Chapter 4. Fundamentals of Material Balance

Transcription

1 Chapter 4 Fundamentals of Material Balance

2 Introduction to Chapter 4 1) In chapter 4 we will present methods for organizing known information about process variables, setting up martial balance equations, and solving these equations for unknown variables. 2) Every chemical process analysis involves writing and solving material balances to account for all process species in feed and product streams. This chapter outlines and illustrates a systematic approach to material balance calculations. The procedure is to draw and label a flowchart, perform a degree-of-freedom analysis to verify that enough equations can be written to solve for all unknown process variables, and write and solve the equations. 3) The general balance equation is: input + generation - output consumption = accumulation

3 4.6 ( Chemical Reaction Stoichiometry) Stoichiometry: Stoichiometry is the theory of the proportions in which chemical species combine with one another. Stoichiometric equation of a chemical reaction : Is a statement of the relative number of molecules or moles of reactants and products that participate in the reaction. Example on stoichiometric equation : 2 SO 2 + O 2 2 SO 3 This equation indicates (shows) that every two molecules of SO 2 reacts with one molecule of O 2 to produce two molecules of SO 3. Stoichiometric coefficient: Is the number before (in-front of) each molecule in the stoichiometric equation. 2 SO 2 + O » 2 SO 3 Example: The stoichiometric coefficient of SO 2 is 2. A valid stoichiometric equation: A valid stoichiometric equation must be balanced, were the number of atoms of each atomic species must be the same on both sides of the equation. (Atoms can neither be created nor destroyed in a chemical reaction.)

4 4.6 ( Chemical Reaction Stoichiometry) Example on balancing an equation: C 4 H 8 + O 2 ---> CO 2 + H 2 O This equation is not valid because the atoms of each atomic specie is not equal in both sides. Balance Carbon: We need 4 carbon atoms in both sides so we need to add 4 to the right side. C 4 H 8 + O 2 ---> 4 CO 2 + H 2 O Balance Hydrogen: We need 8 hydrogen atoms in both sides so we need to add 4 in the right side were 4 x 2 = 8. C 4 H 8 + O 2 ---> 4 CO 2 + 4H 2 O Balance Oxygen: We have 12 oxygen atoms in the right side so that means we need to get the total of oxygen in the left side equals to 12 so we will add 6 were 6 x 2 = 12. C 4 H O 2 ---> 4 CO 2 + 4H 2 O

5 4.6 ( Chemical Reaction Stoichiometry) Stoichiometric ratio: The stoichiometric ratio of two molecular species participating in a reaction is the ratio of their stoichiometric coefficients in the balanced reaction equation. This ratio can be used as a conversion factor to calculate the amount of a particular reactant (or product) that was consumed (or produced), given a quantity of another reactant of product that participated in the reaction. Example: 2 SO 2 + O » 2 SO 3 We can write a stoichiometric ratios: 2 mol SO3 generated 1 mol 02 consumed also we can say 2 lb-moles SO 2 consumed 2 lb-moles SO 3 generated We can use any unit for moles in the stoichiometric ratio but they have to be the same for each ratio alone.

6 4.6 ( Chemical Reaction Stoichiometry) Example on how we can use the stoichiometric ratio: 2 SO 2 + O » 2 SO 3 If you know, for example, that 1600 kg/h of SO 3 is to be produced, you can calculate the amount of oxygen required as : 1. Find the stoichiometric ratio between SO 3 and O 2 : 1 kmol O 2 2 kmol SO 3 2. Convert the mass flow-rate to mole flow- rate because our conversion factor is in moles: n (mole flow-rate) = m (mass flow-rate) / Mwt Picked kmol to make calculations easier later on because the mass was given in kg in the question. mole flow-rate of SO 3 = 1600 (kg/h) / 80 ( kg/kmol) mole flow-rate of SO 3 = 20 (Kmol/h) (generated)

7 4.6 ( Chemical Reaction Stoichiometry) 3. By using the conversion factor (Stoichiometric ratio) and moles of SO 3 generated we can find the mole flow-rate of O 2 consumed : 1 kmol O 2 (consumed) 2 kmol SO 3 (generated) x 20 (Kmol SO 3 /h) (generated) = 10 (kmol O 2 / h) (consumed) 4. Convert the O 2 from mole flow-rate to mass flow-rate because in the question they asked for it in Kg not Kmol: n (mole flow-rate) = m (mass flow-rate) / Mwt 10 (kmol O 2 / h) = m (mass flow-rate) / 32 (kg/kmol) m (mass flow-rate) = 320 kg O 2 /h

8 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Stoichiometric Proportion: If we had two reactants A and B they are said to be present in Stoichiometric proportion if the ratio ( moles A feed (present) / moles B feed (present) ) equals the stoichiometric ratio obtained (got) from the balanced reaction equation. Example: 2 SO 2 + O » 2 SO 3 For the reactants in the reaction to be present in stoichiometric proportion then there must be 2 moles of SO 2 for every mole of O 2 present in the feed of the reactor. n(so 2 )/n(o 2 ) = 2/1 The ratio is 2:1 Stoichiometric ratio: The ratio of coefficients in the stoichiometric equation. Note: If reactants are fed to a chemical reactor in stoichiometric proportion and the reaction proceeds to completion, all of the reactants are consumed.

9 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) limiting reactant: Is the reactant that would run out if a reaction proceeded to completion. Excess reactant: Is the reactant that would not run out if a reaction proceeded to completion. Remember the limiting and excess reactant are only for molecules in the feed not product. Note: 1) A reactant is limiting if it is present in less than its stoichiometric proportion relative to every other reactant. 2) If all reactants are present in stoichiometric proportion, then no reactant is limiting (or they all are, depending on how you choose to look at it). How to find the limiting reactant and the excess reactant: Feed of Molecule Stoichiometric Coefficient of Molecule

10 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example: 2 C 2 H 4 + O 2 ---» 2 C 2 H 4 O If the feed to the reactor contains 100 kmol C 2 H 4 and 100 kmol O 2 ; which reactant is the limiting? C 2 H 4 : Feed of Molecule Stoichiometric Coefficient of Molecule 100Kmol = =50 2 Kmol O 2 : Feed of Molecule Stoichiometric Coefficient of Molecule 100 kmol 1 Kmol = = 100 Therefore the limiting reactant is C 2 H 4.

11 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Stoichiometric requirement: Let say we have n A which is the amount number of moles of an excess reactant (A) that is present in the feed to a reactor. Then (n A ) stoich is the stoichiometric requiment which is the amount needed to react completely with the limiting reactant. Fractional excess: The fractional excess of the reactant is the ratio of the excess to the stoichiometric requiment. Fractional excess of (i) = (n i ) feed - (n i ) stoich (n i ) stoich The percentage excess of (i) is 100 times the fractional excess.

12 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form ethane: C 2 H 2 + 2H » C 2 H 6 and suppose that 20.0 kmol/h of acetylene and 50.0 kmol/h of hydrogen are fed to a reactor. The stoichiometric ratio of hydrogen to acetylene is 2:1. 1 First find which one is the excess: C 2 H 2 : (feed of C 2 H 2 )/(stoichiometric of C 2 H 2 ) = (20/1) =20 H 2 : (feed of H 2 )/(stoichiometric of H 2 ) =(50/2) =25 Therefore the limiting is C 2 H 2 ; and the excess is H 2.

13 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) 2 Fractional excess of H 2 : Fractional excess of (H 2 ) = (n H2 ) feed - (n H2 ) stoich (n H2 ) stoic (n H2 ) feed = 500 kmol/h (given in question) Steps to find the (n i ) stoich : (n H2 ) stoich (From equation: (C 2 H 2 + 2H » C 2 H 6 ) see relation between H 2 and C 2 H 2 ) L.R H 2 (n H2 ) stoich = 1 C 2 H » 2 H 2 Feed of L.R : 20 c 2 H » (n H2 ) stoich Cross Multiply: 20 c 2 H 2 x 2 H 2 = 1 C 2 H 2 x (n H2 ) stoich (n H2 ) stoich = 40 Kmol/h Fractional excess of (H 2 ) = 500 kmol/h 40 kmol/h = kmol/h We say that there is 25% excess hydrogen in the feed.

14 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Fractional conversion: f conv = moles reacted moles Fed = (n i ) feed - (n i ) product (n i ) feed To find the fraction unreacted is accordingly 1 f conv. To find the percentage conversion it is 100 time the conversion. If the reaction proceed to completion that means f conv = 1 (f conv = 100%) Extent of reaction: (n i ) product - (n i ) feed ξ = v i vi = The stoichiometric coefficient of the (ith) species in a chemical reaction; making it negative for reactants and positive for products. The extent of reaction has the same units as n or mole flow-rate. The extent of reaction is a quantity that measures the degree (range or amount) in which the reaction proceeds. If the reaction rate increases the extent will increase with the rate.

15 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Test yourself page 120: The oxidation of ethylene to produce ethylene oxide proceeds according to the equation : 2 C 2 H 4 + O » 2 C 2 H 4 O The feed to a reactor contains 100 kmol C 2 H 4 and 100 kmol O 2. 1 Which reactant is limiting? C 2 H 2 : (feed of C 2 H 4 )/(stoichiometric of C 2 H 4 ) = (100/2) =50 H 2 : (feed of O 2 )/(stoichiometric of O 2 ) =(100/1) =100 Therefore the limiting is C 2 H 4 ; and the excess is O 2.

16 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) 2 What is the percentage excess of the other reactant? O 2 is excess ; so the percentage excess of O 2 is: Fractional excess of (H 2 ) = (n O2 ) feed - (n O2 ) stoich (n O2 ) stoic Fractional excess of (H 2 ) = 100kmol - (n O2 ) stoich x 100% Find (n O2 ) stoich : (n O2 ) stoich L.R O 2 2 C 2 H » 1 O » (n O2 ) stoich By cross multiplying (n O2 ) stoich = 50 kmol. Fractional excess of (O 2 ) = (100 kmol- 50 kmol) / (50 kmol) = 100%

17 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) 3 If the reaction proceeds to completion, how much of the excess reactant will be left; how much C 2 H 4 O will be formed; and what is the extent of reaction? a. If the reaction proceeds to completion that means all of the limiting reactant will react completely with stoichiometric requiment. 2 C 2 H » 1 O » O 2 O 2 = 50 Therefore 50 kmol of O 2 will be left. b. 2 C 2 H » 2 C 2 H 4 O » C 2 H 4 O C 2 H 4 O= 100 Kmol Therefore 100kmol of C 2 H 4 O will be formed.

18 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) c. (n i ) product - (n i ) feed ξ = v i Previously we found out how much C 2 H 4 O will be formed which is the product and we don t have any C 2 H 4 O in the feed and we know its coefficient from the reaction formula which is 2 so that makes it easy to find the extent. ξ = ξ = 50kmol

19 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Chemical Equilibrium: Its when the reaction is reversible which means that reactants form products and products undergo the reverse reactions to reform the reactants. Lets say we have: C 2 H 4 + H 2 O C 2 H 5 OH o If you start with ethylene and water, the forward reaction occurs; then once ethanol is present, the reverse reaction begins to take place. o If the concentrations of C2H4 and H 2 O decrease, the rate of the forward reaction decreases. o As the C 2 H 5 OH concentration increases, the rate of the reverse reaction increases. o When the rates of the forward and reverse reactions are equal there is no further composition change takes place and the reaction mixture is in chemical equilibrium.

20 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Equilibrium constant: Lets say we have a chemical equilibrium equation A+B C+D K eq = P D P C P A P B We know that partial pressure equals to y i P T (were y i is the mole fraction and P T is the total pressure were we can say that: K eq = (y D P t )(y C p T ) (y B P T )(y A P T ) We can also replace y i with n i /n T because we know that y i =n i /n T : K eq = (n D /n T )(n c /n T ) (n B /n T )(n A /n T ) Where P A, P B, P C, P D are partial for each element. We can cancel all the total pressure because it the same in all elements. Therefore we can say that : K eq = n D n C n B n A

21 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) The terms yield and selectivity are used to describe the degree to which a desired reaction predominates(leads) over competing side reactions: Yield: Selectivity: moles of desired product formed Moles that would have been formed if there were no side reactions and the limiting reactant had reacted completely. moles of desired product formed moles of undesired product formed = actual theoretical The yield is defined is always a fraction but it may also be expressed as a percentage by multiplying by 100%. Notes: 1. Mostly when gives to reactions it would ask about yield and selectivity. 2. The yield and selectivity helps find ratio between two elements. 3. When they give you the yield or selectivity amounts in question it will help you either find moles of input or output.

22 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) The concept of extent of reaction can be extended to multiple reactions, only now each independent reaction has its own extent. If a set of reactions takes place in a batch or continuous steady-state reactor and v ij is the stoichiometric coefficient of substance ( i ) in reaction ( j ) (negative for reactants, positive for products). Formula : n i(out) =n i(feed) + v ij ξ j Example: C 2 H 4 +(1/2) O 2 ->> C 2 H 4 O C 2 H O 2 ->> 2 CO H 2 O Desired reaction undesired reaction (side reaction) ξ 1 ξ 2

23 4.7 (Balances on Reactive Processes) Balances on reactive processes: ( Not needed for this course ) 1) Molecular Species Balance: A material balance equation is applied to each chemical compound appearing in the process. Molecular Species Balance equation 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. Note: Mostly for molecular species which are the reactant for the chemical reaction the molecular species balance equation is Input Consumed = Output and for the molecular species which are the product for the chemical reaction the molecular species balance would be Generated = Output.

24 4.7 (Balances on Reactive Processes) Balances on reactive processes: 2) Atomic species Balances: The balance is applied to each element appearing in the process. Atomic Species Balance equation for steady-state reactive processes : Input = Output Balances on atomic species can be written input = output, since atoms can neither be created (generation = 0) nor destroyed (consumption = 0) in a chemical reaction.

25 4.7 (Balances on Reactive Processes) Independent Chemical Reactions: A chemical reaction is independent if it cannot be obtained algebraically from other chemical reactions involved in the same process. We know that the maximum number of material balances you can write for a nonreactive process equals the number of independent species involved in the process. It is time to take a closer look at what that means and to see how to extend the analysis to reactive processes. 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.

26 4.7 (Balances on Reactive Processes) Example: Consider a continuous process in which a stream of liquid carbon tetrachloride (CCl4) is vaporized into a stream of air.. n 1 (mol O 2 /s) 3.76 n 1 (N 2 /s). n 3 (mol O 2 /S) 3.76 n 3 (mol N 2 /s) n 4 (mol CC! 4 (v)/s). n 2 (mol CCI 4 (l)/s). n 5 (mol CCI 4 (l)/s)

27 4.7 (Balances on Reactive Processes) Example (continued): Molecular Species: Number of molecular species Number of Independent species Independent Chemical Reactions Note : N 2 and O 2 have the same ratio. Atomic Species: Number of atomic species 3 (O 2, N 2, CCl 4 ) 4 (O, N, C, Cl) 2 (O 2 or N 2, CCl 4 ) Number of Independent species 2 (O or N, Cl or C) Note: If you write a O and N balance equation you will find out that they have the same equation, the same thing if you write a balance equation for C and CI.

28 4.7 (Balances on Reactive Processes) Degrees of Freedom of Analysis for Reactive Processes : 1) Molecular Species Balance: + No. unknown labeled variables (unknowns) + No. independent chemical reactions No. of independent molecular species No. other equations relating unknown variables = No. degrees of freedom Once a generation or consumption term has been calculated for a species in a given reaction, the generation and consumption terms for all other species in that reaction may be determined directly from the stoichiometric equation.

29 4.7 (Balances on Reactive Processes) 2) Atomic Species Balance: + No. unknown labeled variables (unknowns) No. independent atomic species No. of independent nonreactive molecular species No. other equations relating unknown variables = No. degrees of freedom Note: The number of independent nonreactive molecular species that means the number of how many species that won't react lets say with have air and another molecular specie going in the reactor the N 2 in the air is a independent nonreactive molecular specie. The number of other equations relating unknown variables means that sometimes they give us extra information like ( fractional conversion or yield or selectivity or excess.) and basis they are counted as the number of other equations relating unknown variables.

30 4.7 (Balances on Reactive Processes) Example: 100 kmol C 2 H 6 /min 40 kmol H. 2 /min n 1 (kmol C 2 H 6 /min!. n 2 (kmol C 2 H 4 /min) Degree of Freedom (Molecular species Balance):.. 2 unknown labeled variables (n 1,n 2 ) + 1 independent chemical reaction - 3 independent molecular species balances (C 2 H 6, C 2 H 4, and H 2 ) - 0 other equations relating unknown variables = 0 degrees of freedom

31 4.7 (Balances on Reactive Processes) Example (continued): Degree of Freedom (Atomic species Balance): 2 unknown labeled variables - 2 independent atomic species balances (C and H) - 0 molecular balances on independent nonreactive species - 0 other equations relating unknown variables Use atomic balance to solve the question: = 0 degrees of freedom After we knew that the question could be solved from the degree of freedom because it is equal to zero we can now solve the question using atomic balance: The atomic balance equation is input = output : 1. C-Balance: # of atoms C in molecular formula x moles (input) = # of atoms C in molecular formula x moles (output). 2 (Kmol C / Kmol C 2 H 6 ) X 100 (Kmol C 2 H 6 /s) = 2(Kmol C / Kmol C 2 H 6 ) x n 1 (kmol C 2 H 6 /s) + 2(Kmol. C / Kmol C 2 H 4 ) x n 2 (kmol C 2 H 4 /s) = n 1 + n 2

32 4.7 (Balances on Reactive Processes) Example (continued): 2. H-Balance: # of atoms H in molecular formula x moles (input) = # of atoms H in molecular formula x moles (output) 6 (kmol H/kmol C 2 H 6 ) x 100 (kmol C 2 H 6 /min) = 2 (kmol H/kmol H 2 ) x 40 (kmol H 2 /min) (kmol H/kmol C 2 H 6 ) x n 1 (kmol C 2 H 6 /min) + 4 (kmol H/C 2 H 4 ) x n 2 (kmol C 2 H 4 /min) = 520= 6n 1 + 4n 2 with two equations and two unkowns we can find n 1 and n 2 :. n 1 = 60 kmol C 2 H 6 /min.. n 2 =40 kmol C 2 H 4 /min...

33 4.7 (Balances on Reactive Processes) Material Balances on Reactive Processes: 3) Extent of Reaction: Expressions for each reactive species is written involving the extent of reaction. For single reaction: (n i ) product = (n i ) feed + v i ξ For multiple reaction: n i(out) =n i(feed) + v ij ξ j

34 4.7 (Balances on Reactive Processes) Given that all three methods of carrying out material balances on reactive systems molecular species balances, were they all give you the same results, the question is which one to use for a given process. There are no hard and fast rules but we suggest the following guidelines: Atomic species balances generally lead to the most straightforward solution procedure, especially when more than one reaction is involved. Extents of reaction are convenient for chemical equilibrium problems and when equation solving software is to be used. Molecular species balances require more complex calculations than either of the other two approaches and should be used only for simple systems involving one reaction.

35 4.7 (Balances on Reactive Processes) Degrees of Freedom of Analysis for Reactive Processes: 3) Extent of Reaction: + No. unknown labeled variables + No. independent chemical reactions (one extent of reaction for each) No. of independent reactive molecular species (one equation for each species in terms of extents of reaction) No. of independent nonreactive molecular species (one balance equation for each) No. other equations relating unknown variables = No. degrees of freedom

36 4.7 (Balances on Reactive Processes) Example page 131: Methane is burned with air in a continuous steady-state combustion reactor to yield a mixture of carbon monoxide, carbon dioxide, and water. The reactions taking place are: CH 4 + (3/2) O 2 CO + 2H 2 O CH 4 + 2O 2 CO 2 + 2H 2 O The feed to the reactor contains 7.80 mole% CH 4, 19.4% O 2, and 72.8% N 2. The percentage conversion of methane is 90.0%, and the gas leaving the reactor contains 8 mol CO 2 /mol CO. Carry out a degree-of-freedom analysis on the process. Then calculate the molar composition of the product stream using atomic species balances, and extents of reaction.

37 4.7 (Balances on Reactive Processes) Example page 131 (continued): 1) Draw the flow chart and Basis of 100 mol Feed: 100 mol mol CH 4 /mol mol O 2 /mol mol N 2 /mol a. Atomic species balance: Degree of Freedom: + 5 unknown variables - 3 independent atomic species balances (C, H, O) - 1 nonreactive molecular species balance (N 2 ) - 1 specified methane conversion = 0 degrees of freedom. n ch4 (mol CH 4 ) n co (mol CO) 8n co (mol CO 2 ) n H2O (mol H 2 O) n O2 (mol O 2 ) n N2 (mol N 2 ) Note: N 2 it goes with the O 2 in air but not consumed during the reaction.

38 4.7 (Balances on Reactive Processes) Example page 131 (continued): Always start with the extra given information it will help you determine at least one of the unknowns: 90% CH 4 Conversion: f conv = (n ch4(feed) - n ch4(out) ) / n ch4(feed) 0.90=((100x0.0780)-n ch4(out) ) / (100x0.0780) n ch4(out) =0.780 Balance ( input=output): We can do C, H, and O balance: C-balance: (1 x (100 x )) = (1 x 0.780) + (1 x n co ) +(1 x 8n co (mol CO 2 ) ) n co = mol CO n co2 = 8n Co = (8 X 0.780) mol CO 2 = 6.24 mol CO 2

39 4.7 (Balances on Reactive Processes) Example page 131 (continued): H-Balance: ( 4 X (100X0.0780) ) = (0.78 x 4) + (2 x n H20 ) n H20 = 14.0 mol H 2 O O-Balance: ( 2 x (100 x 0.194) ) = ( 2 x n O2 ) + ( 1 x 0.780) + ( 2 x 6.24) + (1 x 14) n O2 = 5.75 mol O 2 N 2 Balance : N 2 balance is the regular material balance (input=output) because its not reactive. (100 x 0.728) = n N2 n N2 = 72.8 mol N2

40 4.7 (Balances on Reactive Processes) Example page 131 (continued): b. Extents of reaction balance: Degree of Freedom: + 5 unknown labeled variables + 2 independent reactions - 5 expressions for n I (ξ) (i= CH 4, 02, CO, CO:, H 2 O) - 1 nonreactive molecular species balance (N 2 ) - 1 specified methane conversion = 0 degrees of freedom. Same as we did with atomic species balance always start with the extra given information it will help you determine at least one of the unknowns: 90% CH 4 Conversion: f conv = (n ch4(feed) - n ch4(out) ) / n ch4(feed) 0.90=((100x0.0780)-n ch4(out) ) / (100x0.0780) n ch4(out) =0.780 Balance: n i(out) = n i(feed) + v ij ξ j

41 4.7 (Balances on Reactive Processes) Example page 131 (continued): CH 4 + (3/2) O 2 CO + 2H 2 O CH 4 + 2O 2 CO 2 + 2H 2 O ξ 1 ξ 2 CH 4 - balance: = 100 x mol ξ 1 - ξ mol = ξ 1 + ξ 2 CO- balance: n co = ξ 1 CO 2 - balance: n co2 (= 8n co ) = ξ 2 H 2 O balance: n H2O = 2ξ 1 + ξ 2 O 2 balance: n O2 = (100 x 0.194) (3/2 ξ 1 ) ( 2 ξ 2 )

42 4.7 (Balances on Reactive Processes) Example page 131 (continued): Substitute n co = ξ 1 and n co2 (= 8n co ) = ξ 2 in 7.02 mol = ξ 1 + ξ 2 : we get that n co = 0.78 mol CO Then we can find: n co2 = 8 nco = (8 x 0.780) mol CO 2 = 6.24 mol CO 2 n co = ξ 1 =0.78 mol n co2 = ξ 2 =6.24 mol n h2o = 14.0 mol H 2 O n O2 = 5.75 mol O 2 Once again the same flow rates have been calculated, so that the molar composition of the product gas must therefore also be the same. For this problem, atomic species balances provide the least heavy solution.

43 4.7 (Balances on Reactive Processes) Product Separation and Recycle: There is two type of conversion that we can do on a chemical process with product separation and recycle of unconsumed reactants. 1) Overall Conversion: reactant input to process reactant output from process reactant input to process 2) Single-Pass Conversion: reactant input to reactor reactant output from reactor reactant input to reactor As usual, the corresponding percentage conversions are obtained by multiplying these quantities by 100%.

44 4.7 (Balances on Reactive Processes) Example: 75 mol A/min 100 mol A/min 25 mol A/min Reactor 75 mol B/min Product Separation Unit 75 mol B/min 25 mol A/min 1) Overall conversion of A: (green dashed borders) 75 (mol A/min) in - 0 (mol A/min) out x 100% = 100% 75 (mol A/min) in 2) Single-pass conversion of A: (red dashed borders) 100 (mol A/min) in - 25 (mol A/min) out x 100% = 75% 100 (mol A/min) in

45 4.7 (Balances on Reactive Processes) Example page : Propane is dehydrogenated to form propylene in a catalytic reactor: C 3 H 8 C 3 H 6 + H 2 The process is to be designed for a 95% overall conversion of propane. The reaction products are separated into two streams: the first, which contains H 2, C 3 H 6, and 0.555% of the propane that leaves the reactor, is taken off as product; the second stream, which contains the balance of the unreacted propane and 5% of the propylene in the first stream, is recycled to the reactor. Calculate the composition of the product, the ratio (moles recycled)/(mole fresh feed), and the single-pass conversion.

46 4.7 (Balances on Reactive Processes) Example page (continued): 1) Draw flow-chart and put basis for fresh feed: Fresh feed 100 mol C 3 H 8 n 1 (mol C 3 H 8 ) n 2 (mol C 3 H 6 ) Reactor n 3 (mol C 3 H 8 ) n 4 (mol C 3 H 6 ) n 5 (mol H 2 ) Separator Product n 6 (molc 3 H 8 ) (0.555% of n 3 ) n 7 (mol C 3 H 5 ) n 8 (mol H 8 ) Recycle n 3 (mol C 3 H 8 ) n 4 (mol C 3 H 6 ) n 5 (mol H 2 ) n 9 (mol C 3 H 8 ) n 10 {mol C 3 H 6 ) (5% of n 7 )

47 4.7 (Balances on Reactive Processes) Example page (continued): 2) Degree of freedom: Overall system : ( green dashed border) +3 unknown variables (n 6, n 7, n 8 ) -2independent atomic balances (C and H) -1 additional relation (95% overall propane conversion) = 0 degrees of freedom Recycle-fresh feed mixing point : (blue dashed border) + 4 unknown variables (n 9, n 10, n 1, n 2 ) - 2 balances (C3H 8, C 3 H 6 ) = 2 degrees of freedom Reactor: (red dashed border) +5 unknown variables (n 1 through n 5 ) -2 atomic balances (C and H) = 3 degrees of freedom Can t start with these the degree of freedom doesn t equal to zero.

48 4.7 (Balances on Reactive Processes) Example page (continued): Separator: (purple dashed line) +5 unknown variables (n 3, n 4, n 5, n 9, n 10 ) - 3 balances (C3H8, C3H6, H2 ) - 2 additional relations (n 6 = n 3, n 10 = 0.05n 7 ) = 0 degrees of freedom We can start doing balance with the separator or the overall system then we can go back to the recycle-fresh feed mixing point and reactor to find the rest of the unkowns. 3) Solve the extra information given : (95% Overall Propane Conversion) 0.95= (100 mol C 3 H 8 ) in - (n 6 molc 3 H 8 ) out (100 mol C 3 H 8 ) in n 6 = 5 mol C 3 H 8

49 4.7 (Balances on Reactive Processes) Example page (continued): 4) Atomic balance : Overall C Balance (100 mol C 3 H 8 )(3mol C/mol C 3 H 8 ) = (n 6 (mol C 3 H 8 )](3 mol C/mol C 3 H 8 ))+(n 7 ( moi C 3 H 6 ))(3 mol C/mol C 3 H 6 ) sub in n 6 = 5 mol and we get : n 7 = 95 mol C 3 H 6 Overall H Balance (100)(8) = n 6 (8) + n 7 (6) + n 8 (2) sub in n 6 = 5 mol n 7 = 95 mol and we get: n 8 = 95 mol H 2 Given Relations Among Separator Variables: n 6 = n 3 sub in n 6 = 5 mol and we get : n 3 = 900 mol C 3 H 8

50 4.7 (Balances on Reactive Processes) Example page (continued): n 10 = n 7 sub in n 7 = 95 mol and we get : n 10 = 4.75 mol C 3 H 6 Propane Balance About Separation Unit: n 3 = n 6 + n 8 sub in n 3 = 900 mol, n 6 = 5 mol and we get : n 8 = 895 mol C 3 H 8 Propane Balance About Mixing Point: 100 mol + n 9 =n 1 After finding all the variables find the desired quantities: Recycle ratio = (n 9 + n 10 ) mol recycle 100 mol fresh feed sub in n 9 = 895 mol, n 10 = 4.75 mol and we get : 9.00 mol recycle/mol fresh feed

51 4.7 (Balances on Reactive Processes) Example page (continued): Single-pass conversion = n 1 -n 3 x100 n 1 sub in n 1 = 995 mol, n 3 = 900 mol and we get : 9.6%

52 4.7 (Balances on Reactive Processes) Purging: Sometimes we can face some problems with the recycling process, such as a material that enters with the fresh feed or is produced in a reaction remains entirely in a recycle stream, rather than being carried out in a process product. If nothing was done to prevent this situation the substance would continuously enter the process and would have no way of leaving; then making the attainment of steady state impossible. To prevent this buildup, a portion of the recycle stream must be withdrawn as a purge stream to rid the process of the substance in question. Note: When labeling the flowchart the purge stream and the recycle stream before and after the purge takeoff all have the same composition.

53 4.7 (Balances on Reactive Processes) Example page : Methanol is produced in the reaction of carbon dioxide and hydrogen: CO 2 + 3H 2 CH 3 OH + H 2 O The fresh feed to the process contains hydrogen, carbon dioxide, and mole% inerts (I). The reactor effluent passes to a condenser that removes essentially all of the methanol and water formed and none of the reactants or inerts. The latter substances are recycled to the reactor. To avoid buildup of the inerts in the system, a purge stream is withdrawn from the recycle. The feed to the reactor (not the fresh feed to the process) contains 28.0 mole% CO 2, 70.0 mole% H 2, and 2.00 mole% inerts. The single-pass conversion of hydrogen is 60.0%. Calculate the molar flow rates and molar compositions of the fresh feed, the total feed to the reactor, the recycle stream, and the purge stream for a methanol production rate of 155 kmol CH 3 OH/h.

54 4.7 (Balances on Reactive Processes) Example page (continued): Draw and label flow chart and add basis: n r (mol) x 5c (mol CO 2 /mol) x 5H (mol H 2 /mol) (1 -x 5c -x 5H ) (mol l/mol) n p (mol) x 5c (mol CO 2 /mol) x 5H (mol H 2 mol) (1 -x 5c -x 5H ) (mol l/mol) n 5 (mol) x 5c (mol CO 2 /mol) x 5H (mol H 2 /mol) (1 -x 5c -x 5H ) (mol l/mol) n 0 {mol) x oc (mol CO 2 /mol) ( x oc ) (mol H 2/ mol) mol l/mol 100 mol mol CO 2 mol mol H 2 mol mol l/mol Reactor n 1 (mol CO 2 ) n 2 (mol H 2 ) 2.0 mol I n 3 (mol CH 3 OH) n 4 (mol H 2 O) Condenser n 3 (mol CH 3 OH) n 4 (molh 2 O)

55 4.7 (Balances on Reactive Processes) Example page (continued): Note: That inert is something that doesn t react so the moles in will equal to the moles out. Degree of Freedom: overall system- 7 unknowns (n o,x oc,n 3,n 4,n p,x 5C,x 5H ) + 1 reaction - 5 independent balances (CO 2,H 2,,I,CH 3 OH,H 2 O) = 3 degrees of freedom Recycle-fresh feed mixing point- 5 unknowns (n o,x oc,n r,x 5C,x 5H ) - 3 independent balances (CO 2, H 2, I) = 2 degrees of freedom

56 4.7 (Balances on Reactive Processes) Example page (continued): Reactor- 4 unknowns (n 1,n 2,n 3,n 4 ) + 1 reaction - 4 independent balances (CO 2, H 2, CH 3 OH, H 2 O) - 1 single-pass conversion = 0 degrees of freedom Condenser- 3 unknowns (n 5,x 5c,x 5H ) - 3 independent balances (CO 2, H 2, I) = 0 degrees of freedom Purge-recycle splitting point- 2 unknowns (n r,n p ) - 1 independent balance = 1 degree of freedom

57 4.7 (Balances on Reactive Processes) Example page (continued): Recycle-fresh feed mixing point- 3 unknowns (n o,x oc,n r ) - 3 independent balances = 0 degrees of freedom Purge-recycle splitting point- 1 unknown (n p ) - 1 independent balance = 0 degrees of freedom Extra information: 60% Single-Pass H 2 Conversion: f conv = (n in -n out )/n in 0.6= (70-n out )/70 n out =28 mol=n 2

58 4.7 (Balances on Reactive Processes) Example page (continued): Reactor balance: H 2- Balance: n 2 =100x ξ 28=70-3ξ ξ=14 CO 2 -balance: n 1 =100x0.28-ξ n 1 =28-14 n 1 =14 CH 3 OH-balance: n 3 =14 H 2 O-balance: n 4 = 14

59 4.7 (Balances on Reactive Processes) Example page (continued): Condenser Balance: In=out (no reaction) Total-balance: n 1 +n 2 +2+n 3 +n 4 =n p +n 5 +n 3 +n =n n 5 = 44 CO 2 -balance n 1 = n 5 (x 5c ) 14=44(x 5c ) x 5c =0.32 H 2 -balance n 2 =n 5 (x 5H ) x i = 1 - x sc - x 5h = mol I/mol 28=44(x 5H ) x 5H =0.64

60 4.7 (Balances on Reactive Processes) Example page (continued): Feed-Recycle Mixing Point: In=out (no reaction) Total Mole Balance: n 0 + n r = 100 mol I Balancen 0 ( ) + n r ( ) = 2.0 mol CO 2 Balance- 2 equations with 2 unkowns so by solver we get that n o = 61.4 mol nr = 38.6 mol n o x oc + n T x 5c = 28.0 mol CO 2 x oc = mol CO 2 /mol Therefore: x oh = (1 -x 5c -x 5H ) = mol H 2 /mol Recycle-Purge Splitting Point- n 5 = n T + n p n p = 5.4 mol purge

61 4.7 (Balances on Reactive Processes) Example page (continued): To scale the process to a methanol production rate of 155 kmol CH 3 OH/h, we multiply each total and component molar flow rate by the factor: (155 kmol CH 3 OH/h) / (14 mol CH 3 OH) = 11.1 (kmol/h)/mol

62 4.7 (Balances on Reactive Processes) Variable Basis Value Scaled Value Fresh Feed 61.4 mol 681 kmol/h 25.6 mole% CO mole% CO mole% H mole% H mole% I mole% I Feed to reactor 100 mol 1110 kmol/h 28.0 mole% CO mole% CO mole% H mole% H mole% I 2.0 mole% I

63 4.7 (Balances on Reactive Processes) Variable Basis Value Scaled Value Recycle 38.6 mol 428 kmol/h 31.8 mole% CO mole% CO mole%h mole% H mole% I 4.6 mole% I Purge 5.4 mol 59.9 kmol/h 31.8 mole% CO mole% CO mole% H mole% H mole% I 4.6 mole% I

64 4.8 Combustion Reaction Combustion: Combustion is the rapid reaction of a fuel with oxygen. (More important than any other class of industrial chemical reactions, despite the fact that combustion products (CO 2,H 2 O, and possibly CO and SO 2 ) are worth much less than the fuels burned to obtain them.) Combustion Chemistry: When a fuel is burned, carbon in the fuel reacts to form either CO 2 or CO, hydrogen forms H 2 O, and sulfur forms SO 2. A combustion reaction in which CO is formed from a hydrocarbon is referred to as partial combustion or incomplete combustion of the hydrocarbon. (If there is CO in the product stream then incomplete; if complete there is CO 2 in the product stream only.) Example: C 3 H 8 + 5O 2 3 CO H 2 O Complete combustion of propane C 3 H 8 + (7\2)O 2 3 CO + 4 H2O Partial combustion of propane

65 4.8 Combustion Reaction Note: Air is the source of oxygen in most combustion reactors. In most combustion calculations, it is acceptable to simplify this composition to 79% N 2, 21% O 2. Composition on a wet basis: Commonly used to denote the component mole fractions of a gas that contains water. Composition on dry basis: Signifies the component mole fractions of the same gas without the water. Stack gas or Flue gas: The product gas that leaves a combustion furnace. Note: When the flow rate of a gas in a stack is measured, it is the total flow rate of the gas including water. Therefore you should convert a composition on a dry basis to its corresponding composition on a wet basis before writing material balances on the combustion reactor.

66 4.8 Combustion Reaction To convert from wet basis to dry basis: 1) Do table 2) Put basis and get composition for each element (if composition not given) 3) Add composition of all elements except H 2 O 4) Divide each composition by the total composition found from part (3) (composition will get greater)

67 4.8 Combustion Reaction Example: A stack gas contains 60.0 mole% N 2, 15.0 % CO 2, 10.0% O 2, and the balance H 2 O. Calculate the molar composition of the gas on a dry basis. Component Wet Basis Dry Basis Mole % Dry Basis N /85 = O /85 = CO /85 = H 2 O Add compositions except H 2 O: (60 mol N mol CO mol O 2 ) = 85 mol dry

68 4.8 Combustion Reaction To convert from dry basis to wet basis: 1) Do table 2) Put basis and get composition for all elements (if composition not given) 3) Have to give information for H 2 O ( mostly composition) 4) Multiply the information given about H 2 O with basis to get H 2 O mole 5) Subtract the the moles of H 2 O (from part (4)) by basis and it will give us a number lets say (# 1 ). 6) Divide the number ( # 1 ) by the basis to get it in composition. 7) To get the remaining moles we need to take the number (# 1 ) in composition and multiply it by each elements old composition to get the new composition.

69 4.8 Combustion Reaction An Orsat analysis (a technique for stack analysis) yields the following dry basis composition: N 2 65% CO 2 14% CO 11% O 2 10% A humidity measurement shows that the mole fraction of H 2 O in the stack gas is Calculate the stack gas composition on a wet basis. 1) Basis : 100 mole 2) 100 x =7 3) = 93 which equal to (0.93)

70 4.8 Combustion Reaction Component Dry Basis Wet Basis Mole % wet basis N x 0.93 = % CO x 0.93 = % CO x 0.93 = % O x 0.93 = % H 2 O %

71 4.8 Combustion Reaction Theoretical Oxygen: The moles (batch) or molar flow rate (continuous) of O 2 needed for complete combustion of all the fuel fed to the reactor, assuming that all carbon in the fuel is oxidized to CO 2 and all the hydrogen is oxidized to H 2 O. Theoretical Air: The quantity of air that contains the theoretical oxygen. Excess Air: The amount by which the air fed to the reactor exceeds the theoretical air. Percent Excess Air: (moles air) fed (moles air) theoretical (moles air) theoretical x 100%

72 4.8 Combustion Reaction If you know the fuel feed rate and the stoichiometric equation(s) for complete combustion of the fuel, you can calculate the theoretical 02 and air feed rates. If in addition you know the actual feed rate of air, you can calculate the percent excess air. It is also easy to calculate the air feed rate from the theoretical air and a given value of the percentage excess. Example page : One hundred mol/h of butane (C 4 H 10 ) and 5000 mol/h of air are fed into a combustion reactor. Calculate the percent excess air. 1) First find the stoichiometric equation for complete combustion of butane: 2) Theoretical O 2 :. (n o2 ) theoretical = 1 C 4 H 10 (13/2) O C 4 H 10 X C 4 H 10 + (13/2) O 2 4CO H 2 O After doing cross multiplying we will find out that X = 650 mol O 2 /h.

73 3) Theoretical air: 4.8 Combustion Reaction Example page (continued) :. (n air ) theoretical = sum of (n o2 ) theoretical (n air ) theoretical = 650 mol O 2 / 0.21 (mol O 2 /mol air) = 3094 mol air 4) Percent Excess Air: %excess air = (( ) / 3094) x 100 = 61.6 %

74 4.8 Combustion Reaction Material Balances on Combustion Reactors: The procedure for writing and solving material balances for a combustion reactor is the same as that for any other reactive system. But put in your mind that: 1) When you draw and label the flowchart, be sure the outlet stream (the stack gas) includes: a. Unreacted fuel unless you are told that all the fuel is consumed. b. Unreacted oxygen. c. Water and carbon dioxide, as well as carbon monoxide if the problem statement says any is present. d. Nitrogen if the fuel is burned with air and not pure oxygen. 2) To calculate the oxygen feed rate from a specified percent excess oxygen or percent excess air (both percentages have the same value, so it doesn't matter which one is stated).

75 4.8 Combustion Reaction 3) If only one reaction is involved, all two balance methods (atomic species balances, extent of reaction) are equally convenient. If several reactions occur such as combustion of a fuel to form both CO and CO 2 atomic species balances are usually most convenient. Example page : Ethane is burned with 50% excess air. The percentage conversion of the ethane is 90%; of the ethane burned. 25% reacts to form CO and the balance reacts to form CO 2. Calculate the molar composition of the stack gas on a dry basis and the mole ratio of water to dry stack gas. 1) Draw and label flow chart and add basis: 100 mol C 2 H 6 50% excess air n 0 (mol) 0.21 mol O 2 /mol 0.79 mol N 2 /mol n 1 (molc 2 H 6 ) n 2 (mol O 2 ) n 3 (mol N 2 ) n 4 (mol CO) n 5 (mol CO 2 ) n 6 (mol H 2 O)

76 4.8 Combustion Reaction Example page (continued): 2) Write down the stoichiometric equation: C 2 H 6 + (7/2) O 2 2CO 2 + 3H 2 O C 2 H 6 + (5/2) O 2 2CO + 3H 2 O Note: o Since no product stream mole fractions are known, then calculations are easier if we use individual component amounts rather than total amount and mole fractions labeled. o The composition of air is taken to be approximately 21 mole% O 2, 79 mo!e % N 2. o Since excess air is supplied, O 2 must appear in the product stream.

77 4.8 Combustion Reaction Example page (continued): 3) Degree of Freedom: 7 unknowns (n 0, n 1...,n 6 ) - 3 atomic balances (C, H, O) - 1 N 2 balance - 1 excess air specification (relates n 0 to the quantity of fuel fed) - 1 ethane conversion specification - 1 CO/CO 2 ratio specification = 0 degree of freedom 4) 50% Excess Air: 1 C 2 H 6 (7/2) O C 2 H 6 x Theoretical O 2 x= 350 mol O 2 Therefore: ( 350 mol O 2 / 0.21 (mol O 2 /mol air) ) = Theoretical air

78 4.8 Combustion Reaction Example page (continued): 0.50 = ( n ) / n 0 = 2500 mol air fed 5) 90% Ethane conversion: f conv = (n fed -n 1 )/n fed 0.9= ( 100-n 1 )/100 n 1 = 10 mol C 2 H 6 6) 25% conversion to CO: 1 C 2 H 6 2 CO (100-10) n 4 n 4 = 45 CO 7) Balances: Nitrogen balance ( not reacted) (input = output) o.79(2500)=n 3 n 3 = 1975 mol

79 4.8 Combustion Reaction Example page (continued): Atomic Carbon balance: 100(2)=n 1 (2) + n 4 (1)+ n 5 (1) n 5 = 135 Atomic Hydrogen balance: 100(6)=10(6)+n 6 (2) n 6 = 270 Atomic oxygen balance: 525(2)=n 2 (2)+(45)(1)+(135)(2)+(270)(1) n 2 = 232

80 4.8 Combustion Reaction Example page (continued): Stack gas composition on a dry basis: a) Add the moles without including H 2 O mole: n 1 = 10molC 2 H 6 n 2 = 232 mol O 2 n 3 = 1974 mol N 2 n 4 = 45 mol CO n 5 = 135 mol CO b) Divide each compositions by 2396: y 1 = 10/2396= y 2 =232/2396= y 3 =1974/2396=0.824 y 4 =45/2396=0.019 y 5 =135/2396=

81 4.8 Combustion Reaction Example page : A hydrocarbon gas is burned with air. The dry-basis product gas composition is 1.5 mole% CO, 6.0% CO 2, 8.2% O 2, and 84.3% N 2. There is no atomic oxygen in the fuel. Calculate the ratio of hydrogen to carbon in the fuel gas and speculate on what the fuel might be. Then calculate the percent excess air fed to the reactor. 1) Draw and label flow chart and add basis: n c (mol C) n H (mol H) n a( mol air) 0.21 mol O 2 /mol 0.79 mol N 2 /mol 100 mol dry gas mol CO/mol dry gas mol CO 2 /mol dry gas mol O 2 /mol dry gas mol N 2 /mol dry gas n w (mol H 2 O) Note: Since the fuel is a hydrocarbon, water must be one of the combustion products.

82 4.8 Combustion Reaction Example page : C + O 2 CO 2 2C + O 2 2CO 4H + O 2 2H 2 O 2) Degree of Freedom: 4 unknowns (n H, n c, n a, n w ) -3 independent atomic balances (C, H, O) -1 N 2 balance = 0 degree of freedom 3) Balances: N 2 balance- 0.79n a = 100(0.843) n a = mol Atomic C balance: n c =100(0.015)(1) + (100)(0.060)(1) n c = 7.5

83 4.8 Combustion Reaction Example page : Atomic O balance: 0.21(n a )(2)=n w (1)+100(0.015)(1)+100(0.060)(2)+100(0.082)(2) n w =14.9 Atomic H balance: n H = n w (2) n H = 29.8 C/H Ratio in the fuel: n H /n C = (29.8 mol H)/(7.5 mol C) = 3.97 mol H / mol C Percent Excess Air: (n o2 ) theoretical : 1 C 1 O 2 4H 1 O X 29.8 X X= 7.5 X= 7.45 (n o2 ) theoretical = = 14.95

84 4.8 Combustion Reaction (n air ) theoretical = ( 14.95/0.21) =71.2 (n air ) feed =106.7 %excess = (( )/71.2) x 100 % =49.8%