50
If the water content of the last node (j=nc) were equal to that of
the preceding node (j=nc-1), then no water would flow between the two
nodes. This would cause both of the last two nodes to act as one node
during the simulation. The other viewpoint would be that the sum of
the flows into the last node from the preceding node and an imaginary
node following must be zero.
DLnc DLnc
Sz (On+c+l,n nc,n) = -- l(nc-l,n Onc,n) (2-70)
If the distance between nodes nc and nc-1 is the same as the distance
between the nodes nc and the imaginary node nc+1, then it follows that
Onc+l,n = Onc-1,n (2-71)
Substitution of 2-71 into 2-65 yields as the boundary condition for the
node, j=nc, the following
(Onc,n+1- Onc,n) DLnc
dwnc dt 2 d-- (Onc-1,n-nc,n)
(2-72)
nc,n
dwnc wnc
Pwnc
The same approach produces the following as the boundary condition for
the vapor continuity and energy equations, respectively.
(Pvnc,n+l- Pvnc,n) Dvnc
dwnc dt -= 2 (Pvnc-1-Pvnc)
(2-73)
dwnc Enc,n
(S Oj,n)
(Tnc,n+1-Tnc,n) (Tnc-l,n- Tnc,n)
dwnc Cs,nc dt = 2 fnc dznc-1