Page: Flow between two parallel plates one plate is moving and the other is stationary
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Consider two plates kept at a distance ‘t’ apart in which the upper plate is moving and the lower plate is fixed. A viscous fluid is flowing between two plates from left to right.

$\left(P+\dfrac{\delta P}{\delta x}.\Delta x\right)$ . Let ‘$\tau$’ be the shear stress acting on face BC then shear stress on face AD will be

$\left(\tau+\dfrac{\delta \tau}{\delta y}.\Delta y\right)$

If the width of element in the direction perpendicular to the paper is unity then the force acting on the fluid elements are

1) Pressure force, $P\times\Delta y\times 1$ on face AB

2) Pressure force $\left(P+\dfrac{\delta P}{\delta x}.\Delta x\right)\Delta y\times 1$ on face CD

3) Shear force $\tau\times\Delta x\times 1$ on face BC

4) Shear force $\left(\tau+\dfrac{\delta \tau}{\delta y}.\Delta y\right)\Delta x\times 1$ on face AD

For steady and uniform flow, there is no acceleration, hence resultant force in the direction of flow is zero.

$\therefore P\times\Delta y\times 1-\left(P+\dfrac{\delta P}{\delta x}.\Delta x\right)\Delta y\times 1-\tau\times\Delta x\times 1-\left(\tau+\dfrac{\delta \tau}{\delta y}.\Delta y\right)\Delta x\times 1=0$

$\therefore -\dfrac{\delta P}{\delta x}.\Delta x\Delta y+\dfrac{\delta \tau}{\delta y}.\Delta y\Delta x=0$

Dividing by $\Delta x\Delta y$, we get

$\therefore -\dfrac{\delta P}{\delta x}+\dfrac{\delta \tau}{\delta y}=0$

$\therefore \dfrac{\delta P}{\delta x}=\dfrac{\delta \tau}{\delta y}……………..(1)$

To obtain velocity distribution across a section, the value of shear stress $\tau=\mu\dfrac{\delta u}{\delta y}$ from Newton's law of viscosity for laminar flow

$\therefore \dfrac{\delta P}{\delta x}=\dfrac{\delta }{\delta y}\left(\mu\dfrac{\delta u}{\delta y}\right)$

$\therefore \dfrac{\delta P}{\delta x}=\mu\dfrac{\delta^2 u}{\delta y^2}$

$\therefore \dfrac{\delta^2 u}{\delta y^2}=\dfrac1\mu \dfrac{\delta P}{\delta x}$

Integrating the above equation with respect to ‘y’ we get

$\therefore \dfrac{\delta u}{\delta y}=\dfrac1\mu \dfrac{\delta P}{\delta x}y+C$

Integrating again,

$\therefore u=\dfrac1\mu \dfrac{\delta P}{\delta x}\dfrac{y^2}2+C_1y+C_2………….(2)$

where $C_1$ and $C_2$ are constants. There values are obtained by using to boundary conditions i.e.,

(a) at y=0, u=0

(b) y=t, u=V

Therefore, substituting (a)equation values

$0=0+C_1\times0+C_2$

$\therefore C_2=0$

Substituting the values from (b) equations

y=t, and u=V, we get

$\therefore u=\dfrac1\mu \dfrac{\delta P}{\delta x}\dfrac{t^2}2+C_1\times t+0$

$\therefore C_1=\dfrac ut-\dfrac1\mu \dfrac{\delta P}{\delta x}\dfrac{t^2}{2\times t}$

$\therefore C_1=\dfrac ut-\dfrac1{2\mu }\dfrac{\delta P}{\delta x}t$

Substituting the values of $C_1, C_2$ in equation (2), we get

$\therefore u=\dfrac1\mu \dfrac{\delta P}{\delta x}\dfrac{y^2}2+\dfrac ut-\dfrac1{2\mu }\dfrac{\delta P}{\delta x}t \times y+0$

$\therefore u=\dfrac ut .y-\dfrac1{2\mu }\dfrac{\delta P}{\delta x}[t y-y^2]$

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modified 7 months ago  • written 7 months ago by gravatar for Syedahina Mohi Syedahina Mohi0
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