Power Loss in Bearing Lubrication Analysis
Calculating energy dissipation in a rotating shaft bearing due to viscous friction
Problem Statement
The dynamic viscosity of oil used for lubrication between a shaft and sleeve is 6 poise. The shaft has a diameter of 0.4 m and rotates at 190 rpm. Calculate the power lost in the bearing for a sleeve length of 90 mm, given the thickness of the oil film is 1.5 mm.
Given Data
Dynamic Viscosity (μ)
6 poise
Shaft Diameter (D)
0.4 m
Rotational Speed (N)
190 rpm
Sleeve Length (L)
90 mm = 0.09 m
Oil Film Thickness (t)
1.5 mm = 0.0015 m
Schematic Diagram
Fig. 1: Cross-section of shaft, oil film, and sleeve assembly
Solution
Convert Viscosity to SI Units
\( \mu = 6 \times 0.1 = 0.6 \text{ Ns/m}^2 \)
Calculate Tangential Velocity of Shaft
\( u = \frac{\pi \times 0.4 \times 190}{60} \)
\( u = \frac{3.1416 \times 76}{60} = \frac{238.7616}{60} = 3.97936 \approx 3.98 \text{ m/s} \)
Calculate Shear Stress
\( \tau = 0.6 \times \frac{3.97936}{0.0015} \)
\( \tau = 0.6 \times 2652.9067 = 1591.744 \approx 1592 \text{ N/m}^2 \)
Calculate Shear Force on Shaft
\( F = 1592 \times (\pi \times 0.4 \times 0.09) \)
\( F = 1592 \times (3.1416 \times 0.036) = 1592 \times 0.1130976 \approx 180.05 \text{ N} \)
Calculate Torque on Shaft
\( T = 180.05 \times \frac{0.4}{2} = 180.05 \times 0.2 = 36.01 \text{ Nm} \)
Calculate Power Lost
\( P = \frac{2 \times \pi \times 190 \times 36.01}{60} \)
\( P = \frac{2 \times 3.1416 \times 190 \times 36.01}{60} = \frac{42988.9}{60} \approx 716.48 \text{ W} \)
Explanation
1. Viscosity Conversion:
The dynamic viscosity was converted from poise to SI units (Ns/m²) because all other parameters are in SI units. This ensures dimensional consistency in calculations.
2. Tangential Velocity:
The tangential velocity of 3.98 m/s represents the surface speed of the rotating shaft. This is calculated using the shaft diameter and rotational speed.
3. Shear Stress:
The shear stress of 1592 N/m² is determined using Newton's law of viscosity, where the velocity gradient is approximated as the tangential velocity divided by the oil film thickness.
4. Shear Force:
The shear force of 180.05 N is the total viscous drag acting on the shaft surface, calculated as the product of shear stress and the lateral surface area of the shaft within the sleeve.
5. Torque Calculation:
The torque of 36.01 Nm represents the rotational resistance caused by viscous friction, calculated as the product of shear force and shaft radius.
6. Power Dissipation:
The power lost (716.48 W) is the rate at which energy is dissipated as heat due to viscous friction in the oil film. This represents the energy required to overcome friction in the bearing.
Physical Meaning & Applications
1. Power Loss Significance:
The calculated power loss of 716.48 watts represents the energy converted to heat due to viscous friction in the oil film. This energy must be dissipated to prevent overheating of the bearing system.
2. Design Implications:
This calculation helps engineers:
- Select appropriate oil viscosity for efficient lubrication
- Design cooling systems for bearings
- Determine power requirements for rotating machinery
- Optimize bearing dimensions for minimal energy loss
3. Industrial Applications:
Understanding power loss in bearings is crucial for:
- Electric motor and generator design
- Turbine and pump efficiency calculations
- Automotive and aerospace transmission systems
- Industrial machinery maintenance planning
4. Efficiency Considerations:
While some power loss is inevitable, engineers aim to minimize it through:
- Optimal oil viscosity selection
- Proper film thickness maintenance
- Surface finish improvements
- Temperature control systems
