00:01
In our question, we are told that the viscosity of liquids can be measured through the use of two concentric cylinders.
00:06
Now in this device, the fluid is filled in between the cylinder and during the measurement, the outer cylinder is fixed.
00:12
While the inner cylinder is rotated with an angular velocity, the torque required to develop the measured and the viscosity is calculated from these two measurements.
00:23
Now we need to develop an equation relating to the viscosity torque and inner cylinder rotation speed to to determine the viscosity of the fluids, neglecting the effects and assuming the velocity distribution in the gap being linear.
00:39
Now during the measurement, the outer cylinder is kept fixed while the inner cylinder is varied by n number of terms rpm.
00:49
Now let omega be the angular speed of the inner cylinder, mu the viscosity of the fluid that is between the two cylinders, the tangential speed of cylinder that is outside, b equal to 0 and that of the inner cylinder being equal to omega r1.
01:05
Now the velocity gradient in a given problem can be given as d u by ty which will be equal to omega r1 r2 minus r1 and the sheer stress that is acting on the element can be given as mu d u by d y.
01:36
Substering the value we have tau to be equal to mu, omega r 1 upon r2 minus r1.
01:45
Now the shear force is equal to the sheer stress multiplied by the surface area...
