AZAS TEKNIK KIMIANeraca Bahan Dalam Sistem Tanpa & Dengan Reaksi OLEH: Kelompok X (Sepuluh) Andre P. W (130405090) Maria Paula Sihombing (130405094) Panji Umbar (130405096) Ria Eirene (130405098) Irsa Septiawan (130405100) Salwa Jody Gustia (130405104) Siti Maysarah (130405106) Departemen Teknik Kimia Fakultas Teknik Universitas Sumatera Utara Medan 2014 Bab 2 Neraca Bahan Dalam Sistem Tanpa Reaksi 2.8 A gas containing 79.1% N2, 1.7% O2, and 19.2% SO2 is mixed with another containing 50% SO2, 6.53% O2, and 43.47% N2 to produce a product gas containing 21.45% SO2, 2.05% O2, and 76.50% N2. All compositions are in mol %. Determine: (a) How many independent stream variables there are in the problem. (b) How many material balances can be written and how many of them will be independent. (c) the ratio in which streams should be mixed. Solution: 1 = N2 1 2 = O2 3 Mixer 3 = SO2 3 x1 = 0,765 x 32 = 1 1 x = (a) Derajat Kebebasan 0,791 1 1.Variabel kebebasanx 2 = x 21 = 2 2.Neraca Zat 3.Spesifik Komposisi Laju alir Hubungan pembantu 3 0,4347 9 2 x2 = 6 0 (b) Persama an Neraca Massa N2: x 11 . N 1 O2: x2 . N 1 1 SO2:............................... x 3 . N 1 + x 21 . N 2 + x2 . N 2 2 1 N3 = x 31 . N 3 = x2 . N 3 3 + x 23 . N 2 = ........................................................(3) (c) Perbandingan aliran yang harus dicampurkan (1) (2) x 33 . Langkah I : Gunakan basis N mol = 1000 h 1 N1 + N2 = N 3 1000 + N2 = N3 Neraca komponen N2 x 11 . N 1 + x 21 . N 2 0,791.1000 + 0,4347. N = 2 x 31 . N 3 (1) = 0,765.(1000 + N2) 791 + 0,4347N2 = 765 + 0,765N2 2 0,3303 N = 26 N2 = 78,71 1000 + N2 = N3 1000 + 78,71 = N3 1078,71 = N3 Rasio = N1 N2 = 1000 78,71 = 12,704 F2 2.12 The feed to a distilation column contains 36% benzene w12 by weight, the = 0,52 remainder being toluene. The overhead distillate is to contain benzene by w22 =52% 0,48 weight , while the bottoms are to contain 5% benzene by weight. Calculate: 2 (a) The percentage of the benzene feed which is contained in the distilate. (b) The percentage of the total feed which leaves as distillate. F1 w11 = 0,36 w21 = 0,64 1 3 F3 w13 = 0,05 w23 = 0,95 Tabel Derajat Kebebasan : Jumlah Variabel Alur Persamaan Neraca TTSL Spesifik: - Komposisi - Laju Alir Hubungan Pembantu Basis Derajat Kebebasan 6 2 3 -1 1 0 F1 = 1000 lbm/h Neraca Total : F1 = F2 + F3 1000 = F2 + F3.......................................................................(1) Neraca Benzena : F1 w11 = F2 w12 + F3 w13 1000 (0,36) = F2(0,52) + F3(0,05)................................. (2) Neraca O2 : F1 w11 = F2 w12 + F3 w13 2 1000 (0,64) = F (0,48) + F3(0,95)...............................(3) Subtitusi pers (1) dan pers (2) Pers (1) = F2 = 1000 – F3 Pers (2) = 1000 (0,36) = F2(0,52) + F3(0,05) 3 Maka ; 360 = (1000 – F )(0,52) + F3(0,05) 360 = 520 - 0,52 F3 + 0,05 F3 160 =340,43 F3 = 0,47 Sehingga diperoleh: F2 = 1000 – F3 = 1000 – 340,43 = 659,57 a. Persentase benzena umpan dengan yang didistilasi : F 2 w21 ( 659,58 ) (0,52) = x 100 =95,27 F 1 w11 (1000 )(0,36) b. Persen total umpan yang didestilasi : 2 F ( 659,58 ) = x 100 =65,96 F 1 ( 1000 ) 2.17. A slurry consisting of CaC03 precipitate in a solution of NaOH and H 20 is washed with an equal mass of a dilute solution of 5% (wt) NaOH in H 20. The washed and settled slurry which is withdrawn from the unit contains 2 lbm of solution per 1 lb of solid (CaC03). The clear solution withdrawn from the unit can be assumcd. to have the same concentration as the solution withdrawn with the 'solids (sec Figure P2.17). If the feed ·slurry contains equal mass fractions of all Components, calculate the concentration of the clear solution. Penyelesaian : F2 x 2 1 Wash solution 2 = 3 4 x 41 Neraca Zat 3.Variabel kebebasan 2.NaOH = 5 % Feed Slurry F F4 1 1 x1 = 1 x 12 = Clear solution Wash Slurry F3 3 Keterangan: NaOH =1 x1 = x 32 = CaCO3 = 2 H2O = 3 Tabel Derajat Kebebasan : 1.Spesifik Komposisi Laju alir Hubungan pembantu Basis Hubungan Pembantu: 10 3 3 0 3 -1 1 0 . F1 = F2 = 1000 lbm/ jam 2.333 lbm 1000+1000 = F3 + F4 (W3 NaOH + W3 H2O ) = 2/3.100% = 0.1.67.W4 NaOH F4 = 2000 -1000 = 1000 lbm/jam 3. W 3 CaCO3 : (W3 NaOH + W3 H2O ) = 1:2 F 1 + F 2 = F3 + F4 maka W 3 CaCO3 = 1/3.67 lbm 2000 = F3 + F4 ……1) Maka W 3 NaOH = 0.100% = 0. ( ( ( massa NaOH massa H 2 O+massa NaOH massatotal massa H 2 O+massa NaOH F F 3 NaOH 3 3 H2 O NaOH +F ) ( = W ¿ ¿3 ¿ NaOH (¿ H 2 O+W 3¿ X F 3 ¿¿¿) ¿ ¿ 3 W NaOH X F 3 ¿ ¿ ( ) = ) F 3 NaOH F3 ( = massa NaOH massa H 2 O+massa NaOH NaOH ( massa massa total ) ) ) ( x 3 F 3 3 F H O + F NaOH 2 W 3NaOH X F3 F3 = ( ) W ¿ ¿3 ¿ NaOH (¿ H 2 O+W 3¿ X F 3 ¿¿¿) ¿ ¿ F3 ¿ ¿ x w3 NaOH 1−w 3C aCO ) 3 ) * . 333 lbm.3 /0.513 + W4 .W3 H2O + 1000.0.383/1.F3 +0 W3 CaCO3. NaOH = 5 % Pada feed slurry komposisi dari ketiga komponen sama yaitu masing – masing komponen = 1/3 lbm Neraca Komponen CaCO3 W1 CaCO3.77 W3 H2O = 1.333 ) = W 4 NaOH Spesifikasi: Pada wash solution.1000 = W3 NaOH.F2 = W3 H2O.383 W4 NaOH = 0.F3 + W4 NaOH.23 lbm Maka W4 H2O = 1-0.(1-0.F3 = 1000/3 = 333.F1 + W2 CaCO3.77 lbm Neraca Komponen H2O W1 H2O.77 W3 H2O = 0.3 lbm F3 = 333.23 = 0.67 W4 NaOH = 0.1000 = NaOH.F4 1/3.333 = 1000 lbm/jam Neraca Komponen NaOH W1 NaOH.283 – 0.05 = W3 NaOH + W4 NaOH = 0.F4 1000.( 3 w NaOH 1−0.67.F2 = W3 CaCO3.23) 1.1000 1 3 + 0.W4 NaOH + W4 NaOH 0.F2 1 3 .1000 + 5%.1/3 + 1000.283 = W3 H2O +0.F3 = 333.F4 W3 NaOH.F3 + W4 CaCO3.95 = 1000.F1 + W2 H2O.F1 + W2 NaOH.3 lbm 0.1000 + 0 = W3 CaCO3.F3 + W4 H2O. 30% salt and the rest water.157 lbm 2.21.21 A slurry consisting of TiO2 precipitate in a salt water solution is to be washed in three stages as shown in the flowsheet of figure P2.8 F1. If the feed slurry consists of 1000 lbm/h of 20% TiO2 .Maka W3 NaOH = 1-0. W21 F7.8 F10. W28 = 0. The stages are operated so that the slurry leaving contains one third solids c. 80% of the salt fed to each stage leaves in the waste solution b. Penyelesaian : Hubungan pembantu: F8. W26 = 0.513-0. W27 = 0. the salt concentration in its waste solution is the same as the salt concentration of the solution entrained with the slurry leaving that stage. In each stage. W210 W19 = 1/3 . Assume that : a. W29 F6.8 F9.33 = 0. calculate the wash water rate to each stage. 1 Senyawa H2O pada unit I W31F1 + W32 F2 = W38 F8 + W39 F9 500 + W32 F2 = W38 F8 + W39 F9 = 0.0.8 W21 F1 1 1 9 9 300 = 0. W110 = 1/3 W15 = 1/3 W28 = W29 W27 = W210 W26 = W25 Tabel Derajat Kebebasan Unit 1 Variabel alur alir 9 Neraca TTSL 3 Spesifikasi: -Komposisi 2 -Laju Alir 1 -Hub. Pembantu 3 Total 9 DK 0 Unit 2 9 3 Unit 3 9 3 Proses 21 9 Overall 15 3 0 0 4 7 2 0 0 4 7 2 2 1 9 21 0 2 1 3 9 6 Langkah I.2 + 0.9 F8 + 0. selesaikan unit I F = F1 + F2 = F8 + F9 1000 + F2 = F8 + F9 Senyawa TiO2 pada unit I TiO2 = W11F1 = W19 .567.0.9 F8 .9 F8 0.0. F9 200 = 1/3 .9 F8 – 159. 600 2 500 + F = 340. F9 200 = W19 .567 Masukkan nilai F2 kembali ke persamaan awal bahwa : 1000 + F2 = F8 + 600 1000 + 0.F2 = 159.8 – 600 = F8 .9 F8 – 159.9 F8 .8 W2 F + W2 F 300 = 240 + 600 W29 600 W29 = 60 W29 = 0.8 = F8 + 600 1000 – 159. F9 F9 = 600 Senyawa Salt pada unit I W21F1 = W28 F8 + W29 F9 300 = W28 F8 + W29 F9 W28 F8 = 0.33 = 0.8 F2 = 0.8 W39= 1.1. 48 = W210 . F = F9 + F3 = F7 + F10 600 + F3 = F7 + F10 Senyawa TiO2 pada unit II TiO2 = W19F9 = W110 . 600 = 1/3. Maka kita dapat meyelesaikan persamaan pada unit II Penentuan persamaan senyawa dan laju alir pada unit II yaitu : Dimana. F10 1/3. 600 = W110 . F10 F10 = 600 Senyawa Salt pada unit II W29F9 = W27 F7 + W210 F10 0. 600 W27 = W210 W27 = 0.1 F8 F8 = 2402 2 Maka. 600 60 . Pembantu Total DK Unit 1 9 3 Unit 2 9 3 2 1 3 9 0 2 0 4 7 2 240. 600. 0.1 + W210 F10 60 = 48 + W210 F10 60 = 48 + W210 .600 = W27 F7 + W210 F10 W27 F7 = 0.02 Senyawa H2O pada unit II .8.8 F9 W29 + W210 F10 60 = 0.Variabel alur alir Neraca TTSL Spesifikasi: -Komposisi -Laju Alir -Hub. F10 1/3.1.2 = 0. 600 12 = W27. 600 12 = W210 . nilai F : F2 = F8 + F9 – F1 = 2402 + 600 – 1000 2 F = 2002 Setelah menyelesaikan alur F2 maka variabel alur dan derajat kebebasan juga berubah menjadi berikut : Karena nilai F2 dan F8 sudah didapat. maka didapat pertambahan Variabel total dimana pada unit II DK menjadi 0 karena adanya pertambahan variabel komposisi.8 F9 W29 60 = 0. 02 = 0.9 F7 – F3 = -49.0.98 10 W3 = 1-0.98 F7 49.0.8 -Komposisi 2 2 2 -Laju Alir 1 0 0 Subsitusikan ke -Hub. maka didapat pertambahan Variabel total dimana pada unit III DK menjadi 0 karena adanya pertambahan variabel komposisi. F = F10 + F4 = F6 + F5 600 + F4 = F6 + F5 Senyawa TiO2 pada unit III TiO2 = W110F10 = W15 .8 – 600 = F7 .65 Variabel alur alir 0.0.F9 = 2490 + 600 – 600 = 2490 Setelah menyelesaikan alur F3 maka variabel alur dan derajat kebebasan juga berubah menjadi berikut : Karena nilai F3 dan F7 sudah didapat. 600 = 1/3.33 Unit 3 9 3 = 0.8 = F7 + 600 600 – 49. 600 + W33 F3 = W37 F7 + W310 F10 340.8 = 0.22.8 Neraca TTSL F3 = 0.2 + W33 F3 = 0.98 F7 – Spesifikasi: 49. Maka kita dapat meyelesaikan persamaan pada unit III Penentuan persamaan senyawa dan laju alir pada unit III yaitu : Dimana.567.W39 F9 + W33 F3 = W37 F7 + W310 F10 0. F5 1/3.98 F7 – 49. 600 = W15 . F5 1/3. F5 F5 = 600 . Pembantu 3 4 4 persamaan awal : Total 9 9 9 600 + F3 = F7 + F10 DK 0 0 0 600 + 0.02 F7 F7 = 2490 Maka subsitusikan ke persamaan utk mendapat nilai F3 F3 = F7 + F10.98 F7 + 390 Unit 1 9 3 Unit 2 9 3 W37= 1. 0.0.6 / 0.8 F10 W210 + W25 F5 12 = 0.65 W36 = 1-0.996 F6 + 399.6 – 600 = F6 .65.33 = 0.666 .22.996 F6 – F4 = 9.004 Senyawa H2O pada unit III W310 F10 + W34 F4 = W36 F6 + W35 F5 0.9.666 390 + W34 F4 = 0.996 F6 – 9.6 = F6 + 600 600 – 9.0.Senyawa Salt pada unit III W210 F10 = W26 F6 + W25 F5 W210 = W27 6 6 5 5 0.6 Subsitusikan ke persamaan awal : 600 + F4 = F6 + F5 600 + 0. 600. 600 W310 = 1-0.6 + W25 F5 12 = 9.600 = W2 F + W2 F W27 F7 = 0. 600 9.004 F6 F6 = 9.996 F6 + 0. 600 + W34 F4 = W36 F6 + W35 F5 390 + W34 F4 = 0.F10 = 2400 + 600 – 600 4 F = 2400 .996 F6 + 399.996 F6 – 9.6 F4 = 0. 0.8.004 F6 = 2400 Maka subsitusikan ke persamaan utk mendapat nilai F4 : F4 = F6 + F5.33 = 0.996 W35 = 1-0.004 = 0.6 390 + F4 = 0.6 0.4 / 600 W25 = 0.6 + W25 .8 F10 W210 12 = 0.6 W25 = 2.004.996 F6 -9.6 + W25 .02 + W25 F5 12 = 9. 600 = 12 600 W25 = 12.02.6 = 0. sesame. 10% linter. . and 49% oil. For each ton of raw seeds to be processed. copra. hexane. peanut. castor and flax. determine the amount of oil and oil free meal produced.29 Oilseed protein source include soybean. must be used per 1 lb m clean seeds processed. Commonly. During the extraction step. cottonseed. the separation of the oil from the protein meal is performed by solvent extraction. 2 lbm solvent.2. safflower. The analysis of cotton is 4% hull. rapeseed. 37% meal. sunflower. Penyelesaian: KET: H= Hull L= Linter M= Meal O= Oil S= Solvent Hubungan pembantu: F7+ F8 =2 F7 . Pembantu Total 8 DK 0 Solven Extraction Solven Recovery Proses Overall 8 3 4 2 15 9 10 5 1 0 1 5 3 0 0 0 2 2 4 1 1 15 0 4 1 0 10 0 Langkah I : Kerjakan unit Seed Clearning Unit: Seed Clearning .Tabel Derajat Kebebasan Seed Clearnin g Variabel alur alir 8 Neraca TTSL 4 Spesifikasi: -Komposisi 3 -Laju Alir 1 -Hub. 43 W3O = 1.• Sehingga diperoleh w2H F2 = 40 w2H = (40/140) = 0.43= 0.0.w3M = 1.286 w2L F2 = 100 w2L = (100/140) = 0.57 Dari Over all (keseluruhan) diperoleh .714 W3M F3 = 370 w2L = (370/860) = 0. . r1 . can be produced by catalytic dehydrogenation of ethanol. Penyelesaian C2H5OH 2 C2H5OH Keterangan: C2H5OH = 1 CH3CHO = 2 H2 = 3 CH3CHO +H2 CH3COOC2H5 + 2H2 . CH3COOC2H5 2 C2H5OH CH3COOC2H5 + 2H2 Suppose that in a given reactor the conditions are adjusted so that a conversion of 95% ethanol is obtained with an 80% yield of acetaldehyde.3. a parallel reaction producing ethyl acetate. CH3CHO.19 Acetaldehyde. r2 . C2H5OH. Calculate the composition of the reactor product assuming the feed consists of pure ethanol. via the reaction C2H5OH CH3CHO +H2 There is. however. Neraca Zat 3.r1 – 2r2 2 => N22= N21+ r1 3 => N32= N31+ r1+ 2r2 4 => N42= N41+ r2 Basis => N1 = 1000 mol/h 7 4 0 0 2 -1 1 0 N2 N12 N22 N32 N42 .Spesifik Komposisi Laju alir Hubungan pembantu Basis NeracaMassa : 1 => N12= N11 .Variabel kebebasan 2.CH3COOC2H5 = 4 1 N11 Reaktor 2 X1 = 95% Tabel Derajat Kebebasan 1. 95 => X1 N ¿ = 0. maka anggap tidak ada 2 CH3COOC2H5 ( N 4 ) yang dihasilkan di produk.N 1−¿ X1 = 0.95 1 N 21 1 1 1(1000)−N 21 1(1000) = 2 N1 50 mol/h = 2 Agar dapat diperoleh CH3CHO ( N 2 Max ). Neraca 4 Neraca 1 r1 = => N 22=N 12 + R2 = 0 = 0 r2 = 0 => N 21 = + N 11 - r1 = 1000 r1 = 950 mol /h => N 14 r2 r2 + 50 950 mol /h Neraca 2 N 24 - - r1 -2 r2 2(0) subtitusi ke Neraca 2 N 22 = = 0 = 950 mol /h + N 12 + r1 950 ==>> N 22 max . Y 2= R2 R2 max N 22−N 12 Y 2= R2 max 0.950 = 0 + R2 R2 950 mol /h = Hubungan yield.8= 2 N 22−0 950 N 2=760 mol h Neraca 2 Neraca 1 Neraca 3 Neraca 4 => 2 N2 = = 0 r1 = 760 mol/h => 2 N1 = 50 = 1000 r2 = 95 mol /h => => 2 N3 1 N1 -760 - = = 0 = 950 mol /h N 24 + r1 r1 760 + 1 N2 1 N3 - r1 2 r2 + r1 + 760 + 2(95) = N 14 2 r2 - + r2 + 2 r2 . = 0 = 95 mol /h + 95 Maka.laju alir mol di output reaktor adalah N 2 = N 21 + N 22+ N 23+ N 24 = 50+760+950+ 95 = 1855 mol/h . are removed from aromatics by reactions with hydrogen to form the parent aromatic compound. Penyelesaian: C6H5CH3 + H2 C6H6 + CH4 C6H4(CH3)2 + H2 C6H5CH3 + CH4 C6H3(CH3)3 + H2 C6H4(CH3)2 +CH4 2C6H6 .1% of byphenyl. 35% xylene. The products stream is found to contain a small amount. If 5 mol H2 is used per 1 mol feed. 74% conversion of xylene. toluene can be converted to benzene: C6H5CH3 + H2 C6H6 + CH4 Xylene can be converted to toluene: C6H4(CH3)2 + H2 C6H5CH3 + CH4 Pseudocumene and other C9 Hydrocarbons containing three CH3 groups can be converted to xylenes: C6H3(CH3)3 + H2 C6H4(CH3)2 +CH4 In a given application. r4 Keterangan 1 = benzene 5 = Hydrogen 2 = toluene 6 = byphenyl 3 = xylene 7 = CH4 4 = C9 Hydrocarbons . For instance. 0. and 70% conversion of C9 Hydrocarbons are attained.3. 80% conversion of toluene. r1 . 20% toluene. and 40% C 9 Hydrocarbons is reacted with hydrogen. r2 C6H5C6H5 + H2 . a refinery reformate stream consisting of 5 % benzene. Calculate the complete composition of the reactor outlet stream. r3 . consisting of alkyl groups. indicating that the side reaction 2C6H6 C6H5C6H5 + H2 Occurs to some extent.22 Hydrodealkylation is a process in which side chains. 05 N3 X23 =…? X22= 0. 0.Variabel kebebasan 2.20 = 200 . 0.70 DerajatKebebasan 1.74 R2 X2 = 0.35 N3 X53=…? X42= 0.05 = 50 N2 X22 = 1000.N1 X51 1 Reaktor 2 3 N2 N3 X33=…? X12= 0.40 X63= 0.Neraca Zat 3.20 N3 X43=…? X32= 0.001 N3 X73=…? HubunganPembantu: Jika 5 mol H2 digunakan/ 1 mol umpan R1 N1 X51 = 5 N2 R3 X3 = 0.Spesifik Komposisi Laju alir Hubungan pembantu Basis N =1000 4 0 4 -1 1 0 Basis 2 12+4 7 mol h N2 X12 = 1000.80 R4 X4= 0. 2r4 N3 X13 = 749 .74) = 91 N3 X33 = N2 X3 2 (1-X3) N3 X43 = 400 (1-0.70) = 120 N2 X43 = N2 X4 2 (1-X4) Toluena N3 X2 3 40 Xylen N3 X33 91 = N2 X22 – r1 + r2 = 200 – r1 + r2…………………(1) = N2 X32 – r2 + r3 = 350 – r2 + r3……………………(2) C9 Hydrocarbons N3 X43 = N2 X4 2 – r3 120 = 400 – r3 r3 = 280 Maka.8) = 40 N3 X23 = N2 X22 (1-X2) N3 X33 = 350 (1-0. dapat dimasukkan kepersamaan (2) menjadi: 91 = 350 – r2 + r3 91 = 350 – r2 + 280 r2 = 350 + 280 – 91 = 539 dari hasil yang didapat dimasukkan ke persamaan (1) menjadi : 40 = 200 – r1 + r2 40 = 200 – r1 + 539 r1 = 200 + 539 – 40 = 699 CH4N3X7 3 = r1 + r2 + r3 = 280 + 539+ 699 = 1518 C6H6 N3X13 = 50 + r1 .N2 X32 = 1000.40 = 400 Dari hubungan pembantu didapatkan: N1 X51 = 5 N2 N1 X51 = 5 .35 = 350 N2 X4 2 = 100.(3) N3 X53 = 500 – r1 – r2 – r3 + r4 = 5000 – 699 – 539 – 280 + r4 .. 1000 = 5000 N3 X23 = 200 (1-0.2r4……………. 0. N2 = 5.2r4 N3 X13 = 50 + 699 . 0. 001 . a) Benzene: b) Toluena: N3 X13 = 737 737 3 X3 = 6000 N3 X23 = 40 = X33 N3 X33 = 91 91 = 6000 = 0.5813 N3 X73 = 1518 1518 X73 = 6000 = 0.= 3482 + r4……………………. 6000 =6 N 31 = 749 – 2(r4 ¿ 749 – 2(6) = 737 N 35 = 3482 + r4 = 3482 + 6 = 3488 Maka.00667 N3 X43 = 120 120 = 6000 = 0. 3 N6 = 3 N❑ .015 d) Hidrokarbon: X43 f) CH4: 40 6000 X23 c) Xylena: e) H2: = 0.(4) N1 + N2 = 5000 + 1000 = 6000 Jadi. Ntotal = 0.02 N3 X53 = 3488 3488 X53 = 6000 = 0.1228 = 0.253 . calculate all flows in the process for a fresh feed rate of 100mol/h. on a molar basis. . If the reactor outlet stream analyzes 21% P and 7% B. The unreacted R and inerts are separated from the reactor products and recycled.3.28) when a fresh feed containing 1 mol inert 1 per 11 mol R is used.28 Product P is produced from reactant R according the reaction 2R=>P + W With side reactions R=>B + W P=>2B + W Only 50% conversion of R is achieved in the reactor (Figure P3. Some of the unreacted R and inert are purged to limit the inerts level in the combined 7 Recycle Purged R 6 2 1 R Seperator I 3 I 4 R I R 12% P 21% W B 7% 5 P I B W Reactor feed to 12 mol %. maka kolom kedua dikali dengan -2 dan dijumlahkan dengan kolom 3 1 0 0 R 1 0 0 R -1/2 1 -1 P -1/2 1 0 P 0 -2 2 B 0 -2 0 B -1/2 -1 1 W -1/2 -1 0 W .Penyelasaian: N 1R : N 1I =11 :1 Komponen R P B W Reaksi 1 -2 1 0 1 -2 -1 0 R 1 0 -1 P 0 1 2 B 1 1 1 W Reaksi 2 -1 0 1 1 Reaksi 3 0 -1 2 1 Tahap I : Elemen pada kolom pertama dibagi dengan elemen kolom pertama yang paling atas. 1 -1 0 R -1/2 0 -1 P 0 1 2 B -1/2 1 1 W Hasilnya dijumlahkan dengan elemen pada kolom 2 sehingga elemen paling atas kolom 2 bernilai 0. 1 0 0 R -1/2 -1/2 -1 P 0 1 2 B -1/2 1/2 1 W Agar nilai elemen kolom ketiga bernilai 0. Agar baris kedua pada kolom 1 bernilai 0. maka kolom 1 dikali dengan 2 dan hasilnya dijumlahkan dengan kolom 3 2 0 0 R 2 0 0 R -1 1 0 P 0 1 0 P 0 -2 0 B -2 -2 0 B -1 -1 0 W -2 -1 0 W Kolom pertama disederhanakan dengan dibagi dengan 2 1 0 0 R 0 1 0 P -1 -2 0 B -1 1 0 W Maka terdapat 2 persamaan reaksi yang TTSL. yaitu: B + W => R 2B + W => R Variabel alur alir Peneracaan TTSL Spesifika si Mixe Reakto Seperato Splitte Prose Keseluruha r r r n r s 6 9 10 6 20 9 2 5 5 2 14 5 1 3 2 0 3 0 1 0 0 0 0 0 Rasio 1 0 0 0 1 1 Konversi Kendala 0 1 0 0 1 0 0 0 0 1 1 1 5 1 9 0 7 3 3 3 20 0 7 2 Komposi si Laju Alir Hub. Pembant u Pembagi Total Derajat Kebebasan . dan 4.(2) N 4p=r 2 4 3 N B =N B −r 1−2 r 2 B N 4B =−r 1−2 r 2 ………………(3) W 4 3 N W =N W −r 1−r 2 N 4W =−r 1−r 2 N 4I =N 3I I N 4I =X 3I ∙ N 3 ¿ 0.5 N 4R =N 3R ( 1−0. Reaktor N3 100mol/ jam Dari hubungan konversi: XR= 50% = 0.21 …….88∗100 09.50 = 44mol/jam N 4R =N 3R +r 1 …………………. 3.(5) . N 4R =N 3R +r 1 N 4R =X 3R ∙ N 3 +r 1 44=0. 2..12∙ 100 ¿ 12 mol jam Dari persamaan 1.21 N 4=r 2 4 N B =−r 1−2 r 2 4 4 X B ∙ N =−r 1−2 r 2 N4= r2 0.5 ) N 4R =0.88 ∙ 100+r 1 r 1=−44 mol jam 4 4 N p=r 2 4 X P ∙ N =r 2 0.(1) Peneracaan: R P 4 3 N p=N p + r 2 …………………. 857 ) ¿ 6.285 mol jam Dari persamaan 4 .4 0. Persamaan 5 ke persamaan 6 0.(6) N 4W =−r 2 −r 2 …………………………………………………………………………… (7) Subs.21 0.21 r 2=18.857 mol jam Dari persamaan 3 N 4B =−r 1−2 r 2 4 N B =44−2 ( 18.07 r2 =−r 1 −2r 2 0.07 r2 =44−2 r 2 0.857 mol jam N 4P =r 2=18.07 N =−r 1−2 r 2 …………………………. 285+25.4139 4 106.4 N W =−r 1−r 2 ¿ 44−18.143 X = 4= =0.285 N N W4 25.143 ¿ 106.285 N 4 4 I X = NI N 4 = 12 =0.857 = =0.285 Maka.1129 106.285 = =0.143 4 4 4 4 4 mol jam 4 N =N R + N I + N P+ N B + N W ¿ 44 +12+ 18.177 N 4 106.285 X 4P= N 4P 18.23 106.857=25.857+6.285 X 4B= N 4B 6. mol jam X 4R= N 4R 44 = =0.285 N 4 W .0591 4 106. akan diperoleh tabel derajat kebebasan Mixe Reakto Seperato Splitte Prose Keseluruha r r r n Variabel alur alir Peneracaan TTSL Spesifika si r s 6 9 10 6 20 9 2 5 5 2 14 5 1 3 4 0 3 0 0 0 1 0 0 0 Rasio 1 0 0 0 1 1 Konversi Kendala 0 1 0 0 1 0 0 0 0 1 1 1 4 2 9 0 10 0 3 3 20 0 7 2 Komposi si Laju Alir Hub.288=18.813 W: N 4W =N 5W N 5W = X 4W ∙ N 4 mol jam .117 ∙ 106.Dari informasi yang didapatkan. Pembant u Pembagi Total Derajat Kebebasan Seperator Neraca Total: ∑ masuk =∑ keluar 4 5 N =N + N Neraca P: 6 N 4P =N 5P X 4P ∙ N 4 =N 5P N 5P =0. 288=6.23 ∙106.813+24.747 mol jam Neraca N 6R =N 4R R: N 6R =44 Neraca I: N =12 Splitter N 6R 44 = =0.2114 N 56.¿ 0. dari persamaan neraca total 4 5 N =N + N 6 5 5 5 106.0591∙ 106.2816 mol jam Maka.446 mol jam N 4B =N 5B B: 5 4 N B =X B ∙ N 4 ¿ 0.2816 ) ¿ 56.747 6 I .747 N N 6I =N 4I 6 B X 6R= N 6I 12 X = 6= =0.446+6.7753 6 56.288=24.288−( 18.288=N P +N W + N B + N 6 N 6=106. 88 ∙100−0.7753 ∙28.7753 X 7I =X 6I = X 2I =0.3725 mol jam 2 X R= X R =X R=0.2114 Mixer Neraca total: ∑ masuk =∑ keluar 1 2 N + N =N 3 N 1+ 28.Neraca total: 6 7 2 N =N + N 7 N =N 2 N 6=2 N N= 56.6265 Neraca R: mol jam N 1R =N 3R−N 2R ¿ 0.3725 2 jam 7 2 7 6 N =N =28.3735=100 N 1=71.3735 ¿ 66 mol jam N 1R 66 mol N = = =6 11 11 jam 1 I .747 mol =28. 23 N 5W =24.4139 N p=18.288 4 mol jam 4 X R=0.446 mol jam 4 5 N B =6.1129 N 4B =6.2816 mol jam N 6=56.2114 N 4R =44 mol jam X I =0.177 N 4W =25. jawabannya ialah 1 N =71.0591 N P =18.285 N 1I =6 mol jam N 4 =106.143 mol jam X 4B=0.813 mol jam X 4W =0.857 mol jam X 4P=0.Maka.7753 .3735 N 3R =88 mol jam mol jam mol jam X 2R= X 7R =X 6R=0.747 mol jam 5 N 2=N 7 =28.6265 mol jam N 1R =66 mol jam N 4I =12 mol jam X 2I = X 7I = X 6I =0. Tabel Derajat kebebasan Variabel alur alir Neraca TTSL Spesifikasi: -Komposisi -Laju Alir -Hub. and 79% N2. ammonia and air are mixed in 1:10 molar ratio and reacted catalytically in the first reactor stage.29. Assuming all composition are in mol %. and about 10% H2O. Pembantu R1 R2 Total DK Reakto r1 7+2 5 Reaktor 2 Adsorber Proses Overall 8+1 5 11 + 1 5 18 + 4 15 10 + 4 7 1 0 0 0 3 0 4 0 4 0 1 1 8 1 0 0 5 4 0 0 8 4 1 1 21 1 1 0 12 2 .The two reaction are the main reaction occuring are the main reaction • 4 NH3 + 5 O2 4 NO + 6 H2O • 2 NH3 + 3/2 O2 • NO • 2 NO2 + ½ O2 + H2O N2 + ½ O2 + 3 H2O NO2 2HNO3 The waste gas from the proccess contains ½% NO2. In the flowsheet shown on figure P3. calculate the composition of all streams in the proccess.3 N I =12 mol jam 3. The air composition may be taken as 21% O2.29 Most modern proccess from production of nitrid acid are based on the sequential oxidation of ammonia to oxides of nitrogen followed by absorption of these intermediate in water. 95 N1 = 0. N3 = N3O2 + N3H2O + N3NO + N3N2 .79 (10.5r1 – 3/2 r2 = 0.4r1 – 2r2 = 1000 – 4(237. hitung Reaktor I Persamaan neraca REAKTOR I : Basis N1 = 1000 mol/jam atau N1 NH3 • Dari hubungan pembantu N2 = 10 N1 N2 • = 10 (1000) = 10.000) + 25 = 7925 mol/jam Jadi.5) – 3/2 (25) = 875 mol/jam • Neraca H2O NOUTH2O • • = NinH2O + 6r1 + 3r2 = 0 + 6 (237.95 (1000) = 950 mol/jam • Neraca NO NOUTNO = NinNO + 4r1 950 = 0 + 4r1 r1 = 237.5 mol/jam Neraca NH3 NOUTNH3 0 r2 • = NinNH3 .Langkah I.5) + 3 (25) = 1500 mol/jam Neraca N2 NOUTN2 = NinN2 + r2 = 0.000 mol/jam Dari hubungan pembantu • N3 NO = 0.5) – 2r2 = 25 mol/jam Neraca O2 NOUTO2 = NinO2 .21 (10000) – 5(237. = 875 + 1500 + 950 + 7925 = 11.5.5) – 3/2 (25) – ½ (950) – ½ r4 = 400 – ½ r4 Neraca N2 . .7) In = 1. N4 = N4NO2 + N4O2 + N4H2O + N4N2 = 950 + 400 + 1500 + 7925 = 10775 mol/jam Setelah itu. . hitung alur Overall (1.000) – 5 (237. . hitung REAKTOR II • Neraca NO NoutNO 0 r3 • = NinNO – r3 = 950 – r3 = 950 mol/jam Neraca O2 NoutO2 = NinO2 – ½ r3 0 = 875 – ½ (950) = 400 mol/jam • Neraca N2 NoutN2 • Neraca NO2 NoutNO2 • = NinN2 + 0 = 7925 mol/jam = NinNO2 + r3 = 0 + 950 = 950 mol/jam Neraca H2O NoutH2O = NinH2O + 0 = 1500 + 0 = 1500 mol/jam • Jadi.21 (10. . (1) Nout X6O2 • Out = 6.7 = NinO2 – 5r1 – 3/2 r2 – ½ r3 – ½ r4 = X2O2 N2 – 5r1 – 3/2 r2 – ½ r3 – ½ r4 = 0. .2. . .2.6. (2) . .250 mol/jam Langkah II. . .5 • Neraca O2 . . .95mol/jam • Neraca N2 ------>> (0.755 = 452.X O2)N = 7925 ------------>> (0. . (4) NoutNO2 X6NO2 N6 0. N2 + r2 6 6 (1 – 0. . . .005 N6 = 950 – 2 r4 ------------>> 0.0192 . . .37 = 0. .92 = 904. (5) NoutHNO3 = NinHNO3 + 2r4 7 7 X HNO3 N = 0 + 2r4 7 0.895 . .37) N7 = 1507. . .89375 N6 = 8087.79 (10.X6O2)N6 = 8325 – ½ r4 ----------------------------------------.4 N7 + 0. .005 N6 • = NinNO2 + r3 – 2r4 = 0 + r3 – 2r4 = 950 – 2r4 Neraca HNO3 .5) + 3(25) – r4 0.95) = 950 – 2r4 2r4 r4 • Neraca HNO3 ------>> 0. . .6 N = 2r4 Eliminasi neraca O2 dengan N2 .+ 0.895 N6 = 8325 – ½ r4 Eliminasi neraca 4 dengan 6 0.1 N = N5 + 1500 – r4 Neraca NO2 . . (3) • NoutH2O = NinH2O + 6r1 + 3r2 – r4 X7H2O N7 + X6H2O N6 = XH2O5 + 6r1 + 3r2 – r4 0.895 – XO2 )N = 7925 • Neraca H2O . .895 .4 N7 + 0.5 – ½ r4 ------------>> (0.895 – XO26) .6 N7 = 2 (452.005 (9048.95) = 7925 XO26 • Neraca NO2 ------>> 0.000) + 25 6 6 (0. 0.(6) X6O2 N6 = 400 – ½ r4 ------------>> X6O2 N6 = 400 – ½ r4 6 6 (0.00125 N6 = 237.1 N6= 1 (N5) + 6(237.895 .+ 0.Nout = NinN2 + r2 X2N2 N6 = X2N2 . .1 – XO2 )N = 0.5 6 N =9048. . (9048. .X6O2)N6 = 7925 ----------------------------------------.005 . . . Part of gases from the absorber are purged and the remainder is recycled to the reactor. (b) Deduce a calculation order assuming all flows and compositions are to be calculated.1 (9048. The overhead gases from the absorber are found to contain 6 mol C2H4 per 1 mol CO2. The C2H4O-containing liquid (4% C2H4O) from the absorber is sent to a steam stripper. .4 (1507. Explain why these are different. Calculate the fractional yield of C2H4o as it occurs in the reactor itself.95) = N5+1500 – 452.63 N5 = 460. 25% conversion is observed and the reactor outlet stream is found to contain 2 mol CO2 per 1 mol H2O.32 Ethylene oxide is made by the partial oxidation of ethylene with oxygen using a silver catalyst: 2C2H4 + O2 → 2C2H4O An undesirable side reaction also occurs: C2H4 + 3O2 → 2CO2 + 2H2O With a reactor inlet composition of 10% C 2H4.17 + 904. 1% CO2.37 603. Give explanations. analyzes 25% C2H4O. (d) Calculate the overall yield of C2H4O from C2H4 in the process. (a) Construct a degree-of-freedom table and show the process is specified.92) + 0. The ethylene oxide is removed from the stream leaving the reactor by means of an absorber.895 = N5 + 1047. (c) Calculate the diluent N2 required per mole of C2H4 fed to the process. and the rest N2 inert dilute. The product.• Neraca H2O ------>> 0.435 3. 11% O2. stream 12. Gambar 3.32 : Splitter Absorb Mixe Diluent Reaktor Stipper Strippe Steam Steam Penyelesaian : Diketahui : HubunganPembantu : 5 NC 2H 4 = 0.1 x N4 N 5CO2 = 2 x N 5H 2 O KendalaPembagi : (N-1)(S-1) = (4-1)(2-1) .75 x 0. N 4C 2 H 4 – 2r1 – r2 75 = 1000 x 0.2r1 – r2 2r1 – r2= 25…………………(1) b.75 x 0. Persamaan dari Laju Alir dan Komposisi : 1.2r1 – r2 75 = 100 .11 . Pembantu Total DK Strippe Splitte r r Proses Overall 12 33 + 2 11 + 2 3 4 23 6 1 0 2 0 0 0 5 0 1 0 2 2 0 4 6 1 11 1 9 5 5 3 34 1 8 5 a.1 x N4 = 0. Tabel Derajat kebebasan b. =3 Reaksi yang terjadi : 2C2H4 + O2 → 2C2H4O C2H4 + 3O2 → 2CO2 + 2H2O Mixe r Reakto r Absorber 11 10 + 2 14 8 4 6 6 3 0 3 0 0 7 4 Variabel alur alir Neraca TTSL Spesifikasi: -Komposisi -Laju Alir -Hub. N 5O 2 4 = N O2 –r1 – 3r2 = 1000 x 0.75 x 0.1 x 1000 N 5C 2 H 4 = 75 5 N CO2 N 5C 2 H 4 =2 5 NH 2O Persamaan : N 5C 2 H 4 = a.1 .r1 – 3r2 . Reaktor Basis ( N4 ) = 1000 HubunganPembantu N 5C 2 H 4 = 0. 4 = N C 2 H 4 O + 2r1 = 0 + 2r1 = 2r1…………………(5) N 5CO2 = 10 + 2r2………………(3) g..78 x 1... 4 N H 20 + 2r 2 = 0 + 2r2 = 2r2…………………..r1 – 3r2..(2) 5 4 NN 2 = c. Absorber HubunganPembantu : 6 XC 2H 4 Persamaan : N 5C 2 H 4 = a. 2r1 – r2= 25…………………(1) 2r1 = 20 r1= 10 Jadi.(3) = 10+ 2r2 5 N H 20 = e.. NN 2 = 0.000 = 780 N 5CO2 d.... N5 Total = 990 2... 5 N H 2 O = 10+ 2r2 2 2 x 2r2 = 10+ 2r2 4r2= 10+ 2r2 2r2 = 10 r2 = 5 h. 6 6 = 6 XC 2H 4 N 6C 2 H 4 N C 2 H 4 = 75……………………..(1) ..= 110...(4) N 5C 2 H 4 O f.. 4 = N CO2 + 2r2…………. 5 NC 2H 4O N 9H 20 ………. Di Alur 6 : N 6C 2 H 4 = 75 N 6O 2 = 85 6 N N 2 = 780 N 6CO2 = 12.. N 6CO2 + N 9CO2 = 6 9 N CO2 + N CO2 ………….(3) d.. 5 6 NO2 = NO2 N 6O2 = 85……………………….(2) c. XC2H4O 20 = N9x 0. N6= 75 6 N6.5 h.5 + .(4) N 5H 2 O = 9 N 10 H 2 O + N H 20 10 10 + N H 2 O = f.(1) X 6C 2 H 4 .5 N6 = 952.04 N9 = 500 g.(5) = N9.b. N 5CO2 20 = e.6 N CO2 = 75 6 6 N CO2 N 6CO2 6 N CO2 = 75 = 75/6 = 12. N 5N 2 = N 6N 2 N 6N 2 =780………………. 6 N C 2 H 4 = 75……………….. (4) N 9CO2 20 = 12.5 X 9CO 2 = 0.96 .5 X 6O 2 = 0. 5 .5 9 CO2 = 7.0892 6 3.(5) 10 9 10 + N H 2 O = N9. Alur 6 terdiri dari: 6 1.5 + N 9CO2 = 7. X = 7. X O 2 = 85 X 6O 2 = 85 / 952. N6. X H 20 10 9 10 + N H 2 O = ( 0. 10 + 9 N H 20 ………….X CO 2 ) 500 10 10 + N H 2 O = 480 – 500 X 9CO 2 10 10 + N H 2 O = 480 – 500 (0.5 j.015 Maka. X CO2 500 . X C 2 H 4 = 75 X 6C 2 H 4 = 75/952.5 9 N9 . N6.10 NH 2O = i.015) 10 N H 2 O = 462. X N 2 = 780 6 X N 2 = 780 / 952. N6.0787 6 2. 20 = 6 9 N CO2 + N CO2 ………….5 X 6C 2 H 4 = 0. X 9CO 2 = 0. N 12 CO2 = 80 . 0.5 / 952.01312 = 500 3. 0. N6. 0. N9 = 0.5 X 6CO 2 = 0. = 7.09375 12 X CO 2 = 0. X 9H 2O 4.5 4. 6 X CO 2 = 12.09375 N 12 CO2 3.8188 6 X CO 2 = 12. Stripper HubunganPembantu : a.015 3.65625 N 12 H 2O = 12 N H 2 O = 52.04 1.65625 Maka : 1.5 Persamaan : 12 X H 2O .5 80 .6 X N 2 = 0.75 12 N C 2 H 4 O = 20 2. 1 - 12 X CO 2 + 12 XC 2H 4 O = X 12 CO 2 = 1 – 0. 2.25 – 0. N 12 C 2 H 4 O = 80 .01312 Maka Alur 9 terdiri dari : X 9C 2 H 4 = 0. 5 4. XC2H4O N12= 20 / 0.09375 c. N 9H 2 O + 472.5 g.65625 N 12 H 2O = 12 N H 2 O = 52.5 12 X CO 2 + d.5 = N12.5 / 80 XCO2 = 0. 0. XCO2 7. 9 NC 2H 4O 12 NC2H 4O = 20 = N12. XCO2 XCO2 = 7.09375 12 X CO 2 = 0. 0. Mixer Persamaan : a. 9 N CO2 = N12. 4 8 N CO2 = N8. 0.09375 N 12 CO2 = 7.5+ N 11 H 2O = N 12 H 2O + 11 N H 2 O =52. 1 - 12 XC 2H 4 O = X 12 CO 2 = 1 – 0.25 N12= 80 b. 80 .75 N 12 C 2 H 4 O = 20 f.5 + 462.25 – 0. 0.01312 12 X H 2O N 10 H 2O . 12 N CO2 = 80 .5 N 11 H 2O =42. 12 N C 2 H 4 O = 80 .a.65625 e. X CO 2 10 = N8 . ( 0.9 . X 8C 2 H 4 + N 7C 2 H 4 75 =762 . X O 2 = N8. X 8C 2 H 4 = NC 2H 4 = 40 N 4N 2 = N 3N 2 + + 762 (0. N4 . Splitter (Pembagi) : Karena di Splitter ( Pembagi ). X O 2 + 110 = 67. 0291 N 4C 2 H 4 c.8188 d. X 6CO 2=X 8CO 2=X 7CO2 = 0.07874 a.0892 c.07874 ) + N7 ( 0.01312 Persamaan : N 6C 2 H 4 = N8.9709 + N O2 N 2O2 N 2O 2 = 42.07879) N 8N 2 3 780 = N N 2 + 762 (0.5 c. X O 2 =X O 2= X O 2 = 0.0188) 3 N N 2 = 156 5. 1 NC2H 4 100 = 1 d. terdapat kendala pembagi : X 6C 2 H 4 =X C8 2 H 4= X 7C 2 H 4 = 0. X N 2= X N 2=X N 2 = 0. 6 8 7 6 8 7 b.N8 = 762 4 8 2 b. Perbandingan N2dengan C2H4 terhadap umpan yang masuk : N 3N 2 N 1C 2 H 4 = 156 40 = 3. N 1C 2 H 4 + N8.07874) N7= 190. 0.1 .d.1000 X 100% 20 = 25 X 100% = 80% e. N 4 X 100% C 2H 4O = 20 0. Fraksi Yield dari C2H4O : Y = N 5C 2 H 4 O X Fraksi .25.1000 = 20 25 = 80% X 100% X 100% .25.1 . 0. Fraksi Yield dari C2H4O : 5 N C 2H 4O Y = X Fraksi . N 4C 2 H 4 O X 100% = 20 0.