The flow between two parallel, coaxial disks is called

*torsional flow*. In this flow, the bottom disk is fixed, and the top disk rotates at an angular velocity of

\(\varOmega \). The distance between the disks is

*h*. Neglecting the fluid inertia, show that

$$\begin{aligned} \mathbf {u}=\varOmega r \frac{z}{h}\mathbf {e}_{\theta },\;\;\;\dot{\gamma }=\varOmega \frac{r}{h}. \end{aligned}$$

(6.27)

Show that the torque required to turn the top disk is

$$\begin{aligned} M=2\pi \int _{0}^{R}{\dot{\gamma }\eta \left( {\dot{\gamma }}\right) r^{2}dr}, \end{aligned}$$

(6.28)

where

*R* is the radius of the disks. Show that the pressure is

$$\begin{aligned} P\left( r\right) =\int _{\dot{\gamma }}^{\dot{\gamma }_{R}}{\dot{\gamma }\left( { \nu _{1}+\nu _{2}}\right) d\dot{\gamma }}. \end{aligned}$$

(6.29)

From the axial stress, show that the normal force on the top disk is

$$\begin{aligned} F=\pi R^{2}\dot{\gamma }_{R}^{-2}\int _{0}^{\gamma _{R}}{\dot{\gamma }\left( { N_{1}-N_{2}}\right) d\dot{\gamma }}, \end{aligned}$$

(6.30)

where

\(\dot{\gamma }_{R}=\varOmega R/h\) is the shear rate at the rim

\(r=R.\) By normalizing the torque and the force as

$$\begin{aligned} m=\frac{M}{{2\pi R^{3}}},\;\;\;f=\frac{F}{{\pi R^{2}}}, \end{aligned}$$

(6.31)

show that

$$\begin{aligned} \eta \left( {\dot{\gamma }_{R}}\right) =\frac{m}{{\dot{\gamma }_{R}}}\left[ {3+ \frac{{d\ln m}}{{d\ln \dot{\gamma }_{R}}}}\right] , \end{aligned}$$

(6.32)

and

$$\begin{aligned} N_{1}\left( {\dot{\gamma }_{R}}\right) -N_{2}\left( {\dot{\gamma }_{R}}\right) =f\left( {2+\frac{{d\ln f}}{{d\ln \dot{\gamma }_{R}}}}\right) . \end{aligned}$$

(6.33)

Relations (

6.32), (

6.33) are the basis for the operation of the parallel-disk viscometer.