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Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013, Article ID 620238, 11 pageshttp://dx.doi.org/10.1155/2013/620238

Research ArticleAnalysis of Third-Grade Fluid in Helical Screw Rheometer

M. Zeb,1 S. Islam,2 A. M. Siddiqui,3 and T. Haroon1

1 Department of Mathematics, COMSATS Institute of Information Technology, Islamabad, Pakistan2Department of Mathematics, Abdul Wali Khan University, Mardan, Pakistan3 Department of Mathematics, Pennsylvania State University, York Campus, York, PA 17 403, USA

Correspondence should be addressed to M. Zeb; [email protected]

Received 17 October 2012; Revised 4 March 2013; Accepted 20 March 2013

Academic Editor: Juan Torregrosa

Copyright © 2013 M. Zeb et al. This is an open access article distributed under the Creative Commons Attribution License, whichpermits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

The steady flowof an incompressible, third-grade fluid in helical screw rheometer (HSR) is studied by “unwrapping or flattening” thechannel, lands, and the outside rotating barrel. The geometry is approximated as a shallow infinite channel, by assuming that thewidth of the channel is large as compared to the depth. The developed second-order nonlinear coupled differential equations arereduced to single differential equation by using a transformation. Using Adomian decomposition method, analytical expressionsare calculated for the the velocity profiles and volume flow rates.The results have been discussed with the help of graphs as well. Weobserved that the velocity profiles are strongly dependant on non-Newtonian parameter ( ̃𝛽), and with the increase in ̃𝛽, the velocityprofiles increase progressively, which conclude that extrusion process increases with the increase in ̃𝛽. We also observed that theincrease in pressure gradients in x- and z-direction increases the net flow inside the helical screw rheometer, which increases theextrusion process. We noticed that the flow increases as the flight angle increase.

1. Introduction

In real life, there are many materials that exhibit the mechan-ical characteristics of both elasticity and viscosity. Thesematerials are known as non-Newtonian fluids. These fluidscannot be described satisfactorily by the theory of elasticity orviscosity but by a combination of both. Due to the rheologicalbehavior of these fluids, many constitutive equations are pro-posed [1]. In most fluid food products, the shear stress isdependent on the share rate; hence, nonlinear flow curveresults and a unique viscosity are no longer adequate to char-acterize the fluid. Many fluids such as molten plastics, poly-mers, and slurries are non-Newtonian in their flow behav-ior. The basic governing equations for such fluids motionare highly nonlinear differential equations having no generalsolution, and only a limited number of exact solutions havebeen established for particular problems. To solve practi-cal problems in engineering and mathematics, researchersand scientists have developed numerous numerical tech-niques, that is, finite difference method (FDM), finite vol-ume approach, control-volume-based finite element method(CVFEM), lattice Boltzmann method (LBM), and analytical

techniques, that is, variational iteration method (VIM),perturbation method (PM), homotopy perturbation method(HPM), HPM-Pade technique, homotopy analysis method(HAM), optimal homotopy analysis method (OHAM), opti-mal homotopy perturbation method (OHPM), and someother techniques, to overcome nonlinearity and get numer-ical and analytical solutions [2–10]. A brief review on ana-lytical techniques is presented by [11]. In recent years inthe area of series solutions, an iterative technique Adomiandecomposition method [12, 13] has received much attention.A considerable amount of research work has been investedin the application of this method to a wide class of linear,nonlinear, and partial differential equations and integralequations. Many interesting problems in applied science andengineering have been successfully solved by using ADMto their higher degree of accuracy. A useful quality of theADM is that it has proved to be a competitive alternative tothe Taylor series method and other series techniques. Thismethod has been used in obtaining analytic and approximatesolutions to a wide class of linear and nonlinear, differ-ential and integral equations, homogeneous or inhomoge-neous, with constant coefficients or with variable coefficients.

2 Journal of Applied Mathematics

The Adomian decomposition method is comparatively eas-ier to program in engineering problems than other seriesmethods and provides immediate and visible solution termswithout linearization, perturbation, or discretization of theproblem, while the physical behavior of the solution remainsunchanged. It provides analytical solution in the form of aninfinite series in which each term can be easily determined[14–16]. If an exact solution exists for the problem, thenthe obtained series converges very rapidly to the solution.For concrete problems, where a closed-form solution is notobtainable, a truncated number of terms are usually used fornumerical purposes [17].

The Helical Screw Rheometer (HSR) consists of a helicalscrew in a tight fitting cylinder, with the inlet and outlet partsclosing the inner screw. Rotation of screw creates a pressuregradient along the axis of the screw. The HSR is being usedfor rheological measurements of fluid food suspensions. Thegeometry of an HSR is similar to a single-screw extruder [18].Extrusion process is widely used in multigrade oils, liquiddetergents, paints, polymer solutions and polymer melts [19],the injection molding process for polymeric materials, theproduction of pharmaceutical products, food extrusion, andprocessing of plastics [20]. Various food items in daily life,such as cookie dough, sevai, pastas, breakfast cereals, frenchfries, baby food, ready to eat snacks, and dry pet food, aremost commonly manufactured using the extrusion process.

Knowledge of rheological properties is essential in theprocessing of fluid foods since these affect the flow behavior.During processing, physical and chemical changes can occurso it is desirable to monitor the process to achieve excellentoutput and quality control [21]. On-line rheological measure-ments in the food industry have been limited [22].

Bird et al. [23] presented an asymptotic solution andarbitrary values of the flow behavior index, for the power-law fluid in a very thin annulus. A brief discussion is givenbyMohr andMallouk [24] for the same problem consideringNewtonian fluid in a screw extruder. Tamura et al. [18] alsoinvestigated the flow of Newtonian fluid in helical screwrheometer.

The objective of this paper is to study the flow of third-grade fluid in helical screw rheometer (HSR)where the effectsof curvature and also of flights are neglected by assumingthat the helical channel is “unwrapped.” The geometry isapproximated as a shallow infinite channel, with ℎ/𝐵 ≪ 1,where 𝐵 denotes the channel width and ℎ is gap [18]. Theformulation results in second-order nonlinear coupled dif-ferential equations which are reduced to first-order nonlineardifferential equations by integrating and combined in singlefirst-order differential equation using a transformation, thesolution is obtained by using ADM. Analytical expressionsare given for the velocity components in 𝑥-, 𝑧-directionsand in direction of the screw axis. Volume flow rates arealso obtained for all three types of velocities. The paperis organized as follows. Section 2 contains the governingequations of the fluid model. In Section 3, the problemunder consideration is formulated. In Section 4, descriptionof Adomian decompositionmethod is given. In Section 5, thegoverning equation of the problem is solved. In Section 6,results are discussed. Section 7 contains conclusion.

2. Basic Equations

The basic equations governing the motion of an incompress-ible fluid are

divV = 0, (1)

𝜌

𝐷V𝐷𝑡

= 𝜌f + divT, (2)

where 𝜌 is the constant fluid density,V is the velocity vector, fis the body force per unitmass, the𝐷/𝐷𝑡 denotes thematerialtime derivative defined as

𝐷 (∗)

𝐷𝑡

=

𝜕

𝜕𝑡

(∗) + (V ⋅ ∇) (∗) , (3)

and T is the Cauchy stress tensor, given as

T = −𝑃I + S, (4)

where 𝑃 denotes the dynamic pressure, I denotes unit tensor,and S denotes the extra stress tensor. The constitutive equa-tion for third-grade fluid is defined as

S = 𝜇A1+ 𝛼

1A2+ 𝛼

2A21+ 𝛽

1A3

+ 𝛽

2(A1A2+ A2A1) + 𝛽

3(trA21)A1,

(5)

where 𝜇 is the viscosity, 𝛼1, 𝛼2, 𝛽1, 𝛽2, and 𝛽

3are the material

constants, A1, A2, and A

3are the first three Rivlin-Ericksen

tensors defined as [19]

A1= (gradV) + (gradV)𝑇,

A𝑛+1

=

𝐷A𝑛

𝐷𝑡

+ [A𝑛(gradV) + (gradV)𝑇A

𝑛] , (𝑛 = 1, 2) .

(6)

3. Problem Formulation

Consider the steady flow of an isothermal, incompressibleand homogeneous third-grade fluid in helical screw rheome-ter (HSR) in such a way that the curvature of the screw chan-nel is ignored, unrolled and laid out on a flat surface. Thebarrel surface is also flattened. Assume that the screw surface,the lower plate, is stationary, and the barrel surface, the upperplate, is moving across the top of the channel with velocity𝑉 at an angle 𝜙 to the direction of the channel Figure 1. Thephenomena is the same as the barrel held stationary and thescrew rotates. The geometry is approximated as a shallowinfinite channel, by assuming that the width 𝐵 of the channelis large compared with the depth ℎ; edge effects in the fluidat the land are ignored. The coordinate axes are positionedin such a way that the 𝑥-axis is perpendicular to the walland 𝑧-axis is in down channel direction. The liquid wets allthe surfaces and moves by the shear stresses produced by therelative movement of the barrel and channel. For simplicity,the velocity of the barrel relative to the channel is broken upinto two components:𝑈 is along 𝑥-axis and𝑊 is along 𝑧-axis

Journal of Applied Mathematics 3

𝐵

sin 𝜙

Moving barrelsurface

Stationarychannel

Axis of screw

Small gapbetween barrel

and lands

𝑥𝑧

𝑉

𝑊

ℎ

𝐵

𝜙

𝑋

𝑋

𝑈

Figure 1:The geometry of the “unwrapped” screw channel and bar-rel surface.

[24]. Under these assumptions the velocity field and cauchystress tensor can be written as

V = [𝑢 (𝑦) , 0, 𝑤 (𝑦)] , T = T (𝑦) . (7)

On substituting (7) in (4) and (5), we obtain nonzero compo-nents of Cauchy stress T,

𝜏

𝑥𝑥= −𝑃 + 𝛼

2(

𝑑𝑢

𝑑𝑦

)

2

,

𝜏

𝑥𝑦= 𝜏

𝑦𝑥= 𝜇

𝑑𝑢

𝑑𝑦

+ 2 (𝛽

2+ 𝛽

3) {(

𝑑𝑢

𝑑𝑦

)

2

+ (

𝑑𝑤

𝑑𝑦

)

2

}

𝑑𝑢

𝑑𝑦

,

𝜏

𝑥𝑧= 𝜏

𝑧𝑥= 𝛼

2

𝑑𝑢

𝑑𝑦

𝑑𝑤

𝑑𝑦

,

𝜏

𝑦𝑦= −𝑃 + (2𝛼

1+ 𝛼

2) [(

𝑑𝑢

𝑑𝑦

)

2

+ (

𝑑𝑤

𝑑𝑦

)

2

] ,

𝜏

𝑦𝑧= 𝜏

𝑧𝑦= 𝜇

𝑑𝑤

𝑑𝑦

+ 2 (𝛽

2+ 𝛽

3) {(

𝑑𝑢

𝑑𝑦

)

2

+ (

𝑑𝑤

𝑑𝑦

)

2

}

𝑑𝑤

𝑑𝑦

,

𝜏

𝑧𝑧= −𝑃 + 𝛼

2(

𝑑𝑤

𝑑𝑦

)

2

,

(8)

where T = [𝜏𝑖𝑗].

Using (7), (1) is identically satisfied, and (2) in the absenceof body forces results in

0 = −

𝜕𝑃

𝜕𝑥

+

𝜕

𝜕𝑦

× [𝜇

𝑑𝑢

𝑑𝑦

+ 2 (𝛽

2+ 𝛽

3) {(

𝑑𝑢

𝑑𝑦

)

2

+ (

𝑑𝑤

𝑑𝑦

)

2

}

𝑑𝑢

𝑑𝑦

] ,

0 = −

𝜕𝑃

𝜕𝑦

+

1

2

(2𝛼

1+ 𝛼

2)

𝜕

𝜕𝑦

[(

𝑑𝑢

𝑑𝑦

)

2

+ (

𝑑𝑤

𝑑𝑦

)

2

] ,

0 = −

𝜕𝑃

𝜕𝑧

+

𝜕

𝜕𝑦

× [𝜇

𝑑𝑤

𝑑𝑦

+ 2 (𝛽

2+ 𝛽

3) {(

𝑑𝑢

𝑑𝑦

)

2

+ (

𝑑𝑤

𝑑𝑦

)

2

}

𝑑𝑤

𝑑𝑦

] .

(9)

Define the modified pressure ̂𝑃 as

̂

𝑃 = 𝑃 − (𝛼

1+

1

2

𝛼

2){(

𝑑𝑢

𝑑𝑦

)

2

+ (

𝑑𝑤

𝑑𝑦

)

2

} , (10)

which implies that ̂𝑃 = ̂𝑃(𝑥, 𝑧) only; thus, (9) reduce to

𝑑

2𝑢

𝑑𝑦

2+

2 (𝛽

2+ 𝛽

3)

𝜇

𝑑

𝑑𝑦

[{(

𝑑𝑢

𝑑𝑦

)

2

+ (

𝑑𝑤

𝑑𝑦

)

2

}

𝑑𝑢

𝑑𝑦

] =

1

𝜇

𝜕

̂

𝑃

𝜕𝑥

,

𝑑

2𝑤

𝑑𝑦

2+

2 (𝛽

2+ 𝛽

3)

𝜇

𝑑

𝑑𝑦

[{(

𝑑𝑢

𝑑𝑦

)

2

+ (

𝑑𝑤

𝑑𝑦

)

2

}

𝑑𝑤

𝑑𝑦

] =

1

𝜇

𝜕

̂

𝑃

𝜕𝑧

.

(11)

The associated boundary conditions can be taken as (seeFigure 1)

𝑢 = 0, 𝑤 = 0, at 𝑦 = 0,

𝑢 = 𝑈, 𝑤 = 𝑊, at 𝑦 = ℎ,(12)

where

𝑈 = −𝑉 sin𝜙, 𝑊 = 𝑉 cos𝜙. (13)

Introducing nondimensionalized parameters

𝑥

∗=

𝑥

ℎ

, 𝑦

∗=

𝑦

ℎ

, 𝑧

∗=

𝑧

ℎ

, 𝑢

∗=

𝑢

𝑊

, 𝑤

∗=

𝑤

𝑊

,

𝑃

∗=

𝑃

𝜇 (𝑊/ℎ)

,

̃

𝛽

∗=

(𝛽

2+ 𝛽

3)𝑊

2

𝜇ℎ

2,

(14)


in (11)-(12), takes the form

𝑑

2𝑢

∗

𝑑𝑦

∗2+

̃

𝛽

∗ 𝑑

𝑑𝑦

∗[{(

𝑑𝑢

∗

𝑑𝑦

∗)

2

+ (

𝑑𝑤

∗

𝑑𝑦

∗)

2

}

𝑑𝑢

∗

𝑑𝑦

∗] =

𝜕𝑃

∗

𝜕𝑥

∗,

𝑑

2𝑤

∗

𝑑𝑦

∗2+

̃

𝛽

∗ 𝑑∗

𝑑𝑦

∗[{(

𝑑𝑢

∗

𝑑𝑦

∗)

2

+ (

𝑑𝑤

∗

𝑑𝑦

∗)

2

}

𝑑𝑤

∗

𝑑𝑦

∗] =

𝜕𝑃

∗

𝜕𝑧

∗,

𝑢

∗= 0, 𝑤

∗= 0, at 𝑦∗ = 0,

𝑢

∗=

𝑈

𝑊

, 𝑤

∗= 1, at 𝑦∗ = 1.

(15)

Dropping "∗" from (15) onward and defining

𝐹 = 𝑢 + 𝜄𝑤, 𝑉

0=

𝑈

𝑊

+ 𝜄1, 𝐺 = 𝑃

,𝑥+ 𝜄𝑃

,𝑧, (16)

where 𝜕𝑃/𝜕𝑥 = 𝑃,𝑥, 𝜕𝑃/𝜕𝑧 = 𝑃

,𝑧in (15) reduce to

𝑑

2𝐹

𝑑𝑦

2= 𝐺 −

̃

𝛽{(

𝑑𝐹

𝑑𝑦

)

2𝑑

2𝐹

𝑑𝑦

2+ 2

𝑑𝐹

𝑑𝑦

𝑑

2𝐹

𝑑𝑦

2

𝑑𝐹

𝑑𝑦

} , (17)

where 𝐹 is the complex conjugate of 𝐹.The boundary conditions become

𝐹 = 0 at 𝑦 = 0,

𝐹 = 𝑉

0at 𝑦 = 1.

(18)

Equation (17) is second-order nonlinear ordinary differentialequation, and the exact solution seems to be difficult. In thefollowing section, we use Adomian decomposition methodto obtain the approximate solution. To obtain the expressionsfor the velocity components in 𝑥- and 𝑧-directions, (17)together with the boundary conditions (18) is solved up tothe second component approximations by using the symboliccomputation software Wolfram Mathematica 7.

4. Description of AdomianDecomposition Method

Consider equation 𝐽[𝐹(𝑦)] = 𝑔(𝑦), where 𝐽 represents ageneral nonlinear ordinary or partial differential operatorincluding both linear and nonlinear terms. The linear termsare decomposed into 𝐿 + 𝑅, where 𝐿 is invertible. 𝐿 is takenas the highest-order derivative to avoid difficult integrations,and 𝑅 is the remainder of the linear operator. Thus, theequation can be written as

𝐿 (𝐹) + 𝑅 (𝐹) + 𝑁 (𝐹) = 𝑔 (𝑦) , (19)

where𝑁(𝐹) indicates the nonlinear term and 𝑔(𝑦) is forcingfunction. Since 𝐿 is invertible, so 𝐿−1 exist. The above equa-tion can be written as

𝐿

−1𝐿 (𝐹) = 𝐿

−1𝑔 (𝑦) − 𝐿

−1𝑅 (𝐹) − 𝐿

−1𝑁(𝐹) . (20)

If 𝐿(= 𝑑2/𝑑𝑦2) is a second-order operator, 𝐿−1(=∫ ∫(∗)𝑑𝑦 𝑑𝑦) is a twofold indefinite integral. Equation (20)becomes

𝐹 = 𝐶

1+ 𝐶

2𝑦 + 𝐿

−1𝑔 (𝑦) − 𝐿

−1𝑅 (𝐹) − 𝐿

−1𝑁(𝐹) , (21)

where 𝐶1and 𝐶

2are constants of integration and can be

determined by using boundary or initial conditions. ADMassumes that the solution 𝐹 can be expanded into infiniteseries as 𝐹 = ∑∞

𝑛=0𝐹

𝑛; also, the nonlinear term 𝑁(𝐹) will be

written as 𝑁(𝐹) = ∑∞𝑛=0

𝐴

𝑛, where 𝐴

𝑛are special Adomian

polynomials which can be defined as

𝐴

𝑛=

1

𝑛!

[

𝑑

𝑛

𝑑𝜆

𝑛{𝑁(

∞

∑

𝑖=0

𝜆

𝑖𝐹

𝑖)}]

𝜆=0

𝑛 = 0, 1, 2, 3, . . . ,

(22)

finally, the solution can be written as

∞

∑

𝑛=0

𝐹

𝑛= 𝐹

0− 𝐿

−1𝑅(

∞

∑

𝑛=0

𝐹

𝑛) − 𝐿

−1(

∞

∑

𝑛=0

𝐴

𝑛) , (23)

where

𝐹

0= 𝐶

1+ 𝐶

2𝑦 + 𝐿

−1𝑔 (𝑦) (24)

is initial solution and

𝐹

𝑛= −𝐿

−1𝑅 (𝐹

𝑛−1) − 𝐿

−1𝐴

𝑛−1, 𝑛 ≥ 1, (25)

and (25) is 𝑛th-order solution. The practical solution will bethe 𝑛-term approximation

𝜙

𝑛=

𝑛−1

∑

𝑖=0

𝐹

𝑖, (26)

and by definition [25–27],

lim𝑛→∞

𝜙

𝑛=

𝑛−1

∑

𝑖=0

𝐹

𝑖= 𝐹. (27)

5. Solution of the Problem

Adomian decomposition method describes that in the oper-ator form (17) can be written as

𝐿 (𝐹) = 𝐺 −

̃

𝛽{(

𝑑𝐹

𝑑𝑦

)

2𝑑

2𝐹

𝑑𝑦

2+ 2

𝑑𝐹

𝑑𝑦

𝑑

2𝐹

𝑑𝑦

2

𝑑𝐹

𝑑𝑦

} , (28)

where 𝐿 is the differential operator taken as the highest-orderderivative to avoid difficult integrations, assuming that 𝐿 isinvertible, which implies that 𝐿−1 = ∫∫(∗)𝑑𝑦 𝑑𝑦 exist.

On applying 𝐿−1 to both sides of (28) results in

𝐹 = 𝐴 + 𝐵𝑦 + 𝐿

−1(𝐺) −

̃

𝛽𝐿

−1

× {(

𝑑𝐹

𝑑𝑦

)

2𝑑

2𝐹

𝑑𝑦

2+ 2

𝑑𝐹

𝑑𝑦

𝑑

2𝐹

𝑑𝑦

2

𝑑𝐹

𝑑𝑦

} ,

(29)


where 𝐴 and 𝐵 are constants of integration and can be deter-mined by using boundary conditions. According to proce-dure of Adomian decomposition method 𝐹 and 𝐹 can bewritten in component form as:

𝐹 =

∞

∑

𝑛=0

𝐹

𝑛,

𝐹 =

∞

∑

𝑛=0

𝐹

𝑛.

(30)

Thus, (29) takes the form∞

∑

𝑛=0

𝐹

𝑛= 𝐴 + 𝐵𝑦 + 𝐿

−1(𝐺) −

̃

𝛽𝐿

−1

× {(

𝑑

𝑑𝑦

(

∞

∑

𝑛=0

𝐹

𝑛))

2

(

𝑑

2

𝑑𝑦

2(

∞

∑

𝑛=0

𝐹

𝑛))

+ 2(

𝑑

𝑑𝑦

(

∞

∑

𝑛=0

𝐹

𝑛))(

𝑑

2

𝑑𝑦

2(

∞

∑

𝑛=0

𝐹

𝑛))

× (

𝑑

𝑑𝑦

(

∞

∑

𝑛=0

𝐹

𝑛))} .

(31)

Adomian also suggested that the nonlinear terms can beexplored in the form of Adomian polynomials, say, 𝐴

𝑛and

𝐵

𝑛as∞

∑

𝑛=0

𝐴

𝑛= (

𝑑

𝑑𝑦

(

∞

∑

𝑛=0

𝐹

𝑛))

2

(

𝑑

2

𝑑𝑦

2(

∞

∑

𝑛=0

𝐹

𝑛)) ,

∞

∑

𝑛=0

𝐵

𝑛= 2(

𝑑

𝑑𝑦

(

∞

∑

𝑛=0

𝐹

𝑛))(

𝑑

2

𝑑𝑦

2(

∞

∑

𝑛=0

𝐹

𝑛))(

𝑑

𝑑𝑦

(

∞

∑

𝑛=0

𝐹

𝑛)) .

(32)

Equation (31) yields∞

∑

𝑛=0

𝐹

𝑛= 𝐴 + 𝐵𝑦 + 𝐿

−1(𝐺) −

̃

𝛽𝐿

−1(

∞

∑

𝑛=0

𝐴

𝑛+

∞

∑

𝑛=0

𝐵

𝑛) .

(33)

The associated boundary conditions (18) will be∞

∑

𝑛=0

𝐹

𝑛= 0, 𝑦 = 0,

∞

∑

𝑛=0

𝐹

𝑛= 𝑉

0, 𝑦 = 1.

(34)

The recursive relation in (33) and (34) gives the componentproblems

𝐹

0= 𝐴 + 𝐵𝑦 + 𝐿

−1(𝐺) , (35)

along with boundary conditions

𝐹

0= 0, 𝑦 = 0,

𝐹

0= 𝑉

0, 𝑦 = 1,

(36)

𝐹

𝑗+1= −

̃

𝛽𝐿

−1(𝐴

𝑗+ 𝐵

𝑗) , 𝑗 ≥ 0 (37)

together with the boundary conditions

∞

∑

𝑛=1

𝐹

𝑛= 0, 𝑦 = 0,

∞

∑

𝑛=1

𝐹

𝑛= 0, 𝑦 = 1.

(38)

The ADM solution to (33) along with the boundary condi-tions (34) will be

𝐹 =

∞

∑

𝑛=0

𝐹

𝑛. (39)

5.1. Zeroth Component Solution. The relations (35) and (36)give the zeroth component problem

𝐹

0= 𝐴 + 𝐵𝑦 + 𝐿

−1(𝐺) , (40)

and the boundary conditions are

𝐹

0= 0 at 𝑦 = 0,

𝐹

0= 𝑉

0at 𝑦 = 1,

(41)

which gives the solution

𝑢

0=

𝑈

𝑊

𝑦 +

1

2

𝑃

,𝑥(𝑦

2− 𝑦) , (42)

𝑤

0= 𝑦 +

1

2

𝑃

,𝑧(𝑦

2− 𝑦) , (43)

which are the linearly viscous solutions to the problem.

5.2. First Component Solution. Equations (37) and (38) give

𝐹

1= −

̃

𝛽𝐿

−1(𝐴

0+ 𝐵

0) , (44)

𝐹

1= 0 at 𝑦 = 0,

𝐹

1= 0 at 𝑦 = 1,

(45)

where the remainder term 𝑅 of the linear part is zero and

𝐴

0= (

𝑑𝐹

0

𝑑𝑦

)

2𝑑

2𝐹

0

𝑑𝑦

2,

𝐵

0= 2(

𝑑𝐹

0

𝑑𝑦

𝑑

2𝐹

0

𝑑𝑦

2

𝑑𝐹

0

𝑑𝑦

)

(46)

are Adomian polynomials. Using (45)-(46) in (44) results in

𝑢

1= −

̃

𝛽 {𝐿

11(𝑦

2− 𝑦) + 𝐿

12(𝑦

3− 𝑦) + 𝐿

13(𝑦

4− 𝑦)} ,

(47)

𝑤

1= −

̃

𝛽 {𝑇

11(𝑦

2− 𝑦) + 𝑇

12(𝑦

3− 𝑦) + 𝑇

13(𝑦

4− 𝑦)} ,

(48)

where 𝐿11, 𝐿12, 𝐿13, 𝑇11, 𝑇12, and 𝑇

13are constants given in

appendix.


5.3. Second Component Solution. The relations (37) and (38)give

𝐹

2= −

̃

𝛽𝐿

−1(𝐴

1+ 𝐵

1) ,

𝐹

2= 0 at 𝑦 = 0,

𝐹

2= 0 at 𝑦 = ℎ,

(49)

𝐴

1= (

𝑑𝐹

0

𝑑𝑦

)

2𝑑

2𝐹

1

𝑑𝑦

2+ 2

𝑑𝐹

0

𝑑𝑦

𝑑𝐹

1

𝑑𝑦

𝑑

2𝐹

0

𝑑𝑦

2,

𝐵

1= 2(

𝑑𝐹

0

𝑑𝑦

𝑑

2𝐹

0

𝑑𝑦

2

𝑑𝐹

1

𝑑𝑦

+

𝑑𝐹

0

𝑑𝑦

𝑑

2𝐹

1

𝑑𝑦

2

𝑑𝐹

0

𝑑𝑦

+

𝑑𝐹

1

𝑑𝑦

𝑑

2𝐹

0

𝑑𝑦

2

𝑑𝐹

0

𝑑𝑦

) ,

(50)

where 𝐴1and 𝐵

1are Adomian polynomials.

Using (50) in (49), we get

𝑢

2=

̃

𝛽

2{𝐿

14(𝑦

2− 𝑦) + 𝐿

15(𝑦

3− 𝑦) + 𝐿

16(𝑦

4− 𝑦)

+ 𝐿

17(𝑦

5− 𝑦) + 𝐿

18(𝑦

6− 𝑦)} ,

(51)

𝑤

2=

̃

𝛽

2{𝑇

14(𝑦

2− 𝑦) + 𝑇

15(𝑦

3− 𝑦) + 𝑇

16(𝑦

4− 𝑦)

+ 𝑇

17(𝑦

5− 𝑦) + 𝑇

18(𝑦

6− 𝑦)} ,

(52)

where 𝐿14, 𝐿15, 𝐿16, 𝐿17, 𝐿18, 𝑇14, 𝑇15, 𝑇16, 𝑇17, and 𝑇

18are

constants mentioned in appendix.

5.4. Velocity Profiles

5.4.1. Velocity Profile in 𝑥-Direction. Equations (42), (47),and (51) give the ADM solution for the velocity profile in thetransverse plane

𝑢 =

𝑈

𝑊

𝑦 + (

1

2

𝑃

,𝑥+

̃

𝛽𝐿

11+

̃

𝛽

2𝐿

14) (𝑦

2− 𝑦)

+ (

̃

𝛽𝐿

12+

̃

𝛽

2𝐿

15) (𝑦

3− 𝑦) + (

̃

𝛽𝐿

13+

̃

𝛽

2𝐿

16) (𝑦

4− 𝑦)

+

̃

𝛽

2𝐿

17(𝑦

5− 𝑦) +

̃

𝛽

2𝐿

18(𝑦

6− 𝑦) .

(53)

5.4.2. Velocity Profile in 𝑧-Direction. Equations (43), (48) and(52) give the ADM solution for the velocity profile in thedown channel direction

𝑤 = 𝑦 + (

1

2

𝑃

,𝑧+

̃

𝛽𝑇

11+

̃

𝛽

2𝑇

14) (𝑦

2− 𝑦)

+ (

̃

𝛽𝑇

12+

̃

𝛽

2𝑇

15) (𝑦

3− 𝑦) + (

̃

𝛽𝑇

13+

̃

𝛽

2𝑇

16) (𝑦

4− 𝑦)

+

̃

𝛽

2𝑇

17(𝑦

5− 𝑦) +

̃

𝛽

2𝑇

18(𝑦

6− 𝑦) .

(54)

5.4.3. Velocity in the Direction of the Axis of Screw. The veloc-ity in the direction of the axis of the screw at any depth in thechannel can be computed from (53) and (54) as

𝑠 = 𝑤 sin𝜙 + 𝑢 cos𝜙, (55)

𝑠 = {𝑦 + (

1

2

𝑃

,𝑧+

̃

𝛽𝑇

11+

̃

𝛽

2𝑇

14) (𝑦

2− 𝑦)

+ (

̃

𝛽𝑇

12+

̃

𝛽

2𝑇

15) (𝑦

3− 𝑦)

+ (

̃

𝛽𝑇

13+

̃

𝛽

2𝑇

16) (𝑦

4− 𝑦) +

̃

𝛽

2𝑇

17(𝑦

5− 𝑦)

+

̃

𝛽

2𝑇

18(𝑦

6− 𝑦)} sin𝜙

+ {

𝑈

𝑊

𝑦 + (

1

2

𝑃

,𝑥+

̃

𝛽𝐿

11+

̃

𝛽

2𝐿

14) (𝑦

2− 𝑦)

+ (

̃

𝛽𝐿

12+

̃

𝛽

2𝐿

15) (𝑦

3− 𝑦)

+ (

̃

𝛽𝐿

13+

̃

𝛽

2𝐿

16) (𝑦

4− 𝑦) +

̃

𝛽

2𝐿

17(𝑦

5− 𝑦)

+

̃

𝛽

2𝐿

18(𝑦

6− 𝑦) } cos𝜙,

(56)

which shows the resultant velocity of the flow.

5.5. Volume Flow Rates. Volume flow rate in 𝑥-direction perunit width is

𝑄

∗

𝑥= ∫

1

0

𝑢 𝑑𝑦,(57)

where 𝑄∗𝑥= 𝑄

𝑥/𝑊ℎ𝐵, and (57) gives

𝑄

∗

𝑥=

𝑈

2𝑊

−

1

6

(

1

2

𝑃

,𝑥+

̃

𝛽𝐿

11+

̃

𝛽

2𝐿

14) −

1

4

(

̃

𝛽𝐿

12+

̃

𝛽

2𝐿

15)

−

3

10

(

̃

𝛽𝐿

13+

̃

𝛽

2𝐿

16) −

1

3

̃

𝛽

2𝐿

17−

5

14

̃

𝛽

2𝐿

18.

(58)

Volume flow rate in 𝑧-direction per unit width is

𝑄

∗

𝑧= ∫

1

0

𝑤𝑑𝑦,(59)

where 𝑄∗𝑧= 𝑄

𝑧/𝑊ℎ𝐵, and (59) gives

𝑄

∗

𝑧=

1

2

−

1

6

(

1

2

𝑃

,𝑧+

̃

𝛽𝑇

11+

̃

𝛽

2𝑇

14) −

1

4

(

̃

𝛽𝑇

12+

̃

𝛽

2𝑇

15)

−

3

10

(

̃

𝛽𝑇

13+

̃

𝛽

2𝑇

16) −

1

3

̃

𝛽

2𝑇

17−

5

14

̃

𝛽

2𝑇

18.

(60)

Equation (56) gives the resultant volume flow rate forward inthe screw channel, which is the product of the velocity andcross-sectional area integrated from the root of the screw tothe barrel surface

𝑄

∗=

𝑛

sin𝜙∫

1

0

𝑠 𝑑𝑦, (61)


0 0.2 0.4 0.6 0.8 1

0

1

2

3

−1

�̃� = 0.9

�̃� = 0.7

�̃� = 0.5

�̃� = 0.3

�̃� = 0

𝑢(𝑦)

𝑦

Figure 2: Profile of the nondimensional velocity 𝑢(𝑦) for differentvalues of ̃𝛽, keeping 𝑃

,𝑥= −2.0, 𝑃

,𝑧= −2.0, and 𝜙 = 45∘.

where𝑄∗ = 𝑄/𝑊ℎ𝐵 and 𝑛 is the number of parallel flights ina multiflight screw.

Equation (61) gives

𝑄

∗=

𝑛

sin𝜙[{

1

2

−

1

6

(

1

2

𝑃

,𝑧+

̃

𝛽𝑇

11+

̃

𝛽

2𝑇

14)

−

1

4

(

̃

𝛽𝑇

12+

̃

𝛽

2𝑇

15) −

3

10

(

̃

𝛽𝑇

13+

̃

𝛽

2𝑇

16)

−

1

3

̃

𝛽

2𝑇

17−

5

14

̃

𝛽

2𝑇

18} sin𝜙

+ {

𝑈

2𝑊

−

1

6

(

1

2

𝑃

,𝑥+

̃

𝛽𝐿

11+

̃

𝛽

2𝐿

14)

−

1

4

(

̃

𝛽𝐿

12+

̃

𝛽

2𝐿

15) −

3

10

(

̃

𝛽𝐿

13+

̃

𝛽

2𝐿

16)

−

1

3

̃

𝛽

2𝐿

17−

5

14

̃

𝛽

2𝐿

18} cos𝜙] ,

(62)

which can be written as

𝑄

∗=

𝑛

sin𝜙{𝑄

∗

𝑧sin𝜙 + 𝑄∗

𝑥cos𝜙} . (63)

6. Results and Discussion

In the present work, we have considered the steady flowof an incompressible, isothermal, and homogeneous third-grade fluid in helical screw rheometer (HSR). UsingAdomiandecomposition method, solutions are obtained for velocityprofiles in 𝑥-, 𝑧-directions and also in the direction of the axisof the screw 𝑠(𝑦). The volume flow rates are also calculatedby using the velocities in 𝑥, 𝑧 and in the direction of the axisof the screw. Here we discussed the effect of dimensionlessparameters ̃𝛽, 𝑈/𝑊 = − tan𝜙, 𝑃

,𝑥, and 𝑃

,𝑧where 𝜙 = 45∘,

on the velocity profiles given in (53), (54), and (56) with the

0 0.2 0.4 0.6 0.8 10

1

2

3

4

�̃� = 0.9

�̃� = 0.7

�̃� = 0.5

�̃� = 0.3

�̃� = 0

𝑦

𝑤(𝑦)

Figure 3: Profile of the nondimensional velocity 𝑤(𝑦) for differentvalues of ̃𝛽, keeping 𝑃

,𝑥= −2.0, 𝑃

,𝑧= −2.0, and 𝜙 = 45∘.

0 0.2 0.4 0.6 0.8 10

1

2

3

4𝑠(𝑦)

𝑦

�̃� = 0.9

�̃� = 0.7

�̃� = 0.5

�̃� = 0.3

�̃� = 0

Figure 4: Profile of the nondimensional velocity 𝑠(𝑦) for differentvalues of ̃𝛽, keeping 𝑃

,𝑥= −2.0, 𝑃

,𝑧= −2.0, and 𝜙 = 45∘.

help of graphical representation. Figures 2, 3, and 4 for thevelocities in𝑥-direction𝑢(𝑦), 𝑧-direction𝑤(𝑦), and the resul-tant velocity 𝑠(𝑦) are plotted against 𝑦 for different values ofnon-Newtonian parameter ̃𝛽 and constant pressure gradients𝑃

,𝑥= −2.0, 𝑃

,𝑧= −2.0, respectively. From these figures, it is

seen that the velocity profiles are strongly dependant on thenon-Newtonian parameter ̃𝛽, as we increase the value of ̃𝛽 inthe interval 0 to 0.9, the progressive increase in velocities in𝑥-, 𝑧-direction and in the direction of the axis of screw found.It is worthwhile to note that the extrusion process increaseswith the increase of the non-Newtonian parameter ̃𝛽.

Figures 5, 7, and 9 are sketched for the velocity profiles𝑢(𝑦), 𝑤(𝑦), and 𝑠(𝑦) against 𝑦 for different values of 𝑃

,𝑥,

keeping ̃𝛽 = 0.3 and 𝑃,𝑧= −2.0 fixed; enlightened escalation


0 0.2 0.4 0.6 0.8 1

0

1

2

3

−1

𝑃,𝑥 = −4

𝑃,𝑥 = −3.5

𝑃,𝑥 = −3

𝑃,𝑥 = −2.5

𝑃,𝑥 = −2

𝑢(𝑦)

𝑦

Figure 5: Profile of the nondimensional velocity 𝑢(𝑦) for differentvalues of 𝑃

,𝑥, keeping ̃𝛽 = 0.3, 𝑃

,𝑧= −2.0, and 𝜙 = 45∘.

0 0.2 0.4 0.6 0.8 1

0

−1

−0.2

−0.6

−0.4

−0.8

𝑢(𝑦)

𝑦

𝑃,𝑧 = −4

𝑃,𝑧 = −3.5

𝑃,𝑧 = −3

𝑃,𝑧 = −2.5

𝑃,𝑧 = −2

Figure 6: Profile of the nondimensional velocity 𝑢(𝑦) for differentvalues of 𝑃

,𝑧, keeping ̃𝛽 = 0.3, 𝑃

,𝑥= −2.0, and 𝜙 = 45∘.

is noted in the velocity profiles with increase in pressuregradient in 𝑥-direction.

Figures 6, 8, and 10 are sketched for the velocity profiles𝑢(𝑦), 𝑤(𝑦), and 𝑠(𝑦) against 𝑦 for different values of 𝑃

,𝑧,

keeping ̃𝛽 = 0.3 and 𝑃,𝑥

= −2.0 fixed, it is observed thatparabolicity of the velocity profiles increases with increase inpressure gradient in 𝑧-direction.

Figure 11 is plotted for the velocity 𝑠(𝑦) against 𝑦 for dif-ferent values of 𝜙, keeping ̃𝛽 = 0.3,𝑃

,𝑥= −2.0, and𝑃

,𝑧= −2.0.

It is observed that flow increases as the flight angle increasesup to 𝜙 = 45∘.

7. Conclusion

The steady flow of an isothermal, homogeneous and incom-pressible third-grade fluid is investigated in helical screw

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

𝑦

𝑃,𝑥 = −4

𝑃,𝑥 = −3.5

𝑃,𝑥 = −3

𝑃,𝑥 = −2.5

𝑃,𝑥 = −2

𝑤(𝑦)

Figure 7: Profile of the nondimensional velocity 𝑤(𝑦) for differentvalues of 𝑃


,𝑧= −2.0, and 𝜙 = 45∘.

𝑤(𝑦)

0 0.2 0.4 0.6 0.8 10

1

2

3

4

𝑦

𝑃,𝑧 = −4

𝑃,𝑧 = −3.5

𝑃,𝑧 = −3𝑃,𝑧 = −2.5

𝑃,𝑧 = −2

Figure 8: Profile of the nondimensional velocity 𝑤(𝑦) for differentvalues of 𝑃


,𝑥= −2.0, and 𝜙 = 45∘.

rheometer (HSR). The geometry of the problem under con-sideration gives second-order nonlinear coupled differentialequations which are reduced to single differential equationby using a transformation. Adomian decomposition methodis used to obtain analytical expressions for the flow profiles,volume flow rate. It is noticed that the zeroth componentsolution matches with solution of the linearly viscous fluidin HSR, and it is also found that the net velocity of the fluidis due to the pressure gradient as the expression for the netvelocity is free from the drag term. Graphical representationshows that the velocity profiles are strongly dependant onnon-Newtonian parameter ( ̃𝛽) and pressure gradients in 𝑥-and 𝑧-direction.Thus, the extrusion process strongly dependson the involved parameters.


0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

2

2.5

𝑦

𝑃,𝑥 = −4

𝑃,𝑥 = −3.5

𝑃,𝑥 = −3

𝑃,𝑥 = −2.5

𝑃,𝑥 = −2

𝑠(𝑦)

Figure 9: Profile of the nondimensional velocity 𝑠(𝑦) for differentvalues of 𝑃


,𝑧= −2.0, and 𝜙 = 45∘.

0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

2

2.5

𝑦

𝑠(𝑦)

𝑃,𝑧 = −4

𝑃,𝑧 = −3.5

𝑃,𝑧 = −3

𝑃,𝑧 = −2.5

𝑃,𝑧 = −2



,𝑥= −2.0, and 𝜙 = 45∘.

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

𝑦

𝑠(𝑦)

𝜙 = 20∘

𝜙 = 30∘

𝜙 = 45∘



,𝑥= −2.0, and 𝜙 = 45∘.

Appendix

𝐿

11=

1

2

(𝑃

1+

3𝑈

2𝑃

1

𝑊

2−

3𝑈𝑃

2

1

𝑊

+

3𝑃

3

1

4

+

2𝑈𝑃

2

𝑊

− 2𝑃

1𝑃

2−

𝑈𝑃

2

2

𝑊

+

3

4

𝑃

1𝑃

2

2) ,

𝐿

12=

1

6

(

6𝑈𝑃

2

1

𝑊

− 3𝑃

3

1+ 4𝑃

1𝑃

2+

2𝑈𝑃

2

2

𝑊

− 3𝑃

1𝑃

2

2) ,

𝐿

13=

1

12

(3𝑃

3

1+ 3𝑃

1𝑃

2

2) ,

𝐿

14=

1

2

(𝑃

1+

9𝑈

4𝑃

1

𝑊

4+

10𝑈

2𝑃

1

𝑊

2−

27𝑈

3𝑃

2

1

𝑊

3

−

15𝑈𝑃

2

1

𝑊

+

13𝑃

3

1

3

+

21𝑈

2𝑃

3

1

𝑊

2

−

27𝑈𝑃

4

1

4𝑊

+

15𝑃

5

1

16

+

8𝑈

3𝑃

2

𝑊

3+

8𝑈𝑃

2

𝑊

− 10𝑃

1𝑃

2−

34𝑈

2𝑃

1𝑃

2

𝑊

2+

25𝑈𝑃

2

1𝑃

2

𝑊

−

16

3

𝑃

3

1𝑃

2−

5𝑈

3𝑃

2

2

𝑊

3−

17𝑈𝑃

2

2

𝑊

+

38

3

𝑃

1𝑃

2

2

+

38𝑈

2𝑃

1𝑃

2

2

3𝑊

2−

49𝑈𝑃

2

1𝑃

2

2

6𝑊

+

15

8

𝑃

3

1𝑃

2

2

+

25𝑈𝑃

3

2

3𝑊

−

16

3

𝑃

1𝑃

3

2−

17𝑈𝑃

4

2

12𝑊

+

15

16

𝑃

1𝑃

4

2) ,

𝐿

15=

1

6

(

54𝑈

3𝑃

2

1

𝑊

3+

30𝑈𝑃

2

1

𝑊

− 18𝑃

3

1

−

90𝑈

2𝑃

3

1

𝑊

2+

87𝑈𝑃

4

1

2𝑊

−

15𝑃

5

1

2

+

68𝑈

2𝑃

1𝑃

2

𝑊

2−

108𝑈𝑃

2

1𝑃

2

𝑊

+

104

3

𝑃

3

1𝑃

2

+

10𝑈

3𝑃

2

2

𝑊

3+

34𝑈𝑃

2

2

𝑊

− 54𝑃

1𝑃

2

2

−

54𝑈

2𝑃

1𝑃

2

2

𝑊

2+

157𝑈𝑃

2

1𝑃

2

2

3𝑊

− 15𝑃

3

1𝑃

2

2−

36𝑈𝑃

3

2

𝑊

+

104

3

𝑃

1𝑃

3

2+

53𝑈𝑃

4

2

6𝑊

−

15

2

𝑃

1𝑃

4

2+ 20𝑃

1𝑃

2) ,

𝐿

16=

1

12

(18𝑃

3

1+

90𝑈

2𝑃

3

1

𝑊

2−

90𝑈𝑃

4

1

𝑊

+

45𝑃

5

1

2

+

108𝑈𝑃

2

1𝑃

2

𝑊

− 72𝑃

3

1𝑃

2+ 54𝑃

1𝑃

2

2


+

54𝑈

2𝑃

1𝑃

2

2

𝑊

2−

108𝑈𝑃

2

1𝑃

2

2

𝑊

+ 45𝑃

3

1𝑃

2

2+

36𝑈𝑃

3

2

𝑊

− 72𝑃

1𝑃

3

2−

18𝑈𝑃

4

2

𝑊

+

45

2

𝑃

1𝑃

4

2) ,

𝐿

17=

1

20

(

60𝑈𝑃

4

1

𝑊

− 30𝑃

5

1+ 48𝑃

3

1𝑃

2+

72𝑈𝑃

2

1𝑃

2

2

𝑊

− 60𝑃

3

1𝑃

2

2

+ 48𝑃

1𝑃

3

2+

12𝑈𝑃

4

2

𝑊

− 30𝑃

1𝑃

4

2) ,

𝐿

18=

1

30

(15𝑃

5

1+ 30𝑃

3

1𝑃

2

2+ 15𝑃

1𝑃

4

2) ,

𝑇

11=

1

2

(2

𝑈𝑃

1

𝑊

− 𝑃

2

1+ 3𝑃

2+

𝑈

2𝑃

2

𝑊

2

−2

𝑈𝑃

1𝑃

2

𝑊

+

3

4

𝑃

2

1𝑃

2− 3𝑃

2

2+

3

4

𝑃

3

2) ,

𝑇

12=

1

6

(2𝑃

2

1+ 4

𝑈𝑃

1𝑃

2

𝑊

− 3𝑃

2

1𝑃

2+ 6𝑃

2

2− 3𝑃

3

2) ,

𝑇

13=

1

12

(3𝑃

2

1𝑃

2+ 3𝑃

3

2) ,

𝑇

14=

1

2

(

8𝑈

3𝑃

1

𝑊

3+

8𝑈𝑃

1

𝑊

− 5𝑃

2

1−

17𝑈

2𝑃

2

1

𝑊

2

+

25𝑈𝑃

3

1

3𝑊

−

17𝑃

4

1

12

+ 9𝑃

2+

𝑈

4𝑃

2

𝑊

4

+

10𝑈

2𝑃

2

𝑊

2−

10𝑈

3𝑃

1𝑃

2

𝑊

3−

34𝑈𝑃

1𝑃

2

𝑊

+

38

3

𝑃

2

1𝑃

2+

38𝑈

2𝑃

2

1𝑃

2

3𝑊

2−

16𝑈𝑃

3

1𝑃

2

3𝑊

+

15

16

𝑃

4

1𝑃

2− 27𝑃

2

2−

15𝑈

2𝑃

2

2

𝑊

2+

25𝑈𝑃

1𝑃

2

2

𝑊

−

49

6

𝑃

2

1𝑃

2

2+ 21𝑃

3

2+

13𝑈

2𝑃

3

2

3𝑊

2

−

16𝑈𝑃

1𝑃

3

2

3𝑊

+

15

8

𝑃

2

1𝑃

3

2−

27𝑃

4

2

4

+

15𝑃

5

2

16

) ,

𝑇

15=

1

6

(10𝑃

2

1+

34𝑈

2𝑃

2

1

𝑊

2−

36𝑈𝑃

3

1

𝑊

+

53𝑃

4

1

6

+

20𝑈

3𝑃

1𝑃

2

𝑊

3+

68𝑈𝑃

1𝑃

2

𝑊

− 54𝑃

2

1𝑃

2

−

54𝑈

2𝑃

2

1𝑃

2

𝑊

2+

104𝑈𝑃

3

1𝑃

2

3𝑊

−

15

2

𝑃

4

1𝑃

2

+ 54𝑃

2

2+

30𝑈

2𝑃

2

2

𝑊

2−

108𝑈𝑃

1𝑃

2

2

𝑊

+

157

3

𝑃

2

1𝑃

2

2− 90𝑃

3

2−

18𝑈

2𝑃

3

2

𝑊

2

+

104𝑈𝑃

1𝑃

3

2

3𝑊

− 15𝑃

2

1𝑃

3

2+

87𝑃

4

2

2

−

15𝑃

5

2

2

) ,

(A.1)

𝑇

16=

1

12

(

36𝑈𝑃

3

1

𝑊

− 18𝑃

4

1+ 54𝑃

2

1𝑃

2+

54𝑈

2𝑃

2

1𝑃

2

𝑊

2

−

72𝑈𝑃

3

1𝑃

2

𝑊

+

45

2

𝑃

4

1𝑃

2

+

108𝑈𝑃

1𝑃

2

2

𝑊

− 108𝑃

2

1𝑃

2

2+ 90𝑃

3

2+

18𝑈

2𝑃

3

2

𝑊

2

−

72𝑈𝑃

1𝑃

3

2

𝑊

+ 45𝑃

2

1𝑃

3

2− 90𝑃

4

2+

45𝑃

5

2

2

) ,

𝑇

17=

1

20

(12𝑃

4

1+

48𝑈𝑃

3

1𝑃

2

𝑊

− 30𝑃

4

1𝑃

2+ 72𝑃

2

1𝑃

2

2

+

48𝑈𝑃

1𝑃

3

2

𝑊

− 60𝑃

2

1𝑃

3

2+ 60𝑃

4

2− 30𝑃

5

2) ,

𝑇

18=

1

30

(15𝑃

4

1𝑃

2+ 30𝑃

2

1𝑃

3

2+ 15𝑃

5

2) .

(A.2)

Acknowledgment

The first author is very thankful to Higher Education Com-mission (HEC) of Pakistan for funding his higher studiesunder the 5000 indigenous scholarship scheme Batch-IV.

References

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