ChE 400 - Reactive Process Engineering ChE
Multiple Reactions
L11L11-1
• We have largely considered single reactions so far in this class • How many industrially important processes involve a single reaction? • The job of a chemical engineer is therefore to design
ChE 400 - Reactive Process Engineering ChE
Multiple Reactions
L11L11-2
• We need to develop tools that will allow us to quantify how well (or how poorly) we are doing at producing a desired product • There are three concepts that we need to develop for multiple reactions: – Conversion (similar to single reactions) • Xj =
– Selectivity • Sj =
– Yield
• Yj =
ChE 400 - Reactive Process Engineering ChE
Multiple Reactions
L11L11-3
So far, we have exclusively looked at simple system with only one reaction occuring. However, many reaction systems of practical relevance involve many reactions occuring at the same time, either in parallel or in series (i.e. sequentially). Let’s look at steam cracking of ethane (~ 80 bio. to/a world production!): C2H6 -> C2H4 + H2
(I) ethene (ethylene) formation
C2H6 -> C2H2 + 2 H2
(II) acetylene formation
C2H4 -> C2H2 + H2
(III) acetylene formation from ethylene
C2H6 + H2 -> 2 CH4
(IV) methane formation
3 C2H6 -> C6H6 + 6 H2
(V) benzene formation
C2H6 -> 2 C(s) + 3 H2
(VI) coke formation
C2H2 -> 2 C(s) (s) + H2
(VII) coke formation (explosion!)
C(s) (s) + H2O -> CO + H2
(VIII) coke gasification
We distinguish between parallel reactions and series reactions.
ChE 400 - Reactive Process Engineering ChE
Conversion, Selectivity, Yield
L11L11-4
Simple example:
A -> B; A -> C (caveat :
Conversion: X j =
Selectivity:
or:
SB =
SB =
Yield:
YB = XA . SB
Production rate:
FB = YB FA0
note the formulation with mol numbers, not concentrations! -> Why?!)
preferred form !
(remark : you must be sure that you know all products!)
(and similar for C)
Typically, Typically, selectivity selectivity is is the the crucial crucial quantity! quantity!
Why?
ChE 400 - Reactive Process Engineering ChE
Selectivity: Complications…
L11L11-5
Selectivity:
SB =
N B − N B ,0
νB
⋅
νA N A − N A,0
Definition ambiguous if product formed from more than one reactant! Example: Andrussov process (HCN synthesis) (1) (2) (3) Conversions of methane and ammonia can be very different. Extent to which reaction proceeds via (2) or (3) rather than (1) will be different. Hence: SHCN,CH4 ≠ SHCN,NH3
Selectivity Selectivity needs needs aa reference reference point! point!
ChE 400 - Reactive Process Engineering ChE
Si: even more problems…
L11L11-6
Selectivity:
SB =
What if I don’t know the reaction equations (hence no νi!) ? (Often the case for very complex reactant mixtures, networks of many parallel and series reactions!) (Simplified) Example:
Methane Coupling 2 CH4 + 1/2 O2 <=> C2H6 + H2O CH4 + 2 O2 => CO2 + 2 H2O
(1) (2)
Assume equal amounts of methane react along (1) & (2): 1 mol CH4 -> 0.25 mol C2H6 + 0.5 mol CO2. If we did not know the reaction equations we might be tempted to calculate: SC= = nC= / (nC= + nCO2) = 0.25 / 0.75 = 1/3. But we reacted equal amounts of CH4 along both pathways: shouldn’t SC= = 0.5 ?!
“atom selectivity”:
S j ,i =
ChE 400 - Reactive Process Engineering ChE
Selectivity: one last slide… dN B ν A ⋅ dN A ν B
Also called “instantaneous” selectivity in batch reactors… S B′ =
…or “local” selectivity in (plug) flow reactors: S B′ =
And for once we can also use concentrations!
S B′ = ( Why ?? )
L11L11-7
S B′ =
“differential” selectivity:
dN B ν A ⋅ = dN A ν B
Example reaction: A -> B -> C
Let’s look at a reactant, which can form a desired product D, and an undesired side-product U in parallel reactions. A
U
Total Cost
Reactor Cost
Separations Cost 0
L11L11-8
Two problems: • “repair” of low S generally not possible • undesired side-product usually needs to be separated
D
Cost ($$$)
ChE 400 - Reactive Process Engineering ChE
Selectivity: Parallel Reactions
0.1
0.2
0.3
0.4
0.5
Conversion (X)
0.6
0.7
ChE 400 - Reactive Process Engineering ChE
Parallel Reactions: Selectivity II
L11L11-9
A
D
rD = kD CAd = k0,D exp{-ED/RT} CAd
U
rU = kU CAu = k0,U exp{-EU/RT} CAu
Differential Selectivity: S’ = νD rD / (νD rD + νU rU)
(S’)-1 = 1 + νU rU/νD rD
(S’)-1 ~ νD/νU rU/rD = νDU k0U/k0D exp{-(EU-ED)/RT} CA(u-d) (d-u) S’ ~ exp{-(ED-EU)/RT} CA(d-
(I) d > u
S’ increases with
• PFR (BR) > CSTR • no dilution • high pressure
(II) d < u
S’ increases with
• CSTR or PFR w/high recycle • dilution • low pressure
(III) ED < EU
S’ decreases with
(IV) ED > EU
S’ increases with
ChE 400 - Reactive Process Engineering ChE
Differential and Total S
L11L11-10
k1
Parallel reactions, different rctn orders
A
(Reactor) Selectivity:
SD =
A
k2
D
r1 = k1 CAd r2 = k2 CAu
U
Local selectivity: SD’= (or equivalently with Fj!)
Hence we calculate ND: N D =
SD
.
For V = const.
FB 1 = = − FA0 − FAe FA0 − FAe
SD = −
C A0
1 − C Ae
FAe
∫ S′
D
dFA
FA 0
C Ae
∫ S′
D
dC A
CA 0
total totalSS==integral integralaverage averageof oflocal localSS
ChE 400 - Reactive Process Engineering ChE
PFR vs CSTR
L11L11-11
SD = −
C A0
1 − C Ae
C Ae
∫ S′
D
dC A
CA0
SD’=
SD’
dd<
dd>>uu
SD’
1
1
CAe
CA0
CA
CAe
CA0
CA
If rctn order of desired reaction > rctn order of side reaction, PFR better, otherwise CSTR better
Why?
ChE 400 - Reactive Process Engineering ChE
Parallel Reaction Networks
L11L11-12
More interesting case: Network of many parallel reactions,
SD’ 1
some with higher r.o., some with lower r.o. than desired reaction
CAe
CA0
Optimum yield for CSTR first, followed by PFR (Exact shape of S’-CA curve, and hence also precise sizing of reactors is dependent on specific network!)
CA
ChE 400 - Reactive Process Engineering ChE
Conversion in Multireaction Systems
L11L11-13
Let’s again consider the simple parallel reaction system: A -> B,
r1= k1 CA
A -> C,
r2= k2 CA
How can we express the concentrations CA, CB, CC in terms of XA? We can’t – at least not directly! We have to distinguish between XA,1 and XA,2! While we still can define a XA = (NA0 – NA)/NA0 , we have to distinguish between the different pathways to do our ‘book-keeping’ (i.e. mass balances) for B and C: B:
CB= CA0 XA,1
C:
CC= CA0 XA,2
A:
CA= but also: CA=
Everything else remains unchanged – we now simply have one more quantity to keep track of, i.e. the different Xji!
ChE 400 - Reactive Process Engineering ChE
Series Reactions: Selectivity
L11L11-14
A -> B -> C Very common: partial vs total oxidation reactions (Almost) all oxidation reactions follow a sequence of successively “deeper” oxidation. Example: Ethane oxidation C2H6 + ½ O2 -> C2H4 + H2O (oxidative dehydrogenation) C2H4 + O2 -> 2 CO + 2 H2 (syngas formation) CO + H2 + O2 -> CO2 + H2O (combustion)
Let’s assume a PFR for A -> B -> C, with 1st order kinetics: rA =
, rB =
=
Plugging this into the PFR design equation, we get:
C A (τ ) = CC (τ ) =
CB (τ ) =
Check the derivation in the textbook (p. 160ff)! You should be able to derive this yourself.
So, what does this look like?
ChE 400 - Reactive Process Engineering ChE
Series Reactions: CSTR
L11L11-15
What about the same series reaction in a CSTR? (Same procedure: rate laws into design equation, substitute CB0 and CC0 by expressions in terms of CA0)
CA =
C A0
CB =
C A0 k1τ
CC =
C A0 k1k2τ 2
So, what does this look like? How is it different from a PFR? •• For Forseries seriesreactions, reactions,there thereisisalways alwaysan anoptimum optimumresidence residencetime time to toachieve achieveaamaximum maximumyield yieldtowards towardsthe theintermediate intermediateproduct product •• For Forpositive positiveorder orderkinetics, kinetics,the theoptimum optimumyield yieldininthe thePFR PFRisis always alwaysgreater greaterthan thanthat thatininaaCSTR CSTR How can we rationalize this?
How about the BR?
L11L11-16
PFR vs CST R: Series Reaction PFR vs CST R: Series Reaction
Maximum in CB -> optimum in residence time! time How can I calculate the optimum? Look for dCB/dτ = 0! Do it for the PFR!
1
1
0.8 0.8
conc.[a.u.] [a.u.] conc.
ChE 400 - Reactive Process Engineering ChE
Series Rctns - Maximum Yield
A - PFR A - PFR B - PFR B - PFR C - PFR C - PFR A - CSTR A - CSTR B - CSTR B - CSTR C - CSTR C - CSTR
0.6 0.6
0.4 0.4
0.2 0.2
0
0
0
0
2
4
2
4
6
6
time time[a.u.] [a.u.]
PFR:
CSTR:
(log mean rate constant)-1
(geometric mean rate constant)-1
What if C and not B is the desired product?
8
8
10 10
ChE 400 - Reactive Process Engineering ChE
Example:
L11L11-17
We wish to produce a product B from a reactant A in a PFR with V = 4 l/min and CA0 = 2 mol/l. However, another reaction is also occurring, forming an undesired product C. (Both reactions are irreversible, 1st order, with kB= 0.5 min-1 and kC = 0.1 min-1). (a)
Assuming a series reaction A -> B -> C, calculate the maximum achievable yield of B, as well as the necessary reactor volume.
(b) Assuming parallel reactions A -> B and A -> C, calculate the reactor volume necessary to achieve the same conversion of A as in (a). What is the yield of B in this case? Procedure: (a) Calculate τopt, from there CB(τopt) and CA(τopt). From these you obtain SB,max and XA,max. (b) Calculate τ(XA=0.865), from there: V = 13.36 l.l With τ from the equation for CB in a PFR/series reactions…
Check -2 and -3, incl. Check it it in in LDS, LDS, examples examples 44-2 and 44-3, incl. the the CSTR CSTR case! case!
