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Firstly, we can specify the boundary conditions at the symmetric plane z = 0 and the channel wall z = h: ∂ u ∂z z=h =0, u| =0, (10a) z=0 ∂ψ =0, ∂ψ =Es=σ, (10b) ∂z z=0 ∂z z=h ǫ where h is the inverse of internal area density (channel surface area ap per channel volume Vp), i.e., half channel depth for planar channels, σ is the surface charge density, and Es is the electrical field on the walls. In reality, σ comes from surface group dissociation, which is dependent on the local ionic composition [30]. The charged walls attract counter-ions and repel co-ions, and form the Stern layer where ions are immobile and then the diffuse layer where ions are mobile [31]. These space charge layers and the charged wall form the so called EDL. In this paper, we will fix σ for simplicity, and neglect the Stern layer. Next we will simplify the governing equations under the assumption of thin channels. Firstly, we can decompose the electric potential ψ = ψv(x, y) + φ(x, y, z), [29] where φ is the equilibrium part of potential from the Boltzmann distribution: ck = cvk exp(−zkφ ̃), (11) which is essentially the results of Jz,k = 0 and w = 0, and we assume φ ̃ = 0 in the reservoir. Note that zkcvk = 0 since ρe = 0 in the virtual reservoir. We will denote the φ at charged walls as ζ. Plug Eq.(11) into the Nernst-Plank (NP) flux (Eq.(1)), we arrive at Jk = uck − Dk ck ∇xy ln cvk + zk ψ ̃v . (12) Plug Eq.(11) into the water dissociation equation, we get cvNs+1cvNs+2 = Kw. (13) Plug Eq.(11) into the Poisson equation and apply | ∂ | ≫ | ∂ |, | ∂ |, we have the local Poisson-Boltzmann ∂z ∂x∂y (PB) equation: ∂ 2 φ N ǫ∂z2 =−ρe=− zkFcvkexp(−zkφ ̃), (14) k=1 which can be solved with the known cvk and the boundary condition Eq.(10b). It can also be rearranged to λ2d2φ ̃=−N zkcvkexp(−zφ ̃), (15) Ddz2 I k k=1 where I = 1 z2cv is the ionic strength and λD = ǫVT/FI is the Debye length and the characteristic 2kk length for EDL thickness [32, 31]. Then we can simplify the Stokes equations to μ∂2u − ∇xyp − ρe∇xyψ = 0, (16) ∂z e ∂z By substituting Eq.(14) in to Eq.(17) and integrating it, we get ∂z2 − ∂p − ρ ∂φ = 0, (17) N p−pv =VTF[exp(−zkφ ̃)−1]cvk. (18) k=1 5PDF Image | Theory of shock electrodialysis
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