# MHD Thin Film Flows of a Third Grade Fluid on a Vertical Belt with Slip Boundary Conditions

Authors: Taza Gul; Rehan Ali Shah; Saeed Islam; Muhammad Arif
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

The problem of heat transfer analysis is considered in electrically conducting thin film flows with slip boundary conditions. The flow is assumed to be obeying the nonlinear rheological constitutive equation of a third grade fluid. We have solved the governing nonlinear equations of present problems using the traditional Adomian decomposition method (ADM). Particular attention is given to the combined effect of heat and MHD on the velocity field. The results include the profile of velocity, volume flux, skin friction, average velocity, and the temperature distribution across the film. The effects of model parameters on velocity, skin friction and temperature variation have been studied. Optimal homotopy asymptotic method (OHAM) is also used for comparison. The numerical results and absolute errors are derived in tables.


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## Body

## 1. Introduction

The flow and heat transfer inside thin films are ubiquitous in civil, environmental sciences, mechanical engineering, biological sciences, geophysics, and elsewhere. This is due to their several  applications at large scale such as wire and fiber coating, reactor fluidization, paper production, different food stuffs like ketchup, sauce, and honey, transpiration cooling,  gaseous diffusion, drilling mud, oil wills, heat pipes, and fluid cells. The problem of chambers for chemical and biological detection systems like fluid of many chemicals was considered by Lavrik et al. [1]. Heat transfer inside thin film flow with variable pressure was discussed by Khaled and Vafai [2]. Thin film unsteady flow with variable viscosity was investigated by Nadeem and Awais [3]. Munson and Young [4] discussed the thin film flow of Newtonian fluids. Alam et al. investigated the thin film flow of Johnson-Segalman fluids for lifting and drainage problems [5]. The MHD thin film flow was discussed by Hameed and Ellahi for a non-Newtonian fluid on a vertical moving belt [6]. Due to complexity of non-Newtonian fluids, it becomes difficult to suggest a single model which exhibits all properties of non-Newtonian fluids; therefore various empirical and semiempirical models have been imposed. The use of heat transfer together with the MHD and non-Newtonian fluids under the influence of slip boundary conditions is of particular interest in chemical processing. Relevant and interesting work about present work may be found in [7, 8]. One of the established models amongst non-Newtonian fluids is class of third grade fluids which have their constitutive equations based on strong theoretical foundations, where the relation between the stress and strain is not linear. To solve real world problems, different approximate techniques have been used in mathematics, fluid mechanics, and engineering sciences [9]. Some of the common methods are VIM [10–14], DTM [15, 16], HPM [17–19], HAM, and OHAM [20–24]. These methods deal with the nonlinear problems effectively. The work under various configurations on the thin film flows was discussed by Siddiqui et al. [25]. In [26] they examined the thin film flows of Sisko and Oldroyd-6 constant fluids on a moving belt. The heat transfer analysis of thin film is also discussed by Chakraborty and Som [27]. The main aim of the present work is to study heat transfer into a thin film of a third grade fluid on a vertical belt under the influence of transverse magnetic field with slip boundary conditions using ADM. In 1992 Adomian [28, 29] introduced the ADM for the approximate solutions for linear and nonlinear problems. Wazwaz [30, 31] used ADM for the reliable treatment of Bratu-type and Emden-Fowler equations.


## 2. Basic Equations

The governing equations of an incompressible isothermal and electrically conducting third grade fluid are(1)∇·U=0,(2)ρDUDt=∇·T+ρg+J×B,(3)ρcpDΘDt=κ∇2Θ+tr(τ·L),
where ρ is the constant density, g is body force per unit mass, U is velocity vector of the fluid, Θ is temperature, κ is thermal conductivity, cp is specific heat, L=∇U, D/Dt=∂/∂t+(U·∇) denote material time derivative, τ is Cauchy stress tensor, and T is the shear stress. A uniform magnetic field B=(0,B0,0) is imposed transversely on the belt. The Lorentz force per unit volume is given by
(4)J×B=[0,σB02u(x),0].Shear stress tensorT is given by
(5)T=-pI+τ,
where -pI denotes spherical stress and τ is defined as
(6)τ=μA1+α1A2+α2A12+β1A3+β2(A1A2+A2A1)+β3(trA12)A1.Hereαi and βj are the material constants, and A1, A2, and A3 are the kinematical tensors given by
(7)An=DAn-1Dt+An-1(∇u)+(∇u)TAn-1,n≥1,(8)μ≥0,α1≥0,|α1+α2|≤24μβ3,β3≥0.


## 3. Formulation of the Lift Problem

Consider a wide flat belt moving vertically upward at constant speedU0, through a large bath of third grade liquid as shown in Figure 1. The belt carries with it a layer of liquid of constant thickness, δ. For analysis coordinate system is chosen, in which the x-axis is taken parallel to the surface of the belt and y-axis is perpendicular to the belt. Uniform magnetic field is applied transversely to the belt. Assume the flow is steady and laminar after a small distance above the liquid surface layer. The external pressure is atmospheric everywhere.Figure 1
Geometry of the lifting problem.Velocity and temperature fields are(9)u=(0,u(x),0),Θ=Θ(x).Boundary conditions are(10)u=U0-γTxyatx=0,dudx=0,atx=δ,Θ=Θ0,atx=0,Θ=Θ1,atx=δ.Inserting the velocity field given in (9) in (1) and (5)–(7), the continuity equation (1) satisfies identically and (5) gives the following components of stress tensor
(11)Txx=-P+(2α1+α2)(dudx)2,Txy=μdudx+2(β2+β3)(dudx)3,Tyy=-P+α2(dudx),Tzz=-P,Txz=Tyz=0.Using (11), the momentum and energy equations reduce to
(12)0=μd2udx2+6(β2+β3)(dudx)2(d2udx2)-ρg-σB02u(x),0=κd2Θdx2+μ(dudx)2+2(β2+β3)(dudx)4.Introducing the following nondimensional variables:(13)u¯=δνU,x¯=xδ,Θ¯=Θ-Θ0Θ1-Θ0,γ¯=μγδ,Λ=γνδ,λ=μν2k(Θ1-Θ0)δ2,m=δ3gν2,M=σB02δ2μ,β=(β2+β3)ν2μδ4,α=δU0ν,Re=Uδν,ν=μρ,
where m is the gravitational parameter, M is magnetic parameter, β is non-Newtonian effect, Λ is slip parameter, λ is heat dimensionless number, Re is the local Reynolds number, and α is the nondimensional variable using the above dimensionless variables in (10) and in (12) and dropping bars we obtain
(14)d2udx2+6β(dudx)2(d2udx2)-m-Mu(x)=0,(15)d2Θdx2+λ[(dudx)2+2β(dudx)4]=0,(16)un(0)=α-Λ(dundx+2β(dundx)3),un(0)=1α-1dun(1)dx=0,n=0,(17)un(0)=-Λ(dundx+2β(dundx)3),un(0)=-1Λdun(1)dx=0,n>0,(18)Θ(0)=0,Θ(1)=1.


## 4. Adomian Decomposition Method (ADM)

A general description of the method is as follows. Begin with an equationF~(u)=g~(x), where F~ represents a general nonlinear ordinary differential operator involving both linear and nonlinear terms and g~ is a source term. The linear term is decomposed into L~+R~, where L~ is easily invertible and R~ is the remainder of the linear operator. For convenience, L~ may be taken as the highest order derivative which avoids difficult integrations which result when complicated Green’s functions are involved. Thus the equation may be written as follows:
(19)L~u+R~u+N~u=g~.
Solving for L~u,
(20)L~u=g~-R~u-N~u,
where N~ is non-linear operator and g~ is a source term, since L~ is easily invertible.Equation (20) can be written as follows:
(21)L~-1L~u=L~-1g~-L~-1R~u-L~-1N~u.
We use L~-1 depending on the order of the differential equation. If differential equation is an initial-value problem, if it is desired for boundary value problem as well, integrations are used. The constants of integration are evaluated from the given initial and boundary conditions.  L~-1 can also be treated as definite integral from (t~0 to t~). Solving (21) for u we obtained
(22)u=C1+C2t+L~-1g~-L~-1R~u-L~-1N~u.The non-linear termN~u, is defined as follows:
(23)N~u=∑k=0∞A~n.
Here A~n are special polynomials called Adomian polynomials and u~ will be equated to ∑k=0∞u~n. The initial velocity is identified as follows:
(24)u~0=C~1+C~2t+L~-1g~,∑k=0∞un=u0-L~-1R~∑k=0∞un-L~-1∑k=0∞A~n.
Comparison for different components of velocity profile is as follows:
(25)p.u1=-L~-1R~u0-L~-1A~0,p.u2=-L~-1R~u0-L~-1A~1,ppp.⋮un+1=-L~-1R~un-L~-1A~n.
The Adomian polynomials A~n depend on the velocity components u0,u1,u2,…,un, which play a flourishing role in the rapid convergence of the series. In the above series A~0 depends only on u0, A~1 depends on u0 and u1, A~2 depends on u0, u1, and u2, and so forth. Relevant discussion about ADM can be seen in [16, 17].


### 4.1. The ADM Solution of Lifting Problem

Using the inverse operatorL~-1=∬du′ of the Adomian decomposition method on (14), we obtained
(26)u=C1x+C2+mx22+L~-1[Mu]-6βL~-1[(dudx)2d2ud2x].
For series solution we use summation on (26)
(27)∑n=0∞un=C1x+C2+mx22+ML~-1(∑n=0∞un)-6βL~-1×[ddx(∑n=0∞un)2d2dx2(∑n=0∞un)].
Adomian polynomials are defined from (27) as follows:
(28)A~n=(ddx(∑n=0∞un)2)[d2dx2(∑n=0∞un)],A~n=(ddx(∑n=0∞u1n)2)(∑n=0∞un)n≥0.From (28) when n≥0, the Adomian polynomials in components form are
(29)A~0=(du0dx)2d2u0dx2,A~1=(du0dx)2d2u1dx2+2du0dxdu1dx,A~2=(du0dx)2d2u2dx2+(du1dx)2d2u0dx2+2du0dxdu2dxd2u0dx2.
The series solution becomes
(30)u0+u1+u2+⋯=C1x+C2+mx22+ML-1[u0+u1+u2+⋯]-6βL-1[A~0+A~1+A~2+⋯].The velocity components are obtained by comparing both sides of (33).


#### 4.1.1. Zero Component Problem

Consider(31)u0(x)=C1x+C2+mx22.
From (16) the boundary condition, for n=0, is
(32)u0(0)=α-Λ(du0dx+2β(du0dx)3),du0(1)dx=0.
By making use of (32) in (31), after simplification we obtain
(33)u0(x)=α+mx22-mx+Λ(m+2βm3).


#### 4.1.2. First Component Problem

Consider(34)u1(x)=ML-1[u0]-6βL-1[A~0].
From (17) the boundary condition, for n=1, is
(35)u1(0)=-Λ(du1dx+6β(du0dx)2du1dx),du1(1)dx=0.The solution is(36)u1(x)=[(mM24-m3β2)x4(1+6m2β)Λ0×(mM3-Mα+2m3β-mMΛ-2m3MβΛ)0+(mM3-Mα+2m3β-mMΛ-2m3MβΛ)x0+(Mα2-3m3β+mMΛ2+m3MβΛ)x20+(2m3β-mM6)x3+(mM24-m3β2)x4].


#### 4.1.3. Second Component Problem

Consider(37)u2(x)=ML-1[u1]-6βL-1[A~1].
From (17) the boundary condition, for n=2, is
(38)u2(0)=-Λ(du2dx(1+6β(du0dx)2)+6β(du1dx)2du0dx),=-Λ(du2dx(1+6β11(du0dx)2)±-),du2(1)dx=0.The solution is(39)u2(x)=Ψ0-Ψ1x+Ψ2x2-Ψ3x3+Ψ4x4-Ψ5x5+Ψ6x6.The series solution up to the second component is(40)u(x)=u0(x)+u1(x)+u2(x).
Using (33), (36), and (39) in (40), we have
(41)u(x)=Φ0-Φ1x+Φ2x2-Φ3x3+Φ4x4-Φ5x5+Φ6x6.
The constants (Ψ0–Ψ6) and (Φ0–Φ6) are given in the appendix.The dimensionless shear stress is(42)τxy=μνδ2[dudx+2β(dudx)3].
The shear rate becomes
(43)τxy|x=0=μνδ2[-20ΛM2m3β+20Λ2M2m3β)325(90m2Mαβ+90m2M2αβΛ-15Mαvvvvvvvvvvvvv+5M2α+15M2αΛ-180m5β2vvvvvvvvvvvvv+180m2β2M2Λ2-15m+5mMvvvvvvvvvvvvv-2mM2-15ΛmMΛ+15mM2Λ2vvvvvvvvvvvvv+30m3β-36Mm3β+30ΛMm3βvvvvvvvvvvvvv-20ΛM2m3β+20Λ2M2m3β)vvvvvvvvvvv+(90m2Mαβ+90m2M2αβΛ-15Mαvvvvvvvvvvvvvv+5M2α+15M2αΛ-180m5β2vvvvvvvvvvvvvv+180m2β2M2Λ2-15m+5mMvvvvvvvvvvvvvv-2mM2-15ΛmMΛ+15mM2Λ2vvvvvvvvvvvvvv+30m3β-36Mm3β+30ΛMm3βvvvvvvvvvvvvvv-20ΛM2m3β+20Λ2M2m3β)3].The coefficient of skin friction is defined as follows:(44)Cf(0)=τxy|x=0(1/2)ρU2.By use of (43), we have
(45)Cf(0)=2Re2[-20ΛM2m3β+20Λ2M2m3β)325(90m2Mαβ+90m2M2αβΛvvvvvvvvvvvvv-15Mα+5M2α+15M2αΛvvvvvvvvvvvvv-180m5β2+180m2β2M2Λ2-15mvvvvvvvvvvvvv+5mM-2mM2-15ΛmMΛ+15mM2Λ2vvvvvvvvvvvvv+30m3β-36Mm3β+30ΛMm3βvvvvvvvvvvvvv-20ΛM2m3β+20Λ2M2m3β)vvvvvvv+(90m2Mαβ+90m2M2αβΛvvvvvvvvvv-15Mα+5M2α+15M2αΛvvvvvvvvvv-180m5β2+180m2β2M2Λ2vvvvvvvvvv-15m+5mM-2mM2vvvvvvvvvv-15ΛmMΛ+15mM2Λ2+30m3βvvvvvvvvvv-36Mm3β+30ΛMm3βvvvvvvvvvv-20ΛM2m3β+20Λ2M2m3β)3].


## 4.1. The ADM Solution of Lifting Problem

Using the inverse operatorL~-1=∬du′ of the Adomian decomposition method on (14), we obtained
(26)u=C1x+C2+mx22+L~-1[Mu]-6βL~-1[(dudx)2d2ud2x].
For series solution we use summation on (26)
(27)∑n=0∞un=C1x+C2+mx22+ML~-1(∑n=0∞un)-6βL~-1×[ddx(∑n=0∞un)2d2dx2(∑n=0∞un)].
Adomian polynomials are defined from (27) as follows:
(28)A~n=(ddx(∑n=0∞un)2)[d2dx2(∑n=0∞un)],A~n=(ddx(∑n=0∞u1n)2)(∑n=0∞un)n≥0.From (28) when n≥0, the Adomian polynomials in components form are
(29)A~0=(du0dx)2d2u0dx2,A~1=(du0dx)2d2u1dx2+2du0dxdu1dx,A~2=(du0dx)2d2u2dx2+(du1dx)2d2u0dx2+2du0dxdu2dxd2u0dx2.
The series solution becomes
(30)u0+u1+u2+⋯=C1x+C2+mx22+ML-1[u0+u1+u2+⋯]-6βL-1[A~0+A~1+A~2+⋯].The velocity components are obtained by comparing both sides of (33).


### 4.1.1. Zero Component Problem

Consider(31)u0(x)=C1x+C2+mx22.
From (16) the boundary condition, for n=0, is
(32)u0(0)=α-Λ(du0dx+2β(du0dx)3),du0(1)dx=0.
By making use of (32) in (31), after simplification we obtain
(33)u0(x)=α+mx22-mx+Λ(m+2βm3).


### 4.1.2. First Component Problem

Consider(34)u1(x)=ML-1[u0]-6βL-1[A~0].
From (17) the boundary condition, for n=1, is
(35)u1(0)=-Λ(du1dx+6β(du0dx)2du1dx),du1(1)dx=0.The solution is(36)u1(x)=[(mM24-m3β2)x4(1+6m2β)Λ0×(mM3-Mα+2m3β-mMΛ-2m3MβΛ)0+(mM3-Mα+2m3β-mMΛ-2m3MβΛ)x0+(Mα2-3m3β+mMΛ2+m3MβΛ)x20+(2m3β-mM6)x3+(mM24-m3β2)x4].


### 4.1.3. Second Component Problem

Consider(37)u2(x)=ML-1[u1]-6βL-1[A~1].
From (17) the boundary condition, for n=2, is
(38)u2(0)=-Λ(du2dx(1+6β(du0dx)2)+6β(du1dx)2du0dx),=-Λ(du2dx(1+6β11(du0dx)2)±-),du2(1)dx=0.The solution is(39)u2(x)=Ψ0-Ψ1x+Ψ2x2-Ψ3x3+Ψ4x4-Ψ5x5+Ψ6x6.The series solution up to the second component is(40)u(x)=u0(x)+u1(x)+u2(x).
Using (33), (36), and (39) in (40), we have
(41)u(x)=Φ0-Φ1x+Φ2x2-Φ3x3+Φ4x4-Φ5x5+Φ6x6.
The constants (Ψ0–Ψ6) and (Φ0–Φ6) are given in the appendix.The dimensionless shear stress is(42)τxy=μνδ2[dudx+2β(dudx)3].
The shear rate becomes
(43)τxy|x=0=μνδ2[-20ΛM2m3β+20Λ2M2m3β)325(90m2Mαβ+90m2M2αβΛ-15Mαvvvvvvvvvvvvv+5M2α+15M2αΛ-180m5β2vvvvvvvvvvvvv+180m2β2M2Λ2-15m+5mMvvvvvvvvvvvvv-2mM2-15ΛmMΛ+15mM2Λ2vvvvvvvvvvvvv+30m3β-36Mm3β+30ΛMm3βvvvvvvvvvvvvv-20ΛM2m3β+20Λ2M2m3β)vvvvvvvvvvv+(90m2Mαβ+90m2M2αβΛ-15Mαvvvvvvvvvvvvvv+5M2α+15M2αΛ-180m5β2vvvvvvvvvvvvvv+180m2β2M2Λ2-15m+5mMvvvvvvvvvvvvvv-2mM2-15ΛmMΛ+15mM2Λ2vvvvvvvvvvvvvv+30m3β-36Mm3β+30ΛMm3βvvvvvvvvvvvvvv-20ΛM2m3β+20Λ2M2m3β)3].The coefficient of skin friction is defined as follows:(44)Cf(0)=τxy|x=0(1/2)ρU2.By use of (43), we have
(45)Cf(0)=2Re2[-20ΛM2m3β+20Λ2M2m3β)325(90m2Mαβ+90m2M2αβΛvvvvvvvvvvvvv-15Mα+5M2α+15M2αΛvvvvvvvvvvvvv-180m5β2+180m2β2M2Λ2-15mvvvvvvvvvvvvv+5mM-2mM2-15ΛmMΛ+15mM2Λ2vvvvvvvvvvvvv+30m3β-36Mm3β+30ΛMm3βvvvvvvvvvvvvv-20ΛM2m3β+20Λ2M2m3β)vvvvvvv+(90m2Mαβ+90m2M2αβΛvvvvvvvvvv-15Mα+5M2α+15M2αΛvvvvvvvvvv-180m5β2+180m2β2M2Λ2vvvvvvvvvv-15m+5mM-2mM2vvvvvvvvvv-15ΛmMΛ+15mM2Λ2+30m3βvvvvvvvvvv-36Mm3β+30ΛMm3βvvvvvvvvvv-20ΛM2m3β+20Λ2M2m3β)3].


## 4.1.1. Zero Component Problem

Consider(31)u0(x)=C1x+C2+mx22.
From (16) the boundary condition, for n=0, is
(32)u0(0)=α-Λ(du0dx+2β(du0dx)3),du0(1)dx=0.
By making use of (32) in (31), after simplification we obtain
(33)u0(x)=α+mx22-mx+Λ(m+2βm3).


## 4.1.2. First Component Problem

Consider(34)u1(x)=ML-1[u0]-6βL-1[A~0].
From (17) the boundary condition, for n=1, is
(35)u1(0)=-Λ(du1dx+6β(du0dx)2du1dx),du1(1)dx=0.The solution is(36)u1(x)=[(mM24-m3β2)x4(1+6m2β)Λ0×(mM3-Mα+2m3β-mMΛ-2m3MβΛ)0+(mM3-Mα+2m3β-mMΛ-2m3MβΛ)x0+(Mα2-3m3β+mMΛ2+m3MβΛ)x20+(2m3β-mM6)x3+(mM24-m3β2)x4].


## 4.1.3. Second Component Problem

Consider(37)u2(x)=ML-1[u1]-6βL-1[A~1].
From (17) the boundary condition, for n=2, is
(38)u2(0)=-Λ(du2dx(1+6β(du0dx)2)+6β(du1dx)2du0dx),=-Λ(du2dx(1+6β11(du0dx)2)±-),du2(1)dx=0.The solution is(39)u2(x)=Ψ0-Ψ1x+Ψ2x2-Ψ3x3+Ψ4x4-Ψ5x5+Ψ6x6.The series solution up to the second component is(40)u(x)=u0(x)+u1(x)+u2(x).
Using (33), (36), and (39) in (40), we have
(41)u(x)=Φ0-Φ1x+Φ2x2-Φ3x3+Φ4x4-Φ5x5+Φ6x6.
The constants (Ψ0–Ψ6) and (Φ0–Φ6) are given in the appendix.The dimensionless shear stress is(42)τxy=μνδ2[dudx+2β(dudx)3].
The shear rate becomes
(43)τxy|x=0=μνδ2[-20ΛM2m3β+20Λ2M2m3β)325(90m2Mαβ+90m2M2αβΛ-15Mαvvvvvvvvvvvvv+5M2α+15M2αΛ-180m5β2vvvvvvvvvvvvv+180m2β2M2Λ2-15m+5mMvvvvvvvvvvvvv-2mM2-15ΛmMΛ+15mM2Λ2vvvvvvvvvvvvv+30m3β-36Mm3β+30ΛMm3βvvvvvvvvvvvvv-20ΛM2m3β+20Λ2M2m3β)vvvvvvvvvvv+(90m2Mαβ+90m2M2αβΛ-15Mαvvvvvvvvvvvvvv+5M2α+15M2αΛ-180m5β2vvvvvvvvvvvvvv+180m2β2M2Λ2-15m+5mMvvvvvvvvvvvvvv-2mM2-15ΛmMΛ+15mM2Λ2vvvvvvvvvvvvvv+30m3β-36Mm3β+30ΛMm3βvvvvvvvvvvvvvv-20ΛM2m3β+20Λ2M2m3β)3].The coefficient of skin friction is defined as follows:(44)Cf(0)=τxy|x=0(1/2)ρU2.By use of (43), we have
(45)Cf(0)=2Re2[-20ΛM2m3β+20Λ2M2m3β)325(90m2Mαβ+90m2M2αβΛvvvvvvvvvvvvv-15Mα+5M2α+15M2αΛvvvvvvvvvvvvv-180m5β2+180m2β2M2Λ2-15mvvvvvvvvvvvvv+5mM-2mM2-15ΛmMΛ+15mM2Λ2vvvvvvvvvvvvv+30m3β-36Mm3β+30ΛMm3βvvvvvvvvvvvvv-20ΛM2m3β+20Λ2M2m3β)vvvvvvv+(90m2Mαβ+90m2M2αβΛvvvvvvvvvv-15Mα+5M2α+15M2αΛvvvvvvvvvv-180m5β2+180m2β2M2Λ2vvvvvvvvvv-15m+5mM-2mM2vvvvvvvvvv-15ΛmMΛ+15mM2Λ2+30m3βvvvvvvvvvv-36Mm3β+30ΛMm3βvvvvvvvvvv-20ΛM2m3β+20Λ2M2m3β)3].


## 5. Volume Flow Rate and Average Velocity

Volume flow rate in nondimensional form is as follows:(46)Q=∫01u(x)dx.Using velocity field from (41), volume flow rate is obtained as follows.Average velocityu in dimensionless form is given by
(47)u=Q,(48)Q=-m3+2mM15-17mM2315+α-Mα3+2M2α15+2m3β5-64105m3Mβ+65m2Mαβ-12m5β27+mΛ-19mM2Λ-MαΛ+23M2αΛ+4m3βΛ-815m3MβΛ-3320m3M2βΛ+152m2M2αβΛ-6mM2α2βΛ-1435m5Mβ2Λ+78m4Mαβ2Λ-144m7β3Λ-mMΛ2+13mM2Λ2+M2αΛ2-4m3MβΛ2+296m3M2βΛ2+54m5Mβ2Λ2+12m5M2β2Λ2+12m4M2αβ2Λ2+84m7Mβ3Λ2+mM2Λ3+8m3M2βΛ3+36m5M2β2Λ3+48m7M2β3Λ3.


## 6. Temperature Distribution in Case of Lift Problem

On substituting the series solution for velocity field given in (41) in (15) and solving corresponding to the boundary condition given in (18) for fixed values of α=0.1; m=0.5; M=1; Λ=0.4; β=1; λ=100, we obtained
(49)Θ(x)=7.1923x-12.7305x2+9.0310x3-2.4798x4-0.3026x5+0.4717x6-0.2248x7+0.03702x8+0.01515x9-0.0133x10+0.0044x11-0.00032x12-0.00037x13+0.000202x14-0.000048x15+0.00000156x16+0.000003x17-0.0000014x18+3.3326×10-7x19-5.1691×10-8x20+5.0040×10-9x21-2.2745×10-10x22.


## 7. Formulation of Drainage Problem

The geometry and assumptions of the problem are the same as those in the previous problem. Consider a film of non-Newtonian liquid draining at volume flow rateQ down the vertical belt, as shown in Figure 2. The belt is stationary and the fluid drain down the belt due to gravity. The coordinate system is selected in the same way as that in the previous case. Assuming the flow is steady, laminar and external pressure is neglected. Consider that fluid shear forces balance gravity and the film thickness remain constant.Figure 2
Geometry of the drainage problem.Boundary condition for electrically conducting drainage problem is as follows:(50)u=-γTxyatx=0,dudx=0,atx=δ.
Using nondimensional variables the slip boundary conditions for drainage problem become
(51)un(0)=-Λ(dundx+2β(dundx)3),dun(1)dx=0,=-Λ(dundx+2β(dundx)3),dun(1)dx=0n≥0.


### 7.1. Solution of the Drainage Problem by ADM

Using ADM on (14), the Adomian polynomials in (29) for both problems are the same. The different velocity components are obtained as follows.


#### 7.1.1. Zero Component Problem

Consider(52)u0(x)=C1+C2x-mx22,u0(0)=-Λ(du0dx+2β(du0dx)3),du0(1)dx=0.
The solution is
(53)u0(x)=mx-mx22-Λ(m+2βm3).


#### 7.1.2. First Component Problem

Consider(54)u1(x)=ML-1[u0]-6βL-1[A~0],u1(0)=-Λ(du1dx+6β(du0dx)2du1dx),du1(1)dx=0.
For different velocity components, the Adomian polynomials mentioned in (29) are used:
(55)u1(x)=(1+6m2β)Λ×(-mM3-2m3β+mMΛ+2m3MβΛ)+(-mM3-2m3β+mMΛ+2m3MβΛ)x+(3m3β-mMΛ2-m3MβΛ)x2+(mM6-2m3β)x3+(-mM24+m3β2)x4.


#### 7.1.3. Second Component Problem

Consider(56)u2(x)=ML-1[u1]-6βL-1[A~1],u2(0)=-Λ(du2dx(1+6β(du0dx)2)+6β(du1dx)2du0dx),=-Λ(du2dx(16β(du0dx)2)(du1dx)2),du2(1)dx=0.
The solution is
(57)u2(x)=ξ0-ξ1x+ξ2x2-ξ3x3+ξ4x4-ξ5x5+ξ6x6.
The series solution of the velocity field is
(58)u(x)=u0(x)+u1(x)+u2(x).
Substituting (53), (55), and (57) in (58),
(59)u(x)=ω0-ω1x+ω2x2-ω3x3+ω4x4-ω5x5+ω6x6.
The constants (ξ0–ξ6) and (ω0–ω6) are defined and listed in the appendix.Substituting (59) the shear rate is obtained as
(60)τxy|x=0=μνδ2[+43m3M2βΛ-8m3M2βΛ2)312m5β2-12m5β2M2Λ2+m-mM3vvvv+mΛM+2M2m15-M2Λ2m-2m3βvvvv+3615m3Mβ-2m3MβΛ+43m3M2βΛvvvv-8m3M2βΛ2+2βvvvv×(12m5β2-12m5β2M2Λ2+m-mM3vvvvvvv+mΛM+2M2m15-M2Λ2mvvvvvvv-2m3β+3615m3Mβ-2m3MβΛvvvvvvv+43m3M2βΛ-8m3M2βΛ2)3].


## 7.1. Solution of the Drainage Problem by ADM

Using ADM on (14), the Adomian polynomials in (29) for both problems are the same. The different velocity components are obtained as follows.


### 7.1.1. Zero Component Problem

Consider(52)u0(x)=C1+C2x-mx22,u0(0)=-Λ(du0dx+2β(du0dx)3),du0(1)dx=0.
The solution is
(53)u0(x)=mx-mx22-Λ(m+2βm3).


### 7.1.2. First Component Problem

Consider(54)u1(x)=ML-1[u0]-6βL-1[A~0],u1(0)=-Λ(du1dx+6β(du0dx)2du1dx),du1(1)dx=0.
For different velocity components, the Adomian polynomials mentioned in (29) are used:
(55)u1(x)=(1+6m2β)Λ×(-mM3-2m3β+mMΛ+2m3MβΛ)+(-mM3-2m3β+mMΛ+2m3MβΛ)x+(3m3β-mMΛ2-m3MβΛ)x2+(mM6-2m3β)x3+(-mM24+m3β2)x4.


### 7.1.3. Second Component Problem

Consider(56)u2(x)=ML-1[u1]-6βL-1[A~1],u2(0)=-Λ(du2dx(1+6β(du0dx)2)+6β(du1dx)2du0dx),=-Λ(du2dx(16β(du0dx)2)(du1dx)2),du2(1)dx=0.
The solution is
(57)u2(x)=ξ0-ξ1x+ξ2x2-ξ3x3+ξ4x4-ξ5x5+ξ6x6.
The series solution of the velocity field is
(58)u(x)=u0(x)+u1(x)+u2(x).
Substituting (53), (55), and (57) in (58),
(59)u(x)=ω0-ω1x+ω2x2-ω3x3+ω4x4-ω5x5+ω6x6.
The constants (ξ0–ξ6) and (ω0–ω6) are defined and listed in the appendix.Substituting (59) the shear rate is obtained as
(60)τxy|x=0=μνδ2[+43m3M2βΛ-8m3M2βΛ2)312m5β2-12m5β2M2Λ2+m-mM3vvvv+mΛM+2M2m15-M2Λ2m-2m3βvvvv+3615m3Mβ-2m3MβΛ+43m3M2βΛvvvv-8m3M2βΛ2+2βvvvv×(12m5β2-12m5β2M2Λ2+m-mM3vvvvvvv+mΛM+2M2m15-M2Λ2mvvvvvvv-2m3β+3615m3Mβ-2m3MβΛvvvvvvv+43m3M2βΛ-8m3M2βΛ2)3].


## 7.1.1. Zero Component Problem

Consider(52)u0(x)=C1+C2x-mx22,u0(0)=-Λ(du0dx+2β(du0dx)3),du0(1)dx=0.
The solution is
(53)u0(x)=mx-mx22-Λ(m+2βm3).


## 7.1.2. First Component Problem

Consider(54)u1(x)=ML-1[u0]-6βL-1[A~0],u1(0)=-Λ(du1dx+6β(du0dx)2du1dx),du1(1)dx=0.
For different velocity components, the Adomian polynomials mentioned in (29) are used:
(55)u1(x)=(1+6m2β)Λ×(-mM3-2m3β+mMΛ+2m3MβΛ)+(-mM3-2m3β+mMΛ+2m3MβΛ)x+(3m3β-mMΛ2-m3MβΛ)x2+(mM6-2m3β)x3+(-mM24+m3β2)x4.


## 7.1.3. Second Component Problem

Consider(56)u2(x)=ML-1[u1]-6βL-1[A~1],u2(0)=-Λ(du2dx(1+6β(du0dx)2)+6β(du1dx)2du0dx),=-Λ(du2dx(16β(du0dx)2)(du1dx)2),du2(1)dx=0.
The solution is
(57)u2(x)=ξ0-ξ1x+ξ2x2-ξ3x3+ξ4x4-ξ5x5+ξ6x6.
The series solution of the velocity field is
(58)u(x)=u0(x)+u1(x)+u2(x).
Substituting (53), (55), and (57) in (58),
(59)u(x)=ω0-ω1x+ω2x2-ω3x3+ω4x4-ω5x5+ω6x6.
The constants (ξ0–ξ6) and (ω0–ω6) are defined and listed in the appendix.Substituting (59) the shear rate is obtained as
(60)τxy|x=0=μνδ2[+43m3M2βΛ-8m3M2βΛ2)312m5β2-12m5β2M2Λ2+m-mM3vvvv+mΛM+2M2m15-M2Λ2m-2m3βvvvv+3615m3Mβ-2m3MβΛ+43m3M2βΛvvvv-8m3M2βΛ2+2βvvvv×(12m5β2-12m5β2M2Λ2+m-mM3vvvvvvv+mΛM+2M2m15-M2Λ2mvvvvvvv-2m3β+3615m3Mβ-2m3MβΛvvvvvvv+43m3M2βΛ-8m3M2βΛ2)3].


## 8. Volume Flow Rate and Average Velocity of Thin Film Flow

Using (59) in (47) we obtain the following.The average velocity is defined asu-/δ(61)u-=Q,Q=m3-2mM15+17mM2315-2m3β5+64105m3Mβ+12m5β27-mΛ+19mM2Λ-4m3βΛ+815m3MβΛ+3320m3M2βΛ+1435m5Mβ2Λ+144m7β3Λ+mMΛ2-13mM2Λ2+4m3MβΛ2-296m3M2βΛ2-54m5Mβ2Λ2-3m5M2β2Λ2-84m7Mβ3Λ2-mM2Λ3-8m3M2βΛ3-36m5M2β2Λ3-48m7M2β3Λ3.


## 9. Heat Distribution for Drainage Problem

Solving (15) with boundary conditions given in (18), after making use of (59) for fixed values of m=0.5, M=1, Λ=0.4, β=1, and λ=100, we obtain
(62)Θ(x)=7.7356x-14.3046x2+11.1370x3-4.2154x4+0.57907x5+0.241003x6-0.2503x7+0.0947x8-0.01404x9-0.0068x10+0.0053x11-0.0016x12+0.000103x13+0.000127x14-0.0000609x15+0.0000126x16-8.11926×10-8x17-7.8766446×10-7x18+2.651929×10-7x19-4.828779×10-8x20+5.004021×10-9x21-2.274555×10-10x22.


## 10. Results and Discussion

The effects of magnetic parameterM, non-Newtonian parameter β, gravitational parameter m, slip parameter Λ, and the dimensionless number λ, for both lift and drainage velocity profiles, are discussed in Figures 5–14. Figures 1 and 2 show the geometry of lift and drainage velocity profiles. Figures 3 and 4 show the comparison of OHAM and ADM for lift and drainage velocity profiles. Numerical results and absolute error for both problems are shown in Tables 1 and 2, respectively. Figures 5 and 10 show that the rise in the non-Newtonian parameter β increases the speed of flow. For small values of β, the velocity profile differs little from the Newtonian one; however when β is increased, these profiles become more flattened showing the shear-thinning effect. Behavior of the velocity field u, for different values of M, by fixing other physical parameters, is shown in Figures 6 and 11 for lifting and drainage of fluid, respectively. Here, it can be seen that the boundary layer thickness is reciprocal to the transverse magnetic field and the velocity decreases as one progresses towards the surface of the fluid. We note that the velocity of fluid is maximum at the surface of the belt and minimum at the surface. Moreover, it is to be noted that for large values of M, the velocity increases rapidly as compared to small values. When the gravitational parameter m increases, the velocity decreases in lifting flow and increases in case of drainage. This can be seen from Figures 7 and 12, respectively. Due to friction force the gravitational effect seems to be smaller near the belt. It can be seen that there is a point in the domain, where the velocity of the fluid becomes approximately the same for different values of gravitational parameter. The reason is that the friction of the belt becomes negligible at this point. On increasing m after this point in lifting flow, the velocity decreases due to negligible friction but in drainage flow after this point, the velocity of the fluid increases. The effect of slip parameter can be observed from Figures 8 and 13. It is noticed that the speed of the fluid near the belt is greater than the speed at the surface. When we increase the slip parameter, the velocity of the fluid increases and comparatively this increase can be seen more clearly between Λ=1.0 and Λ=1.5 because the friction goes on decreasing. Figures 9 and 14 indicate the dimensionless temperature distribution for different values of λ. It can be seen that temperature distribution increases as the dimensionless parameter λ  increases and becomes more flattened for large values of λ. Figure 15 shows the effect of local Reynolds number versus skin friction. This figure shows that the Reynolds number decreases the skin friction. For large values of Reynolds number, the skin friction vanishes. Figure 16 shows the effect of non-Newtonian parameter β versus skin friction.Table 1
Comparison of (OHAM) and (ADM) for lift velocity profile.
x
OHAM
ADM
Absolut error
0.0
0.105386
0.106257
0.870
×
10
-
3
0.1
0.062934
0.062767
0.167
×
10
-
3
0.2
0.024379
0.023672
0.707
×
10
-
3
0.3
-
0.010137
-
0.011089
0.953
×
10
-
3
0.4
-
0.0404456
-
0.041497
0.105
×
10
-
2
0.5
-
0.066426
-
0.067481
0.106
×
10
-
2
0.6
-
0.087899
-
0.088944
0.105
×
10
-
2
0.7
-
0.104748
0.105779
0.104
×
10
-
2
0.8
-
0.116863
-
0.117886
0.103
×
10
-
2
0.9
-
0.124166
-
0.125185
0.102
×
10
-
2Table 2
Comparison of (OHAM) and (ADM) for drainage velocity profile.
x
OHAM
ADM
Absolut error
0.0
-
0.00096249
-
0.001043
0.804
×
10
-
4
0.1
0.0081824
0.00810405
0.784
×
10
-
4
0.2
0.0163486
0.016272
0.767
×
10
-
4
0.3
0.0235415
0.0234664
0.752
×
10
-
4
0.4
0.0297656
0.0296918
0.739
×
10
-
4
0.5
0.0350252
0.0349525
0.728
×
10
-
4
0.6
0.0393238
0.0392519
0.719
×
10
-
4
0.7
0.0426641
0.0425929
0.713
×
10
-
4
0.8
0.0450485
0.0449777
0.708
×
10
-
4
0.9
0.0464684
0.0464079
0.705
×
10
-
4Figure 3
Comparison of ADM and OHAM methods for lift velocity profile is shown for the given parameters and auxiliary constants.m=0.5, M=0.1, β=0.3, Λ=0.01, α=0.1, C1=-0.6221311197, and C2=-0.0208205048.Figure 4
Comparison of ADM and OHAM methods for drainage velocity profile is shown for the given parameters and auxiliary constants.m=0.1, M=0.1, β=0.3, Λ=0.01, α=0.1, C1=-0.93798569, and C2=-0.000741729.Figure 5
The influence of non-Newtonianβ on velocity profile for lifting problem keeping α=0.1, m=0.5, M=1, and Λ=0.4.Figure 6
The effect of magnetic force “M” on the lift velocity profile keeping other parameters fixed. α=0.1, m=0.2, β=1.2, and Λ=0.4.Figure 7
The figure shows this gravity effect of “m” on velocity profile for lift problem, where α=0.1, β=1, M=1, and Λ=0.2.Figure 8
Variation of velocity for various values of slip parameterΛ, by fixing α=0.1; m=0.2; β=1.2; M=0.6.Figure 9
The influence of the dimensionless “λ” on the temperature distribution for lift problem keeping, α=0.5, β=0.2, M=2, and m=1.Figure 10
The influence of the non-Newtonian effect “β” on the velocity profile u(x) of drainage problem keeping m=0.5, M=1, and Λ=0.2.Figure 11
The influence of the magnetic force “M” on velocity profile shown in this figure for the drainage problem, where m=0.5, β=1, and Λ=0.2.Figure 12
This figure shows the influence of the gravitational parameterm on the velocity profile “u(x)” for drainage problem keeping β=1, M=1, and Λ=0.2.Figure 13
Variation of velocity for various values of slip parameterΛ, by fixing M=0.6, m=0.2, and β=1.2.Figure 14
The influence of dimensionless number “λ” on the temperature distribution u(x), for drainage problem keeping m=0.5, M=1, Λ=0.4, and β=1.Figure 15
Variation of skin friction versus Reynolds number.Figure 16
Variation of skin friction versus non-Newtonian parameterβ.


## 11. Conclusion

The constitutive equation governing the flow of a third grade fluid for lifting and drainage of fluid with slip boundary conditions is solved analytically by using Adomian decomposition method. Expression for velocity field, volume flow rate, skin friction, and temperature distribution is derived and sketched. It is concluded that velocity increases as the gravitational parameterm decreases in lifting case while velocity increases as this parameter increases in drainage. For small values of β, the velocity profile differs little from the Newtonian one; however when β is increased, these profiles become more flattened showing the shear-thinning effect. It can be seen that the boundary layer thickness is reciprocal to the transverse magnetic effect and the velocity decreases as it progresses towards the surface of the fluid. On increasing the slip parameter, the velocity of the fluid increases and comparatively this increase can be seen more clearly between Λ=1.0 and Λ=1.5 because the friction goes on decreasing. According to the best of our knowledge there is no previous literature about discussed problem; this is our first attempt to handle this problem with slip boundary condition. Also this problem is more general when compared to linear viscous model and second grade fluid model.



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