Neraca Massa Prarancangan Anilin

LAMPIRAN A

PERHITUNGAN NERACA MASSA

Kapasitas Produksi = 62500 ton/tahun

= 7891.4141 kg/jam

Hari Kerja = 1 tahun = 330 hari

= 1 hari = 24 jam

Basis Bahan Baku = 100 kg/jam

Spesifikasi bahan baku = Nitrobenzen = 0.9970

= Air = 0.0010

= Di-Nitrobenzen = 0.0010

= Benzen = 0.0010

Spesifikasi Produk = Anilin = 0.9900

= Air = 0.0100

Tabel Data Masing-masing Komponen

Komponen Rumus Molekul BM, kg/kmol

Nitrobenzen 210.9 123

Air 100 18

Anilin 184.1 93

Hidrogen -259.2 2

Di-Nitrobenzen 301 168

Benzen 80.1 78

Metana -164 16

A.1 SEPARATOR

2

1

3

Basis = 100 kg/jam

Asumsi = 80% produk

Arus 2 (produk yang diinginkan):

Nitrobenzen = 79.7600 kg/jam = 0.6485 kmol/jam

= 0.0800 kg/jam = 0.0044 kmol/jam

Di-Nitrobenzen = 0.0800 kg/jam = 0.0005 kmol/jam

Titik Didih, oC

C6H5NO2

H2O

C6H5NH2

H2

C6H4NO2

C6H6

CH4

H2O

Sep

arator

Benzen = 0.0800 kg/jam = 0.0010 kmol/jam

Arus 1 (umpan):


= 0.1000 kg/jam = 0.0056 kmol/jam



Arus 3


= 0.0200 kg/jam = 0.0011 kmol/jam



Tabel Neraca Massa Separator

KomponenMasuk (kg/jam) Keluar (kg/jam)

Arus 1 Arus 2 Arus 3

Nitrobenzen 99.7000 79.7600 19.9400

0.1000 0.0800 0.0200

Di-Nitrobenzen 0.1000 0.0800 0.0200

Benzen 0.1000 0.0800 0.0200

Total80.0000 20.0000

100.0000 100.0000

A.2 REAKTOR

4

2

Konversi = 0.98

Bahan Baku yang Masuk kedalam Reaktor

= 79.7600 kg/jam = 0.6485 kmol/jam

= 0.0800 kg/jam = 0.0044 kmol/jam

= 0.0800 kg/jam = 0.0005 kmol/jam

= 0.0800 kg/jam = 0.0010 kmol/jam

Perbandingan mol Nitrobezen terhadap H2 = 1:3

= 3.8907 kg/jam = 1.9454 kmol/jam

= 0.00004 kg/jam = 0.000002 kmol/jam

H2O

H2O

H2O

C6H5NO2

C6H4NO2

C6H6

H2O

H2

CH4

Reak

tor

Bahan yang Bereaksi

= Konversi x

= 0.98 x 79.7600

= 78.1648 kg/jam = 0.6355 kmol/jam

=

= 0.08 kg/jam = 0.0044 kmol/jam

=

= 0.08 kg/jam = 0.0005 kmol/jam

=

= 0.08 kg/jam = 0.0010256 kmol/jam

=

= 3.8129 kg/jam = 1.9065 kmol/jam

=

= 0.00004 kg/jam = 0.000002 kmol/jam

Reaksi yang Reaksi

+ +

Mula-mula 0.6485 1.9454 - -

Bereaksi 0.6355 1.9065 0.6355 1.2710

Sisa 0.0130 0.0389 0.6355 1.2710

Produk yang Terbentuk

= 0.6355 kmol/jam = 59.1002 kg/jam

= 1.2710 kmol/jam = 22.8775 kg/jam

Neraca Massa Total Reaktor

Komponen Masuk (kg/jam) Keluar (kg/jam)

79.7600 1.5952

0.0800 0.0800

0.0800 0.0800

0.0800 0.0800

3.8907 0.0778

0.00004 0.00004

- 59.1002

- 22.8775

Total 83.8908 83.8908

C6H5NO2 yang bereaksi C6H5NO2 yang masuk

H2O H2O yang masuk

C6H4NO2 C6H4NO2 yang masuk

C6H6 C6H6 yang masuk

H2 yang bereaksi (3/1) x C6H5NO2 yang bereaksi

CH4 CH4 yang masuk

C6H5NO2 3H2 C6H5NH2 2H2O

C6H5NH2

H2O

C6H5NO2

C6H4NO2

H2O

C6H6

H2

CH4

C6H5NH2

H2O

Bahan Yang Masuk

= 1.5952 kg/jam = 0.0130 kmol/jam

= 0.0800 kg/jam = 0.0005 kmol/jam

= 0.0800 kg/jam = 0.0044 kmol/jam

= 0.0800 kg/jam = 0.0010 kmol/jam

= 0.0778 kg/jam = 0.0389 kmol/jam

= 0.00004 kg/jam = 0.000002 kmol/jam

= 59.1002 kg/jam = 0.3210 kmol/jam

= 22.8775 kg/jam = 1.2710 kmol/jam

A.3 Flash Tank

5

4

6

Komposisi Umpan Masuk

Bahan Yang Masuk

= 1.5952 kg/jam = 0.0130 kmol/jam

= 0.0800 kg/jam = 0.0005 kmol/jam

= 0.0800 kg/jam = 0.0044 kmol/jam

= 0.0800 kg/jam = 0.0010 kmol/jam

= 0.0778 kg/jam = 0.0389 kmol/jam

= 0.00004 kg/jam = 0.000002 kmol/jam

= 59.1002 kg/jam = 0.3210 kmol/jam

= 22.8775 kg/jam = 1.2710 kmol/jam

Komposisi Bahan yang Teruapkan

98% x = 1.5633 kg/jam = 0.0127 kmol/jam

99% x = 0.0792 kg/jam = 0.0005 kmol/jam

98% x = 22.4984 kg/jam = 1.2499 kmol/jam

C6H5NO2

C6H4NO2

H2O

C6H6

H2

CH4

C6H5NH2

H2O

C6H5NO2

C6H4NO2

H2O

C6H6

H2

CH4

C6H5NH2

H2O

Nitrobenzen (C6H5NO2)

C6H5NO2 yang masuk

Di-Nitrobenzen (C6H4NO2)

C6H4NO2 yang masuk

Air (H2O)

H2O yang masuk

Flash

T

ank

98% x = 0.0784 kg/jam = 0.0010 kmol/jam

100.0% x = 0.0778146 kg/jam = 0.0389 kmol/jam

100% x = 0.00004 kg/jam = 0.000002 kmol/jam

0.01% x = 0.0059 kg/jam = 0.0001 kmol/jam

Komposisi Bahan Cair

2% x = 0.031904 kg/jam = 0.0003 kmol/jam

1% x = 0.0008 kg/jam = 0.000005 kmol/jam

2% x = 0.45915 kg/jam = 0.0255 kmol/jam

2% x C6H6 yang masuk = 0.0016 kg/jam = 2.051E-05 kmol/jam

99.99% x = 59.094305 kg/jam = 0.6354 kmol/jam

0.00% x = 0.0000 kg/jam = 0.00000 kmol/jam

Komponen Masuk (kg/jam)Keluar (kg/jam)

V L

1.5952 1.5633 0.0319

0.0800 0.0792 0.0008

22.9575 22.4984 0.4592

0.0800 0.0784 0.0016

0.0778 0.0778 -

3.890770615E-05 3.890770615E-05 -

Benzen (C6H6)

C6H6 yang masuk

Gas Hidrogen (H2)

H2 yang masuk

Gas Metana (CH4)

CH4 yang masuk

Anilin (C6H5NH2)

C6H5NH2 yang masuk


C6H5NO2 yang masuk


C6H4NO2 yang masuk

Air (H2O)

H2O yang masuk

Benzen (C6H6)

Anilin (C6H5NH2)

C6H5NH2 yang masuk

Hidrogen (H2)

H2 yang masuk

C6H5NO2

C6H4NO2

H2O

C6H6

H2

CH4

59.1002 0.0059 59.0943

Total 83.890824.3030 59.5878

83.8908

KomponenAntoine

A B C D E

-54.494 -2112.3 29.321 -0.0448 0.0000

-24.246 -4114 16.344 -0.0241 0.0000

18.3036 3816.44 -46.13 - -

15.9008 2788.51 -52.36 - -

13.6333 164.9 3.19 - -

15.2243 597.84 -7.16 - -

16.6748 3857.52 -73.15 - -

(L/V) data = 2.4519

Dengan menggunakan persamaan

Vi =

Fi

Ai =

(L/V)data

ki =

Pi

((L/V)/ki)+1 ki P sistem

T trial = 139.53 C = 412.53 K

Li =

Fi

P sistem = 760 mmHg (1+(L/V)*ki)

logP = A + B/T + ClogT + DT + ET2

Komponen Fi (kmol) Pi (mmHg) ki Ai

0.0130 101.6479 0.1337 18.3321

0.0005 1.1463 0.0015 1625.6623

1.2754 2663.536 3.5047 0.6996

0.0010 3492.8994 4.5959 0.5335

0.0389 560527.74 737.5365 0.0033

2E-06 936103.2 1231.7147 0.0020

0.3210 201.9982 0.2658 9.2249

Total 1.6498

Vi %V Li %Li

0.0007 0.0008 0.0098 0.0289

2.927408E-07 3.561597E-07 0.0005 0.0014

0.7504 0.9130 0.1330 0.3937

0.0007 0.0008 0.0001 0.0002

0.0388 0.0472 2.150358E-05 0.0001

2.426901E-06 2.95266E-06 8.049423E-10 2.383875E-09

C6H5NH2

C6H5NO2

C6H4NO2

H2O

C6H6

H2

CH4

C6H5NH2

C6H5NO2

C6H4NO2

H2O

C6H6

H2

CH4

C6H5NH2

0.0314 0.0382 0.1944 0.5756

0.8219 1.0000 0.3377 1.0000

L = Fi

Vi total

= 2.0072

(L/V)hitung = 2.4421

A.5 MENARA DISTILASI

7

6

8

Arus 6

Komponen Kg/Jam Kmol/Jam

0.0319 0.0003

0.0008 4.761905E-06

0.4592 0.0255

0.0016 2.051282E-05

59.0943 0.6354

Total 59.5878 0.6612

A. Massa Masuk Menara Distilasi pada Kondisi Bubble Point

T = 170.11 °C = 443.11 K ; P = 1 atm = 760 mmHg

Komponen (Kmol/jam) xi Pi ki

0.0003 0.0004 267.6120 0.5177

4.761905E-06 7.201743E-06 5.5490 0.0107

C6H5NO2

C6H4NO2

H2O

C6H6

C6H5NH2

Trial pada T akan dianggap benar apabila Syi = 1

C6H5NO2

C6H4NO2

Men

ara D

istilasi

0.0255 0.0386 5942.6871 11.4957

2.051282E-05 3.102289E-05 6402.8470 12.3859

0.6354 0.9610 516.9484 1.0000

Total 0.6612 1

yi αi

0.0002 1.0000

7.73045E-08 0.0207

0.4435 22.2064

0.0004 23.9259

0.9610 1.9317

1.405

Jadi bisa disimpulkan bahwa suhu pemasukan umpan sebesar 170,11 C

B. Spesifikasi Hasil Yang Diinginkan

B.1 Distilat

• Menentukan Massa Distilat

1. = 1% x Massa = 2.594E-06 kmol/jam

2. = 1% x Massa = 4.762E-08 kmol/jam

3. = 99.00% x Massa = 0.0253 kmol/jam

4. = 1% x Massa = 2.051E-07 kmol/jam

5. = 1% x Massa = 0.0064 kmol/jam

Total = 0.0316 kmol/jam

Menentukan Nilai xdi

1. = Massa Komponen

=

2.594E-06

= 8.205613E-05Massa Total 0.0316

2. = Massa Komponen

=

4.762E-08

= 1.50644E-06Massa Total 0.0316

3. = Massa Komponen

=

0.0253

= 0.7989Massa Total 0.0316

4. = Massa Komponen

=

2.051E-07

= 0Massa Total 0.0316

6. = Massa Komponen

=

0.0064

= 0.2010Massa Total 0.0316103

H2O

C6H6

C6H5NH2

C6H5NO2

C6H4NO2

H2O

C6H6

C6H5NH2

C6H5NO2

C6H4NO2

H2O

C6H6

C6H5NH2

• Menentukan Massa Bottom

1. = 99% x Massa = 0.0003 kmol/jam

2. = 99% x Massa = 4.714E-06 kmol/jam

3. = 1% x Massa = 0.0003 kmol/jam

4. C6H6 = 99% x Massa = 2.03.E-05 kmol/jam

5. = 99% x Massa = 0.6291 kmol/jam


Menentukan Nilai xbi

1. = Massa Komponen

=

0.0003

= 0.0004Massa Total 0.6296

2. = Massa Komponen

=

4.714E-06

= 0.0000Massa Total 0.6296

3. = Massa Komponen

=

0.0003

= 0.0004Massa Total 0.6296

4. C6H6 = Massa Komponen

=

2.0.E-05

= 0.0000Massa Total 0.6296

5. = Massa Komponen

=

0.6290684

= 0.9991Massa Total 0.6296053

KomponenMassa Distilat

(Kmol/jam) % Massa xdi

0.0003 1% 2.593821E-06 8.205613E-05

4.761905E-06 1% 4.761905E-08 1.50644E-06

0.0255 99% 0.0253 0.7989

2.051282E-05 1% 2.051282E-07 0

0.6354 1% 0.0064 0.2010

Total 0.6612 0.0316 1.0000

KomponenMassa Bottom

(Kmol/jam) % Massa xbi

0.0003 99% 0.0003 0.0004

4.761905E-06 99% 4.714286E-06 7.487684E-06

0.0255 1% 0.0003 0.0004

C6H6 2.05.E-05 99% 2.03.E-05 0.0000

0.6354 99% 0.6291 0.9991

Total 0.6612 0.6296 1.0000

C6H5NO2

C6H4NO2

H2O

C6H5NH2

C6H5NO2

C6H4NO2

H2O

C6H5NH2

C6H5NO2

C6H4NO2

H2O

C6H6

C6H5NH2

C6H5NO2

C6H4NO2

H2O

C6H5NH2

C. Perhitungan Suhu Atas (Dew point)

Pt = 1 atm = 760 mmHg

T = 182.7943 °C = 455.9443 K

KomponenMassa yi=

Piki= xi= αi =

(kmol/jam) m/m total Pi/P yi/ki ki/k(HK)

C6H5NO2 2.593821E-06 8.205613E-05 382.33575 0.5031 0.0001631 1.0000

C6H4NO2 4.761905E-08 1.50644E-06 9.9670455 0.0131 0.0001149 0.0261

0.0253 0.7988925214 8030.4177 10.5663 0.0756 21.0036

2.051282E-07 6.489278E-06 8034.002 10.5711 6.139E-07 21.0129

0.0064 0.2010174268 733.28465 0.9648 0.2083 1.9179

Total 0.0316 1 0.284

Jadi dapat disimpulkan bahwa T distilat adalah 182,7943°C

D. Perhitungan Suhu Bawah (Bubble point)


T = 184.0100 °C = 457.15996218 K

KomponenMassa xi=

Piki= yi= αi =

(kmol/jam) m/m total Pi/P xi.ki ki/k(HK)

0.0003 0.0004 394.95015 0.5196713 0.000212 1.0000

4.714286E-06 0.0000 10.513526 0.0138336 1.036E-07 0.0266

0.0003 0.0004 8254.6734 10.861412 0.0044005 20.9005

C6H6 2.03.E-05 0.0000 8202.4461 10.792692 0.0003481 20.7683

0.6291 0.9991 757.05477 0.9961247 0.9952753 1.9168

Total 0.6296 1 1.000

Jadi dapat disimpulkan bahwa T bottom adalah 172,5794°C

KomponenMassa Masuk Massa Keluar

(kg/gr) Distilat Bottom

0.0016 2E-05 2E-03

0.4592 0.4546 0.0046

59.0943 0.5909 58.5034

0.0319 0.0003 0.0316

0.0008 8E-06 0.0008

59.5878 1.0458 58.5419

Total 59.5878 59.5878

Trial pada T akan dianggap benar apabila Sxi = 1

H2O

C6H6

C6H5NH2

Trial pada T dianggap benar apabila Syi = 1

C6H5NO2

C6H4NO2

H2O

C6H5NH2

C6H6

H2O

C6H5NH2

C6H5NO2

C6H4NO2

Produk = 58.5080

Faktor Pengali = 7891.4141

= 134.799458.5419

LAMPIRAN A


kmol/jam

kmol/jam

kmol/jam

8207.0707071

LAMPIRAN A



= 7891.4141 kg/jam


= 1 hari = 24 jam



= Air = 0.0010


= Benzen = 0.0010


= Air = 0.0100




Air 100 18

Anilin 184.1 93

Hidrogen -259.2 2


Benzen 80.1 78

Metana -164 16

A.1 SEPARATOR

2

1

3

Basis = 100 kg/jam

Asumsi = 80% produk



= 10.7840 kg/jam = 0.5991 kmol/jam


Titik Didih, oC

C6H5NO2

H2O

C6H5NH2

H2

C6H4NO2

C6H6

CH4

H2O

Sep

arator


Arus 1 (umpan):


= 13.4799 kg/jam = 0.7489 kmol/jam



Arus 3


= 2.6960 kg/jam = 0.1498 kmol/jam






Nitrobenzen 13439.5002 10751.6001 2687.9000

13.4799 10.7840 2.6960

Di-Nitrobenzen 13.4799 10.7840 2.6960

Benzen 13.4799 10.7840 2.6960

Total10783.9520 2695.9880

13479.9400 13479.9400

A.2 REAKTOR

4

2

Konversi = 0.98


= 10751.6001 kg/jam = 87.4114 kmol/jam

= 10.7840 kg/jam = 0.5991 kmol/jam

= 10.7840 kg/jam = 0.0642 kmol/jam

= 10.7840 kg/jam = 0.1383 kmol/jam


= 524.4683 kg/jam = 262.2341 kmol/jam

= 0.00524 kg/jam = 0.000328 kmol/jam

H2O

H2O

H2O

C6H5NO2

C6H4NO2

C6H6

H2O

H2

CH4

Reak

tor

Bahan yang Bereaksi

= Konversi x

= 0.98 x 10751.6001

= 10536.568 kg/jam = 85.6632 kmol/jam

=

= 10.783952 kg/jam = 0.5991 kmol/jam

=

= 10.783952 kg/jam = 0.0642 kmol/jam

=

= 10.783952 kg/jam = 0.1382558 kmol/jam

=

= 513.9789 kg/jam = 256.9895 kmol/jam

=

= 0.00524 kg/jam = 0.000328 kmol/jam

Reaksi yang Reaksi

+ +

Mula-mula 87.4114 262.2341 - -

Bereaksi 85.6632 256.9895 85.6632 171.3263

Sisa 1.7482 5.2447 85.6632 171.3263


= 85.6632 kmol/jam = 7966.6735 kg/jam

= 171.3263 kmol/jam = 3083.8736 kg/jam



10751.6001 215.0320

10.7840 10.7840

10.7840 10.7840

10.7840 10.7840

524.4683 10.4894

0.00524 0.00524

- 7966.6735

- 3083.8736

Total 11308.4255 11308.4255


H2O H2O yang masuk




CH4 CH4 yang masuk


C6H5NH2

H2O

C6H5NO2

C6H4NO2

H2O

C6H6

H2

CH4

C6H5NH2

H2O

Bahan Yang Masuk

= 215.0320 kg/jam = 1.7482 kmol/jam

= 10.7840 kg/jam = 0.0642 kmol/jam

= 10.7840 kg/jam = 0.5991 kmol/jam

= 10.7840 kg/jam = 0.1383 kmol/jam

= 10.4894 kg/jam = 5.2447 kmol/jam

= 0.00524 kg/jam = 0.000328 kmol/jam

= 7966.6735 kg/jam = 43.2736 kmol/jam

= 3083.8736 kg/jam = 171.3263 kmol/jam

A.3 Flash Tank

5

4

6


Bahan Yang Masuk

= 215.0320 kg/jam = 1.7482 kmol/jam

= 10.7840 kg/jam = 0.0642 kmol/jam

= 10.7840 kg/jam = 0.5991 kmol/jam

= 10.7840 kg/jam = 0.1383 kmol/jam

= 10.4894 kg/jam = 5.2447 kmol/jam

= 0.00524 kg/jam = 0.000328 kmol/jam

= 7966.6735 kg/jam = 43.2736 kmol/jam

= 3083.8736 kg/jam = 171.3263 kmol/jam


98% x = 210.7314 kg/jam = 1.7133 kmol/jam

99% x = 10.676112 kg/jam = 0.0635 kmol/jam

98% x = 3032.7644 kg/jam = 168.4869 kmol/jam

C6H5NO2

C6H4NO2

H2O

C6H6

H2

CH4

C6H5NH2

H2O

C6H5NO2

C6H4NO2

H2O

C6H6

H2

CH4

C6H5NH2

H2O


C6H5NO2 yang masuk


C6H4NO2 yang masuk

Air (H2O)

H2O yang masuk

Flash

T

ank

98% x = 10.568273 kg/jam = 0.1355 kmol/jam

100.0% x = 10.489366 kg/jam = 5.2447 kmol/jam

100% x = 0.00524 kg/jam = 0.000328 kmol/jam

0.01% x = 0.7967 kg/jam = 0.0086 kmol/jam


2% x = 4.3006401 kg/jam = 0.0350 kmol/jam

1% x = 0.1078395 kg/jam = 0.000642 kmol/jam

2% x = 61.893151 kg/jam = 3.4385 kmol/jam

2% x C6H6 yang masuk = 0.215679 kg/jam = 0.0027651 kmol/jam

99.99% x = 7965.8768 kg/jam = 85.6546 kmol/jam

0.00% x = 0.0000 kg/jam = 0.00000 kmol/jam


V L

215.0320 210.7314 4.3006

10.7840 10.6761 0.1078

3094.6576 3032.7644 61.8932

10.7840 10.5683 0.2157

10.4894 10.4894 -

0.0052447354444 0.0052447354444 -

Benzen (C6H6)

C6H6 yang masuk

Gas Hidrogen (H2)

H2 yang masuk

Gas Metana (CH4)

CH4 yang masuk

Anilin (C6H5NH2)

C6H5NH2 yang masuk


C6H5NO2 yang masuk


C6H4NO2 yang masuk

Air (H2O)

H2O yang masuk

Benzen (C6H6)

Anilin (C6H5NH2)

C6H5NH2 yang masuk

Hidrogen (H2)

H2 yang masuk

C6H5NO2

C6H4NO2

H2O

C6H6

H2

CH4

7966.6735 0.7967 7965.8768

Total 11308.42553276.0314 8032.3941

11308.4255

KomponenAntoine

A B C D E

-54.494 -2112.3 29.321 -0.0448 0.0000

-24.246 -4114 16.344 -0.0241 0.0000

18.3036 3816.44 -46.13 - -

15.9008 2788.51 -52.36 - -

13.6333 164.9 3.19 - -

15.2243 597.84 -7.16 - -

16.6748 3857.52 -73.15 - -

(L/V) data = 2.4519


Vi =

Fi

Ai =

(L/V)data

ki =

Pi


T trial = 139.53 C = 412.53 K

Li =

Fi




1.7482 101.6479 0.1337 18.3321

0.0642 1.1463 0.0015 1625.6623

171.9254 2663.536 3.5047 0.6996

0.1383 3492.8994 4.5959 0.5335

5.2447 560527.74 737.5365 0.0033

3E-04 936103.2 1231.7147 0.0020

43.2736 201.9982 0.2658 9.2249

Total 222.3947

Vi %V Li %Li

0.0904 0.0008 1.3165 0.0289

3.946129E-05 3.561597E-07 0.0640 0.0014

101.1562 0.9130 17.9221 0.3937

0.0902 0.0008 0.0113 0.0002

5.2273 0.0472 0.0028986694 0.0001

0.0003271447 2.95266E-06 1.085057E-07 2.383875E-09

C6H5NH2

C6H5NO2

C6H4NO2

H2O

C6H6

H2

CH4

C6H5NH2

C6H5NO2

C6H4NO2

H2O

C6H6

H2

CH4

C6H5NH2

4.2322 0.0382 26.1998 0.5756

110.7966 1.0000 45.5165 1.0000

L = Fi

Vi total

= 2.0072



7

6

8

Arus 6


4.3006 0.0350

0.1078 0.0006419019

61.8932 3.4385

0.2157 0.0027651159

7965.8768 85.6546

Total 8032.3941 89.1315


T = 170.11 °C = 443.11 K ; P = 1 atm = 760 mmHg


0.0350 0.0004 267.6120 0.5177

0.0006419019 7.201743E-06 5.5490 0.0107

C6H5NO2

C6H4NO2

H2O

C6H6

C6H5NH2


C6H5NO2

C6H4NO2

Men

ara D

istilasi

3.4385 0.0386 5942.6871 11.4957

0.0027651159 3.102289E-05 6402.8470 12.3859

85.6546 0.9610 516.9484 1.0000

Total 89.1315 1

yi αi

0.0002 1.0000

7.73045E-08 0.0207

0.4435 22.2064

0.0004 23.9259

0.9610 1.9317

1.405



B.1 Distilat


1. = 1% x Massa = 0.0003496 kmol/jam

2. = 1% x Massa = 6.419E-06 kmol/jam

3. = 99.00% x Massa = 3.4041 kmol/jam

4. = 1% x Massa = 2.765E-05 kmol/jam

5. = 1% x Massa = 0.8565 kmol/jam



1. = Massa Komponen

=

0.0003496

= 8.205613E-05Massa Total 4.2611

2. = Massa Komponen

=

6.419E-06

= 1.50644E-06Massa Total 4.2611

3. = Massa Komponen

=

3.4041

= 0.7989Massa Total 4.2611

4. = Massa Komponen

=

2.765E-05


6. = Massa Komponen

=

0.8565

= 0.2010Massa Total 4.2610529

H2O

C6H6

C6H5NH2

C6H5NO2

C6H4NO2

H2O

C6H6

C6H5NH2

C6H5NO2

C6H4NO2

H2O

C6H6

C6H5NH2


1. = 99% x Massa = 0.0346 kmol/jam

2. = 99% x Massa = 0.0006355 kmol/jam

3. = 1% x Massa = 0.0344 kmol/jam


5. = 99% x Massa = 84.7980 kmol/jam



1. = Massa Komponen

=

0.0346

= 0.0004Massa Total 84.8704

2. = Massa Komponen

=

0.0006355

= 0.0000Massa Total 84.8704

3. = Massa Komponen

=

0.0344

= 0.0004Massa Total 84.8704


=

2.7.E-03

= 0.0000Massa Total 84.8704

5. = Massa Komponen

=

84.798043

= 0.9991Massa Total 84.870416



0.0350 1% 0.0003496455 8.205613E-05

0.0006419019 1% 6.419019E-06 1.50644E-06

3.4385 99% 3.4041 0.7989

0.0027651159 1% 2.765116E-05 0

85.6546 1% 0.8565 0.2010

Total 89.1315 4.2611 1.0000



0.0350 99% 0.0346 0.0004

0.0006419019 99% 0.0006354829 7.487684E-06

3.4385 1% 0.0344 0.0004

C6H6 2.77.E-03 99% 2.74.E-03 0.0000

85.6546 99% 84.7980 0.9991

Total 89.1315 84.8704 1.0000

C6H5NO2

C6H4NO2

H2O

C6H5NH2

C6H5NO2

C6H4NO2

H2O

C6H5NH2

C6H5NO2

C6H4NO2

H2O

C6H6

C6H5NH2

C6H5NO2

C6H4NO2

H2O

C6H5NH2



T = 182.7943 °C = 455.9443 K

KomponenMassa yi=

Piki= xi= αi =


C6H5NO2 0.0003496455 8.205613E-05 382.33575 0.5031 0.0001631 1.0000

C6H4NO2 6.419019E-06 1.50644E-06 9.9670455 0.0131 0.0001149 0.0261

3.4041 0.7988925214 8030.4177 10.5663 0.0756 21.0036

2.765116E-05 6.489278E-06 8034.002 10.5711 6.139E-07 21.0129

0.8565 0.2010174268 733.28465 0.9648 0.2083 1.9179

Total 4.2611 1 0.284




T = 184.0100 °C = 457.15996218 K

KomponenMassa xi=

Piki= yi= αi =


0.0346 0.0004 394.95015 0.5196713 0.000212 1.0000

0.0006354829 0.0000 10.513526 0.0138336 1.036E-07 0.0266

0.0344 0.0004 8254.6734 10.861412 0.0044005 20.9005

C6H6 2.74.E-03 0.0000 8202.4461 10.792692 0.0003481 20.7683

84.7980 0.9991 757.05477 0.9961247 0.9952753 1.9168

Total 84.8704 1 1.000




0.2157 2E-03 2E-01

61.8932 61.2742 0.6189

7965.8768 79.6588 7886.2180

4.3006 0.0430 4.2576

0.1078 1E-03 0.1068

8032.3941 140.9792 7891.4149

Total 8032.3941 8032.3941


H2O

C6H6

C6H5NH2


C6H5NO2

C6H4NO2

H2O

C6H5NH2

C6H6

H2O

C6H5NH2

C6H5NO2

C6H4NO2

LAMPIRAN A


kmol/jam

kmol/jam

kmol/jam

LAMPIRAN B

NERACA PANAS

Sebagai Basis Perhitungan

Kapasitas Produksi = 62500 Ton/Tahun

Hari Kerja = 330 Hari

Produksi Aniline Perja = 7891.4141 Kg/Jam

Basis Waktu = 1 Jam

Satuan Massa = Kg

Satuan Panas = kJ

Satuan Cp = kJ/mol

Suhu Referensi =

Data Data yang Diperlukan

Kapasitas panas gas, cairan dan padatan

Sehingga Cp dT =

keterangan

Cp = kapasitas panas (kJ/kmol K)

A, B, C, D, E= konstanta

T = suhu (K)

Komponen A B C D E

27.1430 9.2730E-03 -1.3800E-05 7.6450E-09

-31.3860 0.4746 -3.11E-04 8.5240E-08 -5.0524E-12

32.243 1.9230E-03 1.0550E-05 -3.5960E-09

-22.0620 0.5731 -4.57E-04 -1.8410E-07 -2.9867E-11

-16.2020 0.5618 -3.93E-04 -1.0040E-07 -1.2252E-12

18.1480 0.5618 -3.93E-04 1.0040E-07 -1.2252E-12

34.9420 -3.9957E-02 -1.9184E-04 -1.5300E-07 3.9321E-11

Sumber Yaws dan Coulson

Komponen A B C D

28.8400 0.00765 3.29E-01 -8.70E-10

-33.6620 0.4743 -0.0036 3.8243E-06

18.2964 0.4721 -1.3388E-03 1.3142E-06

46.9480 0.9896 -2.3583E-03 2.3296E-06

-0.0180 1.1982 -9.8722E-03 3.1670E-05

25oC (298,15K)

Cp = A + BT + CT2 + DT3 + ET4 (kJ/kmol. K)

H2

C6H6

H2O

C6H5NH2

C6H5NO2

C6H4N2O4

CH4

Data kapasitas panas untuk liquid

H2

C6H6

H2O

C6H5NH2

CH4

5432

5432T

ET

DT

CT

BAT

39.4730 0.9128 -0.0021 2.0093E-06

-12.6350 1.5624 -2.9981E-03 2.3171E-06

Sumber Yaws dan Himmelblau

Kapasitas panas untuk Cooper Carbonate dicari dengan pendekatan pada

Perrys 7th Edition tabel 2-393

Elemen Solid ,J/mol C

Cu 26

C 7.5

O 16.7

H 9.6

Sehingga Kapasitas Panas CuCO3.Cu(OH)2 adalah

Cp CuCO3.Cu(OH) = 162.2 J/mol C Treff = 20 C

Data entalpi pembentukan

Komponen ΔHf (Kj/mol)

0

82.9

-241.826

86.86

-74.5

67.6

50.8

Sumber Yaws an Himmelblau

Data berat molekul

Komponen Rumus molekul BM

Hidrogen 2

Benzene 78

Air 18

Anilin 93

Metana 16

Nitrobenzen 123

Di-Nitrobenzen 168

Data bilangan Antoine

C6H5NO2

C6H4N2O4

H2

C6H6

H2O

C6H5NH2

CH4

C6H5NO2

C6H4N2O4

H2

C6H6

H2O

C6H5NH2

CH4

C6H5NO2

C6H4N2O4

KomponenAntoine

A B C D E

-54.494 -2112.3 29.321 -0.0448 0.00002

-24.246 -4114 16.344 -0.0241 0.00001

18.3036 3816.44 -46.13 - -

15.9008 2788.51 -52.36 - -

13.6333 164.9 3.19 - -

15.2243 597.84 -7.16 - -

16.6748 3857.52 -73.15 - -

B.1 Heater (E-114)

Fungsi : Mengubah fase Hidrogen cair menjadi uap Hidrogen

Tujuan :

Q3 Qsteam

Q1 Q2

Neraca Energi

ΔH = Hout - Hin

Hin = Hpendingin + Hout

KomponenMassa n

(kg) (kmol)

524.7727 262.3863

0.0052 0.0003

Tin = 30 = 303.15 K

Tref = 25 = 298.15 K

●

Q = n303.15

298.15

= 262.3863303.15

dT298.15

= 262.3863 27.1430 303.15 - 298.15 +9.27E-03

303.152

- 298.152

2

C6H5NO2

C6H4N2O4

H2O

C6H6

H2

CH4

C6H5NH2

Menaikkan suhu feed menjadi 250oC

1. Menghitung Panas Sensibel Masuk Heater, Q1

H2

CH4

oCoC

H2

Cp dT

27.1430T + 9.2730E-03 T2 - 1.3800E-05 T3 + 7.6450E-09 T4

+-1.3800E-05

303.153

- 298.153

+7.6450E-09

3 4

303.154

- 298.154

= 40129.7403 kJ/jam

●

Q = n303.15

298.15

= 0.000328303.15

34.942 T-3.9957E-02 T2 - 1.9184E-04 T3-1.53E-07 T4+3.9321E-11 T5dT298.15

= 0.000328 34.9420 303.15 - 298.15 +-0.03996

303.152

- 298.152

2

+-1.9184E-04

303.153

- 298.153

+-1.5300E-07

3 4

303.154

- 298.154

+3.9321E-11

303.155

- 298.155

5

= 0.1520 kJ/jam

Q1=

= 40129.7403 + 0.1520 = 40129.8923 kJ/jam

*Hout

Tout = 250 = 523.15 K

Tref = 25 = 298.15 K

●

Q = n523.15

298.15

= 262.3863523.15

dT298.15

= 262.3863 27.1430 523.15 - 298.15 +9.27E-03

523.152

- 298.152

2

+-1.3800E-05

523.153

- 298.153

+7.6450E-09

3 4

523.154

- 298.154

= 1720025.5993 kJ/jam

●

Q = n523.15

298.15

CH4

Cp dT

Q H2 + Q CH4

2. Menghitung Panas Sensibel Keluar Heater, Q2

oCoC

H2

Cp dT

27.1430T + 9.2730E-03 T2 - 1.3800E-05 T3 + 7.6450E-09 T4

CH4

Cp dT

= 0.000328523.15

34.942 T-3.9957E-02 T2 - 1.9184E-04 T3-1.53E-07 T4+3.9321E-11 T5dT298.15

= 0.000328 34.9420 523.15 - 298.15 +-0.03996

523.152

- 298.152

2

+-1.9184E-04

523.153

- 298.153

+-1.5300E-07

3 4

523.154

- 298.154

+3.9321E-11

523.155

- 298.155

5

= 4.75040 kJ/jam

Q2 =

= 1720025.59928 + 4.75040 = 1720030.34968 kJ/jam

3. Menghitung Kebutuhan Pemanas, Q3

Q3= Q2 - Q1

= 1720030.3497 - 40129.8923

= 1679900.4574 kJ/jam

1679900.4574

= 909.99 kJ/kg

Sehingga :

= Q3= 1679900.4574 kJ/mol= 1846.0647 kg/jam

909.99 kJ/kg

KomponenPanas masuk Panas keluar

(kJ/jam) (kJ/jam)

40129.7403 1720025.5993

0.15198 4.7504

Q Steam 1679900.4574 -

Total 1720030.3497 1720030.3497

B.2.

Fungsi :

masuk reaktor

Tujuan : - Menentukan Suhu Keluar (T out) Vaporizer.

Q H2 + Q CH4

Kebutuhan panas Heater sebesar kJ, panas disupply menggunakan

Saturated steam, dengan suhu 300oC dengan tekanan 1 atm, dari properties of saturated

water and saturated steam up to 1 atm, stoichiometry 2004 diperoleh data:

λsteam

Q3 = m.λsteam

Massa steam

λsteam

Neraca Panas Total dalam Heater

Q H2

Q CH4

VAPORIZER (V-130)

Menguapkan bahan baku Nitrobenzene sebelum

- Menghitung kebutuhan pemanas

Arus 3

204.45 250

Qs

Arus 2

30 250

Data Komponen Masuk

KomponenM n

(kg) (kmol)

13447.2999 109.3276

13.5420945 0.0806

13.5420945 0.7523

13.5420945 0.1736

●

Kondisi Operasi P = 1 atm = 760.0 mmHg

T = 210.82 = 483.97 K

Komponenn

xiPi sat Ki yi

kmol/jam mmHg Pi sat/P Ki/xi

0.8073 0.9909 766.9869 1.0092 0.9819

0.0006 0.0007 31.3712 0.0413 0.0177

0.0013 0.0016 12582.7631 16.5563 0.0001

0.0056 0.0068 14575.2847907727 19.1780 0.0004

Total 0.8147 1.0000 1.0000

●

Kondisi Operasi P = 1 atm = 760.0 mmHg

T = 204.45 = 477.60 K

Komponenn

xiPi sat Ki yi

kmol/jam mmHg Pi sat/P Ki.xi

0.8073 0.9909 660.6744 0.8693 0.8614

0.0006 0.0007 24.5323 0.0323 0.0000

oC Steam oC

oC Steam oC

1. Penentuan Kondisi Operasi Vaporizer (E-121)

C6H5NO2

C6H4N2O4

H2O

C6H6

Menentukan Suhu Keluar (T out) Vaporizer

Menentukan Kondisi Dew Point

oC

Tabel Perhitungan Tekanan Dew Point

C6H5NO2

C6H4N2O4

H2O

C6H6

Menentukan Kondisi Bubble Point

oC

Tabel Perhitungan Tekanan Bubble Point

C6H5NO2

C6H4N2O4

0.0013 0.0016 11422.0738 15.0290 0.0236

0.0056 0.0068 12815.3172382983 16.8623 0.1150

Total 0.8147 1.0000 1.0000

Suhu keluar vaporizer merupakan suhu pada kondisi operasinya yaitu

T out = 204.45 = 477.60 K

Perhitungan Neraca Panas pada Vaporizer (V-130)

a.

T Larutan= 30 C T reff = 25 C

= 303.15 K = 298.15 K

●

Q = n303.15

298.15

= 109.328 39.47 303.15 - 298.15 +0.9128

303.152

- 298.152

2

--0.0021

303.153

- 298.153

+2.01E-06

303.154

- 298.154

3 4

= 97189.7934 kJ/jam

●

Q = n303.15

298.15

= 0.0806 -12.6 303.15 - 298.15 +1.56240

303.152

- 298.152

2

+-3E-03

303.153

- 298.153

+2.32E-06

303.154

- 298.154

3 4

= 318.8357 kJ/jam

●

Q = n303.15

298.15

= 0.1736 -33.6620 303.15 - 298.150.4743

303.152

- 298.152

+###

303.153

- 298.153

+3.82E-06

303.154

- 298.153 4

= 467.9 kJ/jam

●

Q = n303.15

298.15

H2O

C6H6

pada kondisi bubble point oC

Panas Sensibel Cairan Masuk, Q1

C6H5NO2

Cp dT

C6H4N2O4

Cp dT

C6H6

Cp dT

H2O

Cp dT

= 0.7523 18.2964 303.15 - 298.15 +0.4721

303.152

- 298.152

2

+###

303.153

- 298.153

+1.31E-06

303.154

- 298.154

3 4

= 281.8968 kJ/jam

Panas sensibel masuk, Q1

Q1= 97189.7934 + 318.8357 + 467.8883 + 281.8968 kJ/jam

Q1= 98258.4142 kJ/jam

b. Panas Laten Penguapan Komponen, Q

Komponen Tb (K) Tc (K) ΔHv (Kj/mol)

483.9500 719.0000 44.0800 Yaws

572.0000 803.0000 61.5600 Yaws

353.2400 562.1600 30.7500 Yaws

373.1500 647.4000 40.6800 Himmelblau

Untuk menghitung entalpi panas penguapan (ΔHv) digunakan Persamaan Watson:

=1 - 0.38 (Pers 4.13 Smith Van Ness, 200)

1 -

=1 - 0.38

1 -

dimana: = Panas laten penguapan pada titik didih normal (kJ/kmol)

=

= (K)

= (K)

= Titik didih normal komponen (K)

= Suhu tertentu

= = 210.82 = 483.97 K

●

= x1 - 0.38

1 -

= 44.08 x1 - 483.97 719.0000 0.38 = 44.0786 kJ/kmol

1 - 483.9500 719.0000

ΔHv = 0.8073 kmol/jam x 44.0786 kJ/kmol = 35.5854 kJ/jam

●

Q1 = Q C6H5NO2 + Q C6H4N2O4 + Q C6H6 + H2O

C6H5NO2

C6H4N2O4

H2O

C6H6

ΔH2 Tr2

ΔH1 Tr1

ΔH2 ΔH1

Tr2

Tr1

ΔH1

ΔH2 Panas laten penguapan pada suhu T2 (kJ/kmol)

Tr2 T2/TC

Tr1 T1/TC

T1

T2

T dew oC

C6H5NO2

ΔH2 ΔH1

Tr2

Tr1

ΔH2

C6H4N2O4

= x1 - 0.38

1 -

= 61.56 x1 - 483.97 803.0000 0.38 = 69.5956 kJ/kmol

1 - 572.0000 803.0000


●

= x1 - 0.38

1 -

= 30.75 x1 - 483.97 562.1600 0.38 = 21.1666 kJ/kmol

1 - 353.2400 562.1600


●

= x1 - 0.38

1 -

= 40.68 x1 - 483.97 647.4000 0.38 = 33.4158 kJ/kmol

1 - 373.1500 647.4000


Q2= 35.5854 + 0.0414 + 0.0271 + 0.1856 kJ/jam

Q2= 35.8396 kJ/jam

c. Panas Sensibel Uap Keluar (Q3)

T uap = 204.45 C T reff = 25 C

= 477.6 K = 298.15 K

●

Q = n477.60

Cp dT298.15

= 109.3276 -16.2 477.60 - 298.15 +0.5618

477.602

- 298.152

2

=-4E-04

477.603

- 298.153

+1.00E-07

477.604

- 298.154

3 4

+###

477.605

- 298.155

ΔH2 ΔH1

Tr2

Tr1

ΔH2

C6H6

ΔH2 ΔH1

Tr2

Tr1

ΔH2

H2O

ΔH2 ΔH1

Tr2

Tr1

ΔH2

Q2= Q C6H5NO2 + Q C6H4N2O4 + Q C6H6 + Q H2O

C6H5NO2

+5

477.60 - 298.15

= 5259818.8569 kJ/jam

●

Q = n477.60

Cp dT298.15

= 0.0806 18.1 477.60 - 298.15 +0.56182

477.602

- 298.152

2

+-4E-04

477.603

- 298.153

+1.00E-07

477.604

- 298.154

3 4

+###

477.605

- 298.155

5

= 2633.8489 kJ/jam

●

Q = n477.60

298.15

= 0.1736 -31.3860 477.60 - 298.150.4746

477.602

- 298.152

+###

477.603

- 298.153

+8.52E-08

477.604

- 298.153 4

+###

477.605

- 298.155

5

= 6410.1414 kJ/jam

●

Q = n477.60

298.15

= 0.7523 32.2430 477.60 - 298.15 +0.0019

477.602

- 298.152

2

+1.06E-05

477.603

- 298.153

+###

477.604

- 298.154

3 4

= 6179.9152 kJ/jam

Q3= 5259818.8569 + 2633.8489 + 6410.1414 + 6179.9152 kJ/jam

Q3= 5275042.7623 kJ/jam

d. Panas yang dibutuhkan oleh pemanas

Q1+Q4 = Q2+Q3

Q4= (Q2+Q3)-Q1

C6H4N2O4

C6H6

Cpg dT

H2O

Cp dT

Q3 = Q C6H5NO2 + Q C6H4N2O4 + Q C6H6 + Q H2O

Q4= 35.8396 + 5275042.7623 - 98258.4142 (kJ/jam)

Q4= 5176820.1877 (kJ/jam)

dan tekanan 1 atm.

λ = 909.99 kJ/kg

Sehingga:

=Q4

λ

=5176820.1877 kJ/jam

= 5688.8759 kg/jam909.99 kJ/kg

(V-130)

Komponen Q in (kJ/jam) Q out (kJ/jam)

97189.7934 -

318.8357 -

467.8883 -

281.8968 -

- 5275042.7623

- 35.8396

5176820.1877 -

Total 5275078.6019 5275078.6019

B.3. SEPARATOR (H-120)

Fungsi : Memisahkan fase gas dan fase cair yang terbentuk

Tujuan : Menghitung panas setiap arus

Arus 4

204

Arus 3

204

Arus 5

204

Neraca Energi

Q in = Q out

Sebagai pemanas digunakan steam yaitu saturated steam pada suhu 300oC dan

Dari properties of saturated water and saturated steam up to 1 atm, stoichiometry 2004

msteam

Tabel Neraca Panas Vaporizer

C6H5NO2

C6H4N2O4

C6H6

H2O

Q preheating

Q vaporizing

Q steam

oC

oC

oC

Sep

arat

or

3 = 4 + 5

dimana: 3 =

4 =

5 =

Untuk menghitug panas masing-masing arus digunakan persamaan:

dimana:

= 5275042.7623 kJ/jam

Komposisi arus 4:

= 10757.8399 kg/jam = 87.4621 kmol/jam

= 10.83368 kg/jam = 0.0645 kmol/jam

= 10.83368 kg/jam = 0.1389 kmol/jam

= 10.83368 kg/jam = 0.6019 kmol/jam

●

Q = n477.60

298.15

= 87.4621477.60

-16.2020 T + 0.5618 T2 - 3.93E-4 T3 - 1.004E-7 T4 - 1.2252E-12 T4 dT298.15

= 87.4621 -16.2020 477.60 - 298.15 +0.5618

477.602

- 298.152

2

+###

477.603

- 298.153

+###

477.604

- 298.154

3 4

+###

477.605

- 298.155

5

= 4207855.0855 kJ/jam

●

Q = n477.60

298.15

= 0.0645477.60

dT298.15

Panas sensibel gas masuk separator, Q1

Panas sensibel gas keluar separator, Q2

Panas sensibel cairan keluar separator, Q3

1. Panas Sensibel Gas Masuk, Q1

Panas sensibel yang masuk Separator sama dengan panas gas yang keluar Vaporizer.

Q1

2. Panas Sensibel Gas Keluar, Q2

C6H5NO2

C6H4N2O4

C6H6

H2O

C6H5NO2

Cpg dT

C6H4N2O4

Cp dT

18.1480 T + 0.5618 T2 - 3.93E-4 T3 + 1.004E-7 T4 - 1.2252E-12 T5

Q=H=∑▒ 〖 n∫_(T_reff)^T▒ 〖 C_(p ) dT 〗〗

(_∫T reff^) T▒〖 C_p dT= [(Ax(T-T_reff ))+(B/2 x(T^2-T_reff^ ))+²(C/3 x(T^3-T_reff )) +(D/4 x(T^4-T_reff ))+(E/5 x(T^5-³ ⁴T_reff ))] ⁵ 〗

= 0.0645 18.1480 477.60 - 298.15 +0.56182

477.602

- 298.152

2

+###

477.603

- 298.153

+1.00E-07

477.604

- 298.154

3 4

+###

477.605

- 298.155

5

= 2118.1309 kJ/jam

●

Q = n477.60

298.15

= 0.1389477.60

-31.3860 T + 0.4746 T2 - 3.11E-4 T3 + 8.524E-8 T4 - 5.0524E-12 T5dT298.15

= 0.1389 -31.3860 477.60 - 298.15 +0.47460

477.602

- 298.152

2

+###

477.603

- 298.153

+8.52E-08

477.604

- 298.154

3 4

+###

477.605

- 298.155

5

= 5128.1131 kJ/jam

●

Q = n477.60

298.15

= 0.6019477.60

32.243 T + 1.923E-3 T2 + 1.055E-5 T3 - 3.596E-9 T4dT298.15

= 0.6019 32.2430 477.60 - 298.15 +0.00192

477.602

- 298.152

2

+1.06E-05

477.603

- 298.153

+###

477.604

- 298.154

3 4

= 3713.5975 kJ/jam

= 4207855.0855 + 2118.1309 + 5128.1131 + 3713.5975

= 4218814.9270 kJ/jam

Sehingga, panas keluar separator:

= +

Q3= Q1- Q2

= 5275042.7623 - 4218814.9270 = 1056227.8353 kkal/jam

C6H6

Cp dT

H2O

Cp dT

Q2

Q1 Q2 Q3

Tabel Neraca Massa Separator (H-120)

KomponenPanas masuk (kJ/jam) Panas keluar (kJ/jam)


5259818.8569 4207855.0855 1051963.7714

2633.8489 2118.1309 515.7179

6410.1414 5128.1131 1282.0283

6179.9152 3713.5975 2466.3177

Total 5275042.76234218814.9270 1056227.8353

5275042.7623

B.4 Heater (E-114)

Fungsi : Menaikkan suhu uap Nitrobenzen

Tujuan :

Q3 Qsteam

Q1 Q2

Neraca Energi

ΔH = Hout - Hin


KomponenM n

(kg) (kmol)

10757.8399 87.4621

10.83368 0.0806

10.83368 0.7523

10.83368 0.1736

*Hin

Tin = 204 = 477.60 K

Tref = 25 = 298.15 K

●

Q = n477.60

298.15

Q C6H5NO2

Q C6H4N2O4

Q C6H6

Q H2O

Menaikkan suhu menjadi 250oC


C6H5NO2

C6H4N2O4

H2O

C6H6

oCoC

C6H5NO2

Cp dT

= 87.4621477.60

-16.2020 T + 0.5618 T2 - 3.93E-4 T3 - 1.004E-7 T4 - 1.2252E-12 T4 dT298.15

= 87.4621 -16.2020 477.60 - 298.15 +0.5618

477.602

- 298.152

2

+###

477.603

- 298.153

+###

477.604

- 298.154

3 4

+###

477.605

- 298.155

5

= 4207855.0855 kJ/jam

●

Q = n477.60

298.15

= 0.0806477.60

dT298.15

= 0.0806 18.1480 477.60 - 298.15 +0.56182

477.602

- 298.152

2

+###

477.603

- 298.153

+1.00E-07

477.604

- 298.154

3 4

+###

477.605

- 298.155

5

= 2633.8489 kJ/jam

●

Q = n477.60

298.15

= 0.1736477.60

-31.3860 T + 0.4746 T2 - 3.11E-4 T3 + 8.524E-8 T4 - 5.0524E-12 T5dT298.15

= 0.1736 -31.3860 477.60 - 298.15 +0.47460

477.602

- 298.152

2

+###

477.603

- 298.153

+8.52E-08

477.604

- 298.154

3 4

+###

477.605

- 298.155

5

= 6410.1414 kJ/jam

●

Q = n477.60

298.15

= 0.7523477.60

32.243 T + 1.923E-3 T2 + 1.055E-5 T3 - 3.596E-9 T4dT

C6H4N2O4

Cp dT

18.1480 T + 0.5618 T2 - 3.93E-4 T3 + 1.004E-7 T4 - 1.2252E-12 T5

C6H6

Cp dT

H2O

Cp dT

= 0.7523298.15

32.243 T + 1.923E-3 T2 + 1.055E-5 T3 - 3.596E-9 T4dT

= 0.7523 32.2430 477.60 - 298.15 +0.00192

477.602

- 298.152

2

+1.06E-05

477.603

- 298.153

+###

477.604

- 298.154

3 4

= 4641.9969 kJ/jam

Q1=

= 4221541.0726 kJ/jam

*Hout

Tin = 250 = 523.15 K

Tref = 25 = 298.15 K

●

Q = n523.15

298.15

= 87.4621523.15

-16.2020 T + 0.5618 T2 - 3.93E-4 T3 - 1.004E-7 T4 - 1.2252E-12 T4 dT298.15

= 87.4621 -16.2020 523.15 - 298.15 +0.5618

523.152

- 298.152

2

+###

523.153

- 298.153

+###

523.154

- 298.154

3 4

+###

523.155

- 298.155

5

= 5706082.1206 kJ/jam

●

Q = n523.15

298.15

= 0.0806523.15

dT298.15

= 0.0806 18.1480 523.15 - 298.15 +0.56182

523.152

- 298.152

2

+###

523.153

- 298.153

+1.00E-07

523.154

- 298.154

3 4

+###

523.155

- 298.155

5

= 3417.6783 kJ/jam

Q C6H5NO2 + Q C6H4N2O4 + Q C6H6 + Q H2O


oCoC

C6H5NO2

Cp dT

C6H4N2O4

Cp dT

18.1480 T + 0.5618 T2 - 3.93E-4 T3 + 1.004E-7 T4 - 1.2252E-12 T5

●

Q = n523.15

298.15

= 0.1736523.15

-31.3860 T + 0.4746 T2 - 3.11E-4 T3 + 8.524E-8 T4 - 5.0524E-12 T5dT298.15

= 0.1736 -31.3860 523.15 - 298.15 +0.47460

523.152

- 298.152

2

+###

523.153

- 298.153

+8.52E-08

523.154

- 298.154

3 4

+###

523.155

- 298.155

5

= 8744.0538 kJ/jam

●

Q = n523.15

298.15

= 0.7523523.15

32.243 T + 1.923E-3 T2 + 1.055E-5 T3 - 3.596E-9 T4dT298.15

= 0.7523 32.2430 523.15 - 298.15 +0.00192

523.152

- 298.152

2

+1.06E-05

523.153

- 298.153

+###

523.154

- 298.154

3 4

= 5855.0194 kJ/jam

Q2=

= 5724098.8721 kJ/jam


Q3= Q2 - Q1

= 5724098.8721 - 4221541.0726

= 1502557.7995 kJ/jam

1502557.7995

water and saturated steam up to 1 atm, stoichiometry 2004

= 909.99 kJ/kg

Sehingga :

C6H6

Cp dT

H2O

Cp dT



saturated steam dengan suhu 300oC dengan tekanan 1 atm. Dari properties of saturated

λsteam

Q3 = m.λsteam

= Q3= 1502557.7995 kJ/jam= 1651.1806 kg/jam

909.99 kJ/kg


(kJ/jam) (kJ/jam)

4207855.0855 5706082.1206

2633.84886 3417.6783

4641.9969 5855.0194

6410.1414 8744.0538

1502557.7995 -

Total 5724098.8721 5724098.8721

B.5. REAKTOR (R-210)

Fungsi : Tempat berlangsungnya reaksi Anilin

Tujuan : - Menghitung suhu keluar reaktor

- Menghitung kebutuhan pendingin

Arus 6 250

Pendingin Qs

28

Pendingin

Qrx 45

Arus 4 250

Neraca Energi

Q in = Q out

4 + Qs = 6 + Qrx

dimana: 4 = Panas Sensibel Gas Reaktan Masuk, Q reaktan

6 = Panas Sensibel Gas Produk Keluar, Q produk

Qs = Jumlah pendingin yang dibutuhkan

Qrx= Panas reaksi

1. Panas Sensibel Gas Reaktan Masuk, Q reaktan

Panas gas umpan masuk reaktor (R-210) besarnya sama dengan panas gas

5724098.872 kJ/jam 1720030.35 kJ/jam

Q reaktan = 7444129.2218 kJ/jam

2. Panas Reaksi, ΔHR

Reaksi:

-

Massa steam

λsteam


Q C6H5NO2

Q C6H4N2O4

Q H2O

Q C6H6

Q Steam

oC

oC

oC

oC

keluar Heater Nitrobenzen dan Heater Hidrogen

C6H5NO2 + H2 C6H5NH2 + H2O

Panas reaksi pada keadaan standar (T = 25oC)

ΔHro298,15 = ∆Hof (produk) - ∆Hof (reaktan)

=

= 86.86 + ( -241.83 ) - ( 67.60 + 0 ) kJ/mol

= -222.57 kJ/kmol

-

= n ( )

= ( n reaktan) ( )

= 0.98 x 88.4687 x -222.57

= -19296.3169 kJ/jam

Q reaksi adalah panas yang diperlukan untuk reaksi di dalam reaktor, dimana

pendingin untuk menjaga suhu optimum sesuai dengan konversi yang diinginkan.

3. Panas Sensibel Gas Produk Keluar, Q produk

Untuk menghitug panas keluar reaktor digunakan persamaan:

dimana:

Komposisi arus 6:

= 215.1568 kg/jam = 1.7492 kmol/jam

= 10.8337 kg/jam = 0.0645 kmol/jam

= 10.8337 kg/jam = 0.6019 kmol/jam

= 10.8337 kg/jam = 0.1389 kmol/jam

= 10.4955 kg/jam = 5.2477 kmol/jam

= 0.0052 kg/jam = 0.0003 kmol/jam

= 7971.2970 kg/jam = 85.7129 kmol/jam

= 3085.6633 kg/jam = 171.4257 kmol/jam

●

Q = n523.15

298.15

= 1.7492523.15

-16.2020 T + 0.5618 T2 - 3.93E-4 T3 - 1.004E-7 T4 - 1.2252E-12 T4 dT298.15

= 1.7492 -16.2020 523.15 - 298.15 +0.5618

523.152

- 298.152

2

+###

523.153

- 298.153

+###

523.154

- 298.154

3 4

+###

523.155

- 298.155

5

= 114121.6424 kJ/jam

(∆H of C6H5NH2 + ∆H o

f H2O) - (∆H of C6H5NO2 + H2)

Panas reaksi, ΔHR

ΔHRo523.15 ΔHro298,15

XA ΔHro298,15

reaksi berupa reaksi eksotermis (ΔHR negatif). Sehingga, perlu ditambahkan

C6H5NO2

C6H4N2O4

H2O

C6H6

H2

CH4

C6H5NH2

H2O

C6H5NO2

Cp dT

Q=H=∑▒ 〖 n∫_(T_reff)^T▒ 〖 Cp_g dT 〗〗(_∫T reff^) T▒〖 Cp_g dT= [(Ax(T-T_reff ))+(B/2 x(T^2-T_reff^ ))+²(C/3 x(T^3-T_reff )) +(D/4 x(T^4-T_reff ))+(E/5 x(T^5-T_reff ))] ³ ⁴ ⁵ 〗

●

Q = n523.15

298.15

= 0.0645523.15

dT298.15

= 0.0645 18.1480 523.15 - 298.15 +0.56182

523.152

- 298.152

2

+###

523.153

- 298.153

+1.00E-07

523.154

- 298.154

3 4

+###

523.155

- 298.155

5

= 2734.1427 kJ/jam

●

Q = n523.15

298.15

= 0.1389523.15

-31.3860 T + 0.4746 T2 - 3.11E-4 T3 + 8.524E-8 T4 - 5.0524E-12 T5dT298.15

= 0.1389 -31.3860 523.15 - 298.15 +0.47460

523.152

- 298.152

2

+###

523.153

- 298.153

+8.52E-08

523.154

- 298.154

3 4

###523.15

5- 298.15

55

= 94198.1462 kJ/jam

●

Q = n523.15

298.15

= 0.6019523.15

32.243 T + 1.923E-3 T2 + 1.055E-5 T3 - 3.596E-9 T4dT298.15

= 0.6019 32.2430 523.15 - 298.15 +0.00192

523.152

- 298.152

2

+1.06E-05

523.153

- 298.153

+###

523.154

- 298.154

3 4

= 4684.0155 kJ/jam

●

Q = n523.15

C6H4N2O4

Cp dT

18.1480 T + 0.5618 T2 - 3.93E-4 T3 + 1.004E-7 T4 - 1.2252E-12 T5

C6H6

Cp dT

H2O

Cp dT

H2

Cp dT

Q = n298.15

= 5.2477523.15

dT298.15

= 5.2477 27.1430 523.15 - 298.15 +9.27E-03

523.152

- 298.152

2

+-1.3800E-05

523.153

- 298.153

+7.6450E-09

3 4

523.154

- 298.154

= 34400.5120 kJ/jam

●

Q = n523.15

298.15

= 0.000328523.15

34.942 T-3.9957E-02 T2 - 1.9184E-04 T3-1.53E-07 T4+3.9321E-11 T5dT298.15

= 0.000328 34.9420 523.15 - 298.15 +###

523.152

- 298.152

2

+-1.9184E-04

523.153

- 298.153

+-1.5300E-07

3 4

523.154

- 298.154

+3.9321E-11

523.155

- 298.155

5

= 4.7504 kJ/jam

●

Q = n523.15

298.15

= 85.7129523.15

-22.0620 T + 0.5731 T2 - 4.57e-4 T3 - 1.841e-7 T4 - 2.9867e-11 T5dT298.15

= 85.7129 -22.0620 523.15 - 298.15 +0.57313

523.152

- 298.152

2

+###

523.153

- 298.153

+###

523.154

- 298.154

3 4

###523.15

5- 298.15

55

= 5918417.6087 kJ/jam

●

Q = n523.15

298.15

= 171.4257523.15

32.243 T + 1.923E-3 T2 + 1.055E-5 T3 - 3.596E-9 T4dT298.15

Cp dT

27.1430T + 9.2730E-03 T2 - 1.3800E-05 T3 + 7.6450E-09 T4

CH4

Cp dT

C6H5NH2

Cp dT

H2O

Cp dT

= 171.4257 32.2430 523.15 - 298.15 +0.00192

523.152

- 298.152

2

+1.06E-05

523.153

- 298.153

+###

523.154

- 298.154

3 4

= 1334108.1608 kJ/jam

Jadi:

Q produk = ### + 2734.1427 + 94198.1462 + 4684.0155 +

34400.5120 + 4.7504 + 5918417.6087 + 1334108.1608

= 7502668.9787 kJ/jam

4. Pendingin yang dibutuhkan oleh Pemanas

Qlepas= (Q produk + Q reaksi ) - Q reaktan

= 7502668.9787 + -19296.3169 - 7444129.2218

= 39243.4401 kJ/jam

Dari App A.2 Geankoplis, 2003 diperoleh Cp air = 0.9987

= 4.1787

Q = m . Cp. ΔT

=Cp. ΔT

=39243.4401

= 469.5695 kg/jam4.1787 ( 45 - 25 )

Tabel Neraca Panas Reaktor (R-210)


5706082.1206 114121.6424

3417.6783 2734.1427

5855.0194 94198.1462

8744.0538 4684.0155

1720025.5993 34400.5120

4.7504 4.7504

- 5918417.6087

- 1334108.1608

39243.4401 -

Q reaksi - -19296.3169

Total 7483372.6618 7483372.6618

B.6.

Fungsi : Mendinginkan gas produk keluar reaktor

Sebagai pendingin digunakan air pada suhu 25oC dan tekanan 1 atm.

Diperkirakan air keluar pada suhu 45oC.

kkal/kg.oC

kJ/kgoC

mc

Qc

Q C6H5NO2

Q C6H4N2O4

Q H2O

Q C6H6

Q H2

Q CH4

Q C6H5NH2

Q H2O

Q pendingin

COOLER 1 (E-221)

Tujuan : Menghitung kebutuhan pendingin

28

250 139.53

Qc 45

Neraca Energi

Q in = Q out

= +

dimana: = Panas sensibel gas keluar reaktor

= Panas yang diserap oleh pendingin

= Panas sensibel cairan keluar cooler 1


dimana:

= 3358498.9810 kJ/jam

Komposisi arus 6:

= 215.1568 kg/jam = 1.7492 kmol/jam

= 10.8337 kg/jam = 0.0645 kmol/jam

= 10.8337 kg/jam = 0.6019 kmol/jam

= 10.8337 kg/jam = 0.1389 kmol/jam

= 10.4955 kg/jam = 5.2477 kmol/jam

= 0.0052 kg/jam = 0.0003 kmol/jam

= 7971.2970 kg/jam = 85.7129 kmol/jam

= 3085.6633 kg/jam = 171.4257 kmol/jam

●

Q = n412.67

298.15

= 1.7492412.67

-16.2020 T + 0.5618 T2 - 3.93E-4 T3 - 1.004E-7 T4 - 1.2252E-12 T4 dT298.15

= 1.7492 -16.2020 412.67 - 298.15 +0.5618

412.672

- 298.152

oC

Q1 Q2

oC oC

oC

Q1 Q2 Q3

Q1

Q3

Q2


Panas sensibel yang masuk Cooler sama dengan panas sensibel yang keluar Condensor

Q1

2. Panas Sensibel gas keluar, Q2

C6H5NO2

C6H4N2O4

H2O

C6H6

H2

CH4

C6H5NH2

H2O

C6H5NO2

Cp dT

Q=H=∑▒ 〖 n∫_(T_reff)^T▒ 〖 Cp_g dT 〗〗(_∫T reff^) T▒〖 Cp_g dT= [(Ax(T-T_reff ))+(B/2 x(T^2-T_reff^ ))+(C/3 x(T^3-T_reff )) +(D/4 x(T^4-T_reff ))+² ³ ⁴(E/5 x(T^5-T_reff ))] ⁵ 〗

= 1.7492 -16.2020 412.67 - 298.15 +2

412.67 - 298.15

+###

412.673

- 298.153

+###

412.674

- 298.154

3 4

+###

412.675

- 298.155

5

= 47718.2680 kJ/jam

●

Q = n412.67

298.15

= 0.0645412.67

dT298.15

= 0.0645 18.1480 412.67 - 298.15 +0.56182

412.672

- 298.152

2

+###

412.673

- 298.153

+1.00E-07

412.674

- 298.154

3 4

+###

412.675

- 298.155

5

= 1273.1909 kJ/jam

●

Q = n412.67

298.15

= 0.1389412.67

-31.3860 T + 0.4746 T2 - 3.11E-4 T3 + 8.524E-8 T4 - 5.0524E-12 T5dT298.15

= 0.1389 -31.3860 412.67 - 298.15 +0.47460

412.672

- 298.152

2

+###

412.673

- 298.153

+8.52E-08

412.674

- 298.154

3 4

+###

412.675

- 298.155

5

= 2878.7365 kJ/jam

●

Q = n412.67

Cp dT298.15

= 0.6019412.67

32.243 T + 1.923E-3 T2 + 1.055E-5 T3 - 3.596E-9 T4dT298.15

= 0.6019 32.2430 412.67 - 298.15 +0.00192

412.672

- 298.152

2

+1.06E-05

412.673

- 298.153

+###

412.674

- 298.154

3 4

C6H4N2O4

Cp dT

18.1480 T + 0.5618 T2 - 3.93E-4 T3 + 1.004E-7 T4 - 1.2252E-12 T5

C6H6

Cp dT

H2O

= 2350.8332 kJ/jam

●

Q = n412.67

298.15

= 5.2477412.67

27.1430T + 9.2730E-03 T2 - 1.3800E-05 T3 + 7.6450E-09 T4 dT298.15

= 5.2477 27.1430 412.67 - 298.15 +0.0093

412.672

- 298.152

2

+###

412.673

- 298.153

+7.65E-09

412.674

- 298.154

3 4

= 17448.4735 kJ/jam

●

Q = n412.67

298.15

= 0.0003412.67

34.942 T-3.9957E-02 T2 - 1.9184E-04 T3-1.53E-07 T4+3.9321E-11 T5dT298.15

= 0.0003 34.9420 412.67 - 298.15 +-0.03996

412.672

- 298.152

2

+###

412.673

- 298.153

+###

412.674

- 298.154

3 4

+3.93E-11

412.675

- 298.155

5

= 1.9867 kJ/jam

●

Q = n412.67

298.15

= 85.7129412.67

-31.3860 T + 0.4746 T2 - 3.11E-4 T3 + 8.524E-8 T4 - 5.0524E-12 T5dT298.15

= 85.7129 -22.0620 412.67 - 298.15 +0.57313

412.672

- 298.152

2

+###

412.673

- 298.153

+###

412.674

- 298.154

3 4

+###

412.675

- 298.155

5

= 1123864.9289 kJ/jam

H2

Cp dT

CH4

Cp dT

C6H5NH2

Cp dT

●

Q = n412.67

298.15

= 171.4257412.67

32.243 T + 1.923E-3 T2 + 1.055E-5 T3 - 3.596E-9 T4dT298.15

= 171.4257 32.2430 412.67 - 298.15 +0.00192

412.672

- 298.152

2

+1.06E-05

412.673

- 298.153

+###

412.674

- 298.154

3 4

= ### kJ/jam

Jadi:

= 47718.2680 + 1273.1909 + 2878.7365 + 2350.8332 + 17448.4735

+ 1.9867 + 1123864.9289 + 669567.7648

= 1865104.1826 kJ/jam

= -

= 3358498.9810 - 1865104.1826 = 1493394.7984 kJ/jam


= 4.1787

Q = m . Cp. ΔT

=Cp. ΔT

=1493394.7984

= 17869.2979 kg/jam4.178661 ( 45 - 25 )

(E-221)


(kJ/jam) (kJ/jam)

54650.8610 47718.2680

1506.9847 1273.1909

3393.9394 671918.5980

871986.3951 2878.7365

19608.8722 17448.4735

783.9192 1.9867

2406568.0094 1123864.9289

Q pendingin - 1493394.7984

Total 3358498.9810 3358498.9810

H2O

Cp dT

Q2

4. Panas yang diserap oleh Pendingin, Qc

Qc Q1 Q2



kkal/kg.oC

kJ/kgoC

mc

Qc

Tabel Neraca Panas Cooler 1

Q C6H5NO2

Q C6H4N2O4

Q H2O

Q C6H6

Q H2

Q CH4

Q C6H5NH2

B.7.

Fungsi: Menguapkan sebagian besar Nitrobenzen dalam campuran produk

keluaran reaktor

Arus 7

139.5

Arus 6

139.5

Arus 8

139.5

Neraca Energi

Q in = Q out

6 = 7 + 8

dimana: 6 =

7 =

8 =


dimana:

a.

Komposisi arus 7:

= 210.8537 kg/jam = 1.7143 kmol/jam

= 10.7253 kg/jam = 0.0638 kmol/jam

= 3034.5671 kg/jam = 168.5871 kmol/jam

= 10.6170 kg/jam = 0.1361 kmol/jam

= 10.4955 kg/jam = 5.2477 kmol/jam

= 0.0052 kg/jam = 0.0003 kmol/jam

= 0.7971 kg/jam = 0.0086 kmol/jam

FLASH TANK (H-220)

oC

oC

oC

Panas sensibel gas masuk flash tank, Q1

Panas sensibel gas keluar, Q2 Q2

Panas sensibel cairan keluar flash tank, Q3

Panas Sensibel gas keluar, Q2

C6H5NO2

C6H4N2O4

H2O

C6H6

H2

CH4

C6H5NH2

Q=H=∑▒ 〖 n∫_(T_reff)^T▒ 〖 C_pl dT 〗〗 (_∫T reff^) T▒〖 Cp_g dT= [(Ax(T-T_reff ))+(B/2 x(T^2-T_reff^ ))+(C/3 x(T^3-T_reff )) +(D/4 x(T^4-T_reff ))+² ³ ⁴(E/5 x(T^5-T_reff ))] ⁵ 〗

Flash Tank

●

Q = n412.67

298.15

= 1.7143412.67

-16.2020 T + 0.5618 T2 - 3.93E-4 T3 - 1.004E-7 T4 - 1.2252E-12 T4 dT298.15

= 1.7143 -16.2020 412.67 - 298.15 +0.5618

412.672

- 298.152

2

+-3.9E-04

412.673

- 298.153

+-1.0E-07

412.674

- 298.154

3 4

+-1.2E-12

412.675

- 298.155

5

= 46763.9027 kJ/jam

●

Q = n412.67

298.15

= 0.0638412.67

dT298.15

= 0.0638 18.1480 412.67 - 298.15 +0.56182

412.672

- 298.152

2

+-3.9E-04

412.673

- 298.153

+1.00E-07

412.674

- 298.154

3 4

+-1.2E-12

412.675

- 298.155

5

= 1260.4590 kJ/jam

●

Q = n412.67

298.15

= 0.1361412.67

-31.3860 T + 0.4746 T2 - 3.11E-4 T3 + 8.524E-8 T4 - 5.0524E-12 T5dT298.15

= 0.1361 -31.3860 412.67 - 298.15 +0.47460

412.672

- 298.152

2

+-3.1E-04

412.673

- 298.153

+8.52E-08

412.674

- 298.154

3 4

+-5.1E-12

412.675

- 298.155

5

= 2821.1618 kJ/jam

●

Q = n412.67

C6H5NO2

Cp dT

C6H4NO2

Cp dT

18.1480 T + 0.5618 T2 - 3.93E-4 T3 + 1.004E-7 T4 - 1.2252E-12 T5

C6H6

Cp dT

H2O

Cp dT

Q = n298.15

= 1.7E+02412.67

32.243 T + 1.923E-3 T2 + 1.055E-5 T3 - 3.596E-9 T4dT298.15

= 1.7E+02 32.2430 412.67 - 298.15 +0.00192

412.672

- 298.152

2

+1.1E-05

412.673

- 298.153

+-3.6E-09

412.674

- 298.154

3 4

= 658480.2261 kJ/jam

●

Q = n412.67

298.15

= 5.2477412.67

27.1430T + 9.2730E-03 T2 - 1.3800E-05 T3 + 7.6450E-09 T4 dT298.15

= 5.2477 27.1430 412.67 - 298.15 +0.0093

412.672

- 298.152

2

+-1.4E-05

412.673

- 298.153

+7.6E-09

412.674

- 298.154

3 4

= 17448.4735 kJ/jam

●

Q = n412.67

298.15

= 0.0003412.67

34.942 T-3.9957E-02 T2 - 1.9184E-04 T3-1.53E-07 T4+3.9321E-11 T5dT298.15

= 0.0003 34.9420 412.67 - 298.15 +-0.03996

412.672

- 298.152

2

+-1.9E-04

412.673

- 298.153

+-1.5E-07

412.674

- 298.154

3 4

+3.9E-11

412.675

- 298.155

5

= 1.9867 kJ/jam

●

Q = n412.67

298.15

= 0.0086412.67

-22.0620 T + 0.5731 T2 - 4.57E-4 T3 - 1.84E-7 T4 - 2.99E-11 T5 dT298.15

= 0.0086 -22.0620 412.67 - 298.15 +0.5731

412.672

- 298.152

2

+-4.6E-04

412.673

- 298.153

+-1.8E-07

412.674

- 298.154

3 4

Cp dT

H2

Cp dT

CH4

Cp dT

C6H5NH2

Cp dT

+-3.0E-11

412.675

- 298.155

5

= 244.2101 kJ/jam

Jadi:

= 46763.9027 + 1260.4590 + 2821.1618 + 658480.2261 +

17448.4735 1.9867 + 244.2101

= 727020.4199 kJ/jam

b.

Komposisi arus 8:

= 4.3031 kg/jam = 0.0350 kmol/jam

= 0.1083 kg/jam = 0.0006 kmol/jam

= 61.9299 kg/jam = 3.4406 kmol/jam

= 0.2167 kg/jam = 0.0028 kmol/jam

= 0.0000 kg/jam = 0.0000 kmol/jam

= 7970.500 kg/jam = 85.7043 kmol/jam

●

Q = n412.67

298.15

= 0.0350412.67

39.4730 T + 0.9128 T2 - 0.00211 T3 + 2.01E-6 T4dT298.15

= 0.0350 39.4730 412.67 - 298.15 +0.9128

412.672

- 298.152

2

+-0.0021

412.673

- 298.153

+2.0E-06

412.674

- 298.154

3 4

= 751.7266 kJ/jam

●

Q = n412.67

298.15

= 0.0006412.67

-12.6350 T + 1.56240 T2 - 3.00E-3 T3 + 2.32E-6 T4 dT298.15

= 0.0006 -12.6350 412.67 - 298.15 +1.56240

412.672

- 298.152

2

+-3.0E-03

412.673

- 298.153

+2.3E-06

412.674

- 298.154

3 4

= 76.1702 kJ/jam

Q2

Panas Sensibel Cairan Keluar, Q3

C6H5NO2

C6H4N2O4

H2O

C6H6

H2

C6H5NH2

C6H5NO2

Cp dT

C6H4N2O4

Cp dT

●

Q = n412.67

298.15

= 0.0028412.67

-33.6620 T + 0.47430 T2 - 3.61E-3 T3 + 3.82E-6 T4 dT298.15

= 0.0028 -33.6620 412.67 - 298.15 +0.47430

412.672

- 298.152

2

+-3.6E-03

412.673

- 298.153

+3.8E-06

412.674

- 298.154

3 4

= 245.1970 kJ/jam

●

Q = n412.67

298.15

= 3.4406412.67

18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT298.15

= 3.4406 18.2964 412.67 - 298.15 +0.47212

412.672

- 298.152

2

+-1.3E-03

412.673

- 298.153

+1.3E-06

412.674

- 298.154

3 4

= 29965.3066 kJ/jam

●

Q = n412.67

298.15

= 0.0E+00412.67

28.8400 T + 0.00765 T2 + 3.29E-1 T3 - 0.8698 T4 dT298.15

= 0.0E+00 28.8400 412.67 - 298.15 +0.00765

412.672

- 298.152

2

+3.3E-01

412.673

- 298.153

+-8.7E-10

412.674

- 298.154

3 4

= 0.0000 kJ/jam

●

Q = n412.67

298.15

= 85.70412.67

46.9480 T + 0.9896 T2 - 2.3583E-3 T3 + 2.3296E-6 T4 dT298.15

= 85.70 46.9480 412.67 - 298.15 +0.9896

412.672

- 298.152

2

+-2.4E-03

412.673

- 298.153

+2.3E-06

412.674

- 298.154

3 4

C6H6

Cp dT

H2O

Cpg dT

H2

Cpg dT

C6H5NH2

Cpg dT

= 2016973.0498 kJ/jam

Jadi:

= 751.7266 + 76.1702 + 245.1970 + 29965.3066 +

0.0000 + 2016973.0498

= 2048011.4502 kJ/jam

Q in = Q out

= +

= 727020.4199 + 2048011.4502 = 2775031.8700 kJ/jam

(H-220)

KomponenQ in (kJ/jam) Q out (kJ/jam)


47515.6293 46763.9027 751.7266

1336.6292 1260.4590 76.1702

688445.5326 658480.2261 29965.3066

3066.3588 2821.1618 245.1970

17448.4735 17448.4735 0.0000

1.9867 1.9867 -

2017217.2599 244.2101 2016973.0498

Total 2775031.8700727020.4199 2048011.4502

2775031.8700

B.4 Heater (E-114)


Tujuan :

Q3 Qsteam

Q1 Q2

Neraca Energi

ΔH = Hout - Hin


KomponenM n

(kg) (kmol)

4.30313 0.0350

Q3

Sehingga, panas yang masuk flash tank:

Q1 Q2 Q3

Tabel Neraca Panas Flash Tank

Q C6H5NO2

Q C6H4NO2

Q H2O

Q C6H6

Q H2

Q CH4

Q C6H5NH2



C6H5NO2

0.10834 0.0006

61.92992 3.4406

0.21667 0.0028

7970.497 85.7043

*Hin

Tin = 139.5 = 412.67 K

Tref = 25 = 298.15 K

●

Q = n412.67

298.15

= 3.4406412.67

18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT298.15

= 3.4406 18.2964 412.67 - 298.15 +0.47212

412.672

- 298.152

2

+###

412.673

- 298.153

+1.31E-06

412.674

- 298.154

3 4

= 29965.2966 kJ/jam

●

Q = n412.67

298.15

= 3E-03412.7

-33.6620 T + 0.47430 T2 - 3.61E-3 T3 + 3.82E-6 T4 dT298.15

= 3E-03 -33.6620 412.67 - 298.15 +0.47430

412.672

- 298.152

2

+###

412.673

- 298.153

+3.82E-06

412.674

- 298.154

3 4

= 245.1970 kJ/jam

●

Q = n412.67

298.15

= 85.70412.67

46.9480 T + 0.9896 T2 - 2.3583E-3 T3 + 2.3296E-6 T4 dT298.15

= 85.70 46.9480 412.67 - 298.15 +0.9896

412.672

- 298.152

2

+###

412.673

- 298.153

+2.33E-06

412.674

- 298.154

3 4

= 2016972.3765 kJ/jam

●

Q = n412.67

C6H4N2O4

H2O

C6H6

C6H5NH2

oCoC

H2O

Cp dT

C6H6

Cp dT

C6H5NH2

Cp dT

C6H5NO2

Cp dT

Q = n298.15

= 3E-02 39.47 412.67 - 298.15 +0.9128

412.672

- 298.152

2

+-0.0021

412.673

- 298.153

+2.01E-06

412.674

- 298.154

3 4

= 751.7263 kJ/jam

●

Q = n412.67

298.15

= 6E-04 -12.6 412.67 - 298.15 +1.56240

412.672

- 298.152

2

+-3E-03

412.673

- 298.153

+2.32E-06

412.674

- 298.154

3 4

= 19.7480 kJ/jam

Q1=

= 2047954.3443 kJ/jam

*Hout

Tin = ### = 444.80 K

Tref = 25 = 298.15 K

●

Q = n444.80

298.15

= 3.4406444.80

18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT298.15

= 3.4406 18.2964 444.80 - 298.15 +0.47212

444.802

- 298.152

2

+###

444.803

- 298.153

+1.31E-06

444.804

- 298.154

3 4

= 38610.6875 kJ/jam

●

Q = n444.80

298.15

= 3E-03444.8

-33.6620 T + 0.47430 T2 - 3.61E-3 T3 + 3.82E-6 T4 dT298.15

= 3E-03 -33.6620 444.80 - 298.15 +0.47430

444.802

- 298.152

2

Cp dT

C6H4NO2

Cp dT

Q C6H5NO2 + Q C6H4N2O4 + Q C6H6 + Q H2O + Q C6H5NH2


oCoC

H2O

Cp dT

C6H6

Cp dT

+###

444.803

- 298.153

+3.82E-06

444.804

- 298.154

3 4

= 346.4803 kJ/jam

●

Q = n444.80

298.15

= 85.70444.80

46.9480 T + 0.9896 T2 - 2.3583E-3 T3 + 2.3296E-6 T4 dT298.15

= 85.70 46.9480 444.80 - 298.15 +0.9896

444.802

- 298.152

2

+###

444.803

- 298.153

+2.33E-06

444.804

- 298.154

3 4

= 2626484.8134 kJ/jam

●

Q = n444.80

298.15

= 0.035 39.47 444.80 - 298.15 +0.9128

444.802

- 298.152

2

--0.0021

444.803

- 298.153

+2.01E-06

444.804

- 298.154

3 4

= 978.0620 kJ/jam

●

Q = n444.80

298.15

= 0.00064 -12.6 444.80 - 298.15 +1.56240

444.802

- 298.152

2

+-3E-03

444.803

- 298.153

+2.32E-06

444.804

- 298.154

3 4

= 104.9953 kJ/jam

Q2=

= 2666525.0385 kJ/jam


Q3= Q2 - Q1

= 2666525.0385 - 2047954.3443

= 618570.6942 kJ/jam

C6H5NH2

Cp dT

C6H5NO2

Cp dT

C6H4N2O4

Cp dT


618570.6942

= 909.99 kJ/kg

Sehingga :

= Q3= 618570.6942 kJ/jam= 679.7555 kg/jam

909.99 kJ/kg


(kJ/jam) (kJ/jam)

751.7263 978.0620

19.74801 104.9953

2016972.3765 2626484.8134

29965.2966 38610.6875

245.1970 346.4803

618570.6942 -

Total 2666525.0385 2666525.0385

B.13. MENARA DISTILASI 1 (D-310)

Fungsi : Memisahkan produk Anilin dan air

Arus 16 138.9

Qc

Arus 15

170.11 Arus 18Arus 17 ###

Qr Arus 21

184.2

Neraca Energi




λsteam

Q3 = m.λsteam

Massa steam

λsteam


Q C6H5NO2

Q C6H4N2O4

Q C6H5NH2

Q H2O

Q C6H6

Q Steam

oC

oCoC

oCMD 2

Q in = Q out

15 + Qr = 18 + 21

16 = 17 + 18 + Qc

dimana: 15 =

16 =

=

17 =

18 =

21 =

Qc = Panas yang diserap oleh pendingin pada kondensor

Qr = Panas yang dibutuhkan oleh pemanas pada reboiler

Menentukan Kondisi Operasi Menara Distilasi

a. Kondisi Operasi Umpan

Kondisi operasi umpan: P = 1 atm = 760.0 mmHg

Fasa = Cair Jenuh

T = 171.65 = 444.80 K

Komponen (lbmol/jam)xi Pi ki yi α

Pi/P ki.xi Ki/KHK

0.0771 0.0004 280.9083 0.3696 0.0001 0.5182

0.0014 0.00001 6.0081 0.0079 0.0000 0.0111

7.5851 0.0386 6189.8280 8.1445 0.3142 11.4194

0.00612 0.00003 6602.6352 8.6877 0.0003 12.1810

188.9453 0.9610 542.0446 0.7132 0.6854 1.0000

Total 196.6151 1.0000 1.00

b. Kondisi Operasi Puncak Menara

● Dew Point

Kondisi operasi: P = 1 atm = 760 mmHg

T = 138.938 = 412.09 K

Komponen (lbmol/jam)xi Pi ki yi α

Pi/P xi/ki Ki/KHK

C6H5NO2 0.0008 0.0001 100.1167 0.1317 0.0006 0.5029

C6H4NO2 1E-05 2E-06 1.1184 0.0015 0.0010 0.0056

7.5093 0.7989 2630.6402 3.4614 0.2308 13.2152

6E-05 7E-06 3460.2298 4.5529 1E-06 17.3827

Panas sensibel cairan umpan masuk MD 2 (D-340), Q1

Panas laten penguapan, Q2

Panas sensibel gas masuk kondensor, Q3

Panas sensibel cairan reflux, Q4

Panas sensibel cairan produk distilat, Q5

Panas sensibel cairan produk bawah, Q6

oC

Trial Kondisi Operasi Umpan

C6H5NO2

C6H4NO2

H2O

C6H6

C6H5NH2

oC

Trial Kondisi Operasi Dew Point Puncak Menara

H2O

C6H6

1.8895 0.2010 199.0620 0.2619 0.7675 1.0000

Total 9.3996 1 6.290.E+03 1.00

● Bubble Point


T = 105.98 = 379.13 K

Komponen (Kmol/jam)xi Pi ki yi α

Pi/P ki.xi Ki/KHK

C6H5NO2 8E-04 0.0001 27.8877 0.0367 3E-06 0.4774

C6H4NO2 1E-05 2E-06 0.1439 0.0002 3E-10 0.0025

7.5093 0.7989 937.0218 1.2329 0.9850 16.0409

6E-05 7E-06 1583.1294 2.0831 0.0000 27.1016

1.8895 0.2010 58.4146 0.0769 0.0155 1.0000

Total 9.3996 1 2.579.E+03 1.00

c. Kondisi Operasi Bawah Menara

● Bubble Point

Kondisi Operasi: P = 1 atm = 760.0 mmHg

T = 184.02 = 457.17 K

Komponen (lbmol/jam)yi Pi ki yi α

Pi/P ki.xi Ki/KHK

0.0759 0.0004 8256.8840 10.8643 0.0044 10.9032

C6H6 0.0061 3E-05 8204.1008 10.7949 0.0003 10.8335

0.0764 0.0004 303.8360 0.3998 0.0002 0.4012

0.0014 8E-06 10.5190 0.0138 0.0000 0.0139

187.0559 0.9991 757.2896 0.9964 0.9956 1.0000

Total 187.2156 1 1.753.E+04 1.00

● Dew Point

Kondisi Operasi: P = 1 atm = 760.0 mmHg

T = 184.18 = 457.33 K

Komponen (lbmol/jam)yi Pi ki xi α

Pi/P yi/ki Ki/KHK

0.0759 0.0004 8.29.E+03 10.9028 0.0000 10.8971

C6H6 0.0061 3E-05 8225.9791 10.8237 0.0000 10.8180

C6H5NH2

oC

Trial Kondisi Operasi Bubble Point Puncak Menara

H2O

C6H6

C6H5NH2

oC

Trial Kondisi Operasi Bubble Point Bawah Menara

H2O

C6H5NO2

C6H4NO2

C6H5NH2

oC

Kondisi Operasi Dew Point Bawah Menara

H2O

0.0764 0.0004 396.7234 0.5220 0.0008 0.5217

0.0014 8E-06 10.5913 0.0139 0.0005 0.0139

187.0559 0.9991 760.3961 1.0005 0.9986 1.0000

Total 187.2156 1 1.77.E+04 1.00

*Condensor

*Penentuan Harga q

Karena umpan masuk pada keadaan titik didih nya (bubble point)

*Menentukan Kebutuhan Reflux Minimum

Dari persamaan Underwood 9.165 :

umpan masuk menara pada keadaan bubble point (titik didih) sehingga q=1

(cair jenuh)

(1-q) = Σ((αF*XF)/(αF-θ))

Komponen αatas αbawah α rata-rata

0.4774 0.4012 0.4377

0.0025 0.0139 0.0058

16.0409 10.9032 13.2249

27.1016 10.8335 17.1349

1.0000 1.0000 1.0000

Total 44.6223 23.1518 31.8033

Trial θ

θ Hasil

2 -0.91561

8.985 0.0000

Perhitungan Reflux minimum (Rmin)

Rmin +1= Σ((αD*XD)/(αD-θ))

= ((13,2221*0,7975)/(13,2221-8,454))+((17,4043*0,0007)/(17,4043-8,454))+

((1315,6591*0,0012)/(1315,6591-8,454))+((1*0,2007)/(1-8,454))

= 2.4706

R min = 2.4706 - 1

= 1.4706

Jadi nilai refluks minimum sebesar

Rm = 1.2 Rm - 1.5 Rm

diambil R operasi = 1.3 = 1.9118

C6H5NO2

C6H4NO2

C6H5NH2

maka harga q = 1

C6H5NO2

C6H4NO2

H2O

C6H6

C6H5NH2

1Rx

mi

idi

menghitung Lo dan V

Lo = R x D V = Lo + D

= 1.9118 x 9.3996 = 17.9699 x 9.3996

= 17.9699 lbmol/jam = 168.9090 lbmol/jam

L' = Lo x F V' = V

= 17.9699 + 196.6151 = 168.9090 lbmol/jam

= 214.5850 lbmol/jam

a. Komposisi Cairan Reflux, Lo

● = yid x Lo

= 0.7989 x 17.9699 = 14.3560 lbmol/jam

= 258.4089 lb/jam

● = yid x Lo

= 0.2010 x 17.9699 = 3.6122 lbmol/jam

= 335.9364 lb/jam

● = yid x Lo

= 0.0001 x 17.9699 = 0.0015 lbmol/jam

= 0.1814 lb/jam

● = yid x Lo

= 2E-06 x 17.9699 = 3E-05 lbmol/jam

= 0.0046 lb/jam

● = yid x Lo

= 7E-06 x 17.9699 = 0.0001 lbmol/jam

= 0.0091 lb/jam

Jadi:

Lo = 594.5403 lb/jam

b. Komposisi Uap Masuk Kondensor, V

● = yid x V

= 0.7989 x 168.9090 = 134.9405 lbmol/jam

= 2428.9287 lb/jam

● = yid x V

= 0.2010 x 168.9090 = 33.9533 lbmol/jam

= 3157.6535 lb/jam

H2O

C6H5NH2

C6H5NO2

C6H4N2O4

C6H6

H2O

C6H5NH2

● = yid x V

= 0.0001 x 168.9090 = 0.0139 lbmol/jam

= 1.7048 lb/jam

● = yid x V

= 2E-06 x 168.9090 = 3E-04 lbmol/jam

= 0.0429 lb/jam

● = yid x V

= 7E-06 x 168.9090 = 0.0011 lbmol/jam

= 0.0858 lb/jam

Jadi:

V = 5588.4157 lb/jam

c. Komposisi Cairan Masuk Reboiler, L' (Arus 19)

● = xib x L'

= 0.0004 x 214.5850 = 0.0869 lbmol/jam

= 1.5649 lb/jam

● = xib x L'

= 0.0004 x 214.5850 = 0.0875 lbmol/jam

= 10.7649 lb/jam

● = xib x L'

= 8E-06 x 214.5850 = 0.0016 lbmol/jam

= 0.2710 lb/jam

● C6H6 = xib x L'

= 3E-05 x 214.5850 = 0.0069 lbmol/jam

= 0.5420 lb/jam

● = xib x L'

= 0.9991 x 214.5850 = 214.4020 lbmol/jam

= 19939.3855 lb/jam

Jadi:

L' = 19952.5284 lb/jam

d. Komposisi Uap yang dikembalikan ke Menara, V'

● = xib x V'

= 0.0004 x 168.9090 = 0.0684 lbmol/jam

= 1.2318 lb/jam

● = xib x V'

= 0.0004 x 168.9090 = 0.0689 lbmol/jam

C6H5NO2

C6H4N2O4

C6H6

H2O

C6H5NO2

C6H4NO2

C6H5NH2

H2O

C6H5NO2

= 8.4735 lb/jam

● = xib x V'

= 8E-06 x 168.9090 = 0.0013 lbmol/jam

= 0.2133 lb/jam

● C6H6 = xib x V'

= 3E-05 x 168.9090 = 0.0055 lbmol/jam

= 0.42666335 lb/jam

● = xib x V'

= 0.9991 x 168.9090 = 168.7649 lbmol/jam

= 15695.1355 lb/jam

Jadi:

V' = 15705.0541 lb/jam

Perhitungan Neraca Panas pada Kondensor

a. Panas Laten Penguapan (ΔHv)



1 -


=

= (K)

= (K)


= Suhu tertentu

= 138.94 = 412.09 K

●

= x1 - 0.38

1 -

= 30.75 x1 - 412.09 562.16 0.38

1 - 353.24 562.16

= 27.1173 kJ/kmol


●

ΔH2 = x1 - 0.38

1 -

= 86.92 x1 - 412.09 972.15 0.38

1 - 457.25 972.15

= 89.7418 kJ/kmol

C6H4NO2

C6H5NH2

ΔH2 Tr2

ΔH1 Tr1

ΔH1


Tr2 T2/TC

Tr1 T1/TC

T1

T2

oC

H2O

ΔH2 ΔH1

Tr2

Tr1

ΔH2

C6H5NH2

ΔH1

Tr2

Tr1

ΔH2


● C6H5NO2

= x1 - 0.38

1 -

= 44.08 x1 - 412.09 719 0.38

1 - 483.95 719

= 48.7827 kJ/kmol


● C6H4N2O4

= x1 - 0.38

1 -

= 61.56 x1 - 412.09 803 0.38

1 - 572 803

= 75.1824 kJ/kmol


● C6H6

= x1 - 0.38

1 -

= 40.68 x1 - 412.09 647.4 0.38

1 - 373.15 647.4

= 38.3804 kJ/kmol


Jadi:

= 6706.9449 kJ/jam

b.

●

Q = n412.09

298.15

= 134.94412.09

32.243 T + 1.923E-3 T2 + 1.055E-5 T3 - 3.596E-9 T4dT298.15

= 134.94 32.2430 412.09 - 298.15 +0.00192

412.092

- 298.152

2

+1.06E-05

412.093

- 298.153

+###

412.094

- 298.154

ΔH2 ΔH1

Tr2

Tr1

ΔH2

ΔH2 ΔH1

Tr2

Tr1

ΔH2

ΔH2 ΔH1

Tr2

Tr1

ΔH2

Q2

Panas Sensibel Gas Masuk Kondensor, Q3

H2O

Cp dT

+3

412.09 - 298.15 +4

412.09 - 298.15

= 524323.7214 kJ/jam

●

Q = n412.09

298.15

= 33.953412.09

-31.3860 T + 0.4746 T2 - 3.11E-4 T3 + 8.524E-8 T4 - 5.0524E-12 T5dT298.15

= 33.953 -22.0620 412.09 - 298.15 +0.57313

412.092

- 298.152

2

+###

412.093

- 298.153

+###

412.094

- 298.154

3 4

+###

412.095

- 298.155

5

= 961295.8875 kJ/jam

●

Q = n412.09

298.15

= 1E-02412.09

-31.3860 T + 0.4746 T2 - 3.11E-4 T3 + 8.524E-8 T4 - 5.0524E-12 T5dT298.15

= 1E-02 -16.2020 412.09 - 298.15 +0.56182

412.092

- 298.152

2

+###

412.093

- 298.153

+###

412.094

- 298.154

3 4

+###

412.095

- 298.155

5

= 375.7357 kJ/jam

●

Q = n412.09

298.15

= 3E-04412.09

-31.3860 T + 0.4746 T2 - 3.11E-4 T3 + 8.524E-8 T4 - 5.0524E-12 T5dT298.15

= 3E-04 18.1480 412.09 - 298.15 +0.56182

412.092

- 298.152

2

+###

412.093

- 298.153

+1.00E-07

412.094

- 298.154

3 4

+###

412.095

- 298.155

5

= 5.0149 kJ/jam

C6H5NH2

Cp dT

C6H5NO2

Cp dT

C6H4N2O4

Cp dT

●

Q = n412.09

298.15

= 1E-03412.09

-31.3860 T + 0.4746 T2 - 3.11E-4 T3 + 8.524E-8 T4 - 5.0524E-12 T5dT298.15

= 1E-03 -31.3860 412.09 - 298.15 +0.47460

412.092

- 298.152

2

+###

412.093

- 298.153

+8.52E-08

412.094

- 298.154

3 4

+###

412.095

- 298.155

5

= 12.7227 kJ/jam

Jadi:

= 1486013.0821 kJ/jam

c.

●

Q = n379.13

298.15

= 14.3560379.13

18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT298.15

= 14.3560 18.2964 379.13 - 298.15 +0.47212

379.132

- 298.152

2

+###

379.133

- 298.153

+1.31E-06

379.134

- 298.154

3 4

= 87977.7282 kJ/jam

●

Q = n379.13

298.15

= 3.61379.13

46.9480 T + 0.9896 T2 - 2.3583E-3 T3 + 2.3296E-6 T4 dT298.15

= 3.61 46.9480 379.13 - 298.15 +0.9896

379.132

- 298.152

2

+###

379.133

- 298.153

+2.33E-06

379.134

- 298.154

3 4

= 59114.1629 kJ/jam

● C6H5NO2

C6H6

Cp dT

Q3

Panas Sensibel Cairan Reflux, Q4

H2O

Cp dT

C6H5NH2

Cp dT

Q = n379.13

298.15

= 0.0015379.13

18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT298.15

= 0.0015 39.4730 379.13 - 298.15 +0.91277

379.132

- 298.152

2

+###

379.133

- 298.153

+2.01E-06

379.134

- 298.154

3 4

= 22.0440 kJ/jam

● C6H4N2O4

Q = n379.13

298.15

= 3E-05379.13

18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT298.15

= 3E-05 -12.6350 379.13 - 298.15 +1.56240

379.132

- 298.152

2

+###

379.133

- 298.153

+2.32E-06

379.134

- 298.154

3 4

= 0.5772 kJ/jam

● C6H6

Q = n379.13

298.15

= 0.0001379.13

18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT298.15

= 0.0001 -33.6620 379.13 - 298.15 +0.47430

379.132

- 298.152

2

+###

379.133

- 298.153

+3.82E-06

379.134

- 298.154

3 4

= 6.5729 kJ/jam

Jadi:

= 147121.0852 kJ/jam

d.

●

Q = n379.13

298.15

Cp dT

Cp dT

Cp dT

Q4

Panas Sensibel Cairan Produk Distilat, Q5

H2O

Cp dT

= 120.5844379.13

18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT298.15

= 120.5844 18.2964 379.13 - 298.15 +0.47212

379.132

- 298.152

2

+###

379.133

- 298.153

+1.31E-06

379.134

- 298.154

3 4

= 738973.9349 kJ/jam

●

Q = n379.13

298.15

= 3.0.E+01379.13

46.9480 T + 0.9896 T2 - 2.3583E-3 T3 + 2.3296E-6 T4 dT298.15

= 3.0.E+01 46.9480 379.13 - 298.15 +0.9896

379.132

- 298.152

2

+###

379.133

- 298.153

+2.33E-06

379.134

- 298.154

3 4

= 496532.7759 kJ/jam

● C6H5NO2

Q = n379.13

298.15

= 0.0124379.13

18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT298.15

= 0.0124 39.4730 379.13 - 298.15 +0.91277

379.132

- 298.152

2

+###

379.133

- 298.153

+2.01E-06

379.134

- 298.154

3 4

= 185.1595 kJ/jam

● C6H4N2O4

Q = n379.13

298.15

= 2E-04379.13

18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT298.15

= 2E-04 -12.6350 379.13 - 298.15 +1.56240

379.132

- 298.152

2

+###

379.133

- 298.153

+2.32E-06

379.134

- 298.154

3 4

= 4.8485 kJ/jam

C6H5NH2

Cp dT

Cp dT

Cp dT

● C6H6

Q = n379.13

298.15

= 0.0010379.13

18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT298.15

= 0.0010 -33.6620 379.13 - 298.15 +0.47430

379.132

- 298.152

2

+###

379.133

- 298.153

+3.82E-06

379.134

- 298.154

3 4

= 55.2094 kJ/jam

Jadi:

= 1235751.9282 kJ/jam

e.

= + - +

= 109847.0136 kJ/jam


= 4.1787

Q = m . Cp. ΔT

=Cp. ΔT

=109847.0136

= 1314.3805 kg/jam4.1787 ( 45 - 25 )

Perhitungan Neraca Panas pada Reboiler

a.

= 2666525.0385 kJ/jam

b.

●

Q = n457.17

298.15

= 0.0759457.17

18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT298.15

= 0.0759 18.2964 457.17 - 298.15 +0.47212

457.172

- 298.152

Cp dT

Q5

Panas yang dilepas oleh Pendingin Kondensor, Qc

Qc Q2 Q3 Q4 Q5



kkal/kgoC

kJ/kgoC

mc

Qc

Panas Sensibel Cairan Masuk MD , Q1

Panas sensibel masuk MD 2 sama dengan panas sensibel cairan keluaran Heater.

Q1

Panas Sensibel Cairan Produk Bawah, Q6

H2O

Cp dT

= 0.0759 18.2964 457.17 - 298.15 +2

457.17 - 298.15

+###

457.173

- 298.153

+1.31E-06

457.174

- 298.154

3 4

= 925.7972 kJ/jam

●

Q = n457.17

298.15

= 187.06457.17

46.9480 T + 0.9896 T2 - 2.3583E-3 T3 + 2.3296E-6 T4 dT298.15

= 187.06 46.9480 457.17 - 298.15 +0.9896

457.172

- 298.152

2

+###

457.173

- 298.153

+2.33E-06

457.174

- 298.154

3 4

= 14486054.3781 kJ/jam

●

Q = n457.17

298.15

= 8E-02 39.47 457.17 - 298.15 +0.9128

457.172

- 298.152

2

+-0.0021

457.173

- 298.153

+2.01E-06

457.174

- 298.154

3 4

= 2329.6045 kJ/jam

●

Q = n457.17

298.15

= 1E-03 -12.6 457.17 - 298.15 +1.56240

457.172

- 298.152

2

+-3E-03

457.173

- 298.153

+2.32E-06

457.174

- 298.154

3 4

= 61.2887 kJ/jam

●

Q = n457.17

298.15

= 6E-03 -33.7 457.17 - 298.15 +0.47430

457.172

- 298.152

2

+-4E-03

457.173

- 298.153

+3.82E-06

457.174

- 298.154

3 4

= 851.1017 kJ/jam

C6H5NH2

Cp dT

C6H5NO2

Cp dT

C6H4NO2

Cp dT

C6H6

Cp dT

Jadi:

= 14490222.1702 kJ/jam

c.

= + + -

= 13169296.0735

13169296.0735

= 909.99 kJ/kg

Sehingga :

= = 13169296.0735 kJ/jam= 14471.9130 kg/jam

909.99 kJ/kg

Tabel Neraca Panas Menara Distil(D-330)

KomponenPanas masuk Panas keluar (kJ/jam)

(kJ/jam) Arus Arus

978.0620 185.1595 2329.6045

104.9953 4.8485 61.2887

38610.6875 738973.9349 925.7972

346.4803 55.2094 851.1017

2626484.8134 496532.7759 14486054.3781

Q pendingin - 109847.0136 -

13169296.07 - -

Total 15835821.11201345598.9418 14490222.1702

15835821.1120

B.14 COOLER



28

184.2 30

Qc 45

Neraca Energi

Q6

Panas yang dibutuhkan oleh Pemanas Reboiler, Qrb

Qrb Q5 Q6 Qc Q1




λsteam

Qr = m.λsteam

Massa steam Qr

λsteam

Q C6H5NO2

Q C6H4N2O4

Q H2O

Q C6H6

Q C6H5NH2

Q steam

oC

Q1 Q2

oC oC

oC

Q in = Q out

= +





dimana:

= 14490222.1702 kJ/jam

Komposisi arus 6:

= 4.2601 kg/jam = 0.0346 kmol/jam

= 0.1073 kg/jam = 0.0006 kmol/jam

= 0.6193 kg/jam = 0.0344 kmol/jam

C6H6 = 0.2145 kg/jam = 0.0028 kmol/jam

= 7890.7922 kg/jam = 84.8472 kmol/jam

●

Q = n457.17

298.15

= 0.0344457.17

18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT298.15

= 0.0344 18.2964 457.17 - 298.15 +0.47212

457.172

- 298.152

2

+###

457.173

- 298.153

+1.31E-06

457.174

- 298.154

3 4

= 1951.0776 kJ/jam

●

Q = n457.17

298.15

= 84.85457.17

46.9480 T + 0.9896 T2 - 2.3583E-3 T3 + 2.3296E-6 T4 dT298.15

= 84.85 46.9480 457.17 - 298.15 +0.9896

457.172

- 298.152

2

Q1 Q2 Q3

Q1

Q3

Q2

1. Panas Sensibel Cairan Masuk, Q1

Panas sensibel yang masuk Cooler sama dengan panas sensibel cairan yang keluar MD

Q1


C6H5NO2

C6H4N2O4

H2O

C6H5NH2

H2O

Cp dT

C6H5NH2

Cp dT

+###

457.173

- 298.153

+2.33E-06

457.174

- 298.154

3 4

= 2838811.5810 kJ/jam

● C6H6

Q = n457.17

298.15

= 3E-03457.17

46.9480 T + 0.9896 T2 - 2.3583E-3 T3 + 2.3296E-6 T4 dT298.15

= 3E-03 -33.6620 457.17 - 298.15 +0.4743

457.172

- 298.152

2

+###

457.173

- 298.153

+3.82E-06

457.174

- 298.154

3 4

= 386.0537 kJ/jam

●

Q = n457.17

298.15

= 3E-02 39.47 457.17 - 298.15 +0.9128

457.172

- 298.152

2

+-0.0021

457.173

- 298.153

+2.01E-06

457.174

- 298.154

3 4

= 1056.6921 kJ/jam

●

Q = n457.17

298.15

= 6E-04 -12.6 457.17 - 298.15 +1.56240

457.172

- 298.152

2

+-3E-03

457.173

- 298.153

+2.32E-06

457.174

- 298.154

3 4

= 115.9067 kJ/jam

Jadi:

= 2842321.3110 kJ/jam

= -

= 14490222.1702 - 2842321.3110 = 11647900.8591 kJ/jam


= 4.1787

Cp dT

C6H5NO2

Cp dT

C6H4N2O4

Cp dT

Q2


Qc Q1 Q2



kkal/kg.oC

kJ/kgoC

Q = m . Cp. ΔT

=Cp. ΔT

=11647900.8591

= ### kg/jam4.178661 ( 45 - 25 )

(E-221)


(kJ/jam) (kJ/jam)

2329.6045 1056.6921

61.2887 115.9067

14486054.3781 2838811.5810

Q C6H6 851.1017 386.0537

925.7972 1951.0776

Q pendingin - 11647900.8591

Total 14490222.1702 14490222.1702

mc

Qc


Q C6H5NO2

Q C6H4N2O4

Q C6H5NH2

Q H2O

LAMPIRAN A



= 7891.4141 kg/jam


= 1 hari = 24 jam



= Air = 0.0010


= Benzen = 0.0010


= Air = 0.0100




Air 100 18

Anilin 184.1 93

Hidrogen -259.2 2


Benzen 80.1 78

Metana -164 16

A.1 SEPARATOR

2

1

3

Basis = 100 kg/jam

Asumsi = 80% produk



= 0.0800 kg/jam = 0.0044 kmol/jam


Titik Didih, oC

C6H5NO2

H2O

C6H5NH2

H2

C6H4NO2

C6H6

CH4

H2O

Sep

arator


Arus 1 (umpan):


= 0.1000 kg/jam = 0.0056 kmol/jam



Arus 3


= 0.0200 kg/jam = 0.0011 kmol/jam






Nitrobenzen 99.3000 79.4400 19.8600

0.1000 0.0800 0.0200

Di-Nitrobenzen 0.1000 0.0800 0.0200

Benzen 0.1000 0.0800 0.0200

Total79.6800 19.9200

99.6000 99.6000

A.2 REAKTOR

4

2

Konversi = 0.98


= 79.4400 kg/jam = 0.6459 kmol/jam

= 0.0800 kg/jam = 0.0044 kmol/jam

= 0.0800 kg/jam = 0.0005 kmol/jam

= 0.0800 kg/jam = 0.0010 kmol/jam


= 3.8751 kg/jam = 1.9376 kmol/jam

= 0.00004 kg/jam = 0.000002 kmol/jam

H2O

H2O

H2O

C6H5NO2

C6H4NO2

C6H6

H2O

H2

CH4

Reak

tor

Bahan yang Bereaksi

= Konversi x

= 0.98 x 79.4400

= 77.8512 kg/jam = 0.6329 kmol/jam

=

= 0.08 kg/jam = 0.0044 kmol/jam

=

= 0.08 kg/jam = 0.0005 kmol/jam

=

= 0.08 kg/jam = 0.0010256 kmol/jam

=

= 3.7976 kg/jam = 1.8988 kmol/jam

=

= 0.00004 kg/jam = 0.000002 kmol/jam

Reaksi yang Reaksi

+ +

Mula-mula 0.6459 1.9376 - -

Bereaksi 0.6329 1.8988 0.6329 1.2659

Sisa 0.0129 0.0388 0.6329 1.2659


= 0.6329 kmol/jam = 58.8631 kg/jam

= 1.2659 kmol/jam = 22.7857 kg/jam



79.4400 1.5888

0.0800 0.0800

0.0800 0.0800

0.0800 0.0800

3.8751 0.0775

0.00004 0.00004

- 58.8631

- 22.7857

Total 83.5552 83.5552


H2O H2O yang masuk




CH4 CH4 yang masuk


C6H5NH2

H2O

C6H5NO2

C6H4NO2

H2O

C6H6

H2

CH4

C6H5NH2

H2O

Bahan Yang Masuk

= 1.5888 kg/jam = 0.0129 kmol/jam

= 0.0800 kg/jam = 0.0005 kmol/jam

= 0.0800 kg/jam = 0.0044 kmol/jam

= 0.0800 kg/jam = 0.0010 kmol/jam

= 0.0775 kg/jam = 0.0388 kmol/jam

= 0.00004 kg/jam = 0.000002 kmol/jam

= 58.8631 kg/jam = 0.3197 kmol/jam

= 22.7857 kg/jam = 1.2659 kmol/jam

A.3 Flash Tank

5

4

6


Bahan Yang Masuk

= 1.5888 kg/jam = 0.0129 kmol/jam

= 0.0800 kg/jam = 0.0005 kmol/jam

= 0.0800 kg/jam = 0.0044 kmol/jam

= 0.0800 kg/jam = 0.0010 kmol/jam

= 0.0775 kg/jam = 0.0388 kmol/jam

= 0.00004 kg/jam = 0.000002 kmol/jam

= 58.8631 kg/jam = 0.3197 kmol/jam

= 22.7857 kg/jam = 1.2659 kmol/jam


98% x = 1.5570 kg/jam = 0.0127 kmol/jam

99% x = 0.0792 kg/jam = 0.0005 kmol/jam

98% x = 22.4084 kg/jam = 1.2449 kmol/jam

C6H5NO2

C6H4NO2

H2O

C6H6

H2

CH4

C6H5NH2

H2O

C6H5NO2

C6H4NO2

H2O

C6H6

H2

CH4

C6H5NH2

H2O


C6H5NO2 yang masuk


C6H4NO2 yang masuk

Air (H2O)

H2O yang masuk

Flash

T

ank

98% x = 0.0784 kg/jam = 0.0010 kmol/jam

100.0% x = 0.0775024 kg/jam = 0.0388 kmol/jam

100% x = 0.00004 kg/jam = 0.000002 kmol/jam

0.01% x = 0.0059 kg/jam = 0.0001 kmol/jam


2% x = 0.031776 kg/jam = 0.0003 kmol/jam

1% x = 0.0008 kg/jam = 0.000005 kmol/jam

2% x = 0.4573143 kg/jam = 0.0254 kmol/jam

2% x C6H6 yang masuk = 0.0016 kg/jam = 2.051E-05 kmol/jam

99.99% x = 58.857216 kg/jam = 0.6329 kmol/jam

0.00% x = 0.0000 kg/jam = 0.00000 kmol/jam


V L

1.5888 1.5570 0.0318

0.0800 0.0792 0.0008

22.8657 22.4084 0.4573

0.0800 0.0784 0.0016

0.0775 0.0775 -

3.8751607028E-05 3.8751607028E-05 -

Benzen (C6H6)

C6H6 yang masuk

Gas Hidrogen (H2)

H2 yang masuk

Gas Metana (CH4)

CH4 yang masuk

Anilin (C6H5NH2)

C6H5NH2 yang masuk


C6H5NO2 yang masuk


C6H4NO2 yang masuk

Air (H2O)

H2O yang masuk

Benzen (C6H6)

Anilin (C6H5NH2)

C6H5NH2 yang masuk

Hidrogen (H2)

H2 yang masuk

C6H5NO2

C6H4NO2

H2O

C6H6

H2

CH4

58.8631 0.0059 58.8572

Total 83.555224.2065 59.3487

83.5552

KomponenAntoine

A B C D E

-54.494 -2112.3 29.321 -0.0448 0.0000

-24.246 -4114 16.344 -0.0241 0.0000

18.3036 3816.44 -46.13 - -

15.9008 2788.51 -52.36 - -

13.6333 164.9 3.19 - -

15.2243 597.84 -7.16 - -

16.6748 3857.52 -73.15 - -

(L/V) data = 2.4518


Vi =

Fi

Ai =

(L/V)data

ki =

Pi


T trial = 139.53 C = 412.53 K

Li =

Fi




0.0129 101.6479 0.1337 18.3314

0.0005 1.1463 0.0015 1625.5992

1.2703 2663.536 3.5047 0.6996

0.0010 3492.8994 4.5959 0.5335

0.0388 560527.74 737.5365 0.0033

2E-06 936103.2 1231.7147 0.0020

0.3197 201.9982 0.2658 9.2246

Total 1.6432

Vi %V Li %Li

0.0007 0.0008 0.0097 0.0289

2.927522E-07 3.575968E-07 0.0005 0.0014

0.7474 0.9130 0.1324 0.3938

0.0007 0.0008 0.0001 0.0002

0.0386 0.0472 2.141814E-05 0.0001

2.417164E-06 2.952565E-06 8.01744E-10 2.383884E-09

C6H5NH2

C6H5NO2

C6H4NO2

H2O

C6H6

H2

CH4

C6H5NH2

C6H5NO2

C6H4NO2

H2O

C6H6

H2

CH4

C6H5NH2

0.0313 0.0382 0.1936 0.5756

0.8187 1.0000 0.3363 1.0000

L = Fi

Vi total

= 2.0072



7

6

8

Arus 6


0.0318 0.0003

0.0008 4.761905E-06

0.4573 0.0254

0.0016 2.051282E-05

58.8572 0.6329

Total 59.3487 0.6586


T = 170.11 °C = 443.11 K ; P = 1 atm = 760 mmHg


0.0003 0.0004 267.6120 0.5177

4.761905E-06 7.230748E-06 5.5490 0.0107

C6H5NO2

C6H4NO2

H2O

C6H6

C6H5NH2


C6H5NO2

C6H4NO2

Men

ara D

istilasi

0.0254 0.0386 5942.6871 11.4957

2.051282E-05 3.114784E-05 6402.8470 12.3859

0.6329 0.9610 516.9484 1.0000

Total 0.6586 1

yi αi

0.0002 1.0000

7.761584E-08 0.0207

0.4435 22.2064

0.0004 23.9259

0.9610 1.9317

1.405



B.1 Distilat


1. = 1% x Massa = 2.583E-06 kmol/jam

2. = 1% x Massa = 4.762E-08 kmol/jam

3. = 99.00% x Massa = 0.0252 kmol/jam

4. = 1% x Massa = 2.051E-07 kmol/jam

5. = 1% x Massa = 0.0063 kmol/jam



1. = Massa Komponen

=

2.583E-06

= 8.205521E-05Massa Total 0.0315

2. = Massa Komponen

=

4.762E-08

= 1.512491E-06Massa Total 0.0315

3. = Massa Komponen

=

0.0252

= 0.7989Massa Total 0.0315

4. = Massa Komponen

=

2.051E-07


6. = Massa Komponen

=

0.0063

= 0.2010Massa Total 0.0314839

H2O

C6H6

C6H5NH2

C6H5NO2

C6H4NO2

H2O

C6H6

C6H5NH2

C6H5NO2

C6H4NO2

H2O

C6H6

C6H5NH2


1. = 99% x Massa = 0.0003 kmol/jam

2. = 99% x Massa = 4.714E-06 kmol/jam

3. = 1% x Massa = 0.0003 kmol/jam


5. = 99% x Massa = 0.6265 kmol/jam



1. = Massa Komponen

=

0.0003

= 0.0004Massa Total 0.6271

2. = Massa Komponen

=

4.714E-06

= 0.0000Massa Total 0.6271

3. = Massa Komponen

=

0.0003

= 0.0004Massa Total 0.6271


=

2.0.E-05

= 0.0000Massa Total 0.6271

5. = Massa Komponen

=

0.6265446

= 0.9991Massa Total 0.6270794



0.0003 1% 2.583415E-06 8.205521E-05

4.761905E-06 1% 4.761905E-08 1.512491E-06

0.0254 99% 0.0252 0.7989

2.051282E-05 1% 2.051282E-07 0

0.6329 1% 0.0063 0.2010

Total 0.6586 0.0315 1.0000



0.0003 99% 0.0003 0.0004

4.761905E-06 99% 4.714286E-06 7.517845E-06

0.0254 1% 0.0003 0.0004

C6H6 2.05.E-05 99% 2.03.E-05 0.0000

0.6329 99% 0.6265 0.9991

Total 0.6586 0.6271 1.0000

C6H5NO2

C6H4NO2

H2O

C6H5NH2

C6H5NO2

C6H4NO2

H2O

C6H5NH2

C6H5NO2

C6H4NO2

H2O

C6H6

C6H5NH2

C6H5NO2

C6H4NO2

H2O

C6H5NH2



T = 182.7943 °C = 455.9443 K

KomponenMassa yi=

Piki= xi= αi =


C6H5NO2 2.583415E-06 8.205521E-05 382.33575 0.5031 0.0001631 1.0000

C6H4NO2 4.761905E-08 1.512491E-06 9.9670455 0.0131 0.0001153 0.0261

0.0252 0.7988947509 8030.4177 10.5663 0.0756 21.0036

2.051282E-07 6.515345E-06 8034.002 10.5711 6.163E-07 21.0129

0.0063 0.2010151661 733.28465 0.9648 0.2083 1.9179

Total 0.0315 1 0.284




T = 184.0100 °C = 457.15996218 K

KomponenMassa xi=

Piki= yi= αi =


0.0003 0.0004 394.95015 0.5196713 0.000212 1.0000

4.714286E-06 0.0000 10.513526 0.0138336 1.04E-07 0.0266

0.0003 0.0004 8254.6734 10.861412 0.0044005 20.9005

C6H6 2.03.E-05 0.0000 8202.4461 10.792692 0.0003495 20.7683

0.6265 0.9991 757.05477 0.9961247 0.9952751 1.9168

Total 0.6271 1 1.000




0.0016 2E-05 2E-03

0.4573 0.4527 0.0046

58.8572 0.5886 58.2686

0.0318 0.0003 0.0315

0.0008 8E-06 0.0008

59.3487 1.0417 58.3071

Total 59.3487 59.3487


H2O

C6H6

C6H5NH2


C6H5NO2

C6H4NO2

H2O

C6H5NH2

C6H6

H2O

C6H5NH2

C6H5NO2

C6H4NO2

Produk = 58.2732

Faktor Pengali = 7891.4141

= 135.420958.2732

135.4209