4. Calculation of the velocity field

Multiplying both sides of Eq. (13) by sin <sup>m</sup>π<sup>x</sup> d � � and sin <sup>n</sup>π<sup>y</sup> h � �, integrating w.r.t x and y over [0, d] � [0, h] and using Eq. (16), we get

$$(1+\lambda)\frac{\partial F\_{mn}(t)}{\partial t} + \nu \lambda\_{mn} \left(1 + \theta \frac{\partial}{\partial t}\right) F\_{mn}(t) + H(1+\lambda)F\_{mn}(t) + \nu K \left(1 + \theta \frac{\partial}{\partial t}\right) F\_{mn}(t)$$

$$= \nu \lambda\_{mn} U \frac{[1 - (-1)^m][1 - (-1)^n]}{\zeta\_m \lambda\_n} (1 + i\nu \theta) e^{iwt},\tag{17}$$

where

Unsteady Magnetohydrodynamic Flow of Jeffrey Fluid through a Porous Oscillating Rectangular Duct http://dx.doi.org/10.5772/intechopen.70891 129

$$
\zeta\_m = \frac{m\pi}{d}, \lambda\_n = \frac{n\pi}{h} \text{ and } \lambda\_{mn} = \zeta\_m^2 + \lambda\_n^2.
$$

The Fourier transform Fmn(t) have to satisfy the initial conditions

w xð Þ¼ ; y; 0 ∂tw xð Þ¼ ; y; 0 0, (9)

F xð Þ¼ ; y; t u xð Þþ ; y; t iv xð Þ ; y; t , (12)

∂t � �

F xð Þ¼ ; y; 0 ∂tF xð Þ¼ ; y; 0 0, (14)

F xð Þ ; y; t dxdy, m, n ¼ 1, 2, 3, :: (16)

� �, integrating w.r.t x and y over

∂t � �

iwt, (17)

Fmnð Þt

<sup>F</sup>ð Þ¼ <sup>0</sup>; <sup>y</sup>; <sup>t</sup> F dð Þ¼ ; <sup>y</sup>; <sup>t</sup> F xð Þ¼ ; <sup>0</sup>; <sup>t</sup> F xð Þ¼ ; <sup>h</sup>; <sup>t</sup> Ueiwt, (15)

F xð Þ� ; y; t Hð Þ 1 þ λ F xð Þ ; y; t ,

(13)

wð Þ¼ 0; y; t w xð Þ¼ ; 0; t w dð Þ¼ ; y; t w xð Þ¼ ; h; t Ucos wt ð Þ, (10)

wð Þ¼ 0; y; t w xð Þ¼ ; 0; t w dð Þ¼ ; y; t w xð Þ¼ ; h; t Usin wt ð Þ, (11)

t > 0, 0 < x < d and 0 < y < h:

The solutions of problems (8)–(10) and (8), (9), (11) are denoted by u(x, y, t) and v(x, y,t)

t > 0, 0 < x < d and 0 < y < h:

The solution of the problem (13)–(15) will be obtained by means of the DFFST and LT.

sin <sup>n</sup>π<sup>y</sup> h � �

d

ζmλ<sup>n</sup>

� � and sin <sup>n</sup>π<sup>y</sup>

h

FmnðÞþ <sup>t</sup> <sup>H</sup>ð Þ <sup>1</sup> <sup>þ</sup> <sup>λ</sup> Fmnð Þþ <sup>t</sup> <sup>ν</sup><sup>K</sup> <sup>1</sup> <sup>þ</sup> <sup>θ</sup> <sup>∂</sup>

ð Þ 1 þ iwθ e

F xð Þ� ; <sup>y</sup>; <sup>t</sup> <sup>ν</sup><sup>K</sup> <sup>1</sup> <sup>þ</sup> <sup>θ</sup> <sup>∂</sup>

respectively. We define the complex velocity field

∂t � �

∂2 <sup>x</sup> <sup>þ</sup> <sup>∂</sup><sup>2</sup> y � �

which is the solution of the problem

128 Porosity - Process, Technologies and Applications

The DFFST of function F(x, y, t) is denoted by

4. Calculation of the velocity field

[0, d] � [0, h] and using Eq. (16), we get

ð Þ <sup>1</sup> <sup>þ</sup> <sup>λ</sup> <sup>∂</sup>Fmnð Þ<sup>t</sup>

where

Multiplying both sides of Eq. (13) by sin <sup>m</sup>π<sup>x</sup>

ðd 0

ðh 0

<sup>∂</sup><sup>t</sup> <sup>þ</sup> νλmn <sup>1</sup> <sup>þ</sup> <sup>θ</sup> <sup>∂</sup>

sin <sup>m</sup>π<sup>x</sup> d � �

∂t � �

<sup>¼</sup> νλmnU <sup>1</sup> � �ð Þ<sup>1</sup> <sup>m</sup> ½ � <sup>1</sup> � �ð Þ<sup>1</sup> <sup>n</sup> ½ �

FmnðÞ¼ t

ð Þ <sup>1</sup> <sup>þ</sup> <sup>λ</sup> <sup>∂</sup>tF xð Þ¼ ; <sup>y</sup>; <sup>t</sup> <sup>ν</sup> <sup>1</sup> <sup>þ</sup> <sup>θ</sup> <sup>∂</sup>

or

$$F\_{mn}(\mathbf{0}) = \partial\_t F\_{mn}(\mathbf{0}) = \mathbf{0}.\tag{18}$$

We apply LT to Eq. (17) and using initial conditions (18) to get

$$\overline{F}\_{nm}(\mathbf{s}) = \frac{\nu \lambda\_{mn} \mathcal{U} [1 - (-1)^{m}] (1 + iw\theta) [1 - (-1)^{n}]}{\zeta\_{m} \lambda\_{n} (\mathbf{s} - iw) [(1 + \lambda)(\mathbf{s} + H) + \nu(1 + \theta \mathbf{s})(\lambda\_{mn} + K)]}. \tag{19}$$

We will apply the discrete inverse LT technique [20] to obtain analytic solution for the velocity fields and to avoid difficult calculations of residues and contour integrals, but first we express Eq. (19) in series form as

$$\begin{split} \overline{F}\_{nm}(s) &= \frac{\nu \lambda\_{nm} \mathcal{U} [1 - (-1)^{m}] (1 + i\nu \mathcal{O}) [1 - (-1)^{n}]}{\zeta\_{m} \lambda\_{n} (s - i\nu)} \sum\_{p=0}^{\infty} \sum\_{q=0}^{\infty} \sum\_{r=0}^{\infty} \sum\_{s=0}^{\infty} \sum\_{l=0}^{\infty} \\ & \times \frac{\nu^{p+1} \lambda^{s} \mathcal{O}^{q} K^{p-r} H^{l} (\lambda\_{nm})^{r+1} \Gamma(q - p) \Gamma(r - p) \Gamma(s + p + 1) \Gamma(l + p + 1)}{(-1)^{-(p + q + r + s + l)} q! r! s! l! \Gamma(p) \Gamma(p) \Gamma(1 + p) \Gamma(1 + p) s^{l - q + p + 1}} . \end{split} \tag{20}$$

We apply the discrete inverse LT to Eq. (20), to obtain

$$\begin{split} &F\_{mn}(t) \\ &= \frac{e^{jwt}\mathcal{U}[1-(-1)^{m}][1-(-1)^{n}]\nu\lambda\_{mn}(1+\text{i}w\theta)}{\zeta\_{m}\lambda\_{n}}\sum\_{p=0}^{\infty}\sum\_{q=0}^{\infty}\sum\_{r=0}^{\infty}\sum\_{s=0}^{\infty}\sum\_{l=0}^{\infty} \\ &\times \frac{\nu^{p+1}\lambda^{s}\theta^{q}\mathcal{K}^{p-r}\mathcal{U}^{l}(\lambda\_{mn})^{r+1}\Gamma(q-p)\Gamma(r-p)\Gamma(s+p+1)\Gamma(l+p+1)t^{l-q+p}}{(-1)^{-(p+q+r+s+l)}q!r!s!l!\Gamma(p)\Gamma(p)\Gamma(1+p)\Gamma(1+p)\Gamma(l-q+p+1)}. \end{split} \tag{21}$$

Taking the inverse Fourier sine transform we get the analytic solution of the velocity field

$$\begin{split} F(\mathbf{x}, y, t) &= \frac{4}{dl} \sum\_{n=1}^{\infty} \sum\_{n=1}^{\infty} \sin \left( \zeta\_{nm} \mathbf{x} \right) \sin \left( \lambda\_{n} y \right) F\_{mn}(\mathbf{x}, y, t) \\ &= \frac{4 e^{jwt} \mathcal{U} (1 + iw\theta)}{dl} \sum\_{n=1}^{\infty} \sum\_{n=1}^{\infty} \frac{[1 - (-1)^{n}][1 - (-1)^{n}] \sin \left( \zeta \mathbf{x} \right) \sin \left( \lambda\_{n} y \right)}{\zeta\_{m} \lambda\_{n}} \\ &\times \sum\_{p=0}^{\infty} \sum\_{q=0}^{\infty} \sum\_{r=0}^{\infty} \sum\_{s=0}^{\infty} \frac{v^{p+1} \lambda^{r} \theta^{q} K^{p-r} H^{l}(\lambda\_{mn})^{r+1} t^{1-q+p}}{(-1)^{-(p+q+r+s+t)} q^{l} r! s! t!} \\ &\times \frac{\Gamma(q-p) \Gamma(r-p) \Gamma(s+p+1) \Gamma(l+p+1)}{\Gamma(p) \Gamma(p) \Gamma(1+p) \Gamma(1+p) \Gamma(l-q+p+1)}. \end{split} \tag{22}$$

To obtain a more compact form of velocity field we write Eq. (22) in terms of Fox H-function,

$$\begin{split} F(x,y,t) &= \frac{4e^{i\nu t}U(1+i\nu\theta)}{dlt} \sum\_{m=1}^{\infty} \sum\_{n=1}^{\infty} \frac{\sin\left(\zeta\_{m}x\right)\sin\left(\lambda\_{n}y\right)[1-(-1)^{m}][1-(-1)^{n}]}{\zeta\_{m}\lambda\_{n}} \\ &\times \sum\_{p=0}^{\infty} \sum\_{q=0}^{\infty} \sum\_{r=0}^{\infty} \sum\_{s=0}^{\infty} \frac{(-1)^{p+q+r+s}\nu^{p+1}\lambda^{s}\theta^{q}K^{p-r}(\lambda\_{mn})^{r+1}t^{-q+p}}{q!r!s!} \\ &\times H\_{4,6}^{1,4} \left[ dt \bigg| \begin{matrix} (1-q+p,0), (1-r+p,0), (-s-p,0), (-p,1) \\ (0,1), (1-p,0), (1-p,0), (-p,0), (-p,0), (q-p,1) \end{matrix} \right]. \end{split} \tag{23}$$

H<sup>1</sup>,p p, <sup>q</sup>þ<sup>1</sup> �<sup>χ</sup>

5. Calculation of the shear stress

If we introduce

in Eq. (7), we get

Eq. (31), we get

τ1ð Þ¼ x; y;s

or

where

τ1ð Þ¼ x; y;s

�

16μð Þ 1 þ θs dhð Þ 1 þ λ

4μð Þ 1 þ θs dhð Þ 1 þ λ

Uνλmnð Þ 1 þ iwθ <sup>λ</sup>nð Þ <sup>s</sup> � iw ,

> X∞ c¼0

X∞ e¼0

X∞ m¼1 X∞ n¼1

and for sine oscillations by τ1s(x, y, t), τ2s(x, y, t).

We apply LT to Eqs. (29) and (30), to obtain

� � � � � �

<sup>¼</sup> <sup>X</sup><sup>∞</sup> k¼0

<sup>τ</sup>1ð Þ¼ <sup>x</sup>; <sup>y</sup>; <sup>t</sup> <sup>μ</sup>

<sup>τ</sup>2ð Þ¼ <sup>x</sup>; <sup>y</sup>; <sup>t</sup> <sup>μ</sup>

2 4

ð Þ 1 � a1; A<sup>1</sup> ,ð Þ 1 � a2; A<sup>2</sup> , …, 1 � ap; Ap

Unsteady Magnetohydrodynamic Flow of Jeffrey Fluid through a Porous Oscillating Rectangular Duct

ð Þ 1; 0 ,ð Þ 1 � b1; B<sup>1</sup> , …, 1 � bq; Bq

<sup>Γ</sup>ð Þ <sup>a</sup><sup>1</sup> <sup>þ</sup> <sup>A</sup>1<sup>k</sup> …<sup>Γ</sup> ap <sup>þ</sup> Apk � � <sup>k</sup>!Γð Þ <sup>b</sup><sup>1</sup> <sup>þ</sup> <sup>B</sup>1<sup>k</sup> …<sup>Γ</sup> bq <sup>þ</sup> Bqk � � <sup>χ</sup><sup>k</sup>

We denote the tangential tensions for the cosine oscillations of the duct by τ1c(x, y, t), τ2c(x, y, t)

ð Þ <sup>1</sup> <sup>þ</sup> <sup>λ</sup> <sup>1</sup> <sup>þ</sup> <sup>θ</sup> <sup>∂</sup>

ð Þ <sup>1</sup> <sup>þ</sup> <sup>λ</sup> <sup>1</sup> <sup>þ</sup> <sup>θ</sup> <sup>∂</sup>

Taking the inverse Fourier transform of Eq. (19) to get F xð Þ ; y;s and then by putting it into

<sup>τ</sup>1ð Þ¼ <sup>x</sup>; <sup>y</sup>;<sup>s</sup> <sup>μ</sup>ð Þ <sup>1</sup> <sup>þ</sup> <sup>θ</sup><sup>s</sup>

<sup>τ</sup>2ð Þ¼ <sup>x</sup>; <sup>y</sup>;<sup>s</sup> <sup>μ</sup>ð Þ <sup>1</sup> <sup>þ</sup> <sup>θ</sup><sup>s</sup>

∂t

∂t

� �

:

τ1ð Þ¼ x; y; t τ1cð Þþ x; y; t iτ1sð Þ x; y; t , (27)

τ2ð Þ¼ x; y; t τ2cð Þþ x; y; t iτ2sð Þ x; y; t , (28)

� �∂xF xð Þ ; <sup>y</sup>; <sup>t</sup> , (29)

� �∂yF xð Þ ; <sup>y</sup>; <sup>t</sup> : (30)

<sup>1</sup> <sup>þ</sup> <sup>λ</sup> <sup>∂</sup>xF xð Þ ; <sup>y</sup>;<sup>s</sup> , (31)

<sup>1</sup> <sup>þ</sup> <sup>λ</sup> <sup>∂</sup>yF xð Þ ; <sup>y</sup>;<sup>s</sup> : (32)

(33)

cosð Þ <sup>ζ</sup>mx sin ð Þ <sup>λ</sup>ny <sup>1</sup> � �ð Þ<sup>1</sup> <sup>m</sup> ½ � <sup>1</sup> � �ð Þ<sup>1</sup> <sup>n</sup> ½ � ½ � ð Þ 1 þ λ ð Þþ s þ H νð Þ 1 þ θs ð Þ λmn þ K

cosð Þ ζcx sin ð Þ λey Uνλceð Þ 1 þ iwθ

<sup>λ</sup>eð Þ <sup>s</sup> � iw ½ � ð Þ <sup>1</sup> <sup>þ</sup> <sup>λ</sup> ð Þþ <sup>s</sup> <sup>þ</sup> <sup>H</sup> <sup>ν</sup>ð Þ <sup>1</sup> <sup>þ</sup> <sup>θ</sup><sup>s</sup> ð Þ <sup>λ</sup>ce <sup>þ</sup> <sup>K</sup> , (34)

3 5

http://dx.doi.org/10.5772/intechopen.70891

131

� �

or

$$\begin{split} F(\mathbf{x}, y, t) &= \frac{16e^{\mathrm{i}\mathbf{r}\cdot\mathbf{t}}U(1 + \mathrm{i}\mathbf{r}\cdot\mathbf{0})}{d\mathbf{h}} \sum\_{c=0}^{\infty} \sum\_{\epsilon=0}^{\infty} \frac{\sin\left(\zeta\_{c}\mathbf{x}\right)\sin\left(\lambda\_{c}y\right)}{\zeta\_{c}\lambda\_{\epsilon}} \\ &\times \sum\_{p=0}^{\infty} \sum\_{q=0}^{\infty} \sum\_{r=0}^{\infty} \sum\_{s=0}^{\infty} \frac{(-1)^{p+q+r+s}\nu^{p+1}\lambda^{s}\theta^{q}K^{p-r}(\lambda\_{c})^{r+1}\mathbf{f}^{-q+p}}{q!r!s!} \\ &\times H\_{4,6}^{1,4}\left[dt\bigg|\bigg|\begin{matrix} (1-q+p, 0), (1-r+p, 0), (1-s-p, 0), (-p, 1) \\ (0, 1), (1-p, 0), (1-p, 0), (-p, 0), (-p, 0), (q-p, 1) \end{matrix}\right],\end{split} \tag{24}$$

where

$$
\zeta\_c = (2m+1)\frac{\pi}{d}, \lambda\_e = (2n+1)\frac{\pi}{h}, c = 2m+1, e = 2n+1.
$$

From Eq. (24), we obtain the velocity field due to cosine oscillations of the duct

$$\begin{split} u(\mathbf{x},y,t) &= \frac{16I l(\cos\left(wt\right) - w\theta\sin\left(wt\right))}{dh} \sum\_{c=0}^{\infty} \sum\_{c=0}^{\infty} \frac{\sin\left(\zeta\_c \mathbf{x}\right) \sin\left(\lambda\_c y\right)}{\zeta\_c \lambda\_c} \\ &\times \sum\_{p=0}^{\infty} \sum\_{q=0}^{\infty} \sum\_{r=0}^{\infty} \sum\_{s=0}^{\infty} \frac{(-1)^{p+q+r+s} \mathbf{v}^{p+1} \lambda^s \theta^q K^{p-r}(\lambda\_c)^{r+1} t^{-q+p}}{q!r!s!} \\ &\times H\_{4,6}^{1,4} \left[ dt \bigg| \begin{matrix} (1-q+p,0), (1-r+p,0), (1-s-p,0), (-p,1) \\ (0,1), (1-p,0), (1-p,0), (-p,0), (-p,0), (q-p,1) \end{matrix} \end{matrix} \right],\end{split} \tag{25}$$

and the velocity field due to sine oscillations of the duct

$$\begin{split} v(\mathbf{x},y,t) &= \frac{16II(\sin\left(wt\right)-w\theta\cos\left(wt\right))}{dh} \sum\_{c=0}^{\infty} \sum\_{c=0}^{\infty} \frac{\sin\left(\zeta\_c \mathbf{x}\right) \sin\left(\lambda\_c y\right)}{\zeta\_c \lambda\_c} \\ &\times \sum\_{p=0}^{\infty} \sum\_{q=0}^{\infty} \sum\_{r=0}^{\infty} \sum\_{s=0}^{\infty} \frac{(-1)^{p+q+r+s} v^{p+1} A^s \theta^q K^{p-r} (\lambda\_{cc})^{r+1} t^{-q+p}}{q! r! s!} \\ &\times H\_{4,6}^{1,4} \left[ Ht\right] \begin{split} &(1-q+p,0),(1-r+p,0),(1-s-p,0),(-p,1) \\ &(0,1),(1-p,0),(1-p,0),(-p,0),(-p,0),(-p,1) \end{split} \tag{26}$$

We use the following property of the Fox H-function [21] in the above equation

Unsteady Magnetohydrodynamic Flow of Jeffrey Fluid through a Porous Oscillating Rectangular Duct http://dx.doi.org/10.5772/intechopen.70891 131

$$H\_{p,q+1}^{1,p}\left[-\chi\middle|\begin{matrix}(1-a\_1,A\_1),(1-a\_2,A\_2),\dots,(1-a\_p,A\_p)\\(1,0),(1-b\_1,B\_1),\dots,\left(1-b\_q,B\_q\right)\end{matrix}\right]$$

$$=\sum\_{k=0}^{\infty}\frac{\Gamma(a\_1+A\_1k)\dots\Gamma(a\_p+A\_pk)}{k!\Gamma(b\_1+B\_1k)\dots\Gamma(b\_q+B\_qk)}\chi^k.$$

### 5. Calculation of the shear stress

We denote the tangential tensions for the cosine oscillations of the duct by τ1c(x, y, t), τ2c(x, y, t) and for sine oscillations by τ1s(x, y, t), τ2s(x, y, t).

If we introduce

To obtain a more compact form of velocity field we write Eq. (22) in terms of Fox H-function,

sin ð Þ <sup>ζ</sup>mx sin ð Þ <sup>λ</sup>ny <sup>1</sup> � �ð Þ<sup>1</sup> <sup>m</sup> ½ � <sup>1</sup> � �ð Þ<sup>1</sup> <sup>n</sup> ½ � ζmλ<sup>n</sup>

> ð Þ <sup>λ</sup>mn <sup>r</sup>þ<sup>1</sup> t �qþp

ð Þ <sup>λ</sup>ce <sup>r</sup>þ<sup>1</sup> t �qþp # :
