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LAMPIRAN A PERHITUNGAN NERACA MASSA Kapasitas Produksi Hari Kerja Basis Bahan Baku
= = = = =
Spesifikasi bahan baku
Spesifikasi Produk
62500 ton/tahun 7891.4141 kg/jam hari 1 tahun = 330 24 jam 1 hari = 100 kg/jam = = = =
Nitrobenzen Air Di-Nitrobenzen Benzen
= Anilin = Air
Tabel Data Masing-masing Komponen Komponen Rumus Molekul C6H5NO2 Nitrobenzen H2 O Air C6H5NH2 Anilin
A.1
0.9930 0.0010 0.0010 0.0010
= =
0.9900 0.0100
Titik Didih, oC 210.9 100 184.1
BM, kg/kmol 123 18 93
-259.2 301 80.1 -164
2 168 78 16
H2 C6H4NO2 C 6 H6 CH4
Hidrogen Di-Nitrobenzen Benzen Metana
= = = =
SEPARATOR 2
Separator
1
3 Basis = 100 kg/jam Asumsi = 80% produk Arus 2 (produk yang diinginkan): 79.4400 Nitrobenzen = H2 O 0.0800 = 0.0800 Di-Nitrobenzen = 0.0800 Benzen =
kg/jam kg/jam kg/jam kg/jam
= = = =
0.6459 0.0044 0.0005 0.0010
kmol/jam kmol/jam kmol/jam kmol/jam
Arus 1 (umpan): Nitrobenzen H2 O Di-Nitrobenzen Benzen Arus 3 Nitrobenzen H2 O Di-Nitrobenzen Benzen
= = = =
99.3000 0.1000 0.1000 0.1000
kg/jam kg/jam kg/jam kg/jam
= = = =
0.8073 0.0056 0.0006 0.0013
kmol/jam kmol/jam kmol/jam kmol/jam
= = = =
19.8600 0.0200 0.0200 0.0200
kg/jam kg/jam kg/jam kg/jam
= = = =
0.1615 0.0011 0.0001 0.0003
kmol/jam kmol/jam kmol/jam kmol/jam
Tabel Neraca Massa Separator Masuk (kg/jam) Komponen Arus 1 Nitrobenzen 99.3000 H2 O 0.1000 0.1000 Di-Nitrobenzen Benzen 0.1000 Total
99.6000
Keluar (kg/jam) Arus 2 Arus 3 79.4400 19.8600 0.0800 0.0200 0.0800 0.0200 0.0800 0.0200 79.6800 19.9200 99.6000
A.2 REAKTOR 4
Reaktor
2
Konversi = 0.98 Bahan Baku yang Masuk kedalam Reaktor C6H5NO2 79.4400 kg/jam = C6H4NO2 0.0800 kg/jam = C 6 H6 0.0800 kg/jam = H2 O 0.0800 kg/jam =
= = = =
Perbandingan mol Nitrobezen terhadap H2 = 1:3 H2 = 3.8751 kg/jam = CH4 = 0.00004 kg/jam = Bahan yang Bereaksi C6H5NO2 yang bereaksi
= =
Konversi 0.98
x
0.6459 0.0044 0.0005 0.0010
1.9376 0.000002
kmol/jam kmol/jam kmol/jam kmol/jam
kmol/jam kmol/jam
x C6H5NO2 yang masuk 79.4400
= H2 O
C6H4NO2
C 6 H6
H2 yang bereaksi
CH4
Reaksi yang Reaksi C6H5NO2 Mula-mula 0.6459 Bereaksi 0.6329 Sisa 0.0129 Produk yang Terbentuk C6H5NH2 = 0.6329 H2 O = 1.2659
kg/jam
=
0.6329
kmol/jam
= H2O yang masuk = 0.08 kg/jam
=
0.0044
= C6H4NO2 yang masuk = 0.08 kg/jam
=
0.0005
kmol/jam
= C6H6 yang masuk = 0.08 kg/jam
=
0.001026
kmol/jam
= (3/1) x C6H5NO2 yang bereaksi = 3.7976 kg/jam = 1.8988 = CH4 yang masuk = 0.00004 kg/jam
+
3H2
=
kmol/jam kmol/jam
kmol/jam
kmol/jam
0.000002
C6H5NH2
1.9376 1.8988 0.0388
Neraca Massa Total Reaktor Komponen Masuk (kg/jam) C6H5NO2 79.4400 C6H4NO2 0.0800 H2 O 0.0800 C 6 H6 0.0800 H2 3.8751 CH4 0.00004 C6H5NH2 H2 O Total 83.5552 Bahan Yang Masuk C6H5NO2 = 1.5888 C6H4NO2 = 0.0800 H2 O = 0.0800 C 6 H6 = 0.0800
77.8512
+
0.6329 0.6329
= =
58.8631 22.7857
= = = =
0.0129 0.0005 0.0044 0.0010
2H2O 1.2659 1.2659
kg/jam kg/jam
Keluar (kg/jam) 1.5888 0.0800 0.0800 0.0800 0.0775 0.00004 58.8631 22.7857 83.5552
kg/jam kg/jam kg/jam kg/jam
kmol/jam
kmol/jam kmol/jam kmol/jam kmol/jam
H2 CH4 C6H5NH2 H2 O
= = = =
0.0775 0.00004 58.8631 22.7857
kg/jam kg/jam kg/jam kg/jam
= = = =
0.0388 0.000002 0.3197 1.2659
kmol/jam kmol/jam kmol/jam kmol/jam
kg/jam kg/jam kg/jam kg/jam kg/jam kg/jam kg/jam kg/jam
= = = = = = = =
0.0129 0.0005 0.0044 0.0010 0.0388 0.000002 0.3197 1.2659
kmol/jam kmol/jam kmol/jam kmol/jam kmol/jam kmol/jam kmol/jam kmol/jam
=
1.5570
A.3 Flash Tank 5
Flash Tank
4
6 Komposisi Umpan Masuk Bahan Yang Masuk C6H5NO2 = 1.5888 C6H4NO2 = 0.0800 H2 O = 0.0800 C 6 H6 = 0.0800 H2 = 0.0775 CH4 = 0.00004 C6H5NH2 = 58.8631 H2 O = 22.7857 Komposisi Bahan yang Teruapkan Nitrobenzen (C6H5NO2) 98% x C6H5NO2 yang masuk Di-Nitrobenzen (C6H4NO2) 99% x C6H4NO2 yang masuk
=
Air (H2O) 98% x H2O yang masuk
=
Benzen (C6H6) 98% x C6H6 yang masuk
=
Gas Hidrogen (H2) 100.0% x H2 yang masuk
=
Gas Metana (CH4)
kg/jam
0.0792
22.4084
0.0784
=
0.0127
kmol/jam
kg/jam
=
0.0005
kmol/jam
kg/jam
=
1.2449
kmol/jam
kg/jam
=
0.0010
kmol/jam
0.077502 kg/jam
=
0.0388
kmol/jam
100%
x CH4 yang masuk
=
0.00004
Anilin (C6H5NH2) 0.01% x C6H5NH2 yang masuk
=
Komposisi Bahan Cair Nitrobenzen (C6H5NO2) 2% x C6H5NO2 yang masuk
=
Di-Nitrobenzen (C6H4NO2) 1% x C6H4NO2 yang masuk
=
Air (H2O) 2% x H2O yang masuk Benzen (C6H6) 2% x C6H6 yang masuk
= =
Anilin (C6H5NH2) 99.99% x C6H5NH2 yang masuk Hidrogen (H2) 0.00% x H2 yang masuk
Komponen
Masuk (kg/jam)
C6H5NO2 C6H4NO2 H2 O C 6 H6 H2 CH4 C6H5NH2
1.5888 0.0800 22.8657 0.0800 0.0775 3.87516E-05 58.8631
Total
83.5552
Komponen C6H5NO2 C6H4NO2 H2 O C 6 H6
A -54.494 -24.246 18.3036 15.9008
0.0059
=
kg/jam
=
0.031776 kg/jam
0.0008
0.0016
0.0000
=
kg/jam
0.457314 kg/jam
=
=
kg/jam
kg/jam
kmol/jam
0.0001
kmol/jam
0.0003
kmol/jam
=
0.000005
0.0254
kmol/jam
=
2.05E-05
kmol/jam
=
=
0.6329
0.00000
V L 1.5570 0.0318 0.0792 0.0008 22.4084 0.4573 0.0784 0.0016 0.0775 3.87516E-05 0.0059 58.8572 24.2065 59.3487 83.5552 Antoine C 29.321 16.344 -46.13 -52.36
kmol/jam
kmol/jam
Keluar (kg/jam)
B -2112.3 -4114 3816.44 2788.51
kmol/jam
=
58.85722 kg/jam
kg/jam
0.000002
D -0.0448 -0.0241 -
E 0.0000 0.0000 -
-
-
H2 CH4 C6H5NH2
13.6333 15.2243 16.6748
164.9 597.84 3857.52
(L/V) data = 2.4518 Dengan menggunakan persamaan Fi Vi = ((L/V)/ki)+1
3.19 -7.16 -73.15
-
(L/V)data Ai = ki
ki =
T trial = 139.5 C = 412.5 K P sistem = 760 mmHg logP = A + B/T + ClogT + DT + ET2
Li =
-
Pi P sistem
Fi (1+(L/V)*ki)
Komponen
Fi (kmol)
Pi (mmHg)
ki
Ai
C6H5NO2 C6H4NO2 H2 O C 6 H6 H2 CH4 C6H5NH2
0.0129 0.0005 1.2703 0.0010 0.0388 2.E-06 0.3197
101.6479 1.1463 2663.536 3492.8994 560527.74 936103.2 201.9982
0.1337 0.0015 3.5047 4.5959 737.5365 1231.7147 0.2658
18.3314 1625.5992 0.6996 0.5335 0.0033 0.0020 9.2246
Total
1.6432
Vi
%V
Li
%Li
0.0007 2.92752E-07 0.7474 0.0007 0.0386 2.41716E-06 0.0313
0.0008 3.57597E-07 0.9130 0.0008 0.0472 2.95257E-06 0.0382
0.0097 0.0005 0.1324 0.0001 2.14181E-05 8.01744E-10 0.1936
0.0289 0.0014 0.3938 0.0002 0.0001 2.38388E-09 0.5756
0.8187
1.0000
0.3363
1.0000
L
=
Fi Vi total
= (L/V)hitung
2.0072 =
2.4518
A.5 MENARA DISTILASI
7
Menara Distilasi
6
8
Arus 6 Komponen
Kg/Jam
Kmol/Jam
C6H5NO2 C6H4NO2 H2 O C 6 H6 C6H5NH2
0.0318 0.0008 0.4573 0.0016 58.8572
0.0003 4.7619E-06 0.0254 2.05128E-05 0.6329
Total
59.3487
0.6586
A. Massa Masuk Menara Distilasi pada Kondisi Bubble Point T = 170.11 °C 443.11 K ; P = = 1 atm Trial pada T akan dianggap benar apabila Syi = 1
=
Komponen
(Kmol/jam)
xi
Pi
ki
C6H5NO2 C6H4NO2
0.0003 4.7619E-06 0.0254 2.05128E-05 0.6329
0.0004 7.23075E-06 0.0386 3.11478E-05 0.9610
267.6120 5.5490 5942.6871 6402.8470 516.9484
0.5177 0.0107 11.4957 12.3859 1.0000
0.6586
1
H2 O C 6 H6 C6H5NH2 Total
yi
αi
0.0002 7.76158E-08 0.4435 0.0004 0.9610 1.405
1.0000 0.0207 22.2064 23.9259 1.9317
760
mmHg
Jadi bisa disimpulkan bahwa suhu pemasukan umpan sebesar 170,11 C B. Spesifikasi Hasil Yang Diinginkan B.1 Distilat • Menentukan Massa Distilat 1. C6H5NO2 = 1% x Massa = 2.58E-06 kmol/jam 2. C6H4NO2 = 1% x Massa = 4.76E-08 kmol/jam H O 3. = 99.00% x Massa = 0.0252 kmol/jam 2 C 6 H6 4. = 1% x Massa = 2.05E-07 kmol/jam C6H5NH2 5. = 1% x Massa = 0.0063 kmol/jam Total = 0.0315 kmol/jam Menentukan Nilai xdi 1. C6H5NO2 = Massa Komponen =
2.58E-06 0.0315
=
8.20552E-05
= Massa Komponen Massa Total
=
4.76E-08 0.0315
=
1.51249E-06
= Massa Komponen Massa Total
=
0.0252 0.0315
=
0.7989
= Massa Komponen Massa Total
=
2.05E-07 0.0315
=
0
= Massa Komponen Massa Total
=
0.0063 0.031484
=
0.2010
Massa Total 2. C6H4NO2
3. H2O
4. C6H6
6.
C6H5NH2
• Menentukan Massa Bottom 1. C6H5NO2 = 99% 2. C6H4NO2 = 99% 3. H2O = 1% 4. C6H6 = 99% C6H5NH2 5. = 99% Total
x x x x x
Menentukan Nilai xbi 1. C6H5NO2 = Massa Komponen Massa Total 2. C6H4NO2
3. H2O
= Massa Komponen Massa Total
Massa Massa Massa Massa Massa
= 0.0003 kmol/jam = 4.71E-06 kmol/jam = 0.0003 kmol/jam = 2.03.E-05 kmol/jam = 0.6265 kmol/jam = 0.6271 kmol/jam
=
0.0003 0.6271
=
0.0004
=
4.71E-06 0.6271
=
0.0000
=
0.0004
= Massa Komponen
0.0003 =
4. C6H6
C6H5NH2
5.
Komponen C6H5NO2 C6H4NO2 H2 O C 6 H6 C6H5NH2 Total
Massa Total
=
0.6271
=
0.0004
= Massa Komponen Massa Total
=
2.0.E-05 0.6271
=
0.0000
= Massa Komponen Massa Total
=
0.626545 0.627079
=
0.9991
Massa (Kmol/jam) 0.0003 4.7619E-06 0.0254 2.05128E-05 0.6329 0.6586
% 1% 1% 99% 1% 1%
Distilat Massa 2.58341E-06 4.7619E-08 0.0252 2.05128E-07 0.0063 0.0315
xdi 8.20552E-05 1.51249E-06 0.7989 0 0.2010 1.0000
Massa (Kmol/jam) 0.0003 4.7619E-06 0.0254 2.05.E-05 0.6329 0.6586
% 99% 99% 1% 99% 99%
Bottom Massa 0.0003 4.71429E-06 0.0003 2.03.E-05 0.6265 0.6271
xbi 0.0004 7.51784E-06 0.0004 0.0000 0.9991 1.0000
Komponen C6H5NO2 C6H4NO2 H2 O C6H6 C6H5NH2 Total
C. Perhitungan Suhu Atas (Dew point) Trial pada T akan dianggap benar apabila Sxi = 1 Pt = T=
1 182.7943
Komponen C6H5NO2 C6H4NO2 H2 O C 6 H6 C6H5NH2 Total
atm °C
Massa (kmol/jam) 2.58341E-06 4.7619E-08 0.0252 2.05128E-07 0.0063 0.0315
= =
760 455.9443
yi= m/m total 8.20552E-05 1.51249E-06 0.798894751 6.51535E-06 0.201015166 1
mmHg K
Pi 382.3358 9.967046 8030.418 8034.002 733.2846
Jadi dapat disimpulkan bahwa T distilat adalah 182,7943°C D. Perhitungan Suhu Bawah (Bubble point)
ki= Pi/P 0.5031 0.0131 10.5663 10.5711 0.9648
xi= yi/ki 0.000163 0.000115 0.0756 6.16E-07 0.2083 0.284
αi = ki/k(HK) 1.0000 0.0261 21.0036 21.0129 1.9179
Trial pada T dianggap benar apabila Syi = 1 Pt = T=
1 184.0100
Komponen C6H5NO2 C6H4NO2 H2 O C6H6 C6H5NH2 Total
atm °C
Massa (kmol/jam) 0.0003 4.71429E-06 0.0003 2.03.E-05 0.6265 0.6271
= =
760 mmHg 457.1599622 K
xi= m/m total 0.0004 0.0000 0.0004 0.0000 0.9991 1
Pi 394.9502 10.51353 8254.673 8202.446 757.0548
ki= Pi/P 0.519671 0.013834 10.86141 10.79269 0.996125
Jadi dapat disimpulkan bahwa T bottom adalah 172,5794°C
Komponen C 6 H6 H2 O C6H5NH2 C6H5NO2 C6H4NO2 Total Produk
=
Faktor Pengali
Massa Masuk (kg/gr) 0.0016 0.4573 58.8572 0.0318 0.0008 59.3487 59.3487
Massa Keluar Distilat Bottom 2.E-05 2.E-03 0.4527 0.0046 0.5886 58.2686 0.0003 0.0315 8.E-06 0.0008 1.0417 58.3071 59.3487
58.2732 =
7891.4141 58.2732
=
135.4209
135.4209
yi= xi.ki 0.000212 1.04E-07 0.004401 0.00035 0.995275 1.000
αi = ki/k(HK) 1.0000 0.0266 20.9005 20.7683 1.9168
A ACA MASSA
kmol/jam
kmol/jam
kmol/jam
kmol/jam
kmol/jam
kmol/jam
kmol/jam
kmol/jam
mmHg
LAMPIRAN A PERHITUNGAN NERACA MASSA Kapasitas Produksi Hari Kerja Basis Bahan Baku
= = = = =
Spesifikasi bahan baku
Spesifikasi Produk
62500 ton/tahun 7891.4141 kg/jam hari 1 tahun = 330 24 jam 1 hari = 100 kg/jam = = = =
Nitrobenzen Air Di-Nitrobenzen Benzen
= Anilin = Air
Tabel Data Masing-masing Komponen Komponen Rumus Molekul C6H5NO2 Nitrobenzen H2 O Air C6H5NH2 Anilin
A.1
0.9930 0.0010 0.0010 0.0010
= =
0.9900 0.0100
Titik Didih, oC 210.9 100 184.1
BM, kg/kmol 123 18 93
-259.2 301 80.1 -164
2 168 78 16
H2 C6H4NO2 C 6 H6 CH4
Hidrogen Di-Nitrobenzen Benzen Metana
= = = =
SEPARATOR 2
Separator
1
3 Basis = 100 kg/jam Asumsi = 80% produk Arus 2 (produk yang diinginkan): Nitrobenzen = 10757.8363 H2 O = 10.8337 Di-Nitrobenzen = 10.8337 Benzen = 10.8337
kg/jam kg/jam kg/jam kg/jam
= = = =
87.4621 0.6019 0.0645 0.1389
kmol/jam kmol/jam kmol/jam kmol/jam
Arus 1 (umpan): Nitrobenzen H2 O Di-Nitrobenzen Benzen Arus 3 Nitrobenzen H2 O Di-Nitrobenzen Benzen
= = = =
13447.2954 13.5421 13.5421 13.5421
kg/jam kg/jam kg/jam kg/jam
= = = =
109.3276 0.7523 0.0806 0.1736
kmol/jam kmol/jam kmol/jam kmol/jam
= = = =
2689.4591 2.7084 2.7084 2.7084
kg/jam kg/jam kg/jam kg/jam
= = = =
21.8655 0.1505 0.0161 0.0347
kmol/jam kmol/jam kmol/jam kmol/jam
Tabel Neraca Massa Separator Masuk (kg/jam) Komponen Arus 1 Nitrobenzen 13447.2954 H2 O 13.5421 13.5421 Di-Nitrobenzen Benzen 13.5421 Total
13487.9216
Keluar (kg/jam) Arus 2 Arus 3 10757.8363 2689.4591 10.8337 2.7084 10.8337 2.7084 10.8337 2.7084 10790.3373 2697.5843 13487.9216
A.2 REAKTOR 4
Reaktor
2
Konversi = 0.98 Bahan Baku yang Masuk kedalam Reaktor C6H5NO2 kg/jam = 10757.8363 C6H4NO2 kg/jam = 10.8337 C 6 H6 10.8337 kg/jam = H2 O kg/jam = 10.8337
= = = =
Perbandingan mol Nitrobezen terhadap H2 = 1:3 H2 = 524.7725 kg/jam = CH4 = 0.00525 kg/jam = Bahan yang Bereaksi C6H5NO2 yang bereaksi
= =
Konversi 0.98
x
87.4621 0.6019 0.0645 0.1389
262.3863 0.000328
kmol/jam kmol/jam kmol/jam kmol/jam
kmol/jam kmol/jam
x C6H5NO2 yang masuk 10757.8363
= H2 O
C6H4NO2
C 6 H6
H2 yang bereaksi
CH4
Reaksi yang Reaksi C6H5NO2 Mula-mula 87.4621 Bereaksi 85.7128 Sisa 1.7492 Produk yang Terbentuk C6H5NH2 = 85.7128 H2 O = 171.4257
kg/jam
=
85.7128
kmol/jam
= H2O yang masuk = 10.83367 kg/jam
=
0.6019
= C6H4NO2 yang masuk = 10.83367 kg/jam
=
0.0645
kmol/jam
= C6H6 yang masuk = 10.83367 kg/jam
=
0.138893
kmol/jam
= (3/1) x C6H5NO2 yang bereaksi = 514.2771 kg/jam = 257.1385
kmol/jam
= CH4 yang masuk = 0.00525 kg/jam
+
3H2
kmol/jam kmol/jam
=
kmol/jam
0.000328
C6H5NH2
262.3863 257.1385 5.2477
Neraca Massa Total Reaktor Komponen Masuk (kg/jam) C6H5NO2 10757.8363 C6H4NO2 10.8337 H2 O 10.8337 C 6 H6 10.8337 H2 524.7725 CH4 0.00525 C6H5NH2 H2 O Total 11315.1151 Bahan Yang Masuk C6H5NO2 = 215.1567 C6H4NO2 = 10.8337 H2 O = 10.8337 C 6 H6 = 10.8337
10542.68
+
85.7128 85.7128
= =
7971.2943 3085.6623
= = = =
1.7492 0.0645 0.6019 0.1389
2H2O 171.4257 171.4257
kg/jam kg/jam
Keluar (kg/jam) 215.1567 10.8337 10.8337 10.8337 10.4955 0.00525 7971.2943 3085.6623 11315.1151
kg/jam kg/jam kg/jam kg/jam
kmol/jam
kmol/jam kmol/jam kmol/jam kmol/jam
H2 CH4 C6H5NH2 H2 O
= = = =
10.4955 0.00525 7971.2943 3085.6623
kg/jam kg/jam kg/jam kg/jam
= = = =
5.2477 0.000328 43.2987 171.4257
kmol/jam kmol/jam kmol/jam kmol/jam
= = = = = = = =
1.7492 0.0645 0.6019 0.1389 5.2477 0.000328 43.2987 171.4257
kmol/jam kmol/jam kmol/jam kmol/jam kmol/jam kmol/jam kmol/jam kmol/jam
A.3 Flash Tank 5
Flash Tank
4
6 Komposisi Umpan Masuk Bahan Yang Masuk C6H5NO2 = 215.1567 C6H4NO2 = 10.8337 H2 O = 10.8337 C 6 H6 = 10.8337 H2 = 10.4955 CH4 = 0.00525 C6H5NH2 = 7971.2943 H2 O = 3085.6623
kg/jam kg/jam kg/jam kg/jam kg/jam kg/jam kg/jam kg/jam
Komposisi Bahan yang Teruapkan Nitrobenzen (C6H5NO2) 98% x C6H5NO2 yang masuk
=
Di-Nitrobenzen (C6H4NO2) 99% x C6H4NO2 yang masuk
210.8536 kg/jam
=
10.72534 kg/jam
Air (H2O) 98% x H2O yang masuk
=
3034.5661
Benzen (C6H6) 98% x C6H6 yang masuk
=
10.617
Gas Hidrogen (H2) 100.0% x H2 yang masuk
=
Gas Metana (CH4)
=
kg/jam
1.7143
kmol/jam
=
0.0638
kmol/jam
=
168.5870
kmol/jam
kg/jam
=
0.1361
kmol/jam
10.49545 kg/jam
=
5.2477
kmol/jam
100%
x CH4 yang masuk
=
0.00525
kg/jam
=
kg/jam
=
Anilin (C6H5NH2) 0.01% x C6H5NH2 yang masuk
=
Komposisi Bahan Cair Nitrobenzen (C6H5NO2) 2% x C6H5NO2 yang masuk
=
4.303135 kg/jam
Di-Nitrobenzen (C6H4NO2) 1% x C6H4NO2 yang masuk
=
0.108337 kg/jam
Air (H2O) 2% x H2O yang masuk Benzen (C6H6) 2% x C6H6 yang masuk
Komponen
Masuk (kg/jam)
C6H5NO2 C6H4NO2 H2 O C 6 H6 H2 CH4 C6H5NH2
215.1567 10.8337 3096.4960 10.8337 10.4955 0.005247778 7971.2943
Total
11315.1151
Komponen C6H5NO2 C6H4NO2 H2 O C 6 H6
A -54.494 -24.246 18.3036 15.9008
=
kmol/jam
0.0086
kmol/jam
0.0350
kmol/jam
=
0.000645
kmol/jam
=
61.92992 kg/jam
=
3.4406
kmol/jam
=
0.216673 kg/jam
=
0.002778
kmol/jam
Anilin (C6H5NH2) 99.99% x C6H5NH2 yang masuk Hidrogen (H2) 0.00% x H2 yang masuk
0.7971
0.000328
=
=
0.0000
7970.497 kg/jam
kg/jam
=
=
85.7043
0.00000
kmol/jam
Keluar (kg/jam) V L 210.8536 4.3031 10.7253 0.1083 3034.5661 61.9299 10.6170 0.2167 10.4955 0.005247778 0.7971 7970.4972 3278.0598 8037.0552 11315.1151
B -2112.3 -4114 3816.44 2788.51
Antoine C 29.321 16.344 -46.13 -52.36
kmol/jam
D -0.0448 -0.0241 -
E 0.0000 0.0000 -
-
-
H2 CH4 C6H5NH2
13.6333 15.2243 16.6748
164.9 597.84 3857.52
(L/V) data = 2.4518 Dengan menggunakan persamaan Fi Vi = ((L/V)/ki)+1
3.19 -7.16 -73.15
-
(L/V)data Ai = ki
ki =
T trial = 139.5 C = 412.5 K P sistem = 760 mmHg logP = A + B/T + ClogT + DT + ET2
Li =
-
Pi P sistem
Fi (1+(L/V)*ki)
Komponen
Fi (kmol)
Pi (mmHg)
ki
Ai
C6H5NO2 C6H4NO2 H2 O C 6 H6 H2 CH4 C6H5NH2
1.7492 0.0645 172.0276 0.1389 5.2477 3.E-04 43.2987
101.6479 1.1463 2663.536 3492.8994 560527.74 936103.2 201.9982
0.1337 0.0015 3.5047 4.5959 737.5365 1231.7147 0.2658
18.3314 1625.5992 0.6996 0.5335 0.0033 0.0020 9.2246
Total
222.5269
Vi
%V
Li
%Li
0.0905 3.96448E-05 101.2179 0.0906 5.2303 0.000327335 4.2348
0.0008 3.57597E-07 0.9130 0.0008 0.0472 2.95257E-06 0.0382
1.3173 0.0642 17.9333 0.0113 0.002900463 1.08573E-07 26.2154
0.0289 0.0014 0.3938 0.0002 0.0001 2.38388E-09 0.5756
110.8644
1.0000
45.5445
1.0000
L
=
Fi Vi total
= (L/V)hitung
2.0072 =
0.0181
A.5 MENARA DISTILASI
7
Menara Distilasi
6
8
Arus 6 Komponen
Kg/Jam
Kmol/Jam
C6H5NO2 C6H4NO2 H2 O C 6 H6 C6H5NH2
4.3031 0.1083 61.9299 0.2167 7970.4972
0.0350 0.000644861 3.4406 0.002777865 85.7043
Total
8037.0552
89.1832
A. Massa Masuk Menara Distilasi pada Kondisi Bubble Point T = 170.11 °C 443.11 K ; P = = 1 atm Trial pada T akan dianggap benar apabila Syi = 1
=
Komponen
(Kmol/jam)
xi
Pi
ki
C6H5NO2 C6H4NO2
0.0350 0.000644861 3.4406 0.002777865 85.7043
0.0004 7.23075E-06 0.0386 3.11478E-05 0.9610
267.6120 5.5490 5942.6871 6402.8470 516.9484
0.5177 0.0107 11.4957 12.3859 1.0000
89.1832
1
H2 O C 6 H6 C6H5NH2 Total
yi
αi
0.0002 7.76158E-08 0.4435 0.0004 0.9610 1.405
1.0000 0.0207 22.2064 23.9259 1.9317
760
mmHg
Jadi bisa disimpulkan bahwa suhu pemasukan umpan sebesar 170,11 C B. Spesifikasi Hasil Yang Diinginkan B.1 Distilat • Menentukan Massa Distilat 1. C6H5NO2 = 1% x Massa = 0.00035 kmol/jam 2. C6H4NO2 = 1% x Massa = 6.45E-06 kmol/jam H O 3. = 99.00% x Massa = 3.4061 kmol/jam 2 C 6 H6 4. = 1% x Massa = 2.78E-05 kmol/jam C6H5NH2 5. = 1% x Massa = 0.8570 kmol/jam Total = 4.2636 kmol/jam Menentukan Nilai xdi 1. C6H5NO2 = Massa Komponen =
0.00035 4.2636
=
8.20552E-05
= Massa Komponen Massa Total
=
6.45E-06 4.2636
=
1.51249E-06
= Massa Komponen Massa Total
=
3.4061 4.2636
=
0.7989
= Massa Komponen Massa Total
=
2.78E-05 4.2636
=
0
= Massa Komponen Massa Total
=
0.8570 4.263572
=
0.2010
Massa Total 2. C6H4NO2
3. H2O
4. C6H6
6.
C6H5NH2
• Menentukan Massa Bottom 1. C6H5NO2 = 99% 2. C6H4NO2 = 99% 3. H2O = 1% 4. C6H6 = 99% C6H5NH2 5. = 99% Total
x x x x x
Menentukan Nilai xbi 1. C6H5NO2 = Massa Komponen Massa Total 2. C6H4NO2
3. H2O
= Massa Komponen Massa Total
Massa Massa Massa Massa Massa
= 0.0346 kmol/jam = 0.000638 kmol/jam = 0.0344 kmol/jam = 2.75.E-03 kmol/jam = 84.8472 kmol/jam = 84.9197 kmol/jam
=
0.0346 84.9197
=
0.0004
=
0.000638 84.9197
=
0.0000
=
0.0004
= Massa Komponen
0.0344 =
4. C6H6
C6H5NH2
5.
Komponen C6H5NO2 C6H4NO2 H2 O C 6 H6 C6H5NH2 Total
Komponen C6H5NO2 C6H4NO2 H2 O C6H6 C6H5NH2 Total
Massa Total
=
84.9197
=
0.0004
= Massa Komponen Massa Total
=
2.8.E-03 84.9197
=
0.0000
= Massa Komponen Massa Total
=
84.84723 84.91966
=
0.9991
Massa (Kmol/jam) 0.0350 0.000644861 3.4406 0.002777865 85.7043 89.1832
% 1% 1% 99% 1% 1%
Distilat Massa 0.000349848 6.44861E-06 3.4061 2.77786E-05 0.8570 4.2636
xdi 8.20552E-05 1.51249E-06 0.7989 0 0.2010 1.0000
Massa (Kmol/jam) 0.0350 0.000644861 3.4406 2.78.E-03 85.7043 89.1832
% 99% 99% 1% 99% 99%
Bottom Massa 0.0346 0.000638413 0.0344 2.75.E-03 84.8472 84.9197
xbi 0.0004 7.51784E-06 0.0004 0.0000 0.9991 1.0000
C. Perhitungan Suhu Atas (Dew point) Trial pada T akan dianggap benar apabila Sxi = 1 Pt = T=
1 182.7943
Komponen C6H5NO2 C6H4NO2 H2 O C 6 H6 C6H5NH2 Total
atm °C
Massa (kmol/jam) 0.000349848 6.44861E-06 3.4061 2.77786E-05 0.8570 4.2636
= =
760 455.9443
yi= m/m total 8.20552E-05 1.51249E-06 0.798894751 6.51535E-06 0.201015166 1
mmHg K
Pi 382.3358 9.967046 8030.418 8034.002 733.2846
Jadi dapat disimpulkan bahwa T distilat adalah 182,7943°C D. Perhitungan Suhu Bawah (Bubble point)
ki= Pi/P 0.5031 0.0131 10.5663 10.5711 0.9648
xi= yi/ki 0.000163 0.000115 0.0756 6.16E-07 0.2083 0.284
αi = ki/k(HK) 1.0000 0.0261 21.0036 21.0129 1.9179
Trial pada T dianggap benar apabila Syi = 1 Pt = T=
1 184.0100
Komponen C6H5NO2 C6H4NO2 H2 O C6H6 C6H5NH2 Total
atm °C
Massa (kmol/jam) 0.0346 0.000638413 0.0344 2.75.E-03 84.8472 84.9197
= =
760 mmHg 457.1599622 K
xi= m/m total 0.0004 0.0000 0.0004 0.0000 0.9991 1
Pi 394.9502 10.51353 8254.673 8202.446 757.0548
ki= Pi/P 0.519671 0.013834 10.86141 10.79269 0.996125
Jadi dapat disimpulkan bahwa T bottom adalah 172,5794°C
Komponen C 6 H6 H2 O C6H5NH2 C6H5NO2 C6H4NO2 Total
Massa Masuk (kg/gr) 0.2167 61.9299 7970.4972 4.3031 0.1083 8037.0552 8037.0552
Massa Keluar Distilat Bottom 2.E-03 2.E-01 61.3106 0.6193 79.7050 7890.7922 0.0430 4.2601 1.E-03 0.1073 141.0619 7895.9934 8037.0552
yi= xi.ki 0.000212 1.04E-07 0.004401 0.00035 0.995275 1.000
αi = ki/k(HK) 1.0000 0.0266 20.9005 20.7683 1.9168
A ACA MASSA
0.98 0.98
yield konversii
kmol/jam
kmol/jam
kmol/jam
kmol/jam
kmol/jam
kmol/jam
kmol/jam
kmol/jam
mmHg
LAMPIRAN B NERACA PANAS Sebagai Basis Perhitungan Kapasitas Produksi = 62500 Ton/Tahun Hari Kerja = 330 Hari Produksi Aniline Perjam= 7891.4141 Kg/Jam Basis Waktu = 1 Jam Satuan Massa = Kg Satuan Panas = kJ Satuan Cp = kJ/mol o
= 25 C (298,15K)
Suhu Referensi
Data Data yang Diperlukan Kapasitas panas gas, cairan dan padatan Cp = A + BT + CT2 + DT3 + ET4 (kJ/kmol. K) Sehingga Cp dT
=
AT
B 2 C 3 D 4 E 5 T T T T 2 3 4 5
keterangan Cp = kapasitas panas (kJ/kmol K) A, B, C, D, E= konstanta T = suhu (K)
Komponen
A
B
C
D
H2 C 6 H6 H2 O C6H5NH2 C6H5NO2 C 6 H4 N2 O4
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 CH4 34.9420 -3.9957E-02 -1.9184E-04 -1.5300E-07 3.9321E-11 Sumber Yaws dan Coulson Data kapasitas panas untuk liquid Komponen
A
B
C
D
H2 C 6 H6 H2 O C6H5NH2 CH4 C6H5NO2 C 6 H4 N2 O4
28.8400 -33.6620 18.2964 46.9480 -0.0180 39.4730 -12.6350
0.00765 0.4743 0.4721 0.9896 1.1982 0.9128 1.5624
3.29E-01 -0.0036 -1.3388E-03 -2.3583E-03 -9.8722E-03 -0.0021 -2.9981E-03
-8.70E-10 3.8243E-06 1.3142E-06 2.3296E-06 3.1670E-05 2.0093E-06 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)2= 162.2 J/mol C Treff =
20
C
Data entalpi pembentukan Komponen ΔHf (Kj/mol) H2 0 C 6 H6 82.9 H2 O -241.826 C6H5NH2 86.86 CH4 -74.5 C6H5NO2 67.6 C 6 H4 N2 O4 50.8 Sumber Yaws an Himmelblau Data berat molekul Komponen Rumus molekul H2 Hidrogen C6 H6 Benzene H2 O Air C6H5NH2 Anilin CH4 Metana C6H5NO2 Nitrobenzen C 6 H4 N 2 O4 Di-Nitrobenzen
BM 2 78 18 93 16 123 168
Data bilangan Antoine Komponen C6H5NO2 C 6 H4 N2 O4
A -54.494 -24.246
B -2112.3 -4114
Antoine C 29.321 16.344
D -0.0448 -0.0241
E 0.00002 0.00001
H2 O C 6 H6 H2 CH4 C6H5NH2
18.3036 15.9008 13.6333 15.2243 16.6748
3816.44 2788.51 164.9 597.84 3857.52
-46.13 -52.36 3.19 -7.16 -73.15
-
-
-
-
B.1 Heater (E-114) Fungsi : Mengubah fase Hidrogen cair menjadi uap Hidrogen Tujuan
:
Menaikkan suhu feed menjadi 250oC Q3
Qsteam
Q1
Q2
Neraca Energi ΔH = Hout - Hin Hin = Hpendingin + Hout 1. Menghitung Panas Sensibel Masuk Heater, Q1 Massa n Komponen (kg) (kmol) H2 524.7727 262.3863 CH4 0.0052 0.0003 Tin
=
Tref
●
=
30 25
o
C=
303.15
K
o
C=
298.15
K
H2 303.15
Cp dT
Q = n
298.15 303.15
= 262.3863
27.1430T + 9.2730E-03 T2 - 1.3800E-05 T3 + 7.6450E-09 T4 dT
298.15
= 262.3863 27.1430 +
303.15 -
298.15
-1.3800E-05 3 303.15 3
298.15
303.15 =
4
-
298.15
40129.7403
kJ/jam
4
+ 3
9.27E-03 2 303.15 2 +
7.6450E-09 4
298.15
2
● CH4 303.15
Cp dT
Q = n
298.15 303.15
34.942 T-3.9957E-02 T2 - 1.9184E-04 T3-1.53E-07 T4+3.9321E-11 T5 dT
= 0.000328
298.15
= 0.000328 34.9420 +
303.15 -
298.15
-1.9184E-04 3 303.15 3
298.15
4
303.15
-
298.15
0.1520
=
4
+
-0.04 2
+ 3
+
3.9321E-11 5
303.15
2
-
298.15
2
-1.5300E-07 4 303.15
5
-
298.15
5
kJ/jam
Q1 = Q H2 + Q CH4 = 40129.7403 + 0.1520 = 40129.8923 kJ/jam 2. Menghitung Panas Sensibel Keluar Heater, Q2 *Hout Tout Tref
= =
250 25
o
C=
523.15
K
o
C=
298.15
K
● H2 523.15
Cp dT
Q = n
298.15 523.15
= 262.3863
27.1430T + 9.2730E-03 T2 - 1.3800E-05 T3 + 7.6450E-09 T4 dT
298.15
= 262.3863 27.1430 +
523.15 -
298.15
-1.3800E-05 3 523.15 3
298.15
4
523.15
-
= 1720025.5993
298.15
+ 3
9.27E-03 2 523.15 2 +
298.15
2
7.6450E-09 4
4
kJ/jam
● CH4 523.15
Q = n
Cp dT
298.15 523.15
= 0.000328
34.942 T-3.9957E-02 T2 - 1.9184E-04 T3-1.53E-07 T4+3.9321E-11 T5dT
298.15
= 0.000328 34.9420 +
523.15 -
298.15
-1.9184E-04 3 523.15 3
298.15
-0.04 2
+ 3
+
523.15
-1.5300E-07 4
2
-
298.15
2
523.15 =
4
-
4.75040
4
298.15
+
3.9321E-11 5
523.15
5
-
298.15
5
kJ/jam
Q2 = Q H2 + Q CH4 = 1720025.59928 +
4.75040
= 1720030.34968 kJ/jam
3. Menghitung Kebutuhan Pemanas, Q3 Q3 = Q2 - Q1 = 1720030.3497 40129.8923 = 1679900.4574 kJ/jam Kebutuhan panas Heater sebesar
1679900.4574 kJ, panas disupply menggunakan o
Saturated steam, dengan suhu 300 C dengan tekanan 1 atm, dari properties of saturated water and saturated steam up to 1 atm, stoichiometry 2004 diperoleh data: λsteam = 909.99 kJ/kg Sehingga : Q3 = m.λsteam Massa steam = Q3 = λsteam
1679900.4574 kJ/mol= 909.99 kJ/kg
1846.0647
Neraca Panas Total dalam Heater Panas masuk Panas keluar Komponen (kJ/jam) (kJ/jam) Q H2 40129.7403 1720025.5993 Q CH4 0.15198 4.7504 Q Steam Total
1679900.4574 1720030.3497
1720030.3497
B.2. VAPORIZER (V-130) Fungsi : Menguapkan bahan baku Nitrobenzene sebelum masuk reaktor Tujuan : - Menentukan Suhu Keluar (T out) Vaporizer. - Menghitung kebutuhan pemanas Arus 3 204.45 oC
Arus 2 o 30 C
Steam 250 Qs
o
C
Steam 250
o
C
kg/jam
1. Penentuan Kondisi Operasi Vaporizer (E-121) Data Komponen Masuk M Komponen (kg) C6H5NO2 13447.2999 C6H4N2O4 13.5420945 13.5420945 H2 O 13.5420945 C 6 H6
n (kmol) 109.3276 0.0806 0.7523 0.1736
Menentukan Suhu Keluar (T out) Vaporizer ● Menentukan Kondisi Dew Point Kondisi Operasi P = T =
1 atm = 760.0 mmHg 210.82 oC = 483.97 K
Tabel Perhitungan Tekanan Dew Point n Pi sat Komponen xi kmol/jam mmHg C6H5NO2 0.8073 0.9909 766.9869 C6H4N2O4 0.0006 0.0007 31.3712 0.0013 0.0016 12582.7631 H2 O 0.0056 0.0068 14575.28479 C 6 H6 Total 0.8147 1.0000
Ki Pi sat/P 1.0092 0.0413 16.5563 19.1780
yi Ki/xi 0.9819 0.0177 0.0001 0.0004 1.0000
● Menentukan Kondisi Bubble Point Kondisi Operasi P = 1 atm = 760.0 mmHg T = 204.45 oC = 477.60 K Tabel Perhitungan Tekanan Bubble Point n Pi sat Komponen xi kmol/jam mmHg C6H5NO2 0.8073 0.9909 660.6744 C6H4N2O4 0.0006 0.0007 24.5323 0.0013 0.0016 11422.0738 H2 O 0.0056 0.0068 12815.31724 C 6 H6 Total 0.8147 1.0000
Ki Pi sat/P 0.8693 0.0323 15.0290 16.8623
Suhu keluar vaporizer merupakan suhu pada kondisi operasinya yaitu pada kondisi bubble point T out = 204.45 oC = 477.60 K Perhitungan Neraca Panas pada Vaporizer (V-130) a. Panas Sensibel Cairan Masuk, Q1 T Larutan = 30 C T reff =
25
C
yi Ki.xi 0.8614 0.0000 0.0236 0.1150 1.0000
= ●
303.15
K
=
298.15
K
C6H5NO2 303.15
Q = n
Cp dT
298.15
= 109.328 39.47 303.15
=
-0.0021 3 303.15 3 97189.7934 kJ/jam
●
C 6 H4 N2 O4
-
303.15
Q = n 298.15
=
0.0806
298.15
-
298.15
0.9128 2
+ 3
303.15
2
-
+
2.01E-06 4 303.15 4
+
1.56240 2
+
2.32E-06 4 303.15 4
298.15 -
2
298.15
4
Cp dT
-12.6
303.15 -
-3.E-03 3 303.15 3 = 318.8357 kJ/jam +
●
-
-
298.15 298.15
3
303.15
2
-
298.15 -
298.15
2 4
C 6 H6 303.15
Q = n 298.15
=
0.1736 -33.6620 +
= ●
Cp dT 303.15 -
####### 3 303.15 3 467.9 kJ/jam
-
0.4743 2
298.15 298.15
3
+
303.15
2
-
3.82E-06 4 303.15 4
298.15 -
298.15
H2 O 303.15
Q = n 298.15
=
0.7523
18.2964
303.15 -
####### 3 303.15 3 281.8968 kJ/jam
+ =
Cp dT
-
298.15 298.15
3
+ +
0.4721 2
2
1.31E-06 4 303.15 4
Panas sensibel masuk, Q1 Q1 = Q C6H5NO2 + Q C6H4N2O4 + Q C6H6 + H2O Q1= 97189.7934 + 318.8357 + 467.8883 Q1= 98258.4142 kJ/jam b. Panas Laten Penguapan Komponen, Q
303.15
+
-
298.15 -
281.8968
298.15
kJ/jam
2 4
Komponen
Tb (K)
Tc (K)
C6H5NO2 483.9500 C6H4N2O4 572.0000 353.2400 H2 O 373.1500 C 6 H6
719.0000 803.0000 562.1600 647.4000
ΔHv (Kj/mol) 44.0800 61.5600 30.7500 40.6800
Yaws Yaws Yaws Himmelblau
Untuk menghitung entalpi panas penguapan (ΔHv) digunakan Persamaan Watson: ΔH2 1 - Tr2 0.38 (Pers 4.13 Smith Van Ness, 200) = ΔH1 1 - Tr1 ΔH2 = ΔH1 ΔH1 ΔH2 Tr2 Tr1 T1 T2
dimana:
●
1 1 = = = = = = =
- Tr2 0.38 - Tr1 Panas laten penguapan pada titik didih normal (kJ/kmol) Panas laten penguapan pada suhu T2 (kJ/kmol) T2/TC (K) T1/TC (K) Titik didih normal komponen (K) Suhu tertentu = 210.82 oC = 483.97 K T dew
C6H5NO2
ΔH2 = ΔH1 x
1 1 1 x 1
Tr2 Tr1
483.97 719.0000 0.38 = - 483.9500 719.0000
ΔH2 =
44.08
ΔHv =
0.8073 kmol/jam x
●
44.0786
kJ/kmol =
44.0786
35.5854
kJ/kmol
kJ/jam
C 6 H4 N2 O4
ΔH2 = ΔH1 x
1 1 1 x 1
Tr2 Tr1
0.38
483.97 803.0000 0.38 = - 572.0000 803.0000
ΔH2 =
61.56
ΔHv =
0.0006 kmol/jam x
●
0.38
69.5956
kJ/kmol =
69.5956
0.0414
kJ/kmol
kJ/jam
C 6 H6
ΔH2 = ΔH1 x
ΔH2 =
30.75
1 1 -
x
1 1
Tr2 Tr1
0.38
483.97 562.1600 0.38 = - 353.2400 562.1600
21.1666
kJ/kmol
ΔHv =
0.0013 kmol/jam x
21.1666
kJ/kmol =
0.0271
kJ/jam
● H2 O ΔH2 = ΔH1 x
Tr2 Tr1
1 1 1 x 1
0.38
483.97 647.4000 0.38 = - 373.1500 647.4000
ΔH2 =
40.68
ΔHv =
0.0056 kmol/jam x
33.4158
kJ/kmol =
Q2= Q C6H5NO2 + Q C6H4N2O4 + Q C6H6 + Q H2O Q2= 35.5854 + 0.0414 + 0.0271 Q2= 35.8396 kJ/jam c. Panas Sensibel Uap Keluar (Q3) T uap = 204.45 C = 477.6 K ●
T reff
= =
25 298.15
33.4158
kJ/kmol
0.1856
+
kJ/jam
0.1856
kJ/jam
C K
C6H5NO2 477.60
Q = n 298.15
Cp dT
= 109.3276 -16.2
=
-4.E-04 3
477.60
3
477.60
-
-
298.15
298.15
3
+
0.5618 2
477.60
2
+
1.00E-07 4 477.60 4
+
0.56182 2
+
1.00E-07 4 477.60 4
-
298.15
-
298.15
2
4
####### 5 5 477.60 - 298.15 5 5259818.8569 kJ/jam = +
●
C 6 H4 N2 O4 477.60
Q = n
Cp dT
298.15
=
0.0806
18.1
+
-4.E-04 3
477.60
477.60 3
-
298.15 298.15
3
####### 5 5 477.60 - 298.15 5 2633.8489 = kJ/jam +
●
C 6 H6
477.60
2
-
298.15 -
298.15
2 4
477.60
Q = n 298.15
=
Cpg dT
0.1736 -31.3860 +
477.60 -
####### 3 477.60 3
####### 5 477.60 5 = 6410.1414 kJ/jam +
●
0.4746 2
298.15
-
298.15
-
298.15
3
+
477.60
2
-
8.52E-08 4 477.60 4
298.15 -
298.15
5
H2 O 477.60
Q = n 298.15
=
0.7523
Cp dT 32.2430
477.60 -
1.06E-05 3 477.60 3 = 6179.9152 kJ/jam +
-
298.15 298.15
3
+ +
0.0019 2
2
####### 4 477.60 4
Q3 = Q C6H5NO2 + Q C6H4N2O4 + Q C6H6 + Q H2O Q3= 5259818.8569 + 2633.8489 + 6410.1414 Q3= 5275042.7623 kJ/jam d. Panas yang dibutuhkan oleh pemanas Q1+Q4 = Q2+Q3 Q4 = (Q2+Q3)-Q1 Q4 = 35.8396 + 5275042.7623 Q4 = 5176820.1877 (kJ/jam)
477.60
98258.4142
+
-
6179.9152
298.15
kJ/jam
(kJ/jam)
Sebagai pemanas digunakan steam yaitu saturated steam pada suhu 300oC dan dan tekanan 1 atm. Dari properties of saturated water and saturated steam up to 1 atm, stoichiometry 2004 λ = 909.99 kJ/kg Sehingga: Q4 λ 5176820.1877 kJ/jam = = 909.99 kJ/kg
msteam =
5688.8759 kg/jam
(V-130) Tabel Neraca Panas Vaporizer Komponen Q in (kJ/jam) Q out (kJ/jam) C6H5NO2 97189.7934 C 6 H4 N2 O4 318.8357 -
2
298.15 4
C 6 H6 H2 O Q preheating Q vaporizing Q steam Total
467.8883 281.8968 5176820.1877 5275078.6019
5275042.7623 35.8396 5275078.6019
B.3. SEPARATOR (H-120) Fungsi : Memisahkan fase gas dan fase cair yang terbentuk :
Menghitung panas setiap arus Arus 4 204 oC
Arus 3 204 oC
Separator
Tujuan
Arus 5 204 oC Neraca Energi Q in = 3 = dimana: 3 = 4= 5=
Q out 4+ 5 Panas sensibel gas masuk separator, Q1 Panas sensibel gas keluar separator, Q2 Panas sensibel cairan keluar separator, Q3
Untuk menghitug panas masing-masing arus digunakan persamaan: dimana: 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^3T_reff ³)) +(D/4 x(T^4-T_reff⁴))+(E/5 x(T^5-T_reff⁵))] 〗
1. Panas Sensibel Gas Masuk, Q1 Panas sensibel yang masuk Separator sama dengan panas gas yang keluar Vaporizer . Q1 = 5275042.7623 kJ/jam 2. Panas Sensibel Gas Keluar, Q2 Komposisi arus 4: C6H5NO2 = 10757.8399 kg/jam C 6 H4 N2 O4 = 10.83368 kg/jam C 6 H6 10.83368 kg/jam =
= = =
87.4621 kmol/jam 0.0645 kmol/jam 0.1389 kmol/jam
H2 O ●
=
10.83368
kg/jam
=
0.6019 kmol/jam
C6H5NO2 477.60
Q = n
Cpg dT
298.15 477.60
= 87.4621
-16.2020 T + 0.5618 T2 - 3.93E-4 T3 - 1.004E-7 T4 - 1.2252E-12 T4
dT
298.15
= 87.4621 -16.2020
+
477.60 -
####### 3 477.60 3
####### 5 477.60 5 = 4207855.0855 kJ/jam ● C 6 H4 N2 O4 +
298.15
-
298.15
-
298.15
3
+
+
0.5618 2
477.60
2
-
####### 4 477.60 4
298.15
-
298.15
2
4
5
477.60
Q = n
Cp dT
298.15
477.60
=
0.0645
18.1480 T + 0.5618 T2 - 3.93E-4 T3 + 1.004E-7 T4 - 1.2252E-12 T5 dT
298.15
=
0.0645 +
18.1480
477.60 -
####### 3 477.60 3
####### 5 477.60 5 2118.1309 kJ/jam
+ = ●
298.15
-
298.15
-
298.15
3
+
0.56182 2
+
1.00E-07 4 477.60 4
477.60
2
-
298.15 -
298.15
2 4
5
C 6 H6 477.60
Q = n
Cp dT
298.15 477.60
=
0.1389
-31.3860 T + 0.4746 T2 - 3.11E-4 T3 + 8.524E-8 T4 - 5.0524E-12 T5dT
298.15
=
0.1389 -31.3860 +
####### 3 477.60 3
####### 5 477.60 5 5128.1131 kJ/jam
+ = ●
477.60 -
H2 O
298.15
-
298.15
-
298.15
3 5
+
0.47460 2
+
8.52E-08 4 477.60 4
477.60
2
-
298.15 -
298.15
2 4
477.60
Cp dT
Q = n
298.15
477.60
=
0.6019 298.15
=
0.6019
32.243 T + 1.923E-3 T2 + 1.055E-5 T3 - 3.596E-9 T4 dT
32.2430
477.60 -
1.06E-05 3 477.60 3 3713.5975 kJ/jam
+ =
Q2 = =
-
298.15 298.15
4207855.0855 + 2118.1309 + 4218814.9270 kJ/jam
Sehingga, panas keluar separator: Q1 = Q2 + Q3 Q3 = Q1 - Q2 = 5275042.7623 - 4218814.9270
3
+
0.00192 2
+
####### 4 477.60 4
5128.1131
477.60
+
Tabel Neraca Massa Separator (H-120) Panas masuk (kJ/jam) Panas keluar (kJ/jam) Komponen Arus 3 Arus 4 Arus 5 Q C6H5NO2 5259818.8569 4207855.0855 1051963.7714 Q C 6 H4 N2 O4 2633.8489 2118.1309 515.7179 Q C 6 H6 6410.1414 5128.1131 1282.0283 Q H2 O 6179.9152 3713.5975 2466.3177 4218814.9270 1056227.8353 Total 5275042.7623 5275042.7623
Tujuan
:
o
Menaikkan suhu menjadi 250 C Q3 Q1
Neraca Energi
Qsteam Q2
-
298.15 -
3713.5975
= 1056227.8353 kkal/jam
B.4 Heater (E-114) Fungsi : Menaikkan suhu uap Nitrobenzen
2
298.15
2 4
ΔH = Hout - Hin Hin = Hpendingin + Hout 1. Menghitung Panas Sensibel Masuk Heater, Q1 M n Komponen (kg) (kmol) C6H5NO2 10757.8399 87.4621 C6H4N2O4 10.83368 0.0806 10.83368 0.7523 H2 O 10.83368 0.1736 C 6 H6 *Hin Tin
=
Tref
=
●
204 25
o
C=
477.60
K
o
C=
298.15
K
C6H5NO2 477.60
Cp dT
Q = n
298.15
477.60
-16.2020 T + 0.5618 T2 - 3.93E-4 T3 - 1.004E-7 T4 - 1.2252E-12 T4 dT
= 87.4621
298.15
= 87.4621 -16.2020 +
477.60 -
####### 3 477.60 3
####### 5 477.60 5 4207855.0855 kJ/jam
+ =
●
298.15
-
298.15
-
298.15
3
+ +
0.5618 2
477.60
2
-
####### 4 477.60 4
298.15 -
298.15
2 4
5
C 6 H4 N2 O4 477.60
Cp dT
Q = n
298.15
477.60
=
0.0806 298.15
=
0.0806 +
●
18.1480
477.60 -
####### 3 477.60 3
####### 5 477.60 5 2633.8489 kJ/jam
+ =
18.1480 T + 0.5618 T2 - 3.93E-4 T3 + 1.004E-7 T4 - 1.2252E-12 T5 dT
C 6 H6
298.15
-
298.15
-
298.15
3 5
+
0.56182 2
+
1.00E-07 4 477.60 4
477.60
2
-
298.15 -
298.15
2 4
477.60
Cp dT
Q = n
298.15
477.60
=
-31.3860 T + 0.4746 T2 - 3.11E-4 T3 + 8.524E-8 T4 - 5.0524E-12 T5dT
0.1736
298.15
=
0.1736 -31.3860 +
477.60 -
####### 3 477.60 3
####### 5 477.60 5 6410.1414 kJ/jam
+ = ●
298.15
-
298.15
-
298.15
3
+
0.47460 2
+
8.52E-08 4 477.60 4
477.60
2
-
298.15 -
298.15
2 4
5
H2 O 477.60
Cp dT
Q = n
298.15
477.60
=
0.7523 298.15
=
0.7523
32.243 T + 1.923E-3 T2 + 1.055E-5 T3 - 3.596E-9 T4 dT
32.2430
477.60 -
1.06E-05 3 477.60 3 4641.9969 kJ/jam
+ =
-
298.15 298.15
3
+
0.00192 2
+
####### 4 477.60 4
477.60
2
-
298.15 -
298.15
2 4
Q1 = Q C6H5NO2 + Q C6H4N2O4 + Q C6H6 + Q H2O = 4221541.0726 kJ/jam 2. Menghitung Panas Sensibel Keluar Heater, Q2 *Hout Tin
=
Tref ●
=
250 25
o
C=
523.15
K
o
C=
298.15
K
C6H5NO2 523.15
Q = n
Cp dT
298.15
523.15
= 87.4621
-16.2020 T + 0.5618 T2 - 3.93E-4 T3 - 1.004E-7 T4 - 1.2252E-12 T4 dT
298.15
= 87.4621 -16.2020 +
####### 3 523.15 3
####### 5 523.15 5 5706082.1206 kJ/jam
+ =
523.15 -
298.15
-
298.15
-
298.15
3 5
+ +
0.5618 2
523.15
2
####### 4 523.15 4
-
298.15 -
298.15
2 4
●
C 6 H4 N2 O4 523.15
Cp dT
Q = n
298.15
523.15
=
18.1480 T + 0.5618 T2 - 3.93E-4 T3 + 1.004E-7 T4 - 1.2252E-12 T5 dT
0.0806
298.15
=
0.0806 +
18.1480
523.15 -
####### 3 523.15 3
####### 5 523.15 5 3417.6783 kJ/jam
+ = ●
298.15
-
298.15
-
298.15
3
+
0.56182 2
+
1.00E-07 4 523.15 4
523.15
2
-
298.15 -
298.15
2 4
5
C 6 H6 523.15
Q = n
Cp dT
298.15 523.15
=
-31.3860 T + 0.4746 T2 - 3.11E-4 T3 + 8.524E-8 T4 - 5.0524E-12 T5dT
0.1736
298.15
=
0.1736 -31.3860 +
523.15 -
####### 3 523.15 3
####### 5 523.15 5 8744.0538 kJ/jam
+ = ●
298.15
-
298.15
-
298.15
3
+
0.47460 2
+
8.52E-08 4 523.15 4
523.15
2
-
298.15 -
298.15
2 4
5
H2 O 523.15
Q = n
Cp dT
298.15
523.15
=
0.7523 298.15
=
0.7523
32.2430
523.15 -
1.06E-05 3 523.15 3 5855.0194 kJ/jam
+ =
32.243 T + 1.923E-3 T2 + 1.055E-5 T3 - 3.596E-9 T4 dT
-
298.15 298.15
3
+
0.00192 2
+
####### 4 523.15 4
Q2 = Q C6H5NO2 + Q C6H4N2O4 + Q C6H6 + Q H2O = 5724098.8721 kJ/jam
3. Menghitung Kebutuhan Pemanas, Q3
523.15
2
-
298.15 -
298.15
2 4
Q3 = Q2 - Q1 = 5724098.8721 4221541.0726 = 1502557.7995 kJ/jam Kebutuhan panas Heater sebesar
1502557.7995 kJ, panas disupply menggunakan o
saturated steam dengan suhu 300 C dengan tekanan 1 atm. Dari properties of saturated water and saturated steam up to 1 atm, stoichiometry 2004 λsteam = 909.99 kJ/kg Sehingga : Q3 = m.λsteam Massa steam = Q3 = λsteam
1502557.7995 kJ/jam= 909.99 kJ/kg
1651.1806
Neraca Panas Total dalam Heater Panas masuk Panas keluar Komponen (kJ/jam) (kJ/jam) Q C6H5NO2 4207855.0855 5706082.1206 Q C6H4N2O4 2633.84886 3417.6783 Q H2 O Q C 6 H6 Q Steam Total
4641.9969 6410.1414 1502557.7995 5724098.8721
5855.0194 8744.0538 5724098.8721
B.5. REAKTOR (R-210) Fungsi : Tempat berlangsungnya reaksi Anilin Tujuan : - Menghitung suhu keluar reaktor - Menghitung kebutuhan pendingin 250 oC Arus 6 Pendingin Qs 28 oC Pendingin 45 Qrx Arus 4 Neraca Energi Q in = 4 + Qs = dimana: 4 = 6 = Qs =
250
o
o
C
C
Q out 6 + Qrx Panas Sensibel Gas Reaktan Masuk, Q reaktan Panas Sensibel Gas Produk Keluar, Q produk Jumlah pendingin yang dibutuhkan
kg/jam
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/jamdan Heater Hidrogen 1720030.35 kJ/jam keluar Heater Nitrobenzen Q reaktan = 7444129.2218 kJ/jam 2. Panas Reaksi, ΔHR Reaksi: C6H5NO2 + H2 C6H5NH2 + H2O -
Panas reaksi pada keadaan standar (T = 25oC) o ΔHr 298,15 = ∆Hof (produk) - ∆Hof (reaktan) = (∆H of C6H5NH2 + ∆H of H2O) - (∆H of C6H5NO2 + H2) 86.86 + ( -241.83 ) - ( 67.60 + 0 = -222.57 kJ/kmol = Panas reaksi, ΔHR
)
kJ/mol
ΔHRo523.15 = n ( ΔHro298,15 ) = XA ( n reaktan) ( ΔHro298,15 ) 88.4687 = 0.98 x x -222.57 -19296.3169 kJ/jam = Q reaksi adalah panas yang diperlukan untuk reaksi di dalam reaktor, dimana reaksi berupa reaksi eksotermis (ΔHR negatif). Sehingga, perlu ditambahkan 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: Q=H=∑▒〖n∫_(T_reff)^T▒〖Cp_g dT 〗〗
dimana:
∫_(T reff )^T▒〖 Cp_g dT= [(Ax(T-T_reff ))+(B/2 x(T^2-T_reff^ ²))+(C/3 x(T^3T_reff ³)) +(D/4 x(T^4-T_reff⁴))+(E/5 x(T^5-T_reff⁵))] 〗
Komposisi arus 6: C6H5NO2 = 215.1568 kg/jam C 6 H4 N2 O4 = 10.8337 kg/jam 10.8337 kg/jam H2 O = 10.8337 kg/jam C 6 H6 = 10.4955 kg/jam H2 = 0.0052 kg/jam CH4 = 7971.2970 kg/jam C6H5NH2 = 3085.6633 kg/jam H2 O = ●
1.7492 = 0.0645 = 0.6019 = 0.1389 = 5.2477 = 0.0003 = = 85.7129 = 171.4257
kmol/jam kmol/jam kmol/jam kmol/jam kmol/jam kmol/jam kmol/jam kmol/jam
C6H5NO2 523.15
Q = n
Cp dT
298.15 523.15
=
1.7492
-16.2020 T + 0.5618 T2 - 3.93E-4 T3 - 1.004E-7 T4 - 1.2252E-12 T4 dT
=
-16.2020 T + 0.5618 T2 - 3.93E-4 T3 - 1.004E-7 T4 - 1.2252E-12 T4 dT
1.7492
298.15
=
1.7492 -16.2020 +
523.15 -
####### 3 523.15 3
####### 5 523.15 5 114121.6424 kJ/jam
+ =
●
298.15
-
298.15
-
298.15
3
+ +
0.5618 2
523.15
2
-
####### 4 523.15 4
298.15 -
298.15
2 4
5
C 6 H4 N2 O4 523.15
Cp dT
Q = n
298.15
523.15
=
18.1480 T + 0.5618 T2 - 3.93E-4 T3 + 1.004E-7 T4 - 1.2252E-12 T5 dT
0.0645
298.15
=
0.0645 +
18.1480
523.15 -
####### 3 523.15 3
####### 5 523.15 5 2734.1427 kJ/jam
+ = ●
298.15
-
298.15
-
298.15
3
+
0.56182 2
+
1.00E-07 4 523.15 4
523.15
2
-
298.15 -
298.15
2 4
5
C 6 H6 523.15
Q = n 298.15
Cp dT 523.15
=
-31.3860 T + 0.4746 T2 - 3.11E-4 T3 + 8.524E-8 T4 - 5.0524E-12 T5dT
0.1389
298.15
=
0.1389 -31.3860 +
523.15 -
####### 3 523.15 3
####### 5 523.15 5 = 94198.1462 kJ/jam ●
298.15
-
298.15
-
298.15
3
+
0.47460 2
+
8.52E-08 4 523.15 4
523.15
2
-
298.15 -
298.15
2 4
5
H2 O 523.15
Q = n
Cp dT
298.15
523.15
=
0.6019
32.243 T + 1.923E-3 T2 + 1.055E-5 T3 - 3.596E-9 T4 dT
298.15
=
0.6019
32.2430
523.15 -
298.15
+
0.00192
523.15
2
-
298.15
2
=
0.6019
32.2430
523.15 -
1.06E-05 3 523.15 3 4684.0155 kJ/jam
+ =
●
-
298.15 298.15
3
+ +
2
523.15
-
####### 4 523.15 4
298.15 -
298.15
4
H2 523.15
Cp dT
Q = n
298.15 523.15
=
5.2477
27.1430T +
9.2730E-03 T2 - 1.3800E-05 T3 + 7.6450E-09 T4 dT
298.15
=
5.2477 +
27.1430
523.15 -
298.15
-1.3800E-05 3 523.15 3
298.15
4
523.15 =
-
298.15
34400.5120
kJ/jam
+ 3
9.27E-03 2 523.15 2 +
298.15
2
7.6450E-09 4
4
● CH4 523.15
Cp dT
Q = n
298.15 523.15
34.942 T-3.9957E-02 T2 - 1.9184E-04 T3-1.53E-07 T4+3.9321E-11 T5 dT
= 0.000328
298.15
= 0.000328 34.9420 +
523.15 -
298.15
-1.9184E-04 3 523.15 3
298.15
4
523.15
298.15
4.7504
= ●
-
4
+
+ 3
####### 2 523.15 2 +
3.9321E-11 5
298.15
2
-1.5300E-07 4 523.15
5
-
298.15
5
kJ/jam
C6H5NH2 523.15
Q = n
Cp dT
298.15
523.15
= 85.7129
-22.0620 T + 0.5731 T2 - 4.57e-4 T3 - 1.841e-7 T4 - 2.9867e-11 T5 dT
298.15
= 85.7129 -22.0620 +
523.15 -
####### 3 523.15 3
####### 5 523.15 5 = 5918417.6087 kJ/jam
298.15
-
298.15
-
298.15
3 5
+
0.57313 2
+
####### 4 523.15 4
523.15
2
-
298.15 -
298.15
2 4
●
H2 O 523.15
Q = n
Cp dT
298.15
523.15
= 171.4257 298.15
32.243 T + 1.923E-3 T2 + 1.055E-5 T3 - 3.596E-9 T4 dT
= 171.4257 32.2430
523.15 -
1.06E-05 3 523.15 3 = 1334108.1608 kJ/jam +
-
298.15 298.15
3
Jadi: Q produk = ########## + 2734.1427 + 34400.5120 + 4.7504 + 7502668.9787 = kJ/jam
+
0.00192 2
+
####### 4 523.15 4
523.15
-
298.15 -
298.15
94198.1462 + 4684.0155 + 5918417.6087 + 1334108.1608
4. Pendingin yang dibutuhkan oleh Pemanas Qlepas = (Q produk + Q reaksi ) - Q reaktan = 7502668.9787 + -19296.3169 - 7444129.2218 39243.4401 = kJ/jam Sebagai pendingin digunakan air pada suhu 25oC dan tekanan 1 atm. Diperkirakan air keluar pada suhu 45oC. Dari App A.2 Geankoplis, 2003 diperoleh Cp air = 0.9987 kkal/kg.oC = 4.1787 kJ/kgoC Q = m . Cp. ΔT Qc mc = Cp. ΔT 39243.4401 = = 469.5695 kg/jam 4.1787 ( 45 - 25 ) Tabel Neraca Panas Reaktor (R-210) Komponen Q in (kJ/jam) Q out (kJ/jam) Q C6H5NO2 5706082.1206 114121.6424 Q C6H4N2O4 3417.6783 2734.1427 5855.0194 94198.1462 Q H2 O 8744.0538 4684.0155 Q C 6 H6 1720025.5993 34400.5120 Q H2 4.7504 4.7504 Q CH4 5918417.6087 Q C6H5NH2 1334108.1608 Q H2 O Q pendingin 39243.4401 Q reaksi -19296.3169 Total 7483372.6618 7483372.6618
2
2 4
B.6. COOLER 1 (E-221) Fungsi : Mendinginkan gas produk keluar reaktor Tujuan : Menghitung kebutuhan pendingin o
28 C Q1 250
o
Q2 139.53
C Qc
45
o
C
o
C Neraca Energi Q in = Q out Q1 = Q2 + Q3 Q1 = Panas sensibel gas keluar reaktor dimana: Q3 = Panas yang diserap oleh pendingin Q2 = Panas sensibel cairan keluar cooler 1 Untuk menghitug panas masing-masing arus digunakan persamaan: Q=H=∑▒〖n∫_(T_reff)^T▒〖Cp_g dT 〗〗
dimana:
∫_(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 ³)) +(D/4 x(T^4-T_reff⁴))+(E/5 x(T^5-T_reff⁵))] 〗
1. Panas Sensibel Gas Masuk, Q1 Panas sensibel yang masuk Cooler sama dengan panas sensibel yang keluar Condensor Q1 = 3358498.9810 kJ/jam
2. Panas Sensibel gas keluar, Q2 Komposisi arus 6: C6H5NO2 = 215.1568 kg/jam C 6 H4 N2 O4 = 10.8337 kg/jam 10.8337 kg/jam H2 O = 10.8337 kg/jam C 6 H6 = 10.4955 kg/jam H2 = 0.0052 CH4 = kg/jam C6H5NH2 = 7971.2970 kg/jam H2 O = 3085.6633 kg/jam ●
= = = = = = = =
1.7492 0.0645 0.6019 0.1389 5.2477 0.0003 85.7129 171.4257
kmol/jam kmol/jam kmol/jam kmol/jam kmol/jam kmol/jam kmol/jam kmol/jam
C6H5NO2 412.68
Q = n
Cp dT
298.15 412.68
=
1.7492
-16.2020 T + 0.5618 T2 - 3.93E-4 T3 - 1.004E-7 T4 - 1.2252E-12 T4 dT
298.15
=
1.7492 -16.2020 + +
= ●
412.68 -
####### 3 412.68 3 ####### 5 412.68 5 47718.2680 kJ/jam
298.15
-
298.15
-
298.15
3
+ +
0.5618 2
412.68
2
-
####### 4 412.68 4
298.15 -
298.15
2 4
5
C 6 H4 N2 O4 412.68
Q = n
Cp dT
298.15
412.68
=
18.1480 T + 0.5618 T2 - 3.93E-4 T3 + 1.004E-7 T4 - 1.2252E-12 T5 dT
0.0645
298.15
=
0.0645 +
18.1480
412.68 -
####### 3 412.68 3
####### 5 412.68 5 1273.1909 kJ/jam
+ = ●
298.15
-
298.15
-
298.15
3
+
0.56182 2
+
1.00E-07 4 412.68 4
412.68
2
-
298.15 -
298.15
2 4
5
C 6 H6 412.68
Cp dT
Q = n
298.15
412.68
=
-31.3860 T + 0.4746 T2 - 3.11E-4 T3 + 8.524E-8 T4 - 5.0524E-12 T5dT
0.1389
298.15
=
0.1389 -31.3860 +
412.68 -
####### 3 412.68 3
####### 5 412.68 5 = 2878.7365 kJ/jam H2 O +
●
298.15
-
298.15
-
298.15
3
+
0.47460 2
+
8.52E-08 4 412.68 4
412.68
2
-
298.15 -
298.15
2 4
5
412.68
Q = n
Cp dT
298.15
32.243 T + 1.923E-3 T2 + 1.055E-5 T3 - 3.596E-9 T4 dT
412.68
=
0.6019
298.15
=
0.6019
412.68 -
1.06E-05 3 412.68 3 2350.8332 kJ/jam
+ =
32.2430
-
298.15 298.15
3
+
0.00192 2
+
####### 4 412.68 4
412.68
2
-
298.15 -
298.15
2 4
●
H2 412.68
Cp dT
Q = n
298.15
412.68
=
27.1430T +
5.2477
9.2730E-03 T2 - 1.3800E-05 T3 + 7.6450E-09 T4 dT
298.15
=
5.2477 +
= ●
27.1430
412.68 -
####### 3 412.68 3 17448.4735 kJ/jam
-
298.15 298.15
3
+ +
0.0093 2
412.68
2
-
7.65E-09 4 412.68 4
298.15 -
298.15
2 4
CH4 412.68
Cp dT
Q = n
298.15
412.68
=
34.942 T-3.9957E-02 T2 - 1.9184E-04 T3-1.53E-07 T4+3.9321E-11 T5 dT
0.0003
298.15
=
0.0003 + +
= ●
34.9420
412.68 -
####### 3 412.68 3 3.93E-11 5 412.68 5 1.9867 kJ/jam
298.15
-
298.15
-
298.15
3
+
-0.03996 2 412.68 2
+
####### 4 412.68 4
298.15 -
298.15
2 4
5
C6H5NH2 412.68
Q = n 298.15
Cp dT 412.68
= 85.7129
-31.3860 T + 0.4746 T2 - 3.11E-4 T3 + 8.524E-8 T4 - 5.0524E-12 T5dT
298.15
= 85.7129 -22.0620 +
412.68 -
####### 3 412.68 3
####### 5 412.68 5 = 1123864.9289 kJ/jam +
●
H2 O 412.68
Q = n
Cp dT
298.15
298.15
-
298.15
-
298.15
3 5
+
0.57313 2
+
####### 4 412.68 4
412.68
2
-
298.15 -
298.15
2 4
412.68
= 171.4257
32.243 T + 1.923E-3 T2 + 1.055E-5 T3 - 3.596E-9 T4 dT
298.15
= 171.4257 32.2430
412.68 -
1.06E-05 3 412.68 3 = ########## kJ/jam +
-
298.15 3
298.15
+
0.00192 2
+
####### 4 412.68 4
412.68
Jadi: Q2 =
2
47718.2680 + 1273.1909 + 2878.7365 + 2350.8332 + + 1.9867 + 1123864.9289 + 669567.7648 = 1865104.1826 kJ/jam
4. Panas yang diserap oleh Pendingin, Qc Qc = Q1 - Q2 = 3358498.9810 - 1865104.1826 =
1493394.7984 kJ/jam
o
Sebagai pendingin digunakan air pada suhu 25 C dan tekanan 1 atm. Diperkirakan air keluar pada suhu 45oC. Dari App A.2 Geankoplis, 2003 diperoleh Cp air = 0.9987 kkal/kg.oC = 4.1787 kJ/kgoC Q = m . Cp. ΔT Qc mc = Cp. ΔT 1493394.7984 = = 17869.2979 kg/jam 4.17866 ( 45 - 25 ) Tabel Neraca Panas Cooler 1 (E-221) Panas masuk Panas keluar Komponen (kJ/jam) (kJ/jam) Q C6H5NO2 54650.8610 47718.2680 Q C 6 H4 N2 O4 1506.9847 1273.1909 3393.9394 671918.5980 Q H2 O 871986.3951 2878.7365 Q C 6 H6 19608.8722 17448.4735 Q H2 783.9192 1.9867 Q CH4 2406568.0094 1123864.9289 Q C6H5NH2 Q pendingin - 1493394.7984 Total 3358498.9810 3358498.9810
-
298.15 -
298.15
17448.4735
2 4
B.7. FLASH TANK (H-220) Fungsi: Menguapkan sebagian besar Nitrobenzen dalam campuran produk keluaran reaktor Arus 7 139.5 oC Arus 6 139.5 oC
Flash Tank
Arus 8 139.5 oC Neraca Energi Q in = 6 = dimana: 6 = 7= 8=
Q out 7+ 8 Panas sensibel gas masuk flash tank , Q1 Panas sensibel gas keluar, QQ2 2 Panas sensibel cairan keluar flash tank , Q3
Untuk menghitug panas masing-masing arus digunakan persamaan: Q=H=∑▒〖n∫_(T_reff)^T▒〖C_pl dT 〗〗
dimana:
∫_(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 ³)) +(D/4 x(T^4-T_reff⁴))+(E/5 x(T^5-T_reff⁵))] 〗
a. Panas Sensibel gas keluar, Q2 Komposisi arus 7: C6H5NO2 = 210.8537 C 6 H4 N2 O4 = 10.7253 3034.5671 H2 O = 10.6170 C 6 H6 = 10.4955 H2 = 0.0052 CH4 = 0.7971 C6H5NH2 = ●
kg/jam kg/jam kg/jam kg/jam kg/jam kg/jam kg/jam
= = = = = = =
1.7143 0.0638 168.5871 0.1361 5.2477 0.0003 0.0086
kmol/jam kmol/jam kmol/jam kmol/jam kmol/jam kmol/jam kmol/jam
C6H5NO2 412.68
Q = n
Cp dT
298.15 412.68
=
1.7143
-16.2020 T + 0.5618 T2 - 3.93E-4 T3 - 1.004E-7 T4 - 1.2252E-12 T4 dT
298.15
=
1.7143 -16.2020
412.68 -
298.15
+
0.5618
412.68
2
-
298.15
2
=
1.7143 -16.2020 + +
= ●
412.68 -
-3.9E-04 3 412.68 3 -1.2E-12 5 412.68 5 46763.9027 kJ/jam
298.15
-
298.15
-
298.15
3
+ +
2
412.68
-
-1.0E-07 4 412.68 4
298.15 -
298.15
4
5
C6H4NO2 412.68
Cp dT
Q = n
298.15
412.68
=
0.0638
18.1480 T + 0.5618 T2 - 3.93E-4 T3 + 1.004E-7 T4 - 1.2252E-12 T5 dT
298.15
=
0.0638 +
18.1480
412.68 -
-3.9E-04 3 412.68 3
-1.2E-12 5 412.68 5 1260.4590 kJ/jam
+ = ●
298.15
-
298.15
-
298.15
3
+
0.56182 2
+
1.00E-07 4 412.68 4
412.68
2
-
298.15 -
298.15
2 4
5
C 6 H6 412.68
Q = n
Cp dT
298.15
412.68
=
0.1361
-31.3860 T + 0.4746 T2 - 3.11E-4 T3 + 8.524E-8 T4 - 5.0524E-12 T5dT
298.15
=
0.1361 -31.3860 +
412.68 -
-3.1E-04 3 412.68 3
-5.1E-12 5 412.68 5 2821.1618 kJ/jam
+ = ●
298.15
-
298.15
-
298.15
3
+
0.47460 2
+
8.52E-08 4 412.68 4
412.68
2
-
298.15 -
298.15
2 4
5
H2 O 412.68
Q = n
Cp dT
298.15 412.68
= 1.7E+02
32.243 T + 1.923E-3 T2 + 1.055E-5 T3 - 3.596E-9 T4 dT
298.15
= 1.7E+02 32.2430
1.1E-05 3 412.68 3 658480.2261 kJ/jam
+ =
412.68 -
298.15 298.15
3
+
0.00192 2
+
-3.6E-09 4 412.68 4
412.68
2
-
298.15 -
298.15
2 4
●
H2 412.68
Cp dT
Q = n
298.15 412.68
=
5.2477
27.1430T +
9.2730E-03 T2 - 1.3800E-05 T3 + 7.6450E-09 T4 dT
298.15
=
5.2477 +
= ●
27.1430
412.68 -
-1.4E-05 3 412.68 3 17448.4735 kJ/jam
-
298.15 298.15
3
+
0.0093 2
412.68
+
7.6E-09 4
412.68
2
-
4
298.15 -
298.15
2 4
CH4 412.68
Q = n
Cp dT
298.15
412.68
=
0.0003
34.942 T-3.9957E-02 T2 - 1.9184E-04 T3-1.53E-07 T4+3.9321E-11 T5 dT
298.15
=
0.0003 + +
= ●
34.9420
412.68 -
-1.9E-04 3 412.68 3 3.9E-11 5 412.68 5 1.9867 kJ/jam
298.15
-
298.15
-
298.15
3
+
-0.03996 2 412.68 2
+
-1.5E-07 4 412.68 4
298.15 -
298.15
2 4
5
C6H5NH2 412.68
Q = n
Cp dT
298.15
412.68
=
0.0086
-22.0620 T + 0.5731 T2 - 4.57E-4 T3 - 1.84E-7 T4 - 2.99E-11 T5
dT
298.15
=
0.0086 -22.0620 + +
= Jadi: Q2 = =
412.68 -
-4.6E-04 3 412.68 3 -3.0E-11 5 412.68 5 244.2101 kJ/jam
298.15
-
298.15
-
298.15
3
+ +
0.5731 2
2
-1.8E-07 4 412.68 4
-
298.15 -
298.15
5
46763.9027 + 1260.4590 + 2821.1618 + 17448.4735 1.9867 + 244.2101 727020.4199 kJ/jam
b. Panas Sensibel Cairan Keluar, Q3 Komposisi arus 8:
412.68
658480.2261 +
2 4
C6H5NO2 C 6 H4 N2 O4 H2 O C 6 H6 H2 C6H5NH2 ●
4.3031 0.1083 61.9299 0.2167 0.0000 7970.500
= = = = = =
kg/jam kg/jam kg/jam kg/jam kg/jam kg/jam
0.0350 0.0006 3.4406 0.0028 0.0000 85.7043
= = = = = =
kmol/jam kmol/jam kmol/jam kmol/jam kmol/jam kmol/jam
C6H5NO2 412.68
Cp dT
Q = n
298.15 412.68
=
39.4730 T + 0.9128 T2 - 0.00211 T3 + 2.01E-6 T4 dT
0.0350
298.15
=
0.0350 +
=
●
39.4730
412.68 -
-0.0021 3 412.68 3 751.7266 kJ/jam
-
298.15 298.15
3
+
0.9128 2
412.68
+
2.0E-06 4
412.68
2
-
4
298.15 -
298.15
2 4
C 6 H4 N2 O4 412.68
Cp dT
Q = n
298.15 412.68
=
0.0006
-12.6350 T + 1.56240 T2 - 3.00E-3 T3 + 2.32E-6 T4 dT
298.15
=
0.0006 -12.6350 +
= ●
412.68 -
-3.0E-03 3 412.68 3 76.1702 kJ/jam
-
298.15 298.15
3
+
1.56240 2
412.68
+
2.3E-06 4
412.68
2
-
4
298.15 -
298.15
2 4
C 6 H6 412.68
Cp dT
Q = n
298.15 412.68
=
0.0028
-33.6620 T + 0.47430 T2 - 3.61E-3 T3 + 3.82E-6 T4 dT
298.15
=
0.0028 -33.6620
-3.6E-03 3 412.68 3 245.1970 kJ/jam
+ = ●
412.68 -
H2 O 412.68
Q = n
Cpg dT
-
298.15 298.15
3
+
0.47430 2
412.68
+
3.8E-06 4
412.68
2 4
-
298.15 -
298.15
2 4
Q = n
Cpg dT
298.15 412.68
=
3.4406
18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT
298.15
=
3.4406 +
= ●
18.2964
412.68 -
-1.3E-03 3 412.68 3 29965.3066 kJ/jam
-
298.15 298.15
3
+
0.47212 2
412.68
+
1.3E-06 4
412.68
2
-
4
298.15 -
298.15
2 4
H2 412.68
Cpg dT
Q = n
298.15
412.68
= 0.0E+00 298.15
28.8400 T + 0.00765 T2 + 3.29E-1 T3 - 0.8698 T4
= 0.0E+00 28.8400 + = ●
412.68 -
3.3E-01 3 412.68 3 0.0000 kJ/jam
298.15 298.15
3
dT
+
0.00765 2
+
-8.7E-10 4 412.68 4
412.68
2
-
298.15 -
298.15
2 4
C6H5NH2 412.68
Q = n
Cpg dT
298.15 412.68
=
85.70
46.9480 T + 0.9896 T2 - 2.3583E-3 T3 + 2.3296E-6 T4
dT
298.15
=
85.70
46.9480
412.68 -
-2.4E-03 3 412.68 3 2016973.0498 kJ/jam
+ = Jadi: Q3 = =
751.7266 + 0.0000 2048011.4502
-
298.15 298.15
3
76.1702
+
0.9896 2
412.68
+
2.3E-06 4
412.68
245.1970 + + 2016973.0498 kJ/jam
Sehingga, panas yang masuk flash tank : Q in = Q out Q1 = Q2 + Q3 = 727020.4199 + 2048011.4502 =
+
4
29965.3066
2775031.8700 kJ/jam
Tabel Neraca Panas Flash Tank (H-220) Q in (kJ/jam) Q out (kJ/jam) Komponen Arus 6 Arus 7 Arus 8 Q C6H5NO2 47515.6293 46763.9027 751.7266
2
-
298.15 -
+
298.15
2 4
Q C6H4NO2 Q H2 O Q C 6 H6 Q H2 Q CH4 Q C6H5NH2
1336.6292 688445.5326 3066.3588 17448.4735 1.9867 2017217.2599
Total
2775031.8700
1260.4590 658480.2261 2821.1618 17448.4735 1.9867 244.2101 727020.4199
76.1702 29965.3066 245.1970 0.0000 2016973.0498 2048011.4502
2775031.8700
B.4 Heater (E-114) Fungsi : Menaikkan suhu uap Nitrobenzen Tujuan
:
o
Menaikkan suhu menjadi 170 C Q3
Qsteam
Q1
Q2
Neraca Energi ΔH = Hout - Hin Hin = Hpendingin + Hout 1. Menghitung Panas Sensibel Masuk Heater, Q1 M n Komponen (kg) (kmol) C6H5NO2 4.30313 0.0350 C 6 H4 N2 O4 0.10834 0.0006 61.92992 3.4406 H2 O 0.21667 0.0028 C 6 H6 C6H5NH2 7970.497 85.7043 *Hin o
Tin
= 139.5 C =
Tref ●
= 25 H2 O
o
C=
412.68
K
298.15
K
412.68
Q = n
Cp dT
298.15
412.68
=
3.4406
18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT
298.15
=
3.4406 +
18.2964
412.68 -
####### 3 412.68 3
-
298.15 298.15
3
+
0.47212 2
+
1.31E-06 4 412.68 4
412.68
2
-
298.15 -
298.15
2 4
29965.2966
= ●
kJ/jam
C 6 H6 412.68
Q = n 298.15
Cp dT 412.7
=
3.E-03
-33.6620 T + 0.47430 T2 - 3.61E-3 T3 + 3.82E-6 T4
dT
298.15
=
3.E-03 -33.6620
412.68 -
####### 3 412.68 3 245.1970 kJ/jam
+ = ●
C6H5NH2
-
298.15 298.15
3
+
0.47430 2
+
3.82E-06 4 412.68 4
412.68
2
-
298.15 -
298.15
2 4
Cp dT
412.68
Q = n
298.15 412.68
=
85.70 298.15
=
85.70
46.9480 T + 0.9896 T2 - 2.3583E-3 T3 + 2.3296E-6 T4
46.9480
412.68 -
####### 3 412.68 3 2016972.3765 kJ/jam
+ = ●
298.15
3
+ +
0.9896 2
412.68
2
-
2.33E-06 4 412.68 4
298.15 -
298.15
C6H5NO2 412.68
Q = n 298.15
=
-
298.15
dT
Cp dT
3.E-02 39.47 412.68
=
-0.0021 3 412.68 3 751.7263 kJ/jam
●
C6H4NO2
+
412.68
Q = n 298.15
=
6.E-04
298.15
-
298.15
+ 3
0.9128 2
412.68
2
-
+
2.01E-06 4 412.68 4
+
1.56240 2
+
2.32E-06 4 412.68 4
298.15 -
2
298.15
4
Cp dT -12.6
412.68 -
-3.E-03 3 412.68 3 19.7480 kJ/jam = +
-
-
298.15 298.15
3
412.68
2
Q1 = Q C6H5NO2 + Q C6H4N2O4 + Q C6H6 + Q H2O + Q C6H5NH2 = 2047954.3443 kJ/jam
-
298.15 -
298.15
2 4
2 4
2. Menghitung Panas Sensibel Keluar Heater, Q2 *Hout o
Tin
= #### C =
Tref
=
●
H2 O
o
25
C=
444.80
K
298.15
K
444.80
Cp dT
Q = n
298.15
444.80
=
3.4406
18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT
298.15
=
3.4406 +
= ●
18.2964
444.80 -
####### 3 444.80 3 38610.6875 kJ/jam
-
298.15 298.15
3
+
0.47212 2
+
1.31E-06 4 444.80 4
444.80
2
-
298.15 -
298.15
2 4
C 6 H6 444.80
Q = n 298.15
Cp dT 444.8
=
3.E-03
-33.6620 T + 0.47430 T2 - 3.61E-3 T3 + 3.82E-6 T4 dT 298.15
=
3.E-03 -33.6620
444.80 -
####### 3 444.80 3 346.4803 kJ/jam
+ = ●
-
298.15 298.15
3
+
0.47430 2
+
3.82E-06 4 444.80 4
444.80
2
-
298.15 -
298.15
2 4
C6H5NH2 444.80
Cp dT
Q = n
298.15
444.80
=
85.70 298.15
=
85.70
46.9480 T + 0.9896 T2 - 2.3583E-3 T3 + 2.3296E-6 T4
46.9480
444.80 -
####### 3 444.80 3 2626484.8134 kJ/jam
+ = ●
C6H5NO2
444.80
Q = n 298.15
=
0.035
-
298.15 298.15
3
+ +
0.9896 2
444.80
2
2.33E-06 4 444.80 4
-
dT
298.15 -
298.15
Cp dT
39.47 444.80
-
298.15
+
0.9128 2
444.80
2
-
298.15
2
2 4
=
-0.0021 3 444.80 3 978.0620 kJ/jam
●
C 6 H4 N2 O4
-
444.80
Q = n 298.15
-
298.15
3
+
2.01E-06 4 444.80 4
+
1.56240 2
+
2.32E-06 4 444.80 4
-
298.15
Cp dT
= 0.00064 -12.6
444.80 -
-3.E-03 3 444.80 3 = 104.9953 kJ/jam +
-
298.15 298.15
3
444.80
2
-
298.15 -
298.15
Q2 = Q C6H5NO2 + Q C6H4N2O4 + Q C6H6 + Q H2O + Q C6H5NH2 = 2666525.0385 kJ/jam 3. Menghitung Kebutuhan Pemanas, Q3 Q3 = Q2 - Q1 = 2666525.0385 2047954.3443 = 618570.6942 kJ/jam Kebutuhan panas Heater sebesar
618570.6942 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 λsteam = 909.99 kJ/kg Sehingga : Q3 = m.λsteam Massa steam = Q3 = λsteam
618570.6942 909.99
kJ/jam= kJ/kg
Neraca Panas Total dalam Heater Panas masuk Panas keluar Komponen (kJ/jam) (kJ/jam) Q C6H5NO2 751.7263 978.0620 Q C 6 H4 N2 O4 19.74801 104.9953 Q C6H5NH2 2016972.3765 2626484.8134 Q H2 O Q C 6 H6 Q Steam
4
29965.2966 245.1970 618570.6942
38610.6875 346.4803 -
679.7555
kg/jam
2 4
Total
2666525.0385
2666525.0385
B.13. MENARA DISTILASI 1 (D-310) Fungsi : Memisahkan produk Anilin dan air Arus 16 138.9 oC Qc Arus 15 170.11
o
C Arus 17
MD 2
Neraca Energi Q in 15 + Qr 16 dimana: 15 = 16 = = 17 = 18 = 21 = Qc = Qr =
Qr
Arus 18 #### oC
Arus 21 184.2 oC
= Q out = 18 + 21 = 17 + 18 + Qc 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 Panas yang diserap oleh pendingin pada kondensor Panas yang dibutuhkan oleh pemanas pada reboiler
Menentukan Kondisi Operasi Menara Distilasi a. Kondisi Operasi Umpan 1 atm = Kondisi operasi umpan: P = Fasa = Cair Jenuh T = 171.65 oC =
760.0
mmHg
444.80 K
Trial Kondisi Operasi Umpan Komponen (Kmol/jam)
xi
Pi
ki Pi/P
yi ki.xi
α Ki/KHK
C6H5NO2 C6H4NO2 H2 O C 6 H6 C6H5NH2
0.0350 0.0006 3.4406 0.00278 85.7043
0.0004 0.00001 0.0386 0.00003 0.9610
Total
89.1832
1.0000
280.9083 6.0081 6189.8280 6602.6352 542.0446
0.0001 0.0000 0.3142 0.0003 0.6854
0.5182 0.0111 11.4194 12.1810 1.0000
1.00
b. Kondisi Operasi Puncak Menara ● Dew Point 1 atm = Kondisi operasi: P = T = 138.938 oC = Trial Kondisi Operasi Dew Point Puncak Menara xi Pi Komponen (Kmol/jam) C6H5NO2
0.0003
C6H4NO2 H2 O C 6 H6 C6H5NH2
6.E-06 3.4061 3.E-05 0.8570
0.0001 2.E-06 0.7989 7.E-06 0.2010
1.1184 2630.6402 3460.2298 199.0620
Total
4.2636
1
6.290.E+03
100.1167
● Bubble Point Kondisi operasi:
1 atm = P = T = 105.98 oC = Trial Kondisi Operasi Bubble Point Puncak Menara xi Pi Komponen (Kmol/jam) C6H5NO2 C6H4NO2 H2 O C 6 H6 C6H5NH2
3.E-04 6.E-06 3.4061 3.E-05 0.8570
0.0001 2.E-06 0.7989 7.E-06 0.2010
27.8877 0.1439 937.0218 1583.1294 58.4146
Total
4.2636
1
2.579.E+03
c. Kondisi Operasi Bawah Menara ● Bubble Point Kondisi Operasi: P = T =
0.3696 0.0079 8.1445 8.6877 0.7132
760 mmHg 412.09 K ki Pi/P 0.1317 0.0015 3.4614 4.5529 0.2619
yi xi/ki 0.0006 0.0010 0.2308 1.E-06 0.7675
α Ki/KHK 0.5029 0.0056 13.2152 17.3827 1.0000
1.00
760 mmHg 379.13 K ki Pi/P
yi ki.xi
0.0367 0.0002 1.2329 2.0831 0.0769
3.E-06 3.E-10 0.9850 0.0000 0.0155
α Ki/KHK 0.4774 0.0025 16.0409 27.1016 1.0000
1.00
1 atm = 760.0 mmHg 184.02 oC = 457.17 K
Trial Kondisi Operasi Bubble Point Bawah Menara yi Pi Komponen (Kmol/jam)
ki
yi
α
Komponen (Kmol/jam) H2 O C6H6 C6H5NO2 C6H4NO2 C6H5NH2
0.0344 0.0028 0.0346 0.0006 84.8472
0.0004 3.E-05 0.0004 8.E-06 0.9991
8256.8840 8204.1008 303.8360 10.5190 757.2896
Total
84.9197
1
1.753.E+04
Pi/P
ki.xi
10.8643 10.7949 0.3998 0.0138 0.9964
0.0044 0.0003 0.0002 0.0000 0.9956
Ki/KHK 10.9032 10.8335 0.4012 0.0139 1.0000
1.00
● Dew Point Kondisi Operasi:
1 atm = 760.0 mmHg P = T = 184.18 oC = 457.33 K Kondisi Operasi Dew Point Bawah Menara yi Pi ki Komponen (Kmol/jam) Pi/P H2 O 0.0344 0.0004 8.29.E+03 10.9028 C6H6 0.0028 3.E-05 8225.9791 10.8237 C6H5NO2 0.0346 0.0004 396.7234 0.5220 C6H4NO2 0.0006 8.E-06 10.5913 0.0139 C6H5NH2 84.8472 0.9991 760.3961 1.0005 Total
84.9197
1
1.77.E+04
*Condensor *Penentuan Harga q Karena umpan masuk pada keadaan titik didih nya (bubble point) maka harga q = 1
xi yi/ki 0.0000 0.0000 0.0008 0.0005 0.9986 1.00
Komponen
αatas
αbawah
α rata-rata
C6H5NO2 C6H4NO2
0.4774 0.0025 16.0409 27.1016 1.0000
0.4012 0.0139 10.9032 10.8335 1.0000
0.4377 0.0058 13.2249 17.1349 1.0000
23.1518
31.8033
H2 O C 6 H6 C6H5NH2 Total
44.6223
a i x id = Rm 1 i -q
åa
*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))
α Ki/KHK 10.8971 10.8180 0.5217 0.0139 1.0000
Trial θ θ 2 8.91
Hasil -0.9156 -0.003
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)) 2.4269 = R min = 2.4269 1 1.4269 = Jadi nilai refluks minimum sebesar Rm = 1.2 Rm - 1.5 Rm 1.3 diambil R operasi = =
1.8549
menghitung Lo dan V Lo = R x D = 1.8549 x 4.2636 = 7.9087 kmol/jam L' = Lo x F = 7.9087 = 97.0919
+ 89.1832 kmol/jam
a. Komposisi Cairan Reflux, Lo H2 O ● = yid x Lo = 0.7989 x ●
●
●
●
V = Lo + D = 7.9087 = 33.7193
x 4.2636 kmol/jam
V' = V = 33.7193
kmol/jam
7.9087
= =
6.3182 kmol/jam 113.7280 kg/jam
C6H5NH2 = =
yid x Lo 0.2010 x
7.9087
= =
1.5898 kmol/jam 147.8485 kg/jam
C6H5NO2 = =
yid x Lo 0.0001 x
7.9087
= =
0.0006 kmol/jam 0.0798 kg/jam
C6H4N2O4 = =
yid x Lo 2.E-06 x
7.9087
= =
1.E-05 kmol/jam 0.0020 kg/jam
yid x Lo 7.E-06 x
7.9087
= =
0.0001 kmol/jam 0.0040 kg/jam
C 6 H6
= =
Jadi: Lo =
261.6624 kg/jam
b. Komposisi Uap Masuk Kondensor, V H2 O ● = yid x V 33.7193 = 0.7989 x
●
●
●
●
= =
26.9382 kmol/jam 484.8875 kg/jam
C6H5NH2 = =
yid x V 0.2010 x
33.7193
= =
6.7781 kmol/jam 630.3630 kg/jam
C6H5NO2 = =
yid x V 0.0001 x
33.7193
= =
0.0028 kmol/jam 0.3403 kg/jam
C6H4N2O4 = =
yid x V 2.E-06 x
33.7193
= =
5.E-05 kmol/jam 0.0086 kg/jam
yid x V 7.E-06 x
33.7193
= =
0.0002 kmol/jam 0.0171 kg/jam
C 6 H6
Jadi: V =
= =
1115.6165
kg/jam
c. Komposisi Cairan Masuk Reboiler, L' (Arus 19) H2 O ● = xib x L' 97.0919 = 0.0393 = 0.0004 x 0.7081 = ● C6H5NO2 = xib x L' 97.0919 = 0.0396 = 0.0004 x 4.8707 = ● C6H4NO2 = xib x L' 97.0919 = 0.0007 = 8.E-06 x 0.1226 = ●
●
C6H6
= =
C6H5NH2 =
xib x L' 3.E-05 x
xib x L'
97.0919
= =
0.0031 0.2453
kmol/jam kg/jam kmol/jam kg/jam kmol/jam kg/jam
kmol/jam kg/jam
= Jadi: L' =
0.9991 x
9027.7950
97.0919
97.0091 kmol/jam 9021.8483 kg/jam
= =
kg/jam
d. Komposisi Uap yang dikembalikan ke Menara, V' H2 O ● = xib x V' 33.7193 = = 0.0004 x = ● C6H5NO2 = xib x V' 33.7193 = = 0.0004 x = ● C6H4NO2 = xib x V' 33.7193 = = 8.E-06 x = ●
●
C6H6
= =
xib x V' 3.E-05 x
33.7193
= =
C6H5NH2 = =
xib x V' 0.9991 x
33.7193
= =
Jadi: V' =
3135.2030
0.0137 kmol/jam 0.2459 kg/jam 0.0138 kmol/jam 1.6916 kg/jam 0.0003 kmol/jam 0.0426 kg/jam
0.0011
kmol/jam 0.08517489 kg/jam
33.6906 kmol/jam 3133.2230 kg/jam
kg/jam
Perhitungan Neraca Panas pada Kondensor a. Panas Laten Penguapan (ΔHv) Untuk menghitung entalpi panas penguapan (ΔHv) digunakan Persamaan Watson: ΔH2 0.38 1 - Tr2 (Pers 4.19 Smith Van Ness, 2005) = ΔH1 1 - Tr1 dimana: ΔH1 = Panas laten penguapan pada titik didih normal (kJ/kmol) ΔH2 = Panas laten penguapan pada suhu T2 (kJ/kmol) Tr2 = T2/TC (K) Tr1 = T1/TC (K) T1 = Titik didih normal komponen (K) T2 = Suhu tertentu = 138.94 oC = 412.09 K H2 O ● 1 - Tr2 0.38 ΔH2 = ΔH1 x 1 - Tr1 412.09 562.16 0.38 1 ΔH2 = 30.75 x 353.24 562.16 1 27.1173 kJ/kmol =
ΔHv = ●
26.9382
kmol/jam x
27.1173
kJ/kmol =
730.4907
kJ/jam
608.2784
kJ/jam
0.1350
kJ/jam
0.0038
kJ/jam
0.0020
kJ/jam
C6H5NH2 1 - Tr2 0.38 1 - Tr1 412.09 972.15 1 86.92 x 457.25 972.15 1 89.7418 kJ/kmol
ΔH2 = ΔH1 x ΔH2 = = ΔHv = ●
6.7781
kmol/jam x
89.7418
0.38
kJ/kmol =
C6H5NO2 1 - Tr2 0.38 1 - Tr1 412.09 719 1 44.08 x 483.95 719 1 48.7827 kJ/kmol
ΔH2 = ΔH1 x ΔH2 = = ΔHv =
0.0028
kmol/jam x
48.7827
0.38
kJ/kmol =
● C6H4N2O4 1 - Tr2 0.38 1 - Tr1 412.09 803 1 61.56 x 572 803 1 75.1824 kJ/kmol
ΔH2 = ΔH1 x ΔH2 = = ΔHv = ●
0.0001
kmol/jam x
75.1824
0.38
kJ/kmol =
C6H6 1 - Tr2 0.38 1 - Tr1 412.09 647.4 1 40.68 x 373.15 647.4 1 38.3804 kJ/kmol
ΔH2 = ΔH1 x ΔH2 = = ΔHv = Jadi: Q2 =
0.0001
1338.9098
kmol/jam x
38.3804
kJ/jam
b. Panas Sensibel Gas Masuk Kondensor, Q3 H2 O ●
0.38
kJ/kmol =
412.09
Cp dT
Q = n
298.15 412.09
=
32.243 T + 1.923E-3 T2 + 1.055E-5 T3 - 3.596E-9 T4 dT
26.94
298.15
=
26.94
32.2430
412.09 -
1.06E-05 3 412.09 3 104670.8476 kJ/jam
+ = ●
-
298.15 298.15
3
+
0.00192 2
+
####### 4 412.09 4
412.09
2
-
298.15 -
298.15
2 4
C6H5NH2 412.09
Cp dT
Q = n
298.15
412.09
=
-31.3860 T + 0.4746 T2 - 3.11E-4 T3 + 8.524E-8 T4 - 5.0524E-12 T5dT
6.778
298.15
=
6.778 +
-22.0620
412.09 -
####### 3 412.09 3
####### 5 412.09 5 191903.6870 kJ/jam
+ = ●
298.15
-
298.15
-
298.15
3
+
0.57313 2
+
####### 4 412.09 4
412.09
2
-
298.15 -
298.15
2 4
5
C6H5NO2 412.09
Cp dT
Q = n
298.15
412.09
=
-31.3860 T + 0.4746 T2 - 3.11E-4 T3 + 8.524E-8 T4 - 5.0524E-12 T5dT
3.E-03
298.15
=
3.E-03 -16.2020 + +
= ●
412.09 -
####### 3 412.09 3 ####### 5 412.09 5 75.0082 kJ/jam
298.15
-
298.15
-
298.15
3
+
0.56182 2
+
####### 4 412.09 4
412.09
2
-
298.15 -
298.15
2 4
5
C6H4N2O4 412.09
Cp dT
Q = n
298.15
412.09
=
-31.3860 T + 0.4746 T2 - 3.11E-4 T3 + 8.524E-8 T4 - 5.0524E-12 T5dT
5.E-05
298.15
=
5.E-05 +
18.1480
#######
412.09
412.09 3
-
298.15 298.15
3
+ +
0.56182 2 1.00E-07
412.09 412.09
2 4
-
298.15 -
298.15
2 4
+ + = ●
412.09
3
####### 5 412.09 5 1.0011 kJ/jam
-
298.15
-
298.15
+
4
412.09
-
298.15
5
C 6 H6 412.09
Cp dT
Q = n
298.15
412.09
=
2.E-04
-31.3860 T + 0.4746 T2 - 3.11E-4 T3 + 8.524E-8 T4 - 5.0524E-12 T5dT
298.15
=
2.E-04 -31.3860 + +
= Jadi: Q3 =
412.09 -
####### 3 412.09 3 ####### 5 412.09 5 2.5398 kJ/jam
296653.0838
298.15
-
298.15
-
298.15
3
+
0.47460 2
+
8.52E-08 4 412.09 4
412.09
2
-
298.15 -
298.15
2 4
5
kJ/jam
c. Panas Sensibel Cairan Reflux, Q4 H2 O ● 379.13
Cp dT
Q = n
298.15
379.13
=
6.3182
18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT
298.15
=
6.3182
+ =
●
18.2964
379.13 -
####### 3 379.13 3 38719.7657 kJ/jam
-
298.15
298.15
3
+
0.47212 2
+
1.31E-06 4 379.13 4
379.13
2
-
298.15
-
298.15
2
4
C6H5NH2 379.13
Q = n 298.15
Cp dT 379.13
=
1.59
46.9480 T + 0.9896 T2
- 2.3583E-3 T3 + 2.3296E-6 T4
dT
298.15
=
1.59 +
46.9480
379.13 -
####### 3 379.13 3
-
298.15 298.15
3
+ +
0.9896 2
379.13
2
2.33E-06 4 379.13 4
-
298.15 -
298.15
2 4
26016.6588
= ●
kJ/jam
C6H5NO2 379.13
Cp dT
Q = n
298.15
379.13
=
0.0006
18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT
298.15
=
0.0006
+ =
39.4730
379.13 -
####### 3 379.13 3 9.7017 kJ/jam
-
298.15
298.15
3
+
0.91277 2
+
2.01E-06 4 379.13 4
379.13
2
-
298.15
-
298.15
2
4
● C6H4N2O4 379.13
Cp dT
Q = n
298.15
379.13
=
1.E-05
18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT
298.15
=
1.E-05 -12.6350
+ = ●
379.13 -
####### 3 379.13 3 0.2540 kJ/jam
-
298.15
298.15
3
+
1.56240 2
+
2.32E-06 4 379.13 4
379.13
2
-
298.15
-
298.15
2
4
C6H6 379.13
Q = n
Cp dT
298.15
379.13
=
0.0001
18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT
298.15
=
0.0001 -33.6620
379.13 -
=
####### 3 379.13 3 2.8928 kJ/jam
Jadi: Q4 =
64749.2731
+
-
298.15
298.15
kJ/jam
d. Panas Sensibel Cairan Produk Distilat, Q5 H2 O ● 379.13
Q = n
Cp dT
3
+
0.47430 2
+
3.82E-06 4 379.13 4
379.13
2
-
298.15
-
298.15
2
4
Q = n
Cp dT
298.15
379.13
= 20.6200
18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT
298.15
= 20.6200 18.2964
379.13 -
####### 3 379.13 3 126364.7574 kJ/jam
+ = ●
-
298.15 298.15
3
+
0.47212 2
+
1.31E-06 4 379.13 4
379.13
2
-
298.15 -
298.15
2 4
C6H5NH2 379.13
Cp dT
Q = n
298.15 379.13
= 5.2.E+00 298.15
46.9480 T + 0.9896 T2 - 2.3583E-3 T3 + 2.3296E-6 T4
= 5.2.E+00 46.9480 + = ●
379.13 -
####### 3 379.13 3 84907.2488 kJ/jam
-
298.15 298.15
3
+ +
0.9896 2
379.13
2
-
2.33E-06 4 379.13 4
dT
298.15 -
298.15
2 4
C6H5NO2 379.13
Q = n
Cp dT
298.15
379.13
=
0.0021
18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT
298.15
=
0.0021
+ =
39.4730
379.13 -
####### 3 379.13 3 31.6623 kJ/jam
-
298.15
298.15
3
+
0.91277 2
+
2.01E-06 4 379.13 4
379.13
2
-
298.15
-
298.15
2
4
● C6H4N2O4 379.13
Q = n
Cp dT
298.15
379.13
=
4.E-05
18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT
298.15
=
4.E-05 -12.6350
+ =
379.13 -
####### 3 379.13 3 0.8291 kJ/jam
-
298.15
298.15
3
+
1.56240 2
+
2.32E-06 4 379.13 4
379.13
2
-
298.15
-
298.15
2
4
●
C6H6 379.13
Q = n
Cp dT
298.15
379.13
=
0.0002
18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT
298.15
=
0.0002 -33.6620
+ = Jadi: Q5 =
379.13 -
####### 3 379.13 3 9.4408 kJ/jam
211313.9385
-
298.15
298.15
3
+
0.47430 2
+
3.82E-06 4 379.13 4
379.13
2
-
298.15
-
298.15
2
4
kJ/jam
e. Panas yang dilepas oleh Pendingin Kondensor, Qc Qc = Q2 + Q3 - Q4 + Q5 21928.7820 kJ/jam = Sebagai pendingin digunakan air pada suhu 25oC dan tekanan 1 atm. Diperkirakan air keluar pada suhu 45oC. Dari App A.2 Geankoplis, 2003 diperoleh Cp air = 0.9987 kkal/kgoC = 4.1787 kJ/kgoC Q = m . Cp. ΔT Qc mc = Cp. ΔT 21928.7820 = 4.1787 ( 45 - 25 )
=
262.3901
kg/jam
Perhitungan Neraca Panas pada Reboiler a. Panas Sensibel Cairan Masuk MD , Q1 Panas sensibel masuk MD 2 sama dengan panas sensibel cairan keluaran Heater . Q1 = 2666525.0385 kJ/jam b. Panas Sensibel Cairan Produk Bawah, Q6 H2 O ● 457.17
Q = n
Cp dT
298.15
457.17
=
0.0344 298.15
=
0.0344 +
18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4
18.2964
457.17 -
####### 3 457.17 3
-
298.15 298.15
3
+
0.47212 2
+
1.31E-06 4 457.17 4
457.17
2
dT -
298.15 -
298.15
2 4
419.9350
= ●
kJ/jam
C6H5NH2 457.17
Cp dT
Q = n
298.15 457.17
=
84.85 298.15
=
84.85
46.9480 T + 0.9896 T2 - 2.3583E-3 T3 + 2.3296E-6 T4
46.9480
457.17 -
####### 3 457.17 3 6570771.5516 kJ/jam
+ = ●
298.15
-
298.15
-
298.15
3
+ +
0.9896 2
457.17
2
2.33E-06 4 457.17 4
-
dT
298.15 -
298.15
C6H5NO2 457.17
Q = n
Cp dT
298.15
= + = ●
3.E-02 39.47
457.17
-0.0021 3 457.17 3 1056.6921 kJ/jam
-
298.15
3
+
0.9128 2
457.17
2
+
2.01E-06 4 457.17 4
+
1.56240 2
+
2.32E-06 4 457.17 4
+
0.47430 2
+
3.82E-06 4 457.17 4
-
298.15 -
298.15
2 4
C6H4NO2 457.17
Q = n
Cp dT
298.15
=
6.E-04
-12.6
457.17 -
-3.E-03 3 457.17 3 27.8001 kJ/jam = +
●
-
298.15 298.15
3
457.17
2
-
298.15 -
298.15
2 4
C 6 H6 457.17
Q = n
Cp dT
298.15
=
3.E-03
-33.7
457.17 -
-4.E-03 3 457.17 3 = 386.0537 kJ/jam +
Jadi: Q6 =
-
6572662.0325 kJ/jam
298.15 298.15
3
457.17
2
-
298.15 -
298.15
2 4
2 4
c. Panas yang dibutuhkan oleh Pemanas Reboiler, Qrb Qrb = Q5 + Q6 + Qc - Q1 4139379.7145 = Kebutuhan panas Heater sebesar 4139379.7145 kJ, panas disupply menggunakan o
saturated steam dengan suhu 300 C dengan tekanan 1 atm. Dari properties of saturated water and saturated steam up to 1 atm, stoichiometry 2004 λsteam = 909.99 kJ/kg Sehingga : Qr = m.λsteam Massa steam = Qr = λsteam
4139379.7145 909.99
kJ/jam= kJ/kg
4548.8189
Tabel Neraca Panas Menara Distilasi (D-330) 2 Panas masuk Panas keluar (kJ/jam) Komponen (kJ/jam) Arus Arus Q C6H5NO2 978.0620 31.6623 1056.6921 Q C 6 H4 N2 O4 104.9953 0.8291 27.8001 38610.6875 126364.7574 419.9350 Q H2 O 346.4803 9.4408 386.0537 Q C 6 H6 84907.2488 6570771.5516 Q C6H5NH2 2626484.8134 Q pendingin 21928.7820 4139379.71 Q steam 233242.7205 6572662.0325 Total 6805904.7530 6805904.7530 B.14 COOLER Fungsi : Mendinginkan gas produk keluar reaktor Tujuan : Menghitung kebutuhan pendingin o
28 C Q1 184.2 oC
Q2 30 Qc
45
o
o
C
C Neraca Energi Q in = Q out Q1 = Q2 + Q3 Q1 = Panas sensibel gas keluar reaktor dimana: Q3 = Panas yang diserap oleh pendingin Q2 = Panas sensibel cairan keluar cooler 1 Untuk menghitug panas masing-masing arus digunakan persamaan:
kg/jam
dimana:
1. Panas Sensibel Cairan Masuk, Q1 Panas sensibel yang masuk Cooler sama dengan panas sensibel cairan yang keluar MD Q1 = 6572662.0325 kJ/jam 2. Panas Sensibel gas keluar, Q2 Komposisi arus 6: C6H5NO2 = 4.2601 kg/jam C 6 H4 N2 O4 = 0.1073 kg/jam 0.6193 H2 O = kg/jam 0.2145 kg/jam = C6H6 C6H5NH2 = 7890.7922 kg/jam ●
0.0346 0.0006 0.0344 0.0028 84.8472
= = = = =
kmol/jam kmol/jam kmol/jam kmol/jam kmol/jam
H2 O 457.17
Cp dT
Q = n
298.15
457.17
=
0.0344
18.2964 T + 0.47212 T2 - 1.34E-3 T3 + 1.31E-6 T4 dT
298.15
=
0.0344 +
= ●
18.2964
457.17 -
####### 3 457.17 3 1951.0776 kJ/jam
-
298.15 298.15
3
+
0.47212 2
+
1.31E-06 4 457.17 4
457.17
2
-
298.15 -
298.15
2 4
C6H5NH2 457.17
Cp dT
Q = n
298.15 457.17
=
84.85 298.15
=
84.85
46.9480
●
457.17 -
####### 3 457.17 3 2838811.5810 kJ/jam
+ =
46.9480 T + 0.9896 T2 - 2.3583E-3 T3 + 2.3296E-6 T4
-
298.15 298.15
3
+ +
0.9896 2
457.17
2
2.33E-06 4 457.17 4
-
dT
298.15 -
298.15
C6H6 457.17
Cp dT
Q = n
298.15 457.17
=
3.E-03
46.9480 T + 0.9896 T2 - 2.3583E-3 T3 + 2.3296E-6 T4
dT
2 4
=
3.E-03 298.15
=
3.E-03 -33.6620 +
= ●
46.9480 T + 0.9896 T2 - 2.3583E-3 T3 + 2.3296E-6 T4 457.17 -
####### 3 457.17 3 386.0537 kJ/jam
298.15
-
298.15
-
298.15
3
+ +
0.4743 2
457.17
2
3.82E-06 4 457.17 4
-
dT
298.15 -
298.15
C6H5NO2 457.17
Q = n
Cp dT
298.15
= + = ●
3.E-02 39.47
457.17
-0.0021 3 457.17 3 1056.6921 kJ/jam
-
298.15
+
3
0.9128 2
457.17
2
+
2.01E-06 4 457.17 4
+
1.56240 2
+
2.32E-06 4 457.17 4
-
298.15 -
298.15
2 4
C 6 H4 N2 O4 457.17
Q = n
Cp dT
298.15
=
6.E-04
-12.6
457.17 -
298.15
-3.E-03 3 3 457.17 - 298.15 3 = 115.9067 kJ/jam Jadi: Q2 = 2842321.3110 kJ/jam +
4. Panas yang diserap oleh Pendingin, Qc Qc = Q1 - Q2 = 6572662.0325 - 2842321.3110 =
457.17
2
3730340.7215 kJ/jam
o
Sebagai pendingin digunakan air pada suhu 25 C dan tekanan 1 atm. Diperkirakan air keluar pada suhu 45oC. Dari App A.2 Geankoplis, 2003 diperoleh Cp air = 0.9987 kkal/kg.oC = 4.1787 kJ/kgoC Q = m . Cp. ΔT Qc mc = Cp. ΔT 3730340.7215 = 4.17866 ( 45 - 25 )
= 44635.5979 kg/jam
Tabel Neraca Panas Cooler 1 (E-221) Panas masuk Panas keluar Komponen (kJ/jam) (kJ/jam) Q C6H5NO2 1056.6921 1056.6921
-
298.15 -
298.15
2 4
2 4
Q C 6 H4 N2 O4 Q C6H5NH2 Q C6H6 Q H2 O Q pendingin Total
27.8001 6570771.5516 386.0537 419.9350 6572662.0325
115.9067 2838811.5810 386.0537 1951.0776 3730340.7215 6572662.0325
kJ/jam
kJ/jam
Neraca Massa (kg/jam) Komponen Arus Arus 1 Arus 2 Arus 3 Arus 4 Arus 5 Arus 6 Arus 7 Arus 8 Arus 9 Arus 10Arus 11Arus 12Arus 13Arus 14Arus 15Arus 16Arus 17 18 C6H5NO2 0 0 ## ## ## ## ## ## ## ## ## ## ## C6H4N2O4 0 0 ## ## ## ## ## ## ## ## ## ## ## H2O 0 0 ## ## ## ## ## ## ## ## ## ## ## C6H6 0 0 ## ## ## ## ## ## ## ## ## ## ## H2 ## ## 0 0 0 0 0 ## ## ## ## 0 0 CH4 ## ## 0 0 0 0 0 ## ## ## ## 0 0 C6H5NH2 0 0 0 0 0 0 0 ## ## ## ## ## Total ## ## ## ## ## ## ## ## ## ## ## ## ##
Nera Komponen C6H5NO2 C6H4N2O4 H2 O C6H6 H2 CH4 C6H5NH2 Total
Arus 1 0 0 0 0 524.7725 0.0052 0 524.7778
Arus 2 Arus 3 Arus 4 Arus 5 Arus 6 Arus 7 0 10757.8363 10757.8363 10757.8363 2689.4591 10757.8363 0 10.8337 10.8337 10.8337 2.7084 10.8337 0 10.8337 10.8337 10.8337 2.7084 10.8337 0 10.8337 10.8337 10.8337 2.7084 10.8337 524.7725 0 0 0 0 0 0.0052 0 0 0 0 0 0 0 0 0 0 0 524.7778 10790.34 10790.337 10790.337 2697.5843 10790.337
Arus 8 10757.8363 10.8337 10.8337 10.8337 524.7725 0.0052 0.0000 11315.115
Neraca Massa (kg/jam) Arus 9 Arus 10 Arus 11 Arus 12 Arus 13 Arus 14 Arus 15 Arus 16 215.1567 215.1568 215.1568 4.3031 210.8536 4.3031 0.0430 1.6916 10.8337 10.8337 10.8337 0.1083 10.7253 0.1083 0.0011 0.0426 3096.4960 3096.4960 3096.4960 61.9299 3034.5661 61.9299 61.3106 0.2459 10.8337 10.8337 10.8337 0.2167 10.6170 0.2167 0.0022 0.0852 10.4955 10.4955 10.4955 0 10 0 0 0 0.0052 0.0052 0.0052 0 0.0052 0 0 0 7971.2943 7971.2970 7971.2970 7970.4972 0.7971 7970.4972 79.7050 3133.2230 11315.115 11315.118 11315.118 8037.0552 3267.5591 8037.0552 141.06187 3135.2882
Arus 17 Arus 18 Arus 19 4.2601 4.8707 4.2601 0.1073 0.1226 0.1073 0.6193 0.7081 0.6193 0.2145 0.2453 0.2145 0 0 0 0 0 0 7890.7922 9021.8483 7890.7922 7895.9934 9027.795 7895.9934
