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Material Balance-Part 2

Chapter 3

Material and Energy Balances

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Stoichiometry

• Stoichiometry – theory of proportions in which chemical species combine with one another.

• Stoichiometric equation of chemical reaction – statement of the relative number of molecules or moles of reactants and products that participate in the reaction.

2 SO2 + O2 ---> 2 SO3

Material Balance Problems With Chemical Reactions

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• Stoichiometric ratio – ratio of species stoichiometry coefficients in

the balanced reaction equation– can be used as a conversion factor to

calculate the amount of particular reactant (or product) that was consumed (produced).

2 mol SO3 generated 2 mol SO2 consumed

2 mol SO2 consumed 1 mol O2 consumed

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C4H8 + 6 O2 --------> 4 CO2 + 4 H2O

1. Is the stochiometric equation balance?Yes

2. What is stochiometric coefficient for CO2

43. What is stochiometric ratio of H2O to O2 including it unit

4 mol H2O generated/ 6 mol O2 consumed4. How many lb-moles of O2 reacted to form 400lb-moles CO2

600 lb-moles O2 reacted5. 100 mol/min C4H8 fed into reactor and 50% is reacted. At what

rate water is formed?200 mol/min water generated

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Limiting Reactant & Excess Reactant

• The reactant that would run out if a reaction proceeded to completion is called the limiting reactant, and the other reactants are termed excess reactants.

• A reactant is limiting if it is present in less than its stoichiometric proportion relative to every other reactant.

• If all reactants are present in stoichiometric proportion, then no reactant is limiting.

100% n

n - n Excess Percentage

n n - n

Excess Fractional

stoich

stoichfeed

stoich

stoichfeed

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C2H2 + 2H2 ------> C2H6

Inlet condition: 20 kmol/h C2H2 and 50 kmol/h H2

What is limiting reactant and fractional excess?

(H2:C2H2) feed = 2.5 : 1

(H2:C2H2) stoich = 2 : 1

H2 is excess reactant and C2H2 is limiting reactant

Fractional excess of H2

n (stoich)= 40n (feed)= 50

= (50-40)/40 =0.25stoich

stoichfeed

n n - n

Excess Fractional

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Fractional Conversion

Chemical reactions do not take place instantaneously. In normal operation, there always will be the unconverted reactant. To calculate the conversion of a chemical reaction;

The fraction unreacted is 1-f.

100%fed mole

reacted moles f ,Conversion Percentage

fed molereacted moles

f ,Conversion Fractional

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Extent of ReactionExtent of Reaction, ξ

ξ = extent of reaction (amount of species react)ni = moles of species i present in the system after the reaction occurrednio = initial moles of species i in system when the reaction startsvi = stoichiometric coefficient for species i in the particular chemical reaction equation. (-ve for reactants, +ve for products.

iioi

iioi

vnn

vnn

or

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Considering C2H2 + 2H2 ------> C2H6 ,

suppose 20 kmol of C2H2, 50 kmol H2 and 50 kmol

C2H6 are charged to the batch reactor to form C2H6.

After sometimes 30 kmol of H2 has reacted. How much of each species will be present in the reactor at this moment?

H2 left = 50 - 30 kmol= 20 kmol H2

30 kmol H2 react => 15 kmol C2H2

C2H2 left= 20 – 15 kmol = 5 kmol C2H2

30 kmol H2 react from 15 kmol C2H6

C2H6 in the reactor = 15 + 50 kmol = 65 kmol C2H6

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N2 + 3H2 ------------> 2NH3

Reactor inlet: 100 mol N2/s; 300 mol H2/s; 1 mol Ar/s. If fractional conversion of H2 is 0.6, calculate extent of reaction and the outlet composition.

From fractional conversion, moles of species H2 present in the system after the reaction occurred nH2

= (1-0.6) 300 = 120 mol H2/sExtent of reaction = (300-120)/3= 60 mol/snN2

= 100- 60= 40 mol/s

nNH3 = 2(60)= 120 mol/s

2

1

100

3300

3

2

2

NH

Ar

N

H

n

n

n

n

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2 C2H4 + O2 ------->2 C2H4O

The feed to a reactors contains 100kmol C2H4 and 100kmol O2.

a) Which is limiting reactant? C2H4

b) Percentage of excess? [(100-50)/50 ]x100%=100%c) O2 out? C2H4 formed? Extent of reaction?

50kmol, 100kmol C2H4, 50kmol

Try This…

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d) If fractional conversion for limiting reactant is 50%, what is outlet composition and extent of reaction?50kmol C2H4, 75 kmol O2, 50 kmol C2H4O

extent of reaction = 25 kmole) If reaction proceed to a point where 60kmol O2 left,

what is fractional conversion for C2H4, fractional conversion of O2 and extent of reaction?

fC2H4=0.8, fO2=0.4, extent of reaction=40 kmol

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Acrylnitrile is produced in the reaction of production of propylene, ammonia and oxygen:

C3H6 + NH3 + 3/2O2 -> C3H3N + 3H2O

The feed contains 10.0 mole% propylene, 12.0mole% ammonia and 78 mole% air. A fractional conversion of 30.0% of the limiting reactant is achieved. Taking 100 mol of feed as a basis, determine which reactant is limiting, the percentage by which each of the other reactant is excess, and the molar amounts of all product gas constituents for 30% conversion of the limiting reactant.

Try This…

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Extent of Reaction for Multiple Reaction

• Concept of extent of reaction can also be applied for multiple reaction.

• Only now each independent reaction has its own extent.

ijjjiioi vnn

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CLASS DISCUSSIONfor

Example 4.6-3

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Chemical Equilibrium

2 fundamental questions might be ask for a give set reactive species and reaction condition, :

1.What will be the final (equilibrium) composition of the reaction mixture? – chemical engineering thermodynamics

2.How long will the system take to reach a specified state short of equilibrium? – chemical kinetics

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• Irreversible reaction– reaction proceeds only in a single direction

(from reactants to products)– the concentration of the limiting reactant

eventually approaches zero.• Reversible reaction

– reactants form products for forward reaction and products undergo the reverse reactions to reform the reactants.

– Equilibrium point is a rate of forward reaction and reverse reaction are equal

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CLASS DISCUSSIONfor

Example 4.6-2

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Multiples Reaction, Yield & Selectivity

• Some of the chemical reaction has a side reaction which is formed undesired product- multiple reaction occurred.

• Effects of this side reaction might be:1. Economic loss2. Less of desired product is obtained for a given

quantity of raw materials3. Greater quantity of raw materials must be fed to

the reactor to obtain a specified product yield.

selectivity =moles of desired product

moles of undesired product

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Yield3 definitions of yield with different definitions:

Yield =

Moles of desired product formed

Moles that would have been formed if there were no side reaction and the limiting

reactant had reacted completely

Yield =Moles of desired product formed

Moles of reactant fed

Yield = Moles of desired product formed

Moles of reactant consumed

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Balance of Reactive Processes

Balance on reactive process can be solved based on 3 methods:

1. Atomic Species Balance

2. Extent of Reaction

3. Molecular Species Balance

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Atomic Species Balance

No. of unknowns variables- No. of independent atomic species balance- No. of molecular balance on indep. nonreactive species- No. of other equation relating the variable=============================

No. of degree of freedom=============================

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Extent of Reaction

No. of unknowns variables+ No. of independent chemical reaction- No. of independent reactive species- No. of independent nonreactive species- No. of other equation relating the variable=============================

No. of degree of freedom=============================

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Molecular Species Balance

No. of unknowns variables

+ No. of independent chemical reaction- No. of independent molecular species balance- No. of other equation relating the variable

=============================No. of degree of freedom

=============================

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Degree-of-Freedom Analysis

Independent EquationIndependent Species Independent Reaction

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Independent Equation

• Algebraic equation are independent if we cannot obtain any one of them by adding and subtracting multiples of any of the others

x + 2y = 4 [1]3x + 6y = 12 [2]

• Only one independent equation because [2]= 3 x [1]x + 2y = 4 [1]2x – z= 2 [2]4y + z= 6 [3]

• Although 3 equations, but only two independent equation exist because [3]=2x[1] –[2]

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• If two MOLECULAR species are in the SAME RATIO to each other wherever they appear in a process, balance on those species will not be independent (i.e. only one independent equation is obtained)

Similarly

• If two ATOMIC species are in the SAME RATIO to each other wherever they appear in a process, balance on those species will not be independent (i.e. only one independent equation is get)

Independent Species

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Independent Molecular Species-cont.

• If two MOLECULAR species are in the SAME RATIO to each other wherever they appear in a process, balance on those species will not be independent (i.e. only one independent equation is obtained)

n3 mol O2

3.76 n3 mol N2

n4 mol CCl4(v)

ProcessUnit

n1 mol O2

3.76 n1 mol N2

n2 mol CCl4(l) n5 mol CCl4(l)

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• Since N2 and O2 have a same ratio wherever they appear on the flowchart (3.76 mol N2 / mol O2), only ONE independent balance can obtain.

• Let make a molecular balance on both species to prove itBalance on O2 :

n1=n3 [1]Balance on N2 :

3.76 n1=3.76n3n1=n3 [2]

• Eq. [1] and [2] are SAME. Only ONE INDEPENDENT EQUATION OBTAINED although two species involved.

Independent Molecular Species-cont.

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Independent Atomic Species

• If two ATOMIC species are in the SAME RATIO to each other wherever they appear in a process, balance on those species will not be independent (i.e. only one independent equation is get)

n3 mol O2

3.76 n3 mol N2

n4 mol CCl4(v)

ProcessUnit

n1 mol O2

3.76 n1 mol N2

n2 mol CCl4(l) n5 mol CCl4(l)

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Independent Atomic Species-cont.• Atomic N and O are always in same proportion to each other

in the process (3.76:1), similar for atom C and Cl which always same ratio too (1:4).

• Although FOUR atomic species exist, only TWO independent equation can obtain for this cases.

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• Prove:Balance on atomic O: 2n1=2n3

n1=n3 [1]Balance on atomic N: 2(3.76)n1=2(3.76)n3

n1=n3 [2]

Balance on atomic C: n2=n4+n5 [3]

Balance on atomic Cl: 4n2=4n4 +4n5 n2=n4+n5 [4]

• Eq. [1]=[2] and [3]=[4], only TWO independent equation obtained

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Independent Reaction• Used when we using either molecular species balancemolecular species balance or

extent of reaction methodextent of reaction method to analyze a balance on reactive process

• Chemical reaction 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

A ------> 2B [1]B ------> C [2]A ------> 2C[3]

• Only TWOTWO independent eqn. can obtained although three equation exist since [3]=[1] + 2[2].

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Balance of Reactive Processes

Balance on reactive process can be solved based on 3 methods:

1.Atomic Species Balance2.Extent of Reaction3.Molecular Species Balance[arrange according to the easiest method(1) to more difficult method(2), but not always true]

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Atomic Species Balance

No. of unknowns variables- No. of independent atomic species balance- No. of molecular balance on indep. nonreactive species- No. of other equation relating the variable

=============================No. of degree of freedom

=============================

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Extent of Reaction

No. of unknowns variables

+ No. of independent chemical reaction- No. of independent reactive species- No. of independent nonreactive species- No. of other equation relating the variable

=============================No. of degree of freedom=============================

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Molecular Species Balance

No. of unknowns variables

+ No. of independent chemical reaction- No. of independent molecular species balance- No. of other equation relating the variable

=============================No. of degree of freedom

=============================

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Application of Method

C2H6 -------> C2H4 + H2

40 kmol H2/min

n1 kmol C2H6/min

n2 kmol C2H4/min

Reactor100 kmol C2H6/min

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Method 1: Atomic Species Balance

• All atomic balance is INPUT=OUTPUT• Degree-of-freedom analysis

2 unknowns variables (n1, n2)- 2 independent atomic species balance (C, H)- 0 molecular balance on indep. nonreactive species- 0 other equation relating the variable

=============================0 No. of degree of freedom

=============================

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Method 1: Atomic Species Balance

• Balance on atomic C (input= output)

200=2n1 + 2n2100=n1 + n2 [1]

• Balance on atomic H (input = output)100(6)=40(2) + 6n1+4n2520 = 6n1 + 4n2 [2]

Solve simultaneous equation, n1= 60 kmol C2H6/min; n2= 40 kmol C2H4/min

100 kmol C2H6

2 knol C

=

n1 kmol C2H6

2 kmol C

+n2(2)1 kmol

C2H61 kmol C2H6

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Method 2: Extent of Reaction

• Degree-of-freedom analysis

2 unknowns variables (n1,n2)+ 1 independent chemical reaction- 3 independent reactive species (C2H6, C2H4, H2)

- 0 independent nonreactive species- 0 other equation relating the variable

=============================0 No. of degree of freedom

=============================

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Method 2: Extent of Reaction

• Write extent of reaction for each species

C2H6 : n1 = 100-ξ

C2H4 : n2= ξ

H2 : 40= ξ

Solve for n1 and n2 (ξ =40)

n1= 60 kmol C2H6/min; n2= 40 kmol C2H4/min

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Method 3: Molecular Species Balance

• Degree-of-freedom analysis

2 unknowns variables (n1, n2)

+1 independent chemical reaction- 3 independent molecular species balance (C2H6, C2H4, H2)

- 0 other equation relating the variable=============================

0 No. of degree of freedom=============================

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Method 3: Molecular Species Balance

H2 balance (Gen=Output):

H2 Gen= 40 kmol H2/min

C2H6 Balance (input=output + cons.):

100 kmol C2H6/min = n1 kmol C2H6/min +

40 kmol H2 gen X (1 kmol C2H4 gen/1 kmol H2 gen)n1= 60 kmol C2H6/min

C2H4 balance (Gen.=Ouput):

40 kmol H2 gen x (1 kmol C2H4 gen./ 1 kmol H2 gen) = n2

n2= 40 kmol C2H4/min

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CLASS DISCUSSION

EXAMPLE 4.7-1

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Product Separation & Recycle

Overall ConversionReactant input to Process – reactant output from Process

Reactant input to Process

Single Pass ConversionReactant input to Reactor – reactant output from Reactor

Reactant input to Reactor

75 mol B/min100 mol A/min75 mol A/minReactor

Product Separation

Unit

25 mol A/min75 mol B/min

25 mol A/min

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Purging

• To prevent any inert or insoluble substance build up and accumulate in the system

• Purge stream and recycle stream before and after the purge have a same composition.

ProductFresh FeedReactor

Product Separation

Unit

Recycle Purge

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CLASS DISCUSSION

EXAMPLE 4.7-2

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CLASS DISCUSSION

EXAMPLE 4.7-3

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COMBUSTION REACTIONS

• Combustion – Rapid reaction of a fuel with oxygen. This reaction releases tremendous quantities of energy that can be manipulated to boil water to produce steam.

• Combustion releases products such as CO, CO2 and SO2 and as chemical engineers, we are tasked to monitor and analyze the production of these noxious gases.

• Combustion fuels could be coal (carbon, some hydrogen, and sulfur and various noncombustible materials), fuel oil (mostly high molecular weight hydrocarbons, some sulfur), gaseous fuel (such as natural gas, which is primarily methane) or it could be liquefied petroleum gas, which is usually propane or butane.

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Combustion Chemistry

From Fuel Perspective• Fuel contains carbonaceous material that will form either

CO2 or CO, Hydrogen forming H2O and Sulfur forming SO2.

From O2 Source Perspective

• For economic reason, AIR AIR is the source of oxygen in most combustion reactions. Dry airDry air has the following average molar composition:

N2 78.03%O2 20.99%Ar 0.94%CO2 0.03%H2, He, Ne, Kr, Xe 0.01%

Average MolecularWeight = 29.0

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• However, in most combustion calculations, it is acceptable to simplify this composition to 79% N2, 21% O2

• For combustion reaction, generally we have TWOTWO type of expressions to express the mole composition of a gas, that is Composition On A Wet BasisComposition On A Wet Basis and Composition On A Composition On A Dry BasisDry Basis.

• Composition On A Wet BasisComposition On A Wet Basis is commonly used to denote mole fractions of a gas that contains water.

• Composition On A Dry BasisComposition On A Dry Basis can also be used to denote mole fractions of the same gas that contains water but by excluding the presence of water in the calculation.

• Example: A gas contains 33.3 mole% CO2, 33.3% N2 and 33.3% H2O on wet basis is deemed to have a composition of 50.0 mol% CO2 and 50.0 mol% H2O on dry basis.

• It is important to have a grasp of knowledge on the correct technique to calculate these two type of compositions or to convert from dry basis to wet basis or vice versa.

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• The MAIN REASON is whenever flow rate of a product gas leaving the stackstack (stack gas or flue gas refers to product gas that stack gas or flue gas refers to product gas that leaves a combustion furnaceleaves a combustion furnace) is measured, the measurement is for the total flow rate that also involved the product H2O, while on the other hand, common techniques for analyzing stack gases provide compositions on a dry basis.

• The procedure to convert Dry Basis to Wet Basis or The procedure to convert Dry Basis to Wet Basis or vice versa follows exactly the same procedure vice versa follows exactly the same procedure outlined earlier on for converting Mass outlined earlier on for converting Mass Compositions to Mole Compositions or vice versa in Compositions to Mole Compositions or vice versa in previous chapterprevious chapter.

EXAMPLE 4.8-1:EXAMPLE 4.8-1:A stack gas contains 60.0 mole% N2, 15.0% CO2, 10.0% O2 and the balance O2. Calculate the molar composition of the gas on a dry basisdry basis.

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SOLUTIONSOLUTION

• Basis: 100 mol Wet Gas 60.0 mol N2

15.0 mol CO2

10.0 mol O2

85.0 mol dry gas (DG)

60.0/85.0 = 0.706 mol N2/mol DG15.0/85.0 = 0.176 mol CO2/mol DG10.0/85.0 = 0.118 mol O2/mol DG

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Theoretical and Excess Air

• Theoretical Oxygen: The moles (batch process) or molar flow rate (continuous process) of O2 needed for complete combustion of all the fuel fed to the reactor, assuming that all the 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: %100

ltheoretica

ltheoreticafed

airmoles

airmolesairmoles

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TEST YOURSELF PAGE 146

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Material Balances on Combustion Process

• The procedure for writing and solving material balances for a combustion reactor is the same as that for any other reactive system. However, be extra cautious with these:

• When you draw and label flow chart, be sure that the outlet stream (the stack gas) includes:

Unreacted fuel unless you are told otherwise

Unreacted O2

H2O and CO2, also CO if the problem statement says so

N2 if the fuel burnedWith AIRAIR and Not PUREPUREOO22

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• To calculate the O2 feed rate from a specified percent oxygen excess, first is to calculate the theoretical O2 from the fuel feed rate and the reaction stoichiometry for COMPLETE COMPLETE COMBUSTIONCOMBUSTION, then calculate the oxygen feed rate by multiplying the theoretical O2 by (1+fractional excess of O2).

• Atomic balances are usually most convenient for use in the calculation.

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EXAMPLE 4.8-3