Chap. 3.4 Mass Balance Reactive System Part B

Chapter 3 Mass Balance

Balance on Reactive Processes System: Part B

Balance of Reactive ProcessesBalance on reactive process can be solved based on three method:

Atomic Species Balance

Extent of Reaction

Molecular Species Balance

Degree-of-Freedom Analysis? Independent EquationIndependent Species Independent Reaction

Independent EquationAlgebraic equation are independent if we cannot obtain any one of them by adding and subtracting multiples of any of the othersx + 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 equation, but only two independent equation exist because [3]=2x[1] [2]

Independent SpeciesIf 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 get)

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 Molecular SpeciesIf 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 get)

Independent Molecular SpeciesSince 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 it

Balance 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 OBTAIN although two species involved.

Independent Atomic SpeciesIf 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 Atomic SpeciesAtomic 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.Prove:Balance on atomic O: 2n1=2n3n1=n3[1]Balance on atomic N:2(3.76)n1=2(3.76)n3n1=n3[2]Balance on atomic C:n2=n4+n5[3]Balance on atomic Cl:4n2=4n4 +4n5n2=n4+n5[4]

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

MUFLIS DALAM HARTA HANYA KEMISKINAN YANG SEMENTARA DI DUNIA, MUFLIS DALAM WATAK ADALAH KEMISKINAN TERBURUK DI DUNIA. IA AKAN DI BAWA BERSAMA KE AKHIRAT

Independent ReactionUsed when we using either molecular species balance or extent of reaction method to analyze a balance on reactive processChemical 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 TWO independent eq. can obtained although three equation exist sice [3]=[1] + 2[2].

Balance of Reactive ProcessesBalance on reactive process can be solved based on three method:

Atomic Species Balance

Extent of Reaction

Molecular Species Balance

[arrange according to the easiest method(1) to more difficult method(2), but not always true]

Atomic Species BalanceNo. 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=============================

Extent of ReactionNo. 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=============================

Molecular Species BalanceNo. 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=============================

Application of Method C2H6 -------> C2H4 + H2

Method 1: Atomic Species BalanceAll atomic balance is INPUT=OUTPUTDegree-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=============================

Method 1: Atomic Species BalanceBalance 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 C2H62 knol C=n1 kmol C2H62 kmol C+n2(2)1 kmol C2H61 kmol C2H6

Method 2: Extent of ReactionDegree-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=============================

Method 2: Extent of ReactionWrite extent of reaction for each species

C2H6:n1 = 100-C2H4:n2= H2:40=

Solve for n1 and n2 ( =40)

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

Method 3: Molecular Species BalanceDegree-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=============================

Method 3: Molecular Species BalanceH2 balance (Gen=Output):H2 Gen= 40 kmol H2/min

C2H6 Balance (input=output + cons.):100 kmol C2H6/min = n1kmol C2H6/min + 40 kmol H2 gen X (1 kmol C2H6 cons/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) = n2n2= 40 kmol C2H4/min

CLASS DISCUSSIONEXAMPLE 4.7-1

Product Separation & Recycle

Overall ConversionReactant input to Process reactant output from ProcessReactant input to Process

Single Pass ConversionReactant input to Reactor reactant output from ReactorReactant input to Reactor

Purging To prevent any inert or insoluble substance build up and accumulate in the systemPurge stream and recycle stream before and after the purge have a same composition.

ANY QUESTION?

20 Sept 2005

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Chap. 3.4 Mass Balance Reactive System Part B