Question

The viscosity of a fluid is to be measured by an viscometer constructed of two 75-cm-long concentric cylinders. The outer diameter of the inner cylinder is 15 cm, and the gap between the two cylinders is 1 mm. The inner cylinder is rotated at 300 rpm, and the torque is measured to be 0.8 N·m. Determine the viscosity of the fluid.

Answer 1

Answer:

Viscosity of the fluid is $\mu=0.013\ \rm kg/m.s$

Explanation:

Given

• Length of the concentric cylinders, L=75 cm
• Gap between two concentric cylinders = 1 mm
• The frequency of inner cylinder = 300 rpm
• Torque acting $\tau=0.8\ \rm N.m$
• Outer diameter of the inner cylinder=15 cm

The Area is given by

$A=\pi dL\\=\pi \times0.15\times0.75\\=0.35\ \rm m^2$

The velocity of the layer is layer is given by

$v=\dfrac{\pi dN}{60}\\=\dfrac{\pi \times0.15\times300}{60}\\\\=2.356\ \rm m/s$

Force acting by one layer of the fluid to other is given by F

$F=\dfrac{\tau}{r}\\=\dfrac{0.8}{0.075}\\=10.7\ \rm N$

Now from the Law of viscosity we have

$F=\mu A\dfrac{du}{dy}\\10.7=\mu \dfrac{0.353\times 2.356}{0.001}\\\mu=0.013\ \rm kg/m.s$

Hence the viscosity of the fluid is $\mu=0.013\ \rm kg/m.s$