Mass Balance on Reactive System

Transkripsi

1 AGUNG ARI WIBOWO,S.T., M.Sc

2 Single Unit Multiple Unit Stoichiometri Recycle with reaction ATK 1 Reaction System Recycle noreaction Mole Balance

3 STOICHIOMETRI Stoichiometric Equation NOT BALANCE The following equation is balanced because the number of atoms is the same on both sides of the equation (C, H, and O)

4 STOICHIOMETRI Stoichiometric Coefficients Component Coeff SO2-2 O2-1 SO3 2 Reactant Product

5 STOICHIOMETRI Limiting Reactant A reactant is limiting if it is present in less than its stoichiometric proportion relatively to every other reactant

6 STOICHIOMETRI Limiting Reactant Which one is the limiting reactant? SO 2 = 3 2 = 1.5 Component Feed (kmol/hr) SO2 3 O2 2 SO3 - O 2 = 2 1 = 2

7 STOICHIOMETRI Conversion Chemical reactions do not occur instantaneously but rather proceed quite slowly. Therefore, it is not practical to design a reactor for the complete conversion of the limiting reactant n A reacteed =

8 STOICHIOMETRI Conversion Calculate % conversion of SO2 from table of data below! Component Feed (kmol/hr) Coeff Reacted Product SO O2 2-1 SO3-2

9 STOIKIOMETRI Conversion Component Feed (kmol/hr) Coeff Reacted Product SO O SO % Conversion SO2 = x 100% = 70%

10 STOICHIOMETRI Excess Reactants All reactants, other than the limiting species, are termed excess reactants Percentage excess of A is 100 times the fractional excess : % Excess = Fraction Excess x 100 Note : n A stoich = Mol A yang dibutuhkan untuk menghabiskan limiting reaktan

11 STOICHIOMETRI Excess Reactants : Case 1 Calculate % Excess of O2? Component Feed (kmol/hr) Coeff Reacted Product SO O2 2-1 SO3-2

12 STOICHIOMETRI Excess Reactants Component Feed (kmol/hr) Coeff Reacted Product SO O SO % Excess of O2 = ((2-1.5)/1.5) x 100% = 33.33%

13 STOICHIOMETRI Excess Reactants : Case 2 Calculate % Excess of O2 if conversion of SO2 is 70%? Component Feed (kmol/hr) Coeff Reacted Product SO2 3-2 O2 2-1 SO3-2

14 STOICHIOMETRI Excess Reactants Component Feed (kmol/hr) Coeff Reaction Product SO O SO % Excess of O2 = ((2-1.5)/1.5) x 100% = 33.33% Angka 1.5 diperoleh dari oksigen yang dibutuhkan untuk menghabiskan 3 mol SO2

15 Yield STOICHIOMETRI Yang Paling sering dipakai adalah yang No :

16 STOICHIOMETRI Problem : Excess of Reactant (A) C2H6 : 13 mol /hr O2 : 20 mol/hr Reaktor PFR (B) C2H6 : 4 mol /hr O2 : mol/hr CH4 : mol/ hr % Conversion and Excess H2? Component Feed (mol/hr) Coeff Reacted Product C2H6 H2 CH4

17 STOICHIOMETRI Problem : Excess of Reactant Component C2H4 O2 CO2 H2O (A) C2H4 : 20 mol /hr O2 : 80 mol/hr Feed (A) (kmol) Reaktor PFR Coeff (B) CO2 :.. mol /hr H2O :.. O2 : mol/hr C2H4 : mol/ hr Reacted If % Conversion of C2H4 : 75%, calculate molar flow rate of product and Excess O2? Product (B) (kmol)

18 Ekstraksi STOICHIOMETRI Problem : Yield Solvent Biji kemiri : Solid + Oil 100 kg / jam 100 kg / jam Oil : 50% wt Produk Oil : 30 kg / jam Solid + Oil Solvent + Oil

19 STOICHIOMETRI Problem : Yield If % Conversion of SO2 : 75%, calculate molar flow rate of product and yield SO3 per SO2 feed? Component Feed (kmol/hr) Coeff Reacted Product SO2 5-2 O2 8-1 SO3-2

20 Mole Balance in Single Reaction

21 Problem 1 Ethylene oxide is produced by the reaction of ethylene with oxygen as per the following reaction. The feed to the reactor contains 5 mol ethylene, 3 mol oxygen, and 2 mol ethylene oxide. Draw and label the process flow sheet and calculate the product s mole and composition leaving reactor, if conversion of ethylene is 90%! A C2H4 : 5 mol O2 : 3 mol C2H4O : 2 mol Conversion C2H4 : 90% Reactor B?

22 Component Feed (A) (kmol) Coeff Reaction Product (B) (kmol) C2H X 5 = = 0.5 O2 3-1 (1/2) X 4.5 = = 0.75 C2H4O 2 2 (2/2) X 4.5 = = 6.5

23 Problem 2 Ammonia is burned to form nitric oxide and water. The fractional conversion of NH3 is 0.5. The inlet molar flow rate is 5 mol/h of NH3 and 10 mol/h of oxygen. Calculate the exit component molar flow and % Excess of O2! Basis : 1 hr operation

24 Fractional Conversion NH3 : 0.5 A NH3 : 5 mol O2 : 10 mol Reactor B? Component Feed (A) (kmol) Coeff Reacted Product (B) (kmol) NH X 5 = = 2.5 O (-5/-4) X -2.5 = = NO 0 4 (4/-4) X -2.5 = = 2.5 H2O 0 6 (6/-4) X -2.5 = = 3.75

25 Component Feed (A) (kmol) NH3 5 O2 10 NO 0 H2O x 5 = 6,25 % Excess of O2 = (( )/6.25) x 100% = 60%

26 Problem 3 The feed contains 10.0 mol% propylene (C3H6), 12.0 mol% ammonia (NH3), and 78.0 mol% oxygen. A fractional conversion of 30.0% of the limiting reactant is achieved. Taking 100 mol of feed as a basis, determine which reactant is limiting, the percentage by which each of the other reactants is in excess, and the molar amounts of all product gas. Acrylonitrile (C3H3N) is produced by the reaction of propylene, ammonia, and oxygen

27 Mole Balance in Multiple Reaction

28 Generally, the synthesis of chemical products does not involve a single reaction but rather multiple reactions. For instance, the goal would be to maximize the production of the desirable product and minimize the production of unwanted by-products. For example, ethylene is produced by the dehydrogenation of ethane: Multiple Reaction

29

30 Component Feed Reaction 1 Reaction 2 Product X mol Mol Coeff Mol Coeff Mol X mol Mol C2H O CO CO H2O Total

31 Problem 4 Ethylene is oxidized to ethylene oxide (desired) and carbon dioxide (undesired). If feed contain 10 kmol C2H4 and 25 kmol O2, with conversion of reaction 1 and 2 are 30%, 20% respectively, Calculate the composition and molar rate of product stream The following reactions are taking place: Reaksi Utama Reaksi Samping

32 Problem 5 Pada pembakaran ethane didapatkan produk samping berupa gas CO dari reaksi pembakaran tidak sempurna seperti yang ditunjukan dibawah. Jika Feed yang mengandung 10 mol gas ethane dibakar dengan excess oksigen 60% berapa komposisi dan molar flow gas buang yang dihasilkan dengan ketentuan konversi reaksi 1 dan 2 adalah 30% dan 20%. Pembakaran Sempurna Pembakaran Tidak Sempurna

33 Pembakaran Sempurna Pembakaran Tidak Sempurna Pada kasus ini, excess hanya dihitung berdasarkan pembakaran sempurna saja %Excess = A = (mol O2 Masuk Mol O2 stoikiometri reaksi sempurna) Mol O2 stoikiometri reaksi sempurna x 100% 0.6 = (mol O2 Masuk 35) / 35 mol O2 Masuk = 56 mol

34 Component Feed Reaction 1 Reaction 2 Product X mol Mol Coeff Mol Coeff Mol X mol Mol C2H O CO CO H2O Total 1 66

35 Combustion Reactions Combustion is the rapid reaction of a fuel with oxygen to produce energy, The product gas is called stack gas or flue gas Fuels include : coal (C, H, S, and others), fuel oil (high Mw hydrocarbons and some S), gaseous fuel (natural gas mostly methane), or liquefed petroleum gas (propane and/or butane) Source of oxygen : For obvious economic reasons, air in mole% (79% N2, 21% O2) is used the in combustion reactions

36 Combustion Reactions : Theoritical and Excess Air Excess air is the amount of air fed to the reactor that exceeds the theoretical air Mol Udara dapat dihitung dengan memanfaatkan perbandingan mole Nitrogen dan Oksigen pada udara sebelum bereaksi Mol N2 : Mol O2 = 79 : 21

37 Problem 6 Pada pembakaran ethane didapatkan produk samping berupa gas CO dari reaksi pembakaran tidak sempurna seperti yang ditunjukan dibawah. Jika Feed yang mengandung 10 mol gas ethane dibakar dengan excess udara 60% berapa komposisi dan molar flow gas buang yang dihasilkan dengan ketentuan konversi reaksi 1 dan 2 adalah 30% dan 20%. Pembakaran Sempurna Pembakaran Tidak Sempurna

38 Problem 7 Ethylene is oxidized to ethylene oxide (desired) and carbon dioxide (undesired). If feed contain 10 kmol C2H4 and 25 kmol O2, with conversion of reaction 1 and 2 are 30%, 20% respectively, Calculate the composition and molar rate of product stream and % Excess of air used in the process. The following reactions are taking place: Reaksi Utama Reaksi Samping

39 Wet and Dry Basis of gases Reaksi Pembakaran menghasilkan H2O, dalam Analisa gas buang, dikenal metode Orsat yang menyatkan komposisi gas buang tanpa menghitung kuantiti dari H2O tersebut Tidak dihitung dalam Analisa orsat, oleh karena itu disebut Dry Basis

40 Problem 8 Air and an unknown amount of methane (CH4) gas are fed to a combustion process. The exhaust gases, leave the top of the combustion process. An orsat analysis of the exhaust gases gives 8.95 mol% CO2. Based on mol/min of exhaust gases (dry basis), determine the % excess air fed to this process. The following combustion reaction takes place:

41 Problem 8 (A) CH4 : mol (B) N2 : 79% mol O2 : 21% mol Reactor (C) N2 O2 CO2 H2O Component Feed (A) Udara (B) Reaction Dry Basis Product Wet Basis Product X mol Mol X mol Mol Coeff Mol X mol Mol X mol Mol CH O N CO H2O Total