: (23)

(24)

(25)

(26)

X∞ m¼1

X∞ s¼0

X∞ n¼1

X∞ c¼0

X∞ e¼0

ð Þ �<sup>1</sup> <sup>p</sup>þqþrþ<sup>s</sup>

<sup>d</sup> , <sup>λ</sup><sup>e</sup> <sup>¼</sup> ð Þ <sup>2</sup><sup>n</sup> <sup>þ</sup> <sup>1</sup> <sup>π</sup>

ð Þ �<sup>1</sup> <sup>p</sup>þqþrþ<sup>s</sup>

ð Þ �<sup>1</sup> <sup>p</sup>þqþrþ<sup>s</sup>

X∞ c¼0

X∞ c¼0

X∞ e¼0

ν<sup>p</sup>þ<sup>1</sup>λ<sup>s</sup> θq K<sup>p</sup>�<sup>r</sup>

q!r!s!

ð Þ 1 � q þ p; 0 ,ð Þ 1 � r þ p; 0 ,ð Þ 1 � s � p; 0 ,ð Þ �p; 1

ð Þ 0; 1 ,ð Þ 1 � p; 0 ,ð Þ 1 � p; 0 ,ð Þ �p; 0 ,ð Þ �p; 0 , qð Þ � p; 1

X∞ e¼0

ν<sup>p</sup>þ<sup>1</sup>λ<sup>s</sup> θq K<sup>p</sup>�<sup>r</sup>

q!r!s!

ð Þ 1 � q þ p; 0 ,ð Þ 1 � r þ p; 0 ,ð Þ 1 � s � p; 0 ,ð Þ �p; 1

ð Þ 0; 1 ,ð Þ 1 � p; 0 ,ð Þ 1 � p; 0 ,ð Þ �p; 0 ,ð Þ �p; 0 , qð Þ � p; 1

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

16U cos wt ð Þ ð Þ� wθ sin wt ð Þ dh

> X∞ s¼0

X∞ r¼0

� � � � �

16U sin wt ð Þ ð Þ� wθ cos wt ð Þ dh

> X∞ s¼0

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

X∞ r¼0

� � � � � ð Þ �<sup>1</sup> <sup>p</sup>þqþrþ<sup>s</sup>

ν<sup>p</sup>þ<sup>1</sup>λ<sup>s</sup> θq K<sup>p</sup>�<sup>r</sup>

q!r!s!

ð Þ 1 � q þ p; 0 ,ð Þ 1 � r þ p; 0 ,ð Þ �s � p; 0 ,ð Þ �p; 1

ð Þ 0; 1 ,ð Þ 1 � p; 0 ,ð Þ 1 � p; 0 ,ð Þ �p; 0 ,ð Þ �p; 0 , qð Þ � p; 1

sin ð Þ ζcx sin ð Þ λey ζcλ<sup>e</sup>

> ν<sup>p</sup>þ<sup>1</sup>λ<sup>s</sup> θq K<sup>p</sup>�<sup>r</sup>

> > q!r!s!

ð Þ 1 � q þ p; 0 ,ð Þ 1 � r þ p; 0 ,ð Þ 1 � s � p; 0 ,ð Þ �p; 1

ð Þ 0; 1 ,ð Þ 1 � p; 0 ,ð Þ 1 � p; 0 ,ð Þ �p; 0 ,ð Þ �p; 0 , qð Þ � p; 1

<sup>h</sup> , c <sup>¼</sup> <sup>2</sup><sup>m</sup> <sup>þ</sup> <sup>1</sup>, e <sup>¼</sup> <sup>2</sup><sup>n</sup> <sup>þ</sup> <sup>1</sup>:

sin ð Þ ζcx sin ð Þ λey ζcλ<sup>e</sup>

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

sin ð Þ ζcx sin ð Þ λey ζcλ<sup>e</sup>

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

F xð Þ¼ ; y; t

130 Porosity - Process, Technologies and Applications

F xð Þ¼ ; y; t

u xð Þ¼ ; y; t

v xð Þ¼ ; y; t

or

where

<sup>4</sup>eiwtUð Þ <sup>1</sup> <sup>þ</sup> iw<sup>θ</sup> dh

> X∞ q¼0

> > "

X∞ r¼0

� � � � �

<sup>16</sup>eiwtUð Þ <sup>1</sup> <sup>þ</sup> iw<sup>θ</sup> dh

> X∞ q¼0

> > � � � � �

X∞ r¼0

X∞ s¼0

�X<sup>∞</sup> p¼0

� <sup>H</sup><sup>1</sup>, <sup>4</sup> <sup>4</sup>, <sup>6</sup> Ht

> �X<sup>∞</sup> p¼0

> > "

<sup>ζ</sup><sup>c</sup> <sup>¼</sup> ð Þ <sup>2</sup><sup>m</sup> <sup>þ</sup> <sup>1</sup> <sup>π</sup>

� <sup>H</sup><sup>1</sup>, <sup>4</sup> <sup>4</sup>, <sup>6</sup> Ht

> �X<sup>∞</sup> p¼0

� <sup>H</sup><sup>1</sup>, <sup>4</sup> <sup>4</sup>, <sup>6</sup> Ht

�X<sup>∞</sup> p¼0

� <sup>H</sup><sup>1</sup>, <sup>4</sup> <sup>4</sup>, <sup>6</sup> Ht

X∞ q¼0

"

and the velocity field due to sine oscillations of the duct

X∞ q¼0

"

$$
\pi\_1(\mathbf{x}, y, t) = \pi\_{1c}(\mathbf{x}, y, t) + i\pi\_{1s}(\mathbf{x}, y, t), \tag{27}
$$

$$
\pi\_2(\mathbf{x}, y, t) = \pi\_{2\mathcal{L}}(\mathbf{x}, y, t) + i\pi\_{2\mathcal{s}}(\mathbf{x}, y, t), \tag{28}
$$

in Eq. (7), we get

$$\tau\_1(\mathbf{x}, y, t) = \frac{\mu}{(1 + \lambda)} \left( 1 + \theta \frac{\partial}{\partial t} \right) \partial\_x F(\mathbf{x}, y, t), \tag{29}$$

$$
\pi\_2(\mathbf{x}, y, t) = \frac{\mu}{(1 + \lambda)} \left( 1 + \theta \frac{\partial}{\partial t} \right) \partial\_y F(\mathbf{x}, y, t). \tag{30}
$$

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

$$
\overline{\tau}\_1(\mathbf{x}, y, \mathbf{s}) = \frac{\mu(1 + \Theta \mathbf{s})}{1 + \lambda} \partial\_\mathbf{x} \overline{F}(\mathbf{x}, y, \mathbf{s}),
\tag{31}
$$

$$
\overline{\tau}\_2(\mathbf{x}, y, \mathbf{s}) = \frac{\mu(1 + \Theta \mathbf{s})}{1 + \lambda} \partial\_y \overline{F}(\mathbf{x}, y, \mathbf{s}). \tag{32}
$$

Taking the inverse Fourier transform of Eq. (19) to get F xð Þ ; y;s and then by putting it into Eq. (31), we get

$$\begin{split} \overline{\tau}\_{1}(\mathbf{x}, \mathbf{y}, \mathbf{s}) &= \frac{4\mu (1 + \Theta \mathbf{s})}{d\hbar (1 + \lambda)} \sum\_{m=1}^{\infty} \sum\_{n=1}^{\infty} \frac{\cos \left( \zeta\_{m} \mathbf{x} \right) \sin \left( \lambda\_{n} y \right) [1 - (-1)^{m}][1 - (-1)^{n}]}{[(1 + \lambda)(\mathbf{s} + H) + \nu (1 + \Theta \mathbf{s})(\lambda\_{m} + K)]} \\ &\times \frac{\mathbf{L} \nu \lambda\_{nm} (1 + iw\theta)}{\lambda\_{n} (\mathbf{s} - iw)}, \end{split} \tag{33}$$

or

$$\nabla\_1(\mathbf{x}, y, s) = \frac{16\mu (1 + \theta \mathbf{s})}{d h (1 + \lambda)} \sum\_{\varepsilon=0}^{\infty} \sum\_{\varepsilon=0}^{\infty} \frac{\cos \left( \zeta\_\varepsilon \mathbf{x} \right) \sin \left( \lambda\_\varepsilon y \right) l l \nu \lambda\_\varepsilon (1 + i w \theta)}{\lambda\_\varepsilon (\mathbf{s} - i w) [(1 + \lambda)(\mathbf{s} + H) + \nu (1 + \theta \mathbf{s})(\lambda\_\varepsilon + K)]},\tag{34}$$

where

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

τ1sð Þ¼ x; y; t

6. Volume flux

lar duct due to cosine oscillations

u xð Þ¼ ; y; t

v xð Þ¼ ; y; t

�X<sup>∞</sup> q¼0

� <sup>H</sup><sup>1</sup>, <sup>4</sup>

The volume flux due to cosine oscillations is given by

�X<sup>∞</sup> p¼0

� <sup>H</sup><sup>1</sup>, <sup>4</sup>

�X<sup>∞</sup> p¼0

� <sup>H</sup><sup>1</sup>, <sup>4</sup>

7. Numerical results and discussion

X∞ q¼0

"

X∞ q¼0

"

X∞ r¼0

"

16Uμð Þ sin wt ð Þ� wθ cos wt ð Þ dh

ð Þ �<sup>1</sup> <sup>p</sup>þqþrþ<sup>s</sup>

In the similar fashion we can find τ2c(x, y, t) and τ2s(x, y, t) from Eqs. (19) and (32).

Qcð Þ¼ x; y; t

64U cos wt ð Þ ð Þ� wθ sin wt ð Þ dh

> X∞ s¼0

ν<sup>p</sup>þ<sup>1</sup>λ<sup>s</sup> θq K<sup>p</sup>�<sup>r</sup>

Similarly, we obtain the volume flux of the rectangular duct due to the sine oscillations

ν<sup>p</sup>þ<sup>1</sup>λ<sup>s</sup> θq K<sup>p</sup>�<sup>r</sup>

X∞ r¼0

� � � � �

64U sin wt ð Þ ð Þ� wθ cos wt ð Þ dh

> X∞ s¼0

X∞ r¼0

� � � � � ðd 0

ðh 0

putting u(x, y,t) from Eq. (25) into the above equation, we obtain the volume flux of the rectangu-

X∞ c¼0

ð Þ �<sup>1</sup> �ð Þ <sup>p</sup>þqþrþ<sup>s</sup> <sup>q</sup>!r!s!

<sup>4</sup>, <sup>6</sup> Ht ð Þ <sup>1</sup> � <sup>q</sup> <sup>þ</sup> <sup>p</sup>; <sup>0</sup> ,ð Þ <sup>1</sup> � <sup>r</sup> <sup>þ</sup> <sup>p</sup>; <sup>0</sup> ,ð Þ <sup>1</sup> � <sup>s</sup> � <sup>p</sup>; <sup>0</sup> ,ð Þ �p; <sup>1</sup>

X∞ e¼0

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

ð Þ 0; 1 ,ð Þ 1 � p; 0 ,ð Þ 1 � p; 0 ,ð Þ �p; 0 ,ð Þ �p; 0 , qð Þ � p; 1

X∞ c¼0

ð Þ �<sup>1</sup> �ð Þ <sup>p</sup>þqþrþ<sup>s</sup> <sup>q</sup>!r!s!

<sup>4</sup>, <sup>6</sup> Ht ð Þ <sup>1</sup> � <sup>q</sup> <sup>þ</sup> <sup>p</sup>; <sup>0</sup> ,ð Þ <sup>1</sup> � <sup>r</sup> <sup>þ</sup> <sup>p</sup>; <sup>0</sup> ,ð Þ <sup>1</sup> � <sup>s</sup> � <sup>p</sup>; <sup>0</sup> ,ð Þ �p; <sup>1</sup>

We have presented flow problem of MHD Jeffrey fluid passing through a porous rectangular duct. Exact analytical solutions are established for such flow problem using DFFST and LT

X∞ e¼0

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

ð Þ 0; 1 ,ð Þ 1 � p; 0 ,ð Þ 1 � p; 0 ,ð Þ �p; 0 ,ð Þ �p; 0 , qð Þ � p; 1

1 ð Þ <sup>ζ</sup>cλ<sup>e</sup> <sup>2</sup>

1 ð Þ <sup>ζ</sup>cλ<sup>e</sup> <sup>2</sup>

X∞ s¼0

� � � � � X∞ c¼0

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

ν<sup>p</sup>þ<sup>1</sup>λ<sup>s</sup> θq K<sup>p</sup>�<sup>r</sup>

X∞ e¼0

q!r!s!

<sup>4</sup>, <sup>6</sup> Ht ð Þ <sup>2</sup> � <sup>q</sup> <sup>þ</sup> <sup>p</sup>; <sup>0</sup> ,ð Þ <sup>1</sup> � <sup>r</sup> <sup>þ</sup> <sup>p</sup>; <sup>0</sup> ,ð Þ �<sup>1</sup> � <sup>s</sup> � <sup>p</sup>; <sup>0</sup> ,ð Þ �p; <sup>1</sup>

ð Þ 0; 1 ,ð Þ �p; 0 ,ð Þ �1 � p; 0 ,ð Þ �p; 0 ,ð Þ �p; 0 , qð Þ � p; 1

cosð Þ ζcx sin ð Þ λey λe

u xð Þ ; y; t dxdy, (40)

ð Þ <sup>λ</sup>mn <sup>r</sup>þ<sup>1</sup> t �qþp X∞ p¼0

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