Starting from Navier-stokes equation for incompressible fluid and laminar flow, derive the equation for velocity profile for couette flow. State the assumptions made.

Couette flow:

Assumptions:

i) Fluid is incompressible (S=constant)

ii) The flow is one dimensional (in x-direction)

$\therefore u \ne 0, v=w=0$

iii) The flow is steady $\therefore \frac{\partial u}{\partial t}=0$

iv) The flow is independent of variation in z-direction

v) The body force per unit mass are zero.

$\therefore F_x=F_y=F_z=0$

From continuity equation

$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0$

$\because v=w=0$

$\therefore \frac{\partial u}{\partial x}=0$

$\therefore u=f(y)$

From Navier-stokes equation

$u\frac{du}{dx}+v\frac{du}{dy}+w\frac{du}{dz}+\frac{du}{dt}=F_x-\frac{1}{S}\frac{\partial P}{\partial x}+\frac{u}{S}[\frac{\partial ^2u}{\partial x^2}+\frac{\partial ^2u}{\partial y^2}+\frac{\partial ^2u}{\partial z^2}]$

$0=-\frac{1}{S}\frac{\partial P}{\partial x}+\frac{u}{S}\frac{\partial ^2u}{\partial y^2}$

$\therefore \frac{\partial ^2u}{\partial y^2}=\frac{1}{u}\frac{\partial P}{\partial x}$

Integrating,

$\frac{\partial u}{\partial y}=\frac{1}{u}\frac{\partial P}{\partial x}\cdot y+c_1$

Again integrating,

$u=\frac{1}{2u}\frac{\partial P}{\partial x}y^2+c_1\cdot y+c_2$............(1)

where $c_1$ & $c_2$ are constants of integration.

Boundary conditions are

i) at y=0,u=0 $\therefore c_2=0$

ii) at y=b,u=U...........Velocity of moving plate

Using boundary conditions

$U=\frac{1}{2u}\cdot \frac{\partial P}{\partial x}b^2 + c_1\cdot b$

$\therefore c_1=\frac{U-\frac{1}{2u}\cdot \frac{\partial P}{\partial x}\cdot b^2}{b}$

$c_1=\frac{U}{b}-\frac{1}{2u}\cdot \frac{\partial P}{\partial x}\cdot b$

Putting $c_1$ and $c_2$ in equation (1)

$U=\frac{1}{2u}\cdot \frac{\partial P}{\partial x}y^2+[\frac{U}{b}-\frac{1}{2u}\cdot \frac{\partial P}{\partial x}\cdot b]y+0$

$u=\frac{U}{b}\cdot y+\frac{1}{2u}(-\frac{\partial P}{\partial x})(by-y^2)$