# 0.7 Laminar flows with dependence on one dimension  (Page 2/6)

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## Couette flow

The flows when the fluid between two parallel surfaces are induced to flow by the motion of one surface relative to the other is called Couette flow . This is the generic shear flow that is used to illustrate Newton's law of viscosity. Pressure and body forces balance each other and at steady state the equation of motion simplify to the divergence of the viscous stress tensor or the Laplacian of velocity in the case of a Newtonian fluid.

Planar Couette flow . (case 7).

$\frac{d\tau }{d{x}_{3}}=-\mu \frac{{d}^{2}{v}_{j}}{d{x}_{3}^{2}}=0,\phantom{\rule{1.em}{0ex}}j=1,\phantom{\rule{0.166667em}{0ex}}2$

The coordinates system can be defined so that $\mathbf{v}=0$ at ${x}_{3}=0$ and the $j$ component of velocity is non-zero at ${x}_{3}=L$ .

$\begin{array}{c}{v}_{j}=0,\phantom{\rule{1.em}{0ex}}{x}_{3}=0\hfill \\ {v}_{j}={U}_{j},\phantom{\rule{1.em}{0ex}}{x}_{3}=L,\phantom{\rule{1.em}{0ex}}j=1,\phantom{\rule{0.166667em}{0ex}}2\hfill \end{array}$

The velocity field is

${v}_{j}=\frac{{U}_{j}\phantom{\rule{0.166667em}{0ex}}{x}_{3}}{L}$

The shear stress can be determined from Newton's law of viscosity.

${\tau }_{j3}=-\mu \frac{d{v}_{j}}{d{x}_{3}}=-\mu \frac{{U}_{j}}{L},\phantom{\rule{1.em}{0ex}}j=1,\phantom{\rule{0.166667em}{0ex}}2$

Cylindrical Couette flow. The above example was the translational movement of two planes relative to each other. Couette flow is also possible in the annular gap between two concentric cylindrical surfaces (cases 8 and 9) if secondary flows do not occur due to centrifugal forces. We use cylindrical polar coordinates rather than Cartesian and assume vanishing Reynolds number. The only independent variable is the radius.

$\begin{array}{c}\frac{1}{r}\frac{\partial }{\partial r}\left(r\phantom{\rule{0.166667em}{0ex}}{v}_{r}\right)=0\hfill \\ {\left[\nabla •\tau \right]}_{r}=\frac{1}{r}\frac{\partial }{\partial r}\left(r\phantom{\rule{0.166667em}{0ex}}{\tau }_{rr}\right)-\frac{{\tau }_{\theta \theta }}{r},\phantom{\rule{1.em}{0ex}}\mathrm{may}\phantom{\rule{0.277778em}{0ex}}\mathrm{not}\phantom{\rule{0.277778em}{0ex}}\mathrm{vanish}\phantom{\rule{0.166667em}{0ex}}\mathrm{if}\phantom{\rule{0.277778em}{0ex}}\mathrm{Reynolds}\phantom{\rule{0.277778em}{0ex}}\mathrm{number}\phantom{\rule{0.277778em}{0ex}}\mathrm{is}\phantom{\rule{0.277778em}{0ex}}\mathrm{high}\hfill \\ {\left[\nabla •\tau \right]}_{\theta }=\frac{1}{{r}^{2}}\frac{\partial }{\partial r}\left({r}^{2}{\tau }_{r\theta }\right)+\frac{{\tau }_{\theta r}-{\tau }_{r\theta }}{r}=0,\phantom{\rule{1.em}{0ex}}\mathrm{may}\phantom{\rule{0.277778em}{0ex}}\mathrm{not}\phantom{\rule{0.277778em}{0ex}}\mathrm{vanish}\phantom{\rule{0.166667em}{0ex}}\mathrm{if}\phantom{\rule{0.277778em}{0ex}}\mathrm{Reynolds}\phantom{\rule{0.277778em}{0ex}}\mathrm{number}\phantom{\rule{0.277778em}{0ex}}\mathrm{is}\phantom{\rule{0.277778em}{0ex}}\mathrm{high}\hfill \\ {\left[\nabla •\tau \right]}_{z}=\frac{1}{r}\frac{\partial }{\partial r}\left(r\phantom{\rule{0.166667em}{0ex}}{\tau }_{rz}\right)=0\hfill \\ \left(\begin{array}{c}\frac{\partial }{\partial r}\left[\frac{1}{r}\frac{\partial }{\partial r}\left(r\phantom{\rule{0.166667em}{0ex}}{v}_{\theta }\right)\right]=0\\ \frac{\partial }{\partial r}\left(r\frac{\partial {v}_{z}}{\partial r}\right)=0\end{array}}\phantom{\rule{1.em}{0ex}}\mathrm{Newtonian}\phantom{\rule{0.277778em}{0ex}}\mathrm{fluid}\hfill \end{array}$

The stress profile can be calculated by integration.

$\begin{array}{c}{r}^{2}{\tau }_{r\theta }={\left({r}^{2}{\tau }_{r\theta }|}_{r={r}_{1}}={\left({r}^{2}{\tau }_{r\theta }|}_{r={r}_{2}}\hfill \\ r{\tau }_{rz}={\left(r{\tau }_{rz}|}_{r={r}_{1}}={\left(r{\tau }_{rz}|}_{r={r}_{2}}\hfill \end{array}$

The boundary conditions on velocity depend on whether the cylindrical surfaces move relative to each other as a result of rotation, axial translation, or both.

$\begin{array}{c}\mathbf{v}=0,\phantom{\rule{1.em}{0ex}}r={r}_{1}\hfill \\ \mathbf{v}=\left(0,{U}_{\theta },{U}_{z}\right),\phantom{\rule{1.em}{0ex}}r={r}_{2}\hfill \end{array}$

The velocity field for cylindrical Couette flow of a Newtonian fluid is .

$\begin{array}{c}{v}_{\theta }=\frac{{U}_{\theta }}{\frac{{r}_{2}}{{r}_{1}}-\frac{{r}_{1}}{{r}_{2}}}\left(\frac{r}{{r}_{1}}-\frac{{r}_{1}}{r}\right)\hfill \\ {v}_{z}=\frac{{U}_{z}\phantom{\rule{0.166667em}{0ex}}}{log\left({r}_{2}/{r}_{1}\right)}\phantom{\rule{0.166667em}{0ex}}log\left(r/{r}_{1}\right)\hfill \end{array}$

Planer rotational Couette flow. The parallel plate viscometer has the configuration shown in case 10. The system is not strictly 1-D because the velocity of one of the surfaces is a function of radius. Also, there is a centrifugal force present near the rotating surface but is absent at the stationary surface. However, if the Reynolds number is small enough that secondary flows do not occur, then the velocity at a given value of the radius may be approximated as a function of only the $z$ distance in the gap. The differential equations at zero Reynolds number are as follows.

$\begin{array}{c}{\left[\nabla •\tau \right]}_{\theta }=\frac{\partial {\tau }_{\theta z}}{\partial z}=0\hfill \\ \frac{{\partial }^{2}{v}_{\theta }}{\partial {z}^{2}}=0,\phantom{\rule{1.em}{0ex}}\mathrm{Newtonian}\phantom{\rule{0.277778em}{0ex}}\mathrm{fluid}\hfill \end{array}$

Suppose the bottom surface is stationary and the top surface is rotating. Then the boundary conditions are as follows.

$\begin{array}{c}{v}_{\theta }=0,\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}z=0\hfill \\ {v}_{\theta }=2\pi \phantom{\rule{0.166667em}{0ex}}r\phantom{\rule{0.166667em}{0ex}}\Omega ,\phantom{\rule{1.em}{0ex}}\phantom{\rule{0.277778em}{0ex}}z=L\hfill \end{array}$

The stress and velocity profiles are as follows.

$\begin{array}{c}{\tau }_{\theta z}={\tau }_{\theta z}\left(r\right)=-2\pi \phantom{\rule{0.166667em}{0ex}}r\phantom{\rule{0.166667em}{0ex}}\Omega \phantom{\rule{0.166667em}{0ex}}\mu \left(r\right)/L\hfill \\ {v}_{\theta }=2\pi \phantom{\rule{0.166667em}{0ex}}r\phantom{\rule{0.166667em}{0ex}}\Omega \phantom{\rule{0.166667em}{0ex}}z/L,\phantom{\rule{1.em}{0ex}}\mathrm{Newtonian}\phantom{\rule{0.277778em}{0ex}}\mathrm{fluid}\hfill \end{array}$

The stress is a function of the radius and if the fluid is non-Newtonian, the viscosity may be changing with radial position.

## Plane-poiseuille and hele-shaw flow

Forced flow between two stationary, parallel plates, case 2, is called plane-Poiseuille flow or if the flow depends on two spatial variables in the plane, it is called Hele-Shaw flow. The flow is forced by a specified flow rate or a specified pressure or gravity potential gradient. The pressure and gravitational potential can be combined into a single variable, $P$ .

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