CHAPTER 4:The Material Balance for Chemical Reactors

1 CHAPTER 4:The Material Balance for Chemical ReactorsCopyright 2022 by Nob Hill Publishing, LLCRjVQ1cj1Q0cj0 Conservation of mass rate ofaccumulationof componentj ={rate of inflowof componentj} {rate of outflowof componentj}+ rate of generationof componentjbychemical reactions ( )1 / 152 General Mole BalanceRjVQ1cj1Q0cj0 Conservation of massddt VcjdV=Q0cj0 Q1cj1+ VRjdV( )2 / 152 General Mole Balanceddt VcjdV=Q0cj0 Q1cj1+ VRjdVEquation applies to every Chemical component in the system,j= 1,2,..,ns,including inerts, which do not take place in any componentjenters and leaves the volume element only by convectionwith the inflow and outflow streams, neglecting diffusional flux through theboundary of the volume element due to a concentration diffusional flux will be considered during the development of the materialbalance for the packed-bed / 152 Rate expressionsTo solve the reactor Material Balance , we require an expression for the productionrates,RjRj= i ijriTherefore we requirerias a function ofcjThis is the subject of Chemical kinetics.

2 CHAPTER 5 Here we use common reaction-rate expressions without derivation4 / 152 The Batch ReactorRjThe batch reactor is assumed well stirredLet the entire reactor contents be the reactor volume element5 / 152 Batch Reactorddt VcjdV=Q0cj0 Q1cj1+ VRjdVBecause the reactor is well stirred, the integrals in Equation are simple toevaluate, VRcjdV=cjVR VRRjdV=RjVRThe inflow and outflow stream flowrates are zero,Q0=Q1= (cjVR)dt=RjVR( )6 / 152 Reactor VolumeEquation applies whether the reactor volume is constant or changes during the reactor volume is constant (liquid-phase reactions)dcjdt=Rj( )Use Equation rather than Equation if the reactor volume changessignificantly during the course of the / 152 Analytical Solutions for Simple Rate LawsIn general the Material Balance must be solved the reactor is isothermal, we have few components, the rate expressions aresimple, then analytical solutions of the Material Balance are next examine derive analytical solutions for some classic / 152 First-order, irreversibleConsider the first-order, irreversible reactionAk B,r=kcAThe Material Balance for a constant-volume reactor givesdcAdt= kcA( )Watch the sign!

3 9 / 152 First-order, irreversibleWe denote the initial concentration of A ascA0,cA(t) =cA0,t= 0 The solution to the differential equation with this boundary condition iscA=cA0e kt( )10 / 152 First-order, 1k= 2k= 5cAcA0t11 / 152 First-order, irreversibleThe A concentration decreases exponentially from its initial value to zero withincreasing rate constant determines the shape of this exponential decrease. RearrangingEquation givesln(cA/cA0) = kt12 / 152 First-order, 1k= 2k= 5cAcA0tOne can get an approximate value of the rate constant from the slope of thestraight procedure is a poor way to determine a rate constant and should be viewedonly as a rough approximation ( CHAPTER 9).13 / 152 First-order, irreversibleThe B concentration is determined from the A the Material Balance for component B,dcBdt=RB=kcA( )with the initial condition for B,cB(0) =cB02 Note that the sum ofcAandcBis (cA+cB)dt=RA+RB= 0 Therefore,cA+cBis a / 152 First-order, reversibleThe value is known att= 0,cA+cB=cA0+cB0So we have an expression forcBcB=cA0+cB0 cAcB=cB0+cA0(1 e kt)( )15 / 152 First-order, reversibleConsider now the same first-order reaction, but assume it is reversibleAk1 k 1 BThe reaction rate isr=k1cA k Material balances for A and B are nowdcAdt= r= k1cA+k 1cBcA(0) =c A0dcBdt=r=k1cA k 1cBcB(0) =cB0 Notice thatcA+cB=cA0+cB0remains constante16 / 152 First-order, reversibleEliminatecBin the Material Balance for A givesdcAdt= k1cA+k 1(cA0+cB0 cA)( )

4 How do we want to solve this one?Particular solution and homogeneous solution (see text)Laplace transforms (control course)Separation!17 / 152 First-order, reversibledcAdt=acA+b cAcA0dcAacA+b= t0dt1aln(acA+b) cAcA0=tcA=cA0eat ba(1 eat)Substitutea= (k1+k 1),b=k 1(cA0+cB0)18 / 152 First-order, reversiblecA=cA0e (k1+k 1)t+k 1k1+k 1(cA0+cB0)[1 e (k1+k 1)t]( )The B concentration can be determined by switching the roles of A and B andk1andk 1in Reaction , yieldingcB=cB0e (k1+k 1)t+k1k1+k 1(cA0+cB0)[1 e (k1+k 1)t]( )19 / 152 First-order, (t)cA(t)cAscBsctFigure : First-order, reversible kinetics in a batch reactor,k1= 1,k 1= ,cA0= 1,cB0= / 152 Nonzero steady stateFor the reversible reaction, the concentration of A does not go to the limitt in Equation givescAs=k 1k1+k 1(cA0+cB0)in whichcAsis the steady-state concentration of / 152 Nonzero steady stateDefiningK1=k1/k 1allows us to rewrite this ascAs=11 +K1(cA0+cB0)Performing the same calculation forcBgivescBs=K11 +K1(cA0+cB0)22 / 152 Second-order, irreversibleConsider the irreversible reactionAk Bin which the rate expression is second order,r= Material Balance and initial condition aredcAdt= kc2A,cA(0) =cA0( )Our first nonlinear differential / 152 Second-order, irreversibleSeparation works heredcAc2A= kdt cAcA0dcAc2A= k t0dt1cA0 1cA= ktSolving forcAgivescA=(1cA0+kt) 1( )

5 Check that this solution satisfies the differential equation and initial condition24 / 152 Second-order, ordersecond ordercAcA0 The second-order reaction decaysmore slowlyto zero than the first-order / 152 Another second-order, irreversibleA+Bk Cr=kcAcBThe Material Balance for components A and B aredcAdt= r= kcAcBdcBdt= r= kcAcBSubtract B s Material Balance from A s to obtaind(cA cB)dt= 026 / 152 Another second-order, irreversibleTherefore,cA cBis constant, andcB=cA cA0+cB0( )Substituting this expression into the Material Balance for A yieldsdcAdt= kcA(cA cA0+cB0)This equation also is separable and can be integrated to give (you should workthrough these steps),cA= (cA0 cB0)[1 cB0cA0e(cB0 cA0)kt] 1,cA06=cB0( )27 / 152 Another second-order, irreversibleComponent B can be computed from Equation , or by switching the roles of Aand B in Reaction , givingcB= (cB0 cA0)[1 cA0cB0e(cA0 cB0)kt] 1 What about component C?

6 C s Material Balance isdcCdt=kcAcBand therefore,d(cA+cC)/dt= 0. The concentration of C is given bycC=cA0 cA+cC028 / 152 Another second-order, irreversibleNotice that ifcA0>cB0(Excess A), the steady statecAs=cA0 cB0cBs= 0cCs=cB0+cC0 ForcB0>cA0(Excess B), the steady state iscAs= 0cBs=cB0 cA0cCs=cA0+cC029 / 152nth-order, irreversibleThenth-order rate expressionr= 3211/20cA30 / 152nth-order, irreversibleAk Br=kcnAdcAdt= r= kcnAThis equation also is separable and can be rearranged todcAcnA= kdtPerforming the integration and solving forcAgivescA=[c n+1A0+ (n 1)kt]1 n+1,n6= 131 / 152nth-order, irreversibleWe can divide both sides bycA0to obtaincAcA0= [1 + (n 1)k0t]1 n+1,n6= 1( )in whichk0=kcn 1A0has units of inverse / 152nth-order, 5cAcA0tThe larger the value ofn, the more slowly the A concentration approaches zero atlarge / 152nth-order, irreversibleExercise care forn<1,cAreaches zero infinite 211/20 1/2 1 2t34 / 152 Negative order, inhibitionForn<0, the rate decreases with increasing reactant concentration.

7 The reactantinhibits the 1/2 1n= 2cA35 / 152 Negative order, inhibitionInhibition reactions are not uncommon, but watch out for small the rate becomes unbounded ascAapproaches zero, which is not using an ODE solver we may modify the right-hand sides of the materialbalancedcAdt={ kcnA,cA>00,cA= 0 Examine the solution carefully if the concentration reaches / 152 Two reactions in seriesConsider the following two irreversible reactions,Ak1 BBk2 CReactant A decomposes to form an intermediate B that can further react to form afinal product the reaction rates be given by simple first-order rate expressions in thecorresponding reactants,r1=k1cAr2=k2cB37 / 152 Two reactions in seriesThe Material balances for the three components aredcAdt=RA= r1= k1cAdcBdt=RB=r1 r2=k1cA k2cBdcCdt=RC=r2=k2cBThe Material Balance for component A can be solved immediately to givecA=cA0e k1tas / 152 Two reactions in series BThe Material Balance for B becomesdcBdt+k2cB=k1cA0e k1tOops, not separable, now what?}

8 Either Laplace transform or particular solution, homogeneous equation approachproducescB=cB0e k2t+cA0k1k2 k1[e k1t e k2t],k16=k2( )39 / 152 Two reactions in series CTo determine the C concentration, notice from the Material balances thatd(cA+cB+cC)/dt= 0. Therefore,cCiscC=cA0+cB0+cC0 cA cB40 / 152 Two reactions in (t)cB(t)cC(t)ctFigure : Two first-order reactions in series in a batch reactor,cA0= 1,cB0=cC0= 0,k1= 2,k2= / 152 Two reactions in parallelConsider next two parallel reactions of A to two different products, B and C,Ak1 BAk2 CAssume the rates of the two irreversible reactions are given byr1=k1cAandr2= / 152 Two reactions in parallelThe Material balances for the components aredcAdt=RA= r1 r2= k1cA k2cAdcBdt=RB=r1=k1cAdcCdt=RC=r2=k2cA43 / 152 Two reactions in parallelThe Material Balance for A can be solved directly to givecA=cA0e (k1+k2)t( )SubstitutingcA(t) into B s Material Balance givesdcBdt=k1cA0e (k1+k2)tThis equation is now separable and can be integrated directly to givecB=cB0+cA0k1k1+k2(1 e (k1+k2)t)( )

9 44 / 152 Two reactions in parallelFinally, component C can be determined from the condition thatcA+cB+cCisconstant or by switching the roles of B and C, andk1andk2in Equation ,cC=cC0+cA0k2k1+k2(1 e (k1+k2)t)( )45 / 152 Two reactions in (t)cB(t)cC(t)ctFigure : Two first-order reactions in parallel in a batch reactor,cA0= 1,cB0=cC0= 0,k1= 1,k2= / 152 Two reactions in parallelNotice that the two parallel reactionscompetefor the same reactant, AThe rate constants determine which product is favoredLarge values ofk1/k2favor the formation of component B compared to C and viceversa47 / 152 Conversion, Yield, SelectivityThere are several ways to define selectivity, yield and conversion, so be clear about thedefinition you selectivity: The point (or instantaneous) selectivity is the ratio of the production rate ofone component to the production rate of another selectivity: The overall selectivity is the ratio of the amount of one component producedto the amount of another component : The yield of componentjis the fraction of a reactant that is converted : Conversion is normally defined to be the fraction of a component that has beenconverted to products by the reaction network.

10 Conversion has severaldefinitions and conventions. It is best to state the definition in the context ofthe problem being / 152 The Continuous-Stirred-Tank Reactor (CSTR)QcjQfcjfRjWriting the Material Balance for this reactor givesd(cjVR)dt=Qfcjf Qcj+RjVR,j= 1,..,ns( )49 / 152 CSTR Constant DensityIf the reactor volume is constant and the volumetric flowrates of the inflow andoutflow streams are the same, Equation reduces todcjdt=1 (cjf cj) +Rj( )The parameter =VR/Qfis called themean residence timeof the refer to this Balance as the constant-density case. It is often a goodapproximation for liquid-phase / 152 CSTR Steady StateThe steady state of the CSTR is described by setting the time derivative inEquation to zero,0 =Qfcjf Qcj+RjVR( )Conversion of reactantjis defined for a steady-state CSTR as followsxj=Qfcjf QcjQfcjf(steady state)( )One can divide Equation through byVRto obtain for the constant-density casecj=cjf+Rj (steady state, constant density)( )