A shaft of diameter 120 mm is rotating inside a journal bearing of diameter 122 mm at a speed of 360 r.p.m. The space between the shaft and the bearing is filled with a lubricating oil of viscosity 6 poise. Find the power absorbed in oil if the length of the bearing is 100 mm.

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Power Absorbed in Journal Bearing

Problem Statement

A shaft of diameter 120 mm is rotating inside a journal bearing of diameter 122 mm at a speed of 360 r.p.m. The space between the shaft and the bearing is filled with a lubricating oil of viscosity 6 poise. Find the power absorbed in oil if the length of the bearing is 100 mm.

Given Data

  • Shaft Diameter, \(D_s = 120 \, \text{mm}\)
  • Bearing Diameter, \(D_b = 122 \, \text{mm}\)
  • Speed, \(N = 360 \, \text{r.p.m.}\)
  • Viscosity, \(\mu = 6 \, \text{poise}\)
  • Bearing Length, \(L = 100 \, \text{mm}\)

Solution

1. Convert All Units to SI

Dimensions to metres:

$$ D_s = 120 \, \text{mm} = 0.120 \, \text{m} $$ $$ D_b = 122 \, \text{mm} = 0.122 \, \text{m} $$ $$ L = 100 \, \text{mm} = 0.100 \, \text{m} $$

Viscosity to N·s/m² (10 poise = 1 N·s/m²):

$$ \mu = 6 \, \text{poise} \times \frac{1 \, \text{N s/m}^2}{10 \, \text{poise}} $$ $$ \mu = 0.6 \, \text{N s/m}^2 $$

2. Calculate Geometric and Kinematic Values

Annular Gap (Oil Film Thickness), \(t\):

$$ t = \frac{D_b – D_s}{2} $$ $$ t = \frac{0.122 \, \text{m} – 0.120 \, \text{m}}{2} $$ $$ t = \frac{0.002 \, \text{m}}{2} = 0.001 \, \text{m} $$

Tangential Velocity of Shaft, \(u\):

$$ u = \frac{\pi D_s N}{60} $$ $$ u = \frac{\pi \times 0.120 \, \text{m} \times 360 \, \text{r.p.m.}}{60} $$ $$ u \approx 2.262 \, \text{m/s} $$

3. Calculate Shear Stress (\(\tau\))

Assuming a linear velocity profile across the small gap:

$$ \tau = \mu \frac{du}{dy} = \mu \frac{u}{t} $$ $$ \tau = 0.6 \, \text{N s/m}^2 \times \frac{2.262 \, \text{m/s}}{0.001 \, \text{m}} $$ $$ \tau = 1357.2 \, \text{N/m}^2 $$

4. Calculate Viscous Drag Force (\(F\))

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

$$ F = \tau \times \text{Area} = \tau \times (\pi D_s L) $$ $$ F = 1357.2 \, \text{N/m}^2 \times (\pi \times 0.120 \, \text{m} \times 0.100 \, \text{m}) $$ $$ F \approx 51.15 \, \text{N} $$

5. Calculate Resisting Torque (\(T\))

Torque is the drag force acting at the radius of the shaft.

$$ T = F \times \text{Radius} = F \times \frac{D_s}{2} $$ $$ T = 51.15 \, \text{N} \times \frac{0.120 \, \text{m}}{2} $$ $$ T \approx 3.069 \, \text{N}\cdot\text{m} $$

6. Calculate Power Absorbed (\(P\))

Power is the product of torque and angular velocity (\(\omega\)).

$$ \omega = \frac{2\pi N}{60} = \frac{2\pi \times 360}{60} = 12\pi \approx 37.7 \, \text{rad/s} $$ $$ P = T \times \omega $$ $$ P = 3.069 \, \text{N}\cdot\text{m} \times 37.7 \, \text{rad/s} $$ $$ P \approx 115.7 \, \text{W} $$
Final Result:

The power absorbed in the oil is approximately \( P \approx 115.7 \, \text{Watts} \).

Explanation of the Physics

1. Viscous Shear: The rotation of the shaft drags the adjacent oil layers, creating a velocity gradient across the annular gap between the shaft and the stationary bearing. The oil’s viscosity resists this shearing motion, resulting in a shear stress on the shaft’s surface.

2. Drag and Torque: This shear stress, acting over the entire surface area of the shaft within the bearing, creates a tangential drag force. This force, in turn, produces a resisting torque that continuously opposes the shaft’s rotation.

3. Power Absorption: To maintain a constant rotational speed, the motor driving the shaft must constantly supply power to overcome this viscous torque. This power is not lost in the sense of being destroyed, but is converted into heat within the lubricating oil. This is why lubricating systems often require cooling.

Physical Meaning

The calculated power absorbed (115.7 W) represents the rate of energy that is continuously converted into waste heat within the lubricant due to fluid friction. This is a direct measure of the energy loss or inefficiency of the bearing.

In designing machinery, engineers must manage this effect. While lubrication is essential to prevent metal-to-metal contact and reduce wear, the choice of lubricant viscosity is a trade-off. A higher viscosity oil provides a stronger protective film but also leads to greater viscous power loss and heat generation. This calculation is fundamental to predicting bearing performance, determining cooling requirements, and optimizing overall machine efficiency.

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