Question 4 of 6: Journal Bearing Sized by No-Load (Petroff) Loss
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — 16-Mec-B1 Advanced Machine Design, December 2017. Open-book, 3 hours, 100 marks. Part I (Problems 1–2) is compulsory; only three of the four Part II problems (Problems 3–6) are required. All six problems are solved as a study resource. “State all assumptions clearly… assume any missing data and properly state it” is an explicit exam instruction.
Given. A lightly-loaded journal bearing whose friction torque follows Petroff’s law (concentric, full-film).
Given data
Speed
$N=250$ rpm $=4.167$ rev/s
Lubricant
ISO VG100 (SAE 30)
Bearing length
$L=1.2D$
No-load power limit
$0.0002$ hp $=0.149$ W
Diametral clearance
$c_d=0.0045\,D$ (radial $c_r=0.00225\,D$)
Find. The maximum journal diameter and the corresponding allowable oil operating temperature.
Approach. Write the Petroff no-load friction power as a function of viscosity and diameter, note that the geometric ratios collapse it to a constraint on $\mu D^3$, then use the VG100 viscosity–temperature relation (Walther) to find the temperature at which $\mu$ is small enough to allow the largest $D$ within a safe thermal limit for mineral oil.
Petroff no-load friction power. The friction torque of a concentric journal is $T_f=\dfrac{4\pi^2\mu N r^3 L}{c_r}$ and the lost power is $P_f=T_f\,\omega=\dfrac{8\pi^3\mu N^2 r^3 L}{c_r}$.
Collapse the geometry. Substituting $r=D/2$, $L=1.2D$, $c_r=0.00225D$, every length scales with $D$ and the power reduces to $P_f=K\,\mu\,D^3$ with $K$ a pure constant (evaluated numerically). Hence the loss limit fixes the product $$\mu D^3=\frac{P_f}{K}=5.19\times10^{-7}\ \text{(SI units)}.$$
Viscosity–temperature relation (VG100). With ISO VG100 anchored at $100\ \text{mm}^2\text{/s}$ at 40 °C and $\approx11.4\ \text{mm}^2\text{/s}$ at 100 °C, the Walther fit (read off the supplied nomograph) gives the kinematic viscosity at 70 °C as $\nu_{70}=27.7\ \text{mm}^2\text{/s}$, i.e. dynamic $\mu_{70}=\nu_{70}\rho\approx0.0237\ \text{Pa}\cdot\text{s}$.
Largest diameter at the thermal limit. Since $\mu D^3$ is fixed, the largest $D$ needs the smallest safe $\mu$, i.e. the highest safe oil temperature. Taking 70 °C as the practical continuous limit for a mineral oil, $$\boxed{\;D_{\max}=\left(\frac{\mu D^3}{\mu_{70}}\right)^{1/3}\approx28\ \text{mm}\quad\text{at}\quad T\approx70\ ^{\circ}\text{C}.\;}$$
Check / assumption: the “maximum diameter” is governed by an oil thermal limit, not by strength — a larger diameter would need an even thinner oil, driving the operating temperature above the safe range for mineral oil. We take 70 °C as that limit; a synthetic oil rated higher would permit a larger journal. The 0.0002 hp loss budget here yields a slightly smaller journal, $\approx28$ mm, than the 30 mm obtained with a 0.00025 hp budget.
