Steady flow of a power law fluid through a tapered non-symmetric stenotic tube

Blood flow through stenotic arteries is a subject of recent investigation from the past few decades. Development of stenosis in the arteries is one of the leading causes of circulatory disorder. In medical terms, "narrowing of anybody passage" is called stenosis. This growth is abnormal and unnatural in the thickness of arterial wall that produces at different positions in the cardiovascular system under particular conditions. It has been observed that stenosis is due to the close association of the conditions of flow in the blood vessels. Rindfleish [1] and Nazemi et al. [2] analyzed that wall shear stress is a normal factor in the growth of atherosclerotic lesions. Caro et al. [3] suggested that atheroma appears in the low shear stress region while in view of Fry [4,5] endothelial damage is caused by high wall shear stress. Once the stenosis is formed, it is quite clear that the flow characteristics in its vicinity may be significantly altered. The knowledge of various flow parameters in the stenotic artery such as the pattern of velocity, the rate of flow and the stresses is quite helpful to develop bio-medical instruments for bio-medical engineers for treating such modalities.

In recent years, the use of non-Newtonian models in studies regarding blood flow through stenotic arteries has considerably increased. In these studies the blood is represented by various non-Newtonian models such as Casson’s model [6], power law model [7, 8, 9], micropolar model [10,11], Sisko model [12] and Herschel–Bulkley model [13] etc. More recently Nadeem et al. [14] performed a study on blood flow through stenotic artery using the power law model. Unfortunately, they did not choose the proper sign preceding the magnitude of the rate of deformation tensor in their study, and thus they provided erroneous results. This motivated us to look into their problem and rectify the results presented by them. The details are included in this article. A detailed derivation of the governing equation is presented in Section 2. The mistake done by Nadeem et al. [14] is also pointed out here. An exact solution of the equation is obtained in Section 3. Section 4 consists of results and discussion. Conclusion of the paper is given in Section 5.

Formulation of the Problem

We consider an incompressible flow of power law fluid through a tube of length L. We choose the cylindrical coordinate system (r,θ, z) such that u, v and w are the velocity components in radial, azimuthal and axial directions, respectively, to analyze this flow. The governing equations of steady incompressible flow of a power law fluid are

(1)

\frac{\partial u}{\partial r} + \frac{u}{r} + \frac{\partial w}{\partial z} = 0,

$$\frac{\partial u}{\partial r}+\frac{u}{r}+\frac{\partial w}{\partial z}=0,$$

(2)

ρ (u \frac{\partial}{\partial r} + w \frac{\partial}{\partial z}) u = - \frac{\partial p}{\partial r} + \frac{1}{r} \frac{\partial}{\partial r} (r S_{r r}) + \frac{\partial}{\partial z} (S_{r z}) - \frac{S_{θ θ}}{r},

$$\rho \left( u\frac{\partial }{\partial r}+w\frac{\partial }{\partial z} \right)u=-\frac{\partial p}{\partial r}+\frac{1}{r}\frac{\partial }{\partial r}\left( r{{S}_{rr}} \right)+\frac{\partial }{\partial z}\left( {{S}_{rz}} \right)-\frac{{{S}_{\theta \theta }}}{r},$$

(3)

ρ (u \frac{\partial}{\partial r} + w \frac{\partial}{\partial z}) w = - \frac{\partial p}{\partial z} + \frac{1}{r} \frac{\partial}{\partial r} (r S_{r z}) + \frac{\partial}{\partial z} (S_{z z}),

$$\rho \left( u\frac{\partial }{\partial r}+w\frac{\partial }{\partial z} \right)w=-\frac{\partial p}{\partial z}+\frac{1}{r}\frac{\partial }{\partial r}\left( r{{S}_{rz}} \right)+\frac{\partial }{\partial z}\left( {{S}_{zz}} \right),$$

where ρ is the density and p is the pressure of the blood.

The extra stress tensor S for power law fluid as per the abovementioned equations is defined in Eq. [10]

(4)

S = μ {(\dot{γ})}^{m - 1} \dot{ϒ},

$$\mathbf{S}=\mu {{\left( {\dot{\gamma }} \right)}^{m-1}}\dot{\Upsilon },$$

where μ is the characteristics of each polymer and

\dot{γ} = \sqrt{\frac{1}{2} \sum_{i} \sum_{j} {\dot{γ}}_{i j} {\dot{γ}}_{j i}},

$$\dot{\gamma }=\sqrt{\frac{1}{2}\sum\limits_{i}{\sum\limits_{j}{{{{\dot{\gamma }}}_{ij}}}{{{\dot{\gamma }}}_{ji}}}},$$

where ${\dot{γ}}_{i j} i, j = 1, 2, 3$ ${{\dot{\gamma }}_{ij}}\,\,i,j=1,2,3$is the rate of strain tensor component and

(5)

\dot{ϒ} = L + L^{T} .

$$\dot{\Upsilon }=\mathbf{L}+{{\mathbf{L}}^{\mathbf{T}}}.$$

The expression that describe the geometry of the constricted portion of the tube is shown in Fig. 1.

Geometry of non-symmetric stenotic artery.

(6)

\begin{array}{l} h (z) = d (z) [1 - η (b^{n - 1} (z - a) - {(z - a)}^{n})], a \leq z \leq a + b, \\ = d (z), O t h e r w i s e \end{array}

$$\begin{align}& \,\,\,\,\,\,\,\,\,h\left( z \right)=d\left( z \right)\left[ 1-\eta \left( {{b}^{n-1}}\left( z-a \right)-{{\left( z-a \right)}^{n}} \right) \right],\,\,\,a\le z\le a+b, \\ & =d\left( z \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Otherwise \\ \end{align}$$

with

(7)

d (z) = d_{0} + ζ^{*} z .

$$d\left( z \right)={{d}_{0}}+{{\zeta }^{*}}z.$$

In the abovementioned equations, ζ ^∗ = tanϕ which is called the tapering parameter, a is the length of the non-stenotic section, b is the length of the stenotic region, d₀ is the radius of the non-tapered artery in the non-stenotic part and n is a parameter that determines the shape of the constriction profile. For n = 2, the stenosis is symmetric. The parameter η is defined as

(8)

η = \frac{δ n^{\frac{n}{(n - 1)}}}{d_{0} b^{n} (n - 1)},

$$\eta =\frac{\delta {{n}^{\frac{n}{\left( n-1 \right)}}}}{{{d}_{0}}{{b}^{n}}\left( n-1 \right)},$$

where δ indicates the maximum height of the stenosis located at

z = a + \frac{b}{n^{\frac{1}{(n - 1)}}} .

$$z=a+\frac{b}{{{n}^{\frac{1}{\left( n-1 \right)}}}}.$$

Now we present a detailed derivation of the governing equation for flow under consideration. For the flow under consideration,

(9)

v = [u (r, z), 0, w (r, z)] .

$$\mathbf{v}=\left[ u\left( r,z \right),0,w\left( r,z \right) \right].$$

Thus,

L = \nabla V = [\begin{matrix} u_{r} & 0 & u_{z} \\ 0 & \frac{u}{r} & 0 \\ w_{r} & 0 & w_{z} \end{matrix}]

$$\mathbf{L}=\nabla \mathbf{V}=\left[ \begin{matrix}{{u}_{r}} & 0 & {{u}_{z}} \\0 & \frac{u}{r} & 0 \\{{w}_{r}} & 0 & {{w}_{z}} \\\end{matrix} \right]$$

and

L^{T} = [\begin{matrix} u_{r} & 0 & w_{r} \\ 0 & \frac{u}{r} & 0 \\ u_{z} & 0 & w_{z} \end{matrix}]

$${{\mathbf{L}}^{T}}=\left[ \begin{matrix}{{u}_{r}} & 0 & {{w}_{r}} \\0 & \frac{u}{r} & 0 \\{{u}_{z}} & 0 & {{w}_{z}} \\\end{matrix} \right]$$

with the subindex denoting differentiation with respect to the indicated variable.

Using the definition of the rate of strain tensor, we can write

(10)

\dot{ϒ} = L + L^{T} = [\begin{matrix} 2 u_{r} & 0 & u_{z} + w_{r} \\ 0 & \frac{2 u}{r} & 0 \\ w_{r} + u_{z} & 0 & 2 w_{z} \end{matrix}] .

$$\dot{\Upsilon }=\mathbf{L}+{{\mathbf{L}}^{T}}=\left[ \begin{matrix}2{{u}_{r}} & 0 & {{u}_{z}}+{{w}_{r}} \\0 & \frac{2u}{r} & 0 \\{{w}_{r}}+{{u}_{z}} & 0 & 2{{w}_{z}} \\\end{matrix} \right].$$

Now

\dot{γ} = | \dot{ϒ} | \sqrt{\frac{1}{2} \sum_{i} \sum_{j} {\dot{γ}}_{i j} {\dot{γ}}_{j i}},

$$\dot{\gamma }=\left| {\dot{\Upsilon }} \right|\sqrt{\frac{1}{2}\sum\limits_{i}{\sum\limits_{j}{{{{\dot{\gamma }}}_{ij}}}{{{\dot{\gamma }}}_{ji}}}},$$

and expanding the double sum we get

\dot{γ} = {[\frac{1}{2} ({\dot{γ}}_{11} {\dot{γ}}_{11} + {\dot{γ}}_{12} {\dot{γ}}_{12} + {\dot{γ}}_{13} {\dot{γ}}_{13} + {\dot{γ}}_{21} {\dot{γ}}_{21} + {\dot{γ}}_{22} {\dot{γ}}_{22} + {\dot{γ}}_{23} {\dot{γ}}_{23} + {\dot{γ}}_{31} {\dot{γ}}_{31} + {\dot{γ}}_{32} {\dot{γ}}_{32} + {\dot{γ}}_{33} {\dot{γ}}_{33})]}^{\frac{1}{2}} .

$$\dot{\gamma }={{\left[ \frac{1}{2}\left( {{{\dot{\gamma }}}_{11}}{{{\dot{\gamma }}}_{11}}+{{{\dot{\gamma }}}_{12}}{{{\dot{\gamma }}}_{12}}+{{{\dot{\gamma }}}_{13}}{{{\dot{\gamma }}}_{13}}+{{{\dot{\gamma }}}_{21}}{{{\dot{\gamma }}}_{21}}+{{{\dot{\gamma }}}_{22}}{{{\dot{\gamma }}}_{22}}+{{{\dot{\gamma }}}_{23}}{{{\dot{\gamma }}}_{23}}+{{{\dot{\gamma }}}_{31}}{{{\dot{\gamma }}}_{31}}+{{{\dot{\gamma }}}_{32}}{{{\dot{\gamma }}}_{32}}+{{{\dot{\gamma }}}_{33}}{{{\dot{\gamma }}}_{33}} \right) \right]}^{\frac{1}{2}}}.$$

Using the values of components of $\dot{γ}$ $\dot{\gamma }$from Eq. (10) results in

(11)

\dot{γ} = {[2 u_{r}^{2} + {(u_{z} + w_{r})}^{2} + 2 \frac{u^{2}}{r^{2}} + 2 w_{z}^{2}]}^{\frac{1}{2}} .

$$\dot{\gamma }={{\left[ 2u_{r}^{2}+{{\left( {{u}_{z}}+{{w}_{r}} \right)}^{2}}+2\frac{{{u}^{2}}}{{{r}^{2}}}+2w_{z}^{2} \right]}^{\frac{1}{2}}}.$$

Let us substitute for convenience

(12)

\dot{γ} = M .

$$\dot{\gamma }=M.$$

From Eqs. (4), (10) and (12) the components of extra stress tensor for power law fluid can be written as

(13)

S_{r r} = 2 μ M u_{r}, S_{r z} = μ M (u_{z} + w_{r}) = S_{z r}, S_{θ θ} = 2 μ M \frac{u}{r} S_{z z} = 2 μ M w_{z} .

$${{S}_{rr}}=2\mu M{{u}_{r}},{{S}_{rz}}=\mu M\left( {{u}_{z}}+{{w}_{r}} \right)={{S}_{zr}},{{S}_{\theta \theta }}=2\mu M\frac{u}{r}{{S}_{zz}}=2\mu M{{w}_{z}}.$$

At this point, we make the assumption that stenosis is mild, i.e.

\frac{δ}{d_{0}} < < 1.

$$\frac{\delta }{{{d}_{0}}}<<1.$$

This assumption enables us to write

(14)

u = 0 a n d \frac{\partial u}{\partial z} = 0.

$$u=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,and\,\,\,\,\,\,\,\,\,\frac{\partial u}{\partial z}=0.$$

In view of Eq. (14), we can write

(15)

\dot{γ} = \sqrt{{(\frac{\partial w}{\partial r})}^{2}} = \pm \frac{\partial w}{\partial r} .

$$\dot{\gamma }=\sqrt{{{\left( \frac{\partial w}{\partial r} \right)}^{2}}}=\pm \frac{\partial w}{\partial r}.$$

The definition of magnitude requires the result to be a positive number. Thus, our choice of the sign that precedes $(\frac{\partial w}{\partial r})$ $\left( \frac{\partial w}{\partial r} \right)$in Eq. (15) depends on whether the derivative $(\frac{\partial w}{\partial r})$ $\left( \frac{\partial w}{\partial r} \right)$is positive or negative. Since the velocity is highest at the centre of the tube, thus, w is a decreasing function of r; therefore, $(\frac{\partial w}{\partial r})$ $\left( \frac{\partial w}{\partial r} \right)$is negative. Thus, the correct sign in Eq. (15) is negative: However Nadeem et al. [14] missed this point and blindly chose a positive sign preceding $(\frac{\partial w}{\partial r})$ $\left( \frac{\partial w}{\partial r} \right)$in Eq. (15) which is given by

(16)

\dot{γ} = - (\frac{\partial w}{\partial r}) > 0.

$$\dot{\gamma }=-\left( \frac{\partial w}{\partial r} \right)>0.$$

With the help of Eqs. (15) and (16), the components of stress and governing equations (1)–(3) reduce to

\begin{array}{l} S_{r r} = 0, S_{θ θ} = 0, S_{z z} = 0, \\ S_{r z} = μ {(- \frac{\partial w}{\partial r})}^{m - 1} (\frac{\partial w}{\partial r}) = S_{z r}, \end{array}

$$\begin{align}& {{S}_{rr}}=0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{S}_{\theta \theta }}=0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{S}_{zz}}=0, \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{S}_{rz}}=\mu {{\left( -\frac{\partial w}{\partial r} \right)}^{m-1}}\left( \frac{\partial w}{\partial r} \right)={{S}_{zr}}, \\ \end{align}$$

(17)

\frac{\partial p}{\partial r} = 0,

$$\frac{\partial p}{\partial r}=0,$$

(18)

\frac{\partial p}{\partial z} = \frac{μ}{r} \frac{\partial}{\partial r} [r {(- \frac{\partial w}{\partial r})}^{m - 1} \frac{\partial w}{\partial r}],

$$\frac{\partial p}{\partial z}=\frac{\mu }{r}\frac{\partial }{\partial r}\left[ r{{\left( -\frac{\partial w}{\partial r} \right)}^{m-1}}\frac{\partial w}{\partial r} \right],$$

(19)

\frac{\partial u}{\partial r} + \frac{u}{r} + \frac{\partial w}{\partial z} = 0.

$$\frac{\partial u}{\partial r}+\frac{u}{r}+\frac{\partial w}{\partial z}=0.$$

To make the governing equations non-dimensional, we introduce the following dimensionless variables

(20)

\begin{array}{l} \bar{r} = \frac{r}{d_{0}}, \bar{z} = \frac{z}{b}, \bar{w} = \frac{w}{u_{0}}, \bar{h} = \frac{h}{d_{0}}, \bar{u} = \frac{b u}{u_{0} δ}, \bar{p} = \frac{d_{0}^{m + 1} p}{u_{0}^{m} b μ}, \\ R_{e} = \frac{ρ d_{0}^{m + 1}}{u_{0}^{m - 2} μ b}, S_{r r}^{-} = \frac{b S_{r r}}{u_{0}^{m} μ}, S_{r z}^{-} = \frac{d_{0}^{m} S_{r z}}{u_{0}^{m} μ}, S_{z z}^{-} = \frac{b S_{z z}}{u_{0}^{m} μ}, S_{θ θ}^{-} = \frac{b S_{θ θ}}{u_{0}^{m} μ}, \end{array}

$$\begin{align} & \overline{r}=\frac{r}{{{d}_{0}}},\,\,\overline{z}=\frac{z}{b},\overline{w}=\frac{w}{{{u}_{0}}},\,\overline{h}=\frac{h}{{{d}_{0}}},\overline{u}=\frac{bu}{{{u}_{0}}\delta },\,\overline{p}=\frac{d_{0}^{m+1}p}{u_{0}^{m}b\mu }, \\ & {{R}_{e}}=\frac{\rho d_{0}^{m+1}}{u_{0}^{m-2}\mu b},\,S_{rr}^{-}=\frac{b{{S}_{rr}}}{u_{0}^{m}\mu },\,S_{rz}^{-}=\frac{d_{0}^{m}{{S}_{rz}}}{u_{0}^{m}\mu },\,S_{zz}^{-}=\frac{b{{S}_{zz}}}{u_{0}^{m}\mu },S_{\theta \,\theta }^{-}=\frac{b{{S}_{\theta \,\theta }}}{u_{0}^{m}\mu }, \\ \end{align}$$

where u₀ is the averaged velocity over the region of the tube with width d₀. Hence, the governing equations in non-dimensional form after dropping the bars read

(21)

\frac{\partial u}{\partial r} + \frac{u}{r} + \frac{\partial w}{\partial z} = 0,

$$\frac{\partial u}{\partial r}+\frac{u}{r}+\frac{\partial w}{\partial z}=0,$$

(22)

\frac{\partial p}{\partial r} = 0,

$$\frac{\partial p}{\partial r}=0,$$

(23)

\frac{\partial p}{\partial z} = \frac{1}{r} \frac{\partial}{\partial r} [r {(\frac{\partial w}{\partial r})}^{m - 1} \frac{\partial w}{\partial r}] .

$$\frac{\partial p}{\partial z}=\frac{1}{r}\frac{\partial }{\partial r}\left[ r{{\left( \frac{\partial w}{\partial r} \right)}^{m-1}}\frac{\partial w}{\partial r} \right].$$

The appropriate boundary conditions are

(24)

w = 0 a t r = h (z),

$$w=0\,\,\,\,\,\,\,\,\,\,\,at\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,r=h\left( z \right),$$

(25)

\frac{\partial w}{\partial r} = 0 a t r = 0,

$$\frac{\partial w}{\partial r}=0\,\,\,\,\,\,\,\,\,\,\,\,at\,\,\,\,\,\,\,\,\,\,r=0,$$

where

\begin{array}{l} h (z) = (1 + ξ z) [1 - η_{1} (z - σ) - {(z - σ)}^{n}], σ \leq z \leq σ + 1, \\ η_{1} = \frac{δ^{*} n^{\frac{n}{n - 1}}}{n - 1}, δ^{*} = \frac{δ}{d_{0}}, σ = \frac{a}{b}, ξ = \frac{ξ^{*} b}{d_{0}} . \end{array}

$$\begin{align}& h\left( z \right)=\left( 1+\xi z \right)\left[ 1-{{\eta }_{1}}\left( z-\sigma \right)-{{\left( z-\sigma \right)}^{n}} \right],\,\,\,\,\,\sigma \le z\le \sigma +1, \\ & \,\,\,\,\,\,\,{{\eta }_{1}}=\frac{{{\delta }^{*}}{{n}^{\frac{n}{n-1}}}}{n-1},\,\,\,\,{{\delta }^{*}}=\frac{\delta }{{{d}_{0}}},\,\,\,\,\sigma =\frac{a}{b},\,\,\,\,\,\xi =\frac{{{\xi }^{*}}b}{{{d}_{0}}}. \\ \end{align}$$

The parameter ξ(= tanϕ) is called the tapering parameter, and ϕ is known as the tapered angle. For the diverging tapering (ϕ > 0), converging tapering (ϕ < 0), non-tapered artery (ϕ = 0) (Fig. 2).

Stenotic tapered artery for different taper angles.

Exact solution of the problem

Now we proceed to solve Eq. (23) subject to boundary conditions (24) and (25). From Eq. (23), we can write

r \frac{\partial p}{\partial z} = \frac{\partial}{\partial r} [r {(\frac{\partial w}{\partial r})}^{m} {(- 1)}^{m - 1}] .

$$r\frac{\partial p}{\partial z}=\frac{\partial }{\partial r}\left[ r{{\left( \frac{\partial w}{\partial r} \right)}^{m}}{{\left( -1 \right)}^{m-1}} \right].$$

Integration of the abovementioned equation yields

\frac{r^{2}}{2} \frac{\partial p}{\partial z} + C = r {(\frac{\partial w}{\partial r})}^{m} {(- 1)}^{m - 1} .

$$\frac{{{r}^{2}}}{2}\frac{\partial p}{\partial z}+C=r{{\left( \frac{\partial w}{\partial r} \right)}^{m}}{{\left( -1 \right)}^{m-1}}.$$

The boundary condition (Eq. 25) gives C = 0.

Thus, the abovementioned equation becomes

(26)

\frac{r^{2}}{2} \frac{\partial p}{\partial z} = r {(\frac{\partial w}{\partial r})}^{m} {(- 1)}^{m} (- 1) .

$$\frac{{{r}^{2}}}{2}\frac{\partial p}{\partial z}=r{{\left( \frac{\partial w}{\partial r} \right)}^{m}}{{\left( -1 \right)}^{m}}\left( -1 \right).$$

Now without giving the detail of algebra involved, we can write

\frac{\partial w}{\partial r} = {(\frac{r}{2})}^{\frac{1}{m}} \frac{\partial p}{\partial z} {| \frac{\partial p}{\partial z} |}^{\frac{1}{m} - 1} .

$$\frac{\partial w}{\partial r}={{\left( \frac{r}{2} \right)}^{\frac{1}{m}}}\frac{\partial p}{\partial z}{{\left| \frac{\partial p}{\partial z} \right|}^{\frac{1}{m}-1}}.$$

Integration of the abovementioned equation gives

(27)

w = \frac{1 r^{\frac{1}{m} + 1}}{2^{\frac{1}{m}} (\frac{1}{m} + 1)} \frac{\partial p}{\partial z} {| \frac{\partial p}{\partial z} |}^{\frac{1}{m} - 1} + C_{1} .

$$w=\frac{1{{r}^{\frac{1}{m}+1}}}{{{2}^{\frac{1}{m}}}\left( \frac{1}{m}+1 \right)}\frac{\partial p}{\partial z}{{\left| \frac{\partial p}{\partial z} \right|}^{\frac{1}{m}-1}}+{{C}_{1}}.$$

From boundary conditions (Eq. 24), we get

(28)

C_{1} = - \frac{1 h^{\frac{1}{m} + 1}}{2^{\frac{1}{m}} (\frac{1}{m} + 1)} \frac{\partial p}{\partial z} {| \frac{\partial p}{\partial z} |}^{\frac{1}{m} - 1},

$${{C}_{1}}=-\frac{1{{h}^{\frac{1}{m}+1}}}{{{2}^{\frac{1}{m}}}\left( \frac{1}{m}+1 \right)}\frac{\partial p}{\partial z}{{\left| \frac{\partial p}{\partial z} \right|}^{\frac{1}{m}-1}},$$

and thus finally

(29)

\begin{array}{l} w = \frac{\partial p}{\partial z} {| \frac{\partial p}{\partial z} |}^{\frac{1 - m}{m}} \frac{m}{2^{\frac{1}{m}} (1 + m)} [r^{\frac{m + 1}{m}} - h^{\frac{m + 1}{m}}], \\ F = 2 π \int_{0}^{h} r w d r . \end{array}

$$\begin{align}& w=\frac{\partial p}{\partial z}{{\left| \frac{\partial p}{\partial z} \right|}^{\frac{1-m}{m}}}\frac{m}{{{2}^{\frac{1}{m}}}\left( 1+m \right)}\left[ {{r}^{\frac{m+1}{m}}}-{{h}^{\frac{m+1}{m}}} \right], \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,F=2\pi \int_{0}^{h}{rwdr.} \\ \end{align}$$

By substituting Eq. (29) in the definition of flow rate, we can write

(30)

F = \int_{0}^{h} A {r . r^{\frac{m + 1}{m}} - r . h^{\frac{m + 1}{m}}} d r,

$$F=\int_{0}^{h}{A\left\{ r.{{r}^{\frac{m+1}{m}}}-r.{{h}^{\frac{m+1}{m}}} \right\}}dr,$$

where

A = 2 π \frac{\partial p}{\partial z} {| \frac{\partial p}{\partial z} |}^{\frac{1 - m}{m}} \frac{m}{2^{\frac{1}{m}} (1 + m)} .

$$A=2\pi \frac{\partial p}{\partial z}{{\left| \frac{\partial p}{\partial z} \right|}^{\frac{1-m}{m}}}\frac{m}{{{2}^{\frac{1}{m}}}\left( 1+m \right)}.$$

From Eq. (30), after some manipulation, we get

(31)

\frac{\partial p}{\partial z} = - (\frac{3 m + 1}{m}) \frac{h^{- (3 m + 1)} 4 F}{{(2 π)}^{m}} (\frac{3 m + 1}{m}) h^{- (3 m + 1)} 2 F |^{m - 1} .

$$\frac{\partial p}{\partial z}=-\left( \frac{3m+1}{m} \right)\frac{{{h}^{-\left( 3m+1 \right)}}4F}{{{\left( 2\pi \right)}^{m}}}\left( \frac{3m+1}{m} \right){{h}^{-\left( 3m+1 \right)}}2F\left| ^{m-1} \right..$$

The pressure drop is given by the following equation [11]:

(32)

Δ p = \int_{0}^{\frac{L}{b}} (- \frac{d p}{d z}) d z .

$$\Delta p=\int_{0}^{\frac{L}{b}}{\left( -\frac{dp}{dz} \right)dz}.$$

Similarly, resistance–impedance is given by the following equation [11]

(33)

λ = \frac{Δ p}{F} = {\int_{0}^{\frac{a}{b}} R (z) |_{h = 1 + ξ z} d z + \int_{\frac{a}{b}}^{\frac{a}{b} + 1} R (z) d z + \int_{\frac{a}{b} + 1}^{\frac{L}{b}} R (z) |_{h = 1 + ξ z} d z},

$$\lambda =\frac{\Delta p}{F}=\left\{ \int_{0}^{\frac{a}{b}}{R\left( z \right)\left| _{h=1+\xi z} \right.dz}+\int_{\frac{a}{b}}^{\frac{a}{b}+1}{R\left( z \right)dz+}\int_{\frac{a}{b}+1}^{\frac{L}{b}}{R\left( z \right)\left| _{h=1+\xi z} \right.dz} \right\},$$

where

R (z) = (\frac{3 m + 1}{m}) \frac{h^{- \frac{3 m + 1}{m}} 4 F}{{(2 π)}^{m}} (2 \frac{3 m + 1}{m}) h^{- \frac{3 m + 1}{m}} F |^{m - 1} .

$$R\left( z \right)=\left( \frac{3m+1}{m} \right)\frac{{{h}^{-\frac{3m+1}{m}}}4F}{{{\left( 2\pi \right)}^{m}}}\left( 2\frac{3m+1}{m} \right){{h}^{-\frac{3m+1}{m}}}F{{|}^{m-1}}.$$

The non-dimensional shear stress is defined as

(34)

S_{r z} = - {(\frac{\partial w}{\partial r})}^{m - 1} \frac{\partial w}{\partial r} .

$${{S}_{rz}}=-{{\left( \frac{\partial w}{\partial r} \right)}^{m-1}}\frac{\partial w}{\partial r}.$$

From Eq. (34), the wall stress can be obtained as

(35)

S_{r z} |_{r = h} = | \frac{\partial w}{\partial r} |^{m - 1} \frac{\partial w}{\partial r} |_{r = h} .

$${{S}_{rz}}\left| _{r=h}= \right|\frac{\partial w}{\partial r}{{|}^{m-1}}\frac{\partial w}{\partial r}{{|}_{r=h}}.$$

Now since

{(\frac{\partial w}{\partial r})}^{m - 1} \frac{\partial w}{\partial r} = \frac{r}{2} (\frac{\partial p}{\partial z}),

$${{\left( \frac{\partial w}{\partial r} \right)}^{m-1}}\frac{\partial w}{\partial r}=\frac{r}{2}\left( \frac{\partial p}{\partial z} \right),$$

therefore

(36)

S_{r z} = \frac{h}{2 {(2 π)}^{m}} [- (\frac{3 m + 1}{m}) h^{- \frac{3 m + 1}{m}} 4 F (\frac{3 m + 1}{m}) h^{- \frac{3 m + 1}{m}} 2 F |^{m - 1}]

$${{S}_{rz}}=\frac{h}{2{{\left( 2\pi \right)}^{m}}}\left[ -\left( \frac{3m+1}{m} \right){{h}^{-\frac{3m+1}{m}}}4F\left( \frac{3m+1}{m} \right){{h}^{-\frac{3m+1}{m}}}2F{{|}^{m-1}} \right]$$

and

(37)

S_{r z} |_{r = h} = \frac{h}{2 {(2 π)}^{m}} [- (\frac{3 m + 1}{m}) h^{- \frac{3 m + 1}{m}} 4 F (\frac{3 m + 1}{m}) h^{- \frac{3 m + 1}{m}} 2 F |^{m - 1}] .

$${{S}_{rz}}{{|}_{r=h}}=\frac{h}{2{{\left( 2\pi \right)}^{m}}}\left[ -\left( \frac{3m+1}{m} \right){{h}^{-\frac{3m+1}{m}}}4F\left( \frac{3m+1}{m} \right){{h}^{-\frac{3m+1}{m}}}2F{{|}^{m-1}} \right].$$

Similarly, wall shear stress at maximum height of the stenosis, i.e. h = 1 −δ is

(38)

τ_{s} = S_{r z} |_{h = (1 - δ)} = \frac{(1 - δ)}{2 {(2 π)}^{m}} [- (\frac{3 m + 1}{m}) {(1 - δ)}^{- \frac{3 m + 1}{m}} 4 F (\frac{3 m + 1}{m}) {(1 - δ)}^{- \frac{3 m + 1}{m}} 2 F |^{m - 1}] .

$${{\tau }_{s}}={{S}_{rz}}{{|}_{h=\left( 1-\delta \right)}}=\frac{\left( 1-\delta \right)}{2{{\left( 2\pi \right)}^{m}}}\left[ -\left( \frac{3m+1}{m} \right){{\left( 1-\delta \right)}^{-\frac{3m+1}{m}}}4F\left( \frac{3m+1}{m} \right){{\left( 1-\delta \right)}^{-\frac{3m+1}{m}}}2F{{|}^{m-1}} \right].$$

Results and discussion

In this section, the behaviour of flow quantities such as velocity profile, wall shear stress, shear stress at stenotic throat and resistance–impedance is displayed graphically for different values of the involved parameters. Figure 3 illustrates the effects of power law index m on velocity profile w in the stenotic region. This figure indicates an increase in w by increasing m. Thus, a fluid that exhibits shear thinning behaviour (m < 1) flows slowly compared to Newtonian (m = 1) and shear-thickening (m > 1) fluids.

Velocity variation for F=2, δ=0.4, n=2, a=0.2, b=0.4.

The effect of the severity of stenosis on the velocity profile w is shown in Fig. 4. We can see from this figure that for a fixed value of prescribed flux F, the velocity profile w increases near the centre, while it decreases near the wall with an increase in the severity of stenosis. The parameter ξ as defined earlier controls the degree of taperness of artery. For non-tapered artery, ξ = 0 while ξ > 0 and ξ < 0 correspond to diverging and converging artery, respectively. The effect of ξ on w can be observed from Fig. 5. Here, we note a decrease in w by increasing ξ . The parameter n that controls the shape of stenosis affects the velocity profile w in a similar way as ξ does. Wall shear stress S_rz profiles for different values of n, δ, m and ξ are shown in Figs. 7–10. Figures 11–13 are plotted to see the influence of n,m and ξ on resistance–impedance, while Figs. 14 and 15 depict the effect of m and F on shear stress at maximum height of the stenosis τ_s.

Velocity variation for F=2, n=2, a=0.2, b=0.4, m=1.1.

Velocity variation for F=2,δ =0.2, n=2, a=0.2, b=0.4, m=1.

Velocity variation for F=2, δ=0.4, a=0.2, b=0.4, m=1.1.

Wall shear stress variation for F=1, δ=0.2, a=0.2, b=0.4, m=1.

Wall shear stress variation for F=1, L=1, a=0.2, b=0.4, m=1.

Wall shear stress variation for F=1, L=1, a=0.2, b=0.4,δ=0.4.

Resistance impedance for F=0.6, L=1, a=0.2, b=0.4, m=2.

Resistance impedance for F=0.6, L=1, a=0.2, b=0.4, n=2.

Resistance impedance for F=0.6, L=1, a=0.2, b=0.4, m=1.2, n=2.

Shear stress at stenotic throat for F=0.2.

Shear stress at stenotic throat for m=2.

The following observations can be made from Figs. 7 to 15.

Wall shear stress is symmetric for n = 2. However, it gets asymmetric otherwise. Its magnitude decreases with an increase in ξ and m. However, the situation becomes opposite with an increase in δ. The resistance impedance λ decreases by increasing m and ξ . The magnitude of shear stress at stenotic throat decreases by increasing m while it shows the opposite trend when m is fixed and F is increased.

Now at the end, we would like to mention that the results presented here for various flow features differ from those of Nadeem et al. [14] in many ways. A complete comparison of these results with ours is not of much interest to the reader and we therefore only present such comparison for results of velocity profile. Nadeem et al. [14] showed in their analysis (Fig. 5 page 7112) that for a prescribed positive flux velocity profile is negative over the whole cross-section [0,h] in the stenotic region which if assumed true suggests that there is no flow in positive axial direction and whole fluid is moving backward. This result is self-contradictory because for prescribed positive flux the velocity profile over the whole cross-section could not be negative. Moreover, the presence of mild stenosis rather altering the direction of flow can alter only the speed of flow. Reverse flow is only possible if there is complete occlusion. On the contrary, our analysis (Fig. 3) shows that the flow speed in a normal artery is less than that of stenotic artery. There is no indication of reverse flow in our results.

Concluding remarks

An exact solution of governing equation for steady flow of a power law fluid through a tapered non-symmetric stenotic tube is obtained. The same problem was already addressed by Nadeem et al. [14], but the results presented by them were erroneous because of an incorrect choice of sign preceding the magnitude of the rate of deformation tensor appearing in the constitutive equation of power law fluid. Using the corrected solution, we have presented the graphical results for velocity profile, resistance–impedance and shear stress. Some discussion about these results is presented. Moreover, the difference between our velocity profile and the one obtained by Nadeem et al. [14] is also highlighted.

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Steady flow of a power law fluid through a tapered non-symmetric stenotic tube

Article Category: Papers dedicated to the memory of Valetnin Afraymovich (1945-2018)

Published Online: Jun 28, 2019

Page range: 255 - 266

Received: Oct 31, 2019

Accepted: Dec 19, 2019

DOI: https://doi.org/10.2478/AMNS.2019.1.00022

Keywords
Stenotic tube, Steady flow, Power law model, Steady flow, Non symmetric

© 2019 Riaz Ahmad et al., published by Sciendo

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.

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Steady flow of a power law fluid through a tapered non-symmetric stenotic tube

Article Category: Papers dedicated to the memory of Valetnin Afraymovich (1945-2018)

Published Online: Jun 28, 2019

Page range: 255 - 266

Received: Oct 31, 2019

Accepted: Dec 19, 2019

DOI: https://doi.org/10.2478/AMNS.2019.1.00022

KeywordsStenotic tube, Steady flow, Power law model, Steady flow, Non symmetric

© 2019 Riaz Ahmad et al., published by Sciendo

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.

Fig. 1

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Keywords
Stenotic tube, Steady flow, Power law model, Steady flow, Non symmetric