1/r plots: a brief reminder

Transcription

1 L10-1 1/r plots: a brief reminder 1/r X target X

2 L10-2 1/r plots: a brief reminder 1/r X target X

3 L10-3 1/r plots: a brief reminder 1/r X target X

4 Special Case: utocatalytic Reactions Let s assume a reaction + B -> 2 B r = k C C B Stoichiometry requires that C B = C B0 + C 0 C, hence: r = k C (C B0 + C 0 C ) or, with no B present in the feed (C B0 = 0): r = k C (C 0 C ) r This is called an autocatalytic reaction. It is characterized by a positive feed-back loop, here given by B! X r -1 X eq X L10-4 Which area is larger?

5 utocatalytic Reactions II Let s calculate the reactor volume (or rather: the residence time) PFR: τ = Partial fractions! C dc kc C C C ( + ) C 0 B0 0 C C How can we integrate this? B 0 0 C B C dc α β = + kc ( C + C C ) kc C + C C dc L10-5

6 utocatalytic Reactions III Let s calculate the reactor volume (or rather: the residence time) PFR: CSTR: C dc τ = = ( + ) C 0 B0 0 kc C C C utocatalytic reaction needs seed concentration to start! τ C C rc (, T) 0 = = Much better! Why no problem here?! Example: Fermentation: yeast cells produce enzymes which catalyze the break-down of sugar, feeding the yeast cells increasing the yeast population and thus producing more enzymes etc. High ethanol concentrations (a toxic by-product of the process) stop the autocatalysis. L10-6

7 Variable Density Reactors L10-7 So far, we have been assuming constant density conditions in all our derivations. While this is an instructive (since simple) case, it is in reality only a good assumption for reactions with no change in mol number or highly diluted reaction mixtures. What if this is not the case? For a reaction with an increase in the number of moles: + B -> > 4 P the volumetric flow rate (CSTR): the linear gas velocity (PFR): will double over the course of the reaction! Obviously, this has to be taken into account in the mass balance! The trick to avoid these problems is to write the mass balance equation in terms of conversion X j rather than the (density dependent!) concentrations C j. v = u =

8 Variable Density CSTR C j V Hence: mass balance: t steady state: dn dt j = F F + V ν r j0 j j 0 = j0 j + ν j F F V r Since the molar flow of j now changes with conversion, we can write for reactant : V CSTR = 0 = 0 + ν = = F F V r Which also yields immediately τ from V = vτ. (This is in fact the alternative form we had written earlier, not knowing what treasure was hidden in there!) L10-8

9 Variable Density PFR z dv z + dz Mass balance for infinitesimal control volume dv! F j (z) F j (z+dz) + ν j r dv = dn j /dt =0 -df j /dz + ν j r = 0 Taking again into account the dependence of F (X) : df = F 0 dx -df /dz + ν r = - F 0 dx /dz + ν r(x ) = 0 Division by (ν ) yields (with dv = dz): F 0 dx /dv = - r(x ) and hence the reactor volume for the variable density PFR is: V PFR = L10-9

10 Space Velocity To circumvent the problems with the (typically unknown) dependence of density on the axial position (i.e. conversion) in the PFR, a new quantity is introduced, which is based only on (known) inlet conditions: space velocity (You will also sometimes find the reciprocal of this number, called the space time ) (The SV can be intuitively regarded as the number of reactor volumes of reactants processed per unit time at inlet conditions.) Common designations of these values in industry are the gas hourly space velocity (GHSV) and the liquid hourly space velocity (LHSV), both given in [h -1 ]. GHSV ( or LHSV ) = SV = volume of gaseous ( or liquid) feed processed per hour reactor volume L10-10 Careful: Careful: Sometimes Sometimes different different reference reference state state used! used! Typically Typically either either inlet inlet conditions conditions or or standard standard conditions conditions (25 (25 C 11 atm) atm)

11 Space Time: Typical Values L Typically: Low T < > < < > T decrease: gas gas -> -> Space time vs (nominal) reactor T for some major industrial processes (L.D. Schmidt)

12 Variable Density: Example Find the reactor volume required to achieve 90% conversion in a first order reaction of the type -> 2 B, assuming the reactants are ideal gases and using the following values: C 0 = 2 mol/l, k= 0.5 min -1, V 0 = 4 l/min. Compare a CSTR with a PFR. Start by writing N j, V, C j and r in terms of conversion X : N 0 (1 X ) 2 N 0 X N tot = N 0 (1 + X ) => V = V 0 (1+X ) [N 0 (1 - X )] / [V 0 (1+X )] = C 0 (1-X )/(1+X ) r = k C = k C 0 (1-X )/(1+X ) Inserting this into the CSTR and PFR design equations: X X = = = ν ( ) 1 rx X kc0 1+ X CSTR: VCSTR F0 F0 PFR: N = N B = V PFR C = N /V = dx = F0 = ν r( X ) L

13 Variable Density, Cont d Comparison: respective reactor volumes for CSTR and PFR for previous example, for -> ν B B: ν B V CSTR V PFR V PFR always ν B large -> When may one assume constant density? no change in number of moles during reaction liquid phase reaction (mass density does not change!) (strongly) diluted gases -> B has diluting effect on r(c )! effect for CSTR much more pronounced -> operates at maximum dilution! L estimations

14 Transients in Continuous Reactors L Not only the semi-batch reactor shows a time-dependent behavior! While continuous reactors are usually operated in steady-state, transients are in practice also very important. Typical transient states in industrial reactors occur due to: start-up of the reactor shut-down of the reactor (intentional or unintentional) changes in operating conditions such as changes in: - feed composition - flow rate - reactor temperature It is therefore important to be able to describe these transients. (lso, these transients tend to be the most accident-prone phases during reactor operation!!)

15 Transients in the CSTR To describe transients, we again take a mass balance over the reactor, this time keeping the time-dependent accumulation term: dn j dt = nd for constant flow rate and constant density, we obtain again: For a simple 1 st order reaction with r = k C, this yields: dc τ dt dc j τ = dt ( ) = C C τ kc C () t = 0 Test: Check the limiting cases of C (t=0) and C (t-> )! constant density, constant flow rate transient CSTR equation L

16 Transients in the PFR s before, we take a mass balance over the reactor keeping the timedependent accumulation term: F j (z) F j (z+dz) + ν j r dv = dn j /dt constant density! -1/ F j / z + ν j r = C j / t and with F j = V C j = u C j : The transient PFR is thus described by a partial different equation (PDE) or rather a set of j PDEs which need to be solved numerically. Test: What equation do we obtain if we set the first derivative term = 0? What if we do the same for the second derivative? L