Page 291 - Numerical Methods for Chemical Engineering

280 6 Boundary value problems

At the reactor inlet and outlet we use Danckwerts’ boundary conditions. At the inlet, the
flux of species j = A, B, C, D entering the reactor is v z c j0 , but once inside the reactor and
in the presence of dispersion, the flux is v z c j − D(dc j /dz). Balancing these two fluxes at
z = 0 yields the inlet boundary condition

dc j
v z [c j (0) − c j0 ] − D = 0 (6.117)
dz
0
When D → 0, c j (0) = c j0 , but when D →∞, this boundary condition correctly enforces
dc j /dz| 0 = 0. If the reaction stops once the stream leaves the reactor, the concentration
profile becomes uniform, and we use the outlet boundary condition

dc j
= 0 (6.118)

dz
L
Solution by upwind finite differences
We solve the coupled set of PDEs (6.116) with the inlet boundary conditions (6.117)
and outlet boundary conditions (6.118) using upwind finite differencing. We place
a grid of N uniformly-spaced points 0 < z 1 < z 2 < ··· < z N < L, z k = k( z), z
= L(N + 1) −1 and write each PDE of (6.116) as
2
dc j d c j
0 =− v z + D 2 + s j [{c m (z)}] (6.119)
dz dz
where {c m (z)} denotes the set of local concentrations, and the source terms for each field
are

s A [{c m (z)}] =−k 1 c A c B s B [{c m (z)}] =−k 1 c A c B − k 2 c B c C
(6.120)
s C [{c m (z)}] = k 1 c A c B − k 2 c B c C s D [{c m (z)}] = k 2 c B c C
Applying upwind finite differences, we obtain for each grid point z k and each species
j = A, B, C, D a nonlinear algebraic equation

c j (z k ) − c j (z k−1 ) c j (z k−1 ) − 2c j (z k ) + c j (z k+1 )
0 =− v z + D 2 + s j [{c m (z k )}]
z ( z)
(6.121)
Collecting terms, we have
0 = α lo c j (z k−1 ) + α mid c j (z k ) + α hi c j (z k+1 ) + s j [{c m (z k )}] (6.122)

with the coefficients
D 2D D
v z v z
α lo = + 2 α mid =− − 2 α hi = 2 (6.123)
z ( z) z ( z) ( z)
We enforce the boundary conditions by removing the nonexistent unknowns c j (z 0 ) and
c j (z N+1 ) through the discretizations of (6.117) and (6.118),

(in) c j (z 1 ) − c j (z 0 )
0 = v z c j (z 0 ) − c − D c j (z N+1 ) = c j (z N ) (6.124)
j0
z