2. Review on Material Balances

Transcription

1 2. Review on Material Balances Objectives After completing this chapter, students should be able to recall the principle of the conservation of mass recall the principle of the stoichiometry of chemical reactions recall such technical terms concerning chemical reactions as limiting and excess reactants, excess percentage, conversion, yield, and selectivity choose an appropriate basis of calculation and system boundary so that further material balance and energy-balance calculations can be done effectively 1

2 perform the degree-of-freedom (dof) analysis in order to select a suitable system boundary perform material-balance calculations for multiple-unit chemical processes, especially with recycle, purge, and by-pass streams 2

3 2.1 The Principle of the Conservation of Mass The law of conservation of mass states that matter can change from one form to another, but the total amount of mass must remain constant [1] For example, for the evaporation of water, liquid water, with the amount of, e.g., 100 kg, can completely be converted to water vapour with the same amount (i.e. 100 kg) In another process, not all 100 kg of liquid water is converted to water vapour; only 40 kg of liquid water is evaporated to water vapour, thus leaving 60 kg of liquid water unchanged 3

4 In the latter case, the initial amount of mass of the system, which contains only liquid water, is 100 kg, while the final state of the system comprises 40 kg of water vapour and 60 kg of liquid water which are added up to 100 kg This can be written in the form of equation as follows 100 kg liquid water = 40 kg water vapour + 60 kg liquid water (2.1) From Eq. 2.1, it implies that the total amount of mass remains unchanged, even though the state (สถานะ) of water changes (totally or partly) from one form to another 4

5 The conversion of mass is also applicable to chemical reactions, in which reactants are decomposed and products are generated For example, for the following chemical reaction: C 2 H 6 + Cl 2 C 2 H 5 Cl + HCl (2.2) C 2 H 6 and Cl 2 (i.e. reactants) can completely disappears, and C 2 H 5 Cl and HCl (i.e. products) are produced If the basis of calculation (will be discussed in detail later) is set as 1 mol of C 2 H 6, the total mass of the reactant side is the mass of 1 mol of C 2 H 6 (= 30 g) and the mass of 1 mol of Cl 2 (= 71 g), which is added up to = 101 g 5

6 the total mass of the product side is the mass of 1 mol of C 2 H 5 Cl (= 64.5 g) and the mass of 1 mol of HCl (36.5 g), which is added up to = 101 g In general, the principle of the conservation of mass can be written as follows [2] m m m m m input generation output consumption accumualation where m m m m acc sys sys sys final initial (2.3) (2.4) Note that Eq. 2.3 is commonly used for batch processes 6

7 For a continuous process, each term in Eq. 2.3 is presented per unit time (i.e. the process time) as follows dm dm in gen dm dm out consump dmacc dt dt dt dt dt (2.5a) or m m m m m in gen out consump acc (2.5b) or Input rate + Generation rate Output rate Consumption rate = Accumulation rate (2.6) Generally, the analysis of chemical processes is performed at a steady-state condition, in which there is no accumulation ( m 0 or no change in mass of the system) acc 7

8 as or Thus, Eq. 2.6 can be written, for a steady state, Input rate + Generation rate = Output rate + Consumption rate m m m m in gen out consump (2.7) Eqs 2.3, 2.5, and 2.7 are the material-balance equations, which can be written for the total mass of the input and output streams (i.e. overall balance) the mass of a single species in the input and output streams (i.e. species balance) The following are the examples the applications of the principle of the law of mass conservation (or the material-balance equation) to chemical processes 8

9 Example A 5 wt% calcium hydroxide [Ca(OH) 2 ] slurry with the amount of 2,000 kg is prepared from the Ca(OH) 2 slurry with the concentration of 20 wt% Determine of the amount of the 20 wt% slurry required for this process Since this is a chemical process without a chemical reaction (i.e. no generation and consumption terms) and no accumulation is normally assumed, Eq. 2.3 is reduced to or m input m (2.8a) output m m (2.8b) input output The input streams comprise 9

10 an unknown x amount of the 20 wt% slurry an unknown amount y of water while the output stream is 2,000 kg of the 5 wt% slurry Hence, the overall-balance equation can be written as follows x y 2, 000 (2.9) Since there are 2 unknowns with only 1 equation, it is NOT possible to solve for the answer; another equation is needed A species-balance equation [for either Ca(OH) 2 or water) is another equation 10

11 A Ca(OH) 2 -balance equation can be written as 20 0 x y 2, (2.10a) [note that water in the amount of y contains no Ca(OH) 2 or 0 wt% Ca(OH) 2 ] or 20 5 x 2, (2.10b) Solving Eq. 2.10b for x [mass of the 20 wt% Ca(OH) 2 slurry] gives x 500 kg. Substituting x = 500 kg into Eq. 2.9 and solving for y (mass of water) yields y 1,500 kg. 11

12 Example A liquid containing cells and broth (soup) with the flow rate and concentration of 1,000 L/h and 500 mg cells/l, respectively, is fed into a separating unit; the density of the feed is 1,200 g/l The output streams of the separating unit are Stream I S I : the stream of the liquid mixture (cells + broth) with the concentration of 50 wt% cells Stream II S II : the stream of the cell-free broth Determine the mass flow rates of Streams I and II This is another example of the process with no chemical reaction Thus, Eq. 2.7 is reduced to m m (2.11a) in 12 out

13 or m m (2.11b) in out From the information of the feed (or the input stream), the mass flow rate of the input stream can be computed as follows L g g m 1, 000 1,200 1,200,000 in h L h Thus, the overall mass balance can be written as follows m m (2.11b) in out g 1,200,000 S S (2.12) h I II The mass flow rate of cells in the feed (input stream) is L mg cells mg cells 1, ,000 h L h 13

14 The output streams comprises Stream I: which is the unknown mass S of the liquid mixture with 50 wt% cells Stream II: which the unknown mass S of the broth containing no cells I II Accordingly, a species balance for cells can be written as mg cells , 000 S S I II h mg cells S 500,000 I 100 h (2.13) Solving Eq for S I results in mg S 1, 000, 000 1,000 I h g h 14

15 Hence, the mass flow rate of Stream I S is 1,000 g/h I Substituting S into Eq and solving for I S yields II 1,200,000 = 1,000 + S II S = 1,200,000 1,000 = 1,199,000 II Thus, the mass flow rate of Stream II S is 1,199,000 g/h II 2.2 Stoichiometry It is common for chemical engineers to have to deal with chemical processes with chemical reactions 15

16 The following is the example of a chemical process with a chemical reaction, for which the principle of stoichiometry is applied Example Dry ice (solid CO 2 ) is produced from the products of the combustion of heptane (C 7 H 16 ) with pure O 2, and 60 wt% of CO 2 from the reaction can be converted to dry ice If 500 kg/h of dry ice is needed, how many kg/h of heptane must be burned The chemical-reaction equation (called stoichiometric equation) of this reaction can be written (and balanced) as follows C 7 H O 2 7CO 2 + 8H 2 O (2.14) 16

17 From Rxn. 2.14, it means that 7 kmol (or 7 kmol 44 kg/kmol = 308 kg) of CO 2 are produced from 1 kmol (or 1 kmol 100 kg/kmol = 100 kg) of C 7 H 6 It is given, the problem statement, that only 60% of CO 2 generated from Rxn can be converted to dry ice; in other words, 100 kg of CO 2 can produce 60 kg of dry ice Thus, if 500 kg of dry ice is to be produced, the amount of C 7 H 16 required can be calculated as follows 100 kg CO kg dry ice kg CO2 60 kg dry ice 100 kg C H kg CO kg C H kg CO2 17

18 Hence, to produce 500 kg/h of dry ice, C 7 H 16 in the amount of kg/h is to be burned 2.3 Limiting & Excess Reactants and Excess Percentage At times, to ensure a complete conversion of a specific reactant, another reactant is fed in excess, which means that it is fed in the amount higher than that required in the stoichiometric reaction The reactant with a complete conversion (or that is used up) is called a limiting reactant, while the reactant fed in excess is called an excess reactant Let s consider Rxn. 2.14: C 7 H O 2 7CO 2 + 8H 2 O once again 18

19 For the reactant side, it is desired that C 7 H 16 is to be completely converted or used up, and to ensure this, O 2 is to be fed excessively Accordingly, in this case, O 2 is an excess reactant C 7 H 16 is a limiting reactant The information from Rxn indicates that, stoichiometrically, 11 kmol of O 2 is required to completely react with 1 kmol of C 7 H 6 From the recent example, for 1 h operation, kg kg or kmol 100 kg/kmol of C 7H 16 is required to produce 500 kg of dry ice 19

20 Accordingly, the amount of O 2 needed is 11 kmol O kmol C H 29.8 kmol O kmol C H 7 16 Thus, in order for O 2 to be an excess reactant, O 2 in the amount of higher than 29.8 kmol is to be fed to react with kmol of C 7 H 16 Assume that O 2 in the amount of 40 kmol (or any amount that is higher than 29.8 kmol) is fed to the reactor, the excess percentage (% excess) can be computed using the following equation: moles of excess reactant actually fed moles of excess reactant required stoichiometrically % excess 100 moles of excess reactant required stoichiometrically as follows 20 (2.15)

21 40 kmol 29.8 kmol % excess of O kmol % excess of O 34.2% 2 It is important to emphasise that the stoichiometric calculations for the reaction with an excess reactant must be based on a limiting reactant, NOT an excess reactant Another example concerning limiting & excess reactants and % excess is as follows Example To ensure a complete combustion of LPG (70 mol% C 3 H mol% C 4 H 10 ), 15% excess O 2 is to be supplied to the combustor Determine the amount of O 2 required per mole of LPG 21

22 The stoichiometric equations concerning the combustion of C 3 H 8 + C 4 H 10 can be written as follows C 3 H 8 + 5O 2 3CO 2 + 4H 2 O (2.16) C 4 H O 2 4CO 2 + 5H 2 O (2.17) If 100 moles of LPG is set as a basis of calculation, it means that 70 mol of C 3 H 8 30 mol of C 4 H 10 is fed into the combustor The amount of O 2 required for Rxn is 5 mol O 70 mol C H 350 mol O mol C H

23 Rxn is 6.5 mol O 30 mol C H 195 mol O mol C H 4 10 Thus, the total amount of O 2 required stoichiometrically is = 545 mol Since 15% excess of O 2 is required, the actual amount of O 2 fed to the reactor can be computed using Eq as follows n n O O 2 actual 2 n O stoich % excess O stoich % excess O 2 n n n O O O 2 stoich 2 actual 2 stoich

24 % excess O n n n O O O actual stoich stoich n 1 n O O actual % excess O stoich (2.18) Note that Eq can also be applied to other species (not only O 2 ) By using Eq. 2.18, the actual amount of O 2 to obtain 15% excess O 2 can be calculated as follows 15 n mol O2 actual 100 n mol O2 actual 24

25 Hence, the amount of O 2 actually fed per mole of LPG is mol O mol LPG Setting a Basis of Calculation When dealing with complicated processes comprising multiple units, in order to make materialbalance calculations in order and as easy to be solved as possible, a proper basis of calculation must be firstly set up A basis of calculation may be a period of time (e.g., 1 h or 1 d), normally the operating time of a process a mass of material, normally mass of a feed with a known composition or mass of a desired product 25

26 Note that, during performing material-balance calculations, the basis of calculation can be changed, and all numbers must be scaled up or down to be in accordance with the new basis of calculation Example A mixture of 60 wt% C 3 H 8 and 40 wt% C 4 H 10 (LPG) is burned to provide energy for the process Determine the amount of CO 2 emitted per 1 kg of the fuel mixture. Since the composition of the mixture (i.e. the feed) is in mass percentage (wt%), the first basis of calculation can be set as 100 g of the mixture Basis: 100 g of the fuel mixture With this basis of calculation, the mixture comprises 26

27 60 g of C 3 H 8 (MW = 44 g/mol) 40 g of C 4 H 10 (MW = 58 g/mol) which can be converted to a molar basis as follows 60 g 1.36 mol 44 g/mol C 3H 8 40 g mol 58 g/mol C 4H 10 The chemical reactions for the combustion of this fuel mixture can be written and balanced as follows C 3 H 8 + 5O 2 3CO 2 + 4H 2 O (2.19) C 4 H O 2 4CO 2 + 5H 2 O (2.20) From Rxns and 2.20, the amounts of CO 2 produced are mol = 4.08 mol CO mol = 2.76 mol CO 2 27

28 Thus, the total amount of CO 2 generated is = 6.84 mol which is equivalent to 6.84 mol 44 g/mol = g Note that the amount of CO 2 of g is based on the basis of calculation of 100 g Since the problem wants to know the amount of CO 2 produced per 1 kg (or 1,000 g) of the feed, the basis of calculation should be changed (from 100 g of the feed) to 1,000 g of the feed It is unnecessary to re-perform material balance all over again; rather, we can scale up the calculation as follows 28

29 Basis (new): 1,000 g of the fuel mixture (feed) Thus, the amount of CO 2 generated is g CO 2 1, 000 g feed 3,010 g 3.01 kg CO2 100 g feed 2.5 Material Balances for Multiple-unit Processes and Degree-of-Freedom (DoF) Analysis In addition to choosing a decent basis of calculation, selecting an appropriate system boundary is also crucial to solving a complicated process It must be very frustrating if you have spent a long period of time (e.g., 1-2 h) solving the problem, but eventually find out that not enough information is available, and you cannot solve the problem any further 29

30 To avoid this unfortunate incident, you should start your problem solving by firstly selecting the system boundary and then determine whether or not enough information is available (to solve for the answer) for the chosen system boundary Such procedure is known as degree-of-freedom (dof) analysis The degree of freedom (dof) is defined as dof # of unknowns # of independent equations (2.21) The principle of the degree-of-freedom analysis is that, in order to be able to solve the problem, dof must be zero (0) or the number of unknowns must be equal to that of independent equations 30

31 The following example shows how to perform a degree-of-freedom analysis [2] Example A 33.3 wt% K 2 CrO 4 solution with the flow rate of 4,500 kg/h (the fresh feed) was joined with a recycle stream containing 36.4% K 2 CrO 4 before flowing into an evaporator The stream leaving the evaporator has the concentration of K 2 CrO 4 of 49.4% This leaving stream is then fed into a crystalliser and a filter, in which the feeding stream is divided into 3 exit streams: 1) a filter cake, consisting of pure K 2 CrO 4 solid crystals; 2) a 12.0 wt% K 2 CrO 4 solution, with the amount of 95 wt% of the filter cake; and 3) a filtrate, a 36.4 wt% K 2 CrO 4 solution, which is recycled back to mix with the freed feed 31

32 follows The flow chart of this process can be drawn as Water K 2 CrO 4 solution (95 wt% m 0 m 2 m 3 m 5 of the filter cake) Feed: 4,500 kg/h 33.3 wt% K 2 CrO 4 m 1 Evaporator 49.4 wt% K 2 CrO 4 solution Filtrate: 36.4 wt% K 2 CrO 4 Crystalliser Filter cake: Solid K 2 CrO 4 crystals m 4 m 6 Determine 1) the rate of evaporation 2) the production rate of the filter cake 3) the recycle ratio (i.e. the ratio of the flow rate of the recycle stream to that of the fresh feed) Basis: 1 h of operation From the flow chart, let (for the basis of calculation of 1 h of operation) 32

33 m = mass of the fresh feed 0 m = mass of the entering stream of the 1 evaporator m = mass of water leaving the evapora- 2 tor m = mass of the 49.4 wt% K 3 2 CrO 4 solu- tion leaving the evaporator m = mass of the filter cake 4 m = mass of the 12.0 wt% K 5 2 CrO 4 solu- tion leaving the crystalliser m = mass of the filtrate (the recycle 6 stream) If the overall process is chosen as a system boundary as shown on the next page 33

34 m 0 m 2 Water m wt% K 2 CrO 4 solution (95 wt% of the filter cake) Feed: 4,500 kg/h 33.3 wt% K2CrO4 m 1 Evaporator 49.4 wt% K 2 CrO 4 solution Filtrate: 36.4 wt% K 2 CrO 4 Crystalliser m 6 Filter cake: Solid K 2 CrO 4 crystals m 4 the unknowns are mass of water m mass of the filter cake m 2 mass of the 12.0 wt% K 2 CrO 4 solution leaving the crystalliser m 5 4 up are The material-balance equations that can be set the overall-balance equation m m m m , 500 m m m (2.22)

35 the species-balance (K 2 CrO 4 : K) equation x m x m x m x m K 0 K 2 K 4 K m m m , 500 m 0.12m (2.23) 4 5 the relationship between m and m (as 4 5 given in the problem statement) 95 m m m 0.95m (2.24) 5 4 As the number of unknowns (= 3) equals that of the equations (= 3), the degree of freedom is 0; thus, this problem can be solved 35

36 Let s consider what would happen if the evaporator is chosen as the system boundary, with the aim to solve for m (i.e. the rate of water eva- 2 poration) Water (m 2 ) Feed (m 0 ) 33.3% K 2 CrO 4 m1?% K2CrO4 Evaporator m % K 2 CrO 4 Filtrate (m 6 ) 36.4% K 2 CrO 4 In this case, the unknowns are m 1 the concentration of K 2 CrO 4 of the stream m 1 m 2 m 3 x K 1 36

37 Note that the streams m (the fresh feed) and 0 m (the recycle stream) are not included in this 6 system boundary The material-balance equations that can be established are the overall-balance equation m m m (2.25) the species-balance (K 2 CrO 4 ) equation x m x m x m K 1 K 2 K x m m m K x m 0.494m (2.26) K

38 The number of unknowns (= 4) is higher than that of the equations (= 2); or the degree of freedom is 4 2 = 2, which means that we cannot solve for m 2 If we chose the mixing point between the fresh feed and the recycle stream as the system boundary, the unknowns are m 1 the concentration of K 2 CrO 4 of the stream m 1 m 6 while the equations that can be set-up are the overall-balance equation m m m , 500 m m (2.27)

39 the species-balance (K 2 CrO 4 ) equation x m x m x m K 0 K 6 K ,500 m K1 x m 6 1 1, m x m (2.28) 6 K 1 1 The dof analysis for this system boundary results in the fact that the dof is 3 2 = 1, which means that this problem cannot be solved How can we solve for m and m? 1 6 After we obtain the values of m and m when 4 5 we select the overall process as our system boundary, we carry out material balance around the crystalliser as follows 39

40 m 5 (95 wt% of m 4 ): 12.0% K 2 CrO 4 m % K 2 CrO 4 Crystalliser m 4 100% K 2 CrO 4 Filtrate (m 6 ) 36.4% K 2 CrO 4 The unknowns are m 3 m 6 (note that m and m are already known) 4 5 The balance equations are the overall-balance equation m m m m (2.29) the species-balance (K 2 CrO 4 ) equation 0.494m m 0.12m 0.364m (2.30) 40

41 As m and m are already known, we can solve 4 5 Eq for m (and m ) 6 3 When m is obtained, we can go back to solve 6 for m using Eqs and 2.28, but, this time, 1 the number of unknowns is reduced to 2 (as m is 6 already known), i.e. m 1 the concentration of K 2 CrO 4 of the stream m 1 x K 1 With 2 unknowns and 2 equations, we can solve for m and 1 x K 1 From this example, it is evident that choosing the appropriate system boundary enables us to solve the problem effectively 41

42 The appropriate system boundary is the system boundary that results in the number of unknowns equals that of the independent equations 2.6 The Processes with Recycle, Purge, and By-pass Streams The example in the preceding section is an example of the process with a recycle stream In that example, with the recycle stream, the amount of desired product (i.e. the solid K 2 CrO 4 crystals) obtained m is 1,347 kg [2] (try doing 4 it yourself) For the process without the recycle stream, shown below [2] 42

43 Water K 2 CrO 4 solution (95 wt% of the filter cake) Feed: 4,500 kg/h 33.3 wt% K 2 CrO 4 Evaporator 49.4 wt% K2CrO4 solution Crystalliser Filter cake: Solid K 2 CrO 4 crystals Filtrate: 36.4 wt% K 2 CrO 4 the amount of the desired product or the solid crystals of K 2 CrO 4 is reduced to 977 kg [2] (again, try performing the calculations yourself) Another example for the processes with recycle streams is as follows [2] 43

44 Example The C 3 H 8 C 3 H 6 + H 2 (A B + C) reaction takes place in the process shown below Fresh feed 100 mol A A: n 1 mol B: n 2 mol Reactor A: n 3 mol B: n 4 mol C: n 5 mol Separation unit A: n 6 mol B: n 7 mol C: n 8 mol A: n 9 mol B: n 10 mol n 6 = 5 mol A = 0.555% n 3 n 10 = 5% n 7 Determine the conversion of A (C 3 H 8 ): 1) at the reactor 2) for the whole process Note that the conversion of A (from the Reaction Engineering course) is defined as Conversion of A n A initial n n A initial A final (2.31) 44

45 The whole process is chosen as a system boundary as follows Fresh feed n0 = 100 mol A A: n 1 mol B: n 2 mol Reactor A: n 3 mol B: n 4 mol C: n 5 mol Separation unit A: n 6 mol B: n 7 mol C: n 8 mol A: n 9 mol B: n 10 mol Since this is a process with a chemical reaction, mole is not conserved Thus, we cannot perform mole balances, but we can carry out atomic balances (as atoms of each species are conserved) as follows C-balance: (100 mol C 3 H 8 )(3 mol C) = (n 6 mol C 3 H 8 )(3 mol C) + (n 7 mol C 3 H 6 )(3 mol C) 300 = 3n 6 + 3n = 3(5) + 3n 7 45

46 n mol H-balance: (100 mol C 3 H 8 )(8 mol H) = (8 mol H)n 6 + (6 mol H)n 7 + (2 mol H)n = 8n 6 + 6n 7 + 2n = 8(5) + 6(95) + 2n 8 n It is given that and Thus, n 5 mol A 0.555% n 6 3 n 5% n mol A 5 mol A n mol A 0.555%

47 and n 5 5 n mol B Performing a C 3 H 8 (or A) balance around the separation unit as shown below yields (note that the separation unit is the unit without a chemical reaction; thus, a mole balance of each species can be carried out) A: n 3 mol B: n 4 mol C: n 5 mol Separation unit A: n 6 mol B: n 7 mol C: n 8 mol A: n 9 mol B: n 10 mol n n n n 9 n mol A 9 47

48 Conducting a C 3 H 8 (or A) balance at the mixing point before entering the reactor (once again, there is no chemical reaction at the mixing point) results in Fresh feed: n 0 = 100 mol A A: n 1 mol B: n 2 mol A: n 9 mol B: n 10 mol n n n n 1 n mol A 1 Hence, the conversion of C 3 H 8 (or A) can be computed using Eq as follows 48

49 at the reactor: Conversion of A Reactor n n A initial n A n 1 3 n initial n A final Reactor Conversion of A % Reactor for the whole process: n n Ainitial Conversion of A Whole Process nainitial n n 0 6 n A final Whole Process Conversion of A % Whole Process 49

50 It is evident that, without recycle, the conversion the reactant (i.e. C 3 H 8 ) is less than 10%, but with the recycle stream, the conversion of 95% can be obtained In the process with the recycle stream as illustrated in the recent examples, the product (or a fraction of the product) is recycled to mix with the fresh feed In the opposite direction, a fraction of the feed is diverted (or by-passed) the process unit to mix with the product as shown below Feed Process Unit Product By-pass Stream 50

51 The by-pass stream is normally used to control the concentration or composition of the final product (or exit stream) and/or to control the temperature of the final product (or exit stream) Example A 100 kg of feed containing 80 wt% n-c 5 H 12 and 20 wt% iso-c 5 H 12 is divided into 2 streams: Stream 1 (y kg) is fed into the iso-pentane tower, in which n-c 5 H 12 and iso-c 5 H 12 are separated from each other Stream 2 (x kg) is by-passed to mix with the stream of pure n-c 5 H 12, and the resulting mixture contains 90% n-c 5 H 12 and 10% iso-c 5 H 12 as depicted in the following flow chart Determine all of the unknowns (i.e. x, y, T, B, and P) [3] 51

52 Top: T kg i-c 5 100% Feed: 100 kg n-c 5 80% i-c 5 20% y kg n-c 5 80% i-c 5 20% iso-pentane Tower By-pass: x kg n-c 5 80% i-c 5 20% Bottom: B kg n-c 5 100% Product: P kg n-c 5 90% i-c 5 10% Performing material balances around the whole process yields Top: T kg i-c 5 100% Feed: 100 kg n-c 5 80% i-c 5 20% iso-pentane Tower Bottom: B kg n-c 5 100% Product: P kg n-c 5 90% i-c 5 10% 52

53 Overall balance 100 T P (2.32) Species (n-c 5 H 12 ) balance T 90 P P (2.33) Solving Eq for P gives 0.80 P kg 0.90 Substituting P into Eq and solving for T results in 100 T T 11.1 kg. 53

54 Mass-balancing around the iso-pentane tower yields Top: T = 11.1 kg i-c 5 100% y kg n-c 5 80% i-c 5 20% iso-pentane Tower Bottom: B kg n-c 5 100% Overall balance y 11.1 B (2.34) Species (i-c 5 H 12 ) balance y B y 11.1 (2.35) 54

55 gives Solving Eq for y results in 11.1 y 55.5 kg 0.20 Substituting y into Eq and solving for B B. B kg Since y = 55.5, the value of x can be solved as follows (conducting the overall balance at the separating point before entering the iso-pentane tower see the Figure on Page 52) 100 = x + y (2.36) 100 = x x = = 45.5 kg Note that x is the mass of the by-pass stream 55

56 In several circumstances, a feed contains an inert (an inert can also be produced via the reaction), when a fraction of the outlet stream is recycled to mix with the feed, the inert remains in the recycle stream The inert is kept adding into the process as it comes with the feed or is generated via the reaction in the reactor, thus resulting an accumulation (or build-up) of the inert in the recycle stream (in other words, the amount of the inert in the recycle stream keeps increasing) until the amount of the inert is too high To avoid an accumulation (a build-up) of the inert in the recycle stream, purging is needed, as exemplified in the following example [4] 56

57 Example In the production of ammonia (NH 3 ) from the reaction of hydrogen (H 2 ) and nitrogen (N 2 ), the conversion is limited to 15% (due to the chemical equilibrium of the reaction). Accordingly, the exit stream containing NH 3 and un-reacted H 2 and N 2, is recycled to mix with the feed, which contains 0.2 mol% argon (an inert gas that comes from the nitrogen separation process). If it is required that argon in the recycle stream be below 5 mol%, calculate the amount of the purge stream per 100 mol of feed. The process flow chart can be drawn as follows Purge Stream: 5 mol% Ar Feed: 0.2 mol% Ar Reactor NH 3, H 2, N 2 57

58 The principle of an inert is that The inlet amount of an inert = The outlet amount of an inert (2.37) If the basis of calculation is set as 100 mol of feed, the mole balance for the inert (i.e. argon: Ar) can be performed as follows P 0 E (2.38) (P = moles of the purge stream and E = moles of the exit stream) Solving Eq for P yields P 4 mol

59 2.7 Conversion, Yield, and Selectivity In the preceding section, a conversion is defined as Eq. 2.31: Conversion of A n A initial n n A initial A final Note that n A initial n is the amount of A (a limiting reactant) consumed in the reaction(s) A final It is desirable that the reactant(s) be converted to the desired product(s) as much as possible 59

60 For example, for Rxn. 2.39: C 2 H 6 C 2 H 4 + H 2 it is desirable that C 2 H 6 be converted to C 2 H 4 (the desired product) as much as possible and that only a single reaction (i.e. Rxn. 2.38) take place such that C 2 H 6 is converted to C 2 H 4 (and H 2 ) only no further reaction of C 2 H 4 and/or H 2 to another product (or other products) Unfortunately, however, such circumstances rarely occur In the real situation, for the dehydrogenation of ethane (i.e. Rxn. 2.39) 60

61 C 2 H 6 C 2 H 4 + H 2 (2.39) once H 2 is formed, it can react with the reactant (C 2 H 6 ) to form another substance as follows C 2 H 6 + H 2 2CH 4 (2.40) Additionally, C 2 H 4 can react with C 2 H 6 to form other products as follows C 2 H 6 + C 2 H 4 C 3 H 6 + CH 4 (2.41) Rxns and 2.41 are the undesired reactions, which is normally called the side reactions or competing reactions When the side reactions take place, the amount of the desired product generated is reduced 61

62 To describe how much a desired product is produced or to describe how much the main reaction proceeds relative to the side (or competing) reaction(s), the following terms are employed The amount of the desired product actually produced Yield The amount of the desired product produced if there were no side reactions and the limiting reactant had reacted completely (2.42) Selectivity The amount of the desired product actually produced The amount of the undesired product actually produced (2.43) 62

63 The following example illustrates how conversion, yield, and selectivity are calculated [2] Example The following reactions: C 2 H 6 C 2 H 4 + H 2 (2.39) C 2 H 6 + H 2 2CH 4 (2.40) take place in a steady-state continuous reactor The feed contains 85 mol% C 2 H 6 and the balance inert (I) The composition of the outlet stream is 30.3 mol% C 2 H 6, 28.6% C 2 H 4, 26.7% H 2, 3.7% CH 4, and 10.7% I Determine the conversion of C 2 H 6 the yield of ethylene (C 2 H 4 ) the selectivity of ethylene to methane (CH 4 ) 63

64 Basis: 100 mol of feed Hence, the feed contains 85 mol C 2 H 6 15 mol inert (I) Let P be the amount of the product, and the amount of P can be computed as follows The amount of I in the inlet stream = The amount of I in the exit stream P P mol 64

65 Therefore, the product comprises C 2 H mol 100 C 2 H mol 100 H mol 100 CH mol 100 Inert 15 mol When considering Rxn. 2.39: C 2 H 6 C 2 H 4 + H 2 (2.39) and the information of the product, in which C 2 H 4 is produced in the amount of 40.0 mol, it means that 65

66 C 2 H 6 in the amount of 40.0 mol is consumed in Rxn H 2 in the amount of 40.0 mol is generated in Rxn When determining Rxn. 2.40: C 2 H 6 + H 2 2CH 4 (2.40) along with the information of the product, in which CH 4 is produced in the amount of 5.2 mol, it implies that C 2 H 6 in the amount of mol is 2 consumed in Rxn H 2 in the amount of mol is also 2 consumed in Rxn

67 Thus, in summary, the final amount of C 2 H 6 (or the remaining amount of C 2 H 6 at the end of the reaction) is = 42.4 mol (40 mol is consumed in Rxn and 2.6 mol is consumed in Rxn. 2.40) Accordingly, the conversion of C 2 H 6 is (using Eq. 2.31) Conversion of C H 2 6 n n C H C H 2 6 initial 2 6 n C H initial final Conversion of C H %

68 Note that the numerator ( ) is, in fact, = 42.6 mol equal to = 42.6 mol which is the amount of C 2 H 6 consumed in both Rxns and 2.40 If there was no side reaction (i.e. Rxn in this example) and all amounts of C 2 H 6 were converted (to the products), the amount of the desired product (i.e. C 2 H 4 ) that would have been produced is 85 mol (see Rxn. 2.39) However, the actual amount of C 2 H 4 produced in Rxn is 40.0 mol Therefore, the yield of C 2 H 4 can be calculated using Eq as follows 68

69 40.0 Yield % 85.0 In this example, the competing (or side) reaction (Rxn. 2.40) produces CH 4, while the desired product is C 2 H 4 (generated in Rxn. 2.39) Hence, selectivity of C 2 H 4 (the desired product) relative to CH 4 (the undesired product) is (using Eq. 2.43) Selectivity The amount of C H produced The amount of CH produced Selectivity

70 References [1] S.S. Zumdahl, Chemical Principles, 6 th ed., Cengage Learning, [2] R.M. Felder and R.W. Rousseau, Elementary Principles of Chemical Processes, Wiley, [3] D.M. Himmelblau and J.B. Riggs, Basic Principles and Calculations in Chemical Engineering, 8 th ed., Pearson, [4] R. Sinnot and G. Towler, Chemical Engineering Design, 5 th ed., Elsevier,