Page 325 - Numerical Methods for Chemical Engineering

314 6 Boundary value problems

the exponential functions, k b T/q e must have the units of electric potential. Therefore, from
dimensional analysis, the characteristic decay length of ϕ(z) satisfies
1/2
q e N av (2c NaCl ) k b T/q e ε 0 ε r k b T
= ⇒ λ = (6.238)
2
ε 0 ε r λ 2 q N av (2c NaCl )
e
We have placed an additional factor of 2 here to agree with the usual definition of λ as
the Debye screening length. Note that in this equation, c NaCl is in units of moles per cubic
meter. We next define the dimensionless quantities
q e z
ϕ = ξ = (6.239)
k b T λ
so that the BVP in dimensionless form is
2
d ϕ 1 −ϕ(ξ) +ϕ(ξ)
− = e − e
dξ 2 2
(6.240)
BC 1 (surface) ϕ(0) = q e 0 /(k b T )
BC 2 (far-field) ϕ(∞) = 0
Make a plot of Debye length (in meters) vs. [NaCl] in moles at room temperature in the
−6
range of 10 –1 M. Does it make sense to have a Debye length less than 10 −10 m? Solve
this BVP numerically using finite differences, and plot ϕ(ξ) for various values of ϕ(0).
Show that when | 0 | (k b T/q e ), the solution is a simple exponential decay. For a further
discussion of screening in ionic solutions, consult Stokes & Evans (1997).
6.B.4. In problem 6.B.3, we assumed a constant surface potential; however, we can modify
the problem to impose a known surface charge density σ 0 . Electroneutrality of the surface
and the neighboring salt solution requires
∞ ∞
d d
' ' 2 ∞
σ 0 =− ρ(z)dz = ε 0 ε r 2 dz = ε 0 ε r (6.241)
0 0 dz dz 0
In the second equality we have used the Poisson equation. Since as z →∞, (z) becomes
uniformly zero, the derivative at z =∞ is zero and the surface charge density is simply
related to the derivative value at the surface:

d
σ 0 =−ε 0 ε r (6.242)
dz
0
Thus, we need only modify the boundary condition at the surface to move from a specified
surface potential value to a specified charge density. Modify your program from problem
6.B.3 to plot the dimensionless solution as a function of dimensionless charge density.
6.B.5. We focus in this text on problems of direct interest to chemical engineers; however,
the solution of BVPs is important in other fields as well. In finance, a common means to
assign a value to a derivative is to solve the Black–Scholes equation. Let S(t) be the spot
price of some asset (e.g. a stock) at time t. We are considering purchasing a European call
option that gives us the opportunity to buy the asset at specified future time T > t for an
exercise price E, yielding a payoff if the future price S(T )isabove E of

payoff = max{S(T ) − E, 0} (6.243)