# Recall Internal Laminar Flows¶

date: 2013/12/18

My memory about the classic flow problems faded in the past several years. The notes here are my redemption.

Consider the steady-state, fully-developed, incompressible, viscous laminar flow between two infinite long parallel plates. $$x, y, z$$ are the axes of the Cartesian coordinate system and all flow properties remain the same in the $$z$$ direction. The flow direction is toward $$+x$$. Let $$u, v$$ be the velocity in the $$x, y$$ directions, respectively, $$p$$ the pressure, and $$\tau_{\xi\eta}$$ the stress. Let the lower plate be at $$y = 0$$ and the upper plate be at $$y = a$$, where $$a$$ is a given constant.

Consider the $$x$$ component of the momentum equation on a infinitesimal square control volume that’s $$\mathrm{d} x$$ wide and $$\mathrm{d} y$$ high:

$\begin{split}& \left[p - \left(p + \frac{\partial p}{\partial x}\right)\right]\mathrm{d} y - \left[\tau_{yx} - \left(\tau_{yx} + \frac{\partial \tau_{yx}}{\partial y} \mathrm{d} y \right) \right] \mathrm{d} x = 0 \\ \Rightarrow & \frac{\partial\tau_{yx}}{\partial y} = \frac{\partial p}{\partial x}\end{split}$

Assume Newtonian fluid:

$\tau_{yx} = \mu\left(\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}\right)$

where $$\mu$$ is the dynamic viscosity coefficient. Because the flow is fully developed, the vertical velocity $$v$$ is void and thus $$\partial v / \partial x = 0$$:

$\tau_{yx} = \mu\frac{\partial u}{\partial y}$

Then we obtain the following equation for $$u$$:

$\mu\frac{\partial^2 u}{\partial y^2} = \frac{\partial p}{\partial x}$

Let $$\partial p / \partial x$$ be constant (this can be justified) and apply the conditions of $$u(0) = u(a) = 0$$:

$u(y) = \frac{1}{2\mu}\frac{\partial p}{\partial x} y \left(y-a\right) = - \frac{a^2}{8\mu}\frac{\partial p}{\partial x} + \frac{1}{2\mu}\frac{\partial p}{\partial x} \left(y - \frac{a}{2}\right)^2$

For $$\partial p / \partial x < 0$$, $$u \ge 0$$ and the peak velocity is $$u(y = a/2) = -(a^2/8\mu)\partial p / \partial x$$.