tugas otk md
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menara distilasi multi komponenTRANSCRIPT
TUGAS
OPERASI TEKNIK KIMIA 2
MENARA DESTILASI MULTI KOMPONEN
DISUSUN OLEH:
NAMA/NIM : SHELLY PUSPITA DEWI (121140082)
LAILA FADHILA (121140088)
ANIDA SARASTIKA (121140112)
KELOMPOK : 2
HARI/TANGGAL : JUMAT/13 MEI 2016
PRODI TEKNIK KIMIA
FAKULTAS TEKNOLOGI INDUSTRI
UNIVERSITAS PEMBANGUNAN NASIONAL “VETERAN”
YOGYAKARTA
2016
1. Jika anda seorang insinyur Teknik Kimia, anda diminta untuk merancang sebuah 10 menara distilasi yang dibagi dikerjakan oleh 10 kelompok insinyur. Masing-masing kelompok dengan umpan dan hasil yang berbeda. Anda diberi kewenangan untuk menentukan:- T dan P pada distilat, umpan dan bottom- Jumlah Menara distilasi- Kemurnian hasil bisa didapat dari produk atau maupun produk bawah
Tuliskan semua asumsi yang anda buat dan berikan penjelasan dalam singkat dalam perancangan anda pada kesimpulan rancangan!
Kelompok 2:
Menara distilasi untuk mendapatkan hasil C3H8 90% dengan komposisi umpan C3H8
30%, C4H10 30%, C5H12 20%, C6H14 20%
Penyelesaian : D
C3H8 : 90%
F C4H10 : 10%
100 kg/jam
C3H8 : 30%
C4H10 : 30%
C5H12 : 20%
C6H14 : 20% B
C3H8
C4H10
C5H12
C6H14
1. Menentukan komponen kunciTd < Lk = C3H8
Td > Hk = C4H10
2. Neraca Massa Atas
Asumsi : 1. Tekanan = 431.7194 mmHg = 8 psia2. Suhu = 798.9998 K
D : 100 kg x 30% = 30 kg
MassaC3H8 : 90% x 30 kg = 27 kg
C4H10 : 10% x 30 kg = 3 kg
Mol
C3H8 : 27 kg / 42 = 0.6428 kgmol
C4H10 : 3 kg / 58 = 0.05172 kgmol
Yi
C3H8 : 0.6428) /(0.6428+0.05172) = 0.9255
C4H10 : 0.05172/(0.6428+0.05172) = 0.0745
Persamaan Antoine
C3H8 C4H10 C5H12 C6H14
A 6.80398 6.80896 6.87632 6.87024
B 803.810 935.860 1075.780 1168.720
C 246.990 238.730 233.00 224.210
Po
C3H8 : ln Po = A – [B /(C+T)]
= 6.80398 – [803.810/(246.990+798.9998)]
Po = 305,5523957 mmHg
C4H10 : ln Po = A – [B /(C+T)]
= 6.80896 – [935.860/(238.730+798.9998)]
Po = 253,4617274 mmHg
Ki
C3H8 : Ki = Po/Pt
Ki = 305,5523957/431.7194
= 0,590839813
C4H10 : Ki = Po/Pt
Ki = 253,4617274/431.7194
= 0,490113256
Xi
C3H8 : Xi = Yi/Ki
Xi = 0.9255/0,590839813
= 0,546822247
C4H10 : Xi = Yi/Ki
Xi = 0.0745/0,490113256
= 0,453599819
α
C3H8 α = ki C3H8/ki C4H10
= 0,590839813/0,490113256
= 1,2055169
C4H10 α= ki C4H10/ki C4H10
= 0,490113256/0,490113256
= 1
Tabel 1. Komponen Atas
Komponen atas A B C Po
C3H8 6,80398 803,81 246,99 305,5523957
C4H10 6,80896 935,86 238,73 253,4617274
Tabel 2. Komponen Atas
Komponen atas Yi Ki Xi Α
C3H8 0,9255 0,590839813 0,546822247 1,2055169
C4H10 0,9255 0,490113256 0,453599819 1
Jumlah (∑Xi) 1,001307863
UmpanAsumsi : 1. Tekanan = 431.7194 mmHg = 8 psia
2. Suhu = 930,8889 K
MassaC3H8 : 30% x 100 kg = 3 kg
C4H10 : 30% x 100 kg = 3 kg
C5H12 : 20% x 100 kg = 2 kg
C6H14 : 20% x 100 kg = 2 kg
Mol
C3H8 : 3 kg / 42 = 0.0714 kgmol
C4H10 : 3 kg / 58 = 0.05172 kgmol
C5H12 : 2 kg/ 72 = 0.0278 kgmol
C6H14 : 2 kg / 86 = 0.02325 kgmol
Yi
C3H8 : 0.0714 /(0.17402) = 0.4102
C4H10 : 0.05172/ (0.17402) = 0.2972
C5H12 : 0.0278/ (0.17402) = 0.1591
C6H14 : 0.02325/ (0.17402) = 0.1333
Po
Persamaan Antoine
C3H8 C4H10 C5H12 C6H14
A 6.80398 6.80896 6.87632 6.87024
B 803.810 935.860 1075.780 1168.720
C 246.990 238.730 233.00 224.210
C3H8 : ln Po = A – [B /(C+T)]
= 6.80398 – [803.810/(246.990+930,8889)]
Po = 455,5741096 mmHg
C4H10 : ln Po = A – [B /(C+T)]
= 6.80896 – [935.860/(238.730+930,8889)]
Po = 253,4617274 mmHg
C5H12 : ln Po = A – [B /(C+T)]
= 6,8763 – [1075,78/(233+930,8889)]
Po = 384,5301336 mmHg
C6H14 : ln Po = A – [B /(C+T)]
= 6,8702 – [1168,72/(224,21+930,8889)]
Po = 350,1801896 mmHg
Ki
C3H8 : Ki = Po/Pt
Ki = 305,5523957/431.7194
= 1.101166901
C4H10 : Ki = Po/Pt
Ki = 253,4617274/431.7194
= 0,983764308
C5H12 : Ki = 384,5301336/431,7194
= 0,929446706
C6H14 : Ki = 350,1801896/431,7194
= 0,846419553
Xi
C3H8 : Xi = Yi/Ki
Xi = 0.4102/0,590839813
= 1,101166901
C4H10 : Xi = Yi/Ki
Xi = 0.2972/0,490113256
= 0,1591
C5H12 : Xi = Yi/Ki
Xi = 0.1591/ 0,929446706
= 0,1591
C6H14 : Xi = Yi/Ki
Xi = 0.1333/0,846419553
=0,1335
α
C3H8 α = ki C3H8/ki C4H10
= 0,590839813/0,983764308
= 1,119340163
C4H10 α= ki C4H10/ki C4H10
= 0,983764308/0,983764308
= 1
C5H12 α = ki C5H12/ki C4H10
= 0,929446706/0,983764308
= 0,94478596C6H14 α = ki C6H14 /ki C4H10
= 0,846419553/0,983764308= 0,860388557
Tabel 1. Komponen Umpan
Komponen
umpanA B C Po
C3H8 6,804 803,81 246,99 455,5741096
C4H10 6,809 935,86 238,73 407,0023792
C5H12 6,8763 1075,78 233 384,5301336
C6H14 6,8702 1168,72 224,21 350,1801896
Tabel 2. Komponen Umpan
Komponen
umpanXi Ki Yi α
C3H8 0,4102 1,101166901 0,451698663 1,119340163
C4H10 0,2972 0,983764308 0,292374752 1
C5H12 0,1591 0,929446706 0,147874971 0,94478596
C6H14 0,1335 0,846419553 0,11299701 0,860388557
Jumlah (∑Yi) 1,004945396