Taylor-Couette Flow

Consider two thin cylindrical shells with the same vertical axis. Let the inner and outer shells be of radius $r_1$ and $r_2$ , respectively. Suppose that the annular region $r_1\leq r\leq r_2$ is filled with fluid of density $\rho$ and viscosity $\mu$ . Let the inner and outer cylinders rotate at the constant angular velocities ${\mit\Omega}_1$ and ${\mit\Omega}_2$ , respectively. We wish to determine the steady flow pattern set up within the fluid. Incidentally, this type of flow is generally known as Taylor-Couette flow, after Maurice Couette and Geoffrey Taylor (1886-1975).

It is convenient to adopt cylindrical coordinates, $r$ , $\theta$ , $z$ , whose symmetry axis coincides with the common axis of the two shells. Thus, the inner and outer shells correspond to $r=r_1$ and $r=r_2$ , respectively. Suppose that the flow velocity within the fluid is written

$${\bf v} = v_\theta(r)\,{\bf e}_\theta = r\,{\mit\Omega}(r)\,{\bf e}_\theta,$$ (10.30)

where ${\mit\Omega}(r)=v_\theta(r)/r$ is the angular velocity profile. Application of the no slip condition at the two shells leads to the boundary conditions

$${\mit\Omega}(r_1) = {\mit\Omega}_1,$$ (10.31)

$${\mit\Omega}(r_2) ={\mit\Omega}_2.$$ (10.32)

It follows from Section 1.19 that $\nabla\cdot{\bf v}=0$ and

$$\frac{D{\bf v}}{Dt} =-r\,{\mit\Omega}^{\,2}\,{\bf e}_r,$$ (10.33)

$$\nabla^{\,2} {\bf v} =\frac{1}{r^{\,2}}\,\frac{d}{dr}\left(r^{\,3}\,\frac{d{\mit\Omega}}{dr}\right)\,{\bf e}_\theta.$$ (10.34)

Hence, Equation (10.2) yields

$$\frac{dP}{dr} = \rho\,r\,{\mit\Omega}^{\,2},$$ (10.35)

and

$$\frac{1}{r^{\,2}}\,\frac{d}{dr}\!\left(r^{\,3}\,\frac{d{\mit\Omega}}{dr}\right) = 0,$$ (10.36)

assuming that $P=P(r)$ . The solution of Equation (10.36) that satisfies the boundary conditions is

$${\mit\Omega}(r) = \frac{1}{r^{\,2}}\left(\frac{{\mit\Omega}_1-{\m... ...mega}_2\,r_1^{\,-2}- {\mit\Omega}_1\,r_2^{\,-2}}{r_1^{\,-2}-r_2^{\,-2}}\right).$$ (10.37)

It can be seen that this angular velocity profile is a combination of the solid body rotation profile ${\mit\Omega}= {\rm constant}$ , and the irrotational rotation profile ${\mit\Omega}\propto r^{\,-2}$ .

From Section 1.19, the only non-zero component of the viscous stress tensor within the fluid is

$$\sigma_{r\theta} = \mu\,r\,\frac{d}{dr}\!\left(\frac{v_\theta}{r}\right) = \mu\,r\,\frac{d{\mit\Omega}}{dr}.$$ (10.38)

Thus, the viscous torque (acting in the $\theta$ -direction) per unit height (in the $z$ -direction) exerted on the inner cylinder is

$$\tau_1 = 2\pi\,r_1^{\,2}\,\sigma_{r\theta}(r_1) = -4\pi\,\mu\left(\frac{{\mit\Omega}_1-{\mit\Omega}_2}{r_1^{\,-2}-r_2^{\,-2}}\right).$$ (10.39)

Likewise, the torque per unit height exerted on the outer cylinder is

$$\tau_2 = -2\pi\,r_2^{\,2}\,\sigma_{r\theta}(r_2) = 4\pi\,\mu\left(\frac{{\mit\Omega}_1-{\mit\Omega}_2}{r_1^{\,-2}-r_2^{\,-2}}\right).$$ (10.40)

As expected, these two torques are equal and opposite, and act to make the two cylinders rotate at the same angular velocity (in which case, the fluid between them rotates as a solid body).