#4#

0!00&

?i7

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4!07++1740* x 55.1668&0&

aliran laminar

4 0!0474

0!0474

1 0!0474

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hal! +

hal! +

hex h3

1!*&+ 14!&

karena 3luida diam dalam tan"ki penampun"an

] " ]P

"C Y

0 #7!+&

44!1&&*

l%3!3t/l%m

erhause[: hal!40

c,

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+0 , = 7 ,

*& , = 07 ,

= *&!00 5 +0!00

+!00 5 +7#!00

= !&&&7

= +!00 5 +7#!00

*&!00 5 +7#!00

= 0!+

5 #

,

,

,

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*#&*+!71 = +7*4!#

17!*+#4

+7*4!#* = 7!& %uah

x 0

= &0#!&*

3t

3t/3t

Btu/jam 3t

!,

nuhi

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E4a$uasi Perpinda&an Panas (Rd"

1`! $en"hitun" Ire pipa

= It x a`

n x 1##

1!& = 1

1## 1

= 0!0&

= $

= 17&4!444

0!0&

= 47#!*

X = 0!01*

= St x di

X x :#

0!0& = 47#!

!# 0!01*

= +7*11!7701&&

3i"!* [Kern[: hal!*+* `! J = 5

+`! $en"itun" har"a koe3isien 3ilm

io = 1400

at

St

at

Iret

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0!*41& = 0!0&4

jamx

1# hari

hari

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7!0&170E500&

0!0011+741#

+!4&4&174E500&

0!0011#*

3t

1#00 %%l

0!1+*&&1*47

?i7

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datar

0!*4

1& = 0!0&4

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jamx

1# hari

hari

!11&4*1#E5004

0!000+*7*

0!000#7##7+1

0!0011#0

3t

?i7

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in

datar

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0!*4

1& = 0!0&4

jamx

1# hari

hari

0!0007*0&11&

0!000&*74#4&

4!*0####*&E5004

#!#40&&+0*E500&

&!14*+477E5004

!#0010#7E500&

0!001&0#

?i7

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3t

1#00 %%l

0!1+7+&14#

1 tan"ki

x

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in

datar

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0!*4

1& = 0!0&4

jamx

1# hari

hari

0!0011&+#**

0!000+*7*&1

0!0000&1

?i7

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0!00174*

3t

1#00 %%l

0!001#+*01

1 tan"ki

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x

in

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datar

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+!17#

= 1

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h3

1!*&+

= 5 Ws

= 5 Ws

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0!#70

1##

CP

0!0&4

!#

a"ian Tube steam

l%/jam!3t

Btu/j!3t!,

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FEED kolom distilasi

Komponen B$

$+ $+

k"/jam k"mol/jam

8oluena !1#**1 110!& &0!+0707+ 0!&4##4+04

Etil%enzena 10&!17& 1+&!1 *0!#*70+ !#1177++7

(EB 1+#!++0* 1*+!4 +01!&&&**++ !410&1*74&1

8EB 1&!*& 17!& &&!4&&4+ 1+!&*

sul3ur Coump! +!07 ###!& 1+!&4##+1*+1# 0!#47&&+7

868'L 141*!+0#7#+4# 1+0!0#&&++7

Dntuk menentukan K dan LK ditentukan dari %e%an distilasi terendah

(iperoleh %e%an destilasi terendah = ( = !11+70017

d8

2ehin""a ditentukan Etil %enzena se%a"ai Li"ht ke dan (EB se%a"ai ea; ke

'sumsi- Etil%enzena li"ht ke an" keluar se%a"ai %ottom se%esar

(ietil%enzena hea; ke an" keluar se%a"ai destilat se%esar

NERACA MASSA OVERALL

F = D + B

F . = D . + B .

Ko%ponen;eed2 ; (M" Desti$at (M:"

k%o$7+a% k%o$7+a%

8oluena 0!&4#4 0!0040 0!&4#4 0!0071

Etil%enzena LK !#1 0!711 1!4&&+ 0!04

(EB K !410& 0!17+1 0!41 0!00#

8EB 1+!+ 0!107# 0!0000 0!0000

sul3ur Coump! 0!#4* 0!00++ 0!0000 0!0000

-.33:9 1.!!!

Menentukan su&u 'u$e Point dan De? Point

Persa%aan Antoine

8 didih o.

F

D

B

@;

@'

lo"10

P = ' B / 8 .

P = ;apor pressure %ar

8 = temperature K

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(ata 'ntoine

Ko%ponen A ' C

8oluena #!4### 17+*!1+0 0!+#0

Etil%enzena #!#04# 1&4!0&0 5+!&*0

(EB #!1+0 147&!&10 57!1#&0

8EB #!70+& 1++!44*0 5+7!#&02ul3ur #!40& 1+7!+4#0 57!4+0

2eader: pa"e 1#

1. ;eed ko$o% desti$asi

$enentukan suhu %u%le point pada 3eed kolom destilasi

P = !+ %ar = 17#* mm"

8 = 1*0!++07 = #4+!#*07 K

Ko%ponen Ki iBKi

8oluena 0!0040 4!1*&1 !4#* 0!011+ 0!0

Etil%enzena 0!711 !*+& 1!4*1 0!*#* 0!7

(EB 0!17+1 0!#0 0!#+ 0!07#* 0!1

8EB 0!107# 0!+74* 0!1&+# 0!0174 0!1

2ul3ur .oump 0!00++ 0!* 0!+1 0!001+ 0!0

Jumlah 1.!!! -.!8! 1.!!! !.>

G = 0!0000

$enentukan suhu de< point pada 3eed kolom destilasi

P = !+ %ar = 17#* mm"

o.

X; i Psat

11 +=

i

i

i

K

zx

101 +=

i

iii

K

Kzy

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8 = 01!00*& = #7#!14*&07#7 K

Ko%ponen X; i Ki i7Ki

8oluena 0!0040 7!&14& +!+111 0!0014 0!0

Etil%enzena 0!711 #!+0 1!0** 0!+7& 0!+

(EB 0!17+1 1!&1* 0!7041 0!#44 0!

8EB 0!107# 0!&47# 0!*4* 0!+747 0!+

2ul3ur 0!00++ 1!#&1# 0!&+4# 0!004 0!0 1.!!! !.>

G = 1!0000

-. Disti$at ko$o% desti$asi

$enentukan suhu %u%le point pada ditilat kolom destilasi

P = %ar = 140 mm"

8 = 1&+!+#+& = #+&!#+& K

Ko%ponen Ki iBKi8oluena 0!0071 +!&*0* 1!*#0# 0!01+0 0!0

Etil%enzena 0!04 1!1# 0!47 0!*& 0!

(EB 0!00# 0!&+7* 0!+1* 0!000* 0!0

8EB 0!0000 0!7+ 0!11+7 0!0000 0!0

2ul3ur .oump 0!0000 0!4*# 0!1 0!0000 0!0

Jumlah 1.!!! 1.3-3 1.!!! !.>

G = 0!0000

$enentukan suhu de< point pada distilat kolom destilasi

P = %ar = 140 mm"

8 = 1&+!&4# = #+&!*0#*7&1**4 K

Ko%ponen Ki i7Ki

8oluena 0!0071 +!70# 1!*4# 0!00+* 0!0

Etil%enzena 0!04 !004& 1!00* 0!*77 0!

(EB 0!00# 0!+ 0!+1& 0!007& 0!0

8EB 0!0000 0!4 0!11#* 0!0000 0!0

2ul3ur 0!0000 0!4*7+ 0!+& 0!0000 0!0

>> !.>

G = 1!0000

. 'otto% ko$o% desti$asi

$enentukan suhu %u%le point pada %ottom kolom destilasi

P = !4 %ar = 100 mm"

8 = +0!&+ = 40+!77+ K

Ko%ponen Ki iBKi

o.

Psat

o.

XD i Psat

o.

XD i Psat

o.

X' i Psat

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8oluena 0!0000 1!## #!* 0!0000 0!0

Etil%enzena 0!0#& 7!#+0 !7 0!07+7 0!0

(EB 0!47 +!01+# 1!04# 0!71## 0!4

8EB 0!+71# 1!+#++ 0!4+7+ 0!1& 0!+

2ul3ur .oump 0!011+ !700 1!0**0 0!01+ 0!0

Jumlah 1.!!! :.313 1.!!! 1.!

G = 0!0000

$enentukan suhu de< point pada %ottom kolom destilasi

P = !4 %ar = 100 mm"

8 = +!#++* = 41!4*+77+&7* K

Ko%ponen X; i Ki i7Ki

8oluena 0!0000 1#!++* 4!74 0!0000 0!0

Etil%enzena 0!0#& *!&7## +!#&* 0!0071 0!0

(EB 0!47 +!4&40 1!#&0 0!#14& 0!#

8EB 0!+71# 1!&++0 0!&4+ 0!4&*& 0!42ul3ur 0!011+ +!#1 1!*& 0!00** 0!0

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= 1!09

= 1!09

'otto% (M="

k%o$7+a%

0!0000 0!0000

0!# 0!0#&

!*44 0!47

1+!+ 0!+71#

0!#4* 0!011+

8.=!!: 1.!!!

@'

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i Fi

40 0!011+

11 0!*#*

+1 0!07#*

7# 0!0174

++ 0!001+

>: !.>99

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i Fi

14 0!0040

& 0!711

44 0!17+1

47 0!107#

4 0!00++>> !.>>:

,- Disti,t /

i Fi71 0!01+0 = = 0.52

04 0!*& 0.318

# 0!000*

00 0!0000 = 3.144368

00 0!0000

> !.>98

,- Disti,t /

i Fi

+* 0!0071 = = 1.0048

77 0!04 0.3416

7& 0!00#

00 0!0000 = 3.118108

00 0!0000

>: !.>>

,- Disti,t /

i Fi

D

$#$

$*$

D

D

$#$

$*$

D

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00 0!0000 = = 4.24

#& 0!07+7 1.405

7 0!71##

1# 0!1& = 4.865354

11+ 0!01+

!! 1.!!!

,- Disti,t /

i Fi

00 0!0000 = = 3.68

71 0!0#& 1.460

14& 0!47

*& 0!+71# = 4.334116** 0!011+

!! 1.!!!

$p&a rata ("N% C dist C ott

1!7#+ #!747

1!0000 1!0000

0!+&4* 0!0&04

0!141+ 0!004

0!+++ 0!0#&+

1.93> 1.11-

3->3> = 3

D

$#$

$*$

D

D

$#$

$*$

D

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'PP B51*4

APPENDIX '

PER0ITNAN NERACA PANAS

Kapasitas produksi Ethlene SlCol = 10101 k"/jam

Basis Perhitun"an = 1 jam operasi

2uhu re3erensi = 4 = *!14 K

1. Pre&eater E11:

Keteran"an-

- Panas %ahan masuk preheater

- Panas %ahan keluar preheater menuju reaktor

- Panas produk keluar re%oiler

- Panas produk keluar preheater menuju stora"e

- Panas an" hilan"

(en"an meman3aatkan panas dari produk keluar re%oiler E51#&: maka direnCanakan-

2uhu pemanas masuk = 10 = #&+!10 K

2uhu pemanas keluar = +4 = +0*!14 K

2uhu %ahan masuk = +0 = +0+!14 K

2uhu produk keluar = #& = +1!#& K

a.

Komponena % C d

#*!1* 0!107+ 5&!&117E500* 51!**+#E500* 1!

1+!0& 0!#*+&7 50!000770& &!0*&E500* 51!

o.

;un)si, Dntuk peman3aatan panas dari produk keluar re%oiler den"an %ahan masuk rea

1

+1

Loss

o.

o.

o.

o.

$en"hitun" panas an" terkandun" pada produk keluar re%oiler dan produkkeluar preheater menuju stora"e

+1

CP: J/mol K

.

&6

.#

106

+

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'PP B51*&

!101 0!7*&4 50!0004&0& !004E5007 5+!

Komponen kmol/jam#&+!10 +0*!14 &!7*#1 +0!0#7

#&+!10 +0*!14 1&!##& 10#40!1+#1

#&+!10 +0*!14 1!0+4 17*0!070*

Jumlah 14+&!7

.

Komponena % C d

!#*& 0!++74 50!00000& !1+0*E5007 5#!++!++ 50!00*#1*& 0!00000& 51!7*4E500* +!

Komponen kmol/jam

+0+!14 +1!#4&* 0!*0+ 1++!+1#

+0+!14 +1!#4&* 100!7*+4 #+*&4!#0+*

Jumlah 3>>9.81=-

c.

= 19 x

= 19 x 1*7774!*7#

= 1*77!74*7 kCal/jam

d. Kesetim%an"an panas o;erall

=

Neraca Panas 0eater



#+*!71& 1*7774!*7#14+&!7 77#!+1

1*77!74*7

8otal 1>8=-8.>3:> 8otal 1>8=-8.>3:>

.&

1#6

#

M(ikutip dari Va

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'PP B51*7

-. Pre&eater E11=

Keteran"an-


- Panas %ahan keluar preheater menuju reaktor- Panas an" hilan"

- Panas an" terkandun" pada steam

(irenCanakan-

2uhu %ahan masuk = #& = +1!#& K

2uhu produk keluar = 40 = ++!14 K

a.

Komponen

a % C d!#*& 0!++74 50!00000& !1+0*E5007 5#!

++!++ 50!00*#1*& 0!00000& 51!7*4E500* +!

Komponen kmol/jam

+1!#4&* ++!14 0!*0+ 4*+!#117

+1!#4&* ++!14 100!7*+4 1*71!#&+

Jumlah 19888:.983>

.

= 19 x

= 19 x 0+&4!40

= 0+!&4& kCal/jam

IeraCa panas total-

=

;un)si, Dntuk pemanasan %ahan se%elum masuk reaktor

+

Loss

2team

o.

o.

$en"hitun" panas an" terkandun" pada %ahan masuk dan %ahan keluar

CP: J/mol K

26

#

6

M(ikutip dari Va

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'PP B51**

1*7774!*7# = 0+&4!40 0+!&

= +#7+!0+&7 kCal/jam

c. $en"hitun" ke%utuhan steam


P = 1444 kPa

= 1+*!& kJ/k"

= #&+!+##* kCal/k"


= +#7+!0+&7

#&+!+##*

= 74!010 k"/jam

Neraca Panas 0eater



1*7774!*7# 0+&4!40

+#7+!0+&7 0+!&440#*+

8otal ---:=9.>11= 8otal ---:=9.>11=

2team

2team

o.

+

2team

Loss

2teamP

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'PP B51*

. Reaktor R11!

Keteran"an-

- Panas %ahan masuk reaktor dari preheater

- Panas "as masuk reaktor dari stora"e

- Panas air masuk dari utilitas

- Panas "as keluar reaktor

- Panas produk keluar reaktor

- Panas reaksi

- Panas an" hilan"- Panas an" diserap air pendin"in

(irenCanakan-

2uhu %ahan masuk = 40 = ++!14 K

2uhu "as masuk = + = +04!14 K

2uhu air masuk = 7 = +00!14 K

2uhu "as keluar = #* = +1!14 K

2uhu produk keluar = 40 = ++!14 K

2uhu air pendin"in masuk = 7 = +00!14 K

2uhu air pendin"in keluar = 40 = ++!14 K

a. $en"hitun" panas dari masin"5masin" %ahan an" masuk reaktor

= *!14 K

'liran dari eater 'liran dari 2tora"e 'liran

;un)si- Dntuk mereaksikan Etilen oksida dan 6

+

#

4

&

7

rxn

Loss

Pendin"in

o.

o.

o.

o.

o.

o.

o.

8re3

[ ]sensi%el3inin //n/ +=

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'PP B510

kmol/jam kmol/jam kmol/ja

0!*0+ ++!14 5 5

100!7*+4 ++!14 5 **!+115 5 1+!*74 +04!14 5

5 5 !1&4 +04!14 5

Jumlah 101!7&+* 1&!0+*# **!+11


a % C d

!#*& 0!++74 50!00000& !1+0*E5007 5#!

++!++ 50!00*#1*& 0!00000& 51!7*4E500* +!

+0!*7 50!007&0#1 0!000++#7 5+!7#7E5007 !

7!#+7 0!0#+14 50!00001444 0!00000000# 5!

KomponenPanas sensi%el Panas Qeaktan $asuk Qeak

1&!+++ 7!#07&

4+!04& 1&7&!7#4

U'LDE

&!70+ 1+!+7*

Jumlah17#!14+4 U'LDE

JA*E

. $en"hitun" panas dari masin"5masin" %ahan an" keluar reaktor

Qeaksi ) = 171!

Qeaksi )) = 17!1

Qeaksi ))) = 1!

8+

8#

26

#f

6

.

#6

.6O

fKatalisN O)mpurities

26

#

6

.#6.6

M(ikutip dari Va

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'PP B511

KomponenQeaksi ) Qeaksi )) Qeaksi ))) $a

51 171!+1 5 17!1+ 5+ 1!0 213.

51 171!+1 51 17!1+ 51 1.90 1*!

1 171!+1

1 17!1+

1 1!0

!1

0!

= +1!14 K

= ++!14 K

= *!14 K


a % C d

+0!*7 50!007&0#1 0!000++#7 5+!7#7E5007 !++!++ 50!00*#1*& 0!00000& 51!7*4E500* +!

#*!1* 0!107+ 5&!&117E500* 51!**+#E500* 1!

1+!0& 0!#*+&7 50!000770& &!0*&E500* 51!

!101 0!7*&4 50!0004&0& !004E5007 5+!

7!#+7 0!0#+14 50!00001444 0!00000000# 5!

!#*& 0!++74 50!00000& !1+0*E5007 5#!

Komponen

Panas 2ensi%el Panas Produk Keluar Qeaktor

J/mol

11&4!#0& 70!7#04

*#!&144 +&44!70&+

&74!#+# 104#*!0**

+#*1!#04 1#4#!#4

#4&!#&* 4#!&7#

i

1 i 2 i 3 n

.

#6

6

.

&6

.#

106

+

.&

1#6

#

.6O

26

#f

fKatalisN O)mpurities

8&

87

8re3

.#6

6

.

&6

.#

106

+

.&

1#6

#

.6

26

#

M(ikutip dari Va

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'PP B51+

= #01*+!7&& 5 U'LDE H 5+0#

= U'LDE kCal/jam

IeraCa panas total-

0 = 5 =

=

= #01*+!7&& 5 U'LDE U'

= U'LDE kCal/jam

= 19 x

= 19 x U'LDE

= U'LDE kCal/jam

U'U'

= m x .p x 8

U'LDE = m x #!1*1& kJ/k" K x ++!14 5 +

m = U'LDE k"/jam

Neraca Panas Reaktor



17#!14+4 11*1!7#40

U'LDE #*00!+1&

1###&!7*7 U'LDE

U'LDE

U'LDE

8otal JA*E 8otal JA*E

keluar

5 masuk

rxn

Loss

Jadi panas an" harus diserap air pendin"in pendin"in adalah

Pendin"in

+ &

#

7

4

Loss

Pendin"in

rxn

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'PP B51#

3. Pre&eater E1-3

Keteran"an-


- Panas %ahan keluar preheater menuju e;aporator

- Panas an" hilan"


(irenCanakan-

2uhu %ahan masuk = +4 = +0*!14 K

2uhu produk keluar = 100 = +7+!14 K

a.

Komponena % C d

++!++ 50!00*#1*& 0!00000& 51!7*4E500* +!

#*!1* 0!107+ 5&!&117E500* 51!**+#E500* 1!

1+!0& 0!#*+&7 50!000770& &!0*&E500* 51!

!101 0!7*&4 50!0004&0& !004E5007 5+!

Komponen kmol/jam

+0*!14 +7+!14 17*!7#4# 1##77!40

+0*!14 +7+!14 171!+1#& #+#4!4&77

;un)si, Dntuk pemanasan a

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'PP B514

+0*!14 +7+!14 17!1+14 4&00!7+0

+0*!14 +7+!14 1!0+4 **4!0*&1

Jumlah 1##4*!*+7

.

= 19 x

= 19 x 1#*700!04

= 1#*70!00& kCal/jam

IeraCa panas total-

=

1##4*!*+7 = 1#*700!04 1#*

= 1+07#1+!0&* kCal/jam



P = 1444 kPa

= 1+*!& kJ/k"

= #&+!+##* kCal/k"


= 1+07#1+!0&*

#&+!+##*= *1!&*41 k"/jam

Neraca Panas 0eater



1##4*!*+7 1#*700!04

1+07#1+!0&* 1#*70!00&

8otal 1:!198-.!9!: 8otal 1:!198-.!9!:

.#

106

+

.&

1#6

#

$en"hitun" panas an" hilan" Loss

Loss

*

2team

Loss

2team

2team

o.

*

2team

Loss

2teamP

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'PP B51&

:. Trip$e E##ect E4aporator J1-!

Keteran"an-


- Panas uap keluar e33eCt 1

- Panas liRuid keluar e33eCt 1

- Panas uap keluar e33eCt ))

- Panas liRuid keluar e33eCt ))

- Panas uap keluar e33eCt )))

- Panas liRuid keluar e33eCt )))

- Panas an" hilan"

- Panas an" terkandun" pada steam(irenCanakan

2uhu %ahan masuk = 100 = +7+!14 K

2aturated steam an" di"unakan = 00 = #7+!14 K

8ekanan pada saturated steam = #7& kPa

8ekanan pada e;aporator e33eCt ))) = +*!44 kPa

;un)si- Dntuk memisahkan kandun"an air dalam produk

10

11

1

1+

1#

14

Loss

2team

o.

o.

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'PP B517

2uhu pada e;aporator e33eCt ))) = &0 = +++!14 K

Komponen B$ kmol/jam k"/jam

&!0&* 171!+1#& 10&++!1*1

10&!11 17!1+14 1*1*!0111

140!17# 1!0+4 *4!*44

1*!014 10#!*&&4 17#!+1

Jumlah -3=1.->:

(en"an men"asumsikan uap an" ter%entuk tiap e33eCt adalah sama: dari perhitun"an ne

massa diperoleh-

, = +#&1!4+ k"/jam

= 1#&4!4007#1 k"/jam

= = = #744!1&& k"jam

= 770&!1*# k"jam

= 40!&14 k"jam

= 1*14!7#& k"jam

sehin""a didapatkan-

= 0!+#

= 0!#47

= 0!444

= 0!7a. $en"hitun" kenaikan titik didih

Qomero: 00*

i = 1 Uan o33 ,aCtor

m = mol solute

k" sole;ent

K% = 0!41

2ehin""a:

=

= 1 x 0!41 x 10+#!4&0# = #!#17#!+1

= 1 x 0!41 x 10+#!4&0# = &!41

1#&!07

= 1 x 0!41 x 10+#!4&0# = !4#

101+!04+

g8 = 5 5

o.

'liran %ahan masuk e;aporator dari tan"ki penukar ion $

.

&6

.#

106

+

.&

1#6

#

6

U1 U

U

+

U1 U

U

+

L1

L

L+

F,

FL1

FL

FL+

o. k"6/mol

BPQ1

o

.

BPQ

o.

BPQ+

o.

8s1

8+

BPQ1 BPQ

BPQ

+

mK%i8% =

mK%i

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'PP B51*

= 00 5 &0 5 0!

= 11!01

2ehin""a asumsi ditetapkan-

= 7**

= +4&

= 170

=

= 11!01 1/7** = +1!*1

1/7** 1/+4& 1/170

= +7!&4

= #!44

g8 = = 11!01

$en"hitun" titik didih aktual

1 = 5

= 00 5 +1!*1 = 1&*!1

= 00 8 kondensat saturated steam pada e33eCt )

= 5 5

= 1&*!1 5 #!# 5 +7!&4 = 14!&0

= 5

= 1&*!1 5 #!#11 = 1&+!4

+ = 5 5

= 14!&0 5 &!41 5 #!44 = &!4#

= 5

= 14!&0 5 &!41 = 11!0

2ehin""a suhu dalam triple e33eCt se%a"ai %erikut-

E33eCt ) E33eCt )) E33eCt )))

00 1&+!4 11!0

1&*!1 14!&0 &!4#

$en"hitun" har"a entalpi pada masin"5masin" e33eCt untuk air

o.

Koe3isien heat trans3er D untuk short tu%e ;ertikal adalah 1105*00 W/mK

D1 W/m!K

D W/m!K

D+ W/m!K

81

8

o.

8+

o.

81

8

8+

81

8s1

81

o.

8s1

o.

8

81

BPQ1

8

o.

8s

81

BPQ1

o.

8+

8

BPQ

8+

o.

8s+

8

BPQ

o.

8s1

o. 8s o. 8s+ o.

81

o. 8 o. 8+ o.

++

+1

1

D/1D/1D/1

D/18

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'PP B51

E33eCt )

= 00 = 1&+!4

= 1&*!1 = #!#

= x

= 7&0!14 1!**# x #!#11

= 7&!414 kJ/k"

= 5

= 70! 5 *4!#

= 1+*!4000 kJ/k"

E33eCt ))

= 1&+!4 = 11!0

= 14!&0 = &!4107

= x

= 70#!&*04 1!**# x &!4107

= 71&!#&& kJ/k"

= entalpi uap E33eCt ) 5

= 7&!414 5 &*!&+*

= 07!**#0 kJ/k"

E33eCt )))

= 11!0 = &0!00

= &!4# = !4#1*

= x

= &0!7000 1!**# x !4#1*

= &7!&7&7 kJ/k"

= entalpi uap E33eCt )) 5

= 71&!#&& 5 #!*++

= 17!11#1 kJ/k"

Kapasitas panas liRuid-

, - .p = 1!**&

- .p = 1!****

- .p = 1!**74

- .p = 1!**4

'liran an" di"unakan untuk men"hitun" air an" teruapkan

= , 5 = +#&1!4+0*1 5

8s1

o. 8s o.

81

o. BPQ1 o.

1

ssatpada 8s

.p steam pada 100 o. BPQ1

s1

s1

entalpi uap pada 8s1 hs

1entalpi liRuid pada 8s

1

8s

o. 8s+ o.

8

o. BPQ o.

s+satpada 8s

+ .p steam pada 100 o. BPQ

s

1

hs

entalpi liRuid pada 8s

8s+

o. 8s# o.

8+

o. BPQ+ o.

+

s#satpada 8s

# .p steam pada 100 o. BPQ+

s+

hs+

entalpi liRuid pada 8s+

L1

L

L+

U1

L1

L1

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'PP B501

Effect I

= 70! kJ/k" = 7&!414 kJ/k"

= *4!# kJ/k" = 1+*!4000 kJ/k"

=

'liran panas masuk -

=

= +#&1!4+ x 1!**& x 100

= 1#&&04#!107 kCal/jam

= 2 x

= 7*1+!77 x 70!

= 4100!7 kCal/jam

= 9 x

= +1!0*# kCal/jam

8otal panas masuk &&7*4#!#*71 kCal/jam

'liran panas keluar -

= x

= ##7+!&* x 7&!414

= &1++#!&7#4 kCal/jam

=

= 7*7!471 x 1!**** x 1&*!1

= 1400+!#00 kCal/jamKondensat = 2 x

= 7*1+!77 x *4!#

= 1411&!#14 kCal/jam

8otal panas keluar &&7*4#!#*71 kCal/jam

Effect II

= 7&!414 kJ/k" = 71&!#&& kJ/k"

= &*!&+74 kJ/k" = 07!**#0 kJ/k"

='liran panas masuk -

= x

= ##7+!&* x 7&!414

= &1++#!&7#4 kCal/jam

=

= 7*7!471 x 1!**** x 1&*!1

U

1

L

s1

, .p 8, 0 2

UL

1.p 8

1 0 U

1

1 2

L

, .p 8, 0

2team

U

Loss

10

U1

1

11

L1.p 8

1 0

L

U

L

s

L1

.p 81

0 U1

1

L

.p 8

0 U

U1

L

10

U1

1

11

L1.p 8

1 0

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'PP B50

= 1400+!#00 kCal/jam

= 9 x

= 4&!&+4 kCal/jam

8otal panas masuk 40*&++*!07#& kCal/jam


= x

= #&&1!77# x 71&!#&&

= +0741!++** kCal/jam

=

= ++4!* x 1!**74 x 14!&0

= 1+1&*4!0&* kCal/jam

Kondensat = 2 x= ##7+!&* x &*!&+74

= 7+7#00!*1 kCal/jam

8otal panas keluar 40*&++*!07#& kCal/jam

Effect III

= 71&!#&& kJ/k" = &7!&7&7 kJ/k"

= #!*+4 kJ/k" = 17!11#1 kJ/k"

=

'liran panas masuk -= x

= #&&1!77# x 71&!#&&

= +0741!++** kCal/jam

=

= ++4!* x 1!**74 x 14!&0

= 1+1&*4!0&* kCal/jam

= 9 x

= &04#4!0&* kCal/jam

8otal panas masuk #+#*+7!#4& kCal/jam


= x

= 41+0!077 x &7!&7&7

= +1*&7!71 kCal/jam

=

Loss

10

1

U

1+

L.p 8

0

L

U

+

L

s+

L.p 8

0 U

L

+.p 8

+ 0 U

+

+ U

L

1

U

1+

L.p 8

0

Loss

1

1#

U+

+

14

L+.p 8

+ 0

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'PP B50+

= 1*14!7#& x 1!**4 x &!4#

= 470140!&7#4 kCal/jam

Kondensat = 2 x

= #&&1!77# x #!*+4

= 44&1*!77 kCal/jam

8otal panas keluar #+#*+7!#4& kCal/jam

$en"hitun" 2team Ekonomi

2team Ekonomi = = 1#&4!4007 = 1!*4

2 7*1+!77

Neraca Panas Trip$e E##ect E4aporatorNeraca Panas E##ect I



1#+&7++!1&4 &1++#!&7#4

4100!7 1400+!#00

Kondensat 1411&!#14

+1!0*#

8otal ==89-:3.3981 8otal ==89-:3.3981

Neraca Panas E##ect II



0107!*10 +0741!++**

1400+!#00 1+1&*4!0&*

Kondensat 7+7#00!*1

4&!&+4

8otal :!9=9.!83= 8otal :!9=9.!83=

Neraca Panas E##ect III


Komponen Ener"i kCal/jam Komponen Ener"i kCal/jam&&70&!+10 +1*&7!71

1+1&*4!0&* 470140!&7#4

Kondensat 44&1*!77

&04#4!0&*

8otal 339>8.-3:= 8otal 339>8.-3:=

L

U1 U

U

+

10

2team

11

Loss

10

1

11

1+

Loss

1

1#

1+

14

Loss

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'PP B50#

=. Pre&eater E11

Keteran"an-

- Panas %ahan masuk preheater- Panas %ahan keluar preheater menuju kolom destilasi

- Panas an" hilan"


(irenCanakan-

2uhu %ahan masuk = &0 = +++!14 K

2uhu produk keluar = 11+ = +*&!4+# K

a.

Komponena % C d

++!++ 50!00*#1*& 0!00000& 51!7*4E500* +!

#*!1* 0!107+ 5&!&117E500* 51!**+#E500* 1!

1+!0& 0!#*+&7 50!000770& &!0*&E500* 51!


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'PP B504

!101 0!7*&4 50!0004&0& !004E5007 5+!

Komponen kmol/jam+++!14 +*&!4+# +0+!007+&** *4404!0&

+++!14 +*&!4+# 171!+1#&0#+77 14#70#!&+#

+++!14 +*&!4+# 17!1+1#&0#+77 010!440*

+++!14 +*&!4+# 1!0+#4&0# +14!**1&

Jumlah &+4&!7

.

= 19 x

= 19 x &***4!7#14

= &**!*47# kCal/jamIeraCa panas total-

=

&+4&!7 = &***4!7#14 &*

= #+147!+7&+ kCal/ja



P = 1444 kPa

= 1+*!& kJ/k"= #&+!+##* kCal/k"


= #+147!+7&+

#&+!+##*

= +1!#*1 k"/jam

Neraca Panas 0eater

'liran Panas $asuk 'liran Panas Keluar Komponen Ener"i kCal/jam Komponen Ener"i kCal/jam

&+4&!7 &***4!7#14

#+147!+7&+ &**!*47#14#+

8otal =>:1=9.:>>! 8otal =>:1=9.:>>!

.&

1#6

#

M(ikutip dari Va

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'PP B50&

8. Ko$o% Desti$asi D1!

Keteran"an-

- Panas %ahan masuk kolom destilasi

- Panas ;apor menuju kondensor

- Panas liRuid keluar kondensor an" re3luks- Panas liRuid keluar kondensor se%a"ai destilat

- Panas liRuid masuk re%oiler

- Panas ;apor keluar re%oiler

- Panas liRuid keluar re%oiler se%a"ai %ottom

- Panas an" hilan"

;un)si- Dntuk memisahkan kandun"an air dalam produk

1&

17

1*

1

0

1

Loss

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'PP B507


- Panas an" terkandun" pada air pendin"in

(irenCanakan-

2uhu %ahan masuk = 11+ = +*&!4+# K

2uhu liRuid keluar re%oiler = 1*7 = #&0!+*+*#& K


2uhu air pendin"in keluar = *0 = +4+!14 K


8emperatur pada 3eed-

Bu%le point = 11+!+*40 = +*&!4+40 K (e< point = 17*!00 = #41!##00 K

8emperatur pada destilat

Bu%le point = 100!14*+ = +7+!+0*+ K

(e< point = 10&!0&0# = +7!10# K

8emperatur pada %ottom

Bu%le point = 1*7!++* = #&0!+*+* K

(e< point = 0#!# = #7*!0 K


Komponen,eed: , (estilat: (


+0+!007# 0!&1# !77+ 0!#+ +!0

171!+1#& 0!+#7 1!71+1 0!0047 1&!

17!1+14 0!0+#7 5 5 17!1

1!0+4 0!00+ 5 5 1!

Jumlah #+!+4& 1 +01!&0# 1 11!




8rial T = 1!4###4

2team

Pendin"in

o.

o.

o.

o.

o

.o.

o.

o.

o.

o.

x,

x(

6

LK

.

&6

K

.#

106

+

.&

1#6

#

=i

i,i xR1

=+ (im

x1Q

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'PP B50*

Dmpan masuk kolom destilasi

Komposisi H

0!&1# +*!4#+& 0!&+*

0!+#7 1!0000 50!+

0!0+#7 0!1+4 50!00++

0!00+ *7!& 0!00+

Jumlah 1 0!0001

(estilat keluar kolom destilasi

Komposisi H

0!#+ #&!&&+0 1!0*+

0!0047 1!0000 50!0104

Jumlah 1 1!017*

= 1!017*

= 0!017*


= 0!07


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'PP B50

kmol/jam

0!#+ *!0&+ *!14 +7+!+1 4#&1!

0!0047 0!0#4* *!14 +7+!+1 10!0Jumlah 1 *!07 Jumlah 44!1

$en"hitun" aliran ;apor masuk kondensor

U = Q 1 x (

= 1!07 x +01!&0#

= +0!7&& kmol/jam

KomposisiPanas ;apor masuk kondensor: U

kmol/jam

0!#+ +0*!00+& *!14 +7!1 ##!

0!0047 1!740 *!14 +7!1 #17!

Jumlah 1 +0!7&& Jumlah *!

! $en"hitun" panas laju alir %a"ian %a

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'PP B510

0!00 +!07&+ *!14 #&0!+* 7+*!&

Jumlah 1 +0!7&& Jumlah 1#7**!



KomposisiPanas liRuid masuk kolom destilasi: ,

kmol/jam

0!&1# +0+!007# *!14 +*&!4+ 1&*!

0!+#7 171!+1#& *!14 +*&!4+ #0*&&0!

0!0+#7 17!1+14 *!14 +*&!4+ 4#0#*!#

0!00+ 1!0+4 *!14 +*&!4+ *47*!+

Jumlah 1 #+!+4& Jumlah &***4!


KomposisiPanas liRuid keluar se%a"ai destilat: (

kmol/jam

0!#+ !77+ *!14 +7+!+1 1*##!

0!0047 1!71+1 *!14 +7+!+1 +#+7!

Jumlah 1 +01!&0# Jumlah 1*4*7!


KomposisiPanas liRuid keluar se%a"ai %ottom: B

kmol/jam

0!014* +!0+01 *!14 #&0!+* #014!1

0!**# 1&!&014 *!14 #&0!+* 7*7&1#!

0!0*# 17!1+14 *!14 #&0!+* 10&4#4!

0!00 1!0+4 *!14 #&0!+* 1&4!

Jumlah 1 11!&&&4 Jumlah 141*!



=

&***4!7#14 = 1*4*7!4404 141*!+

= #17!#4*+

.&

1#6

#

x, 8

re38

1&

1&kCa

6

LK

.

&6

K

.#

106

+

.&

1#6

#

x( 8

re38

1

1kCa

6

LK

.

&6

K

xB 8

re38

kCa

6

LK

.

&6

K

.#

106

+

.&

1#6

#

1&

Q

1

.

Q

Q

.

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'PP B511



=

*!1&11 = 44!114 1*4*7!4404

= +71&!#* kCal/jam


= #17!#4*+

= #17!#4*+ +71&!#*

= ##1*!#74 kCal/jam


= m x .p x 8

+71&!#* = m x #!1*1& kJ/k" K x +4+!14 5

m = 1&7!*+4# k"/jam


Kesetim%an"an panas pada re%oiler-

=

10+##*!7+ = 1#7**!14# 141*!+

= 1+&1&&!0&40

(imana-

= 19

= 19 ? 1#7**!14# 141*!+ @

= +#1!17*0#1*7& kCal/jam

2ehin""a-

= 1+&1&&!0&40

= 1+&1&&!0&40 +#1!17*0

= 1+*4&10!#+0 kCal/jam


Kondisi steam masuk- 8 = 00P = 1444 kPa

= 1+*!& kJ/k"

= #&+!+##* kCal/k"


17

1*

1

.

.

Q

.

Q

Q

.

0

Q

1

Loss

Q

Q

Loss

Loss ? 1 @

Q

Loss

o

.

QP

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'PP B51

= 1+*4&10!#+0

#&+!+##*

= 0!#41* k"/jam




&***4!7#14 1*4*7!4404

##1*!#74 141*!+

+71&!#*

Jumlah 119-!3.=9>! Jumlah 119-!3.=9>!

'liran Panas Kondensor

*!1&11 44!1141*4*7!4404

+71&!#*

Jumlah --9=3!.1=11 Jumlah --9=3!.1=11

'liran Panas Qe%oiler

10+##*!7+ 1#7**!14#

1+*4&10!#+0 141*!+

+#1!17*0

8otal -319!:9.>9-- 8otal -319!:9.>9--

9. Pre&eater E138

Keteran"an-


- Panas %ahan keluar preheater menuju kolom destilasi

- Panas an" hilan"


1&

1

Q

.

17 1*

1

.

0

1

Q

Loss


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'PP B51+

(irenCanakan-

2uhu %ahan masuk = 1*#!+ = #47!+*+*#& K

2uhu produk keluar = 1*7!+ = #&0!+*+*# K

a.

Komponena % C d

++!++ 50!00*#1*& 0!00000& 51!7*4E500* +!

#*!1* 0!107+ 5&!&117E500* 51!**+#E500* 1!

1+!0& 0!#*+&7 50!000770& &!0*&E500* 51!

!101 0!7*&4 50!0004&0& !004E5007 5+!

Komponen kmol/jam

#47!+*+*#& #&0!+*+*# +!0+01 ++!4&&

#47!+*+*#& #&0!+*+*# 1&!&014 771&0!440+

#47!+*+*#& #&0!+*+*# 17!1+14 10#!+4

#47!+*+*#& #&0!+*+*# 1!0+4 1&4+!07Jumlah *&0*4!4&

.

= 19 x

= 19 x 141*!&41

= 141!*&+ kCal/jam

IeraCa panas total-

=

*&0*4!4& = 141*!&41 1= *1#!+#1 kCal/ja



P = 1444 kPa

= 1+*!& kJ/k"

o.

o.

$en"hitun" panas an" terkandun" pada %ahan masuk

dan %ahan keluar

CP: J/mol K

6

.

&6

.#

106

+

.&

1#

6#

M(ikutip dari Va

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'PP B51#

= #&+!+##* kCal/k"


= *1#!+#1

#&+!+##*

= &0!*#& k"/jam

Neraca Panas 0eater



*&0*4!4& 141*!&41

*1#!+#1 141!*&40&**8otal >-3-8>.>11 8otal >-3-8>.>11

>. Ko$o% Desti$asi D13!

Keteran"an-

- Panas %ahan masuk kolom destilasi

- Panas ;apor menuju kondensor

- Panas liRuid keluar kondensor an" re3luks

- Panas liRuid keluar kondensor se%a"ai destilat

- Panas liRuid masuk re%oiler

+

2team

Loss

;un)si- Dntuk memisahkan produk utama dan produk sampin"

+

#

4

&

7

2teamP

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'PP B514

- Panas ;apor keluar re%oiler

- Panas liRuid keluar re%oiler se%a"ai %ottom

- Panas an" hilan"- Panas an" terkandun" pada steam

- Panas an" terkandun" pada air pendin"in

(irenCanakan-

2uhu %ahan masuk = 1*7 = #&0!+*+*# K

2uhu liRuid keluar re%oiler = 10 = #&+!0+0# K




8emperatur pada 3eed-

Bu%le point = 1*7!++* = #&0!+*+* K

(e< point = 0#!4& = #7*!10& K

8emperatur pada destilat

Bu%le point = 1*!#+ = #&+!0+ K

(e< point = 1&!&* = #&!*#* K

8emperatur pada %ottom

Bu%le point = 1&7!1+ = ##0!+&+ K

(e< point = +!#+* = 41!4** K


Komponen,eed: , (estilat: (


+!0+01 0!0 +!0+01 0!01*

1&!&014 0!** 1&!*17# 0!777 &!7

17!1+14 0!0 0!&*4+ 0!00#1 1&!

1!0+4 0!01 5 5 1!

Jumlah 11!&&&4 1 1&&!4+7 1 4!1




*

Loss

2team

Pendin"in

o.

o.

o.

o.

o.

o.

o.

o.

o.o.

x,

x(

6

.

&6

LK

.#

106

+ K

.&

1#6

#

=i

i,i xR1

(i x

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'PP B51&

8rial T = 1!07#70*&

Dmpan masuk kolom destilasi

Komposisi H

0!014* 70!74* 0!01&1

0!**# #!#001 1!170

0!0*# 1!0000 51!1&

0!00 &!4#1 0!0100

Jumlah 1 0!000*

(estilat keluar kolom destilasi

Komposisi H

0!01* &*!1 0!01*&

0!777 #!+44 1!41+7

0!00#1 1!0000 50!007&

Jumlah 1 1!4#*

= 1!4#*

= 0!4#*


= 0!7*7


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'PP B517

KomposisiPanas liRuid an" dire3luks: Lo

kmol/jam

0!01* !+*41 *!14 #&+!10 +1#!4

0!777 1*!1& *!14 #&+!10 &0!

0!00#1 0!4+# *!14 #&+!10 +#1!

Jumlah 1 1+1!0*7# Jumlah &1+0#1!

$en"hitun" aliran ;apor masuk kondensor

U = Q 1 x (

= 1!7*7 x 1&&!4+7

= 7!&0 kmol/jam

KomposisiPanas ;apor masuk kondensor: U

kmol/jam

0!01* 4!#14 *!14 #&!*# 7&0!&

0!777 0!*0+ *!14 #&!*# 1##040*!

0!00#1 1!#7 *!14 #&!*# *1!0

Jumlah 1 7!&0 Jumlah 1#4!

! $en"hitun" panas laju alir %a"ian %a

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'PP B51*

Jumlah 1 7!&0 Jumlah 1*11+!



KomposisiPanas liRuid masuk kolom destilasi: ,

kmol/jam

0!014* +!0+01 *!14 #&0!+* #014!1

0!**# 1&!&014 *!14 #&0!+* 7*7&1#!

0!0*# 17!1+14 *!14 #&0!+* 10&4#4!

0!00 1!0+4 *!14 #&0!+* 1&4!

Jumlah 1 11!&&&4 Jumlah 141*!


Komposisi

Panas liRuid keluar se%a"ai destilat: (

kmol/jam

0!01* +!0+01 *!14 ##0!+7 +411!7

0!777 1&!*17# *!14 ##0!+7 &44*4!

0!00#1 0!&*4+ *!14 ##0!+7 +&&*!0

Jumlah 1 1&&!4+7 Jumlah &47&4!


KomposisiPanas liRuid keluar se%a"ai %ottom: B

kmol/jam

0!& &!7*#1 *!14 #&+!10 +0!0

0!&4#+ 1&!##& *!14 #&+!10 10#40!

0!0747 1!0+4 *!14 #&+!10 17*0!0

Jumlah 1 4!1++* Jumlah 14+&!



=

141*!&41 = &47&4!4* 14+&!7

= 51017+#!1+4&



=

1#4!0+* = &1+0#1!1#0 &47&4!4*

= 1*+#++!&+ kCal/jam

x, 8

re38

+

+kCa

6

.

&6

LK

.#

106

+ K

.&

1#6

#

x( 8re3 8& &kCa

6

.

&6

LK

.#

106

+ K

xB 8

re38

kCa

.

&6

LK

.#106+ K

.&

1#6

#

+

Q

&

.

Q

Q

.

#

4

&

.

.

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'PP B51


= 51017+#!1+4&

= 51017+#!1+4& 1*+#++!&+

= *1&!447& kCal/jam


= m x .p x 8

1*+#++!&+ = m x #!1*1& kJ/k" K x +&+!14 5

m = &&!4 k"/jam


Kesetim%an"an panas pada re%oiler-

=

1&&&0#!11+* = 1*11+!0+# 14+&!7

= +041*!0+

(imana-

= 19

= 19 ? 1*11+!0+# 14+&!7 @

= 17*!++0&* kCal/jam

2ehin""a-

= +041*!0+

= +041*!0+ 17*!+

= +4&!#+4 kCal/jam



P = 1444 kPa

= 1+*!& kJ/k"

= #&+!+##* kCal/k"


= +4&!#+4

#&+!+##*

= 70!*1&7 k"/jam




Q

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Q

Q

.

7

Q

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Q

Q

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? *

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o.

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'PP B50

141*!&41 &47&4!4*

*1&!447& 14+&!7

1*+#++!&+Jumlah >>=9-9.19-= Jumlah >>=9-9.19-=

'liran Panas Kondensor

1#4!0+* &1+0#1!1#0

&47&4!4*

1*+#++!&+

Jumlah 13:=-3!.!>9 Jumlah 13:=-3!.!>9

'liran Panas Qe%oiler

1&&&0#!11+* 1*11+!0+#

+4&!#+4 14+&!717*!+

8otal 1>>-::!.:3= 8otal 1>>-::!.:3=

1!. Coo$er E13

Keteran"an-

- Panas %ahan masuk Cooler

- Panas %ahan keluar Cooler menuju stora"e

- Panas an" hilan"


(irenCanakan-

2uhu %ahan masuk = 1&7 = ##0!+7 K

2uhu produk keluar = +0 = +0+!14 K



+

&

Q

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#

4

&

.

7

*

Q

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;un)si, Dntuk pendin"inan produk se%elum masuk stora"e

&

+0

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Pendin"in

o.

o.

o.

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'PP B51

a.

Komponena % C d

#*!1* 0!107+ 5&!&117E500* 51!**+#E500* 1!

1+!0& 0!#*+&7 50!000770& &!0*&E500* 51!

!101 0!7*&4 50!0004&0& !004E5007 5+!

Komponen kmol/jam

##0!+7 +0+!14 +!0+01 11##!7*+#

##0!+7 +0+!14 1&!*17# *714#!11&

##0!+7 +0+!14 0!&*4+ 4#!77*

Jumlah ***1*!&7*#

.

= 19 x

= 19 x ***1*!&7*#

= ***!1*&* kCal/jam

c.

IeraCa panas total-

0 = 5

= Panas air pendin"in keluar-

=

***1*!&7*# = &1!114 ***!1*&7*#+4

= *4+0+7!4*0 kCal/jam

= m x .p x 8

*4+0+7!4*0 = m x #!1*1& kJ/k" K x +&+!1

m = ++*!0&1* k"/jam

Neraca Panas Coo$er



***1*!&7*# &1!114

*4+0+7!4*0

***!1*&*

$en"hitun" panas an" terkandun" pada %ahan masuk &

dan %ahan keluar

CP: J/mol K

.

&6

.#

106

+

.&

1#6

#

M(ikutip dari Va

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8otal 999>19.=893 8otal 999>19.=893

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e

44E5011 *!14

4+1E501 *!14

tor

8re3

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171!4#4

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4*+!#117

1*71!#&+

19888:.983>

0.0

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kCal/jam

8re3

kCal/jam

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e&*7*E5011 *!14

+#E501 *!14

&**!41+1

1&7&!7#4

--!=:.-:>!

8re3

+kCal/jam

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'PP B5&

440#*+

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'PP B57

ari Dtilitas

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+00!14

e

&*7*E5011

+#E501

71E5011

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or

1###&!7*7

1###&!7*7

1#&

+14

+4

84

4kCal/jam

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suk Keluar

759 !4*7

040 17*!7#4#

171!+1#&

17!1+14

1!0+4

4 !1&4

0+ 0!*0+

e

71E5011+#E501

44E5011

4+1E501

+*&E5011

*7E501+

&*7*E5011

in niout

]d8i

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'PP B5+0

al/jam

&0!7#

7*#!0*1704!11#

1!>.>3:8

al/jam

&!07#

7*!#0*

04*!+0+

9>9.98-8

al/jam

&!#47

1!*+1

7*!+#7+=!.=-1

**&0!&+1

00#

xn 1 *:14

xn )) *:14

n ))) *:14

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'PP B5+1

&!##

DE

DE kCal/jamDE kJ/jam

00!14 K

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e

+#E501 *!14

44E5011 *!14

4+1E501 *!14

+*&E5011 *!14

101&4!777

+#4!14+

8re3

*kCal/jam

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'PP B5++

#4+!704#

717+!#41

1#*700!04

70!00&

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'PP B5+#

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'PP B5+4

raCa

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&0!00

o.

8s#

o.

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'PP B5+7

7/25/2019 excel new 0

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'PP B5+*

!**** x 1&*!1

x 7&!414

!!!!!!!!!!!!!!!!!!!!!!!!1

= x 1!**7471&!#&&

!!!!!!!!!!!!!!!!!!

!7#& x 1!**4

& x &7!&7&7

!!!!!!!!!!!!!!!!!!!!!!!!+

L

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'PP B5+

0.0000

7/25/2019 excel new 0

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'PP B5#0

0.0000

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'PP B5#1

0.0000

7

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e

+#E501 *!14

44E5011 *!14

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1&

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+*&E5011 *!14

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#0*&&0!707*

4#0#*!#+0&

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&***4!7#14

!*47#14#

excel new 0

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