Question

Power required in overcoming the viscous resistance in case of shaft in oiled bearing is: where \(\mu\) = coefficient of viscosity, \(D\) = diameter of shaft, \(N\) = angular velocity in RPM, \(L\) = length of shaft, \(t\) = thickness of oil

• A. \(\frac{\mu \pi^2 D^3 N^2 L}{1800t}\)
• B. \(\frac{\mu \pi^3 D^3 N^2 L}{3600t}\)
• C. \(\frac{\mu \pi^2 N^2 R^4 L}{1800t}\)
• D. \(\frac{\mu \pi^3 N^2 R^4 L}{3600t}\)

Hint:

Always convert RPM to rad/s when calculating power or torque.

Answer 1

The Correct Option is A

Concept: Power lost due to viscous resistance in a journal bearing is derived from shear stress in fluid film. Shear stress: \[ \tau = \mu \frac{du}{dy} \] Velocity gradient: \[ \frac{du}{dy} = \frac{\omega R}{t} \]

Step 1: Express torque.
\[ T = \tau \times \text{Area} \times R \] Area: \[ A = \pi D L \]

Step 2: Substitute all values.
\[ T = \mu \cdot \frac{\omega R}{t} \cdot \pi D L \cdot R \] \[ T = \frac{\mu \pi D L \omega R^2}{t} \]

Step 3: Express in terms of \(D\).
\[ R = \frac{D}{2} \] \[ T = \frac{\mu \pi D L \omega (D/2)^2}{t} \] \[ T = \frac{\mu \pi D^3 L \omega}{4t} \]

Step 4: Convert angular speed.
\[ \omega = \frac{2\pi N}{60} \]

Step 5: Power relation.
\[ P = T \omega \] Substituting: \[ P = \frac{\mu \pi D^3 L}{4t} \cdot \frac{2\pi N}{60} \cdot \frac{2\pi N}{60} \] \[ P = \frac{\mu \pi^2 D^3 N^2 L}{1800t} \] Final Answer: \[ \boxed{\frac{\mu \pi^2 D^3 N^2 L}{1800t}} \]