L2b-1 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. L2b: Reactor Molar Balance Example

L2b-1

Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.

L2b: Reactor Molar Balance Example Problems

reactor

Fj0 Fj

Gj

Today we will use BMB to derive reactor design equations. Your

goal is to learn this process, not to memorize the equations!

L2b-2


Review: Basic Molar Balance BMB

jdN

dt

V

jr dV

System volume

Fj0 FjGj

Rate of flow of jinto the system[moles/time]

Rate of flow of j out of system[moles/time]

Rate of generation of j

by chemical rxn[moles/time]

Rate of accumulation of j

in the system[moles/time]

Fj0 Fj Gj

- + =

L2b-3


Rate of generation of reactant A in reactor due to rxn

Rate of accumulation ofreactant A in reactor =

Review: Batch Reactor Basic Mole Balance

• No material enters or leaves the reactor• In ideal reactor, composition and temperature are spatially

uniform (i.e. perfect mixing)

• No flow in or out of reactor. Fj0 and Fj = 0.

dt

dNdVr

jVj Batch Reactor

Design Equation

dt

dNVr

jj

Ideal Batch Reactor Design Equation

Ideal (perfectly mixed) reactor: spatially uniform

temp, conc, & reaction rate

L2b-4


Review: CSTR Basic Mole Balance

Accumulation = In - Out + Generation by rxn

0 = Fj0 - Fj +

Vrj

V

jdVr

No spatial variation:

0 0 0

j Cj j A A

jj A

F F C CV F V

r r

• Continuously add reactants and remove products• In an ideal reactor, composition and temperature

are spatially uniform (i.e. perfect mixing) • At steady state- no accumulation

Fj0 Fj

Ideal Steady State CSTR Design Equation:

in terms of concentration

in terms of flow

(upsilon)

L2b-5


ΔV

FA0 FA

Review: Mole Balance – PFR

jVjVVj r

V

FF

0V

lim

VrjFj0 Fj dt

dNj+- =

0VrFF jVVjVj

jj r

dV

dF Ideal SS PFR

Design Eq.

• Flow reactor operated at steady state (no accumulation per Δ)• Composition of fluid varies down length of reactor (material

balance for differential element of volume V

L2b-6


• Heterogeneous rxn: reaction occurs at catalyst particle surface • Concentration gradient of reactant and product change down

length of the reactor• Rxn rate based on the mass of catalyst W, not reactor volume V

Review: Mole Balance- Packed Bed Reactor (PBR)

jj r

dV

dF Similar to PFR, but expressed in terms of

catalyst weight instead of reactor volume

Units for the rate of a homogeneous rxn (rj) :

Units for the rate of a catalytic rxn (rj’) : catalyst kgs

mol3ms

mol

So in terms of catalyst weight instead of reactor volume:

catalyst the of weightthe is W where'rdW

dFj

j

L2b-7


Consider a reaction that occurs on a catalyst surface (a heterogeneous rxn). How is the reaction rate r’j that is in terms of the weight of catalyst related to the rate in terms of volume (rj)?

catalyst kgs

molx

ms

mol3

Hint: rj = x r’j What is x?

xmolcatalyst kgs

ms

mol3

xm

catalyst kg3

Rearrange to solve for x

bvolume catalyst weightcatalyst Bulk catalyst

density

rj = b r’j

L2b-8


Use your result from the previous question to derive a reactor design equation for a fluidized CSTR containing catalyst particles. The equation should be in terms of catalyst weight (W) and the reaction rate for an equation that uses solid catalyst. Assume perfect mixing and steady-state operation of the CSTR.

What is the CSTR design equation?

j0 j

j

F FV

r

0 jj j j

dNF F r V

dt

In Out- +Gen = Accumulation

0Rearrange to put in terms of V

L2b-9


Use your result from the previous question to derive a reactor design equation for a fluidized CSTR containing catalyst particles. The equation should be in terms of catalyst weight (W) and the reaction rate for an equation that uses solid catalyst. Assume perfect mixing and steady-state operation of the CSTR.

Step 1: Come up with an equation that relates V to W (V=?W) & substitute this equivalency into the CSTR design equation.

CSTR design equation:j0 j

j

F FV

r

Need an equation that has W instead of V and –rj’ instead of -rj

bWV

b

WV

Substitute W/ρb for V in design eq:

j0 j

b j

F FWr

Step 2: Substitute an expression that relates –rj to –rj’ into the design eq:

j0 j

b jb

F

r '

FW

Simplify:j0 j

j

F FW

r '

Ideal Fluidized CSTR Design Equation

Units for rj: Units for rj’: 3

mol

s m mol

s kg catalyst

From the previous question: rj = b r’j

j0 j

jb

F

r

FW

j0 j

j

F

rV

F

L2b-10


Use basic molar balance to derive a reactor design equation for a fluidized CSTR containing catalyst particles. The equation should be in terms of catalyst weight (W) and the reaction rate for an equation that uses solid catalyst. Assume perfect mixing and steady-state operation of the CSTR.

In Out- +Generation = AccumulationW j

j0 j jdN

F F r 'dW dt

1. Simplify this expression. Things to consider: Is there flow? Accumulation? Is the reaction rate the same everywhere in the reactor?

W jj0 j j

dNF F r 'dW

dt

0

At steady state

mol mol mol d kg mol

s s kg s dt

j0 j jF F r 'W 0 Rearrange to get in terms of W

j0 j jF F r 'W j0 j

j

F FW

r '

Ideal Fluidized CSTR Design Equation

L2b-11


The reaction A→B is to be carried out isothermally in a continuous-flow reactor. Calculate the CSTR volume to consume 99% of A (CA=0.01CA0) when the entering molar flow rate is 5 mol A/h, the volumetric flow rate is constant at 10 dm3/h and the rate is –rA=(3dm3/mol•h)CA

2.0 = 10 dm3/h =

reactorFA0=5 mol A/h FA=CA where CA = 0.01CA0

jjF C

CSTR design eq:

A A

A

0VF

r

F

A0

A02

A230

3dm mol h

0.01C

0.01 C

CV

Substitute in: –rA=(3dm3/mol•h)CA

2 & CA=0.01CA0

Factor numerator

A023 2

A0

C 1 0.01V

3dm mol h 0.01 C

0 jj j j

dNF F r V

dtIn Out- +Gen = Accumulation

0A0

A

ACV

C

r

A023

0.99V=

3dm mol h 0.01 C

We know . What is CA0?

AA0

0

0FC

3A0 310dm

molC 0

mol5.5

hh =

dm=

2

3

33

0.99V

3dm mol h 0.01 0.5

10

mo

dm

m

h

l d

3V 66,000 dm

L2b-12


The reaction A→B is to be carried out isothermally in a continuous-flow reactor. Calculate the PFR volume to consume 99% of A (CA=0.01CA0) when the entering molar flow rate is 5 mol A/h, the volumetric flow rate is constant at 10 dm3/h and the rate is –rA=(3dm3/mol•h)CA

2.0 = 10 dm3/h =

reactorFA0=5 mol A/h FA=CA where CA = 0.01CA0

jjF C

PFR design eq: AA

dFr

dV A

Ad

rd

C

V

Substitute in: –rA=(3dm3/mol•h)CA2 but

not CA=0.01CA0 until after integration!

32

AA dm

3 Cmol h

dC

dV

A0

A0 VA2

0

3C 0A

.01C dCdV

C3dm mol h

3A0A00.01

V3dm mol

1CC

1

h

b

a

b

2a

dxxx

1

3660 dm V Much smaller V required to get same conversion in a PFR than in a CSTR

3

3 31 1V

0.dm

0.10dm

01dm3 mol h5 0.5 m

hol

b bn

na a

1dx x dx

x

b

n 1

ax n 1

REVIEW:for n≠1:

n 1 n 1b an 1 n 1

3A0 310dm

molC 0

5mo= = 5

h

l h.

dm

L2b-13


The gas phase reaction A→B+C will be carried out isothermally in a 20 dm3 constant volume, well-mixed batch reactor. 20 moles of pure A is initially placed in the reactor. If the rate is –rA=kCA and k=0.865 min-1, calculate the time needed to reduce the number of moles of A in the reactor to 0.2 mol. DDDD Batch reactor

design eqA

AdN

r V dt

Need to convert to dCA/dtHow is dCA/dt related to dNA/dt?

AA AA so

NC N V

VC

AAd d

dtV

N

dtC AA

ACd d d

dt dV

Vt d

CN

t

0

A Ad d

dt tV

d

CN Plug into

design eq

AAdC

Vr Vdt

AAdC

t

rd

Plug in rate law A

AdCk

dtC

Rearrange & integrate

0

A

A

tA

A0 C

C dCk dt

C

A

A0kt ln

C

C Convert Cj

back to Nj/VA

A0

N V

Nkt ln

V A

A0

1t ln

N

k N

min

0.86t ln

0.0

5 2

2 Substitute in values for k, NA0, & NA t 5.3 min

0 jj j j

dNF F r V

dtIn Out- +Gen = Accum0 0

REVIEW:for n=1:

b

a

b

n a

1dx ln x

x ln b ln a

aln

b

L2b-14


There are initially 500 rabbits (x) and 200 foxes (y). Use Polymath to plot the number of rabbits and foxes as a function of time for a period of up to 500 days. The predator-prey relationship is given by the following coupled ODEs:

xykxkdt

dx21 ykxyk

dt

dy43

Constant for growth for rabbits k1= 0.02 day-1

Constant for death of rabbits k2=0.00004/(day∙number of foxes)Constant for growth of foxes after eating rabbits k3=0.0004/(day∙number of rabbits)Constant for death of foxes k4= 0.04 day-1

Also, what happens if k3=0.00004/day and t=800 days? Plot the number of foxes vs rabbits.

Polymath example problem 1-17

L2b-15


• Make sure the “Graph” and “Report” buttons are checked above• After typing in the 2 differential equations, conditions for t=0, constants,

and initial and final time pts, press the magenta arrow to solve

Initially 500 rabbits (x) and 200 foxes (y). Predator-prey relationship is given by the following coupled ODEs: xykxk

dt

dx21 ykxyk

dt

dy43

Constant for growth of rabbits: k1= 0.02 day-1 Constant for death of rabbits: k2=0.00004/(day∙number of foxes)Constant for growth of foxes after eating rabbits k3=0.0004/(day∙number of rabbits)Constant for death of foxes k4= 0.04 day-1

t= 0 to 500 days

L2b-16


Polymath report:

Number of rabbits at 500 days

Number of foxes at 500 days

L2b-17


Polymath graph:

L2b-18


What happens if k3=0.00004/day and t=800 days? Plot the number of foxes vs rabbits.

• Make sure the “Graph” and “Report” buttons are checked above• After changing t(f) to 800 and k3 to 0.00004, press the magenta arrow

to solve

L2b-19


Number of rabbits at 800 days

Number of foxes at 800 days

L2b-20


Documents

L2b-1 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. L2b: Reactor Molar Balance Example