196
J. Prakash et al.
∂ p
∂ y
= 0 ,
(31)
1
Pr
+ Rn
∂
2
θ
∂ y 2 + N b
∂σ
∂ y
∂θ
∂ y
+ N t
∂θ
∂ y
2
+ βθ = 0,
(32)
∂
2
σ
∂ y 2 +
N t
N b
∂
2
θ
∂ y 2 = 0,
(33)
in which
S xx = (λ 1 + μ 1 )
∂
2
ψ
∂ y 2 S xy , S yy = (−λ 1 + μ 1 )
∂
2
ψ
∂ y 2 S xy ,
S xy =
∂
2
ψ
∂ y 2
1 + ξ
∂
2
ψ
∂ y 2
2 −1
,
(34)
Employing Eqs. (30) and (34), we have
∂ p
∂ x
=
∂
3
ψ
∂ y 3 − ξ
∂
∂ y
∂
2
ψ
∂ y 2
3
+ U hs
∂
2
∂ y 2 + Grθ + Brσ,
(35)
Equations (35) and (31) yield,
∂
4
ψ
∂ y 4 − ξ
∂
2
∂ y 2
∂
2
ψ
∂ y 2
3
+ U hs
∂
3
∂ y 3 + Gr
∂θ
∂ y
+ Br
∂σ
∂ y
= 0.
(36)
Employing Debye–Hückel linearization the nondimensional Poisson-Boltzmann
equation can also be deduced in Eq. (13) as follows:
∂
2
∂η 2 = κ
2
.
(37)
The instantaneous volume flow rate in the fixed frame is given by
Q(X, t) =
H 2 (X,t)
H 1 (X,t)
U (X, Y, t)dY .
(38)
The above expression in wave frame becomes,
F(x) =
h 2 (x)
h 1 (x)
u (x, y)dy .
(39)