Question 4 of 6: Journal Bearing Sized by No-Load (Petroff) Power 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, May 2018. Open book, 3 hours, 100 marks. Part I (Problems 1 & 2) is compulsory; candidates answer only three of the four Part II problems (3–6). All six problems are solved here as a complete study resource.
Given. A lightly-loaded journal running at 250 rpm in VG100 oil; the friction (no-load) power loss is capped, which through the Petroff relation limits the product $\mu D^3$.
Given data
Speed
$N=250\ \text{rpm} = 4.167\ \text{rev/s}$
Lubricant
ISO VG100 (SAE 30)
Bearing length
$L=1.2D$
No-load power loss
$P_f \le 0.0002\ \text{hp} = 0.149\ \text{W}$
Diametral clearance
$c_d = 0.0045D$ → radial $c_r=0.00225D$
Find. The maximum journal diameter $D$ and the corresponding oil (operating) temperature limit.
ISO VG100 viscosity–temperature curve (schematic): raising the oil temperature lowers $\mu$, which is how a larger journal is made to meet the fixed loss cap. The chosen limit is $70^\circ$C ($\nu\approx27.7$ cSt).
Approach. The no-load (Petroff) friction power is $P_f=8\pi^3\mu N^2 r^3 L/c_r$. Substituting $L=1.2D$ and $c_r=0.00225D$ collapses this to $P_f=K\,\mu D^3$, so the loss cap fixes only the product $\mu D^3$. A larger $D$ then demands a lower $\mu$, obtained by running the oil hotter — up to the practical thermal limit for mineral oil.
Petroff loss in terms of $\mu D^3$. With $r=D/2$, $L=1.2D$, $c_r=0.00225D$ and $N=4.167\ \text{rev/s}$,
$$P_f=\frac{8\pi^3\mu N^2 r^3 L}{c_r}=K\,\mu D^3,\qquad K=2.87\times10^{5}\ \text{W}/(\text{Pa}\cdot\text{s}\cdot\text{m}^3).$$
Setting $P_f=0.149\ \text{W}$ gives
$$\mu D^3=\frac{0.149}{2.87\times10^{5}}=\boxed{5.19\times10^{-7}\ \text{Pa}\cdot\text{s}\cdot\text{m}^3}.$$
Read the viscosity at the thermal limit. Mineral oils are held to roughly $70^\circ$C bulk temperature for long life (oxidation accelerates above this). From the VG100 chart at $70^\circ$C, $\nu\approx27.7\ \text{cSt}$; with $\rho_{70}\approx854\ \text{kg/m}^3$,
$$\mu_{70}=\nu\rho=27.7\times10^{-6}\times854=2.37\times10^{-2}\ \text{Pa}\cdot\text{s}.$$
Maximum diameter. Solve $\mu D^3$ for $D$ at the lowest usable viscosity (hottest allowable oil):
$$D=\left(\frac{\mu D^3}{\mu_{70}}\right)^{1/3}=\left(\frac{5.19\times10^{-7}}{2.37\times10^{-2}}\right)^{1/3}=\boxed{28.0\ \text{mm}}.$$
Running any cooler raises $\mu$ and would force a smaller journal to stay under the loss cap, so 28 mm at the $70^\circ$C limit is the maximum.
$\mu_{70}\approx0.0237\ \text{Pa}\cdot\text{s}$ ($\nu\approx27.7$ cSt at $70^\circ$C)
Maximum journal diameter
$D_{\max}\approx28\ \text{mm}$
Allowable temperature limit
$\approx70^\circ$C (mineral-oil oxidation limit)
Check: the $70^\circ$C ceiling is the standard long-life bulk-temperature limit for mineral oils; a synthetic or an oil cooler would allow a hotter, lower-viscosity film and hence a larger journal. The Petroff (concentric, no-load) model is exact only for the unloaded friction loss, which is precisely what the question caps.
