Lecture 12 f17

Lecture 12—Initial Value Problems
An example of a reactor design.
reactor_eq.m
hexene(X) −−−− 3 H2(H) + Benzene(B)
s = [−1 3 1]
Keq =
C3
H CB
CX
A chemical reaction in a process flow diagram is more than just a ‘‘black
box’’.
To handle the material and energy balances, you are typically told to
assume a specified a constant conversion, or that a reversible reaction
proceeds to equilibrium.
What actually happens inside of a chemical reactor depends on:
The reactor type (batch, CSTR, semi-batch, plug flow, fluidized bed, etc…)
The size/dimensions of the reactor
The phase of the reactants (vapor vs. liquid)
Temperature, pressure, heat of reaction, etc…
Che 310 | Chapra 22 | Initial Value Problems 12 — Initial Value Problems Novenber 14, 2017 1 / 7

Lecture 12—Initial Value Problems
An example of a reactor design.
reactor_eq.m
hexene(X) −−−− 3 H2(H) + Benzene(B)
s = [−1 3 1]
Keq =
C3
H CB
CX
This particular reaction is one example of an industrial process that is used to
produce hydrogen (H) and benzene (B) from hexene (H).
Thermodynamics actually favor the reverse reaction, with Keq 1
The trick that is used to force the forward reaction to good conversion is
based on the reactor design.

The Inert Membrane Reactor With Catalyst Pellets on the Feed
Side (IMRCF)
H2
Hydrogen
Hexene Hexene, Benzene
x = 0 x = L
Hexene enters the inner tube, which is packed with a catalyst that promotes the
reaction.
As hydrogen forms, it escapes through the tube wall, which is a membrane that is
impermeable to benzene.
This keeps the H2 concentration CH small, shifting the equilibrium towards the
continued production of H2 and C6H6.

The Inert Membrane Reactor With Catalyst Pellets on the Feed
Side (IMRCF)
H2
Hydrogen
Hexene Hexene, Benzene
x = 0 x = L
The concentrations of each reactant/product change continuously throughout the
length of the reactor.
The challenge is to calculate what these concentrations are, as a function of T, P, the
reactor geometry, and the inlet concentration of hexene (X), CX,0.

Modeling the IMRCF (with shell balances)
x = 0 x = L
CX,0, mol
m3
v0, m3
s
nX = v0CX,0, mol
s
RH2 = kHCH(x)
V = AcL
To get started, we need to find the rates of formation/consumption of each
component:
X
kX
+
−−−→ 3 H + B (1)
3 H + B
kX
−
−−−→ X (2)
Reversible reactions can be decomposed into two separate reactions,
each with its own rate constant.

x = 0 x = L
CX,0, mol
m3
v0, m3
s
nX = v0CX,0, mol
s
RH2 = kHCH(x)
V = AcL
The rate of hexene (X) consumption
volume
from the forward reaction is:
−r+
X = k+
X CX (consumption of X)
The rate of hexene (X) production
volume
from the reverse reaction is:
r−
X = k−
X C3
HCB (production of X)
The total rate of hexene (X) consumption
volume
is:
−rX = k−
X CX − k−
X C3
HCB (net consumption of X)

x = 0 x = L
CX,0, mol
m3
v0, m3
s
nX = v0CX,0, mol
s
RH2 = kHCH(x)
V = AcL
The rate of hexene (X) consumption
volume
from the forward reaction is:
−r+
X = k+
X CX (consumption of X)
The rate of hexene (X) production
volume
from the reverse reaction is:
r−
X = k−
X C3
HCB (production of X)
The total rate of hexene (X) consumption
volume
is:
−rX = k+
X CX −
C3
HCB
Keq
(net consumption of X)

x = 0 x = L
CX,0, mol
m3
v0, m3
s
nX = v0CX,0, mol
s
RH2 = kHCH(x)
V = AcL
∆x L
The concentrations all vary as a function of position x. Over a short span ∆x,
we can assume the concentration to be constant.
The volume of the reactants in this region is ∆V = Ac∆x
v
The rate of hexene generation in this region is then:
∆nX = rX ∆V

x = 0 x = L
CX,0, mol
m3
v0, m3
s
nX = v0CX,0, mol
s
RH2 = kHCH(x)
V = AcL
∆x L
rX ∆V
The concentrations all vary as a function of position x. Over a short span ∆x,
we can assume the concentration to be constant.
The volume of the reactants in this region is ∆V = Ac∆x
v
The rate of hexene generation in this region is then:
∆nX = rX ∆V

x = 0 x = L
CX,0, mol
m3
v0, m3
s
nX = v0CX,0, mol
s
RH2 = kHCH(x)
V = AcL
∆x L
rX ∆V
nX (x)
Hexene flows into this region at a molar flow rate of nX (x)

x = 0 x = L
CX,0, mol
m3
v0, m3
s
nX = v0CX,0, mol
s
RH2 = kHCH(x)
V = AcL
∆x L
rX ∆V
nX (x) nX (x + ∆x)
Hexene flows out of this region at a molar flow rate of nX (x + ∆x)

x = 0 x = L
CX,0, mol
m3
v0, m3
s
nX = v0CX,0, mol
s
RH2 = kHCH(x)
V = AcL
∆x L
rX ∆V
nX (x) nX (x + ∆x)
The total mole balance on this region is:
IN
by flow
−
OUT
by flow
+ Generation = 0

x = 0 x = L
CX,0, mol
m3
v0, m3
s
nX = v0CX,0, mol
s
RH2 = kHCH(x)
V = AcL
∆x L
rX ∆V
nX (x) nX (x + ∆x)
The total mole balance on this region is:
nX (x) − nX (x + ∆x) + rX ∆V = 0

x = 0 x = L
CX,0, mol
m3
v0, m3
s
nX = v0CX,0, mol
s
RH2 = kHCH(x)
V = AcL
∆x L
rX ∆V
nX (x) nX (x + ∆x)
This becomes
nX (x + ∆x) − nX (x)
∆x
= AcrX

x = 0 x = L
CX,0, mol
m3
v0, m3
s
nX = v0CX,0, mol
s
RH2 = kHCH(x)
V = AcL
∆x L
rX ∆V
nX (x) nX (x + ∆x)
This becomes
nX (x + ∆x) − nX (x)
∆x
= AcrX
Now take the limit as ∆x → 0
dnX
dx
= AcrX nX (x = 0) = nX0

x = 0 x = L
CX,0, mol
m3
v0, m3
s
nX = v0CX,0, mol
s
RH2 = kHCH(x)
V = AcL
∆x L
nX (x) nX (x + ∆x)
This becomes
nX (x + ∆x) − nX (x)
∆x
= AcrX
Now take the limit as ∆x → 0
dnX
dx
= AcrX nX (x = 0) = nX0
This differential equation is the first part of an initial value problem

x = 0 x = L
CX,0, mol
m3
v0, m3
s
nX = v0CX,0, mol
s
RH2 = kHCH(x)
V = AcL
∆x L
nH (x)
We also need a mole balance on the hydrogen (H). There is H2 going in.

x = 0 x = L
CX,0, mol
m3
v0, m3
s
nX = v0CX,0, mol
s
RH2 = kHCH(x)
V = AcL
∆x L
nH (x) nH (x + ∆x)
There is H2 going out by flow.

x = 0 x = L
CX,0, mol
m3
v0, m3
s
nX = v0CX,0, mol
s
RH2 = kHCH(x)
V = AcL
∆x L
nH (x) nH (x + ∆x)
RH (x) = kH CH (x)
There is H2 going out by permeation through the membrane.

x = 0 x = L
CX,0, mol
m3
v0, m3
s
nX = v0CX,0, mol
s
RH2 = kHCH(x)
V = AcL
∆x L
nH (x) nH (x + ∆x)
RH (x) = kH CH (x)
rH =
−3rX ∆V
There is H2 going out by permeation through the membrane.
There is H2 generation by the reaction:
rH = −3rX

x = 0 x = L
CX,0, mol
m3
v0, m3
s
nX = v0CX,0, mol
s
RH2 = kHCH(x)
V = AcL
∆x L
nH (x) nH (x + ∆x)
RH (x) = kH CH (x)
rH =
−3rX ∆V
The total H2 mole balance is:
IN
by flow
−
OUT
by flow
−
OUT
by diffusion
+ Generation = 0

x = 0 x = L
CX,0, mol
m3
v0, m3
s
nX = v0CX,0, mol
s
RH2 = kHCH(x)
V = AcL
∆x L
nH (x) nH (x + ∆x)
RH (x) = kH CH (x)
rH =
−3rX ∆V
IN
by flow
−
OUT
by flow
−
OUT
by diffusion
+ Generation = 0
This is:
nH(x) − nH(x + ∆x) − kHCH∆V − 3rX ∆V = 0

x = 0 x = L
CX,0, mol
m3
v0, m3
s
nX = v0CX,0, mol
s
RH2 = kHCH(x)
V = AcL
∆x L
nH (x) nH (x + ∆x)
RH (x) = kH CH (x)
rH =
−3rX ∆V
IN
by flow
−
OUT
by flow
−
OUT
by diffusion
+ Generation = 0
This is:
dnH
dx
= −3AcrX − kHAcCH nH(x = 0) = 0

The IMRCF: A coupled initial value problem
If we want to know the concentrations of the 3 reactants, then we have to
solve 2 initial value problems:
dnX
dx
= AcrX nX (x = 0) = nX0 (1)
dnH
dx
= −3AcrX − kHAcCH nH(x = 0) = 0 (2)

dnX
dx
= AcrX nX (x = 0) = nX0 (1)
dnH
dx
= −3AcrX − kHAcCH nH(x = 0) = 0 (2)
How do we go about this? First, we need to convert concentrations into
molar flow rates.

dnX
dx
= AcrX nX (x = 0) = nX0 (1)
dnH
dx
= −3AcrX − kHAcCH nH(x = 0) = 0 (2)
molar flow rates.
1 The volumetric flow rate in the reactor is related to the initial flow rate and
concentration:
v0 =
nX0
mol
s
CX0
mol
m3

dnX
dx
= AcrX nX (x = 0) = nX0 (1)
dnH
dx
= −3AcrX − kHAcCH nH(x = 0) = 0 (2)
molar flow rates.
concentration:
v0 =
nX0
mol
s
CX0
mol
m3
2 Since this is a gas phase reaction, the volumetric flow rate at position x is:
v(x) = v0
Current Molar Flow Rate
Initial Molar Flow Rate
=
nX + nH + nB
nX0

molar flow rates.
concentration:
v0 =
nX0
mol
s
CX0
mol
m3
v(x) = v0
= v0
nX (x) + nH(x) + nB(x)
nX0

molar flow rates.
concentration:
v0 =
nX0
mol
s
CX0
mol
m3
v(x) = v0
= v0
nX0
3 Since benzene cannot escape the membrane, we know from stoichiometry that
nB(x) = nX0 − nX (x)

molar flow rates.
concentration:
v0 =
nX0
mol
s
CX0
mol
m3
v(x) = v0
= v0
nX0
3 Since benzene cannot escape the membrane, we know from stoichiometry that
nB(x) = nX0 − nX (x)
4 This leaves us with
v(x) = v0 1 +
nH(x)
nX0

molar flow rates.
v(x) = v0 1 +
nH(x)
nX0

molar flow rates.
v(x) = v0 1 +
nH(x)
nX0
Concentrations are mol
m3 . To express the concentrations in terms of molar flow
rates:
CX =
nX
v
=
nX (x)
v0 1 + nH (x)
nX0
(1)
CH =
nH
v
=
nH(x)
v0 1 + nH (x)
nX0
(2)
CB =
nB
v
=
nX0 − nX
v0 1 + nH (x)
nX0
(3)

Now we can substitute everything into our 2 initial value problems:
dnX
dx
= AcrX nX (x = 0) = nX0 (1)
dnH
dx
= −3AcrX − kHAcCH nH(x = 0) = 0 (2)

Now we can substitute everything into our 2 initial value problems:
dnX
dx
= AcrX nX (x = 0) = nX0 (1)
dnH
dx
= −3AcrX − kHAcCH nH(x = 0) = 0 (2)
Insert the rate law (−rX = CX −
CBC3
H
Keq
) :
dnX
dx
= −Ack+
X CX −
CBC3
H
Keq
nX (x = 0) = nX0 (3)
dnH
dx
= 3Ack+
X CX −
CBC3
H
Keq
− kHAcCH nH(x = 0) = 0 (4)

Insert the rate law (−rX = CX −
CBC3
H
Keq
) :
dnX
dx
= −Ack+
X CX −
CBC3
H
Keq
nX (x = 0) = nX0 (1)
dnH
dx
= 3Ack+
X CX −
CBC3
H
Keq
− kHAcCH nH(x = 0) = 0 (2)
Now convert Ci’s into ni’s:
dnX
dx
= −
Ack+
X
v
nX −
nBn3
H
v3Keq
nX (x = 0) = nX0 (3)
dnH
dx
=
3Ack+
X
v
nX −
nBn3
H
v3Keq
−
kHAc
v
nH nH(x = 0) = 0 (4)

Now convert Ci’s into ni’s:
dnX
dx
= −
Ack+
X
v
nX −
nBn3
H
v3Keq
nX (x = 0) = nX0 (1)
dnH
dx
=
3Ack+
X
v
nX −
nBn3
H
v3Keq
−
kHAc
v
nH nH(x = 0) = 0 (2)
Now use nB = nX0 − nX :
dnX
dx
= −
Ack+
X
v
nX −
(nX0 − nX )n3
H
v3Keq
(3)
dnH
dx
=
3Ack+
X
v
nX −
(nX0 − nX )n3
H
v3Keq
−
kHAc
v
nH (4)

dnX
dx
= −
Ack+
X
v
nX −
(nX0 − nX )n3
H
v3Keq
(1)
dnH
dx
=
3Ack+
X
v
nX −
(nX0 − nX )n3
H
v3Keq
−
kHAc
v
nH (2)
These equations represent all we need to find ni(x).

dnX
dx
= −
Ack+
X
v
nX −
(nX0 − nX )n3
H
v3Keq
(1)
dnH
dx
=
3Ack+
X
v
nX −
(nX0 − nX )n3
H
v3Keq
−
kHAc
v
nH (2)
Unfortunately, there is no analytic solution. Furthermore, they are coupled,
meaning that to know nX we have to know nH, and vice versa.

We do know these quantities at x = 0, at the beginning of the reactor:
nX (x = 0) = nX0 (1)
nH(x = 0) = 0 (2)

nX (x = 0) = nX0 (1)
nH(x = 0) = 0 (2)
We are going to develop numerical methods that start by evaluating the
reaction rates at the beginning of the reactor.

nX (x = 0) = nX0 (1)
nH(x = 0) = 0 (2)
We are going to develop numerical methods that start by evaluating the
reaction rates at the beginning of the reactor.
Using these initial reaction rates, we will cut the reactor up into slices and
work our way from left to right, until we have the outlet concentrations.

Lecture 12 f17

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