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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
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
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
. $enentukan re3luks minimum
(en"an men""unakan persamaan 11:751 dan 11:750 Seankoplis: 1+
(imana 3eed masuk dalam keadaan liRuid pada titik didihna: R = 1 Kusnarjo: 01
8rial T = 404!4++&
Dmpan masuk kolom destilasi u$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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(irenCanakan re3luks rasio = 1!4 x
= 0!000
c. $en"hitun" panas pada laju alir %a"ian atas dan laju alir %a"ian %a
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e. $en"hitun" IeraCa Panas 6;erall
Kesetim%an"an panas o;erall pada kolom destilasi-
=
+0+&47!4+* = #+7*04#!4&+ 1+1!41#4
= 1+&0*+!44+0
1! $en"hitun" panas air keluar kondensor
Kesetim%an"an panas pada kondensor-
1+ =
#0*0*!4**# = 1#!+7* #+7*04#!4&+
= 47*1+! kCal/jam
! $en"hitun" panas steam pada re%oiler
= 1+&0*+!44+0
= 1+&0*+!44+0 47*1+!
= 1***&4+!00 kCal/jam
#. $en"hitun" ke%utuhan air pendin"in
= 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
). $en"hitun" neraCa panas pada re%oiler
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
&. $en"hitun" ke%utuhan steam
Kondisi steam masuk- 8 = 00
P = 1444 kPa
= 1+*!& kJ/k"
= #&+!+##* kCal/k"
$aka steam an" di%utuhkan =
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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-
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-
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K
-
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-
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-
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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
%ahan konstruksi = .2 2' #0 "rade $ tpe +1&
, 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
%ahan konstruksi = .2 2' #0 "rade $ tpe +1&
, 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
+`! $en"itun" har"a koe3isien 3ilm
io = 1400
at
St
at
Iret
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0!*41& = 0!0&4
jamx
1# hari
hari
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7!0&170E500&
0!0011+741#
+!4&4&174E500&
0!0011#*
3t
1#00 %%l
0!1+*&&1*47
?i7
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datar
0!*4
1& = 0!0&4
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jamx
1# hari
hari
!11&4*1#E5004
0!000+*7*
0!000#7##7+1
0!0011#0
3t
?i7
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in
datar
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0!*4
1& = 0!0&4
jamx
1# hari
hari
0!0007*0&11&
0!000&*74#4&
4!*0####*&E5004
#!#40&&+0*E500&
&!14*+477E5004
!#0010#7E500&
0!001&0#
?i7
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3t
1#00 %%l
0!1+7+&14#
1 tan"ki
x
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in
datar
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0!*4
1& = 0!0&4
jamx
1# hari
hari
0!0011&+#**
0!000+*7*&1
0!0000&1
?i7
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0!00174*
3t
1#00 %%l
0!001#+*01
1 tan"ki
-
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x
in
-
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datar
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+!17#
= 1
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h3
1!*&+
= 5 Ws
= 5 Ws
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0!#70
1##
CP
0!0&4
!#
a"ian Tube steam
l%/jam!3t
Btu/j!3t!,
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-
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-
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-
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FEED kolom distilasi
Komponen B$
$+ $+
k"/jam k"mol/jam
8oluena !1#**1 110!& &0!+0707+ 0!&4##4+04
Etil%enzena 10&!17& 1+&!1 *0!#*70+ !#1177++7
(EB 1+#!++0* 1*+!4 +01!&&&**++ !410&1*74&1
8EB 1&!*& 17!& &&!4&&4+ 1+!&*
sul3ur Coump! +!07 ###!& 1+!&4##+1*+1# 0!#47&&+7
868'L 141*!+0#7#+4# 1+0!0#&&++7
Dntuk menentukan K dan LK ditentukan dari %e%an distilasi terendah
(iperoleh %e%an destilasi terendah = ( = !11+70017
d8
2ehin""a ditentukan Etil %enzena se%a"ai Li"ht ke dan (EB se%a"ai ea; ke
'sumsi- Etil%enzena li"ht ke an" keluar se%a"ai %ottom se%esar
(ietil%enzena hea; ke an" keluar se%a"ai destilat se%esar
NERACA MASSA OVERALL
F = D + B
F . = D . + B .
Ko%ponen;eed2 ; (M" Desti$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
'liran Panas $asuk 'liran Panas Keluar
Komponen Ener"i kCal/jam Komponen Ener"i kCal/jam
#+*!71& 1*7774!*7#14+&!7 77#!+1
1*77!74*7
8otal 1>8=-8.>3:> 8otal 1>8=-8.>3:>
.&
1#6
#
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'PP B51*7
-. Pre&eater E11=
Keteran"an-
- Panas %ahan masuk preheater
- 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
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'PP B51**
1*7774!*7# = 0+&4!40 0+!&
= +#7+!0+&7 kCal/jam
c. $en"hitun" ke%utuhan steam
Kondisi steam masuk- 8 = 00
P = 1444 kPa
= 1+*!& kJ/k"
= #&+!+##* kCal/k"
$aka steam an" di%utuhkan =
= +#7+!0+&7
#&+!+##*
= 74!010 k"/jam
Neraca Panas 0eater
'liran Panas $asuk 'liran Panas Keluar
Komponen Ener"i kCal/jam Komponen Ener"i kCal/jam
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
KomponenPanas 2ensi%el
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
KomponenPanas 2ensi%el
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
'liran Panas $asuk 'liran Panas Keluar
Komponen Ener"i kCal/jam Komponen Ener"i kCal/jam
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 masuk preheater
- Panas %ahan keluar preheater menuju e;aporator
- Panas an" hilan"
- Panas an" terkandun" pada steam
(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
c. $en"hitun" ke%utuhan steam
Kondisi steam masuk- 8 = 00
P = 1444 kPa
= 1+*!& kJ/k"
= #&+!+##* kCal/k"
$aka steam an" di%utuhkan =
= 1+07#1+!0&*
#&+!+##*= *1!&*41 k"/jam
Neraca Panas 0eater
'liran Panas $asuk 'liran Panas Keluar
Komponen Ener"i kCal/jam Komponen Ener"i kCal/jam
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 %ahan masuk preheater
- 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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-
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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
'liran panas keluar -
= 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
'liran panas keluar -
= 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
'liran Panas $asuk 'liran Panas Keluar
Komponen Ener"i kCal/jam Komponen Ener"i kCal/jam
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
'liran Panas $asuk 'liran Panas Keluar
Komponen Ener"i kCal/jam Komponen Ener"i kCal/jam
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
'liran Panas $asuk 'liran Panas Keluar
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
-
7/25/2019 excel new 0
204/284
'PP B50#
=. Pre&eater E11
Keteran"an-
- Panas %ahan masuk preheater- Panas %ahan keluar preheater menuju kolom destilasi
- Panas an" hilan"
- Panas an" terkandun" pada steam
(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!
;un)si, Dntuk pemanasan a
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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
c. $en"hitun" ke%utuhan steam
Kondisi steam masuk- 8 = 00
P = 1444 kPa
= 1+*!& kJ/k"= #&+!+##* kCal/k"
$aka steam an" di%utuhkan =
= #+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 steam
- 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 masuk = 7 = +00!14 K
2uhu air pendin"in keluar = *0 = +4+!14 K
(ata an" diperoleh dari perhitun"an temperatur %u%le point dan de< point
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
a. Komposisi masin"5masin" komponen pada tiap aliran %ahan dalam kolom destilasi
Komponen,eed: , (estilat: (
kmol/jam kmol/jam kmo
+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!
. $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 = 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*
(irenCanakan re3luks rasio = 1!4 x
= 0!07
c. $en"hitun" panas pada laju alir %a"ian atas dan laju alir %a"ian %a
-
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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**!
d. $en"hitun" panas 3eed masuk: destilat keluar dan %ottom keluar
1! $en"hitun" panas pada 3eed masuk kolom destilasi
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!
! $en"hitun" panas liRuid keluar se%a"ai destilat
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!
+! $en"hitun" panas liRuid keluar se%a"ai %ottom
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*!
e. $en"hitun" IeraCa Panas 6;erall
Kesetim%an"an panas o;erall pada kolom destilasi-
=
&***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! $en"hitun" panas air keluar kondensor
Kesetim%an"an panas pada kondensor-
=
*!1&11 = 44!114 1*4*7!4404
= +71&!#* kCal/jam
! $en"hitun" panas steam pada re%oiler
= #17!#4*+
= #17!#4*+ +71&!#*
= ##1*!#74 kCal/jam
#. $en"hitun" ke%utuhan air pendin"in
= m x .p x 8
+71&!#* = m x #!1*1& kJ/k" K x +4+!14 5
m = 1&7!*+4# k"/jam
). $en"hitun" neraCa panas pada re%oiler
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
&. $en"hitun" ke%utuhan steam
Kondisi steam masuk- 8 = 00P = 1444 kPa
= 1+*!& kJ/k"
= #&+!+##* kCal/k"
$aka steam an" di%utuhkan =
17
1*
1
.
.
Q
.
Q
Q
.
0
Q
1
Loss
Q
Q
Loss
Loss ? 1 @
Q
Loss
o
.
QP
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'PP B51
= 1+*4&10!#+0
#&+!+##*
= 0!#41* k"/jam
Neraca Panas Ko$o% Desti$asi
'liran Panas $asuk 'liran Panas Keluar
Komponen Ener"i kCal/jam Komponen Ener"i kCal/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 masuk preheater
- Panas %ahan keluar preheater menuju kolom destilasi
- Panas an" hilan"
- Panas an" terkandun" pada steam
1&
1
Q
.
17 1*
1
.
0
1
Q
Loss
;un)si, Dntuk pemanasan a
-
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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
c. $en"hitun" ke%utuhan steam
Kondisi steam masuk- 8 = 00
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"
$aka steam an" di%utuhkan =
= *1#!+#1
#&+!+##*
= &0!*#& k"/jam
Neraca Panas 0eater
'liran Panas $asuk 'liran Panas Keluar
Komponen Ener"i kCal/jam Komponen Ener"i kCal/jam
*&0*4!4& 141*!&41
*1#!+#1 141!*&40&**8otal >-3-8>.>11 8otal >-3-8>.>11
>. Ko$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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- 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
2uhu air pendin"in masuk = 7 = +00!14 K
2uhu air pendin"in keluar = 0 = +&+!14 K
(ata an" diperoleh dari perhitun"an temperatur %u%le point dan de< point
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
a. Komposisi masin"5masin" komponen pada tiap aliran %ahan dalam kolom destilasi
Komponen,eed: , (estilat: (
kmol/jam kmol/jam kmo
+!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
. $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
*
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#*
(irenCanakan re3luks rasio = 1!4 x
= 0!7*7
c. $en"hitun" panas pada laju alir %a"ian atas dan laju alir %a"ian %a
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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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Jumlah 1 7!&0 Jumlah 1*11+!
d. $en"hitun" panas 3eed masuk: destilat keluar dan %ottom keluar
1! $en"hitun" panas pada 3eed masuk kolom destilasi
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*!
! $en"hitun" panas liRuid keluar se%a"ai destilat
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!
+! $en"hitun" panas liRuid keluar se%a"ai %ottom
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+&!
e. $en"hitun" IeraCa Panas 6;erall
Kesetim%an"an panas o;erall pada kolom destilasi-
=
141*!&41 = &47&4!4* 14+&!7
= 51017+#!1+4&
1! $en"hitun" panas air keluar kondensor
Kesetim%an"an panas pada kondensor-
=
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
! $en"hitun" panas steam pada re%oiler
= 51017+#!1+4&
= 51017+#!1+4& 1*+#++!&+
= *1&!447& kCal/jam
#. $en"hitun" ke%utuhan air pendin"in
= m x .p x 8
1*+#++!&+ = m x #!1*1& kJ/k" K x +&+!14 5
m = &&!4 k"/jam
). $en"hitun" neraCa panas pada re%oiler
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
&. $en"hitun" ke%utuhan steam
Kondisi steam masuk- 8 = 00
P = 1444 kPa
= 1+*!& kJ/k"
= #&+!+##* kCal/k"
$aka steam an" di%utuhkan =
= +4&!#+4
#&+!+##*
= 70!*1&7 k"/jam
Neraca Panas Ko$o% Desti$asi
'liran Panas $asuk 'liran Panas Keluar
Komponen Ener"i kCal/jam Komponen Ener"i kCal/jam
Q
.
Q
Q
.
7
Q
*
Loss
Q
Q
Loss
Loss
? *
@
Q
Loss
o.
QP
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'PP B50
141*!&41 &47&4!4*
*1&!447& 14+&!7
1*+#++!&+Jumlah >>=9-9.19-= Jumlah >>=9-9.19-=
'liran Panas Kondensor
1#4!0+* &1+0#1!1#0
&47&4!4*
1*+#++!&+
Jumlah 13:=-3!.!>9 Jumlah 13:=-3!.!>9
'liran Panas Qe%oiler
1&&&0#!11+* 1*11+!0+#
+4&!#+4 14+&!717*!+
8otal 1>>-::!.:3= 8otal 1>>-::!.:3=
1!. Coo$er E13
Keteran"an-
- Panas %ahan masuk Cooler
- Panas %ahan keluar Cooler menuju stora"e
- Panas an" hilan"
- Panas an" terkandun" pada steam
(irenCanakan-
2uhu %ahan masuk = 1&7 = ##0!+7 K
2uhu produk keluar = +0 = +0+!14 K
2uhu air pendin"in masuk = 7 = +00!14 K
2uhu air pendin"in keluar = 0 = +&+!14 K
+
&
Q
.
#
4
&
.
7
*
Q
Loss
;un)si, Dntuk pendin"inan produk se%elum masuk stora"e
&
+0
Loss
Pendin"in
o.
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
'liran Panas $asuk 'liran Panas Keluar
Komponen Ener"i kCal/jam Komponen Ener"i kCal/jam
***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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'PP B5
8otal 999>19.=893 8otal 999>19.=893
-
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'PP B5+
e
44E5011 *!14
4+1E501 *!14
tor
8re3
-
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'PP B5#
+*&E5011 *!14
171!4#4
4+7&!701&
**4!0*&1
77#!+1
e
&*7*E5011 *!14+#E501 *!14
4*+!#117
1*71!#&+
19888:.983>
0.0
+1
kCal/jam
8re3
kCal/jam
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'PP B54
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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'PP B5*
+00!14
e
&*7*E5011
+#E501
71E5011
*7E501+
or
1###&!7*7
1###&!7*7
1#&
+14
+4
84
4kCal/jam
-
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'PP B5
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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'PP B5+
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&
-
7/25/2019 excel new 0
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'PP B5+#
-
7/25/2019 excel new 0
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'PP B5+4
raCa
-
7/25/2019 excel new 0
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'PP B5+&
&0!00
o.
8s#
o.
-
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'PP B5+7
-
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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
-
7/25/2019 excel new 0
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'PP B5#1
0.0000
7
-
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'PP B5#
e
+#E501 *!14
44E5011 *!14
4+1E501 *!14
1&
8re3
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'PP B5#+
+*&E5011 *!14
1&*!+#7
#0*&&0!707*
4#0#*!#+0&
*47*!+&*#
&***4!7#14
!*47#14#