Oil viscosity

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1

A shaft with outside diameter of 18 mm turns at 0.3 RPM inside a stationary journal bearing 60 mm long. A thin film of oil 0.2 mm thick fills the concentric annulus between the shaft and lining. The torue needed to turn the shaft is 0.0036 Nm. Estimate the viscosity ofthe oil that fills the gap.

2

This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook

PK Bot’s answer:

  1. PURPOSE OF THE EXERCISE:
    Demonstrate how to use a viscous Couette flow analysis to predict the required oil viscosity from a measured torque in a journal bearing.

  2. ENGINEERING CONTEXT:
    This problem models a stationary journal bearing with a thin lubricant film; solving it enables one to design or evaluate hydrodynamic bearings in rotating machinery by linking measured torque to fluid viscosity.

  3. PHYSICAL CONTEXT:
    The dominant effect is viscous shear in a narrow gap that produces a retarding torque on the rotating shaft.

  4. ASSUMPTIONS:
    Assume that the flow is steady, incompressible, and unidirectional (i.e. Couette flow), with a constant gap thickness; these assumptions are reasonable because the gap is thin relative to the shaft dimensions and the rotational speed is low.

  5. SOLUTION TECHNIQUE:
    For Couette flow in a journal bearing the shear stress is

\tau = \mu \frac{U}{h},

where the shaft surface speed is

U = \omega R.

The torque on the rotating shaft is obtained by integrating the shear stress over the surface area. For a full circular surface the torque is

T = 2\pi R^2 L \, \mu \frac{U}{h}.

Here, the provided parameters are:

  • Shaft diameter is 18,mm so that R=9\times10^{-3}\,m.
  • Bearing length L=60\times10^{-3}\,m.
  • Gap thickness h=0.2\times10^{-3}\,m.
  • Rotation speed 0.3\,\mathrm{RPM} corresponds to an angular speed
\omega = \frac{0.3\,\mathrm{RPM}\times2\pi}{60} \approx 0.03142\,\mathrm{rad/s},

so that

U = \omega R \approx 0.03142\times9\times10^{-3} \approx 2.8278\times10^{-4}\,m/s.

Given the measured torque T=0.0036\,Nm, solve for the viscosity

\mu = \frac{T}{2\pi R^2 L\,\frac{U}{h}}.

Substitute numerical values:

  • Compute 2\pi R^2 L:
    2\pi (9\times10^{-3})^2(0.06) \approx 2\pi (81\times10^{-6})(0.06) \approx 2\pi (4.86\times10^{-6}) \approx 3.06\times10^{-5}\,m^3.
  • Compute U/h:
    \frac{2.8278\times10^{-4}}{0.2\times10^{-3}} \approx 1.414\,s^{-1}.

Thus,

\mu = \frac{0.0036\,Nm}{(3.06\times10^{-5}\,m^3)(1.414\,s^{-1})} \approx \frac{0.0036}{4.324\times10^{-5}} \approx 83\,Pa\cdot s.
  1. REFLECTION:
    The analysis indicates that to generate a torque of 0.0036,Nm under the given conditions, the oil must have a viscosity of approximately 83\,Pa\cdot s. This value directly ties to the shear-induced torque in a narrow film and highlights how sensitive torque is to both the rotation rate and gap thickness; the approximations used are accurate for low-speed, thin-film, viscous-dominated flows in journal bearings.

In formulating this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked the sections according to their relevance. If they are marked as relevant, I encourage you to explore them:

Textbook sections:

  1. Chapter 11.1: Bearings Part 1 (HIGHLY RELEVANT)
  2. Chapter 11.1: Bearings Part 2 (HIGHLY RELEVANT)
  3. Chapter 11.1: Bearings Part 3 (HIGHLY RELEVANT)
  4. Chapter 10 Fundamentals: Viscosity (RELEVANT)
  5. Chapter 6 Fundamentals: Viscosity (RELEVANT)

Disclaimer:

See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.

3

motor oil viscosity is ~100 fold lower than that value. that viscosity value is like bitumen. I have never heard of using a fluid with such a high viscosity in a bearing, even a low-speed bearing.