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Chapter 4 Fundamentals of Material Balance

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Page 1: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

Chapter 4

Fundamentals of Material Balance

Page 2: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

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

Page 3: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

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 SO2 + O2

This equation indicates (shows) that every two molecules of SO2 reacts with one molecule of O2 to produce two molecules of SO3.

• Stoichiometric coefficient: Is the number before (in-front of) each molecule in the stoichiometric equation. • 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.)

2 SO3

2 SO2 + O2 ----» 2 SO3 Example: The stoichiometric coefficient of SO2 is 2.

Page 4: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

4.6 ( Chemical Reaction Stoichiometry)

Example on balancing an equation: C4H8 + O2 ---> CO2 + H2O

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.

C4H8 + O2 ---> 4 CO2 + H2O 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. C4H8 + O2 ---> 4 CO2 + 4H2O

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. C4H8 + 6 O2 ---> 4 CO2 + 4H2O ✔

Page 5: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

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 SO2 + O2 ----» 2 SO3

We can write a stoichiometric ratios: also we can say 2 lb-moles SO2 consumed 2 lb-moles SO3 generated

2 mol SO3 generated

1 mol 02 consumed

We can use any unit for moles in the stoichiometric ratio but they have to be the same for each ratio alone.

Page 6: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

4.6 ( Chemical Reaction Stoichiometry)

Example on how we can use the stoichiometric ratio: 2 SO2 + O2 ----» 2 SO3

If you know, for example, that 1600 kg/h of SO3 is to be produced, you can calculate the amount of oxygen required as : 1. Find the stoichiometric ratio between SO3 and O2 : 1 kmol O2

2 kmol SO3

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 mole flow-rate of SO3= 1600 (kg/h) / 80 ( kg/kmol) mole flow-rate of SO3= 20 (Kmol/h) (generated)

Picked kmol to make calculations easier later on because the mass was given in kg in the question.

Page 7: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

4.6 ( Chemical Reaction Stoichiometry)

3. By using the conversion factor (Stoichiometric ratio) and moles of SO3 generated we can find the mole flow-rate of O2 consumed :

1 kmol O2 (consumed) 2 kmol SO3 (generated)

= 10 (kmol O2 / h) (consumed) 4. Convert the O2 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 O2 / h) = m (mass flow-rate) / 32 (kg/kmol) m (mass flow-rate) = 320 kg O2/h

x 20 (Kmol SO3 /h) (generated)

Page 8: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

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 SO2 + O2 ----» 2 SO3

For the reactants in the reaction to be present in stoichiometric proportion then there must be 2 moles of SO2 for every mole of O2 present in the feed of the reactor.

n(SO2)/n(O2) = 2/1

The ratio is 2:1 Note: If reactants are fed to a chemical reactor in stoichiometric proportion and the

reaction proceeds to completion, all of the reactants are consumed.

Stoichiometric ratio: The ratio of coefficients in the stoichiometric equation.

Page 9: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

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

Page 10: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant)

Example: 2 C2H4 + O2 ---» 2 C2H4O

If the feed to the reactor contains 100 kmol C2H4 and 100 kmol O2; which reactant is the limiting ? C2H4: Feed of Molecule 100Kmol Stoichiometric Coefficient of Molecule 2 Kmol O2: Feed of Molecule 100 kmol Stoichiometric Coefficient of Molecule 1 Kmol Therefore the limiting reactant is C2H4.

= =50

= = 100

Page 11: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant)

Stoichiometric requirement: Let say we have nA which is the amount number of moles of an excess reactant (A) that is present in the feed to a reactor. Then (nA)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) = (ni)feed - (ni)stoich

(ni)stoich The percentage excess of (i) is 100 times the fractional excess.

Page 12: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant)

Example on Fractional excess: The hydrogenation of acetylene to form ethane:

C2H2 + 2H2 ----» C2H6

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: C2H2: (feed of C2H2)/(stoichiometric of C2H2) = (20/1) =20 H2: (feed of H2)/(stoichiometric of H2) =(50/2) =25

Therefore the limiting is C2H2; and the excess is H2.

Page 13: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant)

② Fractional excess of H2: Fractional excess of (H2) = (nH2)feed - (nH2)stoich

(nH2)stoic

(nH2)feed = 500 kmol/h (given in question) Steps to find the (ni)stoich :

(nH2)stoich (From equation: (C2H2 + 2H2 ----» C2H6 ) see relation between H2 and C2H2)

L.R H2

(nH2)stoich = 1 C2H2 ----» 2 H2

Feed of L.R : 20 c2H2----» (nH2)stoich Fractional excess of (H2) = 500 kmol/h – 40 kmol/h = 0.25 40 kmol/h We say that there is 25% excess hydrogen in the feed.

Cross Multiply: 20 c2H2 x 2 H2 = 1 C2H2 x (nH2)stoich (nH2)stoich = 40 Kmol/h

Page 14: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant)

• Fractional conversion: fconv= moles reacted moles Fed To find the fraction unreacted is accordingly 1 – fconv . To find the percentage conversion it is 100 time the conversion. • Extent of reaction: (ni)product- (ni)feed

vi

The extent of reaction has the same units as n or mole flow-rate.

= (ni)feed - (ni)product

(ni)feed

ξ =

vi = The stoichiometric coefficient of the (ith) species in a chemical reaction; making it negative for reactants and positive for products.

If the reaction proceed to completion that means fconv= 1 (fconv= 100%)

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.

Page 15: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

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 C2H4 + O2 ----» 2 C2H4O The feed to a reactor contains 100 kmol C2H4 and 100 kmol O2 . ① Which reactant is limiting? C2H2: (feed of C2H4)/(stoichiometric of C2H4) = (100/2) =50 H2: (feed of O2)/(stoichiometric of O2) =(100/1) =100

Therefore the limiting is C2H4; and the excess is O2.

Page 16: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant)

② What is the percentage excess of the other reactant? O2 is excess ; so the percentage excess of O2 is: Fractional excess of (H2) = (nO2)feed - (nO2)stoich

(nO2)stoic

Fractional excess of (H2) = 100kmol - (nO2)stoich x 100%

(nO2)stoich Find (nO2)stoich: L.R O2 2 C2H4 ----» 1 O2 100 ----» (nO2)stoich

Fractional excess of (O2) = (100 kmol- 50 kmol) / (50 kmol) = 100%

By cross multiplying (nO2)stoich = 50 kmol.

Page 17: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

③ If the reaction proceeds to completion, how much of the excess reactant will be left; how much C2H4O 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.

O2= 50 Therefore 50 kmol of O2 will be left. b. 2 C2H4 ----» 2 C2H4O 100 ----» C2H4O C2H4O= 100 Kmol Therefore 100kmol of C2H4O will be formed.

4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant)

2 C2H4 ----» 1 O2 100 ----» O2

Page 18: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

c. (ni)product- (ni)feed

vi

Previously we found out how much C2H4O will be formed which is the product and we don’t have any C2H4O in the feed and we know its coefficient from the reaction formula which is 2 so that makes it easy to find the extent. 100 – 0 2 50kmol

ξ =

ξ =

ξ =

4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant)

Page 19: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

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: C2H4 + H2O ⇌C2H5OH 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 H2O decrease, the rate of the forward

reaction decreases. o As the C2H5OH 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.

4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant)

Page 20: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

Equilibrium constant: Lets say we have a chemical equilibrium equation A+B ⇌ C+D Keq= PD PC

PA PB

We know that partial pressure equals to yiPT (were yi is the mole fraction and PT is the total pressure were we can say that: Keq= (yDPt)(yCpT) (yBPT)(yAPT) We can also replace yi with ni/nT because we know that yi=ni/nT: Keq= (nD/nT)(nc/nT) (nB/nT)(nA/nT)

Where PA , PB , PC , PD are partial for each element.

We can cancel all the total pressure because it the same in all elements.

Therefore we can say that : Keq= nD nC nB nA

4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant)

Page 21: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

The terms yield and selectivity are used to describe the degree to which a desired reaction predominates(leads) over competing side reactions: Yield: moles of desired product formed Selectivity: moles of desired product formed moles of undesired product formed

4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant)

Moles that would have been formed if there were no side reactions and the limiting reactant had reacted completely.

= actual theoretical

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.

The yield is defined is always a fraction but it may also be expressed as a percentage by multiplying by 100%.

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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 vij is the stoichiometric coefficient of substance ( i ) in reaction ( j ) (negative for reactants, positive for products).

Formula : ni(out)=ni(feed) + ∑ vijξj

Example: • C2H4 +(1/2) O2 ->> C2H4O • C2H4 + 3 O2 ->> 2 CO2 + 2 H2O

Desired reaction

undesired reaction (side reaction)

ξ 1

ξ 2

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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 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.

The generation and consumption terms in the molecular balance equation is usually obtained from chemical stoichiometry.

Page 24: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

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.

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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. 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.

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.

Page 26: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

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.

n1 (mol O2 /s) 3.76 n1 (N2 /s)

n2(mol CCI4(l)/s)

n3 (mol O2/S) 3.76 n3 (mol N2/s) n4 (mol CC!4(v)/s)

n5 (mol CCI4 (l)/s)

.

.

.

.

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4.7 (Balances on Reactive Processes)

Example (continued): Independent Chemical Reactions

Molecular Species: Number of molecular species Number of Independent species Note : N2 and O2 have the same ratio. Atomic Species: Number of atomic species Number of Independent species 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.

3 (O2, N2, CCl4) 2 (O2 or N2, CCl4)

4 (O, N, C, Cl)

2 (O or N, Cl or C)

Page 28: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

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.

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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 N2 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.

Page 30: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

4.7 (Balances on Reactive Processes)

Example:

Degree of Freedom (Molecular species Balance): 2 unknown labeled variables (n1 ,n2) + 1 independent chemical reaction

- 3 independent molecular species balances (C2H6 , C2H4 , and H2 ) - 0 other equations relating unknown variables

= 0 degrees of freedom

100 kmol C2H6/min 40 kmol H2/min n1 (kmol C2H6/min! n2 (kmol C2H4/min)

.

.

. .

Page 31: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

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

= 0 degrees of freedom Use atomic balance to solve the question: • 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: 2 (Kmol C / Kmol C2H6) X 100 (Kmol C2H6/s) = 2(Kmol C / Kmol C2H6) x n1(kmol C2H6/s) + 2(Kmol C / Kmol C2H4) x n2(kmol C2H4/s)

# of atoms C in molecular formula x moles (input) = # of atoms C in molecular formula x moles (output)

100 = n1 + n2

. . . .

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4.7 (Balances on Reactive Processes)

Example (continued): 2. H-Balance: 6 (kmol H/kmol C2H6) x 100 (kmol C2H6 /min) = 2 (kmol H/kmol H2) x 40 (kmol H2/min) + 6 (kmol H/kmol C2H6) x n1 (kmol C2H6/min) + 4 (kmol H/C2H4) x n2(kmol C2H4/min) = 520= 6n1+ 4n2

with two equations and two unkowns we can find n1 and n2 : n1= 60 kmol C2H6/min n2=40 kmol C2H4/min

# of atoms H in molecular formula x moles (input) = # of atoms H in molecular formula x moles (output)

.

.

. .

. .

. .

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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: (ni)product= (ni)feed + vi ξ For multiple reaction:

ni(out)=ni(feed) + ∑ vijξj

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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.

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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

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4.7 (Balances on Reactive Processes)

Example 4.7-1 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:

CH4 + (3/2) O2 CO + 2H2O CH4 + 2O2 CO2 + 2H2O

The feed to the reactor contains 7.80 mole% CH4, 19.4% O2 , and 72.8% N2 . The percentage conversion of methane is 90.0%, and the gas leaving the reactor contains 8 mol CO2/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.

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4.7 (Balances on Reactive Processes) Example 4.7-1 page 131 (continued): 1) Draw the flow chart and Basis of 100 mol Feed:

a. Atomic species balance: Degree of Freedom: + 5 unknown variables - 3 independent atomic species balances (C, H, O) - 1 nonreactive molecular species balance (N2) - 1 specified methane conversion = 0 degrees of freedom.

100 mol 0.0780 mol CH4/mol

0.194 mol O2/mol 0.728 mol N2/mol

nch4(mol CH4) nco(mol CO) 8nco(mol CO2 ) nH2O(mol H2O) nO2(mol O2 ) nN2(mol N2 )

Note: N2 it goes with the O2 in air but not consumed during the reaction.

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4.7 (Balances on Reactive Processes) Example 4.7-1 page 131 (continued): Always start with the extra given information it will help you determine at least one

of the unknowns: 90% CH4 Conversion: fconv= (nch4(feed)- nch4(out)) / nch4(feed)

0.90=((100x0.0780)-nch4(out)) / (100x0.0780) nch4(out)=0.780 Balance ( input=output): We can do C , H , and O balance: C-balance: (1 x (100 x 0.0780)) = (1 x 0.780) + (1 x nco) +(1 x 8nco(mol CO2 ) ) nco = 0.780 mol CO nco2 = 8nCo = (8 X 0.780) mol CO2 = 6.24 mol CO2

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4.7 (Balances on Reactive Processes)

Example 4.7-1 page 131 (continued): H-Balance: ( 4 X (100X0.0780) ) = (0.78 x 4) + (2 x nH20 ) nH20= 14.0 mol H2O O-Balance: ( 2 x (100 x 0.194) ) = ( 2 x nO2) + ( 1 x 0.780) + ( 2 x 6.24) + (1 x 14) nO2= 5.75 mol O2

N2 – Balance : N2 balance is the regular material balance (input=output) because its not reactive. (100 x 0.728) = nN2 nN2 = 72.8 mol N2

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4.7 (Balances on Reactive Processes)

b. Extents of reaction balance: Degree of Freedom: + 5 unknown labeled variables + 2 independent reactions - 5 expressions for nI(ξ) (i= CH4, 02 , CO, CO: , H2O) - 1 nonreactive molecular species balance (N2 ) - 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% CH4 Conversion: fconv= (nch4(feed)- nch4(out)) / nch4(feed)

0.90=((100x0.0780)-nch4(out)) / (100x0.0780) nch4(out)=0.780 Balance: ni(out)= ni(feed) + ∑ vijξj

Example 4.7-1 page 131 (continued):

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4.7 (Balances on Reactive Processes)

Example 4.7-1 page 131 (continued): CH4 + (3/2) O2 CO + 2H2O CH4 + 2O2 CO2 + 2H2O CH4- balance: 0.780 = 100 x 0.0780 mol – ξ1 - ξ 2 7.02 mol = ξ1 + ξ 2

CO- balance: nco= ξ1 CO2- balance: nco2(= 8nco) = ξ2

H2O balance: nH2O= 2ξ1 + ξ 2

O2 –balance: nO2= (100 x 0.194) – (3/2 ξ1) – ( 2 ξ 2)

ξ1

ξ 2

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4.7 (Balances on Reactive Processes)

Example 4.7-1 page 131 (continued): Substitute nco= ξ1 and nco2(= 8nco) = ξ2 in 7.02 mol = ξ1 + ξ 2: we get that nco = 0.78 mol CO Then we can find: nco2= 8nco= (8 x 0.780) mol CO2 = 6.24 mol CO2

nco= ξ1=0.78 mol nco2= ξ2=6.24 mol nh2o = 14.0 mol H2O nO2 = 5.75 mol O2

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.

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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:

2) Single-Pass Conversion:

reactant input to process – reactant output from process

reactant input to process

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%.

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4.7 (Balances on Reactive Processes)

Example: 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

Reactor Product Separation

Unit

75 mol A/min 100 mol A/min 25 mol A/min 75 mol B/min

75 mol B/min

25 mol A/min

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4.7 (Balances on Reactive Processes)

Example 4.7-2 page 135-137: Propane is dehydrogenated to form propylene in a catalytic reactor:

C3H8C3H6 + H2

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 , 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 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.

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4.7 (Balances on Reactive Processes)

Example 4.7-2 page 135-137 (continued): 1) Draw flow-chart and put basis for fresh feed:

Reactor Separator

Fresh feed 100 mol C3H8 n1(mol C3H8)

n2(mol C3H6) n3(mol C3 H8) n4(mol C3H6 ) n5(mol H2)

Product n6(molC3H8) (0.555% of n3)

n7(mol C3H5) n8(mol H8 )

Recycle n3(mol C3H8) n4(mol C3H6 ) n5(mol H2) n9(mol C3H8 ) n10{mol C3H6) (5% of n7 )

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4.7 (Balances on Reactive Processes) 2) Degree of freedom: Overall system : ( green dashed border) +3 unknown variables (n6 , n7 , n8) -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 (n9 , n10, n1, n2 ) - 2 balances (C3H8, C3H6) = 2 degrees of freedom Reactor: (red dashed border) +5 unknown variables (n1 through n5) -2 atomic balances (C and H) = 3 degrees of freedom

Can’t start with these the degree of freedom doesn’t equal to zero.

Example 4.7-2 page 135-137 (continued):

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4.7 (Balances on Reactive Processes)

Example 4.7-2 page 135-137 (continued): Separator: (purple dashed line) +5 unknown variables (n3 , n4 , n5 , n9 , n10) - 3 balances (C3H8, C3H6, H2 ) - 2 additional relations (n6=0.00555n3, n10 = 0.05n7 ) = 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 C3H8)in- (n6molC3H8)out

(100 mol C3H8)in n6 = 5 mol C3H8

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4.7 (Balances on Reactive Processes)

4) Atomic balance : • Overall C Balance (100 mol C3H8)(3mol C/mol C3H8 ) = (n6(mol C3H8)](3 mol C/mol C3H8 ))+(n7( moI C3H6 ))(3 mol C/mol C3H6 ) sub in n6= 5 mol and we get : n7 = 95 mol C3H6

• Overall H Balance (100)(8) = n6(8) + n7 (6) + n8 (2) sub in n6 = 5 mol n7 = 95 mol and we get: n8 = 95 mol H2

• Given Relations Among Separator Variables: n6 = 0.00555n3

sub in n6= 5 mol and we get : n3= 900 mol C3H8

Example 4.7-2 page 135-137 (continued):

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4.7 (Balances on Reactive Processes)

n10 = 0.0500n7

sub in n7 = 95 mol and we get : n10= 4.75 mol C3H6

• Propane Balance About Separation Unit: n3 = n6 + n8

sub in n3 = 900 mol, n6 = 5 mol and we get : n8= 895 mol C3H8

• Propane Balance About Mixing Point: 100 mol + n9 =n1

After finding all the variables find the desired quantities: Recycle ratio = (n9 + n10) mol recycle 100 mol fresh feed sub in n9 = 895 mol, n10 = 4.75 mol and we get : 9.00 mol recycle/mol fresh feed

Example 4.7-2 page 135-137 (continued):

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4.7 (Balances on Reactive Processes)

Single-pass conversion = n1-n3 x100 n1

sub in n1= 995 mol, n3 = 900 mol and we get : 9.6%

Example 4.7-2 page 135-137 (continued):

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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.

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4.7 (Balances on Reactive Processes)

Example 4.7-3 page 139-142: Methanol is produced in the reaction of carbon dioxide and hydrogen:

CO2 + 3H2 CH3OH + H2O The fresh feed to the process contains hydrogen, carbon dioxide, and 0.400 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% CO2 , 70.0 mole% H2 , 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 CH3OH/h.

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4.7 (Balances on Reactive Processes) Example 4.7-3 page 139-142 (continued): Draw and label flow chart and add basis:

Reactor Condenser n0{mol) xoc(mol CO2/mol) (0.996 -xoc) (mol H2/mol) 0.00400 mol l/mol

100 mol 0.280 mol CO2mol 0.700 mol H2mol 0.020 mol l/mol

n1 (mol CO2) n2(mol H2) 2.0 mol I n3(mol CH3OH) n4(mol H2O)

n3(mol CH3OH) n4(molH2O)

np(mol) x5c(mol CO2 /mol) x5H(mol H2mol) (1 -x5c-x5H) (mol l/mol) n5(mol) x5c(mol CO2/mol) x5H(mol H2/mol) (1 -x5c-x5H) (mol l/mol)

nr(mol) x5c(mol CO2/mol) x5H(mol H2/mol) (1 -x5c-x5H) (mol l/mol)

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4.7 (Balances on Reactive Processes)

Example 4.7-3 page 139-142 (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 (no,xoc,n3,n4,np,x5C,x5H) + 1 reaction - 5 independent balances (CO2 ,H2,,I,CH3OH,H2O) = 3 degrees of freedom Recycle-fresh feed mixing point- 5 unknowns (no,xoc,nr,x5C,x5H) - 3 independent balances (CO2, H2 , I) = 2 degrees of freedom

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4.7 (Balances on Reactive Processes)

Example 4.7-3 page 139-142 (continued): Reactor- 4 unknowns (n1 ,n2 ,n3 ,n4) + 1 reaction - 4 independent balances (CO2 , H2 , CH3OH, H2O) - 1 single-pass conversion = 0 degrees of freedom Condenser- 3 unknowns (n5,x5c,x5H) - 3 independent balances (CO2 , H2 , I) = 0 degrees of freedom Purge-recycle splitting point- 2 unknowns (nr,np ) - 1 independent balance = 1 degree of freedom

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4.7 (Balances on Reactive Processes)

Example 4.7-3 page 139-142 (continued): Recycle-fresh feed mixing point- 3 unknowns (no,xoc,nr) - 3 independent balances = 0 degrees of freedom Purge-recycle splitting point- 1 unknown (np) - 1 independent balance = 0 degrees of freedom Extra information: 60% Single-Pass H2 Conversion: fconv= (nin-nout)/nin

0.6= (70-nout)/70 nout=28 mol=n2

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4.7 (Balances on Reactive Processes)

Example 4.7-3 page 139-142 (continued): Reactor balance: • H2-Balance: n2=100x0.700-3ξ 28=70-3ξ ξ=14 • CO2-balance: n1=100x0.28-ξ n1=28-14 n1=14 • CH3OH-balance: n3=14 • H2O-balance: n4= 14

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4.7 (Balances on Reactive Processes)

Example 4.7-3 page 139-142 (continued): Condenser Balance: In=out (no reaction) Total-balance: n1+n2+2+n3+n4=np+n5+n3+n4 14+28+2+14+14=n5+14+14 n5= 44 CO2-balance n1= n5(x5c) 14=44(x5c) x5c=0.32 H2-balance n2=n5(x5H) 28=44(x5H)

x5H=0.64

xi= 1 - xsc - x5h = 0.04545 mol I/mol

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4.7 (Balances on Reactive Processes)

Example 4.7-3 page 139-142 (continued): Feed-Recycle Mixing Point: In=out (no reaction) Total Mole Balance: n0 + nr = 100 mol I Balance- n0 (0.00400) + nr (0.04545) = 2.0 mol CO2 Balance- noxoc + nTx5c = 28.0 mol CO2

xoc = 0.256 mol CO2/mol Therefore: xoH = (1 -x5c-x5H) = 0.740 mol H2/mol Recycle-Purge Splitting Point- n5 = nT + np

2 equations with 2 unkowns so by solver we get that no = 61.4 mol nr = 38.6 mol

np = 5.4 mol purge

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4.7 (Balances on Reactive Processes)

Example 4.7-3 page 139-142 (continued): To scale the process to a methanol production rate of 155 kmol CH3OH/h, we multiply each total and component molar flow rate by the factor: (155 kmol CH3OH/h) / (14 mol CH3OH) = 11.1 (kmol/h)/mol

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4.7 (Balances on Reactive Processes)

Variable Basis

Value Scaled

Value

Fresh Feed 61.4 mol 681 kmol/h

25.6 mole% CO2 25.6 mole% CO2

74.0 mole% H2 74.0 mole% H2

0.400 mole% I

0.400 mole% I

Feed to reactor 100 mol 1110 kmol/h

28.0 mole% CO2

28.0 mole% CO2

70.0 mole% H2

70.0 mole% H2

2.0 mole% I

2.0 mole% I

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4.7 (Balances on Reactive Processes) Variable Basis Value Scaled

Value

Recycle 38.6 mol 428 kmol/h

31.8 mole% CO2

31.8 mole% CO2

63.6 mole%H2

63.6 mole% H2

4.6 mole% I

4.6 mole% I

Purge 5.4 mol 59.9 kmol/h

31.8 mole% CO2

31.8 mole% CO2

63.6 mole% H2

63.6 mole% H2

4.6 mole% I

4.6 mole% I

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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 (CO2 ,H2O, and possibly CO and SO2) 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 CO2 or CO, hydrogen forms H2O, and sulfur forms SO2 . 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 CO2 in the product stream only.) Example: C3H8 + 5O2 3 CO2 + 4 H2O Complete combustion of propane C3H8 + (7\2)O2 3 CO + 4 H2O Partial combustion of propane

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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% N2 , 21% O2. 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.

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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 H2O 4) Divide each composition by the total composition found from part (3)

(composition will get greater)

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4.8 Combustion Reaction Example: A stack gas contains 60.0 mole% N2 , 15.0 % CO2 , 10.0% O2 , and the balance H2O. Calculate the molar composition of the gas on a dry basis. Add compositions except H2O: (60 mol N2 + 15 mol CO2+ 10 mol O2 ) = 85 mol dry

Component Wet Basis Dry Basis Mole % Dry Basis

N2 60 60/85 = 0.706 70.6

O2 10 10/85 = 0.176 17.6

CO2 15 15/85 = 0.118 11.8

H2O 15 - -

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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 H2O ( mostly composition) 4) Multiply the information given about H2O with basis to get H2O mole 5) Subtract the the moles of H2O (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.

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4.8 Combustion Reaction An Orsat analysis (a technique for stack analysis) yields the following dry basis composition: N2 65% CO2 14% CO 11% O2 10% A humidity measurement shows that the mole fraction of H2O in the stack gas is 0.0700. Calculate the stack gas composition on a wet basis. 1) Basis : 100 mole 2) 100 x 0.0700 =7 3) 100 – 7 = 93 which equal to (0.93)

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4.8 Combustion Reaction

Component Dry Basis Wet Basis Mole % wet basis

N2 65 0.65 x 0.93 = 0.605 60.5%

CO2 14 0.14 x 0.93 = 0.130 13%

CO 11 0.11 x 0.93 =0.102 10.2%

O2 10 0.10 x 0.93 = 0.093 9.3%

H2O - 0.07 7%

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4.8 Combustion Reaction Theoretical Oxygen: The moles (batch) or molar flow rate (continuous) of O2 needed for complete combustion of all the fuel fed to the reactor, assuming that all carbon in the fuel is oxidized to CO2 and all the hydrogen is oxidized to H2O. 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%

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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 4.8-2 page 145-146 : One hundred mol/h of butane (C4H10) 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:

C4H10 + (13/2) O2 4CO2 + 5 H2O 2) Theoretical O2: (no2)theoretical = 1 C4H10 (13/2) O2 100 C4H10 X

After doing cross multiplying we will find out that X = 650 mol O2/h.

.

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4.8 Combustion Reaction 3) Theoretical air: (nair)theoretical = sum of (no2)theoretical

0.21 (nair)theoretical = 650 mol O2 / 0.21 (mol O2/mol air) = 3094 mol air 4) Percent Excess Air: %excess air = (( 5000 – 3094) / 3094) x 100 = 61.6 %

.

.

Example 4.8-2 page 145-146 (continued) :

Page 74: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

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).

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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 CO2 atomic species balances are usually most convenient.

Example 4.8-3 page 147-149: 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. 1) Draw and label flow chart and add basis: 100 mol C2H6

50% excess air n0(mol)

0.21 mol O2/mol 0.79 mol N2/mol

n1 (molC2H6) n2(mol O2 ) n3(mol N2 ) n4(mol CO) n5(mol CO2 ) n6(mol H2O)

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4.8 Combustion Reaction Example 4.8-3 page 147-149 (continued): 2) Write down the stoichiometric equation:

C2H6 + (7/2) O2 2CO2 + 3H2O C2H6 + (5/2) O2 2CO + 3H2O

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% O2 , 79 mo!e % N2. o Since excess air is supplied, O2 must appear in the product stream.

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4.8 Combustion Reaction Example 4.8-3 page 147-149 (continued): 3) Degree of Freedom: 7 unknowns (n0 , n1…......,n6 ) - 3 atomic balances (C, H, O) - 1 N2 balance - 1 excess air specification (relates n0 to the quantity of fuel fed) - 1 ethane conversion specification - 1 CO/CO2 ratio specification = 0 degree of freedom 4) 50% Excess Air: 1 C2H6 (7/2) O2 100 C2H6 x x= 350 mol O2

Therefore: ( 350 mol O2 / 0.21 (mol O2/mol air) ) = 1666.7

Theoretical O2

Theoretical air

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4.8 Combustion Reaction Example 4.8-3 page 147-149 (continued): 0.50 = ( n0 – 1666.7) / 1666.7 n0= 2500 mol air fed 5) 90% Ethane conversion: fconv= (nfed-n1)/nfed 0.9= ( 100-n1)/100 n1= 10 mol C2H6

6) 25% conversion to CO: 1 C2H6 2 CO (100-10) n4 n4= 45 CO 7) Balances: Nitrogen balance ( not reacted) (input = output) o.79(2500)=n3 n3= 1975 mol

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4.8 Combustion Reaction Example 4.8-3 page 147-149 (continued): Atomic Carbon balance: 100(2)=n1(2) + n4(1)+ n5(1) n5= 135 Atomic Hydrogen balance: 100(6)=10(6)+n6(2) n6= 270 Atomic oxygen balance: 525(2)=n2(2)+(45)(1)+(135)(2)+(270)(1) n2= 232

Page 80: Chapter 4 · 2016-06-04 · 4.6 (Limiting and Excess Reactants, Fractional Conversion, and Extent of Reactant) Example on Fractional excess: The hydrogenation of acetylene to form

4.8 Combustion Reaction Example 4.8-3 page 147-149 (continued): Stack gas composition on a dry basis: a) Add the moles without including H2O mole: n1 = 10molC2H6

n2 = 232 mol O2

n3 = 1974 mol N2

n4 = 45 mol CO n5 = 135 mol CO2

2396 b) Divide each compositions by 2396:

y1= 10/2396= 0.00417 y2=232/2396=0.0970 y3=1974/2396=0.824 y4=45/2396=0.019 y5=135/2396= 0.0563

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4.8 Combustion Reaction Example 4.8-4 page 149-150: A hydrocarbon gas is burned with air. The dry-basis product gas composition is 1.5 mole% CO, 6.0% CO2 , 8.2% O2 , and 84.3% N2. 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:

100 mol dry gas 0.015 mol CO/mol dry gas 0.060 mol CO2/mol dry gas 0.082 mol O2/mol dry gas 0.843 mol N2/mol dry gas nw(mol H2O)

nc(mol C) nH (mol H)

na(mol air) 0.21 mol O2/mol 0.79 mol N2/mol

Note: Since the fuel is a hydrocarbon, water must be one of the combustion products.

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4.8 Combustion Reaction Example 4.8-4 page 149-150:

C + O2 CO2

2C + O2 2CO 4H + O2 2H2O

2) Degree of Freedom: 4 unknowns (nH , nc, na, nw) -3 independent atomic balances (C, H, O) -1 N2 balance = 0 degree of freedom 3) Balances: N2balance- 0.79na= 100(0.843) na= 106.7 mol Atomic C balance: nc=100(0.015)(1) + (100)(0.060)(1) nc= 7.5

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4.8 Combustion Reaction Example 4.8-4 page 149-150: Atomic O balance: 0.21(na)(2)=nw(1)+100(0.015)(1)+100(0.060)(2)+100(0.082)(2) nw=14.9 Atomic H balance: nH = nw (2) nH= 29.8 C/H Ratio in the fuel: nH/nC= (29.8 mol H)/(7.5 mol C) = 3.97 mol H / mol C Percent Excess Air: (no2)theoretical: 1 C 1 O2 4H 1 O2 7.5 X 29.8 X X= 7.5 X= 7.45

(no2)theoretical= 7.5 +7.45 = 14.95

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4.8 Combustion Reaction (nair )theoretical= ( 14.95/0.21) =71.2 (nair )feed

=106.7 %excess = ((106.7-71.2)/71.2) x 100 % =49.8%