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10.37 Chemical and Biological Reaction Engineering, Spring 2007 Exam 1 Review In-Out+Production=Accumulation Accumulation=0 at steady state

FA0 − FA + rAV = 0 FA0 = [ A]0 ν 0 FA = [ A]ν

For a liquid phase with constant density:

ν0 =ν

For A → B the reaction moles are the same, so For A → 2 B ,

ξ [ =]

ν0 =ν

ν0 ≠ν

moles (extent of reaction)

(-) for a reactant and (+) for a product

N rxns

N i = N i 0 + ∑ υi ,nξ n n =1

Suppose A → B + C

XA =

N A0 − N A N A0

N A = N A0 (1 − X A )

Thermodynamics Suppose

Ke = e

k1 ZZZ X A + B YZZ Z 2C k −1

− ΔG RT

ΔG = ΔG °f , products − ΔG °f ,reactants Kc =

[ B ][C ] [ A]

has units. You need to use standard states, such as 1M, to make it

dimensionless.

Enzyme Catalysis

Energy

without enzyme with enzyme S

P Reaction Progress Figure 1. Energy diagram for a reaction with and without enzyme. Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].

E + S U ES ES → E + P Pseudo steady state approximation:

d [ ES ]

=0 dt [ ES ] = f (other species)

Cell Growth # cells

N=

volume N = N0e μ t Monod kinetics:

μ=

μmax [ S ]

KS + [S ]

YA = B

ΔA ΔB

Rate Constants k (T ) = Ae − Ea

RT

Given k1 and k2, you can calculate k and a different temperature.

CSTRs V=

FA0 X A −rA Incorporates changing volumetric flow rate

If the reaction is 1st order and it consists of liquids with constant density:

XA k (1 − X A ) V volume τ= =

τ=

ν0

volumetric flow rate

τk 1+ τ k Da = τ k =Damköhler number: ratio of kinetic effect to volumetric effect or XA =

2nd order reaction:

ratio of reaction rate to dilution rate

10.37 Chemical and Biological Reaction Engineering, Spring 2007

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τ=

XA

kC A0 (1 − X A )

2

Da = τ kC A0

XA =

1 + 2 Da − 1 + 4 Da 2 Da

For constant density and 1st order reaction:

dN A dt dC A C A0 − C A + rAτ = τ dt FA0 − FA + rAV =

Let:

C Cˆ A = A C A0

t tˆ =

τ

Nondimesionalize:

dCˆ A + (1 + Da)Cˆ A = 1 dtˆ Solve given Cˆ A = 0 at tˆ = 0 1 ˆ 1 − e− (1+ Da )t Cˆ A = 1 + Da

(

)

Tanks in series: 1st order reaction

C A, n =

C A,0

(1+ Da )

n

Reactor Design Equations FA0 X A −rA dN A Batch: rAV = dt N A = V [ A] If V changes, then V must remain in the differential

CSTR:

PFR:

V=

dX A −rA = Adz FA0

PBR: Pressure drop consideration If

A( g ) → 2 B( g )

Use

ν 0 ρ0 = νρ

(conservation of mass)

10.37 Chemical and Biological Reaction Engineering, Spring 2007

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Introduce ideal gas law

Pm FT RT

ρ=

FT = FT 0 (1 + ε X )

⎛ ⎞ m  = Pν = P = FT RT ⎟ ⎜ nRT ρ ⎝ ⎠

Reactor volume: positive order reactions

VCSTR FAo − rA

VPFR (area under curve)

XA Figure 2. Levenspiel plot for a CSTR and a PFR for positive order reactions.

Selectivity CSTR

φinst Φ overall ΔC A C Af

C A0

CA

Figure 3. Fractional yield versus concentration. Overall yield times concentration difference shown for a CSTR.

CSTR k1 k2 A ⎯⎯ → P ⎯⎯ →C k3 A ⎯⎯ →U ( k1 [ A] + k3 [ A])V = FA0 − [ A]ν

Non-ideal reactors Residence time distribution E(t) 10.37 Chemical and Biological Reaction Engineering, Spring 2007

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reactor δ (t ) ⎯⎯⎯ → C (t )

E(t) must have a pulse trace

E (t ) =

C (t )

∞

∫ C (t )dt

0

∫ E (t ) = 1

∞

0

∞

tm = ∫ tE (t )dt 0

Mean residence time, tm, for an: Ideal CSTR: τ Ideal PFR: τ ∞

σ 2 = ∫ ( t − tm ) E (t )dt 2

0

Variance, σ2, for an: Ideal CSTR: τ2 Ideal PFR: 0

⎛ ⎛ V ⎜ E (t ) = δ ⎜ t − ⎝ ν ⎝

⎞⎞ ⎟⎟ ⎠⎠

∞

∫ f (t )δ (t − t )dt = f (t ) 0

0

Åproperty of a dirac delta function

0

For a CSTR, E (t ) =

e−t τ

τ

Example 1 z L FA0

FA, FB, FC

Figure 4. Schematic of a PFR with inflow of A and outflow of A, B, and C. r1 r2 A ⎯⎯ → B ⎯⎯ →C r1 = k1C A r2 = k2CB

YB =

moles of B produced moles of A in

=

FB ( L) FA0

Mole balance on B

1 dFB = rB = r1 − r2 = k1C A − k2CB Axs dz FB = CBν 0 FA0 = C A0ν 0 10.37 Chemical and Biological Reaction Engineering, Spring 2007

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ν 0 dCB

= k1C A − k2CB Axs dz ν 0 dC A = −k1C A Axs dz dC A −k1 Axs = dz CA ν0 −k A ln C A = 1 xs z + φ

ν0

Initial condition at z=0 gives:

ln C A0 = 0 + φ

⎡ −k A ⎤ C A = C A0 exp ⎢ 1 xs z ⎥ ⎣ ν0 ⎦ ⎡ −k A kA dCB k2 Axs + CB = 1 xs C A0 exp ⎢ 1 xs dz ν0 ν0 ⎣ ν0

Axs

ν0

⎤ z⎥ ⎦

z [ = ] time, call it τ

This is the time it takes for something to flow to the end of the reactor (of length z).

ν0

G = V velocity

Axs dCB + k2CB = k1C A0 e− k1τ dτ kτ Integrating factor: e 2 d ⎡⎣CB ek2τ ⎤⎦ = k1C A0 e( k2 − k1 )τ dτ kC k −k τ CB ek1τ = 1 A0 e( 2 1 ) + φ k2 − k1 Initial condition: z=0, CB=0 B

kC 0 = 1 A0 + φ k2 − k1 kC CB (τ ) = 1 A0 ⎡⎣e− k1τ − e− k2τ ⎤⎦ k2 − k1 1 dFC = + r2 Axs dz 1 dFA = −r1 Axs dz 1 dFB = r1 − r2 Axs dz 10.37 Chemical and Biological Reaction Engineering, Spring 2007

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1 ⎡ dFA dFB dFC ⎤ + + =0 Axs ⎢⎣ dz dz dz ⎥⎦ FA0 = FA + FB + FC

FC = FA0 − FA + FB F

find

A

C

B

τ∗

τ

Figure 5. Graphs of flow rates of A, B, and C as a function of residence time.

(

)

dCB k1C A0 = −k1e− k1τ * + k2 e− k2τ * = 0 dτ k2 − k1

k1e− k1τ * = k2 e− k2τ * ln k1 − k1τ * = ln k2 − k2τ * k ln 1 = ( k1 − k2 )τ * k2 k ln 1 k2 τ* = k1 − k2 dA( x) A( x) 0 dx L’Hopital’s rule: lim = do lim x → x* B ( x ) x → x* dy 0 dB ( x) dx dx Find τ* for k1=k2

1

k1 1 1 = = k1 → k2 k1 → k2 1 k2 k A L A z τ * = xs τ = xs lim τ * = lim

ν0

ν0

10.37 Chemical and Biological Reaction Engineering, Spring 2007

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L=

τ *ν 0 Axs

= length of reactor

Example 2 If A → 2 B (Assume negligible pressure drop)

r = kC A z L FA0 Figure 6. Schematic of a PFR.

dFA = −rA = − kC A dV ν0 ↔ν  PV = nRT ( Ftotal = n , ν = V ) RT −1 ν = Ftotal = Ftotal Ctotal P Ftotal = FA + FB FA = C Aν = y A Ftotal dFA FA P = − ky ACtotal = − k dV FA + FB RT dFB FA P = +2ky ACtotal = +2k dV FA + FB RT If P or T changes, you need other equations.

Derivation of E(t) for a CSTR N 0δ (t )

Figure 7. Schematic of a CSTR.

No reaction: 10.37 Chemical and Biological Reaction Engineering, Spring 2007

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dN = N 0δ (t ) −ν C dt N = N 0δ (t ) −ν V dN ν + N = N 0δ (t ) dt V

⎛ν ⎞ t⎟ ⎝V ⎠ d ⎛ ⎛ν ⎞⎞ ⎛ν ⎞ ⎜ N exp ⎜ t ⎟ ⎟ = exp ⎜ t ⎟ N 0δ (t ) dt ⎝ ⎝V ⎠⎠ ⎝V ⎠ ⎛ν ⎞ ⎛ν ⎞ N exp ⎜ t ⎟ = N 0 ⋅ exp ⎜ 0 ⎟ = N 0 + φ ⎝V ⎠ ⎝V ⎠

Integrating factor: exp ⎜

Initial condition: t=0, N=N0 Æ φ=0

⎛ ν ⎞ N = N 0 exp ⎜ − t ⎟ ⎝ V ⎠ ν 1 = V τ ⎛ −t ⎞ C = C0 exp ⎜ ⎟ ⎝τ ⎠ C E (t ) = ∞ ∫ Cdt 0

∞

∞

∫ Cdt = ∫ C0e 0

−t

τ

0

−t

dt = C0 ⎡ −τ e ⎣⎢

−t

τ

∞

⎤ = C ⎡ −τ ( 0 − 1) ⎤ = C τ 0 ⎣ 0 ⎦ ⎦⎥ 0

−t

Ce τ e τ = E (t ) = 0 τ C0τ

Long-chain approximation k2 1 ZZZ X E + S YZZ →P+ E Z ES ⎯⎯ k k

−1

tRNA ⎯⎯ →E k3

k4 E ⎯⎯ → Deactivate

Enzyme propagates a long time before it is destroyed. LCA: k3 tRNA = k 4 E

[

]

[ ]

st

(assume 1 order) If there are other steps, add them into the equation 10.37 Chemical and Biological Reaction Engineering, Spring 2007

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k5 ES ⎯⎯ → destruction k3 [tRNA] = k4 [ E ] + k5 [ ES ]

Suppose there is a production term k6 C ⎯⎯ →E

Add another term

k6 [C ] + k3 [tRNA] = k4 [ E ] + k5 [ ES ]

F dX = A0 , − rA = kC A Adz −rA This is a single differential equation in terms of X. Use for PFR with gas flow.

ρ0ν 0 = ρν

CA =

C A0 (1 − X ) T0 P

1+ ε X

P0T

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