lecture 2 2012

8/2/2019 Lecture 2 2012

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Introduction to Chemical ReactionEngineering

Lecture 2

General Mole Balance for Ideal Reactors

8/2/2019 Lecture 2 2012


Lecture 2 Plan

• General mole balance equation for ideal reactors

• Assumptions used in ideal reactors

• Design equations for ideal reactors

Learning outcomes:

• Describe the assumptions used in ideal reactors

• Derive the general mole balance equation • Apply the general mole balance equation to the 3 most common

types of reactor

8/2/2019 Lecture 2 2012


General Mole Balance for IdealReactors

In - Out + Generation = Accumulation

=

+

−

e)(moles/tim

jof onaccumulati

of Rate

e)(moles/tim

reaction

chemicalby jof

generation

of Rate

e)(moles/tim

out jof flow

of Rate

e)(moles/tim

in jof flow

of Rate

n j0 - n j + G j =

dt dN j

n j is the number of moles of species j

8/2/2019 Lecture 2 2012


If all the system variables (e.g. temperature, concentration) are spatially uniformthroughout the volume then G is the product of reaction volume, V , and the rate offormation of j, r j :

G j = r j V

volume.

etime.volum

moles

time

moles=

Determine the time (batch) or reactor volume (flow reactor) to convert a specifiedamount of reactants into products.

n j0 n j

V

8/2/2019 Lecture 2 2012


Reaction rate

• Reaction rate, r j : moles of j appearing (or formed)because of the reaction per unit volume of reacting mixture per unit time (mol j /m 3 s)

If j is reacting (disappearing), rate is –ve (i.e. = -r j )

If j is product (appearing), rate is +ve (i.e. = r j )

Several sign conventions – take care!

8/2/2019 Lecture 2 2012


Rate Equation (or Rate Law)

A → products

May be a linear function of concentration: -r A = kC A

A

A A

C k

C k r

2

1

1+=−

The concentration dependence must be determined from experimental

observation

The rate equation is independent of the type of reactor (e.g. batch orcontinuous flow) in which the reaction is carried out.

or it may be some other algebraic function of concentration: -r A = kC A2

or

8/2/2019 Lecture 2 2012


Batch Reactors

V

N A moles of A

Assume:

• Well-mixed

• Often assume constant V and

constant P

• All reactants in at t=0 and out at t=t

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Batch Reactors

8/2/2019 Lecture 2 2012


dt

dN

V r

j

j

1=

dt

V N d j ) / (=

Apply Mole Balance:

No inflow or outflow

n j0 = n j = 0

dt

dN V r

j

j =

For constant V

dt

dC j=

8/2/2019 Lecture 2 2012


Consider A → B

t

N A0

N A

N A1

t 1 t

N B

N B1

t 1

Mole-time trajectories

dt

dN V r A

A = Rearranging:V r

dN dt

A

A=

Integrating with limits at t = 0, N A = N A0 and t = t 1, N A = N A1

∫=1

0

1

A

A

N

N A

A

V r

dN t

8/2/2019 Lecture 2 2012


Substitute the rate law into previous equation:

-r A = kC A (1st-order)

-r AV = kC AV

= k(N A /V)V

= kN A

∫=0

1

1

NA

NA A

A

kN

dN t

So:

1

0

1 ln1

A

A

N

N

k t =

8/2/2019 Lecture 2 2012


CSTR

V

C A

n A0 C A0

ν T0 (m3 /s)

n A

C A

ν T (m3 /s)

Exit – representative of contents

Assume:

• Continuous supply of feed and product removal• Well-mixed

• Steady-state – reaction rate the same everywhere and time independent.

concentration the same everywhere so exit point the same

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Design equation for CSTR

Apply the mole balance with

state)(steady0=dt

dN j

So

n j0 - n j + r j V = 0

j

j j

r

nnV

−

−=

0

This is the reactor volume required to reduce the entering flow rate, n j0 to n j when species j is disappearing at a rate, -r j .

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Molar flow rate:

n j = C j .ν ν = volumetric flow

time

volume

volume

moles

time

moles=

Combining:

j

j j

r

C C V

−

−=

ν ν 00

8/2/2019 Lecture 2 2012


Plug Flow Reactor (PFR)

reactants products

Assumes:

• Plug flow – no radial variations in velocity, concentration or temperature (‘flat’

velocity profile)

• Steady state

• Continuous supply of feed and product removed

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Consider the mole balance on j in a differential segment of reactor volume ∆V

n j0 n j n j|V n j|V+∆V

V V+∆V

∆V

Mole balance in a differential segment/volume ∆V:

In - out + generation = accumulation

n j|V - n j|V+∆V + r j∆V = 0

Divide by ∆V:

j

V jV V jr

V

nn=

∆

−∆+ ||

In the limit ∆V → 0

j

j

r dV

dn

=

8/2/2019 Lecture 2 2012


Consider A → B

n A0

V

n A n A1

V 1

What is the reactor volume V 1 necessary to reduce n A0 to nA?

A

A

r

dndV =

Integrating with limits at V = 0, then n A = n A0 and V = V 1, n A = n A1

∫∫ −==

0

1

1

0

1

A

A

A

A

n

n A

A

n

n A

A

r

dn

r

dnV

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For a 1st – order reaction, -r A = kC A

Also n A = C Aν

∫∫ ==0

1

0

1

1

A

A

A

A

n

n A

A

n

n A

A

n

dn

k kC

dnV

ν

A

A

A

A

C

C

k n

n

k V 00

1lnln ν ν ==

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Design equations for ideal reactors

j

j j

r

C C V −

−= ν ν 00

Batch reactor

Continuously stirredtank reactor (CSTR)

Plug flow reactor (PFR)

dt

dN V r

j

j =

j

j

r dndV =

Derived from mole balance

lecture 2 2012

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