Chapter 3 Mass Balance Balance on Reactive Processes System: Part B

Balance of Reactive Processes Balance on reactive process can be solved based on three method: Atomic Species Balance Extent of Reaction Molecular Species Balance

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

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] 2x – z= 2 [2] 4y + z= 6 [3] Although 3 equation, but only two independent equation exist because [3]=2x[1] –[2]

Independent Species 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 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 Species 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 get) n3 mol O2 3.76 n3 mol N2 n4 mol CCl4(v) Process Unit n1 mol O2 3.76 n1 mol N2 n2 mol CCl4(l) n5 mol CCl4(l)

Independent Molecular Species 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 be obtained. Let’s make a molecular balance on both species to prove it Balance on O2: n1=n3 [1] Balance on N2: 3.76 n1=3.76n3 n1=n3 [2] Eq. [1] and [2] are SAME. Only ONE INDEPENDENT EQUATION OBTAINED although two species are involved.

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 obtained) n3 mol O2 3.76 n3 mol N2 n4 mol CCl4(v) Process Unit n1 mol O2 3.76 n1 mol N2 n2 mol CCl4(l) n5 mol CCl4(l)

Independent Atomic Species 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 have the same ratio too (1:4). Although FOUR atomic species exist, only TWO independent equation can be obtained for this case. 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

Independent Reaction Used when we are using either molecular species balance or extent 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 TWO independent eq. can obtained although three equation exist since [3]=[1] + 2[2].

Balance of Reactive Processes Balance on reactive process can be solved based on three method: Atomic Species Balance Extent of Reaction Molecular Species Balance [arranged according to the easiest method(1) to more difficult method(2), but not always true]

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

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

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

Application of Method C2H6 -------> C2H4 + H2 Reactor 40 kmol H2/min n1 kmol C2H6/min n2 kmol C2H4/min Reactor 100 kmol C2H6/min

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

Method 1: Atomic Species Balance Balance on atomic C (input= output) 200=2n1 + 2n2 100=n1 + n2 [1] Balance on atomic H (input = output) 100(6)=40(2) + 6n1+4n2 520 = 6n1 + 4n2 [2] Solve simultaneous equation, n1= 60 kmol C2H6/min; n2= 40 kmol C2H4/min 100 kmol C2H6 2 kmol C = n1 kmol C2H6 + n2(2) 1 kmol C2H6

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

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

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

Method 3: Molecular Species Balance H2 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 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

CLASS DISCUSSION EXAMPLE 4.7-1

Product Separation & Recycle 75 mol B/min 100 mol A/min 75 mol A/min Reactor Product Separation Unit 25 mol A/min Overall Conversion Reactant input to Process – reactant output from Process Reactant input to Process Single Pass Conversion Reactant input to Reactor – reactant output from Reactor Reactant input to Reactor

Product Separation Unit 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. Product Fresh Feed Reactor Product Separation Unit Recycle Purge

CLASS DISCUSSION EXAMPLE 4.7-2

CLASS DISCUSSION EXAMPLE 4.7-3

ANY QUESTION?