Couette Flow and Poiseuille Flow
- Fluid Mechanics 2.3
Plane Couette Flow
Consider a viscous incompressible fluid between two parallel planes. Let the lower plane be $y=0$, the upper plane be $y=h$, and take the $x$-axis in the direction of motion. Assume a steady one-dimensional velocity field
\[ \b{v}=v(y)\hat{\b{x}}. \]
The equation of continuity is identically satisfied. The incompressible Navier–Stokes equation reduces to
\[ \mu \odvn{2}{v}{y}=\odv{p}{x}, \qquad \pdv{p}{y}=0. \]
For pure Couette flow, there is no pressure gradient:
\[ \odv{p}{x}=0. \]
Hence
\[ \odvn{2}{v}{y}=0, \]
so that $v$ is linear in $y$. If the lower plane is fixed and the upper plane moves with velocity $u$, then
\[ v(0)=0,\qquad v(h)=u. \]
Therefore
\[ v(y)=\frac{u}{h}y \]
and the mean velocity is
\[ \bar{v} =\frac{1}{h}\int_0^h v(y)\,dy =\frac{u}{2}. \]
The shear stress is constant:
\[ \varepsilon_{xy} =\mu\odv{v}{y} =\mu\frac{u}{h}. \]
Thus the tangential force per unit area on the lower plane is $\mu u/h$, while the force on the upper plane has the opposite sign.
Plane Poiseuille Flow
Now suppose both planes are fixed and the motion is caused by a pressure gradient. The boundary conditions are
\[ v(0)=v(h)=0. \]
Since $\pdv{p}{y}=0$, the pressure is constant across the gap. Also the equation
\[ \mu \odvn{2}{v}{y}=\odv{p}{x} \]
has a left-hand side depending only on $y$ and a right-hand side depending only on $x$. Hence $\odv{p}{x}$ must be constant. Let
\[ \Delta p=p(0)-p(l)>0 \]
be the pressure drop over a length $l$ in the positive $x$-direction. Then
\[ \odv{p}{x}=-\frac{\Delta p}{l}. \]
Solving the differential equation gives
\[ v(y)=\frac{\Delta p}{2\mu l}y(h-y) \]
which is parabolic and reaches its maximum at $y=h/2$:
\[ v_{\max}=\frac{\Delta p}{8\mu l}h^2. \]
The mean velocity is
\[ \bar{v} =\frac{1}{h}\int_0^h v(y)\,dy =\frac{\Delta p}{12\mu l}h^2 =\frac{2}{3}v_{\max}. \]
Per unit width in the $z$-direction, the volume discharge is
\[ q=\int_0^h v(y)\,dy =\frac{\Delta p}{12\mu l}h^3. \]
The shear stress distribution is
\[ \varepsilon_{xy} =\mu\odv{v}{y} =\frac{\Delta p}{2l}(h-2y). \]
It changes sign at the middle of the channel. The wall shear stress has magnitude
\[ \abs{\varepsilon_{xy}}_{y=0,h} =\frac{\Delta p}{2l}h. \]
If the upper plane also moves with velocity $u$, the two effects superpose:
\[ v(y)=\frac{u}{h}y+\frac{\Delta p}{2\mu l}y(h-y). \]
This is often called Couette–Poiseuille flow.
Flow in a Circular Pipe
Consider steady flow in a circular pipe of radius $R$ and length $l$. Take the $x$-axis along the pipe. By symmetry,
\[ \b{v}=v(r)\hat{\b{x}}, \]
where $r$ is the distance from the axis. The pressure is constant over each cross-section, and
\[ \odv{p}{x}=-\frac{\Delta p}{l}. \]
The $x$-component of the Navier–Stokes equation becomes
\[ \mu \frac{1}{r}\odv{}{r}\left(r\odv{v}{r}\right) =-\frac{\Delta p}{l}. \]
Integrating once,
\[ r\odv{v}{r} =-\frac{\Delta p}{2\mu l}r^2+C. \]
The velocity must remain finite at $r=0$, so $C=0$. Integrating again gives
\[ v(r)=-\frac{\Delta p}{4\mu l}r^2+C_1. \]
The no-slip condition $v(R)=0$ gives
\[ v(r)=\frac{\Delta p}{4\mu l}(R^2-r^2) \]
This is again a parabolic profile. The maximum velocity is
\[ v_{\max}=v(0)=\frac{\Delta p}{4\mu l}R^2. \]
The volume discharge is
\[ \begin{align*} Q &=2\pi\int_0^R v(r)r\,dr \nl &=\frac{\pi\Delta p}{8\mu l}R^4. \end{align*} \]
Thus
\[ Q=\frac{\pi R^4}{8\mu l}\Delta p \]
which is the Hagen–Poiseuille law. The mean velocity in the pipe is
\[ \bar{v} =\frac{Q}{\pi R^2} =\frac{\Delta p}{8\mu l}R^2 =\frac{1}{2}v_{\max}. \]
The mass discharge is $\rho Q$. In particular, the dependence on the fourth power of $R$ shows why the radius of a pipe is the dominant geometric factor in viscous laminar flow.
