Process Classification Fundamentals of Material Balances Process Classification

Balances on Reactive Systems Material balance no longer takes the form INPUT = OUTPUT Must account for the disappearance of reactants and appearance of products through stoichiometry. العناصر المتفاعلة

Stoichiometric Equations The stoichiometric equation of a chemical reaction is a statement of the relative amounts of reactants and products that participate in the reaction. 2 SO2 + O2 → 2 SO3 A stoichiometric equation is valid only if the number of atoms of each atomic species is balanced. 2 S → 2 S 4 O + 2 O → 6 O

Stoichiometric Equations The stoichiometric equation of a chemical reaction is a statement of the relative amounts of reactants and products that participate in the reaction. 2 SO2 + O2 → 2 SO3 A stoichiometric ratio of two molecular species participating in a reaction is the ratio of their stoichiometric coefficients: 2 mol SO3 generated /1 mol O2 consumed 2 mol SO3 generated /2 mol SO2 consumed

Stoichiometric Equations C4H8 + 6 O2 → 4 CO2 + 4 H2O Is this stoichiometric equation balanced? What is the stoichiometric coefficient of CO2? What is the stoichiometric ratio of H2O to O2? How many lb-mol O2 react to form 400 lb-mol CO2? 100 lbmol/min C4H8 is fed and 50% reacts. At what rate is water formed?

Limiting and Excess Reactants Two reactants are said to be in stoichiometric proportion if the ratio (moles A present/moles B present) equals the stoichiometric ratio from the balanced reaction equation. 2 SO2 + O2 → 2 SO3 The feed ratio that would represent stoichiometric proportion is nSO2/nO2 = 2:1 If reactants are fed in stoichiometric proportion, and the reaction proceeds to completion, all reactants are consumed. الفائض

Limiting and Excess Reactants Stoichiometric Proportion – Reactants are present in a ratio equivalent to the ratio of the stoichiometric coefficients. A + 2B → 2C

Limiting and Excess Reactants Limiting reactant – A reactant is limiting if it is present in less than stoichiometric proportion relative to every other reactant. A + 2B → 2C Excess reactant – All other reactants besides the limiting reactant.

Limiting and Excess Reactants fractional excess (fXS) – ratio of the excess to the stoichiometric proportion. A + 2B → 2C الكسر او النسبة الفائضة

Limiting and Excess Reactants Fractional conversion (f) – ratio of the amount of a reactant reacted, to the amount fed. A + 2B → 2C

Limiting and Excess Reactants Fractional conversion (f) – ratio of the amount of a reactant reacted, to the amount fed. A + 2B → 2C

Limiting and Excess Reactants Fractional conversion (f) – ratio of the amount of a reactant reacted, to the amount fed. A + 2B → 2C

Limiting and Excess Reactants Fractional conversion (f) – ratio of the amount of a reactant reacted, to the amount fed. A + 2B → 2C

Limiting and Excess Reactants Fractional conversion (f) – ratio of the amount of a reactant reacted, to the amount fed. A + 2B → 2C

Extent of Reaction A + 2B → 2C extent of reaction (ξ) – an extensive quantity describing the progress of a chemical reaction . ν = stoichiometric coefficients: νA = –1, νB = –2, νC = 2 A + 2B → 2C مدى التفاعل

Extent of Reaction A + 2B → 2C extent of reaction (ξ) – an extensive quantity describing the progress of a chemical reaction . ν = stoichiometric coefficients: νA = –1, νB = –2, νC = 2 A + 2B → 2C مدى التفاعل

Extent of Reaction A + 2B → 2C extent of reaction (ξ) – an extensive quantity describing the progress of a chemical reaction . ν = stoichiometric coefficients: νA = –1, νB = –2, νC = 2 A + 2B → 2C

Extent of Reaction A + 2B → 2C extent of reaction (ξ) – an extensive quantity describing the progress of a chemical reaction . ν = stoichiometric coefficients: νA = –1, νB = –2, νC = 2 A + 2B → 2C

Extent of Reaction A + 2B → 2C extent of reaction (ξ) – an extensive quantity describing the progress of a chemical reaction . ν – stoichiometric coefficients: νA = -1, νB = -2, νC = 2 A + 2B → 2C

ethylene Example Assume an equimolar reactant feed of 100 kmol: What is the limiting reactant? A reactant is limiting if it is present in less than stoichiometric proportion relative to every other reactant. ethylene

Example Assume an equimolar reactant feed of 100 kmol: What is the percentage excess of each reactant?

Example Assume an equimolar reactant feed of 100 kmol: If the reaction proceeds to completion: (a) How much of the excess reactant will be left? (b) How much C2H4O will be formed? (c) What is the extent of reaction?

Assume an equimolar reactant feed of 100 kmol: If the reaction proceeds to a point where the fractional conversion of the limiting reactant is 50%, how much of each reactant and product is present at the end? What is ξ?

Assume an equimolar reactant feed of 100 kmol: If the reaction proceeds to a point where 60 mol of O2 is left, what is the fractional conversion of C2H4? What is ξ?

Assume an equimolar reactant feed of 100 kmol: If the reaction proceeds to a point where 60 mol of O2 is left, what is the fractional conversion of C2H4? What is ξ?

Incomplete Reaction From this reaction, suppose that 0.6 kg of stibnite and 0.25 kg of iron are heated together to give 0.2 kg of Sb metal. Determine:

= 600/339.7 = 250/55.85 = 200/121.8

1 3 x 4.48 2 3 1.64 x

2 1 1.64 x

Reaction Stoichiometry Acrylonitrile produced by reaction of ammonia, propylene, and O2 at 30% conversion of limiting reactant: Determine limiting reactant limiting

Reaction Stoichiometry Acrylonitrile produced by reaction of ammonia, propylene, and O2 at 30% conversion of limiting reactant: Determine fractional excesses limiting

Reaction Stoichiometry Acrylonitrile produced by reaction of ammonia, propylene, and O2 at 30% conversion of limiting reactant: Determine the amount of propylene that leaves the reactor limiting use fractional conversion to determine amount of propylene that leaves the reactor.

Reaction Stoichiometry Acrylonitrile produced by reaction of ammonia, propylene, and O2 at 30% conversion of limiting reactant: limiting Determine extent of reaction by applying mole balance to propylene

Reaction Stoichiometry Acrylonitrile produced by reaction of ammonia, propylene, and O2 at 30% conversion of limiting reactant: limiting Apply mole balance to all remaining species

Example Fuel for motor vehicles other than gasoline are being eyed because they generate lower levels of pollutants than does gasoline. Compressed propane has been suggested as a source of economic power for vehicles. Suppose that in a test 22 kg of C3H8 is burned with 400 kg of air to produce 44 kg of CO2 and 12 kg of CO. a. What was the percent excess air? b. The components in the product stream.

C3H8 + 5O2 3CO2 + 4H2O C3H8 + 5O2 3CO2 + 4H2O Component Amount (kg) Molecular weight Input (kg-mol) Output C3H8 22 44 0.50 ? Air 400 29 13.80 O2 0.21*13.80=2.90 N2 0.79*13.80=10.9 CO2 - 1 CO 12 28 0.43 H2O 18 Total 29.87 C3H8 + 5O2 3CO2 + 4H2O 1 5 0.5 x(stoich) Percent excess air = (O2Fed – O2stoich)/(O2stoich) = (2.90 – 5*0.5)/(5*0.5) = 16%

C3H8 + 5O2 3CO2 + 4H2O C3H8 + 7/2O2 3CO + 4H2O [1] [2] y 3 C3H8 + 5O2 3CO2 + 4H2O [1] C3H8 + 7/2O2 3CO + 4H2O [2] Component Input (kg-mol) Consumed [1] Consumed [2] Generated [1] Generated [2] Total Output C3H8 0.50 0.33 0.43/3=0.14 - 0.03 O2 0.21*13.80=2.90 5*0.33= 1.65 3.5*0.14= 0.49 0.76 N2 0.79*13.80=10.9 10.9 CO2 1 CO 0.43 H2O 4*0.33= 1.32 4*0.14= 0.56 1.88 Total 29.87 15 =0.55-(0.33+0.14) =2.90-(1.65+0.49) C3H8 + 7/2O2 3CO + 4H2O [2] 1 z 3 0.43

Example A gas stream containing 80 mol% C2H6 and 20 mol% O2 is burned in an engine with 200% excess air. If 80% of the ethane goes to CO2, 10% goes to CO and the remaining unburned, determine the amount of the excess air per 100 moles of the feed gas and the components in the product stream. C2H6 + 7/2O2 2CO2 + 3H2O [1] C2H6 + 5/2O2 2CO + 3H2O [2]

A gas stream containing 80 mol% C2H6 and 20 mol% O2 is burned in an engine with 200% excess air. If 80% of the ethane goes to CO2, 10% goes to CO and the remaining unburned, determine the amount of the excess air per 100 moles of the feed gas. Basis=100 mole feed gas C2H6 + 7/2O2 2CO2 + 3H2O [1] Percent excess air (based on air): (O2 in – O2used )/(O2used) = 2 O2 theory (Eq 1) = 3.5*80 = 280 O2 used (air) = O2 theory (Eq 1) – 20 = 260 moles (O2 in (air) – 260)/(260) = 2 O2 in (air) = 260 + (2*260) = 780 moles N2 in = 3.76*780 = 2,933.1 Air stream = 4.76*780 = 3,712.8 Component Input Gas Stream Output (kg-mol) C2H6 80 Air Gair= 4.76*780 = 3,712.8 O2 20 + 0.21*Gair = 800 N2 0.79*Gair= 2,933.1 CO2 CO H2O. Total (200%)

C2H6 + 7/2O2 2CO2 + 3H2O [1] C2H6 + 5/2O2 2CO + 3H2O [2] A gas stream containing 80 mol% C2H6 and 20 mol% O2 is burned in an engine with 200% excess air. If 80% of the ethane goes to CO2, 10% goes to CO and the remaining unburned, determine the amount of the excess air per 100 moles of the feed gas. 2 1 1 2 8 Generated CO 64 Generated CO2 C2H6 + 7/2O2 2CO2 + 3H2O [1] C2H6 + 5/2O2 2CO + 3H2O [2] Component Input (mol) Consumed [1] Consumed [2] Generated [1] Generated [2] Total Output C2H6 80 64 8 - O2 800 224 20 556 N2 2933.1 CO2 128 CO 16 H2O 192 24 216 Total 3812.8 3,505.1 =80%C2H6 10% goes to CO =64×7/2 =8×5/2 =3×64=192 =3×8=24

Equilibrium Composition Given: a set of reactive species, and reaction conditions Determine: the final (equilibrium) composition of the reaction mixture. how long the system takes to reach a specified state short of equilibrium.

Equilibrium Composition Irreversible reaction reaction proceeds only in a single direction A → B concentration of the limiting reactant eventually approaches zero (time duration can vary widely). Equilibrium composition of an irreversible reaction is that which corresponds to complete conversion.

Equilibrium Composition Reversible reaction reaction proceeds in both directions A B net rate (forward – backward) eventually approaches zero (again, time can vary widely) Equilibrium composition of a reversible reaction is that which corresponds to the equilibrium conversion. ↔

Equilibrium Composition An equilibrium reaction proceeds to an extent at temperature T based on the equilibrium constant, K(T). where yi is the mole fraction of species i Example: Water-gas shift reaction: Assume 1 mole CO and 2 mole H2O K(1105 K) = 1.00

Equilibrium Composition Example: If the water-gas shift reaction, proceeds to equilibrium at a temperature T (K), the mole fractions of the four reactive species satisfy the relation where K(T) is the reaction equilibrium constant. At T = 1105 K, K (T)= 1.00. Suppose the feed to a reactor contains 1.00 mol of CO, 2.00 mol of H20, and no CO2 or H2 , and the reaction mixture comes to equilibrium at 1105 K. Calculate the equilibrium composition and the fractional conversion of the limiting reactant.

Equilibrium Composition Water-gas shift reaction: Assume 1 mole CO and 2 mole H2O K(1105 K) = 1.00

Equilibrium Composition Water-gas shift reaction: Assume 1 mole CO and 2 mole H2O K(1105 K) = 1.00

Equilibrium Composition Water-gas shift reaction: Assume 1 mole CO and 2 mole H2O K(1105 K) = 1.00

Water-gas shift reaction: Assume 1 mole CO and 2 mole H2O limiting reactant is CO limiting reactant is CO limiting reactant is CO limiting reactant is CO limiting reactant is CO limiting reactant is CO at equilibrium, fractional conversion at equilibrium: Water-gas shift reaction: Assume 1 mole CO and 2 mole H2O K(1105 K) = 1.00

Multiple Reactions For j reactions of i species, mole balance becomes:

Multiple Reactions for j reactions of i species, mole balance becomes:

Multiple Reactions 100 moles A fed to a batch reactor product composition: 10 mol A, 160 B, 10 C What is: fA? YB? SB/C? ξ1, ξ2 EXAMPLE: Consider the following pair of reactions. A → 2B (desired) A → C (undesired) 100 moles of A are fed to a batch reactor and the final product contains 10 mol of A, 160 mol of B and 10 mol of C. Calculate (1) fractional conversion for A fA (2) percentage yield of B, (3) the selectivity of B relative to C (4) the extents of the reactions.

Multiple Reactions EXAMPLE: Consider the following pair of reactions. A → 2B (desired) A → C (undesired) 100 moles of A are fed to a batch reactor and the final product contains 10 mol of A, 160 mol of B and 10 mol of C. Calculate: (1) fractional conversion for A (fA), (2) percentage yield of B, (3) the selectivity of B relative to C, (4) the extents of the reactions.

Multiple Reactions EXAMPLE: Consider the following pair of reactions. A → 2B (desired) A → C (undesired) 100 moles of A are fed to a batch reactor and the final product contains 10 mol of A, 160 mol of B and 10 mol of C. Calculate: (1) fractional conversion for A (fA), (2) percentage yield of B, (3) the selectivity of B relative to C, (4) the extents of the reactions. YB : moles of desired product (B) formed = 160 moles of desired product formed if there were no side reactions and the limiting reactant:

Multiple Reactions SB/C: EXAMPLE: Consider the following pair of reactions. A → 2B (desired) A → C (undesired) 100 moles of A are fed to a batch reactor and the final product contains 10 mol of A, 160 mol of B and 10 mol of C. Calculate: (1) fractional conversion for A fA (2) percentage yield of B, (3) the selectivity of B relative to C (4) the extents of the reactions. SB/C: moles of desired product (B) formed = 160 moles of undesired product (C) formed = 10

Multiple Reactions EXAMPLE: Consider the following pair of reactions. A → 2B (desired) A → C (undesired) 100 moles of A are fed to a batch reactor and the final product contains 10 mol of A, 160 mol of B and 10 mol of C. Calculate: (1) fractional conversion for A fA (2) percentage yield of B, (3) the selectivity of B relative to C (4) the extents of the reactions. Extent of Reactions (ξ1, ξ2):

Balances on Reactive Processes Continuous, steady-state dehydrogenation of ethane Total mass balance still has INPUT = OUTPUT form Molecular balances contain consumption/generation Atomic balances (H and C) also have simple form

Balances on Reactive Processes Continuous, steady-state dehydrogenation of ethane First consider molecular balances: Molecular H2 balance: generation = output generation H2 = 40 kmol H2/min C2H6 balance: input = output + consumption 100 kmol C2H6/min = n1 (kmol C2H6/min )+ (C2H6 consumed) C2H4 balance: generation = output generated C2H4 = n2

Balances on Reactive Processes Continuous, steady-state dehydrogenation of ethane Atomic C balance: input = output Atomic H balance: input = output Atomic balances

Independent Equations To understand the number of independent species balances in a reacting system requires an understanding of independent algebraic equations. Algebraic equations are independent if you cannot obtain any of them by adding/subtracting multiples of the others. 3×[1] = [2] 2×[3] – [4] = [5]

Independent Equations To understand the number of independent species balances in a reacting system requires an understanding of independent algebraic equations. Algebraic equations are independent if you cannot obtain any of them by adding/subtracting multiples of the others.

Independent Species If two molecular or atomic species are in the same ratio to each other where ever they appear in a process and this ratio is incorporated in the flowchart labeling, balances on those species will not be independent equations.

Independent Chemical Reactions When using molecular species balances or extents of reaction to analyze a reactive system, the degree of freedom analysis must account for the number of independent chemical reactions among the species entering and leaving the system.

Independent Chemical Reactions Chemical reactions are independent if the stoichiometric equation of any one of them cannot be obtained by adding and subtracting multiples of the stoichiometric equations of the others. 2×[2] + [1] = [3]

Solving Reactive Systems There are 3 possible methods for solving balances around a reactive system: Molecular species balances require more complex calculations than the other methods and should be used only for simple (single reaction) systems. 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.

Molecular Species Balances To use molecular species balances to analyze a reactive system, the balances must contain generation and/or consumption terms. The degree-of-freedom analysis is as follows: # unknown labeled variables + # independent chemical reactions - # independent molecular species balances - # other equations relating unknown variables # of degrees of freedom

Molecular Species Balances Example: 2 unknown labeled variables + 1 independent chemical reactions - 3 independent molecular species balances - 0 other equations relating unknown variables 0 degrees of freedom at steady state

Molecular Species Balances At steady state H2 Balance: generation = output C2H6 Balance: input = output + consumption 1 1 Consumn 40 kmol C2H4 Balance: generation = output 1 1 Output 40 kmol

Atomic Species Balances All atomic species balances take the form: INPUT = OUTPUT Degree-of-freedom analysis, ndf = # unknown labeled variables - # independent atomic species balances - # molecular balances on independent nonreactive species - # other equations relating unknown variables

Atomic Species Balances Example: All atomic species balances take the form INPUT = OUTPUT Degree-of-freedom analysis, ndf = 0 = 2 unknown labeled variables - 2 independent atomic species balances - 0 molecular balances on independent nonreactive species - 0 other equations relating unknown variables

Atomic Species Balances C Balance: input = output H Balance: input = output

Atomic Species Balances Solve simultaneously

Extent of Reaction The 3rd method by which to determine molar flows in a reactive system is using expressions for each species flow rate in terms of extents of reaction (ξ). Degree-of-freedom analysis for such an approach: ndf = # of unknown labeled variables + # independent reactions - # independent nonreactive species - # other relationships or specifications

Incomplete Combustion of CH4 Example Methane is burned with air in a continuous steady-state reactor to yield a mixture of carbon monoxide, carbon dioxide, and water. The feed to the reactor contains 7.8 mol% CH4, 19.4 mol% O2, 72.8 mol% N2. Methane undergoes 90% conversion, and the effluent gas contains 8 mol CO2 per mole CO.

fCH4 = 0.9 N2 balance: nonreactive species, INPUT = OUTPUT CH4 conversion specification: O2 Fed:

Extent of reaction mole balances: The feed to the reactor contains 7.80 mol% CH4, 19.4 mol% O2, 72.8 mol% N2. Methane undergoes 90.0% conversion, and the effluent gas contains 8 mol CO2 per mole CO.

Product Separation and Recycle Two definitions of reactant conversion are used in the analysis of chemical reactors with product separation and recycle of unconsumed reactants.

Catalytic Propane Dehydrogenation Example: 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 H2 gas, propylene (C3H6), and 0.555% of the propane that leaves the reactor, is taken off as product; the second stream, which contains the balance of the un-reacted propane, 5% of the propylene in the product stream, and no H2, is recycled to the reactor. Calculate the flow rates and compositions of all streams and the single-pass conversion of propane in the reactor. 95% overall conversion

Catalytic Propane Dehydrogenation 95% overall conversion Overall Process ndf = 3 unknowns (n6, n7, n8) – 2 independent atomic balances (C and H) – 1 relation (overall conversion) = 0 Consider n6, n7, n8 known for further DOF analyses

Catalytic Propane Dehydrogenation 95% overall conversion ndf = 2 Mixing point ndf = 4 unknowns (n9, n10, n1, n2) – 2 balances (C3H8 and C3H6) = 2

Reactor ndf = 5 unknowns (n3, n4, n5, n1, n2) – 2 balances (C and H) 95% overall conversion ndf = 3 ndf = 2 Reactor ndf = 5 unknowns (n3, n4, n5, n1, n2) – 2 balances (C and H) = 3

Separator ndf = 5 unknowns (n3, n4, n5, n9, n10) 95% overall conversion ndf = 3 ndf = 0 ndf = 2 Separator ndf = 5 unknowns (n3, n4, n5, n9, n10) – 3 balances (C3H8, C3H6, and H2) – 2 relations (reactant and product recovery fractions) = 0

95% overall conversion Overall conversion relationship

95% overall conversion overall C atomic balance

95% overall conversion Overall H atomic balance

95% overall conversion Separator Given Relations

95% overall conversion Separator Propane balance

95% overall conversion mixer Propane balance

95% overall conversion mixer Propylene balance

95% overall conversion Reactor C atomic balance

95% overall conversion Reactor H atomic balance

95% overall conversion Single-pass conversion

95% overall conversion Recycle ratio

Purging Necessary with recycle to prevent accumulation of a species that is both present in the fresh feed and is recycled rather than separated with the product. mixed fresh feed and recycle is a convenient basis selection

Methanol Synthesis ndf = 7 unknowns (n0, x0C, np, x5C, x5H, n3, n4) + 1 rxn - 5 independent species balances = 3

Methanol Synthesis ndf = 4 unknowns (n1, n2, n3, n4) + 1 rxn – 4 independent species balances – 1 single pass conversion = 0 REACTOR

– 3 independent species balances = 0 ndf = 3 unknowns (n5, x5C, x5H) – 3 independent species balances = 0 CONDENSER

ndf = 3 unknowns (n0, x0C, nr) – 3 independent species balances = 0

– 1 independent species balance = 0 Class work… without support Class work… without support Class work… without support Class work… without support Class work… without support Class work… without support Class work… without support Class work… without support Class work… without support Class work… without support Class work… without support ndf = 1 unknowns (np) – 1 independent species balance = 0 investigate mole balances and their solution in the text

Combustion Reactions Combustion - rapid reaction of a fuel with oxygen. Valuable class of reactions due to the tremendous amount of heat liberated, subsequently used to produce steam used to drive turbines which generates most of the world’s electrical power. Common fuels used in power plants: Coal Fuel oil (high MW hydrocarbons) Gaseous fuel (natural gas) Liquified petroleum gas (propane and/or butane) هائل tremendous

Combustion Chemistry When a fuel is burned C forms CO2 (complete) or CO (partial combustion) H forms H2O S forms SO2 N forms NO2 (above 1800°C) Air is used as the source of oxygen. DRY air analysis: 78.03 mol% N2 20.99 mol% O2 0.94 mol% Ar 0.03 mol% CO2 0.01 mol% H2, He, Ne, Kr, Xe Usually safe to assume: 79 mol% N2 21 mol% O2

Combustion Chemistry Stack (flue) gas – product gas that leaves a furnace. Composition analysis: wet basis – water is included in mole fractions dry basis – does not include water in mole fractions Stack gas contains (mol) on a wet basis: 60.0% N2, 15.0% CO2, 10.0% O2, 15.0% H2O Dry basis analysis: 60/(60+15+10) = 0.706 mol N2/mol 15/(60+15+10) = 0.176 mol CO2/mol 10/(60+15+10) = 0.118 mol O2/mol

Combustion Chemistry Assume 100 mole dry gas basis 7.53 mole H2O Stack gas contains (mol) on a dry basis: 65% N2, 14% CO2, 10% O2, 11% CO Assume 100 mole dry gas basis 7.53 mole H2O 65 mole N2 14 mole CO2 10 mole O2 11 mole CO total = 107.5 mole

Theoretical and Excess Air The less expensive reactant is commonly fed in excess of stoichiometric ratio relative to the more valuable reactant, thereby increasing conversion of the more expensive reactant at the expense of increased use of excess reactant. In a combustion reaction, the less expensive reactant is oxygen, obtained from the air. Consequently, air is fed in excess to the fuel. Theoretical oxygen is the exact amount of O2 needed to completely combust the fuel to CO2 and H2O. Theoretical air is that amount of air that contains the amount of theoretical oxygen.

Theoretical and Excess Air Excess air is the amount by which the air fed to the reactor exceeds the theoretical air.

Theoretical and Excess Air A 100 mol/h of butane (C4H10) and 5000 mol/h of air are fed into a combustion reactor. Calculate the percent excess air. nC4H10 = 100 mol/hr; nair = 5000 mol/hr

Combustion Reactors Procedure for writing/solving material balances for a combustion reactor: When you draw and label the flowchart, be sure the outlet stream (the stack gas) includes: un-reacted fuel (unless the fuel is completely consumed). un-reacted oxygen. water and carbon dioxide (and CO if combustion is incomplete). nitrogen (if air is used as the oxygen source). Calculate the O2 feed rate from the specified percent excess oxygen or air. If multiple reactions, use atomic balances.

Combustion Reactors Example 100 mol of 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 CO2. Calculate the molar composition of the stack gas on a dry basis and the mole ratio of water to dry stack gas.

Combustion of Ethane Degree-Of-Freedom analysis: fC2H6 = 0.9 +7 unknowns - 3 atomic balances - 1 nitrogen balance - 1 excess air specification - 1 ethane conversion specification - 1 CO/CO2 ratio specification ndf = 0 25% of the ethane burned forms CO

Excess air specification fC2H6 = 0.9 = 1+ 0.5 = 1.5 25% of the ethane burned forms CO Ethane conversion specification

CO/CO2 ratio specification fC2H6 = 0.9 25% of the ethane burned forms CO Nitrogen balance C Atomic balance

fC2H6 = 0.9 atomic H balance

fC2H6 = 0.9 25% of the ethane burned forms CO atomic O balance

stack gas composition (dry basis) fC2H6 = 0.9 25% of the ethane burned forms CO stack gas composition (dry basis)

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