# Three-Dimensional Rotating Flow of MHD Jeffrey Fluid Flow between Two Parallel Plates with Impact of Hall Current

Authors: Mehreen Fiza; Abdelaziz Alsubie; Hakeem Ullah; Nawaf N. Hamadneh; Saeed Islam; Ilyas Khan
Journal: Mathematical Problems in Engineering
 (2021)

---

## Abstract

This article deals with three-dimensional non-Newtonian Jeffrey fluid in rotating frame in the presence of magnetic field. The flow is studied in the application of Hall current, where the flow is assumed in steady states. The upper plate is considered fixed, and the lower is kept stretched. The fundamental equations are transformed into a set of ordinary differential equations (ODEs). A homotopy technique is practiced for a solution. The variation in the skin friction and its effects on the velocity fields have been examined numerically. The effects of physical parameters are discussed in various plots.


---

## Body

## 1. Introduction

The rotation of fluid exists in nature due to the fact that the fluid particles rotate internally and rises with fluid movement. Due to engineering and industrial applications, the scientist considers the rotational fluid coupled with various features. Rotational fluids have many applications in engineering. Taylor and Geoffrey introduced the motion of viscous fluid in the rotating system [1]. The detailed study of fluid in rotating system is done by Greenspan [2] and Goodman [3]. The effects of MHD in a rotating system and stretched and porous mediums have been studied by Attia and Kotb [4], Borkakoti and Bharali [5], and Vajravelu and Kumar [6]. This work has been magnified along with the temperature effects by Mehmood and Ali [7], Das et al. [8], and Tauseef et al. [9].The non-Newtonian fluid is used in many industry and technology appliances. Hayat et al. studied the non-Newtonian fluid in a rotating frame, considering the effects of MHD for micropolar nanofluids [11, 12]. Jeffrey’s model was presented by Jeffrey as a subclass of non-Newtonian fluid and studied with convection term [13, 14].Most of the physical problems are nonlinear and have rare exact solutions. The numerical methods (NMs) and analytical methods (AMs) are used to get the results. The NMs required discretization techniques which can affect the results. Among the AMs, HAM proposed by Liao is the most powerful and fast convergent [15–19]. Hall introduced Hall current and proves that, in case of strong magnetic field, the Hall current effects cannot be ignored [20]. Similar other interesting studies are provided in [21–32] for different fluid models. This article aims to elaborate the non-Newtonian nanofluid in the rotating frame with Hall effect. Hall effect is produced due to the potential difference across an electrical conductor when a magnetic field is acting in a direction vertical to that of the flow of current. So, for this aim, Jeffrey fluid flow is considered. For the proposed model, HAM is used.


## 2. Problem Formulation

Assume the Jeffrey fluid between two parallel plates havingd separation. The plate and fluid rotate about y axis with Ω. The lower plate is stretched by two opposite and equal forces. A uniform magnetic field B0 is applied perpendicularly with a steady-state condition (Figure 1).Figure 1
Geometrical structure of the flow problem.The fundamental identities are(1)u⌢x+v⌢y+w⌢z=0,(2)ρuu⌢x+vu⌢y+2wΩ=−p⌢x+μ1+γ1u⌢xx+u⌢yy−σB021+m2u+mw+μγ21+γ1⋅2u⌢xu⌢xx+2v⌢xu⌢xy+uu⌢xxx+u⌢xyy+vu⌢xxy+u⌢yyy+u⌢y⋅v⌢xx+u⌢yx+u⌢yy+v⌢xyv⌢y,(3)ρuv⌢x+vv⌢y=−p⌢y+μ1+γ1v⌢xx+v⌢yy+μγ21+γ1⋅2v⌢yv⌢yy+2u⌢yv⌢yx+v⌢xxx+v⌢xyyu+v⌢xxy+v⌢xyyv+u⌢xy+v⌢xxu⌢x+u⌢yy+v⌢xyv⌢y,(4)ρuw⌢x+vw⌢y−2Ωu=μ1+γ1w⌢xx+w⌢yy−σB021+m2mu−w+μγ21+γ1w⌢yyy+w⌢xxyv+w⌢xyu⌢y+u⌢xw⌢xx+uw⌢xxx+w⌢xyy.The BCs are(5)u⌢0=ax,v⌢0=0=w⌢0,u⌢d=v⌢d=w⌢d=0.The similarity transformation used is(6)u⌢=axf′η,v⌢=−adfη,w⌢=axgη,η=yd.Using equation (6) in (1)–(4), we get(7)f″″+1+γ1Rff‴−f′f″−2Krg′=M1+m2f″+mg′+β2f″f‴−ff″″−f′f‴″,(8)g″+1+γ1Rfg′−gf′+2Krf′=M1+m2g−mf′+βfg‴−g′f″.Substituting equation (8) in (7), we get(9)f0=0,f′0=1,g0=0,f1=0,f′1=0,g1=0,where(10)Kr=2Ωd2ν,R=ad2ν,M=σB02d2ρυ,β=aγ2.Cf is given as(11)cf=μ/1+γ1∂2v/∂x2+∂2v/∂y2+γ2u∂2u/∂x∂y+v∂2v/∂x∂y+u∂2v/∂x2+v∂2u/∂y2y=0ρuw2.The dimensionless form ofcf is(12)cfRex=1+γ1−1f″0+βf″0.


## 3. Solution Procedure

HAM was introduced by Liao. LetΨ1,Ψ2 are two continuous functions defined on topological spaces Χ˜, Y˜, then(13)ψ:Χ˜⌢×0,1⟶Y˜⌢,such that x˜∈Χ˜:(14)ψ0=Ψ1,ψ1=Ψ2.The initial guesses are(15)f0=η3−2η3+η,g0=0.The linear terms are(16)Llf=fηηηη,Llg=gηη,with differential operator(17)LlD1+D2η+D3η2+D4η3=0,LllD5+D6η=0,where Dn represents arbitrary constants, where n=1,2,3,…,6.


### 3.1. Zeroth-Order Problem

Expressq∈01 as an embedding parameter with hf and hg, where h≠0. Then,(18)1−qLif˜η,q−f˜0η=phfΝff⌢,g⌢,1−qLgg˜.−g˜0η=phgΝgf˜,g˜.The BCs are(19)f˜o,q.=g⋅⋅˜o,q.=f˜′o,q.=0,f˜1,q.=g˜1,q=f˜′1,q=0,Νl=f^ηηηη+1+γ1Rf^ηηηf^−f^ηηf^η−2krg^ηη;q−M1+m2f^ηη+mg^ηη;q+β2f^ηηf^ηηη−f^f^ηηηη−f^ηf^ηηηηη,Νll=g^ηη+1+γ1Rf⌢g'⌢η−g^f^η+2krf^η+M1+m2mf^η−g^+βg^ηf^ηη−g^ηηηf^,where(20)flη=1l!f^ηq=0,glη=1l!g^ηq=0.


### 3.2. lth-Order Deformation Problem

(21)Llflη−∏lfl−1η=hfℜlfη,Lllglη−∏igl−1η=hgℜlgη,f^l=f^l′=g^l=0,atη=0,f^l=f′^=g^l=0,atη=1,ℜlfη=fl−1iv+2krgl−1+1+γ1R∑j=0l−1fl−1−jfj″′−fl−1−j′fj″−M1+m2fl−1″+mgl−1′+β∑j=0l−12fl−1.−j″fj″′−fl−1−j.fjiv−fl−.1−j′f.jv,ℜlgη=gl−1″−1+γ1R∑j=0k−1fl−1−jgj′−gl−1−jfj′+Kr⋅fl−1′−M1+m2m.fl−1′−gl−1+β∑j=0l−1gl−1−j′fj″−fl−1−jgj″′,where(22)ζ=1,ifq>1,0,ifq≤1.


## 3.1. Zeroth-Order Problem

Expressq∈01 as an embedding parameter with hf and hg, where h≠0. Then,(18)1−qLif˜η,q−f˜0η=phfΝff⌢,g⌢,1−qLgg˜.−g˜0η=phgΝgf˜,g˜.The BCs are(19)f˜o,q.=g⋅⋅˜o,q.=f˜′o,q.=0,f˜1,q.=g˜1,q=f˜′1,q=0,Νl=f^ηηηη+1+γ1Rf^ηηηf^−f^ηηf^η−2krg^ηη;q−M1+m2f^ηη+mg^ηη;q+β2f^ηηf^ηηη−f^f^ηηηη−f^ηf^ηηηηη,Νll=g^ηη+1+γ1Rf⌢g'⌢η−g^f^η+2krf^η+M1+m2mf^η−g^+βg^ηf^ηη−g^ηηηf^,where(20)flη=1l!f^ηq=0,glη=1l!g^ηq=0.


## 3.2. lth-Order Deformation Problem

(21)Llflη−∏lfl−1η=hfℜlfη,Lllglη−∏igl−1η=hgℜlgη,f^l=f^l′=g^l=0,atη=0,f^l=f′^=g^l=0,atη=1,ℜlfη=fl−1iv+2krgl−1+1+γ1R∑j=0l−1fl−1−jfj″′−fl−1−j′fj″−M1+m2fl−1″+mgl−1′+β∑j=0l−12fl−1.−j″fj″′−fl−1−j.fjiv−fl−.1−j′f.jv,ℜlgη=gl−1″−1+γ1R∑j=0k−1fl−1−jgj′−gl−1−jfj′+Kr⋅fl−1′−M1+m2m.fl−1′−gl−1+β∑j=0l−1gl−1−j′fj″−fl−1−jgj″′,where(22)ζ=1,ifq>1,0,ifq≤1.


## 4. Convergence of HAM

With the help of assisting constraintshf and hg, the convergence region is achieved. The possible region of convergence for the proposed model is given in Figure 2 and Table 1.Figure 2
Combinedh curves of function and velocity, at the 15th-order approximation.Table 1
Convergence table of HAM up to the 25th-order approximations whenR=β=γ1=m=M=0.01.
Order of approximationf″0g′013.188860.220792133.176500.238706463.175840.2396453113.175830.2396677153.175830.2396682203.175830.2396682


## 5. Results and Discussion

The effect ofR on fη and gη is given in Figures 3 and 4. An increase in R decreases fη and gη. The large amounts of viscous energy reduction produce large inertial forces, which decreases fη and gη. The effect of kr on the fη and gη is shown in Figures 5 and 6. It is evident that an increase in kr increases fluid flow due to increase in Cariolis force. This fluid rotation increases kinetic energy which also increases the flow rate. The influence of m and γ1 on fη and gη is given in Figures 7–10, respectively. Both reduce velocity profile. The effect β is given in Figures 11 and 12, showing that the velocity profile increases by increasing β. The relaxation time gets smaller by enhancing γ1. The effects of Μ on fη and gη are presented in Figures 13 and 14, respectively. β and Μ oppose the flow due to large relaxation time and magnetic effects. The magnetic field opposes the flow in the y direction and enhance in the z direction.Figure 3
Effect ofR on fη when m=0.5,γ1=0.7,M=1,β=0.4,andkr=0.6.Figure 4
Effect ofR on gηwhen m=0.5,γ1=0.7,M=1,β=0.4,andkr=0.6.Figure 5
Effect ofkr on fη when m=γ1=0.8,M=0.4,R=1,andβ=0.4.Figure 6
Effect ofkr on gη when m=γ1=0.8,M=0.4,R=1,andβ=0.4.Figure 7
Effect ofm onfη when R=1,γ1=0.7,β=0.4,M=1,andkr=0.6.Figure 8
Effect ofm on gη when R=1,γ1=0.7,β=0.4,M=1,andkr=0.6.Figure 9
Effect ofγ1 onfη when.R=0.1,m=0.8,β=0.4,andM=kr=1.Figure 10
Effect ofγ1 on gη when R=0.1,m=0.8,β=0.4,andM=kr=1.Figure 11
Effect ofβ onfη when R=1,m=γ1=0.8,M=1,andkr=0.6.Figure 12
Effect ofβ on gη when R=1,m=γ1=0.8,M=1,andkr=0.6.Figure 13
Effect ofM on fη and gη when R=1,m=γ1=0.8,β=1,andkr=0.6.Figure 14
Effect ofM on fη and gη when R=1,m=γ1=0.8,β=1,andkr=0.6.The numerical values ofR,γ1,β, and kr on Cf are presented in Table 2. We see that Cf has inverse relations with R,γ1,β and decreases Cf while on direct relation with kr.Table 2
Variation in skin friction coefficient for dissimilar values ofR,Kr,β, and γ1 when m=0.1andM=0.5.
RΚrγ1βcfShehzad et al. [15] resultsPresent results0.010.51.00.52.633123.864160.12.651332.948820.52.639952.642081.00.01.312174.339990.51.259174.341570.91.216944.368970.50.02.385085.642270.51.259175.445760.91.033994.899111.00.00.619112.22743


## 6. Conclusion

The following conclusion is observed:(i)
A rise inR causes to decline cf.(ii)
The mass flux decreases at a lower plate and increases at upper plate.(iii)
R,γ1,m resist the velocity profile.(iv)
β,Kr assist the velocity profile.(v)
M resists the flow along the y direction and assists the flow along the z direction.



---

*Source: 6626411-2021-03-19.xml*