Lecture 10 Solving Material Balances Problems Involving Reactive Processes

Transcription

1 CHE 31. INTRODUCTION TO CHEMICAL ENGINEERING CALCULATIONS Lecture 10 Solving Material Balances Problems Involving Reactive Processes

2 Material Balances on Reactive Processes Material balances on processes involving chemical reactions may be solved by applying: 1. Molecular Species Balance a material balance equation is applied to each chemical compound appearing in the process. 2. Atomic Species Balance the balance is applied to each element appearing in the process. 3. Extent of Reaction expressions for each reactive species is written involving the extent of reaction. 2

3 Molecular and Elemental Balances For steady-state reactive processes, Input + Generation = Output + Consumption The generation and consumption terms in the molecular balance equation is usually obtained from chemical stoichiometry. But for an atomic balance, for all cases Input = Output 3

4 Dehydrogenation of Ethane Consider the dehydrogenation of ethane in a steady-state continuous reactor, 4

5 Dehydrogenation of Ethane Total Balance: Input = Output Molecular Species Balance: C 2 H 6 : C 2 H 4 : H 2 : Input Consumed = Output Generated = Output Generated = Output Atomic (Elemental) Species Balance: C-Balance: H-Balance: Input = Output Input = Output 5

6 Degrees of Freedom of Analysis for Reactive Processes Molecular Species Balance + No. identified/labeled unknowns + No. independent chemical reactions No. of independent molecular species No. other equations relating unknown variables = No. degrees of freedom 6

7 Degrees of Freedom of Analysis for Reactive Processes Atomic Species Balance + No. identified/labeled unknowns No. independent atomic species No. of independent nonreactive molecular species No. other equations relating unknown variables = No. degrees of freedom 7

8 Degrees of Freedom of Analysis for Reactive Processes Extent of Reaction + No. identified/labeled unknowns + No. independent chemical reactions No. of independent reactive molecular species No. of independent nonreactive molecular species No. other equations relating unknown variables = No. degrees of freedom 8

9 Independent Chemical Reactions, Molecular and Atomic Species Chemical reaction: A chemical reaction is independent if it cannot be obtained algebraically from other chemical reactions involved in the same process. Molecular Species: If two molecular species are in the same ratio to each other wherever they appear in a process, then these molecular species are not independent. Atomic Species: If two atomic species occur in the same ration wherever they appear in a process, balances on those species will not be independent equations. 9

10 Independent Chemical Reactions, Molecular and Atomic Species Consider the following reactions: A =======> 2B B =======> C A =======> 2C Are these chemical reactions independent? 10

11 Independent Chemical Reactions, Molecular and Atomic Species Consider a continuous process in which a stream of liquid carbon tetrachloride (CCl 4 ) is vaporized into a stream of air. 11

12 Independent Chemical Reactions, Molecular and Atomic Species Molecular Species Analysis Total: 3 (O 2, N 2, CCl 4 ) Independent: 2 (O 2 or N 2, CCl 4 ) Atomic Species Analysis Total: 4 (O, N, C, Cl) Independent 2 (O or N, Cl or C) 12

13 Example Production of Chlorine (Deacon Process) In the Deacon process for the manufacture of chlorine, HCl and O 2 react to form Cl 2 and H 2 O. Sufficient air (21 mole% O 2, 79% N 2 ) is fed to provide 35% excess oxygen and the fractional conversion of HCl is 85%. Determine the amount of air required per mole of HCl fed into the process.calculate the mole fractions of the product stream components using: a. molecular species balances b. atomic species balances c. extent of reaction 13

14 Example Production of Chlorine (Deacon Process) Identify the components of the product stream: HCl O 2 N 2 Cl 2 H 2 O since not all will be converted (based on fractional conversion) since it is supplied in excess it goes with the O 2 in air but not consumed during the reaction produced during the process produced during the process 14

15 Example Production of Chlorine (Deacon Process) 15

16 Example Production of Chlorine (Deacon Process) To get mole fractions of components in the product stream: y i = n i /n t For the identified components: y HCl = n 2 /n t y O2 = n 3 /n t y N2 = n 4 /n t y Cl2 = n 5 /n t y H2O = n 6 /n t where n t = n 2 + n 3 + n 4 + n 5 + n 6 16

17 Example Production of Chlorine (Deacon Process) DEGREES OF FREEDOM ANALYSIS: Molecular Balance Unit: Reactor unknowns (n 1,n 2,n 3,n 4,n 5,n 6 ) +6 independent chemical reaction +1 independent molecular species 5 other equations: 35% excess O 2 & fractional HCl conversion Degrees of freedom

18 Example Production of Chlorine (Deacon Process) Method I: Molecular Species Balance 35% excess O 2 : 0.5 molo (O ) 100mol HCl 25molO 2molHCl (O ) 25mol O molO 2 2 T 2 2 A 2 2 1molair n mol O molair 0.21mol O mol air molair Required air molHCl molhcl 18

19 Example Production of Chlorine (Deacon Process) HCl Balance: Input Consumed Output = 0 (100 mol) 0.85(100 mol) n 2 = 0 n 2 = 15 mol HCl O 2 Balance: Input Consumed Output = 0 (33.75 mol) 85 mol HCl react (0.5/2) n 3 = 0 n 3 = 12.5 mol O 2 N 2 Balance: Output = Input n 4 = mol air (0.79 mol N 2 /1 mol air) n 4 = 127 mol N 2 19

20 Example Production of Chlorine (Deacon Process) Cl 2 Balance: Generated Output = 0 85 mol HCl react (1/2) n 5 = 0 n 5 = 42.5 mol Cl 2 H 2 O Balance: Generated Output = 0 85 mol HCl react (1/2) n 6 = 0 n 6 = 42.5 mol H 2 O 20

21 Example Production of Chlorine (Deacon Process) Calculation for mole fractions: Component i n i (moles) y HCl 15.0 (15.0/239.5) = O (12.5/239.5) = N (127.0/239.5) = Cl (42.5/239.5) = H 2 O 42.5 (42.5/239.5) = Total

22 Example Production of Chlorine (Deacon Process) DEGREES OF FREEDOM ANALYSIS: Atomic Balance Unit: Reactor unknowns (n 1,n 2,n 3,n 4,n 5,n 6 ) +6 independent atomic specie(s) 3 independent nonreactive molecular specie(s) 1 other equations: 35% excess O 2 & fractional HCl conversion Degrees of freedom

23 Example Production of Chlorine (Deacon Process) From % excess O 2 ======> n 1 From fractional conversion ======> n 2 Atomic Species Balance: H-Balance: 100(1) = n 2 + 2n 6 O-Balance: n 1 (0.21)(2) = 2n 3 + n 6 Cl-Balance: 100(1) = n 2 + 2n 5 N-Balance: n 1 (0.79)(2) = 2n 4 23

24 Example Production of Chlorine (Deacon Process) DEGREES OF FREEDOM ANALYSIS: Extent of Reaction Unit: Reactor unknowns (n 1,n 2,n 3,n 4,n 5,n 6 ) +6 independent chemical reaction(s) +1 independent reactive molecular species 4 independent nonreactive molecular species 1 other equations: 35% excess O 2 & fractional HCl conversion Degrees of freedom

25 Example Production of Chlorine (Deacon Process) From % excess O 2 ======> n 1 From fractional conversion ======> n 2 Extent of Reaction: HCl: Cl 2 : H 2 O: N 2 : O 2 : n 2 = 100 (2) n 5 = 0 + (1) n 6 = 0 + (1) n 4 = 0.79n 1 ± (0) n 3 = 0.21n 1 (0.5) 25

26 Example Production of Ethyl Bromide The reaction between ethylene and hydrogen bromide to form ethyl bromide is carried out in a continuous reactor. C 2 H 4 + HBr =====> C 2 H 5 Br The product stream is analyzed and found to contain 51.7 mole% C 2 H 5 Br and 17.3% HBr. The feed to the reactor contains only ethylene and hydrogen bromide. Calculate the fractional conversion of the limiting reactant and the percentage by which the other reactant is in excess. If the molar flow rate of the feed stream is 165 mol/s, what is the extent of reaction? 26

27 Example Production of Ethyl Bromide 27

28 Example Production of Ethyl Bromide DEGREES OF FREEDOM OF ANALYSIS: Atomic Species Unit: Reactor unknowns (x and n 2 ) +2 independent atomic specie(s) 2 independent nonreactive molecular specie(s) 0 other equations 0 Degrees of freedom 0 28

29 Example Production of Ethyl Bromide Determine the limiting reactant: C H HBr 2 4 Stoichiometric Ratio : 1.0 Actual Ratio : S 2 4 A C H x 165mol / s x HBr (1 x)(165mol / s) 1 x Solve x and n 2 using any 2 of the 3 atomic species balances: C-Balance H-Balance Br-Balance 29

30 Example Production of Ethyl Bromide C-Balance: mol x molc H 2mol C n n s mol 1molC2H 4 330x 1.654n Br-Balance: 2 mol 1 x mol HBr 1mol Br 165 n2 s mol 1mol HBr n (1 x) 0.69n 2 30

31 Example Production of Ethyl Bromide Solving simulateneously, x = mol C 2 H 4 /mol ; n 2 = mol/s Solving for the actual ratio of C 2 H 4 and HBr in the feed: C2H HBr A 1.0 Therefore, HBr is limiting. actual stoichiometric % excessc2h4 100 actual 31

32 Example Production of Ethyl Bromide Actual feed for C 2 H 4 : (165 mol/s)(0.545) = mol/s Theoretical requirement for C 2 H 4 based on stoichiometry: mol HBr 1molC2H4 mol mol s mol 1mol HBr s % excess C2H %

33 Example Production of Ethyl Bromide Fractional conversion of HBr: X HBr HBr amount reacted input output amount fed input X The can be determined based on C 2 H 4, HBr, C 2 H 5 Br: C 2 H 4 : HBr: C2H5Br: Solving for : 0.310(108.77) = (165)(0.545) 0.173(108.77) = (165)( ) 0.517(108.77) = 0 = 56.2 mol/s 33