# Analysis of Unsteady Axisymmetric Squeezing Fluid Flow with Slip and No-Slip Boundaries Using OHAM

Authors: Mubashir Qayyum; Hamid Khan; M. Tariq Rahim; Inayat Ullah
Journal: Mathematical Problems in Engineering
 (2015)

---

## Abstract

In this manuscript, An unsteady axisymmetric flow of nonconducting, Newtonian fluid squeezed between two circular plates is studied with slip and no-slip boundaries. Using similarity transformation, the system of nonlinear partial differential equations is reduced to a single fourth order ordinary differential equation. The resulting boundary value problems are solved by optimal homotopy asymptotic method (OHAM) and fourth order explicit Runge-Kutta method (RK4). It is observed that the results obtained from OHAM are in good agreement with numerical results by means of residuals. Furthermore, the effects of various dimensionless parameters on the velocity profiles are investigated graphically.


---

## Body

## 1. Introduction

The squeezing of an incompressible viscous fluid between two parallel plates is an essential type of flow that is frequently observed in many hydro dynamical tools and machines. In food industries, squeezing flows have several applications particularly in chemical engineering [1, 2]. Compression and injection molding, polymer processing, and modeling of lubrication system are some practical examples of squeezing flows. The modeling and analysis of squeezing flow has started in nineteenth century and continues to receive significant attention due to its vast applications in biophysical and physical sciences. The initial work in squeezing flows has been done by Stefan [3], who developed an ad hoc asymptotic solution of Newtonian fluids. An explicit solution of the squeeze flow, considering inertial terms, has been established by Thorpe and Shaw [4]. However, P. S. Gupta and A. S. Gupta [5] proved that the solution set up by [4] fails to satisfy the boundary conditions. Considering fluid inertia effects, Elkouh [6] studied the squeeze film between two plane annuli. Verma [7] and Singh et al. [8] have conducted Numerical solutions of the squeezing flow between parallel plates. Leider and Bird [9] performed theoretical analysis for squeezing flow of power-law fluid between parallel disks. Analytic solution for the squeezing flow of viscous fluid between two parallel disks with suction or blowing effect has been proposed by Domairry and Aziz [10]. Islam et al. [11] studied Newtonian squeezing fluid flow in a porous medium channel. Ullah et al. [12] discussed the Newtonian fluid flow with slip boundary condition keeping MHD effect into account. Siddiqui et al. [13] investigated the unsteady squeezing flow of viscous fluid with magnetic field. Apart from the mentioned researchers, other prominent scholars have conducted various theoretical and experimental studies of squeezing flows [14–17].The difference between fluid and boundary velocity is proportional to the shear stress at the boundary. The dimension of proportionality constant is length, which is known as slip parameter. In fluids with elastic character, slip condition has great importance [18]. It has many applications in medical sciences, for instance, polishing artificial heart valves [19]. There are various situations in which no-slip boundary condition is inappropriate. Some of these situations include polymeric liquids when the weight of molecule is high, flow on multiple interfaces, fluids containing concerted suspensions, and thin film problems. The general boundary condition which shows the fluid slip at the wall was initially proposed by Navier [20]. Recently, Ebaid [21] studied the effect of magnetic field in Newtonian fluid in an asymmetric channel with wall slip conditions.Most of scientific incidents are modeled by nonlinear partial or ordinary differential equations. In literature, we have variety of perturbation techniques which can solve nonlinear boundary value problems analytically. But the limitations of these techniques are based on the assumption of small parameters. Detailed review of these methods is given by He [22]. In recent times, the ideas of Homotopy and Perturbation have been combined together. Liao [23] and He [24, 25] have done the primary work in this regard. In series of papers, Marinca with various scholars used OHAM to find the approximate solution of nonlinear differential equations arising in heat transfer, steady flow of a fourth-grade fluid, and thin film flow [26–28].In this work, OHAM is used to analyze an unsteady squeezing fluid flow between two circular nonrotating disks with slip and no-slip boundary conditions. In addition, movement of the circular plates is considered to be symmetric about the axial line and the fluid is considered to be Newtonian, incompressible and viscous. Sections2 and 3 include the description and mathematical formulation of the problem. Sections 4, 5, and 6 present the basic theory of OHAM and its application in case of no-slip and slip boundaries. Results and discussions are given in Section 7 while conclusions are mentioned in Section 8.


## 2. Description of the Problem

The unsteady axisymmetric squeezing flow of incompressible Newtonian fluid with densityρ, viscosity μ and kinematic viscosity ν, squeezed between two circular plates having speed v
w
(
t
) is considered. It is assumed that at any time t, the distance between two circular plates is 2
s
(
t
). It is also assumed that r-axis is the central axis of the channel while z-axis is taken normal to it. Plates move symmetrically with respect to the central axis z
=
0 while the flow is axisymmetric about r
=
0. The longitudinal and normal velocity components in radial and axial directions are u
r
(
r
,
z
,
t
) and u
z
(
r
,
z
,
t
), respectively. The geometrical interpretation of the problem is given in Figure 1.Figure 1
Geometrical interpretation of the problem.


## 3. Mathematical Formulation

The governing equations of motion are(1)
∇
·
U
=
0
,
ρ
∂
U
∂
t
+
U
·
∇
U
=
ρ
f
+
∇
·
T
,
where
(2)
T
=
-
p
I
+
μ
A
1
,
A
1
=
∇
U
+
∇
U
t
,
and U is the velocity vector, p is the pressure, f is the body force, T is the Cauchy stress tensor, A
1 is the Rivlin-Ericksen tensor, and μ is the coefficient of viscosity. Now we formulate the unsteady two-dimensional flow. Let us assume that
(3)
U
=
u
r
(
r
,
z
,
t
)
,
0
,
u
z
(
r
,
z
,
t
)
,
and introduce the vorticity function Ω
(
r
,
z
,
t
) and generalized pressure h
(
r
,
z
,
t
) as
(4)
Ω
r
,
z
,
t
=
∂
u
z
∂
r
-
∂
u
r
∂
z
,
(5)
h
r
,
z
,
t
=
ρ
2
u
r
2
+
u
z
2
+
p
.
Equations (1) are reduced to
(6)
∂
u
r
∂
r
+
u
r
r
+
∂
u
z
∂
z
=
0
,
∂
h
∂
r
+
ρ
∂
u
r
∂
t
-
u
z
Ω
=
-
μ
∂
Ω
∂
z
,
∂
h
∂
z
+
ρ
∂
u
z
∂
t
+
u
r
Ω
=
μ
r
∂
∂
r
r
Ω
.
The boundary conditions on u
r
(
r
,
z
,
t
) and u
z
(
r
,
z
,
t
) are
(7)
u
r
r
,
z
,
t
=
0
,
u
z
r
,
z
,
t
=
v
w
t
at
z
=
s
∂
∂
z
u
r
r
,
z
,
t
=
0
,
u
z
r
,
z
,
t
=
0
at
z
=
0
,
where v
w
(
t
)
=
d
s
/
d
t is the velocity of the plates. The boundary conditions (7) are due to symmetry at z
=
0 and no-slip at the upper plate when z
=
s. If we introduce the dimensionless parameter
(8)
η
=
z
s
(
t
)
.
Equations (4) and (6) transforms to
(9)
Ω
r
,
z
,
t
=
∂
u
z
∂
r
-
1
s
∂
u
r
∂
η
,
(10)
∂
u
r
∂
r
+
u
r
r
+
1
s
∂
u
z
∂
η
=
0
,
(11)
∂
h
∂
r
+
ρ
∂
u
r
∂
t
-
u
z
Ω
=
-
μ
s
∂
Ω
∂
η
,
(12)
1
s
∂
h
∂
η
+
ρ
∂
u
z
∂
t
+
u
r
Ω
=
μ
r
∂
∂
r
r
Ω
.
The boundary conditions on u
r and u
z are
(13)
u
r
=
0
,
u
z
=
v
w
t
at
η
=
1
∂
u
r
∂
η
=
0
,
u
z
=
0
at
η
=
0
.
After eliminating the generalized pressure between (11) and (12), we obtained
(14)
ρ
∂
Ω
∂
t
+
u
r
∂
Ω
∂
r
+
u
z
s
∂
Ω
∂
η
-
u
r
r
Ω
=
μ
∇
2
Ω
-
Ω
r
2
,
where ∇
2 is the Laplacian operator.Defining velocity components as [5]
(15)
u
r
=
-
r
2
s
t
v
w
t
g
′
η
,
u
z
=
v
w
(
t
)
g
(
η
)
we see that (10) is identically satisfied and (14) becomes
(16)
d
4
g
d
η
4
+
R
η
-
g
d
3
g
d
η
3
+
2
d
2
g
d
η
2
-
Q
d
2
g
d
η
2
=
0
,
where
(17)
R
=
s
v
w
υ
,
Q
=
s
2
υ
v
w
d
v
w
d
t
.
Here both R and Q are functions of t but we consider R and Q constants for similarity solution. Since v
w
=
d
s
/
d
t, Integrate first equation of (17), we get
(18)
s
t
=
W
t
+
S
1
/
2
,
where W and S are constants. The plates move away from each other symmetrically with respect to η when W
>
0 and S
>
0. Also the plates approach to each other and squeezing flow exists with similar velocity profiles when W
<
0, S
>
0, and s
(
t
)
>
0. From (17) and (18) it follows that Q
=
-
R. Then (16) becomes
(19)
d
4
g
d
η
4
+
R
η
-
g
d
3
g
d
η
3
+
3
d
2
g
d
η
2
=
0
.
Using (13) and (15) we determine the boundary conditions in case of no-slip and slip at the upper plate as follows:
(20)
g
1
=
1
,
g
′
1
=
0
g
0
=
0
,
g
′′
0
=
0
(
No-slip
at
the
wall
)
(21)
g
(
1
)
=
1
,
g
′
(
1
)
=
γ
g
′′
(
1
)
g
(
0
)
=
0
,
g
′′
(
0
)
=
0
0000
.
(
Slip
at
the
wall
)
.


## 4. Basic Theory of OHAM [26, 29–32]

Let us apply OHAM to the following differential equation:(22)
I
v
(
x
)
+
f
(
x
)
+
ℵ
v
x
=
0
,
B
v
,
d
v
d
x
=
0
,
where x represents an independent variable, v
(
x
) is unknown function and f
(
x
) is known function. B
,
ℵ
,
I are boundary, nonlinear, and linear operators, respectively.According to OHAM, we construct HomotopyΦ
(
x
,
p
)
:
R
×
0,1
→
R which satisfies
(23)
1
-
p
I
Φ
x
,
p
+
f
x
=
h
p
I
Φ
x
,
p
+
f
x
+
ℵ
Φ
x
,
p
,
B
Φ
x
,
p
,
∂
Φ
x
,
p
∂
x
=
0
,
where x
∈
R and p
∈
0,1 is an embedding parameter, h
(
p
) is a nonlinear auxiliary function for p
≠
0, h
(
0
)
=
0 and Φ
(
x
,
p
) is an unknown function. Clearly, when p
=
0 and p
=
1, it holds that Φ
(
x
,
0
)
=
v
0
(
x
) and Φ
(
x
,
1
)
=
v
~
(
x
), respectively.Thus, asp varies from 0 to 1, the solution Φ
(
x
,
p
) approaches from v
0
(
x
) to v
~
(
x
).We choose the auxiliary functionh
(
p
) in the form of
(24)
h
p
=
∑
k
=
0
m
p
k
C
k
,
where C
k are convergence controlling constants to be determined.To obtain an approximate solution, we expandΦ
(
x
,
p
,
C
i
) in a Taylor series about p as follows:
(25)
Φ
x
,
p
,
C
i
=
v
0
x
+
∑
k
=
1
m
v
k
C
1
,
C
2
,
…
,
C
k
p
k
.
Substituting (25) into (23) and equating the coefficients of like powers of p, we obtain the following equations.The zeroth-order problem is(26)
I
v
0
(
x
)
+
f
(
x
)
=
0
,
B
v
0
,
d
v
0
d
x
=
0
.
First-order problem is
(27)
I
v
1
(
x
)
+
f
(
x
)
=
C
1
ℵ
0
v
0
x
,
B
v
1
,
d
v
1
d
x
=
0
.
Second-order problem is
(28)
I
v
2
x
+
I
v
1
x
=
C
2
ℵ
0
v
0
x
+
C
1
I
v
1
x
+
ℵ
1
v
0
x
,
v
1
x
,
B
v
2
,
d
v
2
d
x
=
0
.
The general equations for v
k
(
x
) are given by
(29)
I
v
k
x
-
I
v
k
-
1
x
=
C
1
ℵ
0
v
0
x
+
∑
i
=
1
k
-
1
C
i
I
v
k
-
i
x
0000000000.0
+
ℵ
k
-
i
v
0
x
,
v
1
x
,
…
,
v
k
-
1
x
,
B
v
k
,
d
v
k
d
x
=
0
,
1111
k
=
2,3
,
…
,
where the coefficient of p
m in the expansion of ℵ
Φ
x
,
p about p is ℵ
m
v
0
(
x
)
,
v
1
(
x
)
,
…
,
v
m
-
1
(
x
). 						(30)
ℵ
Φ
x
,
p
,
C
i
=
ℵ
0
v
0
x
+
∑
m
=
1
m
ℵ
m
v
0
x
,
v
1
x
,
…
,
v
m
-
1
x
p
m
.It is noted that the convergence of the series (25) depends upon C
k. For convergence atp
=
1, the mth order approximation v
~ is
(31)
v
~
x
,
C
1
,
C
2
,
…
C
m
=
v
0
x
+
∑
j
=
1
m
v
j
x
,
C
1
,
C
2
,
…
C
j
.
Substituting (31) in (22), the expression for residual is
(32)
R
x
,
C
1
,
C
2
,
…
,
C
m
=
I
v
~
x
,
C
1
,
C
2
,
…
,
C
m
+
f
(
x
)
+
ℵ
v
~
x
,
C
1
,
C
2
,
…
,
C
m
.
If R
=
0, then v
~ will be the exact solution but usually this does not happen in nonlinear problems.There are various methods to find the optimal values ofC
i, i
=
1,2
,
…. We apply the method of least square and Galerkin’s method in the following manner:In method of least square(33)
J
x
,
C
1
,
C
2
,
…
,
C
m
=
∫
a
b
R
2
x
,
C
1
,
C
2
,
…
,
C
m
d
x
.
Minimizing J
x
,
C
1
,
C
2
,
…
,
C
m, we have
(34)
∂
J
∂
C
i
=
0
,
i
=
1,2
,
…
,
m
.
In Galerkin’s method, we solve the following system for C
i
i
=
1,2
,
…
,
m:
(35)
∫
a
b
R
∂
v
~
∂
C
i
d
x
=
0
,
i
=
1,2
,
…
,
m
.
To find appropriate C
i
i
=
1,2
,
…
,
m, we choose a and b in the domain of the problem. Approximate solution of order m is well-determined with these known constants.


## 5. Application of OHAM in Case of No-Slip Boundary

Using (19) and (20) various order problems are as follows:Zeroth-order problem(36)
v
0
i
v
η
=
0
,
v
0
(
0
)
=
0
,
v
0
′
′
(
0
)
=
0
,
v
0
(
1
)
=
1
,
v
0
′
(
1
)
=
0
.
First-order problem
(37)
v
1
i
v
η
=
3
C
1
R
v
0
′
′
η
+
C
1
R
η
v
0
′
′
′
η
-
C
1
R
v
0
η
v
0
′
′
′
η
+
v
0
i
v
η
+
C
1
v
0
i
v
η
,
v
1
(
0
)
=
0
,
v
1
′
′
(
0
)
=
0
,
v
1
(
1
)
=
0
,
v
1
′
(
1
)
=
0
.
Second-order problem
(38)
v
2
i
v
η
=
3
C
1
R
v
1
′
′
η
-
C
1
R
v
1
η
v
0
′
′
′
η
+
C
1
R
η
v
1
′
′
′
η
-
C
1
R
v
0
η
v
1
′
′
′
η
+
v
1
i
v
η
+
C
1
v
1
i
v
η
,
v
2
(
0
)
=
0
,
v
2
′
′
(
0
)
=
0
,
v
2
(
1
)
=
0
,
v
2
′
(
1
)
=
0
.
Third-order problem
(39)
v
3
i
v
η
=
3
C
1
R
v
2
′
′
η
-
C
1
R
v
2
η
v
0
′′′
η
-
C
1
R
v
1
η
v
1
′
′
′
η
+
C
1
R
η
v
2
′
′
′
η
-
C
1
R
v
0
η
v
2
′
′
′
η
+
v
2
i
v
η
+
C
1
v
2
i
v
η
,
v
3
(
0
)
=
0
,
v
3
′
′
(
0
)
=
0
,
v
3
(
1
)
=
0
,
v
3
′
(
1
)
=
0
.
Fourth-order problem
(40)
v
4
i
v
η
=
3
C
1
R
v
3
′
′
η
-
C
1
R
v
3
η
v
0
′
′
′
η
-
C
1
R
v
2
η
v
1
′
′
′
η
-
C
1
R
v
1
η
v
2
′
′
′
η
+
C
1
R
η
v
3
′
′
′
η
-
C
1
R
v
0
η
v
3
′
′
′
η
+
v
3
(
i
v
)
η
+
C
1
v
3
(
i
v
)
η
,
v
4
(
0
)
=
0
,
v
4
′
′
(
0
)
=
0
,
v
4
(
1
)
=
0
,
v
4
′
(
1
)
=
0
.
By considering fourth-order solution, we have
(41)
v
~
η
=
v
0
η
+
v
1
η
+
v
2
η
+
v
3
η
+
v
4
η
.
The residual of the problem is
(42)
R
=
d
4
v
~
η
d
η
4
+
R
η
-
v
~
η
d
3
v
~
η
d
η
3
+
3
d
2
v
~
η
d
η
2
.
We apply Galerkin’s method to find constant C
1 as follows:
(43)
∫
0
1
R
∂
v
~
∂
C
1
d
x
=
0
.
Solving (43) and keeping R
=
0.1, we get
(44)
C
1
=
1.01328
.
Using above value of C
1, the approximate solution is
(45)
v
~
η
=
1
2
3
η
-
η
3
+
1
560
3.74914
η
-
7.39694
η
3
+
3.54648
η
5
000000000
+
0.101328
η
7
+
1
776160
95.6941
η
-
430.222
η
3
+
514.14
η
5
000000000000
-
122.33
η
7
-
55.3411
η
9
-
1.94053
η
11
+
1
565045
2544.14
η
+
14364.4
η
3
-
37116.4
η
5
000000000000
+
25023.2
η
7
-
8133.38
η
9
+
2599.01
η
11
000000000000
+
696.10
η
13
+
22.962
η
15
+
1
722737
1.53702
η
-
1.87989
×
1
0
8
η
3
000000000000
+
1.93991
×
1
0
8
η
5
+
8.28837
×
1
0
7
η
7
000000000000
-
8.03587
×
1
0
7
η
9
-
2.92844
×
1
0
7
η
11
000000000000
+
1.67506
×
1
0
7
η
13
-
9.66217
×
1
0
6
η
15
000000000000
-
1.6529
×
1
0
6
η
17
-
48038.9
η
19
.


## 6. Application of OHAM in Case of Slip Boundary

Using (19) and (21) different order problems are as follows.Zeroth-order problem(46)
u
0
(
i
v
)
η
=
0
,
u
0
0
=
0
,
u
0
′
′
0
=
0
,
u
0
1
=
1
,
u
0
′
1
=
γ
u
0
′
′
1
.
First-order problem
(47)
u
1
i
v
η
=
3
C
1
R
u
0
′
′
η
+
C
1
R
η
u
0
′
′
′
η
-
C
1
R
u
0
η
u
0
′
′
′
η
+
u
0
(
i
v
)
η
+
C
1
u
0
(
i
v
)
η
,
u
1
0
=
0
,
u
1
′
′
0
=
0
,
u
1
1
=
0
,
u
1
′
1
=
γ
u
1
′
′
1
.
Second-order problem
(48)
u
2
i
v
η
=
3
C
1
R
u
1
′
′
η
-
C
1
R
u
1
η
u
0
′
′
′
η
+
C
1
R
η
u
1
′
′
′
η
-
C
1
R
u
0
η
u
1
′
′
′
η
+
u
1
(
i
v
)
η
+
C
1
u
1
(
i
v
)
η
u
2
0
=
0
,
u
2
′
′
0
=
0
,
u
2
1
=
0
,
u
2
′
1
=
γ
u
2
′
′
1
.
Third-order problem
(49)
u
3
i
v
η
=
3
C
1
R
u
2
′
′
η
-
C
1
R
u
2
η
u
0
′
′
′
η
-
C
1
R
u
1
η
u
1
′
′
′
η
+
C
1
R
η
u
2
′
′
′
η
-
C
1
R
u
0
η
u
2
′
′
′
η
+
u
2
i
v
η
+
C
1
u
2
i
v
η
,
u
3
0
=
0
,
u
3
′
′
0
=
0
,
u
3
1
=
0
,
u
3
′
1
=
γ
u
3
′
′
1
.
Fourth-order problem
(50)
u
4
i
v
η
=
3
C
1
R
u
3
′
′
η
-
C
1
R
u
3
η
u
0
′
′
′
η
-
C
1
R
u
2
η
u
1
′
′
′
η
-
C
1
R
u
1
η
u
2
′
′
′
η
+
C
1
R
η
u
3
′
′
′
η
-
C
1
R
u
0
η
u
3
′
′
′
η
+
u
3
(
i
v
)
η
+
C
1
u
3
(
i
v
)
η
,
u
4
0
=
0
,
u
4
′
′
0
=
0
,
u
4
1
=
0
,
u
4
′
1
=
γ
u
4
′
′
1
.
By considering fourth-order solution, we have
(51)
u
~
η
=
∑
i
=
0
4
u
i
η
,
C
1
.
The residual of the problem is
(52)
ζ
=
d
4
u
~
η
d
η
4
+
R
η
-
u
~
η
d
3
u
~
η
d
η
3
+
3
d
2
u
~
η
d
η
2
.
We apply Galerkin’s method to find constant C
1 as follows:
(53)
∫
0
1
ζ
∂
u
~
∂
C
1
d
x
=
0
.
Solving (53) and taking R
=
0.2 and γ
=
1, we get
(54)
C
1
=
-
1.06872
.
Using above value of C
1, the approximate solution is
(55)
u
~
η
=
1
2
3
η
+
η
3
+
1
4480
-
113.284
η
+
151.758
η
3
-
38.9012
η
5
0000000000
+
0.427486
η
7
+
1
248372
-
156189
.
η
+
273825
.
η
3
-
160808
.
η
5
000000000000
+
45952.6
η
7
-
2814.27
η
9
+
34.5387
η
11
+
1
723257
-
9.83844
η
+
1.149
×
1
0
8
η
3
000000000000
-
3.35388
×
1
0
7
η
5
+
4.04185
×
1
0
7
η
7
000000000000
-
2.68964
×
1
0
7
η
9
+
3.65673
×
1
0
6
η
11
000000000000
-
157538
.
η
13
+
1724.2
η
15
+
1
370042
-
1.97195
×
1
0
13
η
+
4.25372
×
1
0
13
η
3
000000000000
-
2.94213
×
1
0
13
η
5
+
6.06765
×
1
0
12
η
7
000000000000
-
1.39445
×
1
0
12
η
9
+
2.47846
×
1
0
12
η
11
000000000000
-
5.97188
×
1
0
11
η
13
+
5.08263
×
1
0
10
η
15
000000000000
-
1.62422
×
1
0
9
η
17
+
1.52182
×
1
0
7
η
19
.


## 7. Results and Discussions

In this article we considered the unsteady axisymmetric flow of nonconducting, incompressible Newtonian fluid between two circular plates. The resulting nonlinear boundary value problems are solved with OHAM and fourth-order Runge-Kutta method using Mathematica 7.0.Tables1, 3, and 5 reflect OHAM solutions along with residuals in case of no-sip and slip boundaries for various values of Reynolds number R and slip parameter γ. Also, Tables 2, 4, and 6 represent RK4 solutions along with residuals in case of no-slip and slip boundaries for various values of R and γ. All the tables demonstrate that results obtained using OHAM are in agreement with RK4 by means of residuals. In addition to above mentioned tables, Table 7 shows the comparison of solutions obtained from OHAM and RK4 for various values of Reynolds number R.Table 1
OHAM solutions along with residuals for variousR in case of no-slip boundary.
η
R
=
0.1
R
=
0.3
R
=
0.5
Solution
Residual
Solution
Residual
Solution
Residual
0.0
0.
0.
0.
0.
0.
0.
0.1
0.150158
5.37334 × 10−11
0.151534
−6.69283 × 10−9
0.152999
−6.23412 × 10−8
0.2
0.297237
3.14675 × 10−10
0.299827
1.66025 × 10−9
0.302582
6.58388 × 10−8
0.3
0.438170
8.51847 × 10−10
0.441661
2.33504 × 10−8
0.445373
3.42092 × 10−7
0.4
0.569900
1.51431 × 10−9
0.573869
3.93945 × 10−8
0.578082
4.93942 × 10−7
0.5
0.689397
1.90824 × 10−9
0.693354
2.81614 × 10−8
0.697548
2.42092 × 10−7
0.6
0.793661
1.33301 × 10−9
0.797122
−3.19654 × 10−8
0.800780
−6.53046 × 10−7
0.7
0.879734
−1.59255 × 10−9
0.882300
−2.21911 × 10−7
0.885004
−3.23415 × 10−6
0.8
0.944705
−1.02814 × 10−8
0.946167
−8.84707 × 10−7
0.947702
−1.2206 × 10−5
0.9
0.985722
−3.446 × 10−8
0.986180
−3.21268 × 10−6
0.986659
−4.41008 × 10−5
1.0
1.
−1.03509 × 10−7
1.
−1.11826 × 10−5
1.
−1.54022 × 10−4Table 2
RK4 solutions along with residuals for variousR in case of no-slip boundary.
η
R
=
0.1
R
=
0.3
R
=
0.5
Solution
Residual
Solution
Residual
Solution
Residual
0.0
0.
2.88535 × 10−7
0.
9.66906 × 10−5
0.
5.30549 × 10−4
0.1
0.150158
3.88278 × 10−8
0.151534
5.71393 × 10−6
0.152999
3.04600 × 10−5
0.2
0.297237
−9.97789 × 10−9
0.299827
−1.41434 × 10−6
0.302582
−7.52772 × 10−6
0.3
0.438170
2.93767 × 10−9
0.441661
3.95997 × 10−7
0.445373
2.10296 × 10−6
0.4
0.569900
−7.51436 × 10−10
0.573869
−1.15203 × 10−7
0.578082
−6.18081 × 10−7
0.5
0.689397
−4.71705 × 10−10
0.693354
−9.6761 × 10−9
0.697548
−2.95916 × 10−8
0.6
0.793661
2.23004 × 10−9
0.797122
1.4593 × 10−7
0.800780
7.14133 × 10−7
0.7
0.879734
−7.18245 × 10−9
0.882300
−4.86272 × 10−7
0.885004
−2.39245 × 10−6
0.8
0.944705
2.68744 × 10−8
0.946167
1.77796 × 10−6
0.947702
8.70664 × 10−6
0.9
0.985722
−1.11599 × 10−7
0.986180
−7.28952 × 10−6
0.986659
3.56157 × 10−5
1.0
1.
−2.71683 × 10−6
1.
−1.47695 × 10−4
1.
6.90088 × 10−4Table 3
OHAM solutions along with residuals for various values ofR in case of slip boundary.
η
R
=
0.2
R
=
0.3
R
=
0.4
Solution
Residual
Solution
Residual
Solution
Residual
0.0
0.
0.
0.
0.
0.
0.
0.1
0.072692
3.18345 × 10−8
0.071220
3.15621 × 10−7
0.069593
1.74626 × 10−6
0.2
0.147091
−1.72789 × 10−8
0.144268
−9.14579 × 10−8
0.141147
−1.13148 × 10−7
0.3
0.224893
−1.96995 × 10−7
0.220953
−1.66461 × 10−6
0.216599
−7.79061 × 10−6
0.4
0.307773
−4.94482 × 10−7
0.303048
−4.29173 × 10−6
0.297832
−2.07401 × 10−5
0.5
0.397370
−7.99437 × 10−7
0.392277
−6.99917 × 10−6
0.386657
−3.41643 × 10−5
0.6
0.495283
−8.5784 × 10−7
0.490289
−7.5499 × 10−6
0.484784
−3.70689 × 10−5
0.7
0.603056
−2.1053 × 10−7
0.598654
−1.91141 × 10−6
0.593805
−9.6958 × 10−6
0.8
0.722172
1.87065 × 10−6
0.718841
1.62936 × 10−5
0.715176
7.91082 × 10−5
0.9
0.854042
6.40576 × 10−6
0.852211
5.59717 × 10−5
0.850196
2.72708 × 10−4
1.0
1.
1.46076 × 10−5
1.
1.27671 × 10−4
1.
6.22242 × 10−4Table 4
RK4 solutions along with residuals for various values ofR in case of slip boundary.
η
R
=
0.2
R
=
0.3
R
=
0.4
Solution
Residual
Solution
Residual
Solution
Residual
0.0
0.
4.12179 × 10−6
0.
7.56765 × 10−6
0.
3.62897 × 10−6
0.1
0.072692
1.74999 × 10−7
0.071220
2.16125 × 10−7
0.069593
−3.49994 × 10−7
0.2
0.147091
−4.21503 × 10−8
0.144268
−4.97423 × 10−8
0.141147
9.53266 × 10−8
0.3
0.224893
1.12883 × 10−8
0.220953
1.22294 × 10−8
0.216598
−3.0735 × 10−8
0.4
0.307773
−3.52272 × 10−9
0.303048
−4.30187 × 10−9
0.297832
7.2493 × 10−9
0.5
0.397370
7.80684 × 10−10
0.392276
3.06668 × 10−9
0.386657
8.46409 × 10−9
0.6
0.495283
1.10391 × 10−9
0.490289
−5.28586 × 10−9
0.484783
−3.36828 × 10−8
0.7
0.603056
−4.35033 × 10−9
0.598654
1.54159 × 10−8
0.593805
1.06816 × 10−7
0.8
0.722172
1.44198 × 10−8
0.718841
−6.05873 × 10−8
0.715175
−3.98519 × 10−7
0.9
0.854042
−5.53933 × 10−8
0.852211
2.59233 × 10−7
0.850196
1.65518 × 10−6
1.0
1.
−1.97434 × 10−7
1.
7.96168 × 10−6
1.
3.89126 × 10−5Table 5
OHAM solutions along with residuals for various values ofγ in case of slip boundary.
η
γ
=
0.5
γ
=
0.6
γ
=
0.7
Solution
Residual
Solution
Residual
Solution
Residual
0.0
0.
0.
0.
0.
0.
0.
0.1
−0.00880011
−1.07933 × 10−6
0.0336596
−7.54858 × 10−8
0.0522402
−7.74445 × 10−9
0.2
−0.010881
−2.2445 × 10−6
0.0714091
−1.73419 × 10−7
0.1074223
−2.51485 × 10−8
0.3
0.000449012
−3.45224 × 10−6
0.117324
−3.00407 × 10−7
0.168479
−5.68899 × 10−8
0.4
0.0318278
−4.41058 × 10−6
0.175449
−4.32397 × 10−7
0.238322
−9.8056 × 10−8
0.5
0.0898133
−4.47787 × 10−6
0.249785
−5.00627 × 10−7
0.319832
−1.29675 × 10−7
0.6
0.18086
−2.60114 × 10−6
0.344278
−3.77978 × 10−7
0.415852
−1.13733 × 10−7
0.7
0.311298
2.65136 × 10−6
0.462801
1.30706 × 10−7
0.529175
1.20785 × 10−8
0.8
0.487312
1.28583 × 10−5
0.609145
1.28166 × 10−6
0.662538
3.36724 × 10−7
0.9
0.714929
2.91822 × 10−5
0.787012
3.34515 × 10−6
0.818613
9.67763 × 10−7
1.0
1.
5.14019 × 10−5
1.
6.49074 × 10−6
1.
2.00176 × 10−6Table 6
RK4 solutions along with residuals for various values ofγ in case of slip boundary.
η
γ
=
0.5
γ
=
0.6
γ
=
0.7
Solution
Residual
Solution
Residual
Solution
Residual
0.0
0.
2.04594 × 10−5
0.
7.34794 × 10−6
0.
3.56291 × 10−6
0.1
−0.00880012
7.72953 × 10−7
0.0336596
3.21066 × 10−7
0.0522402
1.63598 × 10−7
0.2
−0.0108811
−1.8406 × 10−7
0.0714091
−7.75305 × 10−8
0.1074223
−3.96814 × 10−8
0.3
0.000448983
4.82747 × 10−8
0.117324
2.08509 × 10−8
0.168479
1.0753 × 10−8
0.4
0.0318277
−1.54235 × 10−8
0.175449
−6.43999 × 10−9
0.238322
−3.28063 × 10−9
0.5
0.0898132
5.14031 × 10−9
0.249785
1.21029 × 10−9
0.319832
4.50154 × 10−10
0.6
0.18086
−6.51866 × 10−10
0.344278
2.64494 × 10−9
0.415852
1.84079 × 10−9
0.7
0.311298
−1.86837 × 10−9
0.462801
−9.88382 × 10−9
0.529175
−6.51979 × 10−9
0.8
0.487312
−1.33003 × 10−9
0.609145
3.37336 × 10−8
0.662538
2.27322 × 10−8
0.9
0.714929
2.66400 × 10−8
0.787012
−1.32232 × 10−7
0.818613
−9.05714 × 10−8
1.0
1.
5.55277 × 10−6
1.
−1.1671 × 10−6
1.
−1.1767 × 10−6Table 7
Comparison of OHAM and RK4 solutions for variousR in case of slip and no-slip boundary.
η
In case of no-slip boundary
In case of slip boundary
|
RK4 Solution
-
OHAM Solution
|
|
RK4 Solution
-
OHAM Solution
|
R
=
0.1
R
=
0.3
R
=
0.5
R
=
0.2
R
=
0.3
R
=
0.4
0.0
0.
0.
0.
0.
0.
0.
0.1
2.33147 × 10−15
1.62249 × 10−10
2.41725 × 10−9
8.44379 × 10−10
7.90896 × 10−9
4.1479 × 10−8
0.2
9.49241 × 10−14
3.14228 × 10−10
4.67809 × 10−9
1.6672 × 10−9
1.55832 × 10−8
8.15527 × 10−8
0.3
3.82361 × 10−13
4.46143 × 10−10
6.63065 × 10−9
2.44889 × 10−9
2.28102 × 10−8
1.18952 × 10−7
0.4
9.13492 × 10−13
5.47932 × 10−10
8.11705 × 10−9
3.1663 × 10−9
2.93539 × 10−8
1.52339 × 10−7
0.5
1.65451 × 10−12
6.07476 × 10−10
8.95202 × 10−9
3.77514 × 10−9
3.48026 × 10−8
1.79574 × 10−7
0.6
2.42317 × 10−12
6.09434 × 10−10
8.91445 × 10−9
4.18241 × 10−9
3.8327 × 10−8
1.96532 × 10−7
0.7
2.86218 × 10−12
5.36833 × 10−10
7.77928 × 10−9
4.22088 × 10−9
3.84569 × 10−8
1.96009 × 10−7
0.8
2.52039 × 10−12
3.78563 × 10−10
5.4278 × 10−9
3.65208 × 10−9
3.31085 × 10−8
1.67861 × 10−7
0.9
1.20232 × 10−12
1.54202 × 10−10
2.18716 × 10−9
2.24454 × 10−9
2.02732 × 10−8
1.02385 × 10−7
1.0
1.26281 × 10−16
2.53545 × 10−16
2.93337 × 10−17
1.7436 × 10−14
4.92715 × 10−14
3.8179 × 10−14Furthermore, Figures2, 3, and 4 indicate the OHAM residuals in case of no-slip and slip boundaries for various values of R and γ.Figure 2
OHAM residuals at various values ofR in case of no-slip boundary.Figure 3
OHAM residuals at various values ofR in case of slip boundary.Figure 4
OHAM residuals at various values ofγ in case of slip boundary.The effect of Reynolds numberR on velocity profiles in case of no-slip boundary is shown in Figure 5. In these profiles we varied R as R
=
0.1,1
,
2,3 and observed that the normal velocity is increased with the increase of Reynolds number (Figure 5(a)). It is also noted that the normal velocity monotonically increases from η
=
0 to η
=
1 for fixed positive value of R at a given time. Figure 5(b) describes the impact of R on the longitudinal velocity in case of no-slip boundary. It is experienced that this component of velocity deceases near the wall but increases near the central axis of the channel.Velocity profiles for various values ofR
=
0.1,1
,
2,3 in case of no-slip boundary.
(a)
The effect ofR on the Normal velocity profiles
(b)
The effect ofR on the longitudinal velocity profilesThe effect of Reynolds numberR on velocity profiles in case of slip boundary is depicted in Figure 6. In these profiles, we fixed slip parameter γ
=
1 and varied Reynolds number R as R
=
0.2,0.6,1
,
1.5. It is noted that the normal velocity decreases as the Reynolds number increases (Figure 6(a)). It is also observed that longitudinal velocity decreases near the central axis of the channel but increases near the walls when R increases (Figure 6(b)).Velocity profiles for various values ofR
=
0.2,0.6,1
,
1.5 fixing γ
=
1 in case of slip boundary.
(a)
The effect ofR on the Normal velocity profiles
(b)
The effect ofR on the longitudinal velocity profilesFigure7 demonstrates the effect of slip parameter γ on the velocity profiles. After fixing Reynolds number R
=
0.3 we varied γ as γ
=
0.6,0.7,0.8,1. We find that normal velocity increases as γ increases. It is also noted that longitudinal velocity decreases near the walls but increases near central axis of the channel.Velocity profiles for various values ofγ
=
0.6,0.7,0.8,1 fixing R
=
0.3 in case of slip boundary.
(a)
The effect ofγ on the Normal velocity profiles
(b)
The effect ofγ on the longitudinal velocity profiles


## 8. Conclusions

In this article, we find the similarity solution for unsteady axisymmetric squeezing flow of incompressible Newtonian fluid between two circular plates. We observed that the similarity solution exists only when distance between the plates varies asW
t
+
S
1
/
2, and squeezing flow occurs when W
<
0, S
>
0 and W
t
+
S
>
0. The key findings of the present analysis are as follows:In case of no-slip at boundary;(i)
It has been found that increase in Reynolds numberR increases the normal velocity.
(ii)
It has been observed that normal velocity increases monotonically fromη
=
0 to η
=
1 for fixed positive value of R at a given time.
(iii)
It has been seen that longitudinal velocity deceases near the walls and increases near the central axis of the channel.In case of slip at boundary;(i)
It has been noted that after fixing slip parameterγ and varying the Reynolds number R, the normal velocity profile decreases with the increase in R. Also the longitudinal velocity increases near the walls but decreases near the central axis of the channel.
(ii)
It has been examined that for a fixed Reynolds numberR when we vary slip parameter γ, the normal velocity increases with the increase in γ. Also the longitudinal velocity decreases near the walls and increases near the central axis of the channel.
(iii)
It has been investigated that Reynolds numberR and slip parameter γ have opposite effects on the normal and longitudinal velocity components.In case of slip versus no-slip boundary;(i)
It has been observed that Reynolds numberR has opposite behavior on the normal velocity in case of slip and no-slip boundaries.
(ii)
It has been also noticed that Reynolds numberR has opposite effect on the longitudinal velocity near the central axis of the channel, while near the wall longitudinal velocity increases in case of slip boundary and decease in no-slip boundary. This is in conformance to [33].



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