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## Mass Transpiration in Nonlinear MHD Flow Due to Porous Stretching Sheet

### Abstract

Motivated from numerous practical applications, the present theoretical and numerical work investigates the nonlinear magnetohydrodynamic (MHD) laminar boundary layer flow of an incompressible, viscous fluid over a porous stretching sheet in the presence of suction/injection (mass transpiration). The flow characteristics are obtained by solving the underlying highly nonlinear ordinary differential equation using homotopy analysis method. The effect of parameters corresponding to suction/injection (mass transpiration), applied magnetic field, and porous stretching sheet parameters on the nonlinear flow is investigated. The asymptotic limits of the parameters regarding the flow characteristics are obtained mathematically, which compare very well with those obtained using the homotopy analysis technique. A detailed numerical study of the laminar boundary layer flow in the vicinity of the porous stretching sheet in MHD and offers a particular choice of the parametric values to be taken in order to practically model a particular type of the event among suction and injection at the sheet surface.

### Introduction

The steady, laminar MHD boundary layer flows driven by moving boundaries are among the classical problems of theoretical fluid mechanics (see Schlichting1). The usual stretching sheet problems arise in polymer extrusion processes that involve the cooling of continuous strips extruded from a dye by drawing them horizontally through a stagnant cooling fluid2.

The phenomena of momentum transfer in steady, laminar boundary layer flows have received much attention due to their wide applications which include drawing of plastics and elastic sheets, metal and polymer expulsion forms, paper creation, and cooling of metallic sheets etc.2,3.

Viscous fluid flow past a linear stretching sheet is a classical problem of laminar boundary layer flow. Blasius4 first discovered the boundary layer flow on a flat plate using similarity transformations. Sakiadis5,6 investigated the steady laminar boundary layer flow on a moving plate in a quiescent liquid and obtained both closed form as well as approximate solutions. Crane7 considered the flow due to stretching of plastic sheets in the polymer industry and obtained an analytical solution of the laminar boundary layer equations. Recently, Al-Housseiny and Stone8 have investigated the laminar boundary layer flow due to motion of stretching sheets by taking into account both the fluid motion as well as the motion of the sheet.

The control of the boundary layer flow due to stretching sheet can be enhanced by introducing magnetohydrodynamic (MHD) effects. This can be done by taking an electrically conducting fluid above the sheet and applying a magnetic field perpendicular to the plane of the sheet. In this connection, Pavlov9 was the first to investigate the MHD flow past a stretching sheet using the Hartman formulation. He found that the applied magnetic field and permeability cause depletion of the boundary layer thickness near the sheet surface. Chakrabarti and Gupta10 extended the classical work of Crane7 to include the effect of a transverse magnetic field and obtained the analytical solutions to the MHD flow over a stretching sheet.

Siddheshwar and Mahabaleshwar3 investigated MHD viscoelastic fluid flow and heat transfer over a stretching sheet in the presence of radiations using Chandrasekhar formulation11. Thereafter, many authors have investigated the MHD boundary layer flows past a stretching/shrinking sheet with different control parameters and conditions12,13,14,15,16,17,18,19,20,21,22,23.

Boundary layer flows through saturated porous media and MHD has gained significant attention in the recent times because of their promising engineering applications, such as in moisture transport in thermal insulation, ceramic processing, extraction of geothermal energy, nuclear reactor cooling systems, underground nuclear waste disposal, ground water pollution control, and filtration processes.

The objective of the present study is to investigate the boundary layer fluid flow over a permeable stretching sheet. The fluid under consideration is taken as electrically conducting and subjected to an externally imposed magnetic field normal to the stretching surface. The flow, mass transfer, and MHD effects are examined by applying the well known homotopy analysis method1Footnote 124 (hereafter referred to as HAM)13,25,26 to the highly nonlinear ordinary differential equations governing the flow. The HAM was originally proposed and developed by Liao24,25,27,28,29,30,31. This method is suitable for investigations of the present type as is evident from its use in the earlier studies13 regarding ease of its applicability and accuracy for obtaining the solution of nonlinear ordinary differential equations. There is another powerful variant of HAM known as optimal homotopy perturbation method (OHPM)32,33 available in literature to obtain analytic solutions of nonlinear differential equations. However, we will not use OHPM in the present work since HAM is sufficient for our purpose. The present study focuses on obtaining nonlinear numerical calculations relevant for visualizing the changes happening in the boundary layer flow generated due to stretching of the sheet by application of the external magnetic field, the porosity of the medium around the sheet, and suction/injection of the fluid at the sheet surface.

### Mathematical Model

We consider the steady two dimensional flow of an electrically conducting, incompressible Newtonian fluid through a porous medium over a stretching sheet issuing from a slit at the origin of the rectangular coordinates as shown in Fig. 1. The sheet is assumed to be horizontal and coincides with the plane y = 0. Two equal and opposite forces are applied to the sheet to stretch it along the x-axis. The velocity of the stretching is (U(x),0), where

$$U(x)={U}_{0}{(x/L)}^{n}$$

such that n > −1 is the stretching parameter; U0 and L are the characteristic scales for measuring the velocity of the stretching and distance, respectively. The porous medium is assumed to have permeability

$$\kappa (x)={\kappa }_{0}{(x/L)}^{1-n}$$

and is subjected to an external vertical magnetic field

$${{\bf{H}}}_{0}(x)=(0,{H}_{0}{(x/L)}^{(n-1)/2}),$$

where κ0 and H0 denote the characteristic scales for measuring the permeability of the porous medium and magnetic field, respectively. The electrical conductivity σ of the fluid is assumed to be small so that the induced magnetic field in the fluid is weak as compared to the applied magnetic field. The sheet is permeable and subjected to suction velocity (0,V(x)) (see [1, Ch. 11, pp. ]), where

$$V(x)=-{V}_{0}\sqrt{\frac{n+1}{2}}{(x/L)}^{(n-\mathrm{1)/2}};\,n >$$

As a convention, V(x) > 0 implies suction while V(x) < 0 implies injection of the fluid at y = 0.

The stretching of the sheet induces a fluid velocity field (u(x, y), v(x, y)) which satisfies the equation of continuity

$$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$$

and the boundary layer approximation of the momentum equation

$$u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=\nu \frac{{\partial }^{2}u}{\partial {y}^{2}}-[\frac{\sigma H{(x)}^{2}}{\rho }+\frac{\nu }{\kappa (x)}]u,$$

where ν is the kinematic viscosity and ρ is the fluid density. The boundary conditions for the flow are given by

$$u(x,0)=U(x);\,v(x,0)=V(x);\,\mathop{\mathrm{lim}}\limits_{y\to \infty }u(x,y)=0,$$

where U(x) and V(x) are as in (1) and (4).

We use similarity transformations to convert the system (5) and (6) in simpler form. Let ψ be the stream function for the flow, such that $$u=\frac{\partial \psi }{\partial y}$$ and $$v=-\,\frac{\partial \psi }{\partial x}$$. Introducing the Reynolds number $${\mathcal R} e$$ and the similarity variable η, such that

$${\mathcal R} e={U}_{0}L/\nu > 0;\,\eta =\sqrt{ {\mathcal R} e(n+1)/2}{(x/L)}^{\frac{n-1}{2}}y/L > 0,$$

and

$$\psi =\nu \sqrt{2 {\mathcal R} e/(n+1)}{(x/L)}^{\frac{n+1}{2}}f(\eta ),$$

where f(η) denotes the dimensionless form of the stream function, we have

$$u=U(x)f^{\prime} (\eta );\,v=(V(x)/{V}_{c})[f(\eta )+\frac{n-1}{n+1}\eta \,f^{\prime} (\eta )],$$

where

$${V}_{c}=({V}_{0}L/\nu )/\sqrt{ {\mathcal R} e}$$

is the dimensionless measure of suction/injection known as the mass transpiration parameter22.

Using (9) and (10) in (5) and (6), we have the following nonlinear third order system

$$\begin{array}{c}{f}^{^{\prime\prime\prime} }+f{f}^{{\rm{^{\prime} }}{\rm{^{\prime} }}}-\frac{2n}{n+1}{f}^{{}^{{\rm{^{\prime} }}}2}-\frac{2 {\mathcal M} }{n+1}{f}^{{\rm{^{\prime} }}}=0;\,f(0)={V}_{c};\,{f}^{{\rm{^{\prime} }}}\,(0)=1;\,\mathop{{\rm{l}}{\rm{i}}{\rm{m}}}\limits_{\eta \to {\rm{\infty }}}{f}^{{\rm{^{\prime} }}}\,(\eta )=0,\end{array}$$

where 0 ≤ η < ∞. The stretching parameter n, the magnetic parameter $${\mathcal M} ={\mathscr Q}+\kappa$$ appear in (12), where

$${\mathscr Q}=\sigma {L}^{2}{H}_{0}^{2}/(\rho \nu {\mathcal R} e);\,\kappa ={L}^{2}/({\kappa }_{0} {\mathcal R} e\mathrm{)}.$$

The symbol $$\kappa$$ is the dimensionless form of the porosity of the medium so that $${\kappa }^{-1}$$ is the dimensionless measure of permeability. Also, the parameter $${\mathscr Q}$$ is related to the Chandrasekhar number Q by $${\mathscr Q}=Q/ {\mathcal R} e$$, where Q = σL2H02/(ρν). Observe that Q ≥ 0 and $$\kappa > 0$$. If no magnetic field is applied, $${\mathscr Q}=0$$ so that $${\mathcal M} =\kappa$$. Increasing the permeability of the porous medium lowers the value of $$\kappa$$ and hence of $${\mathcal M}$$. If the me0dium is not porous, the term containing porosity is absent which correspond to $$\kappa \to 0$$ and $${\mathcal M} ={\mathscr Q}$$. On the other hand, if the medium is not porous and no magnetic field is applied then $${\mathcal M} \to 0$$. Also note that Vc < 0 for suction and Vc > 0 for injection. Finally, recall that n > −1.

### Solution

To solve the nonlinear system (12), we use the well known HAM25,31 which is described as follows. For p∈ [0,1] as the homotopy embedding parameter, we consider the following boundary value problem

$$\begin{array}{c}{f}^{{\rm{^{\prime} }}{\rm{^{\prime} }}{\rm{^{\prime} }}}-{\alpha }^{2}{f}^{{\rm{^{\prime} }}}+p\{f{f}^{{\rm{^{\prime} }}{\rm{^{\prime} }}}-\frac{2n}{n+1}{f}^{{}^{{\rm{^{\prime} }}}2}+({\alpha }^{2}-\frac{2 {\mathcal M} }{n+1})\,{f}^{{\rm{^{\prime} }}}\}=0,\,\\ f(0)={V}_{c},\,\\ {f}^{{\rm{^{\prime} }}}(0)=1,\,\\ \mathop{{\rm{l}}{\rm{i}}{\rm{m}}}\limits_{\eta \to {\rm{\infty }}}{f}^{{\rm{^{\prime} }}}(\eta )=0.\end{array}$$

where α ≠ 0 is the unknown scalar to be determined. For p = 0, the system (14) gives the linear system f ′′′−α2f ′ = 0 and for p = 1, it is the nonlinear system (12). Now assume a solution of (14) in the form

$$f={f}_{0}+p{f}_{1}+{p}^{2}{f}_{2}+\cdots +{p}^{k}{f}_{k}+\cdots$$

and compare the like powers of p to obtain the following sequence of boundary value problems

$${{f}^{{\rm{^{\prime} }}{\rm{^{\prime} }}{\rm{^{\prime} }}}}_{0}-{\alpha }^{2}{{f}^{{\rm{^{\prime} }}}}_{0}=0;\,{f}_{0}(0)={V}_{c};\,{{f}^{{\rm{^{\prime} }}}}_{0}(0)=1;\,\mathop{{\rm{l}}{\rm{i}}{\rm{m}}}\limits_{\eta \to {\rm{\infty }}}{{f}^{{\rm{^{\prime} }}}}_{0}(\eta )=0,$$
$$\begin{array}{c}{{f}^{{\rm{^{\prime} }}{\rm{^{\prime} }}{\rm{^{\prime} }}}}_{k}-{\alpha }^{2}{{f}^{{\rm{^{\prime} }}}}_{k}+({\alpha }^{2}-\frac{2 {\mathcal M} }{n+1}){{f}^{{\rm{^{\prime} }}}}_{k-1}+\mathop{\sum }\limits_{m=0}^{k-1}\,({f}_{m}{{f}^{{\rm{^{\prime} }}{\rm{^{\prime} }}}}_{km}-\frac{2n}{n+1}{{f}^{{\rm{^{\prime} }}}}_{m}{{f}^{{\rm{^{\prime} }}}}_{km})=0,\\ {f}_{k}(0)={{f}^{{\rm{^{\prime} }}}}_{k}(0)=\mathop{{\rm{l}}{\rm{i}}{\rm{m}}}\limits_{\eta \to {\rm{\infty }}}{{f}^{{\rm{^{\prime} }}}}_{k}(\eta )=0,\,k=1,\,2,\,\ldots .\end{array}$$

The system (15) and (16) can be solved recursively starting with k = 0, where at each later stage, one needs to solve a linear boundary value problem using the solutions obtained from all of the previous stages. We have obtained the solutions of first four of these problems in closed form as follows.

$${f}_{0}(\eta )={V}_{c}+\frac{1-\exp \{\,-\,\alpha \eta \}}{\alpha },$$
$$\begin{array}{ccc}{f}_{1}(\eta ) & = & \frac{1}{6{\alpha }^{3}}(3\alpha (\alpha -{V}_{c})\beta -3 {\mathcal M} (2-\beta ))\\ & & +\,\frac{1}{6{\alpha }^{3}}(1+2\beta +3 {\mathcal M} (2-\beta )+3\alpha ({V}_{c}-\alpha ))\exp \{-\alpha \eta \}\\ & & +\,\frac{1}{2{\alpha }^{2}}(1+ {\mathcal M} (2-\beta )+\alpha ({V}_{c}-\alpha ))\,\eta \,\exp \{-\alpha \eta \}+\frac{(1-\beta )}{6{\alpha }^{3}}\exp \{-2\alpha \eta \},\end{array}$$
$$\begin{array}{rcl}{f}_{2}(\eta ) & = & \frac{{q}_{0}}{72{\alpha }^{5}}-\frac{1}{{\alpha }^{5}}({q}_{1}+6{q}_{2}\alpha \eta +6{q}_{3}{\alpha }^{2}{\eta }^{2})\\ & & \times \exp \{\,-\,\alpha \eta \}-\frac{\mathrm{(1}-\beta )}{36{\alpha }^{5}}({q}_{4}+{q}_{5}\alpha \eta )\exp \{\,-\,2\alpha \eta \}\\ & & +\,\frac{\mathrm{(1}-\beta \mathrm{)(4}\beta -\mathrm{5)}}{{\alpha }^{5}}\exp \{\,-\,3\alpha \eta \},\end{array}$$
$$\begin{array}{rcl}{f}_{3}(\eta ) & = & \frac{{r}_{0}}{{\alpha }^{7}}+\frac{1}{{\alpha }^{7}}({r}_{1}+30{r}_{2}\alpha \eta +{r}_{3}{\alpha }^{2}{\eta }^{2}+{r}_{4}{\alpha }^{3}{\eta }^{3})\\ & & \times \exp \{\,-\,\alpha \eta \}-\frac{\mathrm{(1}-\beta )}{{\alpha }^{7}}({r}_{5}+{r}_{6}\alpha \eta +{r}_{7}{\alpha }^{2}{\eta }^{2})\exp \{\,-\,2\alpha \eta \}\\ & & +\,\frac{\mathrm{(1}-\beta \mathrm{)(5}-4\beta )}{{\alpha }^{7}}({r}_{8}+18{r}_{9}\alpha \eta )\exp \{\,-\,3\alpha \eta \}\\ & & +\,\frac{\mathrm{(1}-\beta \mathrm{)(33}\beta +20{\beta }^{2})}{{\alpha }^{7}}\exp \{\,-\,4\alpha \eta \},\end{array}$$

where

$$\beta =\frac{2n}{1+n},$$
$$\begin{array}{ccc}{q}_{0} & = & 10+2\beta (2\beta +9 {\mathcal M} (2-\beta ))+13\beta \\ & & +\,9 {\mathcal M} (2-\beta )(4+3 {\mathcal M} (2-\beta ))+2(11+7\beta +18 {\mathcal M} (2-\beta )){V}_{c}\alpha \\ & & -\,9(4+2\beta +6 {\mathcal M} (2-\beta )-{V}_{c}^{2}){\alpha }^{2}{\alpha }^{3}+27{\alpha }^{4},\end{array}$$
$$\begin{array}{ccc}{q}_{1} & = & 3+4\beta (5\beta +18 {\mathcal M} (2-\beta ))+31\beta \\ & & +\,18 {\mathcal M} (2-\beta )(2+3 {\mathcal M} (2-\beta ))+8(2+7\beta +9 {\mathcal M} (2-\beta )){V}_{c}\alpha \\ & & -\,18(2+4\beta +6 {\mathcal M} (2-\beta )-{V}_{c}^{2}){\alpha }^{2}{\alpha }^{3}+54{\alpha }^{4},\end{array}$$
$$\begin{array}{ccc}{q}_{2} & = & 3+2\beta (3+2 {\mathcal M} (2-\beta ))\\ & & +\,14 {\mathcal M} (2-\beta )(1+9 {\mathcal M} (2-\beta ))+4(2+\beta +3 {\mathcal M} (2-\beta )){V}_{c}\alpha \\ & & -\,(2(7+2\beta +9 {\mathcal M} (2-\beta ))-3{V}_{c}^{2}){\alpha }^{2}{\alpha }^{3}+9{\alpha }^{4},\end{array}$$
$$\begin{array}{c}{q}_{3}=3(1+ {\mathcal M} (2-\beta {))}^{2}+2(1+ {\mathcal M} (2-\beta )){V}_{c}\alpha \\ \,\,\,\,\,-\,3(2(1+ {\mathcal M} (2-\beta ))-{V}_{c}^{2}){\alpha }^{2}-6{\alpha }^{3}+3{\alpha }^{4},\end{array}$$
$${q}_{4}=3+4\beta +9 {\mathcal M} (2-\beta )+7{V}_{c}\alpha -9{\alpha }^{2};\,{q}_{5}=6(1+ {\mathcal M} (2-\beta )+{V}_{c}\alpha -{\alpha }^{2}),$$
$$\begin{array}{ccc}{r}_{0} & = & -\,\beta (+\beta +{\beta }^{2}) {\mathcal M} \{(2+\beta )(5+4\beta +9 {\mathcal M} )+9{ {\mathcal M} }^{2}\}\\ & & -\,5(+\beta +84{\beta }^{2}+ {\mathcal M} +\beta {\mathcal M} +{ {\mathcal M} }^{2}){V}_{c}\alpha \\ & & +\,10(+\beta +60{\beta }^{2}+ {\mathcal M} +\beta {\mathcal M} +{ {\mathcal M} }^{2}{V}_{c}^{2}\beta {V}_{c}^{2} {\mathcal M} {V}_{c}^{2}){\alpha }^{2}\\ & & +\,(11+7\beta +18 {\mathcal M} ){V}_{c}{\alpha }^{3}(10+5\beta +15 {\mathcal M} -3{V}_{c}^{2}){\alpha }^{4}{V}_{c}{\alpha }^{5}+{\alpha }^{6},\end{array}$$
$$\begin{array}{ccc}{r}_{1} & = & -\,9+\beta +{\beta }^{2}+{\beta }^{3}+ {\mathcal M} (3+31\beta +20{\beta }^{2}+18 {\mathcal M} +36\beta {\mathcal M} +18{ {\mathcal M} }^{2})\\ & & +\,5\{(-9+\beta +{\beta }^{2}+ {\mathcal M} +\beta {\mathcal M} +{ {\mathcal M} }^{2}){V}_{c}\}\alpha \\ & & +\,10\{(3+31\beta +20{\beta }^{2}) {\mathcal M} (2+4\beta +3 {\mathcal M} )+2(-1+82\beta +81 {\mathcal M} ){V}_{c}^{2}\}{\alpha }^{2}\\ & & -\,(2+7\beta +9 {\mathcal M} ){V}_{c}{\alpha }^{3}\{-5(1+2\beta +3 {\mathcal M} )+3{V}_{c}^{2}\}{\alpha }^{4}+{V}_{c}{\alpha }^{5}{\alpha }^{6},\end{array}$$
$$\begin{array}{ccc}{r}_{2} & = & 1+53\beta +36{\beta }^{2}+5 {\mathcal M} (15+35\beta +4{\beta }^{2})+6{ {\mathcal M} }^{2}(31+14\beta +15 {\mathcal M} )\\ & & +\,\{9+\beta +20{\beta }^{2}+16 {\mathcal M} (10+8\beta +9 {\mathcal M} )\}{V}_{c}\alpha \\ & & -\,\{75+\beta +20{\beta }^{2}+6 {\mathcal M} (62+28\beta +45 {\mathcal M} )-2(5+22\beta +27 {\mathcal M} ){V}_{c}^{2}\}{\alpha }^{2}\\ & & -\,32(5+4\beta +9 {\mathcal M} ){V}_{c}{\alpha }^{3}+6(31+14\beta +45 {\mathcal M} -9{V}_{c}^{2}){\alpha }^{4}+{V}_{c}{\alpha }^{5}{\alpha }^{6},\end{array}$$
$$\begin{array}{ccc}{r}_{3} & = & (1+\beta )(1+2\beta +5 {\mathcal M} +\beta {\mathcal M} +3{\beta }^{2})+(3+3\beta +10 {\mathcal M} +2\beta {\mathcal M} +6{ {\mathcal M} }^{2}){V}_{c}\alpha \\ & & -\,(6+3\beta + {\mathcal M} (16+2\beta +9 {\mathcal M} )-(2+\beta +3 {\mathcal M} ){V}_{c}^{2}){\alpha }^{2}-2(5+\beta +6 {\mathcal M} ){V}_{c}{\alpha }^{3}\\ & & +\,(8+\beta +9 {\mathcal M} -3{V}_{c}^{2}){\alpha }^{4}+6{V}_{c}{\alpha }^{5}-3{\alpha }^{6},\end{array}$$
$$\begin{array}{ccc}{r}_{4} & = & {(1+ {\mathcal M} )}^{3}+3(1+ {\mathcal M} {)}^{2}{V}_{c}\alpha -3(1+ {\mathcal M} )(1+ {\mathcal M} -{V}_{c}^{2}){\alpha }^{2}\\ & & -\,\{6(1+ {\mathcal M} )-{V}_{c}^{2}\}{V}_{c}{\alpha }^{3}+\,3(1+ {\mathcal M} -{V}_{c}^{2}){\alpha }^{4}+3{V}_{c}{\alpha }^{5}-{\alpha }^{6},\end{array}$$
$$\begin{array}{ccc}{r}_{5} & = & 6+49\beta +28{\beta }^{2}+15 {\mathcal M} (6+8\beta +9 {\mathcal M} )\\ & & +\,4(9+22\beta +42 {\mathcal M} ){V}_{c}\alpha -(90+\beta + {\mathcal M} {V}_{c}^{2}){\alpha }^{2}-\,{V}_{c}{\alpha }^{3}+{\alpha }^{4},\\ & \end{array}$$
$$\begin{array}{ccc}{r}_{6} & = & 2(21+30\beta +90 {\mathcal M} +24\beta {\mathcal M} +63{ {\mathcal M} }^{2})+(+48\beta + {\mathcal M} ){V}_{c}\alpha \\ & & -\,6(30+8\beta +42 {\mathcal M} {V}_{c}^{2}){\alpha }^{2}{V}_{c}{\alpha }^{3}+{\alpha }^{4},\end{array}$$
$${r}_{7}=36+72 {\mathcal M} +36{ {\mathcal M} }^{2}+72(1+ {\mathcal M} ){V}_{c}\alpha (2(1+ {\mathcal M} )-{V}_{c}^{2}){\alpha }^{2}{V}_{c}{\alpha }^{3}+36{\alpha }^{4},$$
$${r}_{8}=11+12\beta +30 {\mathcal M} +23{V}_{c}\alpha {\alpha }^{2};{r}_{9}=1+ {\mathcal M} +\alpha ({V}_{c}-\alpha ).$$

The expressions for the functions fk for k > 3 are lengthy; so, we omit them. The order of approximation of f using HAM is the integer N such that $$f\approx {\sum }_{k=0}^{N}\,{f}_{k}$$. In the present work, we have obtained up to seventh order approximation of f. All numerical computations have been done using computer programming in MATLAB.

The HAM allows us to choose the scalar α appropriately. We take

$$\alpha =-\,f^{\prime\prime} \mathrm{(0)}\approx -\mathop{\sum }\limits_{m\mathrm{=0}}^{N}\,{f^{\prime\prime} }_{m}\mathrm{(0)}=\alpha -\mathop{\sum }\limits_{m\mathrm{=1}}^{N}\,{f^{\prime\prime} }_{m}\mathrm{(0),}$$

where we have f'0(0) = −α from (16a). From (32), we have $${\sum }_{m=1}^{N}\,{f^{\prime\prime} }_{m}(0)\approx 0$$, which can be solved numerically to obtain an approximate value of α, depending upon N and the dimensionless parameters. For N = 1, (32) gives the following closed form expression

$$\alpha \approx \frac{{V}_{c}}{2}+\frac{1}{2}\sqrt{{V}_{c}^{2}+\frac{4}{3}\{1+2\beta +3 {\mathcal M} (2-\beta )\}},$$

which is valid for all n ≥ 0 as $${V}_{c}^{2}+\frac{4}{3}\{1+2\beta +3 {\mathcal M} (2-\beta )\}\ge 0$$ holds since $${\mathcal M} > 0$$ and 0 ≤ β < 2 for all n ≥ 0. For −1 < n < 0, β∈ (−∞,0). In this case, (33) holds for $${\mathcal M} =2/3$$. For $${\mathcal M} \ne 2/3$$, (33) holds either for all β∈ (−∞,0) and $${\mathcal M} > 2/3$$ or for β ≥ β0 and $${\mathcal M} < 2/3$$, where

$${\beta }_{0}=-\frac{3{V}_{c}^{2}+4+24 {\mathcal M} }{4( {\mathcal M} )}.$$

The present formula (33) recovers the value of f ′′(0) as in Crane7 on taking $${V}_{c}\to 0= {\mathcal M}$$ and β = 1, Pavlov9 on taking Vc→0 and β = 1, Gupta and Gupta34 for β = 1, and Hayat et al.14 and Rashidi15 for Vc→0.

To compare the first order approximation of f ′′(0) by (33) with the higher order approximations, we have obtained Fig. 2(a) which shows the variation of $$f^{\prime\prime} \mathrm{(0)}\approx {\sum }_{k\mathrm{=0}}^{N}\,{f^{\prime\prime} }_{k}\mathrm{(0)}$$ with Vc for $${\mathcal M} =0$$ and n = 1, 5, The solid thick curves and the thin dashed curves have been obtained for N = 6 and N = 1, respectively. Clearly for each n, the formula (33) gives good approximation to f ′′(0). To quantify this approximation, we define

$${E}_{R}^{N}=|\mathop{\sum }\limits_{k=0}^{N}\,{{f}^{{\rm{^{\prime} }}{\rm{^{\prime} }}}}_{k}(0)+\frac{{V}_{c}}{2}+\frac{1}{2}\sqrt{{V}_{c}^{2}+\frac{4}{3}\{1+2\beta +3 {\mathcal M} (2-\beta )\}}|/|\mathop{\sum }\limits_{k=0}^{N}\,{{f}^{{\rm{^{\prime} }}{\rm{^{\prime} }}}}_{k}(0)|,$$

which is the relative difference in the values of f ′′(0) = −α obtained using first order approximation with respect to the N-th order approximation. Figure 2(b) shows the variation of ER6 with Vc for the same parametric values as in Fig. 2(a). It is clear from the Fig. 2(a) that the relative difference ER6 = 0 for n = 1, and it remains less than 5% for the other two values of n.

Table 1 shows the numerical value of α obtained with the higher order approximations for Vc → 0, $${\mathcal M} =0$$, and n =  Clearly the method converges for N = 3 within the tolerance of 10−2. For N = 7 and Vc → 0, we have f ′′(0) ≈ −, which is close to the corresponding numerical value − obtained by Vajravelu and Cannon35 and Cortell36.

To compare the third order approximation (N = 3) of HAM to the solution with the higher order approximations, we have plotted f(η) with respect to η in Fig. 3 for n = 10 and $${\mathcal M} =0$$. We have taken $${\mathcal M} =0$$ since the convergence of the method is comparatively rapid for  $${\mathcal M}$$ > 0. The other parametric values are chosen to test the extreme case where the error can possibly be maximum. The three curves in each subfigure correspond to N = 3, 4, and 5, respectively as shown in the legend. Clearly, the three curves are indistinguishable for each value of Vc which shows that the method converges for N = 3 and justifies that the third order analytic approximation to f using HAM are sufficient to describe the solution correctly. For rest of the numerical calculations, we have taken 4 ≤ N ≤ 7.

### Asymptotic Analysis

To understand the full parametric dependence of the present boundary layer flow, we obtain approximate analytic solution f for the following extreme cases.

Case I: $${\mathcal M} \gg 1$$

Let $$\eta ={ {\mathcal M} }^{r}\hat{\eta }$$ for some nonzero real number r when $${\mathcal M}$$ is large, where $$\hat{\eta }=O\mathrm{(1)}$$. Assume

$$f={V}_{c}+{ {\mathcal M} }^{g}F(\hat{\eta })+O({ {\mathcal M} }^{2g}),$$

where g < 0. The boundary condition f(0) = Vc gives F(0) = 0, and f ′(0) = 1 implies $${ {\mathcal M} }^{g-r}{F}^{{\rm{^{\prime} }}}(0)=1$$, so that g = r and F′(0) = 1. Now using (12), we have at the leading order $${ {\mathcal M} }^{2r}$$

$$\begin{array}{c}{ {\mathcal M} }^{-2r-1}{F}^{{\rm{^{\prime} }}{\rm{^{\prime} }}{\rm{^{\prime} }}}+{V}_{c}{ {\mathcal M} }^{-r-1}{F}^{{\rm{^{\prime} }}{\rm{^{\prime} }}}+{ {\mathcal M} }^{-1}(F{F}^{{\rm{^{\prime} }}{\rm{^{\prime} }}}-\frac{2n}{n+1}{F}^{{}^{{\rm{^{\prime} }}}2})-\frac{2}{n+1}{F}^{{\rm{^{\prime} }}}=0,\\ F(0)=0,\,{F}^{{\rm{^{\prime} }}}(0)=1,\,\mathop{{\rm{l}}{\rm{i}}{\rm{m}}}\limits_{\eta \to {\rm{\infty }}}\,{F}^{{\rm{^{\prime} }}}(\eta )=0,\end{array}$$

where r < 0, which for sufficiently large $${\mathcal M}$$ requires −1− 2r = 0 or r = −1/2. So, for $${\mathcal M} \to {\rm{\infty }}$$, we get

$${F}^{{\rm{^{\prime} }}{\rm{^{\prime} }}{\rm{^{\prime} }}}-\frac{2}{n+1}{F}^{{\rm{^{\prime} }}}=0,\,F(0)=0,\,{F}^{{\rm{^{\prime} }}}(0)=1,\,\mathop{\mathrm{lim}}\limits_{\eta \to {\rm{\infty }}}{F}^{{\rm{^{\prime} }}}(\eta )=0.$$

The solution of (36) is given by $$F=\sqrt{\frac{n+1}{2}}(1-\exp \{-\sqrt{\frac{2}{n+1}}\hat{\eta }\})$$. Thus, we have the following asymptotic solution for large $${\mathcal M}$$.

$$f={V}_{c}+\sqrt{\frac{n+1}{2 {\mathcal M} }}(1-\exp \{-\sqrt{\frac{2 {\mathcal M} }{n+1}}\eta \})+O({ {\mathcal M} }^{-1}),\, {\mathcal M} \gg 1.$$

The function f − Vc as obtained from (37) and (12) for large $${\mathcal M}$$ is shown in Fig. 4(a) for Vc =  and n =  In each case, the points marked * correspond to the asymptotic solution (37), while the solid curves correspond to the numerical solution of (12). The asymptotic solution is in a maximum relative error of % for $${\mathcal M} ={10}^{2}$$, which decreases rapidly on increasing $${\mathcal M}$$. Thus the two solutions are in good agreement for $${\mathcal M} \ge {10}^{2}$$.

Case II: Vc >> 1

For Vc >> 1, we take η = Vchη*, where $${\eta }^{\ast }=O\mathrm{(1)}$$ and h is a nonzero scalar. Assume

$$f={V}_{c}+{V}_{c}^{q}G({\eta }^{\ast })+O({V}_{c}^{2q})$$

for large Vc, where q < 0. Using the condition f(0) = Vc implies G(0) = 0 and f′(0) = 1 gives G′(0) = Vchq, so that G′(0) = 1 and h = q. Now from (12), we have

$${G}^{{\rm{^{\prime} }}{\rm{^{\prime} }}{\rm{^{\prime} }}}+{V}_{c}^{1+q}{G}^{{\rm{^{\prime} }}{\rm{^{\prime} }}}-{V}_{c}^{2q}(\frac{2n}{n+1}{G}^{{}^{{\rm{^{\prime} }}}2}-\frac{2 {\mathcal M} }{n+1}{G}^{{\rm{^{\prime} }}})=0,\,G(0)=0,\,{G}^{{\rm{^{\prime} }}}(0)=1,\,\mathop{{\rm{l}}{\rm{i}}{\rm{m}}}\limits_{\eta \to {\rm{\infty }}}{G}^{{\rm{^{\prime} }}}(\eta )=0,$$

where q < 0, which for Vc >> 1 requires for a balance, 1 + q = 0 or q = −1. Thus, for Vc → ∞, we have

$${G}^{{\rm{^{\prime} }}{\rm{^{\prime} }}{\rm{^{\prime} }}}+{G}^{{\rm{^{\prime} }}{\rm{^{\prime} }}}=0,\,G(0)=0,\,{G}^{{\rm{^{\prime} }}}(0)=1,\,\mathop{{\rm{l}}{\rm{i}}{\rm{m}}}\limits_{\eta \to {\rm{\infty }}}{G}^{{\rm{^{\prime} }}}(\eta )=0,$$

which solves to G = 1 − eη* = 1 − eVcη. Thus, in this case the asymptotic solution is given by

$$f={V}_{c}+\frac{1}{{V}_{c}}(1-\exp \{-{V}_{c}\eta \})+O({V}_{c}^{-2}),\,{V}_{c}\gg 1.$$

Figure 4b demonstrates a comparison between the asymptotic solution f − Vc as obtained from (39) marked * and the numerical solution of (12) (solid and dashed curves). The fixed parametric values are n =  and $${\mathcal M} =1$$. Clearly, the two solutions are in very good agreement for Vc ≥ 4 within a relative error not exceeding %.

### Results and Discussion

The numerical results have been obtained for a wide range of parameters. For most of the numerical calculations, we have taken −2 ≤ Vc ≤ 2, 0 ≤ n ≤ 5, and $$0 < {\mathcal M} \le$$. Even higher values of $${\mathcal M}$$ are permissible since the method converges faster as $${\mathcal M}$$ becomes larger, which can be seen from (33) through (16a)–(16d), since α = |f ′′(0)| which in turn rises with $${\mathcal M}$$ for large $${\mathcal M}$$. Also, we have taken at least the fourth order approximation of f, that is N ≥ 4 for the rest of the numerical calculations to meet the convergence issues using HAM. For obtaining the streamline patterns, we have taken 7th order approximation (N = 7) to f.

### Skin friction coefficient at sheet wall

If τ denotes the shear stress near the stretching sheet due to the fluid flow, we have

$$\tau =\mu \mathop{\mathrm{lim}}\limits_{y\to 0}(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x})=\mu (f^{\prime\prime} \mathrm{(0)}U\frac{\partial \eta }{\partial y}+\frac{dV}{dx}).$$

Then the coefficient Cf of the frictional drag which is also known as the skin friction parameter is defined for x > 0 as

$${C}_{f}=-\frac{\tau }{\frac{1}{2}\rho \{U{(x)}^{2}+V{(x)}^{2}\}}=\sqrt{\frac{\mathrm{2(}n+\mathrm{1)}}{ {\mathcal R} e}}\frac{\{-f^{\prime\prime} \mathrm{(0)}+\frac{(n-\mathrm{1)}{V}_{c}}{2 {\mathcal R} e}{(x/L)}^{-(n+\mathrm{1)/2}}\}}{\{{(x/L)}^{n+1}+\frac{(n+\mathrm{1)}{V}_{c}^{2}}{2 {\mathcal R} e}\}}.$$

Note that the skin friction coefficient Cf depends upon f ′′(0), n, $${\mathcal R} e$$, Vc, and the horizontal distance x from the slit, where f ′′(0) further depends upon $${\mathcal M}$$, n, $${\mathcal R} e$$, and Vc.

In the literature, f ′′(0) is generally taken as a measure of the skin friction for a fixed x, which in the present situation holds only if there is no suction/injection, that is, Vc ≈ 0. If this is the case, we have

$$\mathop{\mathrm{lim}}\limits_{{V}_{c}\to 0}{C}_{f}=-\sqrt{\frac{\mathrm{2(}n+\mathrm{1)}}{ {\mathcal R} e}}{(x/L)}^{-(n+\mathrm{1)}}f^{\prime\prime} \mathrm{(0)}.$$

We first discuss the variation of Cf with x in the following three cases.

Case I: n ≥ 1. Figure 5 depicts the variation of |Cf| with x/L for $${\mathcal M} =1$$ and $${\mathcal R} e=2$$. Each subfigure corresponds to one value of Vc among $$\mp$$, $$\mp$$, $$\mp$$, $$\mp$$, and $$\mp$$. The different curves (labeled with different color and style as shown in the legend) in each subfigure correspond to different values of n ≥ 1 among 1, , , and

Источник: [totalfon.net]