L2b: Reactor Molar Balance
Example Problems
Fj0
L2b-1
Fj
Gj
reactor
Today we will use BMB to derive
reactor design equations. Your
goal is to learn this process, not
to memorize the equations!
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L2b-2
Review: Basic Molar Balance BMB
Rate of flow of j
Rate of flow of
into the system - j out of system +
[moles/time]
[moles/time]
Fj0
Rate of
Rate of
generation of j
accumulation of j
=
by chemical rxn
in the system
[moles/time]
[moles/time]
Fj
dN j
Gj
dt
V
 rjdV
Fj0
Gj
Fj
System volume
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
Review: Batch Reactor Basic Mole
Balance
L2b-3
• 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.
Rate of generation of reactant
Rate of accumulation of
=
A in reactor due to rxn
reactant A in reactor
V
 rjdV 
dN j
dt
Ideal (perfectly mixed)
dN j
reactor: spatially uniform r V 
j
temp, conc, & reaction rate
dt
Batch Reactor
Design Equation
Ideal Batch Reactor
Design Equation
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L2b-4
Review: CSTR Basic Mole Balance
• Continuously add reactants and remove products Fj0
• In an ideal reactor, composition and temperature
are spatially uniform (i.e. perfect mixing)
• At steady state- no accumulation
Accumulation =
In
- Out + Generation
by rxn
V
0
=
Fj0 Fj
+
 rjdV
Fj
No spatial variation:
rj V
Ideal Steady State CSTR Design Equation:
in terms V 
of flow
Fj 0 Fj
r j
0C A 0  C A
in terms of
F j  C j    V 
r A
concentration
 
 (upsilon)
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L2b-5
Review: Mole Balance – PFR
• 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
ΔV
FA0
FA
Fj0
Fj
V
 Fj
V  V
Fj
 rjV  0
dFj
dV
 rj
rjV
+

lim
=
Fj
V  0
dN j
V  V Fj V
V
dt
 rj
Ideal SS PFR
Design Eq.
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L2b-6
Review: Mole Balance- Packed Bed
Reactor (PBR)
• 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
dFj
dV
 rj
Similar to PFR, but expressed in terms of
catalyst weight instead of reactor volume
Units for the rate of a mol
homogeneous rxn (rj) : s  m3
Units for the rate of
mol
a catalytic rxn (rj’) : s  kg catalyst
So in terms of catalyst weight instead of reactor volume:
dFj
dW
 rj ' where W is the weight of the catalyst
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
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)?
Hint: rj = x r’j What is x?
mol

 Rearrange to

x

 solve for x
3
sm
 s  kg catalyst 
 mol  s  kg catalyst 

x




mol

 s  m3 
mol
catalyst weight
 kg catalyst 


x

 b

3
catalyst volume


m
Bulk catalyst
density
rj = b r’j
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
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? In - Out + Gen = Accumulation
dN j
Fj0  F j rj V 
Rearrange to put in terms of V
V
Fj0  Fj
dt
0
rj
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
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.
Need an equation that has
Fj0  Fj
V
W instead of V and –rj’
CSTR design equation:
rj
instead of -rj
Step 1: Come up with an equation that relates V to W (V=?W) &
substitute this equivalency into the CSTR design equation.
F  Fj
W  V  W Substitute W/ρb for V V  Fj0  Fj  W  j0
b 
b
rj
rj
b in design eq:
V
Step 2: Substitute an expression that relates –rj to –rj’ into the design eq:
mol
mol
Units for rj:
Units for rj’:
3
s  kg catalyst
sm
From the previous question: rj = b r’j
W
b

Fj0  Fj
rj

W
b

Fj0  Fj
b  rj ' 
Simplify:  W 
Fj0  Fj
rj '
Ideal Fluidized
CSTR Design
Equation
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
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 = Accumulation
dN j
W
Fj0  F j   rj ' dW 
 mol   mol   mol 
kg

 
 

 s   s   kg s 
dt
d
mol 
dt
1. Simplify this expression. Things to consider: Is there flow?
Accumulation? Is the reaction rate the same everywhere in the reactor?
dN j
W
Fj0  F j   rj ' dW 
dt
At steady state
0
 Fj0  F j rj ' W  0
 Fj0  Fj  rj ' W
Rearrange to get in terms of W

Fj0  Fj
rj '
Ideal Fluidized
 W CSTR Design
Equation
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
The reaction A→B is to be carried out isothermally in a continuous-flow L2b-11
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)CA2.
Fj  C j   0 = 10 dm3/h = 
 
reactor
FA0=5 mol A/h
FA=CA where CA = 0.01CA0
dN
j
In - Out + Gen = Accumulation Fj0  F j rj V 
dt
0
C A0  C A  Substitute in:
FA 0  F A
V
V
–rA=(3dm3/mol•h)CA2 & CA=0.01CA0
r
r
CSTR design eq:
A
A
V

C A0 1  0.01
Factor
V
2
numerator
3 dm3 mol h  0.01 C A02
C A0  0.01C A 0


3 dm3 mol h  0.01 C A02
2
0.99

mol
5
FA 0
We know .
h =0.5 mol
 V=
C
=
C

2
A0
3 dm3 mol h  0.01 C A0 What is CA0? A0 0
10 dm3 h
dm3

V


0.99 10 dm3 h

3 dm3 mol h 0.012 0.5 mol dm3 
 V  66,000 dm3
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
The reaction A→B is to be carried out isothermally in a continuous-flow L2b-12
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)CA2. CA0 = 5mol h =0.5 mol
10 dm3 h
dm3
3/h = 

=
10
dm
F  C  
0
j
 j
reactor
FA0=5 mol A/h
FA=CA where CA = 0.01CA0
3/mol•h)C 2 but
dC A
Substitute
in:
–r
=(3dm
dF
A
A
A
PFR design eq:
 rA
 rA 
not CA=0.01CA0 until after integration!
dV
dV
0.01C A0
dC A V

dC A 
dm3  2
  dV

  3

 C A 
2
3

dV
3 dm mol h CA0 C A
0
 mol h 

REVIEW: b
b
1

n1
dx   x n dx   x

n
ax
a

for n≠1:

  n  1
b
a

n1
n1
b
a

n  1 n  1
b
1

 2
x  a
ax
b dx

10 dm3 h
1
1  dm3
 1
1 

V


V




3
 3 dm
  0.01 0.5  0.5  m ol
3 dm3 mol h  0.01C A0 C A0 

mol h 




 660 dm3  V
Much smaller V required to get same conversion in a PFR
than in a CSTR
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
The gas phase reaction A→B+C will be carried out isothermally in a 20 dm3L2b-13
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
dN j
dNA Batch reactor
 rA V 
Fj0  F j rj V 
In - Out + Gen = Accum
dt design eq
dt
0 0
NA
Need to convert to dCA/dt
C

so NA  C A V
A
How is dCA/dt related to dNA/dt?
V
dNA
dC A
dV
dNA
dNA
dC A Plug into
d


V

C

  CA V 

V
A
dt
dt
dt 0
dt
dt
dt
dt design eq
 rA V  V
dC A
dC A
 rA 
dt
dt
REVIEW:
for n=1:
b
1
Plug in
dC A Rearrange


k
C

A
rate law
dt & integrate
b
b


dx

ln
x

ln
b

ln
a



ln
 n
    a

a
 
ax
dC A  kt  ln C A Convert Cj  kt  ln NA V  t  1 ln NA
 k  dt  
C A0 back to Nj/V
k NA0
NA0 V
0
CA 0 C A
min 0.2

t


ln
 t  5.3 min
Substitute in values for k, NA0, & NA
0.865 20
t
CA
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L2b-14
Polymath example problem 1-17
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:
dx
 k1x  k 2 xy
dt
dy
 k 3 xy  k 4 y
dt
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.
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
Initially 500 rabbits (x) and 200 foxes (y). Predator-prey relationship is given by the
dx
dy
following coupled ODEs:
 k1x  k 2 xy
 k 3 xy  k 4 y
dt
L2b-15
dt
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
• 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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L2b-16
Polymath report:
Number of rabbits at
500 days
Number of foxes at
500 days
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L2b-17
Polymath graph:
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
What happens if k3=0.00004/day and t=800
days? Plot the number of foxes vs rabbits.
L2b-18
• 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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L2b-19
Number of rabbits
at 800 days
Number of foxes at
800 days
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
L2b-20
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.