Problem Statement
A 150 mm diameter vertical cylinder rotates concentrically inside another cylinder of diameter 151 mm. Both the cylinders are of 250 mm height. The space between the cylinders is filled with a liquid of viscosity 10 poise. Determine the torque required to rotate the inner cylinder at 100 r.p.m.
Given Data
- Inner Cylinder Diameter, \(D_1 = 150 \, \text{mm}\)
- Outer Cylinder Diameter, \(D_2 = 151 \, \text{mm}\)
- Cylinder Height, \(L = 250 \, \text{mm}\)
- Viscosity, \(\mu = 10 \, \text{poise}\)
- Rotational Speed, \(N = 100 \, \text{r.p.m.}\)
Solution
1. Convert All Units to SI
We must convert all given measurements to base SI units (meters, seconds, etc.).
Convert rotational speed (r.p.m.) to angular velocity (rad/s).
2. Calculate Shear Stress (\(\tau\))
The shear stress on the surface of the inner cylinder is given by Newton’s law of viscosity. For a small gap, we can assume a linear velocity profile, so \(\frac{du}{dy} \approx \frac{u}{dy}\).
First, find the tangential velocity (\(u\)) and the gap width (\(dy\)).
Now, calculate the shear stress:
3. Calculate Shear Force (\(F\))
First, calculate the surface area (\(A\)) of the inner cylinder in contact with the fluid.
The shear force is the shear stress acting over this area.
4. Calculate Torque (\(T\))
Torque is the shear force applied at the radius of the inner cylinder.
The torque required to rotate the inner cylinder is approximately \( T = 13.88 \, \text{N·m} \).
Explanation of Viscous Torque
Viscosity is a measure of a fluid’s resistance to shear or angular deformation. When the inner cylinder rotates, it drags the adjacent layer of fluid with it. This motion is transferred through successive fluid layers to the stationary outer cylinder. This creates a velocity gradient across the fluid-filled gap.
According to Newton’s law of viscosity, this velocity gradient results in a shear stress (\(\tau\)) within the fluid. This stress, acting on the surface area of the inner cylinder, produces a shear force (\(F\)) that opposes the rotation. To maintain a constant rotational speed, an external torque (\(T\)) must be applied to overcome this resistive force.
Physical Meaning
The calculated value of 13.88 N·m is the amount of rotational force (torque) needed to continuously overcome the internal friction of the 10-poise liquid and keep the inner cylinder spinning at 100 r.p.m.
This demonstrates several key relationships in fluid dynamics:
- Higher Viscosity: If a more viscous fluid were used, the shear stress would be higher, and a greater torque would be required.
- Higher Speed: Increasing the rotational speed would increase the velocity gradient, also requiring more torque.
- Larger Gap: Increasing the gap between the cylinders (\(dy\)) would decrease the velocity gradient and the shear stress, thus reducing the required torque.
