Problem Statement
A disk of radius \( R \) rotates at an angular velocity \( \omega \) inside an oil bath of viscosity \( \mu \). Assuming a linear velocity profile and neglecting shear on the outer disk edges, derive an expression for the viscous torque on the disk.
- Radius of the disk: \( R \)
- Clearance on both sides: \( h \)
- Torque: \( T \)
- Angular velocity: \( \omega = \frac{2N\pi}{60} \)
- Oil viscosity: \( \mu \)
Solution
Step-by-Step Derivation:
\(\tau = \mu \frac{u}{h} = \mu \frac{\omega r}{h}\)
The shear stress depends on the velocity gradient (\( \frac{u}{h} \)) and the distance \( r \) from the disk’s center.
Consider a ring of radius \( r \) and width \( dr \):
Area of the ring: \( dA = 2\pi r \, dr \)
Shear force: \( dF = \tau \cdot dA = \mu \frac{\omega r}{h} \cdot (2\pi r \, dr) \)
Torque on the ring: \( dT = dF \cdot r = \frac{2\pi\mu\omega}{h} r^3 \, dr \)
Integrate \( dT \) over the radius of the disk:
\( T = \int_0^R dT = \int_0^R \frac{2\pi\mu\omega}{h} r^3 \, dr \)
Solving the integral:
\( T = \frac{2\pi\mu\omega}{h} \cdot \frac{R^4}{4} \)
Simplify:
\( T = \frac{\pi\mu\omega R^4}{h} \)
Final Result:
The viscous torque on the disk is: \( T = \frac{\pi\mu\omega R^4}{h} \)
Explanation
This derivation is based on the following concepts:
- Shear stress calculation: The shear stress is proportional to the velocity gradient and depends on the viscosity of the fluid and the rotational speed of the disk.
- Elementary ring: By considering an infinitesimally small ring on the disk, the local contributions to torque are computed. These contributions are summed by integration.
- Integration: The total torque is found by integrating over the entire radius of the disk, from \( r = 0 \) to \( r = R \).
- Physical interpretation: The resulting formula highlights how the torque is influenced by the viscosity of the oil (\( \mu \)), angular velocity (\( \omega \)), disk radius (\( R \)), and clearance (\( h \)).
