Page 105 - Numerical methods for chemical engineering

Example. Steady-state modeling 91

that should be read as “A chain with m monomer units reacts reversibly with a chain
containing n monomer units, to produce a combined chain containing m+n monomer units
and a single condensate (water) molecule.” We then combine the contributions from each
individual reaction in the forward and reverse directions to calculate the net rate of change
of each species’ concentration and those of the leading moments.

Forward condensation reaction

P m + P n → P m+n + W (2.119)
The net rate of change of m-mer from forward condensation is
∞ m−1

r P m (fc) =−2k fc [P m ] [P n ] + k fc [P n ][P m−n ] (2.120)
n=1 n=1
The first term is the rate of disappearance of m-mer from reaction with all other species, and
the second term is the rate of m-mer creation through the reaction of two smaller species.
The rate of change of the kth moment is
∞
k

r λ k (fc) = m r P m (fc) (2.121)
m=1
After some mathematics, we get for the three leading moments of interest,
r λ 0 (fc) =−k fc λ 2 0 r λ 1 (fc) = 0 r λ 2 (fc) = 2k fc λ 2 1 (2.122)

Reverse condensation reaction
The reverse reaction

P m+n + W → P m + P n (2.123)
has a rate constant k fc /K eq , where K eq is the equilibrium constant. The net rate of change
of m-mer concentration is
(
∞

r P m (rc) = k fc K −1 [W] 2 [P n ] − (m − 1)[P m ] (2.124)
eq
n=m+1
The first term is the rate of production of m-mer through the scission of larger molecules, and
the second term is the rate at which m-mer disappears when one of its linkages is attacked
by water. The rates of change of the three leading moments from reverse condensation are

r λ 0 (rc) = k fc K −1 [W](λ 1 − λ 0 ) r λ 1 (rc) = 0
eq
1
−1

r λ 2 (rc) = k fc K eq [W] λ 1 − λ 3 (2.125)
3
From these results, we have the following rates of change of each leading moment due to
forward and reverse condensation:
2
=−k fc λ + k fc K −1 [W](λ 1 − λ 0 ) = 0
0 eq
r λ 0 r λ 1
2
= 2k fc λ + k fc K −1 1 (2.126)
1 eq 3
r λ 2 [W] λ 1 − λ 3