Applying the quadratic formula, we find that
c
A,n
=

1 +
p
1 + 4
k τ
PFR
c
A,n

1
/n
2
k τ
PFR
/n
Hence, we obtain the relation
c
A,n
c
A,
0
=

1 +
p
1 + 4
k τ
PFR
c
A,n

1
/n
2
k τ
PFR
c
A,
0
/n
Since
k τ
PFR
c
A,
0
= 1, we arrive at the following recursion relation:
c
A,n
c
A,
0
=

n
+
p
n
2
+ 4
n c
A,n

1
/c
A,
0
2
In a PFR, we can use the
n
th order batch reactor formula from last week’s set with
n
= 2 by simply replacing
t
with
τ
PFR
. This yields
c
A
=
k τ
PFR
+
c

1
A,
0

1
=
c
A,
0
1 +
k τ
PFR
c
A,
0
= 0
.
5
c
A,
0
We then require that
1

c
A,n
c
A,
0
≥
0
.
95 (1

0
.
5) = 0
.
4750
,
or equivalently,
c
A,n
c
A,
0
≤
0
.
5250
Since we do not have a closedform solution for
c
A,n
like for the first order case, we can
find the required value of
n
using the following Matlab script:
n
=
0 ;
ratio
=
1 ;
while
ratio
>
0.5250
n
=
n
+
1 ;
ratio
=
1 ;
f o r
i
=
1 :
n
ratio
= (

n
+
sq rt
(
n
ˆ2 + 4
*
n
*
ratio
) )
/
2 ;
end
end
s p r i n t f
(
num2str
(
n
) )
This calculation yields a requirement of 7 CSTRs in series to achieve the desired con
version.
2.
(a) Writing out the reaction stoichiometry in terms of the conversion of A, we have
c
A
=
c
A,
0
(1

x
)
c
Q
=
c
Q,
0
+
c
A,
0
x
2