: (41)

(42)

(39)

133

We express Eq. (34) in series form in order to obtain a more suitable form of τ1,

$$\begin{split} \overline{\tau}\_{1}(\mathbf{x},y,s) &= \frac{16\mu\mathcal{U}(1+i\nu\theta)}{d\hbar(s-iw)} \sum\_{\epsilon=0}^{\nu} \sum\_{\epsilon=0}^{\infty} \frac{\cos\left(\zeta\_{c}x\right)\sin\left(\lambda\_{c}y\right)}{\lambda\_{\epsilon}} \sum\_{p=0}^{\infty} \sum\_{q=0}^{\infty} \sum\_{r=0}^{\infty} \sum\_{s=0}^{\infty} \sum\_{l=0}^{\infty} \\ &\times \frac{\nu^{p+1}\lambda^{s}\theta^{q}\mathcal{K}^{p-r}H^{l}(\lambda\_{c\epsilon})^{r+1}\Gamma(q-p-1)\Gamma(r-p)\Gamma(s+p+2)\Gamma(l+p+1)}{(-1)^{-(p+q+r+s+l)}q!r!s!!\mathcal{U}\Gamma(p)\Gamma(p+1)\Gamma(2+p)\Gamma(1+p)s^{l-q+p+1}}. \end{split} \tag{35}$$

Using the inverse LT of the last equation, we obtain

$$\begin{split} \tau\_{1}(\mathbf{x}, y, t) &= \frac{16\mu e^{i\nu t} U(1 + i\nu \theta)}{d\hbar} \sum\_{c=0}^{\infty} \sum\_{\ell=0}^{\infty} \frac{\cos\left(\zeta\_{c} \mathbf{x}\right) \sin\left(\lambda\_{c} y\right)}{\lambda\_{c}} \sum\_{p=0}^{\infty} \sum\_{q=0}^{\infty} \sum\_{r=0}^{\infty} \sum\_{s=0}^{\infty} \\ & \times \sum\_{l=0}^{\infty} \frac{\nu^{p+1} \lambda^{s} \theta^{q} K^{p-r} H^{l}(\lambda\_{c})^{r+1} t^{-q+p} \Gamma(q-p-1) \Gamma(s+p+2)}{(-1)^{-(p+q+r+s+l)} q! r! s! l! \Gamma(p) \Gamma(p+1) \Gamma(2+p) \Gamma(1+p)} \\ & \times \frac{\Gamma(l+p+1)}{\Gamma(l-q+p+1)}. \end{split} \tag{36}$$

Lastly, we write the stress field in a more compact form by using Fox H-function

$$\tau\_1(x, y, t) \quad = \frac{16\mu \epsilon^{\mu\nu t} U(1 + iw\theta)}{dh} \sum\_{c=0}^{\infty} \sum\_{c=0}^{\infty} \frac{\cos\left(\zeta\_c x\right) \sin\left(\lambda\_c y\right)}{\lambda\_c} \sum\_{p=0}^{\infty}$$

$$\times \sum\_{q=0}^{\infty} \sum\_{r=0}^{\infty} \sum\_{s=0}^{\infty} \frac{\nu^{p+1} \lambda^s \theta^q K^{p-r} (\lambda\_{um})^{r+1} t^{-q+p}}{(-1)^{-(p+q+r+s)} q! r! s!} \tag{37}$$

$$\times H\_{4,b}^{1,4} \left[ Ht \bigg| \begin{array}{c} (2-q+p, 0), (1-r+p, 0), (-1-s-p, 0), (-p, 1) \\\\ (0, 1), (-p, 0), (-1-p, 0), (-p, 0), (-p, 0), (q-p, 1) \end{array} \right].$$

From Eq. (37), we obtain the tangential tension due to cosine oscillations of the duct

$$\begin{split} \tau\_{1\epsilon}(x,y,t) &= \frac{16L\mu(\cos(wt) - w\theta \sin(wt))}{dh} \sum\_{c=0}^{\infty} \sum\_{c=0}^{\infty} \frac{\cos\left(\zeta\_c x\right) \sin\left(\lambda\_c y\right)}{\lambda\_c} \\ &\times \sum\_{p=0}^{\infty} \sum\_{q=0}^{\infty} \sum\_{r=0}^{\infty} \sum\_{s=0}^{\infty} \frac{(-1)^{p+q+r+s} \nu^{p+1} \lambda^s \theta^q K^{p-r} (\lambda\_{um})^{r+1} t^{-q+p}}{q! r! s!} \\ &\times H\_{4,6}^{1,4} \left[ Ht \bigg| \begin{matrix} (2-q+p,0), (1-r+p,0), (-1-s-p,0), (-p,1) \\ (0,1), (-p,0), (-1-p,0), (-p,0), (-p,0), (q-p,1) \end{matrix} \right], \end{split} \tag{38}$$

and the tangential tension corresponding to sine oscillations of the duct

Unsteady Magnetohydrodynamic Flow of Jeffrey Fluid through a Porous Oscillating Rectangular Duct http://dx.doi.org/10.5772/intechopen.70891 133

$$\begin{split} \tau\_{1s}(x,y,t) &= \frac{16L\mu(\sin\left(wt\right)-w\theta\cos\left(wt\right))}{dh} \sum\_{c=0}^{m} \sum\_{\varepsilon=0}^{\infty} \frac{\cos\left(\zeta\_c x\right)\sin\left(\lambda\_c y\right)}{\lambda\_\varepsilon} \sum\_{p=0}^{\infty} \\ &\times \sum\_{q=0}^{m} \sum\_{r=0}^{\infty} \sum\_{s=0}^{\infty} \frac{(-1)^{p+q+r+s} v^{p+1} \lambda^s \theta^q K^{p-r} (\lambda\_{mn})^{r+1} t^{-q+p}}{q!r!s!} \\ &\times H\_{4,6}^{1,4} \left[ Ht \bigg| \begin{matrix} (2-q+p,0), (1-r+p,0), (-1-s-p,0), (-p,1) \\ (0,1), (-p,0), (-1-p,0), (-p,0), (-p,0), (q-p,1) \end{matrix} \right]. \end{split} \tag{39}$$

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

### 6. Volume flux

<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>

16μUð Þ 1 þ iwθ dh sð Þ � iw

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

Using the inverse LT of the last equation, we obtain

�X<sup>∞</sup> l¼0

�

�X<sup>∞</sup> q¼0

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

> �X<sup>∞</sup> p¼0

> > "

�H<sup>1</sup>, <sup>4</sup>

X∞ q¼0

> � � � � �

�

132 Porosity - Process, Technologies and Applications

τ1ð Þ¼ x; y; t

τ1ð Þ¼ x; y; t

τ1cð Þ¼ x; y; t

τ1ð Þ¼ x; y;s

<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>

cosð Þ ζcx sin ð Þ λey λe

We express Eq. (34) in series form in order to obtain a more suitable form of τ1,

X∞ e¼0

ð Þ <sup>λ</sup>ce <sup>r</sup>þ<sup>1</sup>

X∞ c¼0

X∞ e¼0

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

X∞ c¼0

ð Þ �<sup>1</sup> �ð Þ <sup>p</sup>þqþrþsþ<sup>l</sup>

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

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

Γð Þ l þ p þ 1 <sup>Γ</sup>ð Þ <sup>l</sup> � <sup>q</sup> <sup>þ</sup> <sup>p</sup> <sup>þ</sup> <sup>1</sup> :

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

> X∞ r¼0

2 4 X∞ s¼0

> � � � � � �

Lastly, we write the stress field in a more compact form by using Fox H-function

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

X∞ c¼0

From Eq. (37), we obtain the tangential tension due to cosine oscillations of the duct

16Uμð Þ cos wt ð Þ� wθ sin wt ð Þ dh

> X∞ s¼0

X∞ r¼0

and the tangential tension corresponding to sine oscillations of the duct

X∞ e¼0

<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>:

X∞ q¼0

Γð Þ q � p � 1 Γð Þ r � p Γð Þ s þ p þ 2 Γð Þ l þ p þ 1

<sup>q</sup>!r!s!l!Γð Þ<sup>p</sup> <sup>Γ</sup>ð Þ <sup>p</sup> <sup>þ</sup> <sup>1</sup> <sup>Γ</sup>ð Þ <sup>2</sup> <sup>þ</sup> <sup>p</sup> <sup>Γ</sup>ð Þ <sup>1</sup> <sup>þ</sup> <sup>p</sup> sl�qþpþ<sup>1</sup> :

X∞ r¼0

X∞ s¼0

X∞ p¼0

<sup>l</sup>�qþ<sup>p</sup>Γð Þ <sup>q</sup> � <sup>p</sup> � <sup>1</sup> <sup>Γ</sup>ð Þ <sup>s</sup> <sup>þ</sup> <sup>p</sup> <sup>þ</sup> <sup>2</sup>

X∞ p¼0

cosð Þ ζcx sin ð Þ λey λe

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

X∞ q¼0

X∞ r¼0

X∞ s¼0

> 3 5:
