Mole balance and design equations for batch and continuous mode ...

General algorithm of Chemical Reaction Engineering• Mole balance• Rate laws• Stoichiometry• Energy balance• Combine and Solve

Rates of chemical reactionsA + 2B⎯⎯→ 3C+D• Usually we interested in the concentration of one particular reagent, say A.• The reaction rate in terms of reagent A is the number of moles of Areacting per unit time, per unit volume (mol·m -3·s -1 )In the case of heterogeneous reaction the rate will be defined perunit area of catalyst as mol/m 2·s or per unit mass of catalystmol/kg·s− r = d[ A]/dtAOk for closed well stirredsystem but,this definition is inconvenient and can be misleading because• the concentration of A is varying with time and position insidethe reactor,• for a continuous reactor in steady state, the concentration ata given point is constant in time

The general mole balance equation• Generally, the rate of reaction varies frompoint to point in the reactor:GiV= ∫rdVi• The general mole balance equation:Fj0− Fj + ∫ rdVi=VdNdtj• From here, design equation for differenttypes of the reactors can be developed

Types of Chemical Reactors• Depending on loading/unloading of the reactorBatchSemi-BatchContinuosFlowCSTR (Continuous-Stirred Tank Reactor)Tubular reactorPacked-bed reactor

Batch reactors• for small-scale operation;• testing new processes• manufacturing expensiveproducts• processes difficult to convert tocontinuous operation

Batch reactorsFj0− Fj + ∫ rdVi=VdNdtjV∫rdVi=dNdtj0• assuming perfect mixing, reaction rate the same through the volumedNdtj=rV• integrating the equation we can get N j vs t –“mole-time trajectory”jt1=∫NNA0A1dNA−rVAtime to reachconcentration N A1 in thereactor can be found

Batch reactorsPfaudler’s Batch reactor

Continuous Flow Reactors• CSTR (Continuous-Stirred Tank Reactor)InOutPfaudler’s CSTR reactor

Continuous Flow Reactors• CSTR (Continuous-Stirred Tank Reactor)Fj0− Fj + ∫ rdVi=VdNdtjF j0F j=0, operation in a steady mode• assuming perfect mixing, so• Reaction rate is the same through the volume• Conditions of exit stream are the same as in the reactorF − F =−rVj0j jV =Fj0−−rjFDesign equation of CSTRjorV =vC0 A0−r− vCAA

Continuous Flow Reactors• Tubular reactor• usually operates in steady state• primarly used for gas reactions• easy to maintain, no moving parts• produce highest yield• temperature could be difficult to control, hot spots might occur

Continuous Flow Reactors• Tubular reactor– Reaction continuously progresses along the length of the reactor, so theconcentration and consequently the reaction rate varies in axial direaction– in the model of Plug Flow Reactor (PFR) the velocity is considered uniform andthere are no variation of concentration (and reaction rate) in the radialdirection– If it cannot be neglected we have a model of Laminar Flow Reactor.

Continuous Flow Reactors• PFR (plug flow reactor) – useful approximation of a tubular reactorV∫F − F + rdV =j0j idNdtj• For every slice of volume:F| +Δ− F|Fj0 − Fj + riΔ V = 0 ri=ΔVj V V jVNo accumulationri=dFdVj0• From here, a volume required to produce given molar flow rate of productcan be determined

Continuos Flow Reactors• Design equation for PFRrj=dFdVjdV=dFrjjj j= ∫ =Fj0• If we know a profile of molar flow rate vs. Volume we can calculate therequired volume to produce given molar flow rate at the outlet.VFdFrj∫FFjj0dF−rjj

Continuous Flow Reactors• Packed-Bed reactor – here the reaction takes placeon the surface of catalyst• reaction rate defined per unitarea (or mass) of catalyst( mol A reacted ) /s ( g catalyst )− r′A= ⋅

Continuous Flow ReactorsW – catalyst weight coordinate• as in the PFR case, we can calculate design equation now interms of catalyst weight coordinateFFAW |FAW | Wr′| +Δ− F|−+Δ+AΔ W = 0 r′ A=ΔWAW W AWr′ =AdFdWA

Reactors Mole Balance: Summary

How we can use design equations: Example 1-1• ProblemAn isomerization reaction A->B (first order reaction, k=0.23 min -1 ) is run ina tubular reactor with a constant volumetric flow rate v 0 = 10 l/min.Derive design equation, sketch concentration profile and determine thereactor volume required to achive 10% of A at the exit.• SolutionFrom the mole balance for PFR:Reaction rate law:− r = kCAs the volumetric flow is kept constant:AAdFAAdV = rdFdVA=v0dCdVAv0dCA1Combining with the rate law:AdV =−kCAVCv0 dCA100dmk C=− = ∫CA0A3

Sizing of reactorsHere we’ll find how to find the size of a reactorif the relation between the reaction rate andconversion factor is known

Conversion in the reactorsaA + bB ⎯⎯→ cC + dD• if we are interested in species A we can define thereactant A as the basis of calculation• conversion:b c dA+ B⎯⎯→ C+Da a aXA=Moles of A reactedMoles of A fed• maximum conversion for reversible reactions is theequilibrium conversion X e .

Batch reactor design equations[ ] [ N ]Moles of A reacted =A0⋅ XA[ ] [ ] [ ]Moles of A in reactor, NA = NA0 − NA0⋅XAdNA− = ( −rA) VdtdNAdX=−NA0dt dtNAdX0= ( −rA)VdtDesign equation for Batch Reactor• the equation can be integrated to find the time necessary toachieve required conversion• the longer reactants spend in the chamber the higher is thedegree of conversion

Design equations for flow reactors[ ][ ]F X =A0[ ][ ]F X =A0Molar flow ratefed to the system[ Moles of A fed][ Moles of A reacted][ time][ Moles of A fed][ F ] −[ F ] ⋅ X = [ F ]A0 A0A• molar flow rate can be expressed asconcentration * volume rate[ ] [ ][ Moles of A reacted][ time]F = F (1 − X) = C v (1 −X)A A0 A0 0Molar flow rate of the consumptionof A in the systemMolar flow rate ofA leaving the system

Design equations for flow reactors• CSTR:[ ] [ ]F (1 )A= FA0 −XVFA0 −FA FA0⋅X= =−r−r( )A A exit• Because the reactor is perfectly mixed, the exit composition isidentical to the composition inside the reactor

Design equations for flow reactors• Tubular Flow Reactor (PFR):− r =A−dFdVA[ ] [ ]F (1 )A= FA0 −X− r =AF dXA0dVV=FA0X∫0dX−rA• to integrate we need to know r A depends on the concentration(and therefore on conversion)

Design equations for flow reactors• Packed-Bed Reactor: similar derivation, but W instead of V− r′=A−dFdWA[ ] [ ]F (1 )A= FA0 −X− r ′ =AF dXA0dWW=FA0X∫0dX−rA′• from this equation we can find weight of catalyst W required toachieve the conversion X

Accounting for the reaction rate law• As we see, to find the volume of the reactor weneed to know the 1/r A dependence on X• Let’s consider a first-order reaction:− r (1 )A= kCA = kCA0 −X1 1 1=−r kC (1 −X)AA0

Levenspiel plot• reactor volume required is always reciprocal inr A and proportional to X.PFR:V=FX∫A00dX−r• Levenspiel plot:ACSTR:VFA0⋅ X= − rA

Example (2.2, p.48)• Reaction A→B described by the data below and the species Aenter the reactor at a molar flow rate of 0.4 mol/s:• Calculate the volume necessary for 80% conversion

Example (2.2, p.48)• Solution:– Based on the table the Levenspiel plot can be constructed– The design equation for the CSTR:V3Fmol m ⋅ s= X V = 0.4 20 0.8 = 6.4m− s molA( 0r )A1exit3

Example (2.3, p.50)• Calculate based on the same data the volume of PFR:– Again, we construct the Levenspiel plot– The design equation for the PFR:F0.8A03V = ∫ dX = 2.165m0−rA1

• CSTR in seriesReactors in series–1 st reactorF − F + r V =A0 A1 A1 1A1 A0 A0 101V = F X−rF = F −F X1 A0 1–2 nd reactorF − F + r V =A1 A2 A2 2F = F −F XA2 A0 A0 20A11V = F X −X( )2 A0 1 2−rA2

Mean residence time (Space Time)• space time is defined as:τ =Vv0equal to the mean rresidence time inthe absence of dispersion

Reactor design equations: Summary

Problems• Class problem:– P1-6b (p.30): Calculate the volume of CSTR forthe conditions of example 1-1.– P2-7b (p.74)• Home problems:– P2-5b– P2-6a ”Hippopotamus stomack”http://www.engin.umich.edu/~cre/web_mod/hippo/index.htm