Question 4 of 6: Journal Bearing — Petroff No-Load Sizing
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Exams, May 2014 — 07-Mec-B1 Advanced Machine Design. Open book, 3 hours, 100 marks. Part I (Problems 1–2) compulsory; answer three of the four Part II problems (3–6). All six problems are solved here as a complete study resource.
Reference texts: R. G. Budynas & J. K. Nisbett, Shigley’s Mechanical Engineering Design, 10th ed. (shafts & fatigue §6–7, journal bearings §12, clutches & brakes §16); R. C. Juvinall & K. M. Marshek, Fundamentals of Machine Component Design (lubrication, brakes); R. C. Hibbeler, Mechanics of Materials (beam deflection, impact).
Problem 4: Journal Bearing — Petroff No-Load Sizing (20 marks)
Given. A lightly-loaded journal to be sized from its no-load (Petroff) friction loss.
Given data
Quantity
Value
Rotational speed $N$
250 rpm (4.167 rev/s)
Lubricant
ISO VG100 (SAE 30)
Bearing length $L$
$1.2D$
Diametral clearance $c_d$
$0.0045D$ (so $c_r = 0.00225D$)
No-load power loss limit
$2.5\times10^{-4}$ hp $= 0.186$ W
Find. The maximum journal diameter and the corresponding allowable oil temperature.
Approach. Write Petroff’s no-load friction power, substitute the geometric ratios so the loss depends only on the product $\mu D^3$, then push the diameter to its maximum by taking the lowest viscosity the oil may reach at its practical thermal limit.
Petroff no-load loss. Petroff friction torque $T_f = 4\pi^2\mu N r^3 L/c_r$; the dissipated power is $P_f = T_f\,\omega = 8\pi^3\mu N^2 r^3 L/c_r$.
Collapse the geometry. With $r=D/2$, $L=1.2D$ and $c_r = 0.00225D$, all the $D$-dependence gathers into $D^3$:
$$P_f = \pi^3 (533.3)\,\mu N^2 D^3 \equiv K\,\mu D^3 .$$
Setting $P_f = 0.186$ W fixes only the product $\boxed{\,\mu D^3 = 6.49\times10^{-7}\ \text{(SI)}\,}$.
Maximise the diameter. Since $\mu D^3$ is fixed, the largest $D$ corresponds to the smallest viscosity — i.e. the highest temperature the oil may safely run at. For a mineral oil that practical limit is about $70\,{}^\circ\text{C}$ (above which oxidation life falls sharply).
Viscosity at the limit. From the Walther (ASTM D341) chart for VG100, the kinematic viscosity at $70\,{}^\circ\text{C}$ is $\nu \approx 27.7\ \text{cSt}$; with $\rho \approx 854\ \text{kg/m}^3$ this is $\mu \approx 0.0237\ \text{Pa}\!\cdot\!\text{s}$.
Maximum diameter. $D_{\max} = \left(\mu D^3/\mu\right)^{1/3} = \left(6.49\times10^{-7}/0.0237\right)^{1/3} = 0.0302\ \text{m}$, so
$$\boxed{\,D_{\max} \approx 30\ \text{mm at an allowable oil temperature of about }70\,{}^\circ\text{C}\,}$$
Check / assumptions: “Maximum diameter” is bounded by the oil’s thermal limit, taken as $70\,{}^\circ\text{C}$ for mineral VG100 (a standard bulk-oil ceiling); a synthetic or a lower service temperature would change $D_{\max}$. VG100 viscosity is read from the Walther interpolation between its $40\,{}^\circ\text{C}$ (100 cSt) and $100\,{}^\circ\text{C}$ (11.4 cSt) grade points.
