Problem Statement
An oil of viscosity 5 poise is used for lubrication between a shaft and sleeve. The diameter of the shaft is 0.5 m and it rotates at 200 r.p.m. Calculate the power lost in oil for a sleeve length of 100 mm. The thickness of the oil film is 1.0 mm.
Given Data
- Viscosity, \(\mu = 5 \, \text{poise}\)
- Shaft Diameter, \(D = 0.5 \, \text{m}\)
- Shaft Speed, \(N = 200 \, \text{r.p.m.}\)
- Sleeve Length, \(L = 100 \, \text{mm} = 0.1 \, \text{m}\)
- Oil Film Thickness, \(t = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m}\)
Solution
1. Convert Viscosity to SI Units
The viscosity is given in poise. We convert it to N·s/m² (10 poise = 1 N·s/m²).
2. Calculate Tangential Velocity of the Shaft
The linear velocity (\(u\)) at the surface of the rotating shaft is calculated from its angular velocity.
3. Calculate Shear Stress in the Oil
Assuming a linear velocity profile across the thin oil film, the shear stress (\(\tau\)) is given by Newton's law of viscosity.
Here, \(du = u - 0 = u\) and the film thickness \(dy = t\).
$$ \tau = 0.5 \, \text{N s/m}^2 \times \frac{5.236 \, \text{m/s}}{1.0 \times 10^{-3} \, \text{m}} = 2618 \, \text{N/m}^2 $$4. Calculate Viscous Drag Force
The shear force (\(F\)) is the shear stress acting over the surface area of the shaft inside the sleeve.
5. Calculate the Resisting Torque
The torque (\(T\)) is the drag force multiplied by the shaft radius (\(D/2\)).
6. Calculate the Power Lost
Power (\(P\)) is the product of torque and angular velocity (\(\omega\)).
The power lost in the oil is approximately \( P \approx 2153 \, \text{W} = 2.15 \, \text{kW} \).
Explanation of the Process
1. Viscous Shear:
The rotation of the shaft drags the adjacent layer of oil with it. This motion is transmitted through the oil film to the stationary sleeve, creating a velocity gradient. The oil's internal friction (viscosity) resists this shearing motion, resulting in a shear stress.
2. Drag Force and Torque:
The shear stress, when integrated over the entire lubricated surface area of the shaft, produces a total drag force. This force acts tangentially to the shaft's surface, creating a resisting torque that opposes the rotation.
3. Power Loss:
To overcome this viscous torque and maintain a constant rotational speed, continuous power must be supplied. This power is dissipated as heat within the oil, representing an energy loss in the system. The formula \(P = T \omega\) directly calculates this required power.
Physical Meaning
This calculation is fundamental to the design of all rotating machinery that uses fluid lubrication, such as journal bearings in engines, turbines, and generators.
The power lost represents a direct loss of efficiency, as energy that could be doing useful work is instead converted into waste heat in the lubricant. Engineers must balance the need for lubrication (to reduce friction and wear between solid parts) with the inevitable power loss due to viscous shearing in the lubricant itself.
The result shows that power loss is highly dependent on viscosity, rotational speed, and the geometry of the components. A higher viscosity oil, a faster speed, or a larger surface area would all lead to greater power loss.
