Question 3 [40 marks] - Rotating Cylinder Viscometer The...

Question 3 [40 marks] - Rotating Cylinder Viscometer The viscosity of liquids can be determined with a rotating cylinder viscometer, shown alongside. In such a device, the outer cylinder is fixed and the inner cylinder is rotated with an angular velocity (omega). The torque (T) required to develop (omega) is measured and the viscosity is calculated from these two measurements. (See sketch below.) 3.1. Apply the Navier-Stokes and Continuity equations for this viscometer and determine the fluid's velocity profile, assuming laminar flow. 3.2. Develop an equation relating (mu) to (omega), (T), (l), (R_o), and (R_i).

Transcript

00:01 So here we are given that the viscosity of liquid can be determined with the rotation slender visco viscometer which is shown here in such a device.

00:12 So this device is shown here.

00:14 Here liquid is present in all this direction.

00:20 Here we are given the value of the r.

00:21 This is ri, this is ro, inner radius and outer radius.

00:26 Here we are given this shaft which is moving at the t with the angular velocity omega.

00:33 This is a fixed outer cylindrical and this is the liquid which is filled here.

00:40 So this value is given.

00:42 We have to apply the vector, stroke and continuity equation for this viscometer.

00:48 So here we have to apply the equation.

00:51 So here firstly we are about the assumption that there must be the 3d flow.

00:59 Next is that there can be incompressible flow.

01:06 Flow will be incompressible.

01:08 Next is that no flow or variation of properties in two directions.

01:16 So we are considering the value of v2 is 0 and sigma divided by sigma z that too is equal to 0.

01:22 Next from here is symmetry in the circumcenter that is sigma divided by sigma theta that too is equal to 0.

01:30 So we are considering about the navier -stokes and the continuity equation.

01:38 So from here we are considering about the continuity.

01:44 So equation will be 1 divided by r, sigma divided by the sigma r, r multiplied by the vn plus 1 by r, sigma divided by sigma theta multiplied by the v theta plus sigma divided by sigma z, vz that is equal to 0.

01:58 This term from here is u and this term from here is 3.

02:03 So the value of sigma divided by the sigma r multiplied by the r of vn is equal to 0.

02:08 So the value of r of vn from here is constant...

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