Question 4 of 6: Journal Bearing Sized by No-Load (Petroff) Power Loss

Question 4: Journal Bearing Sized by No-Load (Petroff) Power Loss (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Journal bearing running unloaded (concentric) at $N=250$ rpm on ISO VG100 oil, with $L=1.2D$ and diametral clearance $c_d=0.0045D$; the friction (no-load) power must not exceed $2.5\times10^{-4}$ hp.

Find. The maximum journal diameter $D$ and the oil-temperature limit that keeps the no-load loss at or below the allowance.

Approach. With no radial load the journal runs concentric, so Petroff’s equation gives the friction torque and power. Because $L$ and $c_r$ both scale with $D$, the power collapses to a product $\mu D^{3}$; the loss limit therefore fixes only $\mu D^{3}$. Larger $D$ demands lower $\mu$, i.e. a hotter (thinner) oil, so the “maximum diameter” is reached at the highest safe oil temperature.

1. Petroff friction power. For a concentric journal the friction torque is $T=4\pi^{2}\mu N' r^{3}L/c_r$ and the power $P_f=2\pi N'\,T$. Substituting $r=D/2$, $L=1.2D$, $c_r=0.00225D$ shows every geometric factor scales with $D^{3}$: $$P_f=K\,\mu D^{3},\qquad K=8\pi^{4}N'^{2}\frac{(1.2)}{(0.00225)\,2^{3}}\ \text{(consistent SI units)}.$$
2. Solve for the required $\mu D^{3}$. Setting $P_f=0.1864$ W, $$\boxed{\mu D^{3}=\frac{P_f}{K}=6.49\times10^{-7}\ \text{(SI: Pa\,s\,m}^3).}$$ This single group is all the loss limit constrains—diameter and viscosity trade off directly.
3. Viscosity of VG100 versus temperature. ISO VG100 has $\nu=100$ cSt at $40\,{}^\circ\text{C}$ and $\approx11.4$ cSt at $100\,{}^\circ\text{C}$. Fitting Walther’s equation $\log_{10}\log_{10}(\nu+0.7)=A-B\log_{10}T$ through these points gives, at $70\,{}^\circ\text{C}$, $\nu_{70}\approx27.7$ cSt; with $\rho_{70}\approx854$ kg/m$^3$ the dynamic viscosity is $\mu_{70}=\nu\rho\approx0.0237$ Pa·s.
4. Maximum diameter at the temperature limit. Mineral oil should not run much above $\approx70\,{}^\circ\text{C}$ in service (oxidation / film loss), so this is the practical thermal limit and gives the lowest safe $\mu$. Hence $$\boxed{D_{\max}=\left(\frac{\mu D^{3}}{\mu_{70}}\right)^{1/3}\approx30\text{ mm}\quad(\text{30.2 mm}),\qquad T_{\text{limit}}\approx70\,{}^\circ\text{C}.}$$
5. Resulting proportions. With $D=30$ mm, $L=1.2D=36$ mm and $c_r=0.00225(30)=0.068$ mm—a conventional slow-speed journal.