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The document describes a process for producing cumene from benzene and propylene. Key details include: - Benzene and propylene react in a reactor at 400°F to produce cumene. The conversion of propylene is 80%. - Streams exiting the reactor include unreacted benzene, propylene, butane, and cumene. These streams enter a heat exchanger and separator. - Recycled benzene is mixed with fresh benzene feedstock. Energy balances are used to calculate the temperature of one stream entering the heat exchanger.

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Angga Widya Putra
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81 views7 pages

Atk2

The document describes a process for producing cumene from benzene and propylene. Key details include: - Benzene and propylene react in a reactor at 400°F to produce cumene. The conversion of propylene is 80%. - Streams exiting the reactor include unreacted benzene, propylene, butane, and cumene. These streams enter a heat exchanger and separator. - Recycled benzene is mixed with fresh benzene feedstock. Energy balances are used to calculate the temperature of one stream entering the heat exchanger.

Uploaded by

Angga Widya Putra
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Propylene

Benzene Butane
8
77 F 2 150 F 9
Benzene
3

HE 6
5
400 F
D1 D2
Reactor

75% Propylene 4
25% Butane 77 F Cumene
1 7 10

Reaksi : Komponen MP/K BP/K


C6H6 + C3H6 C9H12 C6H6 278,693 353,26
Mole ratio Recycle Benzene : C3H6 87,91 255,46
1 New = 3 Recycle C9H12 177,125 425,56
Konversi Propylene dalam reactor = 80 % n-C4H10 134,83 272,66
Basis 100 mol Fresh Feed

Ditanya
a. Komposisi & Molar Flow Rate all stream
b. T stream <3>, <4>, <6>
c. Q/mol cumene

Unit Reaktor

5 C9H12 Misal, jumlah benzene yang masuk dalam


C6H6 reaktor adalah A mol
C3H6 Reaksi
Reactor
C6H6 + C3H6 C9H12
75 mol C3H6 M A 75
25 mol n-C4H10 C6H R 60 60 60
1 S A-60 15 60
4

Aliran <4> Aliran <5>


C6H6 = A mol C9H12 = 60 mol
C3H6 = 15 mol
C6H6 = A-60 mol
n-C4H10 = 25 mol
Unit HE
3
Aliran <3> Aliran <6>
C6H6 = A mol C9H12 = 60 mol
5 6
HE C3H6 = 15 mol
C6H6 = A-60 mol
n-C4H10 = 25 mol
4

Unit Pemisahan

C3H6 Aliran <7> Aliran <9>


n-C4H10 C6H6 C9H12 = 60 mol C6H6 = A-60 mol
8 9
C6H6 = A-60 mol

Aliran <8> Aliran <10>


6 C3H6 = 15 mol C9H12 = 60 mol
D1 D2 n-C4H10 = 25 mol

7 10 C9H12

Mixing Point
Aliran <2>
= Aliran <3> - Aliran <9> Aliran <3>
New
= A - ( A - 60 ) C6H6 = 80 mol
C6H = 60 mol
2 Aliran <9>
Mole ratio Recycle (R) Benzene : C6H6 = 20 mol
9 Recycle 1 New = 3 Recycle
C6H 1*(60) = 3 R
R = 20 mol
3
C6H
Mencari nilai A
R = A- 60
20 = A - 60
A = 80 mol

Komposisi dan Molar Flow Rate All Stream


Aliran/mol
Komponen
<1> <2> <3> <4> <5> <6> <7> <8> <9> <10>
C3H6 75 0 0 0 15 15 0 15 0 0
n-C4H10 25 0 0 0 25 25 0 25 0 0
C6H6 0 60 80 80 20 20 20 0 20 0
C9H12 0 0 0 0 60 60 60 0 0 60
Total 100 60 80 80 120 120 80 40 20 60
Neraca Energy
E = Q + W - (H + P + K)
Asumsi steady state
E = 0 ; W=0; P = 0 ; K = 0
sehingga,
Q = H

Asumsi, operasi pada keadaan vakum


shg seluruh komponen berfasa gas

Tref = 25 C = 298,15 K
Data Cp untuk tiap komponen (Himmleblau Ap. E)
Komponen state T a b.102 c.105 d.109
C3H6 g C 59,58 17,71 -10,17 24,6
n-C4H10 g C 92,3 27,88 -15,47 34,98
C6H6 l K -7,27329 77,054 -164,82 1897,9
g C 74,06 32,95 -25,2 77,57
C9H12 g C 139,2 53,76 -39,79 120,5
Form 1
2 3
Cp = a + bT + cT + dT
H = m Cp dT
2 3 4 T
= m [aT + (bT )/2 + (cT )/3 + (dT )/4] Tref

Data Hf pada 25C Data Hvap pada Boiling Point


Komponen Hf (kJ/gmol) Komponen Hvap (kJ/gmol)
C3H6 20,41 C6H6 30,76
C6H6 48,66
C9H12 3,93

Mixing Point

C6H6 Asumsi sistem Adiabatis


77 F 2 shg, Q = H
(25 C)
0 = H
9
0 = Houtput - Hinput
C6H6 Hinput = Houtput
150 F
3 (65.6 C) H2 + H9 = H3
C6H

H2
HC6H6 = 60 [aT + (bT2)/2 + (cT3)/3 + (dT4)/4]2525 = 0 J
H9
HC6H6 = 20 [aT + (bT2)/2 + (cT3)/3 + (dT4)/4]65.5625 = 71740,247 J

H3
HC6H6 = 80 [aT + (bT2)/2 + (cT3)/3 + (dT4)/4]T25
= 5924,8 T+ 13,18 T2 + ( -0,00672 ) T3 + 1,551E-06 T4 - 156253,11

H2 + H9 = H3
71740,247 = 5924,8 T+ 13,18 T2 + ( -0,00672 ) T3 + 1,551E-06 T4 - 156253,11
T = 35,7 C

Jadi, suhu pada aliran <3> = 35,7 C


= 96,26 F

Propylene
Benzene Butane
8
77 F 2 150 F 9
Benzene
3

HE 6
5
400 F
D1 D2
Reactor

75% Propylene 4
25% Butane 77 F Cumene
1 7 10

Asumsi pada proses pemisahan tidak terjadi perubahan suhu


shg
T aliran <6> = T aliran <9>
= 150 F

Unit HE
C6H6
95.9 F Asumsi sistem Adiabatis
3 C9H12
C9H12 (35.5 C) shg, Q = H
C6H6 C6H6 0 = H
C3H6 5 6 C3H6 0 = Houtput - Hinput
n-C4H10 HE n-C4H10
150 F Hinput = Houtput
400 F
(204.44 C) (65.6 C) H3 + H5 = H4 + H6
4
C6H6

H3
HC6H6 = 80 [aT + (bT2)/2 + (cT3)/3 + (dT4)/4]308.65298.15= 107348,04 J
H4
HC6H6 = 80 [aT + (bT2)/2 + (cT3)/3 + (dT4)/4]T25
2 3 4
= 5924,8 T+ 13,18 T + ( -0,00672 ) T + 1,551E-06 T - 156253,11
H5
HC3H6 = 15 [aT + (bT2)/2 + (cT3)/3 + (dT4)/4]204.4425 = 210874,6 J
2 3 4 204.44
Hn-C4H10 = 25 [aT + (bT )/2 + (cT )/3 + (dT )/4] 25 = 546924,14 J
2 3 4 204.44
HC6H6 = 20 [aT + (bT )/2 + (cT )/3 + (dT )/4] 25 = 387792,53 J
2 3 4 204.44
HC9H12 = 60 [aT + (bT )/2 + (cT )/3 + (dT )/4] 25 = 2097966,4 J +
3243557,7 J
H6
HC3H6 = 15 [aT + (bT2)/2 + (cT3)/3 + (dT4)/4]65.625 = 210874,6 J
2 3 4 65.6
Hn-C4H10 = 25 [aT + (bT )/2 + (cT )/3 + (dT )/4] 25 = 546924,14 J
2 3 4 65.6
HC6H6 = 20 [aT + (bT )/2 + (cT )/3 + (dT )/4] 25 = 387792,53 J
2 3 4 65.6
HC9H12 = 60 [aT + (bT )/2 + (cT )/3 + (dT )/4] 25 = 2097966,4 J +
3243557,7 J

H3 + H5 = H6 + H4
107348,04 + 3243557,7 = 3243557,7 + ( 5924,8 T+ 13,18 T2 + ( -0,00672 ) T3
4
+ 1,551E-06 T - ( 156253,11 ))
T = 40,86 C
Jadi, suhu pada aliran <4> = 40,86 C
= 105,548 F

Propylene
Benzene Butane
8 150 F
77 F 2 150 F 9
Benzene
3

HE 6
5
400 F
D1 D2
Reactor

75% Propylene 4 Cumene


25% Butane 77 F 150 F

1 7 10

Q = H
= Houtput - Hinput
= (H8 + H10 ) - (H1 + H2)

H1
HC3H6 = 75 [aT + (bT2)/2 + (cT3)/3 + (dT4)/4]2525 = 0 J
2 3 4 25
Hn-C4H10 = 25 [aT + (bT )/2 + (cT )/3 + (dT )/4] 25 = 0 J +
0 J
H2
HC6H6 = 60 [aT + (bT2)/2 + (cT3)/3 + (dT4)/4]2525 = 0 J
H8
HC3H6 = 15 [aT + (bT2)/2 + (cT3)/3 + (dT4)/4]65.625 = 210874,6 J
2 3 4 65.6
Hn-C4H10 = 25 [aT + (bT )/2 + (cT )/3 + (dT )/4] 25 = 546924,14 J +
757798,74 J
H10
HC9H12 = 60 [aT + (bT2)/2 + (cT3)/3 + (dT4)/4]65.625 = 2097966,4 J

Q = H
= Houtput - Hinput
= (H8 + H10 ) - (H1 + H2)
= 2855765,2 J

Untuk 60 mol produk cumene dibutuhkan panas sebesar = 2855765,2 J


shg untuk 1 mol cumene dibutuhkan panas sebesar = 47596,086 J/mol cumene
= 47,6 kJ/mol cumene
Perhitungan T aliran <3>
2 3 4
71740,247 = 5924,8 T + 13,18 T + -0,00672 T + 1,551E-06 T - 156253,11
T f(T)
35,697581 7,537E-06

Perhitungan T aliran <4>


2 3
107348,04 + 3243557,7 = 3243557,7 + ( 5924,8 T + 13,18 T + ( -0,00672 ) T
4
+ 1,551E-06 T - ( 156253,11 ))
T f(T)
40,854745 0,0002128

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