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 the oil for a sleeve length of 100 mm. The thickness of the oil film is 1.0 mm.

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Power Loss in Lubrication

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 the oil for a sleeve length of 100 mm. The thickness of the oil film is 1.0 mm.

Given Data

  • Dynamic Viscosity, \(\mu = 5 \, \text{poise}\)
  • Shaft Diameter, \(D = 0.5 \, \text{m}\)
  • Shaft Speed, \(N = 200 \, \text{r.p.m.}\)
  • Sleeve Length, \(L = 100 \, \text{mm}\)
  • Oil Film Thickness, \(dy = 1.0 \, \text{mm}\)

Solution

1. Convert All Units to SI

We convert all given values to standard SI units for consistency in calculations.

$$ \mu = 5 \, \text{poise} \times \frac{0.1 \, \text{N·s/m}^2}{1 \, \text{poise}} = 0.5 \, \text{N·s/m}^2 $$ $$ L = 100 \, \text{mm} = 0.1 \, \text{m} $$ $$ dy = 1.0 \, \text{mm} = 0.001 \, \text{m} $$

2. Calculate Tangential Velocity of the Shaft (u)

The rotational speed is converted to the linear velocity at the shaft's surface.

$$ u = \frac{\pi D N}{60} $$ $$ u = \frac{\pi \times 0.5 \, \text{m} \times 200 \, \text{r.p.m.}}{60 \, \text{s/min}} $$ $$ u \approx 5.236 \, \text{m/s} $$

3. Calculate Shear Stress (τ)

Using Newton's law of viscosity, where \(du\) is the tangential velocity \(u\) and \(dy\) is the film thickness.

$$ \tau = \mu \frac{du}{dy} $$ $$ \tau = 0.5 \, \text{N·s/m}^2 \times \frac{5.236 \, \text{m/s}}{0.001 \, \text{m}} $$ $$ \tau = 2618 \, \text{N/m}^2 $$

4. Calculate Shear Force (F)

The shear force is the shear stress acting over the surface area of the shaft within the sleeve.

$$ A = \pi D L $$ $$ A = \pi \times 0.5 \, \text{m} \times 0.1 \, \text{m} \approx 0.157 \, \text{m}^2 $$ $$ F = \tau \times A $$ $$ F = 2618 \, \text{N/m}^2 \times 0.157 \, \text{m}^2 \approx 411.03 \, \text{N} $$

5. Calculate Torque (T)

Torque is the rotational force, calculated as the shear force acting at the shaft's radius.

$$ T = F \times R = F \times \frac{D}{2} $$ $$ T = 411.03 \, \text{N} \times \frac{0.5 \, \text{m}}{2} $$ $$ T \approx 102.76 \, \text{N·m} $$

6. Calculate Power Lost (P)

Power is the rate at which work is done. It can be calculated from torque and angular velocity.

$$ \omega = \frac{2 \pi N}{60} = \frac{2 \pi \times 200}{60} \approx 20.944 \, \text{rad/s} $$ $$ P = T \times \omega $$ $$ P = 102.76 \, \text{N·m} \times 20.944 \, \text{rad/s} \approx 2152 \, \text{W} $$
Final Result:

The power lost in the oil is approximately \( P = 2152 \, \text{Watts} \) or \( 2.152 \, \text{kW} \).

Explanation of Key Terms

  • Shear Force (F): The total frictional drag force exerted by the oil on the surface of the rotating shaft.
  • Torque (T): The rotational equivalent of a linear force. It is the twisting force that the motor must overcome to keep the shaft rotating at a constant speed against the oil's resistance.
  • Power (P): The rate at which energy is consumed or work is done. In this context, it's the energy per second that is converted into heat within the oil due to viscous friction.

Physical Meaning

The calculated power loss of 2.152 kW represents the continuous energy that is being converted into heat within the lubricating oil due to internal friction (viscosity). This is a significant amount of energy, equivalent to running more than twenty 100-watt light bulbs.

This calculation is critical for engineers because:

  • Energy Efficiency: It quantifies the energy being wasted as friction. Minimizing this loss is key to designing efficient machinery.
  • Thermal Management: The 2.152 kW of power becomes heat that raises the oil's temperature. A cooling system must be designed to dissipate this heat effectively, otherwise the oil could break down and the bearing could fail.
  • Component Sizing: The motor driving the shaft must be powerful enough to supply the useful work plus this 2.152 kW lost to friction.

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