Unsteady flow of second grade fluid problem is examined between two horizontal oscillating parallel plates. Plates are oscillating with same periods.
J. Appl. Environ. Biol. Sci., 5(2)52-62, 2015 © 2015, TextRoad Publication
ISSN: 2090-4274 Journal of Applied Environmental and Biological Sciences www.textroad.com
Unsteady Second order Fluid Flow between Two Oscillating Plates Taza Gul 1, Imran Khan2, Muhammad Altaf Khan1, S.Islam1, Tanveer Akhtar 3, S.Nasir 1 1
1
Department of mathematics, Abdul Wali K.U Mardan, KPK Pakistan Department of mathematics, Is PAR, Bacha Khan University Charsadda Pakistan 3 Centre of Excellence in Solid State Physics PU, Lahore Pakistan Received: November 19, 2014 Accepted: January 27, 2015
ABSTRACT Unsteady flow of second grade fluid problem is examined between two horizontal oscillating parallel plates. Plates are oscillating with same periods. We consider three unsteady fluid flow problems. We use the OHAM method to solve the non-linear problems arising from the modeling. The physical effect of various modeled parameters on the velocity and temperature fields is discussed and their numerical results are presented. KEYWORDS: Unsteady Second Grade Fluid, Parallel horizontal plates, Optimal Homotopy Asymptotic method (OHAM). I
INTRODUCTION
The industrial and technological applications of non-Newtonian fluids greatly change from the Newtonian fluids due to their rheological characteristics. These type liquids do not follow the Newton's law of viscosity. Actually, such fluids are significantly used in geophysics, petroleum, chemical, nuclear industries and many others scientific technologies. Since non-Newtonian fluids cannot be totally defined by one of constitutive equation. One of the very important models suggested for non-Newtonian fluids is called the second grade fluid. In past scholars gives the considerable attention to the second grade fluid which is a simplest subclass of differential type non-Newtonian fluids. Second grade fluid was employed to study many problems due to their comparatively simple model. Few recent articles discussed the flow of second grade fluid have been proposed in literature [1-6]. The limited study exists in literature the problem of unsteady fluid flow. Certain progress and related work on unsteady flow of second grade fluid have been discussed by Gul et al. in [7]. They investigated the analytical solution of the model by using ADM and OHAM techniques. The exact solution of unsteady second grade fluid on vertical oscillating palate by using Laplace transformation and Fourier sine transforms have been investigated in [8-9]. In problem under discussion the flow of fluid consider between two horizontal parallel plates. Chen and Zhu [10-11] studied Couette and Poiseuille flow of Bingham fluids between two porous parallel plates and concentric annuli with slip conditions. They calculated the analytical solution of eight different form of velocity profile depending on three model parameters Bingham parameter, axial Couette numbers and the radius ratio. Francisca et al. [12] examined the analytical solution of laminar flow of Newtonian fluid between two parallel plates where the upper plate moving and lower plate is kept in rest. The problem associated to the present work MHD unsteady second grade fluid flow between two vertical and horizontal moving and oscillating parallel plates studied [13-14] Joseph et al. [15] investigated the unsteady MHD couette flow between two parallel plates. The fluid motion of two dimensional steady flow of a viscous and incompressible fluid flowing between two parallel plates under the constant pressure gradient investigated in [16-18]. The fundamental theme of present article is to study the three different unsteady flows of second grade fluid namely (i) couette flow, (ii) Poiseuille flow and (iii) plug flow between two horizontal parallel plates . Optimal Homotopy Asymptotic Method (OHAM) has been used to obtain the analytical solution of the above problems. OHAM is a new and powerful method and has effectively applied to a number of nonlinear problems producing in the science and engineering by several researchers. Some new investigations about the implementation of OHAM on nonlinear problems are mentioned in [19-25]. In the remaining paper the basic equations for second grade fluid are given in section II. Formulation of first problem presented in section III. In section IV, the basic concepts of OHAM are discussed. The formulation and solution of second and third problem have been discussed in section V and VI. The graphical results are given in section VII. The paper ends with some conclusions in section VIII. Fundamental equations The constitutive equations of second grade fluid model is given by 𝐓 = −𝑝𝐈 + 𝜇𝐀1 + 𝛼1 𝐀𝟐 + 𝛼2 𝐀𝟐1 , *Corresponding Author: Taza Gul, Department of mathematics, Abdul Wali K.U Mardan, KPK Pakistan
52
(1)
Gul et al.,2015
Where 𝐓is the Cauchy stress tensor,𝐈 is identity tensor, 𝑝 is fluid pressure, 𝛼1 𝑎𝑛𝑑 𝛼2 are the material constants, 𝐀1 𝑎𝑛𝑑 𝐀2 are the Rivlin-Ericksen tensor given by 𝐀0 = 𝐈, 𝐀1 = (∇𝐯) + (∇𝐯)𝑇 , 𝐷𝐀 𝐀𝑛 = 𝒏−𝟏 + 𝐀𝑛−1(∇𝐯) + (∇𝐯)𝑇 𝐀𝑛−1 , 𝑛 = 2,3,4, …, 𝐷𝑡 The basic equation of continuity and motion for incompressible fluid are given by ∇. 𝐯 = 0, 𝐷𝐯 𝜌 = div𝐓 + 𝜌g, 𝐷𝑡
(2) (3) (4) (5)
𝐷
Where 𝜌is the fluid density,𝐯is the velocity vector of the fluid,gis gravity,the material time derivative define 𝐷𝑡 as 𝐷(∗) 𝜕(∗) = + (𝐯. ∇) ∗, (6) 𝐷𝑡 𝜕𝑡 II Couette flow problem Formulation of problem when one plate moving with oscillation Consider two horizontal and parallel plates such that the upward plate oscillating and moving with constant velocity 𝑉 and the lower plate kept oscillating. The total thickness of the fluid between the plates assumed to be 2ℎ. Moving and oscillating plate caries with itself a liquid of width ℎ.The configuration of fluid flow is along the x-axis and perpendicular to y-axis. The velocity field is of the form 𝐯 = (𝑣(𝑦, 𝑡), 0,0), (7) Boundary conditions are: 𝐯(ℎ, 𝑡) = 𝑉 + 𝑉𝑐𝑜𝑠𝜔𝑡 , 𝐯(−ℎ, 𝑡) = 0 (8) Here 𝜔is the frequency of the oscillating belt. Using equation (7) the continuity equation (4) is satisfied identically, while equation (5)reduces to the form 𝜕𝐯 𝜕 𝜌 = 𝑇𝑥𝑦, (9) 𝜕𝑡
𝜕𝑦
Expending equation (1) components of the Cauchy stress tensor T as
𝜕𝐯 2
𝑇𝑥𝑥 = −𝑝 + 2𝛼1 ( ) , 𝑇𝑥𝑦 = 𝜇
𝜕𝐯 𝜕𝑦
+ 𝛼1
Inserting equation (11) in equation (9), we get 𝜌
𝜕𝐯 𝜕𝑡
=𝜇
𝜕2 𝐯 𝜕𝑦 2
𝜕
(10)
𝜕𝑦 𝜕𝐯
( ) = 𝑇𝑦𝑥,
+ 𝛼1
𝜕
𝜕2 𝐯
(
𝜕𝑡 𝜕𝑦 2
),
Introducing the following non-dimensional physical quantities 𝑣̅ =
𝐯 𝑉
, 𝑦̅ =
𝑦 ℎ
, 𝑡̅ =
𝑡𝜇 2
𝛿 𝜌
,𝛼 =
𝛼1
2
𝜌𝛿
,𝛺 =
Where,𝛼 is the non-dimensional variable and 𝛺 is pressure gradient. The dimensionless form of equation (12) using equation (13) and dropping the bars is 𝜕𝑣 𝜕𝑡
Boundary conditions
=
𝜕2 𝑣
𝜕𝑦 2
𝑣 (1, 𝑡) = 1 + 𝑐𝑜𝑠𝜔𝑡 ,
+𝛼
𝜕
𝜕2 𝑣
(
𝜕𝑡 𝜕𝑦 2
),
𝑣(−1,𝑡) = 0 ,
Figure 1: Flow Configuration of Poiseuille Problem III Optimal Homotopy Asymptotic Method (OHAM) Basic idea For the basic study of OHAM, we consider the boundary value problem as
53
(11)
𝜕𝑡 𝜕𝑦
(12) ℎ2 𝜕𝑝
,
𝜇𝑉 𝜕𝑦
(13)
(14) (15)
J. Appl. Environ. Biol. Sci., 5(2)52-62, 2015
𝜕𝑢, ̃ ̃(𝑢̃(𝑦)) + 𝐺̃ (𝑦 ) = 0, 𝐵̃ (𝑢, 𝐿̃(𝑢̃(𝑦)) + 𝑁 ̃ ) = 0, (16) 𝜕𝑦 ̃ ̃ Where𝐿a linear operator in the differential equation is, 𝑁 is a non-linear term,𝑦 ∈ 𝑅 is an independent variable, 𝐵̃ is a boundary operator and 𝐺̃is a source term.We construct a set of equation as: ̃𝜑̃(𝑦, 𝑝)], 𝐵̃ (𝜑̃(𝑦, 𝑝), 𝜕𝜑̃(𝑦,𝑝)) = 0, [1 − 𝑝][𝐿̃𝜑̃(𝑦, 𝑝) + 𝐺̃ (𝑦)] = 𝐻 (𝑝)[𝐿̃𝜑̃ (𝑦, 𝑝) + 𝐺̃ (𝑦) + 𝑁 (17) 𝜕𝑥 where𝑝 ∈ [0,1] is an embedding parameter, 𝐻 (𝑝), is a non-zero auxiliary function for 𝑝 ≠ 0 and 𝐻 (0) = 0. 𝜑̃(𝑦,𝑝) is an unknown function. Obviously, when 𝑝 = 0and 𝑝 = 1, it holds that: 𝜑̃(𝑦,0) = 𝑢̃0 (𝑦), 𝜑̃(𝑦, 1) = 𝑢̃(𝑦). (18) Note that, when 𝑝 varies from 0 to 1 then 𝜑̃(𝑦, 𝑝) also varies from 𝑢̃0 (𝑦)to 𝑢̃(𝑦). The zero component solution 𝑢̃0 (𝑥) is obtained from equation (17) when 𝑝 = 0 i.e 𝜕𝑢 ̃ (𝑦 ) 𝐿̃(𝑢̃0 (𝑦)) + 𝐺̃ (𝑦) = 0, 𝐵̃ (𝑢̃0 (𝑦), 0 ) = 0. (19) 𝜕𝑥 Auxiliary function 𝐻(𝑝) is chosen as: 𝐻 (𝑝) = 𝑝𝑐1 + 𝑝 2 𝑐2 + ⋯ (20) where𝑐1,𝑐2 are auxiliary constants. Expand 𝜑̃ (𝑦,𝑝) with respect to 𝑝 by using Taylor series. 𝜑̃(𝑦,𝑝, 𝑐𝑖 ) = 𝑢̃0 (𝑦) + ∑𝑘≥1 𝑢̃𝑘 (𝑦, 𝑐𝑖 )𝑝 𝑘 ,𝑖 = 1,2, … . (21) Inserting Eq. (32) into Eq. (28), collecting the same powers of 𝑝, and equating each coefficient of 𝑝, the zero order problem is given in equation (19) and the first and second order are given in equations (22, 23). ̃0(𝑢̃0 (𝑦)), 𝐵̃ (𝑢̃1(𝑦), 𝜕𝑢̃1(𝑦)) = 0, L̃(ũ1 (𝑦)) + ̃ G (y) = c1𝑁 (22) 𝜕𝑦 ̃ ̃ ̃ ̃ ̃ 𝐿 (𝑢̃2(𝑦)) − 𝐿(𝑢̃1(𝑦)) = 𝑐2 𝑁0 (𝑢̃0(𝑦 )) + 𝑐1[𝐿(𝑢̃1 (𝑦)) + 𝑁1(𝑢̃0 (𝑦), 𝑢̃1(𝑦))], 𝜕𝑢 ̃ (𝑦 ) 𝐵̃ (𝑢̃2 (𝑦), 2 ) = 0. 𝜕𝑥 The general governing equations for 𝑢𝑘 (𝑥) are given by
(23)
𝑘−1
̃0 (𝑢̃0(𝑦 )) + ∑ 𝑐𝑖 [𝐿̃(𝑢̃𝑘−𝑖 (𝑦)) + 𝑁 ̃𝑘−1(𝑢̃0 (𝑦),𝑢̃1 (𝑦). . 𝑢̃𝑘−𝑖 (𝑦))] 𝐿̃(𝑢̃𝑘 (𝑦)) − 𝐿̃(𝑢̃𝑘−1 (𝑦)) = 𝑐𝑘 𝑁 𝑖=1
𝜕𝑢 ̃ (𝑦 ) 𝑘 = 2,3, … , 𝐵̃ (𝑢̃𝑘 (𝑦 ), 𝑘 ) = 0
(24)
𝜕𝑥
̃𝑚 (𝑢̃0 (𝑦),𝑢̃1 (𝑦). . 𝑢̃𝑚−1(𝑦)) is the coefficient of 𝑝 𝑚, in the expansion of 𝑁 ̃𝜑̃(𝑦,𝑝). Here 𝑁 ∞ ̃(𝜑̃(𝑦,𝑝, 𝑐𝑖 )) = 𝑁 ̃0 (𝑢̃0(𝑦)) + ∑𝑚=1 𝑁 ̃𝑚 (𝑢̃0(𝑦),𝑢̃1 (𝑦).. 𝑢̃𝑚 (𝑦))𝑝𝑚. 𝑁 The convergence of the series in equation (21) depend upon the auxiliary constants 𝑐1,𝑐2 ,… If it converges at 𝑝 = 1, then the 𝑚𝑡ℎ order approximation 𝑢̃ is 𝑢̃(𝑦, 𝑐1,𝑐2 … . 𝑐𝑚 ) = 𝑢̃0 (𝑦) + ∑𝑚 ̃ 𝑖 (𝑦,𝑐1, 𝑐2 … . 𝑐𝑖 ). 𝑖=1 𝑢 Inserting Eq. (25) into Eq. (16), the residual is obtained as: ̃(𝑢̃(𝑦,𝑐𝑖 )),𝑖 = 1,2 … 𝑚. 𝑅̃(𝑦, 𝑐𝑖 ) = 𝐿̃(𝑢̃(𝑦,𝑐𝑖 )) + 𝐺̃ (𝑦) + 𝑁 Numerous methods like Ritz Method, Method of Least Squares, Galerkin Method and Collocation Method are used to find the optimal values of 𝑐𝑖 , 𝑖 = 1,2,3,4 …. We apply the Method of Least Squares in our problem as given below: 𝑏 𝐽 (𝑐1 ,𝑐2 ,… . 𝑐𝑛) = ∫𝑎 𝑅̃2 (𝑦,𝑐1, 𝑐2,… . 𝑐𝑚 )𝑑𝑦, where𝑎 and 𝑏 are the constant values taking from domain of the problem. Auxiliary constants (𝑐1, 𝑐2, … . 𝑐𝑛 ) can be identified from: 𝜕𝐽 𝜕𝐽 = = ⋯ = 0. 𝜕𝑐1
𝜕𝑐2
(25)
(26) (27)
(28)
(29)
Finally, from these auxiliary constants, the approximate solution is well determined. The OHAM Solution of Poiseuille flow Problem Here we apply the OHAM technique on Eq. (14) with boundary condition in Eq. (15) and as a result we, obtain zero, first and second order problems. Zero, first and second order problems of velocity profile are: 𝑝0 ∶ 𝑝1 : 𝜕2 𝑣2(𝑦,𝑡) 𝑝2 : 𝜕𝑦 2
= −𝑐2
∂𝑣0 ∂t
− 𝑐1
∂𝑣1 ∂t
𝜕2 𝑣1(𝑦,𝑡)
+ 𝑐2
𝜕𝑦 2 ∂2 𝑣0 ∂𝑦 2
= −𝑐1
+ 𝛼𝑐2
∂ ∂t
∂𝑣0
+
∂t ∂2𝑣0
(
∂𝑦 2
∂2𝑣0
𝜕𝑦 2
(1 + 𝑐1) + 𝛼𝑐1
∂𝑦 2 ∂2 𝑣1
)+
𝜕2𝑣0(𝑦,𝑡)
∂𝑦 2
= 0, ∂
∂t
(
),
∂𝑦 2 ∂ ∂ 2𝑣 1
(1 + 𝑐1) + 𝛼𝑐1 ( ∂t
Solutions of Eq. (31, 32 and 33) using boundary conditions in Eq. (15)are: 𝑣0 (𝑦, 𝑡) = [0.5 + Cos[𝑡𝜔]]𝑦 + 0.5𝑦 2,
54
(31)
∂2 𝑣0
∂𝑦 2
(32) ),
(33)
(34)
Gul et al.,2015
1
1
2
2
𝑣1 (𝑦, 𝑡) = 𝑐1 [− 𝜔Sin [𝑡𝜔] + 0𝑦 + 𝜔Sin[𝑡𝜔]𝑦2 ],
(35)
𝑣2 (𝑦, 𝑡) = 5 1 1 1 1 [− 𝜔2 Cos[𝑡𝜔]𝑐12 − 𝜔Sin[𝑡𝜔]𝑐2 ] − 𝜔2 Cos[𝑡𝜔]𝑐1(1 + 𝑐1 )𝑦 + [ 𝜔2 Cos[𝑡𝜔]𝑐12 + 𝜔Sin[𝑡𝜔]𝑐2 ] 𝑦 2 + 1 6
24
2
6
1
4
2
𝜔2 Cos[𝑡𝜔]𝑐1 (1 + 𝑐1)𝑦3 − [ 𝜔2 Cos[𝑡𝜔]𝑐12 ] 𝑦 4,
(36)
24
The series solution of first problem velocity profile up to second order is: 𝑣 (𝑦,𝑡) = 𝑣0 (𝑦, 𝑡) + 𝑣1 (𝑦,𝑡) + 𝑣2 (𝑦,𝑡), (37) Substituting Eq. (34-35) in Eq. (37), we have 𝑣 (𝑦,𝑡) = 1 5 1 1 1 1 [ + Cos[𝑡𝜔] (1 − 𝜔2 𝑐12 ) − 𝜔Sin[𝑡𝜔](𝑐1 + 𝑐2)] + [ − 𝜔2 Cos[𝑡𝜔](1 + 𝑐1 )]𝑦 + [ 𝜔Sin[𝑡𝜔](𝑐1 + 𝑐2) + 2 24 2 2 6 1 2 1 1 𝜔 Cos[𝑡𝜔]𝑐12] 𝑦 2 + 𝜔2 Cos[𝑡𝜔]𝑐1 (1+ 𝑐1 )𝑦 3 − 𝜔2 Cos[𝑡𝜔]𝑐12𝑦 4, 4 6 24
2
(38)
The value of 𝑐𝑖 for thevelocity components offirst problem are 𝑐1 = −1.0134436970417653, 𝑐2 = 0.022008523038264727 , IV Formulation of Poiseuille flow Problem: Here in this section, we consider the second grade fluid is between two horizontal parallel plates𝑦 = ℎ and 𝑦 = −ℎ. We assumed that both plates are oscillating. The flow of fluid is due to the constant pressure gradient in xdirection. The equation governing this flow of incompressible second grade fluid with boundary condition is 𝜌
𝜕𝐯
=−
𝜕𝑡
𝜕𝑝 𝜕𝑦
𝜕2 𝐯
+𝜇
𝜕𝑦 2
+ 𝛼1
𝜕
𝜕2 𝐯
(
𝜕𝑡 𝜕𝑦 2
),
(39)
Boundary conditions
𝐯(ℎ, 𝑡) = 𝑉𝑐𝑜𝑠𝜔𝑡 , 𝐯(−ℎ, 𝑡) = 0 , Using the dimensionless physical parameters define in Eq. (13) the momentum equations become 𝜕𝑣 𝜕𝑡
=
And Boundary conditions are
𝜕2 𝑣
𝜕
𝜕2 𝑣
) − 𝛺,
(41)
𝑣 (1, 𝑡) = 𝑐𝑜𝑠𝜔𝑡, 𝑣(−1,𝑡) = 0,
(42)
𝜕𝑦 2
+𝛼
(
(40)
𝜕𝑡 𝜕𝑦 2
Figure 2: Flow Configuration of Poiseuille Problem The OHAM Solution of Poiseuille Problem: Here we apply the OHAM technique on Eq. (41) with boundary condition in Eq. (42) and as a result we, obtain zero, first and second order problems. Zero, first and second order problems velocity profiles are 𝜕2 𝑣0(𝑦,𝑡)
𝑝0 : 𝑝1 : 𝜕2 𝑣
2(𝑦,𝑡) 𝜕𝑦 2
𝑝2 :
𝜕2 𝑣1(𝑦,𝑡)
= −𝛺𝑐2 − 𝑐2
𝜕𝑦 2 ∂𝑣0 ∂t
= −𝛺 − 𝛺𝑐1 − 𝑐1
− 𝑐1
∂𝑣1 ∂t
+ 𝑐2
∂2 𝑣
0 ∂𝑦 2
∂𝑣0 ∂t
+ 𝛼𝑐2
+
∂2𝑣0
𝜕𝑦 2
(1 + 𝑐1) + 𝛼𝑐1
∂𝑦 2 ∂ ∂2𝑣0 ∂t
(
∂𝑦 2
= 𝛺,
)+
∂2 𝑣1 ∂𝑦 2
(43) ∂ ∂t
∂2 𝑣0
(
∂𝑦 2
), ∂
(44) ∂ 2𝑣 1
(1 + 𝑐1) + 𝛼𝑐1 ( ∂t
∂𝑦 2
),
Solutions of Eq. (43, 44 and 45) using boundary conditions in Eq. (42) are: 𝛺 𝛺 𝑣0 (𝑦, 𝑡) = − + Cos[𝑡𝜔] + 0𝑦 + y 2, 1
2
1
2
𝑣1 (𝑦, 𝑡) = − 𝜔Sin[𝑡𝜔]𝑐1 + 0𝑦 + 𝜔Sin[𝑡𝜔]𝑐1 𝑣2 (𝑦, 𝑡) = [− 1
5 24
1
2
1
2
(46) y2,
(47) 1
𝜔2 Cos[𝑡𝜔]𝑐12 − 𝛺𝑐2 − 𝜔Sin[𝑡𝜔]𝑐2 ] − [ 𝜔2 Cos[𝑡𝜔]𝑐1(1 + 𝑐1 )] 𝑦 + [ 𝜔2 Cos[𝑡𝜔]𝑐12 + 1
2
6 1
𝛺𝑐2 + 𝜔Sin[𝑡𝜔]𝑐2 ]y 2 + [ 𝜔2 Cos[𝑡𝜔]𝑐1 (1 + 𝑐1)] y 3 − [ 𝜔2 Cos[𝑡𝜔]𝑐12] y 4, 2
(45)
6
24
55
4
(48)
J. Appl. Environ. Biol. Sci., 5(2)52-62, 2015
The series solution of second problem velocity profile up to second order is: 𝑣 (𝑦,𝑡) = 𝑣0 (𝑦, 𝑡) + 𝑣1 (𝑦,𝑡) + 𝑣2 (𝑦,𝑡), Substituting Eq. (34-35) in Eq. (37), we have 5 1 𝛺 1 𝑣 (𝑦,𝑡) = [Cos[𝑡𝜔] (1 − 𝜔2 𝑐12 ) − 𝜔Sin[𝑡𝜔](𝑐1 + 𝑐2 ) − 𝛺𝑐2 − ] − [ 𝜔2 Cos[𝑡𝜔]𝑐1(1 − 𝑐1 )]𝑦 + 1
1
24
2
𝛺
2
1
6
(49)
1
[ 𝜔Sin[𝑡𝜔](𝑐1 + 𝑐2) + 𝜔2 Cos[𝑡𝜔]𝑐12 + + 𝛺𝑐2 ]y 2 + [ 𝜔2 Cos[𝑡𝜔]𝑐1 (1 + 𝑐1)] y 3 − [ 𝜔2 Cos[𝑡𝜔]𝑐12] y 4, 2 4 2 6 24 (50) The values of 𝑐𝑖 for the second problem of velocity profile are 𝑐1 = −0.993732155929142 , 𝑐2 = 0.0006997010832539563 , V Formulation of Couette Poiseuille flow Problem In this problem the second grade fluid is restricted between two horizontal parallel plates with𝑦 = ℎand 𝑦 = −ℎ. The physical model in this case is similar to that of first problem but additionally a constant pressure gradient is applied in x-direction. The governing equation of motion and boundary conditions are 𝜌
𝜕𝐯 𝜕𝑡
=−
𝜕𝑝 𝜕𝑦
+𝜇
𝜕2 𝐯
𝜕𝑦 2
+ 𝛼1
𝜕
𝜕2 𝐯
(
𝜕𝑡 𝜕𝑦 2
),
(51)
Boundary conditions
𝐯(ℎ, 𝑡) = 𝑉 + 𝑉𝑐𝑜𝑠𝜔𝑡 , 𝐯(−ℎ,𝑡) = 0 , Using the dimensionless physical parameters define in Eq. (13) the momentum equations (51) become 𝜕𝑣 𝜕𝑡
=
𝜕2 𝑣
𝜕𝑦 2
+𝛼
Boundary conditions
𝜕
𝜕2 𝑣
(
𝜕𝑡 𝜕𝑦 2
) − 𝛺,
(52) (53)
𝑣 (1, 𝑡) = 1 + 𝑐𝑜𝑠𝜔𝑡, 𝑣(−1,𝑡) = 0,
(54)
Figure 3: Flow Configuration of Couette Poiseuille flow Problem The OHAM Solution of Couette Poiseuille flow Problem: Here we apply the OHAM technique on Eq. (53) with boundary condition in Eq. (54) and as a result we, obtain zero, first and second order problems. Zero, first and second order problems velocity profiles are 𝜕2 𝑣0(𝑦,𝑡)
𝑝0 : 𝑝1 :
𝜕2 𝑣
𝑝2 :
2(𝑦,𝑡)
𝜕𝑦 2
=
𝜕2 𝑣1(𝑦,𝑡)
𝜕𝑦 2 ∂𝑣0 ∂𝑣 −𝛺𝑐2 − 𝑐2 − 𝑐1 1 ∂t ∂t
=
𝜕𝑦 2 ∂𝑣 −𝛺 − 𝛺𝑐1 − 𝑐1 0 ∂t
+ 𝑐2
∂2 𝑣
0
∂𝑦 2
+ 𝛼𝑐2
∂ ∂t
∂2𝑣
(
0
∂𝑦 2
= 𝛺,
+
∂2𝑣0
(55) (1 + 𝑐1) + 𝛼𝑐1
∂
∂𝑦 2 ∂2 𝑣1
)+
∂2 𝑣0
(
),
∂t ∂𝑦 2 ∂ ∂ 2𝑣 (1 + 𝑐1) + 𝛼𝑐1 ( 21), 2 ∂𝑦 ∂t ∂𝑦
(57) Solutions of Eq. (43, 44 and 45) using boundary conditions in Eq. (42) are: 1 𝛺 1 𝛺 𝑣0 (𝑦, 𝑡) = [ − + Cos[𝑡𝜔]] + 𝑦 + y 2, 2
1
2
2
1
2
𝑣1 (𝑦, 𝑡) = − 𝜔Sin[𝑡𝜔]𝑐1 + 0𝑦 + 𝜔Sin[𝑡𝜔]𝑐1 𝑣2 (𝑦, 𝑡) = [− 1
5 24
1
2
1
2
(56)
(58) y2,
(59) 1
𝜔2 Cos[𝑡𝜔]𝑐12 − 𝛺𝑐2 − 𝜔Sin[𝑡𝜔]𝑐2 ] − [ 𝑐1𝜔2 Cos[𝑡𝜔](1 − 𝑐1 )] 𝑦 + [ 𝜔2 Cos[𝑡𝜔]𝑐12 + 1
2
6 1
𝛺𝑐2 + 𝜔Sin[𝑡𝜔]𝑐2 ]y 2 + [ 𝜔2 Cos[𝑡𝜔]𝑐1 (1 + 𝑐1)] y 3 − [ 𝜔2 Cos[𝑡𝜔]𝑐12] y 4, 2 6 24 (60) The series solution of second problem velocity profile up to second order is: 𝑣 (𝑦,𝑡) = 𝑣0 (𝑦, 𝑡) + 𝑣1 (𝑦, 𝑡) + 𝑣2 (𝑦,𝑡), Substituting Eq. (58-60) in Eq. (61), we have
56
4
(61)
Gul et al.,2015
1
𝛺
5
2 1
2 2
24
𝑣 (𝑥, 𝑡) = [ − − 𝛺𝑐2 + Cos[𝑡𝜔] (1 − 𝛺
1
1
1
1
2 2
2
6
𝜔2 𝑐12) − 𝜔Sin[𝑡𝜔](𝑐1 + 𝑐2 )] + [ − 𝜔2 Cos[𝑡𝜔]𝑐1(1 − 𝑐1 )]𝑦 + 1
1
[ + +𝛺𝑐2 𝜔 Cos[𝑡𝜔]𝑐12 + 𝜔Sin[𝑡𝜔](𝑐1 + 𝑐2 )] y + [ 𝜔2 Cos[𝑡𝜔]𝑐1(1 + 𝑐1)] y 3 − [𝜔2 Cos[𝑡𝜔]𝑐12 ]y4, 2 4 2 6 24 (62) The values of 𝑐𝑖 for the third problem of velocity profile are c1=-0.993732155929142, c2=0.0006997010832539563. Table 1: Different time variations in velocity profile of first problem. y 0.0
y=0.2
y=0.4
y=0.6
y=0.8
1.591831748
1.693010067
1.794881818
1.8972870308
0.1
1.590970052
1.689761270
1.787656356
1.8844985269
0.2
1.550601507
1.647053815
1.741056151
1.8844985269
0.3 0.4
1.472335480 1.359292191
1.566590313 1.451578589
1.656939007 1.538658410
1.8324611734 1.7432495355
0.5
1.215978318
1.306603799
1.390929832
1.4688699839
0.6
1.048107333
1.137445630
1.219642747
1.2946407170
0.7
0.862371722
0.950847884
1.031625829
1.1046783701
0.8
0.666176179
0.754249624
0.834374721
0.9065561425
0.9
0.467342401
0.555488603
0.635753201
0.7081725422
1.0
0.273797263
0.362488795
0.443679683
0.5174364972
Table 2: Different time variations in velocity profile of second problem. y
y=0.2
y=0.4
y=0.6
y=0.8
0.0
0.7037681161
0.7409699782
0.8028597754
0.889277513
0.1
0.7029301188
0.7377416068
0.7956497056
0.876497611
0.2
0.6625707973
0.6950413017
0.7490544472
0.824462869
0.3
0.5842991503
0.6145713896
0.6649316060
0.735247751
0.4
0.4712356214
0.4995399517
0.5466348942
0.612408980
0.5 0.6 0.7 0.8
0.3278876968 0.1599702058 −0.0258225110 −0.2220834845
0.3545329286 0.1853312926 −0.0013194209 −0.1979780371
0.3988804285 0.2275587131 0.0394998041 −0.157798983
0.460843751 0.286594492 0.09660797 4 −0.10154165
0.9
−0.42098840889
−0.3968043975
−0.356471968
−0.29995477
1.0
−0.61460757254
−0.5898719225
−0.548598687
−0.49072128
Table 3: Time variations in velocity profile of third problem. y 0.0 0.1
y=0.1 1.194888607 1.194648398
y=0.3 1.323711851 1.321877732
y=0.5 1.481284925 1.476264450
y=0.7
0.2
1.154874899
1.280542610
1.431810412
1.608560048
0.3
1.077153754
1.201354384
1.349695053
1.522069793
0.4
0.964583459
1.087470040
1.233192053
1.401660046
0.5
0.821651838
0.943429787
1.086946019
1.252131164
0.6
0.654057122
0.774976055
0.916787320
1.079444391
0.7
0.468480784
0.588824564
0.729499645
0.890484204
0.8
0.272321168
0.392396585
0.532549562
0.692783849
0.9
0.073398538
0.193523082
0.333788851
0.494225016
1.0
−0.12035668
0.000132515
0.141141474
0.302723619
1.667479435 1.657682717
Figure 4: Influence of different time level on velocity profile of first problem.
57
J. Appl. Environ. Biol. Sci., 5(2)52-62, 2015
Figure 5: Influence of different time level on velocity profile of second problem.
Figure 6: Influence of different time level on velocity profile of third problem.
Figure 7: Velocity distribution of fluid on first problem by taking 𝜔 = 0.2 , 𝛺 = 0.4
58
Gul et al.,2015
Figure 8: Velocity distribution of fluid on second problem by taking 𝜔 = 0.2 , 𝛺 = 0.5
Figure 9: Velocity distribution of fluid on third problem by taking 𝜔 = 0.2 , 𝛺 = 0.7
Fig 10: Effect of the time
t on first problem velocity profile when 𝜔 = 0.2, 𝛺 = 0.4
Fig 11: Effect of frequency 𝜔 on first problem velocity profile when 𝑡 = 5, 𝛺 = 0.3
59
J. Appl. Environ. Biol. Sci., 5(2)52-62, 2015
Fig 12: Effect of time 𝑡 on second problem velocity profile when 𝜔 = 0.02, 𝛺 = 0.4
Fig 13: Effect of frequency 𝜔 on second problem velocity profile when 𝑡 = 5, 𝛺 = 0.3
Fig 14: Effect of pressure 𝛺 on second problem velocity profile when 𝑡 = 5, 𝜔 = 0.02
Fig 15: Effect of time 𝑡 on third problem velocity profile when 𝛺 = 0.5, 𝜔 = 0.02
60
Gul et al.,2015
Fig 16: Effect of frequency 𝜔 on third problem velocity profile when 𝑡 = 6, 𝛺 = 2
Fig 17: Effect of pressure 𝛺 on third problem velocity profile when 𝑡 = 5, 𝜔 = 0.02 I
RESULTS AND DISCUSSION
In this paper, the analytical solutions of velocity field of unsteady second grade fluid between two horizontal plates have been established. OHAM technique has been used for the solution of these problems. The geometrical configurations of the problems are shown in Figure1to 3. The different numerical result of velocity profile of these three problems shown in Table 1-3.The influence of different time level on velocity profiles presented in Figures 4-6. The graphical illustration of velocity distribution of fluid on first, second and third problem shows in Figures7-9 respectively. The effects of some physical parameters on three different velocity problems have been deliberated graphically. Fig 10-11 shows the effect of physical parameters 𝒕, 𝝎 on velocity field of first problem. From the graphical results we observed that velocity decrease by increasing the value of 𝒕,𝝎. Figures 12-14 indicate the effect of 𝒕, 𝝎 𝐚𝐧𝐝 𝜴 on second velocity profile. The velocity of second problem is also decrease by raising the value of 𝒕, 𝝎 𝐚𝐧𝐝 𝜴. Figures 15-17 indicate the effect of 𝒕, 𝝎 𝐚𝐧𝐝 𝜴 on third velocity profile. The velocity of third problem is also decreases by increasing the value of 𝒕, 𝝎 𝐚𝐧𝐝 𝜴. II
CONCLUSION
In this paper we develop an approximate analytical solution for the problem of second grade fluid flow between two parallel horizontal plates. The paper contained three different problems (i) when one plate is moving and oscillating (ii) plates are not moving but only oscillating the flow is due to the pressure gradient (iii) the flow is due to pressure gradient and one plate is moving as well as oscillating. The solution for velocity distributions have been obtained by using OHAM method. Effects of some parameters 𝑡, 𝜔 and 𝛺 on velocity distribution are also presented graphically. REFERENCES [1]
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