15: Laminar Flow

Smoothly flowing fluids, with the fluid flowing in undisrupted layers, are called laminar flows. There are several iconic laminar flows, whose velocity fields are readily found by solving the continuity and Navier-Stokes equations.

Plane Couette flow

Plane Couette flow consists of a fluid flowing between two infinite plates separated by a distance $d$. The lower plate is stationary, and the upper plate is moving to the right with velocity $U$. The pressure $p$ is constant and the fluid is incompressible.

We look for a steady solution for the velocity field of the form

$$\nonumber \mathbf{u}(x, y, z)=(u(y), 0,0) . \nonumber$$

The incompressibility condition is automatically satisfied, and the first-component of the Navier-Stokes equation reduces to

$$\nonumber v \frac{\partial^{2} u}{\partial y^{2}}=0 \nonumber$$

Applying the boundary conditions $u(0)=0$ and $u(d)=U$ on the lower and upper plates, the laminar flow solution is given by

$$\nonumber u(y)=\frac{U y}{d} \nonumber$$

Channel flow

Channel flow, or Poiseuille flow, also consists of a fluid flowing between two infinite plates separated by a distance $d$, but with both plates stationary. Here, there is a constant pressure gradient along the $x$-direction in which the fluid flows. Again, we look for a steady solution for the velocity field of the form

$$\nonumber \mathbf{u}(x, y, z)=(u(y), 0,0) \nonumber$$

and with

$$\nonumber p=p(x) \nonumber$$

and

$$\frac{d p}{d x}=-G \nonumber$$

with $G$ a positive constant. The first-component of the Navier-Stokes equation becomes

$$-\frac{1}{\rho} \frac{d p}{d x}+v \frac{d^{2} u}{d y^{2}}=0 . \nonumber$$

Using (15.1) in (15.2) leads to

$$\frac{d^{2} u}{d y^{2}}=-\frac{G}{v \rho} \nonumber$$

which can be solved using the no-slip boundary conditions $u(0)=u(d)=0$. We find

$$u(y)=\frac{G d^{2}}{2 v \rho}\left(\frac{y}{d}\right)\left(1-\frac{y}{d}\right) . \nonumber$$

The maximum velocity of the fluid occurs at the midline, $y=d / 2$, and is given by

$$u_{\max }=\frac{G d^{2}}{8 v \rho} \nonumber$$

Pipe flow

Pipe flow consists of flow through a pipe of circular cross-section radius $R$, with a constant pressure gradient along the pipe length. With the pressure gradient along the $x$-direction, we look for a steady solution of the velocity field of the the form

$$\mathbf{u}=(u(y, z), 0,0) . \nonumber$$

With the constant pressure gradient defined as in $(15.1)$, the Navier-Stokes equation reduces to

$$\frac{\partial^{2} u}{\partial y^{2}}+\frac{\partial^{2} u}{\partial z^{2}}=-\frac{G}{v \rho} . \nonumber$$

The use of polar coordinates in the $y-z$ plane can aid in solving (15.7). With

$$u=u(r) \nonumber$$

we have

$$\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}}=\frac{1}{r} \frac{d}{d r}\left(r \frac{d}{d r}\right), \nonumber$$

so that (15.7) becomes the differential equation

$$\frac{d}{d r}\left(r \frac{d u}{d r}\right)=-\frac{G r}{v \rho} \nonumber$$

with no-slip boundary condition $u(R)=0 .$ The first integration from 0 to $r$ yields

$$r \frac{d u}{d r}=-\frac{G r^{2}}{2 v \rho} \nonumber$$

and after division by $r$, the second integration from $r$ to $R$ yields

$$u(r)=\frac{G R^{2}}{4 v \rho}\left(1-\left(\frac{r}{R}\right)^{2}\right) \nonumber$$

The maximum velocity occurs at the pipe midline, $r=0$, and is given by

$$u_{\max }=\frac{G R^{2}}{4 v \rho} \nonumber$$