Answer to Question #169587 in Mechanical Engineering for nikko

Question

The viscosity of a fluid is to be measured by a viscometer constructed of two 40-cm-long concentric cylinders, such that in the figure. The outer diameter of the inner cylinder is 12 cm, and the gap between the two cylinders is 0.15 cm. The inner cylinder is rotated at 300 rpm, and the torque is measured to be 1.8 N - m. Determine the absolute viscosity, Pa-s, of the fluid

Answer 1

Gap between two concentric cylinders = $0.15 cm=0.0015m$

The frequency of inner cylinder = $300 rpm=\dfrac{300}{60}s^{-1}=5s^{-1}$

Length of the concentric cylinders, $L=40 cm=0.4m$

Torque acting $1.8 N m.$

Outer diameter of the inner cylinder $=12 cm=0.12m$

$$\mu=\frac{Te}{4\pi^2R^3\dot{n}L} =\frac{(1.8N.m)(0.0015m)}{4\pi^2(0.06m)^3(5s^{-1})(0.4m)}$$

=$0.158Pa.s$