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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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
• 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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
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
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 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
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 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
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 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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
• 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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
Polymath report:
Number of rabbits at 500 days
Number of foxes at 500 days
L2b-17
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
Polymath graph:
L2b-18
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.
• 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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.
Number of rabbits at 800 days
Number of foxes at 800 days
L2b-20
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.